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https://www.youtube.com/watch?v=FWRBoUASFY0
How to Construct a Perpendicular Line through a Point on the Line MathSux 2330 subscribers 61 likes Description 12786 views Posted: 21 Jul 2021 In this video, we are going to go over how to construct a perpendicular line through a point on the line. You will need a compass and a ruler or a straight edge in order to do this construction. What is a Perpendicular Line? A perpendicular line is a line that will intersect our given point already placed on the line to create a 90º angle (or 4 90º angles) How to Construct a Perpendicular line through a Point: Step 1: Place the compass point on the point on our line. Step 2: Draw a semi-circle around our point (any size) using the compass. Step 3: Open up the compass (any size) and take the point of the compass to the intersection of our semi-circle and given line. Then swing our compass above the line. Step 4: Keeping that same length of the compass, go to the other side of our point, where the given line and semi-circle connect. Swing the compass above the line so it intersects with the arc we made in the previous step. Intro 0:00 Perpendicular Lines 0:13 How to Construct a Perpendicular line through a Point 0:19 Step 5: Mark the point of intersection created by these two intersecting arcs we just made and draw a perpendicular line! Practice Questions and more: Twitter: Facebook: TikTok: Patreon: If you have any questions, please don’t hesitate to comment below. And if this video has helped you, please hit the like button and subscribe for more math videos! Thank you and happy calculating! :) Music: Whenever by LiQWYD Creative Commons — Attribution 3.0 Unported — CC BY 3.0 Free Download / Stream: Music promoted by Audio Library mathsux 4 comments Transcript: Intro hi everyone and welcome to math sucks this video is going to help you pass geometry in this video we're going to go over how to construct a perpendicular line through a point on the line so just a little reminder about what a perpendicular line is it's a line that Perpendicular Lines will intersect our given point which is point b once it does it will then create a 90 degree angle so notice here we have How to Construct a Perpendicular line through a Point points a b and c on this line and we want to construct a perpendicular line right through point b so notice point b isn't in the middle so that's why we need to use a different method for constructing this perpendicular line that we've seen before so for this we will need a ruler or a straight edge and our compass so i'm going to open up the compass to something kind of smallish any size we'll do but something smaller so if we can we can fit it on this line so we're going to take the point of our compass put it on point b the line we're trying to intersect and then just swing this around [Music] so we swing this around and make a little bit of a semi-circle making sure that we intersect with our original line now we'll open up the compass a little bit more and take the point to the point of intersection to the line we just made and the original line and then make a little arc above the point and we're going to do the same thing keeping the same distance bringing the point to the other side and doing the same thing making a little arc so notice that we made a point of intersection up here right here so i'm going to mark that with a point and now all we need to do is take our ruler or straight edge and line it up with point b and draw our line and notice we made our perpendicular line so all so this is equal to 90 degrees so it's a super easy construction and if you're looking for more check out the videos right here for more constructions and more videos on geometry i post new videos every week if this video helped you please give it a like and subscribe thanks so much for stopping by and happy calculating need more practice check out mathsucks.org for more questions link below also don't forget to subscribe happy calculating
8401
https://pmc.ncbi.nlm.nih.gov/articles/PMC1083931/
Mapping and identification of essential gene functions on the X chromosome of Drosophila - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer | PMC Copyright Notice EMBO Rep . 2002 Jan 15;3(1):34–38. doi: 10.1093/embo-reports/kvf012 Search in PMC Search in PubMed View in NLM Catalog Add to search Mapping and identification of essential gene functions on the X chromosome of Drosophila Annette Peter Annette Peter Find articles by Annette Peter , Petra Schöttler Petra Schöttler Find articles by Petra Schöttler , Meike Werner Meike Werner Find articles by Meike Werner , Nicole Beinert Nicole Beinert Find articles by Nicole Beinert , Gordon Dowe Gordon Dowe Find articles by Gordon Dowe , Peter Burkert Peter Burkert Find articles by Peter Burkert , Foteini Mourkioti Foteini Mourkioti Find articles by Foteini Mourkioti , Lore Dentzer Lore Dentzer Find articles by Lore Dentzer , Yuchun He Yuchun He Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Yuchun He 1, Peter Deak Peter Deak Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Peter Deak 2,3, Panayiotis V Benos Panayiotis V Benos Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Panayiotis V Benos 4,a, Melanie K Gatt Melanie K Gatt Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Melanie K Gatt 2,3, Lee Murphy Lee Murphy Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Lee Murphy 5, David Harris David Harris Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by David Harris 5, Bart Barrell Bart Barrell Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Bart Barrell 5, Concepcion Ferraz Concepcion Ferraz Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Concepcion Ferraz 6, Sophie Vidal Sophie Vidal Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Sophie Vidal 6, Christine Brun Christine Brun Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Christine Brun 6, Jacques Demaille Jacques Demaille Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Jacques Demaille 6, Edouard Cadieu Edouard Cadieu Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Edouard Cadieu 7, Stephane Dreano Stephane Dreano Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Stephane Dreano 7, Stéphanie Gloux Stéphanie Gloux Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Stéphanie Gloux 7, Valerie Lelaure Valerie Lelaure Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Valerie Lelaure 7, Stephanie Mottier Stephanie Mottier Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Stephanie Mottier 7, Francis Galibert Francis Galibert Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Francis Galibert 7, Dana Borkova Dana Borkova Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Dana Borkova 8, Belen Miñana Belen Miñana Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Belen Miñana 8, Fotis C Kafatos Fotis C Kafatos Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Fotis C Kafatos 8, Slava Bolshakov Slava Bolshakov Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Slava Bolshakov 8,9, Inga Sidén-Kiamos Inga Sidén-Kiamos Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Inga Sidén-Kiamos 9, George Papagiannakis George Papagiannakis Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by George Papagiannakis 9, Lefteris Spanos Lefteris Spanos Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Lefteris Spanos 9, Christos Louis Christos Louis Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Christos Louis 9,10, Encarnación Madueño Encarnación Madueño Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Encarnación Madueño 11, Beatriz de Pablos Beatriz de Pablos Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Beatriz de Pablos 11, Juan Modolell Juan Modolell Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Juan Modolell 11, Alain Bucheton Alain Bucheton Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Alain Bucheton 6, Debbie Callister Debbie Callister Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Debbie Callister 3, Lorna Campbell Lorna Campbell Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Lorna Campbell 3, Nadine S Henderson Nadine S Henderson Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Nadine S Henderson 3, Paul J McMillan Paul J McMillan Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Paul J McMillan 3, Cathy Salles Cathy Salles Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Cathy Salles 3, Evelyn Tait Evelyn Tait Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Evelyn Tait 3, Phillipe Valenti Phillipe Valenti Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Phillipe Valenti 3, Robert DC Saunders Robert DC Saunders Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Robert DC Saunders 3,12, Alain Billaud Alain Billaud Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Alain Billaud 13, Lior Pachter Lior Pachter Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Lior Pachter 14, Robert Klapper Robert Klapper Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Robert Klapper 15, Wilfried Janning Wilfried Janning Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Wilfried Janning 15, David M Glover David M Glover Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by David M Glover 2,3, Michael Ashburner Michael Ashburner Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Michael Ashburner 2,4, Hugo J Bellen Hugo J Bellen Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain Find articles by Hugo J Bellen 1, Herbert Jäckle Herbert Jäckle Find articles by Herbert Jäckle b, Ulrich Schäfer Ulrich Schäfer Find articles by Ulrich Schäfer c Author information Article notes Copyright and License information Max-Planck-Institut für Biophysikalische Chemie, Abt. Molekulare Entwicklungsbiologie, Am Fassberg, 37077 Göttingen, 8 European Molecular Biology Laboratory, 69012 Heidelberg,15 Institut für Allgemeine Zoologie und Genetik, Westfälische Wilhelms-Universität, Schlossplatz 5, 48149 Münster, Germany,1 Howard Hughes Medical Institute, Department of Molecular and Human Genetics, Baylor College of Medicine, Houston, TX 77030,14 Department of Mathematics, University of California at Berkeley, Berkeley, CA, USA,2 Department of Genetics, University of Cambridge, Cambridge CB2 3EH,4 EMBL Outstation—The European Bioinformatics Institute, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SD,5 Sanger Centre, Wellcome Trust Genome Campus, Hinxton, Cambridge CB10 1SA,3 Department of Anatomy and Physiology, CRC Cell Cycle Genetics Group, University of Dundee, Dundee DD1 4HN,12 Department of Biological Sciences, The Open University, Milton Keynes MK7 6AA, UK,6 Montpellier University Medical School, Institut de Génétique Humaine, CNRS, 114 rue de la Cardonille, 34396 Montpellier Cedex 5,7 UPR 41, CNRS, Recombinaisons Génétiques, Faculté de Médecine, 2 Avenue du Pr Leon Bernard, 35043 Rennes Cedex,13 Fondation Jean Dausset—CEPH (Centre d’Etude du Polymorphisme Humain), 27 rue Juliette Dodu, 75010 Paris, France,9 Institute of Molecular Biology and Biotechnology, FORTH, Heraklion,10 Department of Biology, University of Crete, Heraklion, Greece and 11 Centro de Biología Molecular Severo Ochoa, CSIC and Universidad Autónoma de Madrid, 28049 Madrid, Spain a Present address: Department of Genetics, School of Medicine, Washington University, 4566 Scott Avenue, St Louis, MO 63110, USA b Corresponding authors. Tel: +49 551 201 1482; Fax: +49 551 201 1755; E-mail: uschaef@gwdg.de or hjaeckl@gwdg.de c Corresponding authors. Tel: +49 551 201 1482; Fax: +49 551 201 1755; E-mail: uschaef@gwdg.de or hjaeckl@gwdg.de Received 2001 Oct 11; Revised 2001 Nov 22; Accepted 2001 Nov 22. Copyright © 2002 European Molecular Biology Organization PMC Copyright notice PMCID: PMC1083931 PMID: 11751581 Abstract The Drosophila melanogaster genome consists of four chromosomes that contain 165 Mb of DNA, 120 Mb of which are euchromatic. The two Drosophila Genome Projects, in collaboration with Celera Genomics Systems, have sequenced the genome, complementing the previously established physical and genetic maps. In addition, the Berkeley Drosophila Genome Project has undertaken large-scale functional analysis based on mutagenesis by transposable P element insertions into autosomes. Here, we present a large-scale P element insertion screen for vital gene functions and a BAC tiling map for the X chromosome. A collection of 501 X-chromosomal P element insertion lines was used to map essential genes cytogenetically and to establish short sequence tags (STSs) linking the insertion sites to the genome. The distribution of the P element integration sites, the identified genes and transcription units as well as the expression patterns of the P-element-tagged enhancers is described and discussed. INTRODUCTION Drosophila melanogaster has been a model organism for genome research for almost 90 years, starting with Sturtevant’s demonstration that recombination frequencies could be used to map genes in a linear order along the chromosome (Sturtevant, 1913). Physical maps that relate genes to physical sites on the chromosomes (Bridges, 1935), together with the precise mapping of cloned DNA fragments by in situ hybridization to polytene chromosomes, enabled the first chromosomal maps of cloned genomic DNA segments (Wensink et al., 1974). These achievements, the development of procedures for molecular screening (Grunstein and Hogness, 1975) and the assembly of large chromosomal contigs (Bender et al., 1983) have facilitated systematic attempts to construct P1-based physical maps of the Drosophila genome. These maps cover ∼90% of the euchromatic genome and are punctuated by short sequence tag (STS) sites that are, on average, 50 kb apart (Kimmerly et al., 1996). More recently, a whole genome shotgun sequencing project, a collaborative effort between Celera Genomics and the Berkeley Drosophila Genome Project (BDGP), has provided a 95% coverage of the Drosophila genome (Adams et al., 2000; Rubin et al., 2000b). In addition, expressed sequence tag (EST) projects have generated sequence information from >100 000 transcripts (Rubin et al., 2000a). To assign function to the DNA sequences, the BDGP has undertaken large-scale functional analysis of the genome through gene-disruption projects that are based on mutagenesis by transposable P element insertion. These projects have focused on the isolation of essential genes on autosomes. The current collection of strains disrupt at least 1200 different genes, representing ∼30% of the estimated 3600 genes on the autosomes that are mutable to easily scorable phenotypes, mainly sterility or lethality (Spradling et al., 1995, 1999). Here, we report a large-scale P element insertion screen to identify essential genes on the X chromosome, as well as the construction of an X-chromosomal BAC tiling map, which complements other ongoing studies of the Drosophila genome. RESULTS AND DISCUSSION Tiling map of the X chromosome A physical map of the X chromosome was constructed by hybridizing 521 probes to high-density filters bearing both the EDGP and BDGP BAC libraries. 5771 BAC clones were identified by hybridization. Eleven contigs were assembled using a combination of manual analysis of hybridization data and in situ hybridization. These span the X chromosome from the subtelomeric region of 1A to 20C near the base of the chromosome arm (Figure 1; supplementary information available on the website The longest contig, contig 1, spans divisions 1–6 and comprises both cosmid clones (Siden-Kiamos et al., 1990) and BAC clones. Divisions 1–3 were mapped to yield template clones for sequencing (Benos et al., 2000, 2001), and in this region the BAC libraries were screened principally to link cosmid contigs. Open in a new tab Fig. 1. Diagram showing the contig distribution along the X chromosome. Contigs are indicated by the boxes above the schematic representation of the X chromosome. Contig 1 is composed of cosmids and BACs. Probe sequences were selected from several sources. The EDGP cosmid mapping project determined many sequence tags from the termini of cosmid clone inserts. These sequence tags correspond to cosmid clones of known cytological location, with apparently single-copy inserts that are members of contigs (Madueno et al., 1995). Some additional tags were determined from isolated cosmid clones of known cytological location. Additional probes were derived from sequences reported in databases, generally corresponding to genes of known cytological map position and from sequences of P element insertions, again of known cytological position. Probes used were selected partly on the basis of cytological position, in order to enhance the distribution of probe sequences along the chromosome. Example for inter-contig gaps Two probes (ES00262, mapping to 7B1-2, and ES00263, mapping to polytene chromosome bands 7B1-6) fail to hybridize to BAC clones from either library. Both probes are derived from mapped P1 clones. No BAC clones hybridizing to probes on either side of ES00262 and ES00263 were observed. The explanation may be that the sequences from which these probes were derived are incorrect or that there may be polymorphism between the Drosophila strains used in the construction of the BAC and P1 libraries. This gap represents a separation between contig 1 (1A to 7A) and contig 2 (7B to 10A). Other gaps generally correspond to intensely staining polytene chromosome bands (e.g. those in 10B, 11CD, 13A and 17A) and presumably represent long regions with few probes. In a number of cases, clones repeatedly failed to give signals with probes to which, on the basis of cytological location and other hybridization data, they might be expected to hybridize. Possible explanations include clones being absent on certain filters, cloned DNA segments bearing segmental deletion, and other failures in hybridization. Generation of P element insertions The euchromatin of the Drosophila X chromosome comprises ∼22 Mb of DNA, containing 2182 predicted protein coding genes (Adams et al., 2000). As an estimated two-thirds of all Drosophila genes show no obvious loss-of-function phenotype (Miklos and Rubin, 1996), ∼800 of the X-chromosomal genes are mutable to a scorable phenotype such as sterility or lethality. In order to assess such genes in a systematic manner, we generated a collection of lethal P element insertions on the X chromosome. The crossing scheme for generating X-chromosomal P{lacW} insertion lines is depicted in Figure 2. Females with a reinsertion of the P{lacW} element, as shown by the absence of the dominant second chromosome marker Curly and the white + eye colour from P{lacW}, were individually mated to FM7c males. 57 105 such crosses were initially set and 39 900 (69.9%) produced offspring. From these crosses, 501 strains (1.3%) were derived that were either hemizygously lethal or semilethal (defined as <20% viability compared to the sibling balancer males). Taking into account that the X chromosome represents roughly one-seventh of the genomic target size, 1.3% corresponds to a frequency of ∼9% lethal P element insertions, a rate similar to those reported for autosomal insertions with comparable P element vector (Cooley et al., 1988; Bellen, 1999). In addition, 73 lines were isolated that carried a visible mutation or that were sterile in one sex. Open in a new tab Fig. 2. Crossing scheme for isolating lethal X-chromosomal insertion lines. The sex chromosomes (left pair in each genotype) and the two large autosomes are shown schematically with females on the left. Relevant mutations are labelled accordingly. The P{lacW} insertion is indicated by the triangle and its white + selectable marker is represented by the darker chromosome. The transposase source is indicated by the double-headed arrow. Balancer chromosomes are hatched. Absence of the crossed-out males in the F 3 generation indicates a potential lethal P{lacW} insertion. For further details, see Methods. Distribution of P element insertions The insertion sites of 401 strains were determined by in situ hybridization on polytene chromosome squashes. In addition, DNA from the insertion sites was isolated either by plasmid rescue or by inverse PCR to generate an STS for all the insertion lines. Figure 3 summarizes the distribution of the P integrations along the X chromosome, showing a higher frequency of P{lacW} insertions in regions closer to the telomere. An observation that remains valid even when considering only non-redundant insertions. This skewed distribution is not mirrored by the relatively even distribution of transcription units along the X chromosome. Hence, it is likely that an inherent insertion site preference of P elements (Liao et al., 2000) is responsible for the observed distribution. Open in a new tab Fig. 3.P{lacW} distribution along the X chromosome. The localization is based on the molecularly defined integration site. Open bars represent all insertions in a given polytene region whereas shaded bars indicate the number of genes affected. The quality of a P element collection depends largely on the proportion of strains that harbour single P{lacW} insertions. In situ hybridizations to polytene chromosomes were carried out for 401 strains. 89.5% (359 lines) contain single insertions, 9.7% (39 lines) contain two insertions and 0.7% (three lines) contain three P{lacW} insertions. However, for those 14 strains that showed a second hybridization signal in polytene region 3A, we were unable to confirm the second insertion site by molecular means. Conversely, molecular analysis identified eight strains that contain a second integration site and two containing three integrations where only one in situ hybridization signal was observed. In eight such cases, the insertion sites are <50 kb apart, and thus likely to be below the limit of resolution of the in situ hybridization technique. In four of these closely linked strains, the same transcription unit was hit by two P elements, indicative of local hopping during the remobilization of the starting P{lacW}. Collectively, these results indicate that >90% of the lines contain single P element insertions. Hence, this collection of P element insertion strains is comparable with the best P element collections generated to date (Spradling et al., 1999). P{lacW} insertion and an associated lethal phenotype can be causally linked if excision of the P{lacW} restores wild type function. We tested a total of 206 lines by combining a transposase source P{ry + Δ 2-3}(99B) with the the P{lacW} element (Bellen et al., 1989). 134 strains (65.0%) produced fully viable male progeny that lost the white + gene, suggesting that the P element excised precisely. The 65% value should be regarded as a minimum, since the reversion tests were performed with only 10 females bearing both the P{lacW} insertion and the transposase source. Some P elements are difficult to excise and hence exhibit low mobilization frequencies and will almost certainly have been missed in this type of protocol. Characterization of the lines As the β-galactosidase gene in P{lacw} can report the expression pattern of adjacent genes, we immunohistochemically stained embryos with anti-β-gal antibody. 398 of the 501 strains (79.4%) expressed the enzyme. This is considerably higher than the 64% reported after random autosomal insertion of the same P element (Bier et al., 1989) and may result from the fact that the X-chromosomal lines are selected for insertions affecting essential genes. Representative patterns for each line were deposited in the FlyView database, (Janning, 1997), where they can be searched by various criteria. The lethal phase was determined in 497 of the 501 lines. The most frequent stage where development stops is the larval stage, with 29.2% of the analyzed lines (see Supplementary data). This is not unexpected, since homozygous mutants can often survive up to this stage by relying on maternal contribution of gene products deposited by the heterozygous mother. In addition, the majority of the lines probably do not carry a null allele, but rather hypomorphic mutations. We have generated an STS flanking the P element insertion site for 496 of the 501 lines by either plasmid rescue or by inverse PCR (the sequences were submitted to the EMBL nucleotide database; see Supplementary data). The integration sites were determined by BLAST search to the published Drosophila genomic sequence (Adams et al., 2000). In cases where the gene annotation is supported by EST data, we can unambiguously identify the affected gene. For predicted genes whose open reading frame is only annotated we assumed that the insertion is integrated in the vicinity of a putative transcription start site upstream of the nearest translational start codon, since there is a strong preference for P integrations in the 5′-end of genes (Spradling et al., 1995). Since the value of this collection is, in part ,determined by the number of new genes that are affected, we will briefly discuss our mapping data. A total of 513 STS sequences were generated from 496 strains. In five strains, repeated attempts to isolate flanking sequences failed, and in 16 the identification of the integration site was uninformative, since the P element is inserted in repetitive DNA. In 11 of these lines, the insertion is within a yoyo retrotransposon (Whalen and Grigliatti, 1998). The other 497 STSs were generated from the remaining 480 lines and are derived from unique X-chromosomal sequences. Five insertions were >7 kb away from any annotated gene and it is unclear whether the P integration is responsible for the observed lethal phenotype. Three additional P insertions occurred in the vicinity of known EP insertions but were some distance from the next annotated gene, suggesting that the the P elements have inserted into a putative promotor region of an unknown gene. Fourteen P elements, representing five gene pairs, may affect two genes. In each of these cases, one gene is localized within a large intron of another gene. 183 P{lacW} insertions affect 52 different genes that were previously characterized at both the molecular and genetic level, e.g. Notch, pebbled and short gastrulation. Phenotypically well characterized mutations of previously unidentified genes form a second group, consisting of two loci. These are trol, formerly known as zw1, with 12 insertions and stardust with one insertion. The largest group of insertions disrupts genes for which no mutation has previously been reported. In most cases, some molecular information is available or the genes have been predicted by computer algorithms (Adams et al., 2000). In summary, 301 insertions affect 130 genes (including the five nested gene pairs) that remain to be characterized. Among this class are transcription units that code for proteins putatively involved in signaling cascades, e.g. Rala, protein degradation, or Rpt4, intracellular transport, or Ntf-2, as well as many other cellular and developmental processes. To determine the frequency by which genes were hit, we relied on molecular information. Indeed, allelism cannot easily be determined by genetic criteria, as complementation tests between X chromsome lethals are not possible without first introducing a duplication that rescues the lethality. Genes on the X chromosome that are hit only once represent 96 out of 490 insertions (19.6%). This compares to a rate of 25.8 and 28.8% in the second and third chromosome collection, respectively (Spradling et al., 1999). The 60 so-called warm spots (2–5 insertions per gene; Spradling et al., 1999) correspond to 169 X-chromosomal insertions (34.5%), similar to what was observed for the second and third chromosomes (35.8 and 34.2%, respectively). There are 23 hot spots on the X chromosome that are hit at least six times, resulting in a total of 225 insertions (45.9%). The frequencies for the second and third chromosome hot spots are 39.1 and 38.1%, respectively. The six genes that were hit most frequently are: Notch, 22 insertions; Trf2, 17; inx2, 15; ras, 14; and act5C, ctp and trol 12 each. In total, these six hot spots contained 104 insertions, more than all the transcription units that were hit once. In summary, in this X-chromosomal collection, single hits are under-represented and hot spots are over-represented, when compared to the autosomal P insertion collections. Taken together, of the X-chromosomal genes mutated in our screen each is hit, on average, 2.7 times, whereas the figures for the autosomal collection are 2.3 and 2.2, respectively. Conclusions Saturation mutagenesis studies give a ratio of lethal complementation groups to polytene chromosomes bands of 0.81 (Ashburner et al., 1999). This estimate predicts ∼820 essential genes for the X chromosome (1012 bands). Hence, our collection represents mutations in slightly <25% of the X-chromosomal essential genes and is similar to the data obtained for the autosomal collection of lethal P insertions (Spradling et al., 1999). The P element screen for X-chromosomal genes required for adult viability and the BAC tiling map reported here represent a valuable tool for the Drosophila research community. The gene disruption collection provides the first opportunity to link ∼130 X-chromosomal genes with a phenotype. The majority of the lines are available from the Bloomington stock center, together with their in situ hybridization data, and the enhancer trap expression patterns of these strains have been deposited in the FlyView database. Using these resources, the Drosophila research community can utilize the collection for the functional and molecular characterization of affected genes. The BAC tiling map can, on the other hand, serve as a reference point for mapping sequences and, more importantly, it will be essential as a source for a particular genomic fragment, e.g. for germ line transformation. The combined efforts of the community will not only increase our knowledge of the model organism Drosophila but will also, by virtue of the evolutionary conservation of many genes and processes, shed light on human gene function. METHODS Drosophilastrains and embryos. The following fly strains (Lindsley and Zimm, 1992; FlyBase, 1999) were used in the experiments. The starting element, P{lacW} (Bier et al., 1989) originated from two male sterile insertion strains, 1040B and 1260A, which each contain a single transposon on chromosome 2. The strain used to supply transposase activity was w; CyO/wg Sp; TM6/Sb P{ry + Δ 2-3}(99B) (Robertson et al., 1988). The female fertile FM6 and the female sterile FM7c chromosomes were employed as X chromosome balancers. In addition to being fertile in both sexes, FM6 has the advantage of carrying a white null allele, thereby facilitating the detection of low expression of the white + marker gene. Crossing scheme (see also Figure 2) P: w/w; P{lacW}/P{lacW} × w/Y; wg Sp/CyO; ry Sb P{ry+ Δ2-3}(99B)/TM6 B F 1: FM6/FM6 × w/Y; P{lacW}/CyO; ry Sb P{ry+ Δ2-3}(99B)/+ F 2: w/FM6; CyO/+ (with P{lacW} somewhere) × FM7c/Y The absence of non-balancer bearing sons in the F 3 generation indicates a potential lethal insertion in the X chromosome. Immunohistochemistry. Enhancer trap expression pattern was analyzed by immunohistochemistry. Embryos were collected on apple juice agar plates, fixed and stained with anti-β-galactosidase antibody (Cappel). The secondary peroxidase-coupled antibody was a biotinylated anti-rabbit Ig (Organon Teknika). Molecular analysis. DNA was isolated from adult flies with QIAGEN DNeasy according to the instructions of the manufacturer. 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https://www.gauthmath.com/solution/1836436957399122/6-Locate-the-absolute-extrema-of-the-function-fx-x3-3x-on-the-closed-interval-0-
Solved: Locate the absolute extrema of the function f(x)=x^3-3x on the closed interval [0,3]. a. A [Calculus] Drag Image or Click Here to upload Command+to paste Upgrade Sign in Homework Homework Assignment Solver Assignment Calculator Calculator Resources Resources Blog Blog App App Gauth Unlimited answers Gauth AI Pro Start Free Trial Homework Helper Study Resources Calculus Questions Question Locate the absolute extrema of the function f(x)=x^3-3x on the closed interval [0,3]. a. Absolute max: f(3)=18; Absolute min: f(1)=-2 b. Absolute max: f(1)=-2; Absolute min: f(3)=18 c. Absolute max: f(3)=18; No absolute min d. No absolute max; Absolute min: f(3)=18 e. No absolute max or min Show transcript Expert Verified Solution 93%(13 rated) Answer Absolute max: f(3) = 18; Absolute min: f(1) = -2. Explanation Find the first derivative of the function f(x) = x³ - 3x. f'(x) = 3x² - 3 Set the first derivative equal to zero and solve for x to find critical points. 3x² - 3 = 0 3(x² - 1) = 0 x² = 1 x = ±1 Since we are considering the interval [0, 3], we only consider x = 1. Evaluate the function at the critical point and endpoints of the interval. f(0) = 0³ - 3(0) = 0 f(1) = 1³ - 3(1) = -2 f(3) = 3³ - 3(3) = 27 - 9 = 18 Compare the values of the function at the critical point and endpoints to determine the absolute maximum and minimum. The absolute maximum is f(3) = 18. The absolute minimum is f(1) = -2. Helpful Not Helpful Explain Simplify this solution Gauth AI Pro Back-to-School 3 Day Free Trial Limited offer! Enjoy unlimited answers for free. Join Gauth PLUS for $0 Previous questionNext question Related Locate the absolute extrema of the function fx=x3-3x on the closed interval [0,5]. a. absolute max: 1,-2 ; absolute min: 5,110 b. no absolute max; absolute min: 5,110 c. absolute max: 5,110 ; absolute min: 1,-2 d. absolute max: 5,110 ; no absolute min e. no absolute max or min a b c d e 98% (221 rated) Locate the absolute extrema of the function fx=x3-3x on the closed interval [0,3]. absolute max: 3,18 ; no absolute min absolute max: 1,-2 ; absolute min: 3,18 absolute max: 3,18 ; absolute min: 1,-2 no absolute max or min no absolute max; absolute min: 3,18 100% (3 rated) Locate the absolute extrema of the function fx=x3-3x on the closed interval [0,3]. absolute max: 3,18 ; no absolute min no absolute max; absolute min: 3,18 no absolute max or min absolute max: 3,18 ; absolute min: 1,-2 absolute max: 1,-2 ; absolute min: 3,18 95% (17 rated) Locate the absolute extrema of the function fx=x3-3x on the closed interval [0,5]. no absolute max or min absolute max: 1,-2 ; absolute min: 5,110 no absolute max; absolute min: 5,110 absolute max: 5,110 ; absolute min: 1,-2 absolute max: 5,110 ; no absolute min 100% (17 rated) Locate the absolute extrema of the function fx=x3-3x on the closed interval [0,2]. no absolute max; absolute min: 2,2 absolute max: 2,2 ; no absolute min absolute max: 2,2 ; absolute min: 1,-2 no absolute max or min absolute max: 1,-2 ; absolute min: 2,2 Previous 100% (4 rated) Locate the absolute extrema of the function fx=x3-3x on the closed interval [0,5]. no absolute max; absolute min: 5,110 absolute max: 5,110 ; no absolute min absolute max: 5,110 ; absolute min: 1,-2 absolute max: 1,-2 ; absolute min: 5,110 no absolute may or min 94% (15 rated) Locate the absolute extrema of the function fx=x3-3x on the closed interval [0,5]. absolute max: 5,110 ; absolute m in: 1,-2 no absolute max or min no absolute max; absolute min: 5,110 absolute max: 5,110 ; no absolute min absolute max: 1,-2 ; absolute min: 5,110 100% (5 rated) Locate the absolute extrema of the function fx=x3-3x on the closed interval [0,4]. 100% (3 rated) Locate the absolute extrema of the function fx=x3-48x on the closed interval [0,8] 100% (1 rated) Locate the absolute extrema of the function fx=x3-48x on the closed interval [0,7]. 93% (770 rated) Gauth it, Ace it! contact@gauthmath.com Company About UsExpertsWriting Examples Legal Honor CodePrivacy PolicyTerms of Service Download App
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https://physics.stackexchange.com/questions/508927/when-to-use-e-t-rc-capacitor-discharge-and-charge
Skip to main content When to use e−t/RC - Capacitor discharge and charge Ask Question Asked Modified 5 years, 9 months ago Viewed 1k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Respected Team Members, I am learning how to calculate the amount of time it takes to charge/discharge a capacitor. The Formula given in the text book is Vf=Vs(1−e−t/RC) But we are also required to use Vf=Vs(e−t/RC) Can you please guide me when to use which ? I have requested some fellow students and also my professor, and the answer is somewhat ambiguous. For example if t=0 and charging or discharging use the former, if t is a higher value say 40s and counting down then use the latter. Can you please guide me. Thank you. electric-circuits charge electrical-resistance capacitance voltage Share CC BY-SA 4.0 Improve this question Follow this question to receive notifications edited Oct 18, 2019 at 18:53 BioPhysicist 59.1k1919 gold badges117117 silver badges196196 bronze badges asked Oct 18, 2019 at 18:08 KudmiSubbaKudmiSubba 2511 gold badge11 silver badge77 bronze badges 1 2 Don't just memorize an answer. Investigate the math behind this problem. Assume a value for starting voltage, and see what happens for each equation when t=0. Then, choose small values of t and see what happens. Then choose successively larger values for t, and see what happens. My point? Take the time to learn the concept behind the math, and also learn a way to diagnose problems whereby you can answer some of your own questions. – David White Commented Oct 18, 2019 at 18:19 Add a comment | 2 Answers 2 Reset to default This answer is useful 2 Save this answer. Show activity on this post. They are both in essence the same formula. The claim is that you have a capacitor with a voltage V(t) as a function of time, starting at some value V0, that is in series with a resistor and in parallel with a battery at some voltage V1. The physics of this system is that the current through the capacitor must charge the capacitor and must also induce a voltage drop along the resistor, so that Kirchoff's voltage law says that, V1−QC−I R=0, or taking a derivative with respect to time, IC+R dIdt=0. This is an extremely common form of equation, one where the rate-of-change of a quantity, in this case the current, is linear in the value of that quantity. You see it for example if you are losing weight, where when you are heavier your body needs to burn more calories to simply exist and when you lose weight your body goes into a hibernation mode where it burns fewer calories to simply exist. You see it when a sink or bathtub is draining through a clog and the water is being forced through that clog by pressure but the lower the water level in the sink, the lower the pressure forcing water out through the clog. On the flip side, when you see the rate growing, you have the phenomenon of compound interest where the amount of money you earn monthly by saving money is proportional to the amount of money in your savings account. And it is solved by exponential growth or decay, I(t)=α1 exp(−tRC). Note that the equations do not say what this arbitrary parameter α1 is, that is determined by what we call boundary conditions—you choose it to match whatever that current was at t=0, say, because I(0)=α1. Or you can choose a different time τ if you compute I(τ) eτ/(RC). But the dynamics of the system do not care how exactly it started out; they tell you for any starting state how this circuit will continue. In turn this means that the voltage Q/C on the capacitor is V(t)=QC=V1−IR=V1−α1 R exp(−tRC). The battery voltage V1 specifies where the capacitor eventually “wants to go,” the arbitrary initial constant α1 specifies something about where it has “started.” One of your equations corresponds to the case where there is no battery in the loop, which you can think of as V1=0. It also corresponds to the case where α1R=−Vs for some Vs. In other words the battery capacitor has been charged initially to the value Vs but it wants to go to zero because there is no battery in the loop, so it exponentially decays from Vs down to 0. In your other case, you have a battery with voltage V1=Vs in parallel but the capacitor is starting out with absolutely zero charge on it V(0)=0, and it then exponentially decays upwards to Vs. You can also consider situations that do not match either of these two. For example, maybe your capacitor starts with voltage 2V1 and decays down to voltage V1. That requires a new choice of α1, namely V(t)=V1+V1e−t/(RC). All of these are valid, it just depends on what the battery voltage V1 is and what the initial condition voltage V0 is. You can always write this as V(t)=V1+(V0−V1)e−t/(RC). Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Oct 18, 2019 at 19:01 CR DrostCR Drost 39.6k33 gold badges4444 silver badges119119 bronze badges Add a comment | This answer is useful 2 Save this answer. Show activity on this post. You can easily determine which equation to use based on how you expect the circuit to behave as well as the shape of each of your functions. The first equation starts at 0 for t=0 and then grows towards a maximum value of Vs. The second equation starts at a maximum of Vs and then decays towards 0. So, just look at the circuit element in question and ask yourself what it is doing. For example, in a charging RC circuit the capacitor starts with no potential difference and then grows to a maximum potential difference. Therefore, the first equation describes a charging capacitor. For a discharging capacitor the capacitor starts at a maximum potential difference and decays to no potential difference. Therefore, the second equation describes a discharging capacitor. The main point though is don't rely on memorization. Rely on reasoning like what I outline above. By knowing how the equations and circuits behave, you can easily determine how to tackle the problem at hand. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Oct 18, 2019 at 18:59 BioPhysicistBioPhysicist 59.1k1919 gold badges117117 silver badges196196 bronze badges 0 Add a comment | Start asking to get answers Find the answer to your question by asking. 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https://sites.ualberta.ca/~vbouchar/MATH315/section_physical_gcd.html
Skip to main content Section 4.5 Physical interpretation of grad, curl, div 🔗 In this section we pause for a moment and explore further the physical interpretation of the vector calculus operations of gradient, curl, and divergence. 🔗 Objectives You should be able to: Interpret the gradient of a vector field as giving the direction and magnitude of fastest increase. Interpret the curl of a vector field in terms of a rotational motion in a fluid. Interpret the divergence of a vector field in terms of expansion and contraction of a fluid. 🔗 Subsection 4.5.1 The gradient of a function 🔗 Let be a smooth function with . We start by reviewing the interpretation of the gradient , 🔗 which as we saw is the vector field associated to the exterior derivative . At a point : The direction of the gradient is the direction in which the function increases most quickly at ; The magnitude of the gradient is the rate of fastest increase at . 🔗 This may be most easily understood with an example. 🔗 Example 4.5.1. The direction of steepest slope. In this example we work with a function on instead of , but the interpretation is the same. Suppose that the function gives the height above sea level at point on a surface. The height is highest at the origin, where , and decreases as we move away from the origin, until it reaches sea level on the circle with radius . So we can think of this altitude function as representing in a circular mountain centred at the origin. The level curves of are the curves of constant elevation, i.e. for constant elevations . Those would correspond to the level curves of elevation on a topographical map. In this case, they are all circles centred at the origin. The gradient of is For any point on the surface, this is a vector that points towards the origin, which says that the direction of steepest increasing slope is towards the origin, as expected from a circular mountain centred at the origin. Moreover, we see that the magnitude of the gradient vector is This says that the slope is steeper far away from the origin, and becomes less and less steep as we get closer to the origin. Here is below the contour map for this function (the graph of its level curves), and also a 3D plot (both produced using Mathematica). 🔗 Subsection 4.5.2 The curl of a vector field 🔗 Let be a smooth vector field with , with component functions . Its curl is given by the vector field 🔗 What is it actually computing? How can we interpret the vector field ? 🔗 It is easiest to understand the meaning of the curl of a vector field by thinking of as the velocity field of a moving fluid in three dimensions. So let us call it instead. The curl is related to rotational motion induced by the fluid. 🔗 Consider an infinitesimally small ball (sphere) located within the fluid, and centered at a point . Assume that the ball has a rough surface. The fluid moving around the ball will generally make it rotate. Then: The direction of the curl at gives the axis of rotation (according to the right hand rule);1 Half the magnitude of the curl at gives the angular speed of rotation. The “right hand rule” means that if you align the thumb of your right hand along the vector, the fingers of your right hand will “curl” around the axis of rotation in the direction of rotation. 🔗 In particular, if at a point , the fluid does not cause the sphere centered at to rotate. Because of this interpretation, we say that the velocity field of a fluid is irrotational if it is curl-free at all points on , that is on . 🔗 Example 4.5.4. The curl of the velocity field of a moving fluid. Suppose that a moving fluid has velocity field given by Sketching the vector field (see Exercise 2.1.3.2), one sees that at any point, the fluid is rotating counterclockwise around the -axis. Using the right-hand-rule, we thus expect the curl to point in the positive -direction, since it is the axis of rotation. We calculate: Indeed, it points in the positive -direction, as expected. Furthermore, half the magnitude of the curl is which means that the angular speed of rotation of the ball would be radian per unit of time. 🔗 Example 4.5.5. An irrotational velocity field. Conside a moving fluid with velocity field given by Sketching the vector field, we see that this would be a fluid in expansion. As the fluid is expanding, regardless of where the small sphere is located, it should not cause it to rotate (try to visualize this yourself). So we expect the velocity field to be curl-free. From the definition of the curl, we calculate: Thus it is curl-free, as expected, and this is an example of an irrotational velocity field. 🔗 Example 4.5.6. Another irrotational velocity field. Irrotational fluids do not have to be necessarily spherically symmetric. Consider for instance a moving fluid with velocity field This would be a fluid that is moving uniformly in the positive -direction. If you think about it for a little bit, it should not induce any rotation either, regardless of where the small sphere is located. So we expect the velocity field to be curl-free again. Indeed, from the definition we , which says that the velocity field is irrotational. 🔗 Subsection 4.5.3 The divergence of a vector field 🔗 Let be a smooth vector field with , with component functions . Its divergence is the function: 🔗 What is its intepretation? 🔗 As for the curl, it is easiest to interpret the divergence by thinking of the vector field as being the velocity field of a moving fluid or gas. The divergence is then related to expansion and compression of the fluid. 🔗 More precisely, consider an infinitesimally small sphere around a point : The divergence at measures the rate at which the fluid is exiting the small sphere at (per unit of time and unit of volume). 🔗 The key here is that if the divergence at a point is positive, it means that more fluid is exiting the small sphere centered at this point then entering the sphere, and vice-versa if the divergence is negative. If it is zero, this means that the fluid may still be moving, but there is exactly the same amount of fluid entering and existing the sphere. 🔗 For instance, if a gas is heated, it will expand. This means that for any small sphere in the gas, there will be more gas exiting then entering. So the divergence will be positive everywhere. Conversely, if a gas is cooled, it will contract, and its divergence will be negative everywhere. 🔗 Motivated by this interpretation, we say that the velocity field of a fluid is incompressible if it is divergence-free everywhere, i.e. everywhere on . We also call such vector fields solenoidal. 🔗 Example 4.5.7. The divergence of the velocity field of an expanding fluid. Consider a fluid/gas with velocity field The velocity field is pointing outwards in all directions. This corresponds to an expanding fluid. If we first think of a small sphere centered at the origin, then the fluid is moving outwards in all directions, exiting the sphere, and thus we expect the divergence to be positive. In fact, even if the small sphere is located elsewhere, we still expect the divergence to be positive, as there will be more fluid exiting the sphere than entering the sphere. From the definition, we calculate which is indeed positive everywhere, as expected. 🔗 Example 4.5.8. An imcompressible velocity field. Consider a fluid with velocity field As we have seen (see Exercise 2.1.3.2), the fluid is rotating counterclockwise around the -axis. It is not so obvious to see whether more fluid is existing or entering small spheres in the fluid. It is easiest to consider first a sphere centered around the origin. Because of the rotational motion, we see that actually no fluid is entering or leaving the sphere at all. So we expect the divergence to be zero, at least at the origin. It is not so obvious to see why the same should be true for all spheres not centered at the origin, but you can try to visualize it. In the end, through direct calculation, we get that , and hence the velocity field is divergence-free (that is, incompressible). 🔗 Example 4.5.9. Another incompressible velocity field. Consider the fluid with velocity field which has uniform velocity in the positive -direction. Here, if you pick a small sphere centered anywhere, there is certainly fluid entering and exiting the sphere because of the linear motion of the fluid, but the exact same amount of fluid will enter and exit the sphere. Therefore we expect the divergence to be zero everywhere, and indeed, . Another example of an incompressible velocity field. Comparing with Example 4.5.6, we see that this velocity field is both irrotational and incompressible, since it is both curl-free and divergence-free. 🔗 Exercises 4.5.4 Exercises 🔗 1. Show that any vector field of the form is irrotational. Solution. To show that is irrotational, we need to show that We can use directly the formula for the curl, or we can use the language of differential forms. In the latter, we associate a one-form to the vector field : We want to show that . But this is clear true, as taking the exterior derivative gives since are only functions of respectively. Therefore is irrotational. 🔗 2. Show that any vector field of the form is incompressible. Solution. To show that is incompressible, we need to show that This is the statement that But each of those partial derivatives vanishes, since does not depend on , does not depend on , and does not depend on . Therefore is incompressible. 🔗 3. Consider a vector field on ; it is independendent of , and its -component is zero. A sketch of the vector field in the -plane is shown in the figure below; as it is independent of , it looks the same in all other horizontal planes. Is positive, negative, or zero at the origin? What about at the point ? And ? Is at the origin? If not, in what direction does it point? Solution. (a) First we see that all arrows point in the positive -direction. The lengths of the arrows appear to be symmetric about the -axis. The arrows get longer and longer as you move away from the -axis. If you picture a small sphere around the origin, then there is fluid entering and leaving the sphere, and the arrows of the vectors entering and exiting have the same length. So you expect the divergence to be zero at the origin. In fact, the same is true for all points on the -axis, i.e. with . At the point , there are arrows entering and leaving the sphere as well. But the arrows that leave the sphere are longer than those that enter, so there should be more fluid exiting than entering. So we expect the divergence to be positive. This should be the case for all points with . As for , the opposite happens; the arrows entering the sphere are longer than those leaving. So there should be more fluid entering than leaving, and we expect the divergence to be negative. The same is true for all points with . (b) At the origin, the arrows are all pointing horizontally. As such, it will not induce any rotation on a sphere centered at the origin, and we expect the curl to be zero at the origin. In fact, this will be the case everywhere, so the curl should be zero everywhere. For your interest, this is the vector field . Its divergence is . We see that it is zero when , positive for all , and negative for all , as expected. The curl is which vanishes as expected (the vector field is irrotational). 🔗 4. Consider a vector field on ; it is independendent of , and its -component is zero. A sketch of the vector field in the -plane is shown in the figure below; as it is independent of , it looks the same in all other horizontal planes. Is positive, negative, or zero at the origin? What about at ? Is at the origin? If not, in what direction does it point? Solution. (a) We look at a small sphere around the origin, and we want to know whether there is more or less fluid entering the sphere versus exiting the sphere. Around the origin, we see that the arrows appear to be the same length on both sides of the -axis, and on both sides of the -axis. The arrows are all pointing the positive -direction. Furthermore, the arrows above the -axis point downwards, while the arrows below the -axis point upwards, in a symmetric way. If you think about it, this means that there is more fluid entering the sphere than exiting (because the arrows above and below the -axis are pointing towards the -axis). We thus expect the divergence to be negative at the origin. At the point , the fluid is moving downwards, and the arrows appear to be bigger for than for . So it looks like there is still more fluid entering the sphere than exiting the sphere. So we expect the divergence to be negative at this point as well. (b) At the origin, all arrows point in the positive -direction. Furthermore, the arrows above the -axis point downwards, while the arrows below the -axis point upwards, in a symmetric way. Therefore, if you imagine a small sphere around the origin, the motion of the fluid would not induce any rotation. So we expect its curl to vanish. For your interest, this is the vector field . Its divergence is , which is negative everywhere. As for the curl, we get , so it is zero everywhere (the vector field is irrotational). 🔗 5. Consider a vector field on ; it is independendent of , and its -component is zero. A sketch of the vector field in the -plane is shown in the figure below; as it is independent of , it looks the same in all other horizontal planes. Is positive, negative, or zero at the origin? What about at ? Is at the origin? If not, in what direction does it point? Solution. (a) Arrows that are diametrically opposite around the origin appear to have the same length but point in opposite directions. As a result, the same amount of fluid should be entering and exiting a small sphere around the origin, and hence we expect the divergence to be zero at the origin. As for the point , we see that at this point (just like for any point with ) all arrows are pointing horizontally. Furthermore, the length of the arrows on both sides of the -axis appear to be the same, so we expect the same amount of fluid entering and exiting the sphere. Again, the divergence should be zero at this point. (b) It is clear from the figure that the moving fluid would make a sphere centered at the origin rotate clockwise. We thus expect the curl to be non-zero at the origin. Using the right hand rule, we expect it to point in the negative -direction. In fact, looking at the figure, we expect this to be true at all points. So we expect the divergence to be zero everywhere. For your interest, this is the vector field . Its divergence is , which is zero everywhere, as expected (the vector field is incompressible). Its curl is We see that it points in the negative -direction for all values of , as expected. 🔗 6. Consider a vector field on ; it is independendent of , and its -component is zero. A sketch of the vector field in the -plane is shown in the figure below; as it is independent of , it looks the same in all other horizontal planes. Is positive, negative, or zero at the origin? What about at ? Is at the origin? If not, in what direction does it point? Solution. (a) We see that all arrows point away from the origin, in a way that is spherically symmetric (the lengths of all arrows on a circle of a fixed radius about the origin appear to be the same). As such, the fluid is all exiting the sphere, so we expect the divergence to be positive at the origin. At the point , there are arrows coming in and arrows coming out, but the arrows coming out are longer than the arrows coming out, so we expect again the divergence to be positive since more fluid is exiting the sphere than entering. In fact, this will be the case at all points in the figure. So we expect the divergence to be always positive. (b) At the origin, the fluid is all pushing outwards in spherically symmetric way, so it will induce no rotation on a sphere centered at the origin. We thus expect the curl to be zero at the origin. While it may not be as obvious, you can probably convince yourself that this should be true at all points in the figure, so the curl should be zero everywhere. For your interest, this is the vector field . Its divergence is , which is positive everywhere, as expected. Its curl is , which is zero everywhere as expected (the fuild is irrotational).
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http://www.cccg.ca/proceedings/1999/fp16.pdf
When Can a Net F old to a P olyhedron? Therese Biedl y Anna Lubiw y Julie Sun y  In tro duction In this pap er, w e study the problem of whether a p olyhedron can b e obtained from a net , i.e., a p olygon and a set of creases, b y folding along the creases. W e consider t w o cases, dep ending on whether w e are giv en the dihedral angle at eac h crease. If these dihedral angles are giv en the problem can b e solv ed in p olynomial time b y the simple exp edien t of p erforming the folding. If the dihedral angles are not giv en the problem is NP-complete, at least for orthogonal p olyhedra. W e then turn to the actual folding pro cess, and sho w an example of a net with rigid faces that can, in the sense ab o v e, b e folded to form an orthogonal p olyhedron, but only b y allo wing faces to in tersect eac h other during the folding pro cess. In the existing literature, a few related problems ha v e b een studied. Shephard in v estigated when a con v ex p olyhedron has a con v ex net [She]. Lubiw and O'Rourk e sho w ed ho w to test in O (n ) time whether an n-v ertex p olygon (with unkno wn creases) can b e folded in to a con v ex p olyhedron [LO ]. Another problem is the rev erse of ours: Giv en a p olyhedron, can one obtain a net? This can b e done for all con v ex p olyhedra [A O ], as w ell as for some classes of orthogonal p olyhedra [BDD + ]. A fundamen tal op en problem in this area is whether for an y con v ex p olyhedron there exists a net suc h that the edges of the net are also edges of the con v ex p olyhedron, a so-called unfolding with edge cuts only . Related to our problem of folding a rigid net is the problem of straigh tening rigid link ages in D, whic h has b een studied in [BDD + ]. In fact, our pro of that some rigid net cannot b e folded without in tersecting faces is based on the fact that there exists a link age in D that cannot b e straigh tened without ha ving links in tersect [CJ , BDD + ]. De nitions A p olygonal chain is a sequence of line segmen ts [a i ; b i ], i = 0; : : : ; n  that are m utually disjoin t, except that b i = a i+ for i = 0; : : : ; n  (addition mo dulo n). The segmen ts are called e dges and the endp oin ts of the edges are called vertic es. A nite region in the plane b ounded b y a p olygonal c hain is called a p olygon. A chor d of a p olygon is a line segmen t inside the p olygon where b oth endp oin ts are v ertices of the p olygon. A cr e ase-set of a p olygon is a set of c hords of the p olygon that do not in tersect eac h other except p ossibly at endp oin ts. A net is a p olygon together with a crease-set. An ortho gonal net is a net where all edges of the p olygon and all creases are parallel to a co ordinate axis. These results are part of the third author's Master's Thesis at the Univ ersit y of W aterlo o. Researc h supp orted b y NSER C. y Departmen t of Computer Science, Univ ersit y of W aterlo o, W aterlo o, ON NL G, fbiedl, alubiw, jsung@uwate rlo o.c a  A net can also b e view ed as a graph; in fact, it is an outer-planar gr aph since no t w o edges cross and all v ertices are on the un b ounded face (the outer fac e). It is kno wn that an n-v ertex outer-planar graph has at most n edges. The faces that are not the outer face are called interior fac es. F or an outer-planar graph, the incidences b et w een in terior faces form a tree. In a net, for eac h crease w e ma y or ma y not sp ecify the dihe dr al angle, i.e., the angle that the t w o faces inciden t to the crease will form inside the nished p olyhedron. W e imp ose the condition that ev ery crease m ust indeed b e folded, so the dihedral angle cannot b e  . Lik ewise, the dihedral angle cannot b e 0 or  , b ecause faces are not allo w ed to o v erlap. F or an orthogonal net, w e stipulate that all dihedral angles m ust b e  = or  =. T o b e able to test whether a net folds in to a p olyhedron, w e m ust establish a clear de nition of a p olyhedron. This is a non-trivial task (see [Cro  ] for a history of attempts). W e use the follo wing de nition, based on Co xeter [Co x ]: A p olyhe dr on is a nite connected set of plane p olygons, called fac es, suc h that () if t w o faces in tersect, it is only at a common v ertex or a common edge, () ev ery edge of ev ery face is an edge of exactly one other face, and () the faces surrounding eac h v ertex form a single circuit (to exclude anomalies suc h as t w o p yramids with a common ap ex). An ortho gonal p olyhe dr on is a p olyhedron eac h of whose faces is p erp endicular to a co ordinate axis. F or suc h a p olyhedron, w e classify eac h face as an xy -fac e, a y z -fac e or an xz -fac e, dep ending on whic h plane the face is parallel to. Kno wn Dihedral Angles In this section, w e sho w that if dihedral angles are giv en, w e can determine in p olynomial time whether folding the net yields a p olyhedron. Our computation mo del here is the real RAM; for the case of orthogonal nets, whic h is our main in terest, basic arithmetic suces. First, w e sho w ho w to nd the co ordinates of the v ertices in D after the creases ha v e b een folded. Compute the tree of adjacencies b et w een the in terior faces of the net. T ra v erse this tree T in depth- rst-searc h order, starting at an arbitrary leaf in an arbitrary p osition. F or eac h face, the p ositions of v ertices of this face can then b e computed using the p ositions of the v ertices of the paren t of this face in T , and the dihedral angle that connects the t w o faces. This tak es O (n) time, where n is the n um b er of v ertices of the net, b ecause the n um b er of edges and faces of the net is prop ortional to the n um b er of v ertices. No w w e m ust v erify the three prop erties of p olyhedra. W e do so in four steps, not for eciency , but for clarit y of presen tation: () W e reject the input if w e can nd t w o faces that in tersect in a p oin t that is in terior to one or b oth of the faces; () W e add additional v ertices along the edges of the faces, so that an y v ertex of the p olyhedron is a v ertex of all its inciden t faces; () F or eac h edge of eac h face w e nd all iden tical edges of other faces, sim ultaneously building up the the incidenc e gr aph, a data structure to store p olyhedra [Ede ]; () w e use the incidence graph to test that the faces surrounding eac h v ertex form a circuit. F or step  w e consider eac h pair of faces of the net. The running time of our algorithm is then already (n ), so w e will not w orry ab out making other steps of the algorithm faster than this. Let F  ; : : : ; F f b e the in terior faces of the net, and let m i b e the n um b er of edge of F i . F or an y i < j , if F i and F j lie in the same plane, w e can test in O (m i m j ) time ho w they in tersect b y testing ev ery pair of edges, and doing a nal inclusion test. If F i and F j lie in di eren t planes, w e then compute the line of in tersection of these planes and compute the in tersections of this line with F i and F j , forming t w o sets of disjoin t in terv als. W e can certainly test in O (m i m j ) time ho w these in terv als in tersect. The total running time of this step is prop ortional to X i  P n i= x i is big enough to prev en t an y of the links of the jagged sequence from reac hing the link of length v . Con v ersely , assume that S 0 can b e folded in to an orthogonal p olygon, and after p ossible rotation, assume that the link of length v is v ertical. Let H l b e the lengths of the edges p oin ting left and H r b e the lengths of the edges p oin ting righ t. Since all creases ha v e dihedral angle  = or  =, w e kno w that H l [ H r = fx  ; : : : ; x n ; L; Lg. Since S 0 folds to a closed p olygon, the lengths in H l sum to the same as the lengths in H r . Since L + L > P n i= x i , the t w o L's cannot b e in the same subset, so eac h of H l and H r m ust con tain exactly one L. Remo ving b oth L's, w e obtain a partition of fx  ; x ; : : : ; x n g with the t w o parts ha ving equal sums. . D Orthogonal F oldings Theorem The D ortho gonal folding pr oblem is NP-c omplete.  2 3 4 5 6 7 1 0 -3 -2 -1 L =  x = v =  x  = x = x  =  Figure : Constructing S 0 from the instance f,,,g of P ar tition. Pro of: If w e are giv en the assignmen t of dihedral angles of  = or  =, then w e can v erify in p olynomial time whether this net folds in to a p olyhedron (Section ), so the problem is in NP . T o sho w that it is NP-hard, let S = fx  ; : : : ; x n g b e an instance of P ar tition. W e describ e the construction of an instance of the D orthogonal folding problem in successiv ely more correct re nemen ts. W e illustrate these using the P ar tition instance S = f; ; g. The rst approac h is to extrude the p olygon formed in Section . in the z -direction b y units, see Figure . The sequence of links b ecomes in the net a sequence of rectangles all of heigh t . Ho w ev er, the problem then arises of ho w to co v er the fron t face. Note that the righ t b oundary of the fron t face abuts the (extruded) jagged sequence, and th us the shap e of this face dep ends up on the partition of S in to S  and S . Ev en cutting the fron t face in to strips attac hed to the v ertical unit length edges of the jagged sequence as sho wn in Figure do es not resolv e the problem, b ecause the lengths of the strips still dep end on the partition. x y z jagged sequence bottom top Figure : The construction in D extruded, and its net. Our next idea is to mak e the strips that form the fron t face equally long, whic h can b e ac hiev ed b y replicating the jagged sequence. The lengths in the x-direction of the top and b ottom faces are set to P n i= x i + . The resulting p olyhedron lo oks lik e a staggered pile of bric ks. Its net is indep enden t of an y particular partition of S . Unfortunately , this construction is to o general: for an y input the resulting net folds in to a p olyhedron. The pro of from the t w o-dimensional case do es not transfer b ecause w e cannot force the top and the b ottom face to b e aligne d, i.e., to ha v e the same xz -pro jection. See Figure . T o force an alignmen t of the top and the b ottom face, w e add an extruded rectangle with x-dimension  and z -dimension  do wn the middle of the fron t face. W e call this extruded rectangle the midd le notch. See Figure .  x y z top bottom Figure : Replicate the jagged sequence to form a pile of bric ks. No w the top and b ottom face need not align. x y z bottom top Figure : In tro ducing a middle notc h. Unfortunately , no w w e ha v e a problem similar to the one w e had originally: Eac h strip no w consists of v e pieces, the middle three forming part of the middle notc h. The lengths of the rst and last pieces dep end up on the partition of S . W e resolv e this problem b y sco oping out rectangular notc hes along the fron t side of the p oly-hedron (see Figure ). More precisely , starting   units to the righ t of the left end of the top face, w e sco op out notc hes of length  in the x-dimension and length  in the z -dimension at  unit in terv als in the x-direction, un til w e reac h the middle notc h. W e pro ceed in a similar w a y from the righ t end of the top face. Call the resulting net S 00 . x y z backbone bottom top x  = x = x = Figure : Cutting notc hes along the fron t face. The e ect of these notc hes is that no w the strips used to co v er the fron t face of the p olyhedron ha v e man y creases an y of whic h can form part of the middle notc h. Therefore, w e need not kno w the partition to construct the net. On the other hand, w e can sho w that the middle notc h con tin ues  to force alignmen t of the top and the b ottom faces, and th us that this construction w orks: Claim: S has a solution S = S  [ S if and only if S 00 folds in an orthogonal p olyhedron. Pro of: F ollo wing the steps of our construction, one can v erify that if S has a solution, then S 00 folds in to an orthogonal p olyhedron. F or the other direction, assume S 00 folds in to an orthogonal p olyhedron. W e will not sho w that the p olyhedron is lik e the one sho wn in Figure  (ev en though this is p ossible), but only use prop erties of the p olyhedron to extract a partition of S . Rotate the p olyhedron so that the face mark ed \top" in the net is an xz -face; w e refer to this face as T . Also rotate suc h that the edge of length P n i= x i +  of T is the b ack e dge of T , i.e., it is parallel to the x-axis and has the smallest z -co ordinate of all p oin ts of T . Let the b ackb one b e the in tersection of the net with the line through the bac k edge of T in the net, see Figure . By construction of our net, the edges of the bac kb one lie in one xz -plane P in the folded p olyhedron. As a rst step, w e sho w that face T m ust b e aligned with the face mark ed \b ottom" in the net; w e refer to this face as B . Since all folds m ust b e folded at righ t angles, w e can read from the net that the set of xz -faces consists exactly of the faces T and B and the faces corresp onding to x  ; : : : ; x n . Notice that b y construction of our net these faces ha v e their bac k edge on the bac kb one and th us in plane P . Note ho w ev er that T and B stic k farther forw ard in the z -direction than an y other part of an y other xz -face, b ecause of the middle notc h. This implies that T and B m ust b e aligned, for otherwise there w ould b e a line parallel to the y -axis in tersecting the top of the middle notc h and no other xz -face, a con tradiction to the ev en parit y required for the n um b er of in tersections b et w een a p olyhedron and a line. In the p olyhedron, the bac kb one forms an orthogonal c hain in plane P . Let C b e the sub c hain joining the bac k edges of T and B ; th us C consists of the bac k edges of one extruded jagged sequence. Let H l b e the lengths of the edges of C p oin ting left, and H r b e the lengths of the edges of C p oin ting righ t. Since all creases ha v e dihedral angle  = or  =, w e kno w that H l [ H r = fx  ; : : : ; x n g. Because T and B are aligned, c hain C b egins and ends with the same x-co ordinate. Therefore the lengths in H l sum to the same as the lengths in H r , and th us H l and H r form the required partition. This ends the pro of of the claim, and therefore the pro of of the theorem.  Rigid Nets In this section, w e sho w that if the net is made from sti material, i.e., its faces are rigid, then w e cannot alw a ys execute the folding pro cess|from net to p olyhedron|while k eeping faces disjoin t. Theorem Ther e exists an ortho gonal net and an ortho gonal p olyhe dr on it c an b e folde d to with the pr op erty that the folding c annot b e p erforme d while ke eping fac es rigid and disjoint. Pro of: Consider the net sho wn in Figure ; its folded-up v ersion is sho wn in Figure . The ends of this \extruded c hain" are supp osed to b e v ery long. Imagine placing an orthogonal c hain on the faces that are shaded in Figure , with v ertices exactly on the creases that the c hain crosses. No w, if w e could fold this net of rigid faces without self-in tersections, then w e could also fold the c hain without self-in tersections in to the p osition that  a long end a long end Figure : This rigid net cannot b e folded without in tersections. it tak es on the p olyhedron. Ho w ev er, using a pro of v ery similar to the one in [BDD + ], w e can sho w that if the end-links of the c hain are sucien tly long, then this c hain cannot b e str aightene d, i.e., transformed in to a straigh t c hain without self-in tersections, ev en allo wing arbitrary rotations of links, rather than just the one degree of freedom rotations p ossible for the c hain on the net. Since the c hain can b e straigh tened when lying on the net (b ecause it has a planar pro jection [BDD + ]), the net of rigid faces cannot b e folded in to the p olyhedron without self-in tersections. Figure : The p olyhedron for the rigid net, and the c hain em b edded on it.  Conclusions In this pap er, w e studied the problem of determining whether a net folds in to a p olyhedron. W e sho w ed that if the dihedral angle at eac h crease is giv en, the question can b e answ ered in p olynomial time. Ho w ev er, for orthogonal p olyhedra, if the dihedral angle is not giv en, the problem b ecomes NP-complete. Finally , w e examined the dicult y of folding a net (with dihedral angles giv en) made  of sti material in whic h faces cannot in tersect. Here w e sho w ed a net of an orthogonal p olyhedron that cannot fold in to the p olyhedron without an y in tersection of faces. Our NP-completeness result only holds for orthogonal p olyhedra. In particular, it is conceiv able that the nets w e construct (see Figure ) can fold to non-orthogonal p olyhedra that no longer yield the partition. This leads to a question in teresting in its o wn righ t: can an orthogonal net ev er fold to a non-orthogonal p olyhedron? W e men tion one other problem concerning rigidit y: can an y feasible folding pro cess b e accom-plished with rigid non-in tersecting faces if extra creases (of dihedral angle  ) ma y b e in tro duced? References [A O ] B. Arono v and J. O'Rourk e. Nono v erlap of the star unfolding. Discr ete Comput. Ge om., : {0,  . [BDD + ] T. Biedl, E. Demaine, M. Demaine, A. Lubiw, J. O'Rourk e, M. Ov ermars, S. Robbins, and S. Whitesides. Unfolding some classes of orthogonal p olyhedra. In 0th Canadian Confer enc e on Computational Ge ometry, pages 0{,  . [BDD + ] T. Biedl, E. Demaine, M. Demaine, S. Lazard, A. Lubiw, J. O'Rourk e, M. Ov ermars, S. Robbins, I. Strein u, G. T oussain t, and S. Whitesides. Lo c k ed and unlo c k ed p olygonal c hains in D. In SIAM-A CM Confer enc e on Discr ete A lgorithms, pages {,  . [CJ ] J. Can tarella and H. Johnston. Non-trivial em b eddings of p olygonal in terv als and un-knots in -space. Journal of Knot The ory and its R ami c ations, ():0{ 0 ,  . [Co x] H.S.M. Co xeter. R e gular Polytop es. The Macmillan Compan y ,  . [Cro ] P eter R. Crom w ell. Polyhe dr a. Cam bridge Univ ersit y Press,  . [Ede] H. Edelsbrunner. A lgorithms in Combinatorial Ge ometry. Springer-V erlag,  . [GJ ] M. R. Garey and D. S. Johnson. Computers and Intr actability: A Guide to the The ory of NP-Completeness. F reeman,   . [LO ] A. Lubiw and J. O'Rourk e. When can a p olygon fold to a p olytop e? T ec hnical Rep ort 0, Smith College,  . [L W ] W.J. Lenhart and S.H. Whitesides. Recon guring closed p olygonal c hains in euclidean d-space. Discr ete and Computational Ge ometry, :{0,  . [She] G. C. Shephard. Con v ex p olytop es with con v ex nets. Math. Pr o c. Camb. Phil. So c., : {0,  .
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https://math.stackexchange.com/questions/228841/how-do-i-calculate-the-intersections-of-a-straight-line-and-a-circle
Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams How do I calculate the intersection(s) of a straight line and a circle? Ask Question Asked Modified 6 years, 6 months ago Viewed 110k times 23 $\begingroup$ The basic equation for a straight line is $y = mx + b$, where $b$ is the height of the line at $x = 0$ and $m$ is the gradient. The basic equation for a circle is $(x - c)^2 + (y - d)^2 = r^2$, where $r$ is the radius and $c$ and $d$ are the $x$ and $y$ shifts of the center of the circle away from $(0,\ 0)$. I'm trying to come up with an equation for determining the intersection points for a straight line through a circle. I've started by substituting the "y" value in the circle equation with the straight line equation, seeing as at the intersection points, the y values of both equations must be identical. This is my work so far: $$(x - c)^2 + (mx + b - d)^2 = r^2$$ $$x^2 + c^2 - 2xc + m^2x^2 + (b - d)^2 + 2mx(b - d) = r^2$$ I'll shift all the constants to one side $$x^2 - 2xc + m^2x^2 + 2mx(b - d) = r^2 - c^2 - (b - d)^2$$ $$(m^2 - 1)x^2 + 2x(m(b - d) - c) = r^2 - c^2 - (b - d)^2$$ That's as far as I can get. From what I've gathered so far, I ought to be able to break down the left side of this equation into a set of double brackets like so: $$(ex + f)(gx + h)$$ where $e,\ f,\ g$ and $h$ are all constants. Then I simply have to solve each bracket for a result of $0$, and I have my $x$ coordinates for the intersections of my two equations. Unfortunately, I can't figure out how to break this equation down. Any help would be appreciated. circles Share edited Nov 4, 2012 at 12:25 EuYu 42.8k99 gold badges9696 silver badges142142 bronze badges asked Nov 4, 2012 at 12:20 CambotCambot 38322 gold badges33 silver badges1010 bronze badges $\endgroup$ 2 1 $\begingroup$ Why don't you just solve for $x$? You know the constans, right? $\endgroup$ M.B. – M.B. 2012-11-04 12:22:20 +00:00 Commented Nov 4, 2012 at 12:22 3 $\begingroup$ I'm trying to develop an equation that will deduce what x is for any set of constants, not just a specific set. $\endgroup$ Cambot – Cambot 2012-11-04 12:41:35 +00:00 Commented Nov 4, 2012 at 12:41 Add a comment | 4 Answers 4 Reset to default 38 $\begingroup$ Let's say you have the line $y = mx + c$ and the circle $(x-p)^2 + (y-q)^2 = r^2$. First, substitute $y = mx + c$ into $(x-p)^2 + (y-q)^2 = r^2$ to give $$(x-p)^2 + (mx+c-q)^2 = r^2 \, . $$ Next, expand out both brackets, bring the $r^2$ over to the left, and collect like terms: $$(m^2+1)x^2 + 2(mc-mq-p)x + (q^2-r^2+p^2-2cq+c^2) = 0 \, .$$ This is a quadratic in $x$ and can be solved using the quadratic formula. Let us relabel the coefficients to give $Ax^2 + Bx + C = 0$, then we have $$x = \frac{-B \pm \sqrt{B^2-4AC}}{2A} \, . $$ If $B^2-4AC < 0$ then the line misses the circle. If $B^2-4AC=0$ then the line is tangent to the circle. If $B^2-4AC > 0$ then the line meets the circle in two distinct points. Since $y=mx+c$, we can see that if $x$ is as above then $$y = m\left(\frac{-B \pm \sqrt{B^2-4AC}}{2A}\right) + c \, . $$ EDIT: The lines $y=mx+c$ do not cover the vertical lines $x=k$. In that case, substitute $x=k$ into $(x-p)^2+(y-q)^2=r^2$ to give $$y^2 - 2qy + (p^2+q^2-r^2 - 2kp+k^2) = 0$$ This gives a quadratic in $y$, namely $y^2+By+C=0$, where $B=-2q$ and $C=p^2+q^2-r^2 - 2kp+k^2$. Solve using the Quadratic Formula. The solutions are $(k,y_1)$ and $(k,y_2)$, where the $y_i$ are solutions to $y^2+By+C=0$. Share edited Mar 12, 2019 at 17:47 answered Nov 4, 2012 at 12:51 Fly by NightFly by Night 32.9k55 gold badges5757 silver badges103103 bronze badges $\endgroup$ 3 $\begingroup$ Ah! Thank you! This certainly looks like it'll do what I'm trying to do! $\endgroup$ Cambot – Cambot 2012-11-04 13:20:39 +00:00 Commented Nov 4, 2012 at 13:20 5 $\begingroup$ Be careful to manage vertical line !!! $\endgroup$ Eric Ouellet – Eric Ouellet 2016-05-13 19:26:59 +00:00 Commented May 13, 2016 at 19:26 1 $\begingroup$ @EricOuellet Thank you, my friend. I think I've fixed it. $\endgroup$ Fly by Night – Fly by Night 2019-03-12 17:51:21 +00:00 Commented Mar 12, 2019 at 17:51 Add a comment | 4 $\begingroup$ Here's an equation which works even if the line is vertical. This is useful from a computer programming perspective. Let's set up the system of equations using standard formulae: $$ \begin{cases} (x - x_0)^2 + (y - y_0)^2 = r^2 \ Ax + By + C = 0 \end{cases} $$ where the circle with radius $r$ is centered at $(x_0, y_0)$; the line contains the points $(x_1, y_1)$ and $(x_2, y_2)$. From the line equation, we know: $$ y = \frac{-Ax-C}{B} = - \frac{Ax+C}{B} $$ ($B \neq 0$, we'll get into this in a second) Therefore, $$ \begin{align} (x - x_0)^2 + \left( - \frac{Ax+C}{B} - y_0 \right)^2 &= r^2 \ x^2 - 2x_0x + {x_0}^2 + \frac{A^2 x^2 + 2ACx + C^2}{B^2} + 2y_0 \cdot \frac{Ax+C}{B} + {y_0}^2 &= r^2 \ x^2 - 2x_0x + \frac{A^2 x^2 + 2ACx + C^2}{B^2} + 2y_0 \cdot \frac{Ax+C}{B} &= r^2 - {x_0}^2 - {y_0}^2 \end{align} $$ Multiply both sides by $B^2$. $$ \begin{align} B^2 x^2 - (2B^2 x_0)x + A^2 x^2 + (2AC)x + C^2 + 2By_0(Ax+C) &= B^2(r^2 - {x_0}^2 - {y_0}^2) \ (A^2 + B^2)x^2 + (2AC - 2B^2 x_0)x + (2AB y_0)x &= -C^2 - 2BCy_0 + B^2(r^2 - {x_0}^2 - {y_0}^2) \end{align} $$ Therefore, we have a quadratic equation $ax^2 + bx + c = 0$ with: $$ \begin{cases} a = A^2 + B^2 \ b = 2AC + 2AB y_0 - 2B^2 x_0 \ c = C^2 + 2BC y_0 - B^2 (r^2 - {x_0}^2 - {y_0}^2) \end{cases} $$ where $$ \begin{cases} A = y_2 - y_1 \ B = x_1 - x_2 \ C = x_2 y_1 - x_1 y_2 \end{cases} $$ Now you can use ${ x = \frac{-b \pm \sqrt{\Delta}}{2a}, y= -\frac{Ax+C}{B} }$ to solve for the points. The above equation doesn't work for $B=0$. Besides this, if $|B|$ is too small, then floating point computation gets inaccurate. If $B=0$, or is too small, then we let $x = -\frac{By+C}{A}$, and sub it into the circle's formula: $$ \begin{align} \left( - \frac{By+C}{A} - x_0 \right)^2 + (y - y_0)^2 &= r^2 \ \frac{B^2 y^2 + 2BCy + C^2}{A^2} + 2x_0 \cdot \frac{By+C}{A} + y^2 - 2y_0 y &= r^2 - {x_0}^2 - {y_0}^2 \ B^2 y^2 + 2BCy + C^2 + 2Ax_0 (By+C) + A^2y^2 - 2A^2 y_0 y &= A^2(r^2 - {x_0}^2 - {y_0}^2) \ (A^2 + B^2)y^2 + (2BC + 2ABx_0 - 2A^2 y_0)y &= -C^2 - 2ACx_0 + A^2(r^2 - {x_0}^2 - {y_0}^2) \end{align} $$ Thus, we have a quadratic equation $ay^2 + by + c = 0$ where $$ \begin{cases} a = A^2 + B^2 \ b = 2BC + 2ABx_0 -2A^2 y_0 \ c = C^2 + 2ACx_0 -A^2(r^2 - {x_0}^2 - {y_0}^2) \end{cases} $$ Share edited Mar 28, 2019 at 23:28 answered Mar 28, 2019 at 19:19 user658746user658746 $\endgroup$ Add a comment | 2 $\begingroup$ You've got a quadratic in $x$. Use the discriminant to see if it has real solutions, and if so how many. Then solve the quadratic for $x$, and substitute the solution(s), if any, back into the equation for the line. Share answered Nov 4, 2012 at 12:49 Cameron BuieCameron Buie 105k1010 gold badges106106 silver badges245245 bronze badges $\endgroup$ Add a comment | 1 $\begingroup$ Have you checked the general solution on wolframalpha? You've already made your quadratic equation and so, you can get the roots of $x$ as the solutions that can then be used to get the values of $y$. Share answered Nov 4, 2012 at 12:50 TariqTariq 63511 gold badge44 silver badges1717 bronze badges $\endgroup$ Add a comment | You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions circles See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 2 Choosing a value so a line and circle intersect a one, two, and no points 3 Newbie: Find the intersection of a line and a circle and interpret geometrically 2 Find a line parallel to a given line with a specific offset Intersection point between circle and plane in 3d 0 Soving intersection point for a sphere and sphere 0 Distance from point to circle perimeter Related 3 Newbie: Find the intersection of a line and a circle and interpret geometrically 2 General solution for intersection of line and circle 1 Finding the points of intersection of a circle and a line 0 Calculate angle of the next point on a circle How can I find the equation of a circle given two points and a tangent line through one of the points? 0 I calculated the intersection of a line with a circle, got projection instead. 4 Find the center of a circle using straight-edge and compass with given conditions. 1 A line goes through a point outside a circle and a point on the circle. 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https://www.bilibili.com/video/BV1HQ4y1y7BJ/
逻辑推理 列表法 （1）_哔哩哔哩_bilibili 首页 番剧 直播 游戏中心 会员购 漫画 赛事 投稿 逻辑推理 列表法 （1） 97 0 2021-09-12 18:13:51 未经作者授权，禁止转载 1 投币 1 分享 奥数 小学数学 汪汪队学大习发消息 关注 138 那我帮你挠一下 西西里人 接下来播放 自动连播 【全421集】《高斯数学1-6年级动画》一年级奥数思维提升 小初高能量补给站 1.9万 15 【五年级必看】《小学奥数举一反三》1~6年级全集，小学奥数的敲门砖和自学神器，全套视频建议收藏集 资料库-分享版 4307 7 全40集【2025必看一年级数学】 奥数思维1-6年级奥数全套 小学数学培优提升学习，构建完整的小学奥数体系！ o_小葱拌面_o 2966 4 小学奥数经典毕克定理！ 图解数学 1402 0 【全184集】《DK图解数学动画》风靡家长圈的数学启蒙，涵盖小学数学各个核心知识点，锻炼数学思维能力 丁一知识宝库 4005 3 【小学数学全196集＋讲义】这绝对是小破站上最简单易懂的小学数学六年级奥数思维课程（内附配套资料pdf） 中小学资源站 1439 13 【清华附小数学动画】1—6年级 | 405集全 柚子优学社 2586 0 【全1000集】高斯数学1-6年级动画版小学1~6年级奥数动画，奥数思维提升 牛娃学习加油站 1.1万 12 【全760集】高斯数学1-6年级动画版 奥数思维提升一课一练 （配套习题PDF） 清北无忧资料站 3.2万 34 【405集全】清华出品：小学数学启蒙动画，1-6年级数学全打通！让孩子学得更主动更透彻！（下） 小宝妈的资料库 1.4万 1 【全集430集】小学奥数思维启蒙课程《高斯数学动画》，1-6年级全覆盖，孩子看动画轻松掌握数学知识点，逻辑思维全面提升，学习成绩快速提高 儿童教育视频 3053 0 （无偿分享）小学数学六年级上下册资料合集～知识点汇总&随堂测试&基础测评&期中期末测试～持续更新～电子版～可打印～ 学海灯塔计划 377 0 一年级的奥数附加题，99%的孩子一个数都不会填。 #一年级数学 #小学数学 #数学思维 #思维训练 #一学就会 小刘老师数学思维 8017 6 全780集 2025最新版【小学数学思维 小学奥数动画】人教版1-6年级上下册小学数学思维提升 小学数学思维训练 同步小学数学教材基础课＋进阶课 牛娃学习加油站 9385 53 【答疑】我从未见过如此“赏心悦目”的数学题 汪汪队学大习 94 0 【全50集【2025最新版 小学数学速算巧算技巧课程】(巧算 速算简便运算技巧)提高数学计算能力 把计算变得更简单 数学学不好都难 牛娃学习加油站 4888 3 全300集【2025必备五年级数学 奥数思维】1-6年级奥数全套 小学数学培优提升学习，构建完整的奥数体系(知识点讲解+经典例题解析) 清北无忧资料站 4323 16 [全83集]三年级奥数举一反三|轻松开启数学思维之门(视频+PDF)轻松攻克奥数难点 孩子三年级学奥数没思路？这套《三年级奥数举一反三》83集系统课来帮忙！ 橙子资料学习库 4720 31 小学经典平移法——通过周长算面积！ 图解数学 1954 1 【全集184集】小学数学启蒙《DK图解数学动画》，家长圈热推课程，涵盖所有核心知识点，孩子看动画就能轻松掌握数学思维与解题技巧，学习兴趣全面提升 爱学习学院 1797 0 展开 手办模型周边聚集地>> 小窗 客服 顶部 赛事库课堂2021拜年纪
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https://euclidsmuse.com/embedded?id=98
N Gon Inscribed in Circle N Gon Inscribed in Circle An n-gon is a regular polygon with n sides. This figure demonstrates that an n gon becomes colser to a circle as its number of sides increases. This is because there are more, smaller sides, with bigger angles, in regular polygons of more sides. A circle is essentially an n-gon with infinity sides. This document requires an HTML5-compliant browser. n Use the textbox to change n, the number of sides on the n-gon (note: only whole numbers will display properly). Values less than 3 will result in figures that are not regular polygons, by definition. Values of n greater than 12 will show 12 points of the figure, with angles and side lengths positioned as if n sides were present. App generated by Geometry Expressions
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https://www.youtube.com/watch?v=mPDGs_dDK-Q
Engineering Statics: Frame-Pulley Systems Serhan Guner 2930 subscribers 16 likes Description 1415 views Posted: 12 Nov 2021 Dr. Guner solves three questions on frame-pulley systems. 0:00 - Intro 0:37 - Example 1 (Pulleys only) 5:55 - Example 2 (Frame and Pulley) 12:40 - Example 3 (Frame and Pulley - More advanced) 18:29 - Study Tips 4 comments Transcript: Intro hi everyone today i'm going to cover frames with pulleys these are a little bit more advanced systems which we see it in statics from time to time i'm going to solve a couple of questions along with a couple of strategies or tips to help you develop your understanding after watching this session and after solving these questions yourself you should be able to tackle other questions and apply these strategies for solving more questions with or without police if you find these videos useful please don't forget to subscribe as well as like the video let's get into it and start solving Example 1 (Pulleys only) questions i would like to start with a pulley system only this would illustrate how to disintegrate them and how to draw free body diagrams for pulleys only in my second example i would move on to a combined system it would involve a frame and pulley together so let's get into this example you can see we have three police with the cable system and the purpose is to lift a weight as you can imagine if you pull this cable the weight will go up each pulley has a weight of q pulleys are usually very light as compared to the weight that we are lifting or the force that we are applying so in general you will see in many questions the weight of the pulley is neglected but i want to show you how you would incorporate it determine tension forces in each cable and support reactions first of all these type of questions require a strategy when i look at this i see two cables or even three if you count this little one the first cable is this long one starts from a it goes like this if it's the same cable the tension would be the same considering that we are neglecting friction even though you might think that oh there are other cables and the force would change same cable same tension so that is the principle the other cable is this one it's a shorter one it is going to have the same tension which we don't know at this point the third cable is this small one which is actually the support reaction so it's going to be the same as support reaction we are going to have two support reactions one at a one at b and also even though the question is not asking we are able to determine the key relationship it is going to be less than the weight because we are doing a mechanical system and we need to make it more efficient so if this is w let's see is it going to be half of it is it going to be one third of it we would determine that the first step would be free body diagram probably this is first and the last step i'm showing this pulley this one which is this it's going to be call this t1 i'm not writing any equations but i'm just labeling the unknowns two that's the self weight of the pulley here what's moving this up is this tension call this t2 this corresponds to that one this corresponds to that one and i'm missing another one again it's going to be t2 this one and then call this one t2 there is going to be sulfate and from this one you would have called this t3 this cable tension here again it's going to be t3 support reactions this is point b call this b y it is y direction and this is a call this a y the next step would be equations of equilibrium in this problem everything is taking place vertically so i'm only going to be using fy equal to zero when i look at this this seems easier to start this is part one if you write vertical equilibrium equation you will see that t2 would be equal to 2p plus q let's look at this one make this number 2 t3 equal to 2 t2 plus 2. you know t2 you can just substitute that back in from this part i can easily write down t3 is the same as by that is what the question is asking which is support reaction and then let's move on to parts three two p plus t 1 equal to w plus q we didn't need to use two unknowns you will see that it is the same force p1 equal to t2 if you sub t2 into this one uh 2p plus 2p4p plus q would be equal to w plus q these will cancel out p equal to 0.25 w meaning that i have to apply a quarter of this v in terms of force and also from part three i can show that p is the same as a y if you want um just to clarify for p plus 3 2 answer right now i'm done with the question cable tensions i found it but i just would like to write them a little bit better t 2 2 p plus q that's it for this problem let's move on to the next one in the second example i Example 2 (Frame and Pulley) would like to consider a frame and pulley together this is one level above the previous question this is a conceptual problem there is a force applied we know the force we know a distance we know the distance we know radius force p is applied to the system determine the cable force and support reactions so that is the scope that's what we are expected to do in this question looking at this don't start don't do anything before you develop your strategy that's really important let's just think about reactions point a is a pin you would have blue reactions it's just a cable because of that you would have only one reaction which is going to be tension i'm guessing this would be a compression and this would be up given that this force is going down so that's my prediction we should know the angle as well that should be given in the problem so we should be able to find the x that is also given that is also given in the system so if i look at this system that's relatively easy question because you only have three support reactions i should be able to solve this without disintegrating the system from the pins only three unknowns no need to separate the system use three equations of equilibrium the question do not say anything about the weight of the pulley so don't worry about it it's not given it's neglected in many cases it's neglected it doesn't say anything about friction the cable tension will be the same without any friction it is a relatively straightforward problem this support reaction horizontal one call this ax vertical one call this a y call this d x start from y because y is the easiest one there is really only one unknown which is a y t y we know it f x equal to zero call this cable tension equal to t and i can easily say dx the same as t ax would be dx minus px i would need the moment equation doesn't really matter what point either a or d e x will not show up in that equation only two forces will show up dx times a would be p y times b that is answer because we know all these three variables that's given all right that concludes the problem we were able to solve this without disintegrating it from the pin but i'm going to solve it using this integration this is not something i recommend you should always go with the simplest the shortest solution this is longer way i'm just going to say separate from pins and of course you need to draw a three body diagram you shouldn't really forget this tension force because there is a cable whenever you cut a member you are exposing the internal force and you have to deal with it contrast it with the original the easier solution we didn't really consider this cable tension because whatever inside the system remains inside the pulley i would need to consider it separately so let me sketch in terms of reactions at this point you would get two reactions inverted versions should be applied to the other part once you assemble it they would cancel out that's the reason why we don't deal with internal forces because they are in equilibrium meaning that they cancel out each other you have to start from the pulley part fx equal to zero cx would come out to be t then you would do f y equal to 0 c y would come out as can write these down this is going to be t and this is going to be t apply that point c you can also show it this is the same t and that is going to be the same as the x the x equal to t looking at this part you have to be careful with the dimensions this is r so let's start with f x equal to zero a x would be the same as t minus p x t is the same as d x looking at this we got the same equation coming from this and then let's do summation of y forces equal to zero as you see t t would cancel out and a y would be the same as p y this is also answer and it really checks out with the original solution so now let's move on to the interesting part because you would think that the situation for a moment would be radically different because we have so many forces to consider in the original one because we didn't cut we didn't expose cable tensions but here we have a lot going on so it would be a little bit hard to visualize how come it would be the same as the easier solution so let's do that i'm taking moment about this point what i would like to look at this cable tension because that's what's new it's going to be t times that distance from here to that joint minus t times that distance plus r so you know what's going to remain would be t times r and it's going to be this way the horizontal one is also rotating the same way it is going to be t times a minus r if you add these up you will see that tr components will cancel out and what i'm going to be ending up would be t times a here is the same as dx and if you look at this it is again dx times a would be equal to py times b it is going to be the identical equation to the first one so that's why i call this a more complicated solution sometimes you have to do this if you have more than three unknowns in the original system that's the way you have to do in many cases you have to do this unless it's an easy question as this one i will write this down you can again write the x t y times b divided by a so definitely it makes sense it is the same as before that concludes my second example this is my final question we are going Example 3 (Frame and Pulley - More advanced) to be looking at a frame pulley system it's a little bit more involved but we should be able to develop a strategy and then get this solved this is the system as you can see we have a pulley one cable pulley is connected to member abc the question is determine all force components on member abc remember abc is this l-shaped membership since we are doing this i would like to add another part determine all support reactions relatively complicated looking problem let's first of all talk about strategy let's talk about how many unknowns do we have to get support reactions two reactions here to here supports are not aligned they are at a different elevation meaning that when you take moment about one of those two unknowns will show up in that equation meaning that you won't be able to find anything from that equation we cannot proceed with this system we have to separate it from the pins we need to look at each part individually and then try to come up with a strategy for unknowns which is must this integrate or call this disassemble the system of course we do this from the pins i would like to show unknowns in terms of directions you can make assumptions it doesn't matter we would figure them out you would like to disassemble the pulley but let's do this quicker look at the previous example so what happened these two tensions applied opposite direction reactions and when we move on to the original joint it became the same direction as the cable tension here what i'm going to do in this case this is going to be five kilonewtons and it would be five so i'm just saving one little step dimensions would be needed also point b we would have two reactions so that is pretty much it for this part i do have to look at the other part as well five cable tension you know at point b it's going to be the opposite direction a lot of unknowns i expose but typically my experience is that if the question looks really complicated it ends up being easier than some other easy looking questions make sure that you don't forget any force or any reaction i would start with this one this is definitely easier there are less forces let's make this part one let's look at one these two points are aligned that's a big advantage if you take moment about point a you would find b y once you find b y you can do vertical equilibrium and get a y so four unknowns we should be able to find two of them and then a relationship between ax and dx you can look at how this rotates it's rotating this way this is the opposite way 5 is also the opposite way right so b y times 2.5 minus 5 times distance 3 and then minus 5 times distance 4.5 that is going to be equal to zero b y will come out to be 15.0 kilo newtons that's a very good start the second expression is f y equal to zero a y plus five would be equal to b y i know b y which is 15 a y would be 15 minus 5 and then i would do f x 0 this would give me an equation value f x 0 b x minus a x would be equal to five now i can do part two because i figured out here b y which is 15. now i only have three unknowns and i should be able to solve this let's start with the moment take moment about point f dx will rotate the system this way this would rotate this way 3 would rotate this way so the x and 3 rotating the same way also b y rotating all this i mean you have to be careful with this one the x times 1 plus three times five these are clockwise let's go with minus in terms of pluses which is counter clockwise five times three another five times three plus 15 times 4.5 put a check mark this is to enable that we are not forgetting any of the terms if you solve this equation sub bx into this equation 62.5 kilo newton they came out to be positive meaning that the assumed direction was correct let's do f x 0. ax also answers this is also a reaction a y is answer this is also reaction right b x b y also answers that is pretty much it i have a little bit space left i really would like to show the answers very nicely done thank you very much for Study Tips your attention i tried to solve you a couple of questions please solve them at home on a piece of paper but also you have solved the examples you should definitely look and pick at least two three even four questions from these copy the question on your paper and solve it part by part clears easy to understand free body diagrams really half of the solution so i would encourage you to improve your sketches as well as free body diagrams that's it from me today thank you very much for attention i hope you found these solutions useful please don't forget to subscribe as well as like the video if you would like to see more videos in statics thank you and bye bye
8410
https://www.bu.edu/photonics/files/2013/08/RET-2011-Polarization-of-Light_9th-grade-Physics.pdf
Northeastern University Research Experience for Teachers – Partners Linking Urban Schools Claire Duggan / Program Director Lesson Plan Title Polarization of Light Primary Subject Area Physics Grade Level 9 Overview Students will learn about polarization of light, including linear, circular, and elliptical polarization, and different ways of obtaining polarized light. They will conduct an experiment to explore how Brewster’s angle works. Students will also explore real-life applications of polarized light such as polarizing filters and ellipsometry. Approximate Duration Six 45-min classes MA Frameworks Content Skills 4.4 Describe qualitatively the basic principles of reflection and refraction of waves. 6.1 Recognize that electromagnetic waves are transverse waves and travel at the speed of light through a vacuum. Scientific inquiry skills SIS2. Design and conduct scientific investigations. • Articulate and explain the major concepts being investigated and the purpose of an investigation. • Properly use instruments, equipment, and materials (e.g., scales, probeware, meter sticks, microscopes, computers) including set-up, calibration (if required), technique, maintenance, and storage. • Follow safety guidelines. SIS3. Analyze and interpret results of scientific investigations. • Assess the reliability of data and identify reasons for inconsistent results, such as sources of error or uncontrolled conditions. • Use results of an experiment to develop a conclusion to an investigation that addresses the initial questions and supports or refutes the stated hypothesis. Interdisciplinary Connections Chemistry, mathematics Lesson Objectives 1. Differentiate between polarized and nonpolarized light. 2. Explain how light can be polarized by absorption and by reflection. 3. Differentiate between linear, circular, and elliptical polarization. 4. Explain how a polarizing filter works. 5. Describe what an ellipsometer is and what it is used for. Lesson Materials and Resources Hewitt, P. Conceptual Physics Tipler, P., Mosca, G. Physics for scientists and engineers Smith, H. Measuring Brewster’s angle between classes. Phys. Teach. 17, 109 (1979) www.physicsclassroom.com The Brewster’s angle lab from The Pasco Science Workshop. www.pasco.com The Bogus barrier: Brewster’s angle: Brewster’s angle: Northeastern University Research Experience for Teachers – Partners Linking Urban Schools Claire Duggan / Program Director Technology Tools and Materials Polarizing filters Shoeboxes Computer Pasco: Brewster’s Angle accessory, optics bench, high sensitivity light sensor (2), rotary motion sensor, aperture bracket for light sensor (2), red diode laser, spectrophotometer accessory kit Vernier ULI interface or Pasco interface Logger Pro (or DataStudio) software Background Information Light is an electromagnetic wave that consists of electric and magnetic fields oscillating perpendicular to each other and, at the same time, perpendicular to the direction of propagation of light. Since the electric and magnetic components of an electromagnetic wave are interconnected, it is enough to consider only the electric component. Common light coming from a light bulb or from the Sun contains various waves oscillating in different directions. It is called nonpolarized light. When the electric field of a light wave oscillates in one direction only, the wave is linearly polarized. When the electric field vector traces a circle during its oscillations, the light has circular polarization. When the electric field makes an ellipse, the light is elliptically polarized. Some long-chain hydrocarbon molecules absorb the component of the electric field of incident light that is perpendicular to the chains and transmit the component that is parallel to the chains. Polarizing filters made of such molecules produce linearly polarized light. Consider a nonpolarized light beam shining at a non-conducting (dielectric surface). We can resolve the electric field of the incident light into two components: parallel to the plane of incidence and perpendicular to the plane of incidence. See the picture below ( The reflected light mostly contains the perpendicular component of the incident light. At a certain angle of incidence (Brewster’s angle), the reflected light is all linearly polarized perpendicular to the plane of incidence. The value of Brewster’s angle depends on the indices of refraction of the first and the second medium. The polarization effect is used in ellipsometry. Ellipsometry is a method used for determining the thickness of thin films and their optical constants, such as the index of refraction, using polarized light. An ellipsometer sends a beam of linearly polarized light to a sample. The incident beam of light is reflected off the sample and changes its state of polarization during reflection. The change in polarization state is measured and then analyzed by a computer. The obtained information is used to calculate the thickness of the sample and its index of refraction. Useful Vocabulary New Vocabulary Word Meaning polarized light light that consists of electromagnetic waves vibrating in a specific direction polarizer a device that changes a beam of nonpolarized light into a polarized light Brewster’s angle the angle of incidence that makes the reflected beam completely polarized ellipsometry a method of determining the thickness of thin films and their optical constants using polarization of light Northeastern University Research Experience for Teachers – Partners Linking Urban Schools Claire Duggan / Program Director Essential Questions to be answered; Grand Challenges What is the difference between polarized light and nonpolarized light? What is polarized light? How can polarized light be created? How does polarization by reflection work? What is Brewster’s angle? Where do we use polarization in real-life? Misconceptions Light travels as rays in straight lines. Light reflects only from mirrors or objects with polished surface. Lesson Procedures Part 1: Introduction to polarization of light. (45 min) • Pre-lesson homework: Read pages 542- 545 of Conceptual Physics. In your notebook, answer questions 20-22 from page 549 • In class: all class discussion (facilitated by the teacher) of the homework. • The bogus barrier activity – students work in pairs. ( • Post-lesson homework: read the online tutorial on the Brewster angle ( Part 2: Brewster’s angle. (45 min) • All class discussion (facilitated by the teacher) of the homework. • Demonstration: • Activity: Determining Brewster’s angle in the hallway. (See Smith: Measuring Brewster’s angle between classes) • Homework: Read the Brewster’s lab procedure. Part 3. The Brewster’s angle lab. (three 45-min classes) • The purpose of this lab is to measure Brewster’s angle for a light beam reflected off an acrylic semi-circular lens. The intensity of the reflected light is measured by a light sensor. The angle of reflection is measured by a rotary sensor. The Brewster angle is the angle at which the intensity of the reflected light is the lowest. • The experimental set up and calibration are complicated and take a lot of time. Therefore, to save time and make sure that the stations are set up correctly, the lab setup will be performed by a teacher before the lab. • Homework: 1. The Brewster’s angle lab report. 2. Go to www.youtube.com. Search for a movie on ellipsometry or ellipsometer. Pick a movie that you think is the most helpful in understanding what ellipsometry is or how an ellipsometer works. E-mail me the title of the movie. Part 4. Applications of polarized light: ellipsometry. (45 min) • Students will report about the movies on ellipsometry. • We will watch 2-3 movies selected by the students. • Then, I will give a short presentation about my experience of working with the ellipsometer during the RET program at BU. • We will watch the remaining movies. • Homework: Study for the quiz on polarization. Assessment Procedures Observation of students as they work in groups. Homework. Lab report. Quiz on polarization of light. Accommodations/Modifications Visual learners: movies and simulations; white board diagrams Auditory learners: discussions in groups and as a class Reproducible Materials The Bogus barrier activity. Northeastern University Research Experience for Teachers – Partners Linking Urban Schools Claire Duggan / Program Director The Brewster’s angle lab. Explorations and Extensions Research on FTIR: what is FTIR, what it is used for, and what physics concepts are involved. Lesson Development Resources Hewitt, P. Conceptual Physics Tipler, Mosca Physics for scientists and engineers Hastings A. Smith. Measuring Brewster’s angle between classes. Phys. Teach. 17, 109 (1979) www.physicsclassroom.com The Brewster’s angle lab from The Pasco Science Workshop. www.pasco.com Reflections Contact Information Valentina Sountsova Bancroft School 110 Shore Drive Worcester, MA vsountsova@bancroftschool.org
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https://www.quora.com/What-is-the-L-C-M-of-3-and-6
What is the L.C.M of 3 and 6? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics LCM and GCF Least Prime Factor Mathematics Homework Ques... Arithmetical Calculations Lcm of 8 Number and Arithmetic Number Theory Least Common Multiple 5 What is the L.C.M of 3 and 6? All related (45) Sort Recommended Charles Holmes Studied Financial Markets&Mathematics (Graduated 1990) · Author has 16.6K answers and 13.6M answer views ·5y y=6 PREMISES y=the lowest common multiple LCM of 3 and 6 CALCULATIONS The lowest common multiple LCM of a set of “n” numbers is the smallest number that divides all of the numbers with no remainder The factorization of 3: 1×3=3, 2×3=6, 3×3=9,… The factorization of 6: 1×6=6, 2×6=12, 3×6=18,… Since the factorization of 3 and 6 shows 6 to be the smallest number evenly divided by both 3 and 6, 6 is the LCM. y=6 C.H. Upvote · 9 3 Related questions More answers below What is the LCM of 12, 15 and 18? How do you find the L.C.M. of 18, 24, 36, and 48? What is the answer to the L.C.M. of 20, 30 and 40? How can we obtain an L.C.M. of 2 and 11/4? Assistant Bot · 1y The least common multiple (L.C.M.) of 3 and 6 is 6. To find the L.C.M., you can list the multiples of each number: Multiples of 3: 3, 6, 9, 12, ... Multiples of 6: 6, 12, 18, ... The smallest multiple that appears in both lists is 6. Alternatively, you can use the formula: L.C.M(a,b)=|a×b|G.C.D(a,b)L.C.M(a,b)=|a×b|G.C.D(a,b) Where G.C.D. is the greatest common divisor. For 3 and 6: G.C.D.(3, 6) = 3 So: L.C.M(3,6)=|3×6|3=18 3=6 L.C.M(3,6)=|3×6|3=18 3=6 Thus, the L.C.M. of 3 and 6 is 6. Upvote · Aaron Douglas 7y 6. Multiples of 3 are 3,6,9,12,15,18… Multiples of 6 are 6,12,18… Therefore the lowest common multiple is 6. Upvote · 9 1 The Chosen One Service Desk Analyst at Samsung SDS (2007–present) · Author has 14.7K answers and 5.6M answer views ·6y Originally Answered: What is the LCM of 3, 5 & 6? · 3 and 5 are prime 6 = 3 2 So the LCM is 3 5 2 = 30 Upvote · 9 1 Related questions More answers below Teresa G. Colwell Former /Retired-Transportation Spec./Information Spec. (1980–2006) · Author has 123 answers and 191.1K answer views ·6y Originally Answered: What is the LCM of 3, 5 & 6? · LCM of 3, 5 & 6 would be as follows: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 5, 10, 15, 20, 25, 30 6, 12, 18, 24, 30 The lowest number common to all of the multiples of each number would be 30. Answer: 30 Upvote · 9 3 Ryan Bon Former English Teacher at Abies English Room (2017–2018) ·6y Originally Answered: What is the LCM of 3, 5 & 6? · Here’s a pretty cool trick (for me at least): Step 1. Factor out the numbers into prime numbers. 3=3x1 5=5x1 6=3x2 Step 2. Once you break the numbers into prime numbers (numbers that cannot be formed by multiplying two smaller natural numbers), multiply the factors altogether BUT consider the repeating factor as only one digit. LCM= 1 x 2 x 3 x 5 = 60 Since 1 and 3 can be found on the three numbers two times, they may only be considered as one digit. And this applies for all cases of getting the LCM. Hope this helps, goodluck! Upvote · 9 3 9 3 John Maiko High School Teacher at Teachers Service Commission (2007–present) · Author has 5.9K answers and 3.5M answer views ·5y 3=1×3 6=2×3 LCM=1×2×3=6 Upvote · 9 1 Ran Some Where Keen interest in entrepreneurship & artificial intelligence ·7y Related What is the LCM of 3, 6, and 7? Wow another homework question, oh well LCM - Lowest Common Multiple Continue Reading Wow another homework question, oh well LCM - Lowest Common Multiple Upvote · 9 3 Jose Ruiz Bachelor's Science, EE. in Industrial Electrical Engineering, Universidad Central "Marta Abreu" de Las Villas (Graduated 1986) · Author has 76 answers and 53.7K answer views ·5y By dividing the larger by the smaller. Procedure: 6/3= 2, and if the result is an integer, (or exact division= 2 this case), then LCM of the two is 6. Number 6 is the first multiple of 6, and the second of 3. Upvote · Sinnathamby Mahesan Works at University of Jaffna, Sri Lanka · Author has 70 answers and 141K answer views ·7y Related What is the LCM of 6, 4 and 3? A little different from the methods in other answers - it will give you the insight of other methods. Multiply all three numbers together : you get 72 . Prime factors are 2 and 3. Divide 72 by a prime factor, say, 2. You get 36 Check whether 6, 4, 3, divide it , Yes. So divide by a prime factor , 2, again You get, 18 Check whether 6, 4, 3 divide it : No, 4 cannot divide it. Go back to 36: Divide by another prime factor, say, 3: You get 12: 3, 4, 6 divide it and if you divide 12 by any prime factor 2 or 3, the result will not be divided by one or more of the numbers 3, 4, 6. So 12 will be the answer. (Least co Continue Reading A little different from the methods in other answers - it will give you the insight of other methods. Multiply all three numbers together : you get 72 . Prime factors are 2 and 3. Divide 72 by a prime factor, say, 2. You get 36 Check whether 6, 4, 3, divide it , Yes. So divide by a prime factor , 2, again You get, 18 Check whether 6, 4, 3 divide it : No, 4 cannot divide it. Go back to 36: Divide by another prime factor, say, 3: You get 12: 3, 4, 6 divide it and if you divide 12 by any prime factor 2 or 3, the result will not be divided by one or more of the numbers 3, 4, 6. So 12 will be the answer. (Least common multiple!) Try for 12, 16, 27. (Prime factors 2 and 3) :-) Upvote · Laura Kay Posey Education, K-6th from SWT (Graduated 1966) · Author has 6.5K answers and 2M answer views ·2y Originally Answered: What is the LCM of 3, 5 & 6? · ANSWER: 30 ( 5× 6 = 30) LCM LCM of 3, 5, 6 = = 3, 6, 9, 12, 15, 18, 2q, 24, 27, 30 5 = 5, 10, 15, 20, 25, 30 = 6, 12, 18, 24, 30 Upvote · Related questions What is the LCM of 12, 15 and 18? How do you find the L.C.M. of 18, 24, 36, and 48? What is the answer to the L.C.M. of 20, 30 and 40? How can we obtain an L.C.M. of 2 and 11/4? Related questions What is the LCM of 12, 15 and 18? How do you find the L.C.M. of 18, 24, 36, and 48? What is the answer to the L.C.M. of 20, 30 and 40? How can we obtain an L.C.M. of 2 and 11/4? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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https://www.chegg.com/homework-help/questions-and-answers/normal-random-variable-x-sim-n-left-mu-sigma-2-right-mu-15-sigma-125-compute-following-qua-q111581663
Solved For the normal random variable. X∼N(μ,σ2) with μ=15 | Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Math Statistics and Probability Statistics and Probability questions and answers For the normal random variable. X∼N(μ,σ2) with μ=15 and σ=1.25, compute the following quantities. (a) P(∣X−μ∣≥3) (b) The 33rd percentile of X. (c) Find a and b such that P(a≤X≤b)=0.95. (d) Find c such that P(c≤X)=0.01. Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: For the normal random variable. X∼N(μ,σ2) with μ=15 and σ=1.25, compute the following quantities. (a) P(∣X−μ∣≥3) (b) The 33rd percentile of X. (c) Find a and b such that P(a≤X≤b)=0.95. (d) Find c such that P(c≤X)=0.01. Show transcribed image text There are 2 steps to solve this one.Solution 100%(1 rating) Share Share Share done loading Copy link Step 1 (a) We can standardize X to obtain a standard normal variable Z, P(|X−μ|≥3)=P(|Z|≥3 σ)=2 P(Z≥3 1.25)=2 P(Z≥2.4) View the full answer Step 2 UnlockAnswer Unlock Previous questionNext question Transcribed image text: For the normal random variable. X∼N(μ,σ 2) with μ=15 and σ=1.25, compute the following quantities. (a) P(∣X−μ∣≥3) (b) The 33rd percentile of X. (c) Find a and b such that P(a≤X≤b)=0.95. (d) Find c such that P(c≤X)=0.01. Not the question you’re looking for? Post any question and get expert help quickly. 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https://www.opentextbookstore.com/precalc/2.0/Chapter%209.pdf
This chapter is part of Precalculus: An Investigation of Functions © Lippman & Rasmussen 2017. This material is licensed under a Creative Commons CC-BY-SA license. This chapter contains content remixed from work by Lara Michaels and work from OpenStax Precalculus (OpenStax.org), CC-BY 3.0. Chapter 9: Conics Section 9.1 Ellipses ..................................................................................................... 579 Section 9.2 Hyperbolas ............................................................................................... 597 Section 9.3 Parabolas and Non-Linear Systems ......................................................... 617 Section 9.4 Conics in Polar Coordinates..................................................................... 630 In this chapter, we will explore a set of shapes defined by a common characteristic: they can all be formed by slicing a cone with a plane. These families of curves have a broad range of applications in physics and astronomy, from describing the shape of your car headlight reflectors to describing the orbits of planets and comets. Section 9.1 Ellipses The National Statuary Hall1 in Washington, D.C. is an oval-shaped room called a whispering chamber because the shape makes it possible for sound to reflect from the walls in a special way. Two people standing in specific places are able to hear each other whispering even though they are far apart. To determine where they should stand, we will need to better understand ellipses. An ellipse is a type of conic section, a shape resulting from intersecting a plane with a cone and looking at the curve where they intersect. They were discovered by the Greek mathematician Menaechmus over two millennia ago. The figure below2 shows two types of conic sections. When a plane is perpendicular to the axis of the cone, the shape of the intersection is a circle. A slightly titled plane creates an oval-shaped conic section called an ellipse. 1 Photo by Gary Palmer, Flickr, CC-BY, 2 Pbroks13 ( “Conic sections with plane”, cropped to show only ellipse and circle by L Michaels, CC BY 3.0 580 Chapter 9 An ellipse can be drawn by placing two thumbtacks in a piece of cardboard then cutting a piece of string longer than the distance between the thumbtacks. Tack each end of the string to the cardboard, and trace a curve with a pencil held taught against the string. An ellipse is the set of all points where the sum of the distances from two fixed points is constant. The length of the string is the constant, and the two thumbtacks are the fixed points, called foci. Ellipse Definition and Vocabulary An ellipse is the set of all points ( ) y x Q , for which the sum of the distance to two fixed points ( ) 1 1 1 , y x F and ( ) 2 2 2 , y x F , called the foci (plural of focus), is a constant k: ( ) ( ) k F Q d F Q d = + 2 1 , , . The major axis is the line passing through the foci. Vertices are the points on the ellipse which intersect the major axis. The major axis length is the length of the line segment between the vertices. The center is the midpoint between the vertices (or the midpoint between the foci). The minor axis is the line perpendicular to the minor axis passing through the center. Minor axis endpoints are the points on the ellipse which intersect the minor axis. The minor axis endpoints are also sometimes called co-vertices. The minor axis length is the length of the line segment between minor axis endpoints. Note that which axis is major and which is minor will depend on the orientation of the ellipse. In the ellipse shown at right, the foci lie on the y axis, so that is the major axis, and the x axis is the minor axis. Because of this, the vertices are the endpoints of the ellipse on the y axis, and the minor axis endpoints (co-vertices) are the endpoints on the x axis. x y d(Q,F1) d(Q,F2) Q F1 F2 y x Vertices Minor axis endpoints Foci Major axis Minor axis Section 9.1 Ellipses 581 Ellipses Centered at the Origin From the definition above we can find an equation for an ellipse. We will find it for a ellipse centered at the origin ( ) 0 , 0 C with foci at ( ) 0 , 1 c F and ( ) 0 , 2 c F − where c > 0. Suppose ( ) y x Q , is some point on the ellipse. The distance from F1 to Q is ( ) ( ) ( ) ( ) 0 , 2 2 2 2 1 y c x y c x F Q d + − = − + − = Likewise, the distance from F2 to Q is ( ) ( ) ( ) ( ) ( ) 0 , 2 2 2 2 2 y c x y c x F Q d + + = − + − − = From the definition of the ellipse, the sum of these distances should be constant: ( ) ( ) k F Q d F Q d = + 2 1 , , so that ( ) ( ) k y c x y c x = + + + + − 2 2 2 2 If we label one of the vertices ( ) 0 , a , it should satisfy the equation above since it is a point on the ellipse. This allows us to write k in terms of a. ( ) ( ) k c a c a = + + + + − 0 0 2 2 2 2 k c a c a = + + − Since a > c, these will be positive k c a c a = + + − ) ( ) ( k a = 2 Substituting that into our equation, we will now try to rewrite the equation in a friendlier form. ( ) ( ) a y c x y c x 2 2 2 2 2 = + + + + − Move one radical ( ) ( ) 2 2 2 2 2 y c x a y c x + + − = + − Square both sides ( ) ( ) 2 2 2 2 2 2 2       + + − =       + − y c x a y c x Expand ( ) ( ) ( ) 2 2 2 2 2 2 2 4 4 y c x y c x a a y c x + + + + + − = + − Expand more ( ) 2 2 2 2 2 2 2 2 2 2 4 4 2 y c xc x y c x a a y c xc x + + + + + + − = + + − Combining like terms and isolating the radical leaves ( ) xc a y c x a 4 4 4 2 2 2 + = + + Divide by 4 ( ) xc a y c x a + = + + 2 2 2 Square both sides again ( ) ( ) 2 2 2 4 2 2 2 2 c x xc a a y c x a + + = + + Expand ( ) 2 2 2 4 2 2 2 2 2 2 c x xc a a y c xc x a + + = + + + Distribute 582 Chapter 9 2 2 2 4 2 2 2 2 2 2 2 2 2 c x xc a a y a c a xc a x a + + = + + + Combine like terms 2 2 4 2 2 2 2 2 2 c a a y a c x x a − = + − Factor common terms ( ) ( ) 2 2 2 2 2 2 2 2 c a a y a x c a − = + − Let 2 2 2 c a b − = . Since a > c, we know b > 0. Substituting 2 b for 2 2 c a − leaves 2 2 2 2 2 2 b a y a x b = + Divide both sides by 2 2b a 1 2 2 2 2 = + b y a x This is the standard equation for an ellipse. We typically swap a and b when the major axis of the ellipse is vertical. Equation of an Ellipse Centered at the Origin in Standard Form The standard form of an equation of an ellipse centered at the origin ( ) 0 , 0 C depends on whether the major axis is horizontal or vertical. The table below gives the standard equation, vertices, minor axis endpoints, foci, and graph for each. Major Axis Horizontal Vertical Standard Equation 1 2 2 2 2 = + b y a x 1 2 2 2 2 = + a y b x Vertices (−a, 0) and (a, 0) (0, −a) and (0, a) Minor Axis Endpoints (0, −b) and (0, b) (−b, 0) and (b, 0) Foci (−c, 0) and (c, 0) where 2 2 2 c a b − = (0, −c) and (0, c) where 2 2 2 c a b − = Graph x y (a,0) (-a,0) (0,b) (0,-b) (c,0) (-c,0) x y (b,0) (-b,0) (0,a) (0,-a) (0,c) (0,-c) Section 9.1 Ellipses 583 Example 1 Put the equation of the ellipse 9 9 2 2 = + y x in standard form. Find the vertices, minor axis endpoints, length of the major axis, and length of the minor axis. Sketch the graph, then check using a graphing utility. The standard equation has a 1 on the right side, so this equation can be put in standard form by dividing by 9: 1 9 1 2 2 = + y x Since the y-denominator is greater than the x-denominator, the ellipse has a vertical major axis. Comparing to the general standard form equation 1 2 2 2 2 = + a y b x , we see the value of 3 9 = = a and the value of 1 1 = = b . The vertices lie on the y-axis at (0,±a) = (0, ±3). The minor axis endpoints lie on the x-axis at (±b, 0) = (±1, 0). The length of the major axis is ( ) ( ) 6 3 2 2 = = a . The length of the minor axis is ( ) ( ) 2 1 2 2 = = b . To sketch the graph we plot the vertices and the minor axis endpoints. Then we sketch the ellipse, rounding at the vertices and the minor axis endpoints. To check on a graphing utility, we must solve the equation for y. Isolating 2 y gives us ( ) 2 2 1 9 x y − = Taking the square root of both sides we get 2 1 3 x y − ± = Under Y= on your graphing utility enter the two halves of the ellipse as 2 1 3 x y − = and 2 1 3 x y − − = . Set the window to a comparable scale to the sketch with xmin = -5, xmax = 5, ymin= -5, and ymax = 5. 584 Chapter 9 Here’s an example output on a TI-84 calculator: Sometimes we are given the equation. Sometimes we need to find the equation from a graph or other information. Example 2 Find the standard form of the equation for an ellipse centered at (0,0) with horizontal major axis length 28 and minor axis length 16. Since the center is at (0,0) and the major axis is horizontal, the ellipse equation has the standard form 1 2 2 2 2 = + b y a x . The major axis has length 28 2 = a or a = 14. The minor axis has length 16 2 = b or b = 8. Substituting gives 1 8 16 2 2 2 2 = + y x or 1 . 64 256 2 2 = + y x . Try it Now 1. Find the standard form of the equation for an ellipse with horizontal major axis length 20 and minor axis length 6. Example 3 Find the standard form of the equation for the ellipse graphed here. The center is at (0,0) and the major axis is vertical, so the standard form of the equation will be 1 2 2 2 2 = + a y b x . From the graph we can see the vertices are (0,4) and (0,-4), giving a = 4. The minor-axis endpoints are (2,0) and (-2,0), giving b = 2. The equation will be 1 4 2 2 2 2 2 = + y x or 1 16 4 2 2 = + y x . Section 9.1 Ellipses 585 Ellipses Not Centered at the Origin Not all ellipses are centered at the origin. The graph of such an ellipse is a shift of the graph centered at the origin, so the standard equation for one centered at (h, k) is slightly different. We can shift the graph right h units and up k units by replacing x with x – h and y with y – k, similar to what we did when we learned transformations. Equation of an Ellipse Centered at (h, k) in Standard Form The standard form of an equation of an ellipse centered at the point C( ) k h, depends on whether the major axis is horizontal or vertical. The table below gives the standard equation, vertices, minor axis endpoints, foci, and graph for each. Major Axis Horizontal Vertical Standard Equation ( ) ( ) 1 2 2 2 2 = − + − b k y a h x ( ) ( ) 1 2 2 2 2 = − + − a k y b h x Vertices ( h ± a, k ) (h, k ± a) Minor Axis Endpoints ( h, k ± b ) ( h ± b, k ) Foci ( h ± c, k ) where b2 = a2 – c2 (h, k ± c) where b2 = a2 – c2 Graph x y (h+a,k) (h-a,k) (h,k+b) (h,k-b) (h-c,k) (h,k) (h+c,k) x y (h+b,k) (h-b,k) (h,k+a) (h, k-a) (h,k+c) (h, k-c) (h,k) 586 Chapter 9 Example 4 Put the equation of the ellipse 33 24 4 2 2 2 − = − + + y y x x in standard form. Find the vertices, minor axis endpoints, length of the major axis, and length of the minor axis. Sketch the graph. To rewrite this in standard form, we will need to complete the square, twice. Looking at the x terms, x x 2 2 + , we like to have something of the form 2 ) ( n x + . Notice that if we were to expand this, we’d get 2 2 2 n nx x + + , so in order for the coefficient on x to match, we’ll need 1 2 ) 1 ( 2 2 + + = + x x x . However, we don’t have a +1 on the left side of the equation to allow this factoring. To accommodate this, we will add 1 to both sides of the equation, which then allows us to factor the left side as a perfect square: 1 33 24 4 1 2 2 2 + − = − + + + y y x x 32 24 4 ) 1 ( 2 2 − = − + + y y x Repeating the same approach with the y terms, first we’ll factor out the 4. ) 6 ( 4 24 4 2 2 y y y y − = − Now we want to be able to write ( ) y y 6 4 2 − as ( ) 2 2 2 2 4 ) ( 4 n ny y n y + + = + . For the coefficient of y to match, n will have to -3, giving ( ) 36 24 4 9 6 4 ) 3 ( 4 2 2 2 + − = + − = − y y y y y . To allow this factoring, we can add 36 to both sides of the equation. 36 32 36 24 4 ) 1 ( 2 2 + − = + − + + y y x ( ) 4 9 6 4 ) 1 ( 2 2 = + − + + y y x ( ) 4 3 4 ) 1 ( 2 2 = − + + y x Dividing by 4 gives the standard form of the equation for the ellipse ( ) ( ) 1 1 3 4 1 2 2 = − + + y x Since the x-denominator is greater than the y-denominator, the ellipse has a horizontal major axis. From the general standard equation ( ) ( ) 1 2 2 2 2 = − + − b k h a h x we see the value of 2 4 = = a and the value of 1 1 = = b . The center is at (h, k) = (-1, 3). The vertices are at (h±a, k) or (-3, 3) and (1,3). The minor axis endpoints are at (h, k±b) or (-1, 2) and (-1,4). Section 9.1 Ellipses 587 The length of the major axis is ( ) ( ) 4 2 2 2 = = a . The length of the minor axis is ( ) ( ) 2 1 2 2 = = b . To sketch the graph we plot the vertices and the minor axis endpoints. Then we sketch the ellipse, rounding at the vertices and the minor axis endpoints. Example 5 Find the standard form of the equation for an ellipse centered at (-2,1), a vertex at (-2,4) and passing through the point (0,1). The center at (-2,1) and vertex at (-2,4) means the major axis is vertical since the x-values are the same. The ellipse equation has the standard form ( ) ( ) 1 2 2 2 2 = − + − a k y b h x . The value of a = 4-1=3. Substituting a = 3, h = -2, and k = 1 gives ( ) ( ) 1 3 1 2 2 2 2 2 = − + + y b x . Substituting for x and y using the point (0,1) gives ( ) ( ) 1 3 1 1 2 0 2 2 2 2 = − + + b . Solving for b gives b=2. The equation of the ellipse in standard form is ( ) ( ) 1 3 1 2 2 2 2 2 2 = − + + y x or ( ) ( ) 1 9 1 4 2 2 2 = − + + y x . Try it Now 2. Find the center, vertices, minor axis endpoints, length of the major axis, and length of the minor axis for the ellipse ( ) ( ) 1 4 2 4 2 2 = + + − y x . 588 Chapter 9 Bridges with Semielliptical Arches Arches have been used to build bridges for centuries, like in the Skerton Bridge in England which uses five semielliptical arches for support3. Semielliptical arches can have engineering benefits such as allowing for longer spans between supports. Example 6 A bridge over a river is supported by a single semielliptical arch. The river is 50 feet wide. At the center, the arch rises 20 feet above the river. The roadway is 4 feet above the center of the arch. What is the vertical distance between the roadway and the arch 15 feet from the center? Put the center of the ellipse at (0,0) and make the span of the river the major axis. Since the major axis is horizontal, the equation has the form 1 2 2 2 2 = + b y a x . The value of 25 ) 50 ( 2 1 = = a and the value of b = 20, giving 1 15 25 2 2 2 2 = + y x . Substituting x = 15 gives 1 20 25 15 2 2 2 2 = + y . Solving for y, 16 625 225 1 20 = − = y . The roadway is 20 + 4 = 24 feet above the river. The vertical distance between the roadway and the arch 15 feet from the center is 24 − 16 = 8 feet. 3 Maxine Armstrong ( “Skerton Bridge, Lancaster, England”, CC BY-SA x y 50ft 20ft 4ft Section 9.1 Ellipses 589 Ellipse Foci The location of the foci can play a key role in ellipse application problems. Standing on a focus in a whispering gallery allows you to hear someone whispering at the other focus. To find the foci, we need to find the length from the center to the foci, c, using the equation 2 2 2 c a b − = . It looks similar to, but is not the same as, the Pythagorean Theorem. Example 7 The National Statuary Hall whispering chamber is an elliptical room 46 feet wide and 96 feet long. To hear each other whispering, two people need to stand at the foci of the ellipse. Where should they stand? We could represent the hall with a horizontal ellipse centered at the origin. The major axis length would be 96 feet, so 48 ) 96 ( 2 1 = = a , and the minor axis length would be 46 feet, so 23 ) 46 ( 2 1 = = b . To find the foci, we can use the equation 2 2 2 c a b − = . 2 2 2 48 23 c − = 2 2 2 23 48 − = c 42 1775 ± ≈ = c ft. To hear each other whisper, two people would need to stand 2(42) = 84 feet apart along the major axis, each about 48 – 42 = 6 feet from the wall. Example 8 Find the foci of the ellipse ( ) ( ) 1 29 3 4 2 2 2 = + + − y x . The ellipse is vertical with an equation of the form ( ) ( ) 1 2 2 2 2 = − + − a k y b h x . The center is at (h, k) = (2, −3). The foci are at (h, k ± c). To find length c we use 2 2 2 c a b − = . Substituting gives 2 29 4 c − = or 5 25 = = c . The ellipse has foci (2, −3 ± 5), or (2, −8) and (2, 2). 590 Chapter 9 Example 9 Find the standard form of the equation for an ellipse with foci (-1,4) and (3,4) and major axis length 10. Since the foci differ in the x -coordinates, the ellipse is horizontal with an equation of the form ( ) ( ) 1 2 2 2 2 = − + − b k h a h x . The center is at the midpoint of the foci ( ) ( ) 4 , 1 2 4 4 , 2 3 1 2 , 2 2 1 2 1 =       + + − =       + + y y x x . The value of a is half the major axis length: 5 ) 10 ( 2 1 = = a . The value of c is half the distance between the foci: 2 ) 4 ( 2 1 )) 1 ( 3 ( 2 1 = = − − = c . To find length b we use 2 2 2 c a b − = . Substituting a and c gives 2 2 2 2 5 − = b = 21. The equation of the ellipse in standard form is ( ) ( ) 1 21 4 5 1 2 2 2 = − + − y x or ( ) ( ) 1 21 4 25 1 2 2 = − + − y x . Try it Now 3. Find the standard form of the equation for an ellipse with focus (2,4), vertex (2,6), and center (2,1). Planetary Orbits It was long thought that planetary orbits around the sun were circular. Around 1600, Johannes Kepler discovered they were actually elliptical4. His first law of planetary motion says that planets travel around the sun in an elliptical orbit with the sun as one of the foci. The length of the major axis can be found by measuring the planet’s aphelion, its greatest distance from the sun, and perihelion, its shortest distance from the sun, and summing them together. 4 Technically, they’re approximately elliptical. The orbits of the planets are not exactly elliptical because of interactions with each other and other celestial bodies. Section 9.1 Ellipses 591 Example 10 Mercury’s aphelion is 35.98 million miles and its perihelion is 28.58 million miles. Write an equation for Mercury’s orbit. Let the center of the ellipse be (0,0) and its major axis be horizontal so the equation will have form 1 2 2 2 2 = + b y a x . The length of the major axis is 56 . 64 58 . 28 98 . 35 2 = + = a giving 28 . 32 = a and 9984 . 1041 2 = a . Since the perihelion is the distance from the focus to one vertex, we can find the distance between the foci by subtracting twice the perihelion from the major axis length: ( ) 4 . 7 58 . 28 2 56 . 64 2 = − = c giving 7 . 3 = c . Substitution of a and c into 2 2 2 c a b − = yields 3084 . 1028 7 . 3 28 . 32 2 2 2 = − = b . The equation is 1 3084 . 1028 9984 . 1041 2 2 = + y x . Important Topics of This Section Ellipse Definition Ellipse Equations in Standard Form Ellipse Foci Applications of Ellipses Try it Now Answers 1. 2a = 20, so a =10. 2b = 6, so b = 3. 1 9 100 2 2 = + y x 2. Center (4, -2). Vertical ellipse with a = 2, b = 1. Vertices at (4, -2±2) = (4,0) and (4,-4), minor axis endpoints at (4±1, -2) = (3,-2) and (5,-2), major axis length 4, minor axis length 2 3. Vertex, center, and focus have the same x-value, so it’s a vertical ellipse. Using the vertex and center, a = 6 – 1 = 5 Using the center and focus, c = 4 – 1 = 3 2 2 2 3 5 − = b . b = 4. ( ) ( ) 1 25 1 16 2 2 2 = − + − y x 592 Chapter 9 Section 9.1 Exercises In problems 1–4, match each graph with one of the equations A–D. A. 1 9 4 2 2 = + y x B. 1 4 9 2 2 = + y x C. 1 9 2 2 = + y x D. 1 9 2 2 = + y x 1. 2. 3. 4. In problems 5–14, find the vertices, the minor axis endpoints, length of the major axis, and length of the minor axis. Sketch the graph. Check using a graphing utility. 5. 1 25 4 2 2 = + y x 6. 1 4 16 2 2 = + y x 7. 1 4 2 2 = + y x 8. 1 25 2 2 = + y x 9. 25 25 2 2 = + y x 10. 16 16 2 2 = + y x 11. 144 9 16 2 2 = + y x 12. 400 25 16 2 2 = + y x 13. 18 9 2 2 = + y x 14. 12 4 2 2 = + y x In problems 15–16, write an equation for the graph. 15. 16. In problems 17–20, find the standard form of the equation for an ellipse satisfying the given conditions. 17. Center (0,0), horizontal major axis length 64, minor axis length 14 18. Center (0,0), vertical major axis length 36, minor axis length 18 19. Center (0,0), vertex (0,3), 2 = b 20. Center (0,0), vertex (4,0), 3 = b Section 9.1 Ellipses 593 In problems 21–28, match each graph to equations A-H. A. ( ) 1 9 ) 1 ( 4 2 2 2 = − + − y x E. ( ) 1 9 ) 1 ( 4 2 2 2 = + + + y x B. ( ) 1 16 ) 1 ( 4 2 2 2 = − + − y x F. ( ) 1 16 ) 1 ( 4 2 2 2 = + + + y x C. ( ) 1 4 ) 1 ( 16 2 2 2 = − + − y x G. ( ) 1 4 ) 1 ( 16 2 2 2 = + + + y x D. ( ) 1 4 ) 1 ( 9 2 2 2 = − + − y x H. ( ) 1 4 ) 1 ( 9 2 2 2 = + + + y x 21. 22. 23. 24. 25. 26. 27. 28. In problems 29–38, find the vertices, the minor axis endpoints, length of the major axis, and length of the minor axis. Sketch the graph. Check using a graphing utility. 29. 1 4 ) 2 ( 25 ) 1 ( 2 2 = + + − y x 30. 1 36 ) 3 ( 16 ) 5 ( 2 2 = − + + y x 31. 1 25 ) 3 ( ) 2 ( 2 2 = − + + y x 32. 1 ) 6 ( 25 ) 1 ( 2 2 = − + − y x 33. 16 4 8 4 2 2 = + + + y x x 34. 36 16 16 4 2 2 = + + + y y x 35. 1 16 4 2 2 2 − = + + + y y x x 36. 4 8 16 4 2 2 = − + + y y x x 37. 104 8 4 36 9 2 2 = + + − y y x x 38. 4 36 9 8 4 2 2 − = + + + y y x x 594 Chapter 9 In problems 39–40, write an equation for the graph. 39. 40. In problems 41–42, find the standard form of the equation for an ellipse satisfying the given conditions. 41. Center (-4,3), vertex(-4,8), point on the graph (0,3) 42. Center (1,-2), vertex(-5,-2), point on the graph (1,0) 43. Window A window in the shape of a semiellipse is 12 feet wide and 4 feet high. What is the height of the window above the base 5 feet from the center ? 44. Window A window in the shape of a semiellipse is 16 feet wide and 7 feet high. What is the height of the window above the base 4 feet from the center? 45. Bridge A bridge over a river is supported by a semielliptical arch. The river is 150 feet wide. At the center, the arch rises 60 feet above the river. The roadway is 5 feet above the center of the arch. What is the vertical distance between the roadway and the arch 45 feet from the center? 46. Bridge A bridge over a river is supported by a semielliptical arch. The river is 1250 feet wide. At the center, the arch rises 175 feet above the river. The roadway is 3 feet above the center of the arch. What is the vertical distance between the roadway and the arch 600 feet from the center? 47. Racetrack An elliptical racetrack is 100 feet long and 90 feet wide. What is the width of the racetrack 20 feet from a vertex on the major axis? 48. Racetrack An elliptical racetrack is 250 feet long and 150 feet wide. What is the width of the racetrack 25 feet from a vertex on the major axis? Section 9.1 Ellipses 595 In problems 49-52, find the foci. 49. 1 3 19 2 2 = + y x 50. 1 38 2 2 2 = + y x 51. 1 26 ) 1 ( ) 6 ( 2 2 = + − + + y x 52. 1 ) 5 ( 10 ) 3 ( 2 2 = + + − y x In problems 53-72, find the standard form of the equation for an ellipse satisfying the given conditions. 53. Major axis vertices (±3,0), c=2 54. Major axis vertices (0,±7), c=4 55. Foci (0,±5) and major axis length 12 56. Foci (±3,0) and major axis length 8 57. Foci (±5,0), vertices (±7,0) 58. Foci (0,±2), vertices (0,±3) 59. Foci (0,±4) and x-intercepts (±2,0) 60. Foci (±3,0) and y-intercepts (0,±1) 61. Center (0,0), major axis length 8, foci on x-axis, passes through point ( ) 6 , 2 62. Center (0,0), major axis length 12, foci on y-axis, passes through point ( ) 4 , 10 63. Center (-2,1), vertex (-2,5), focus (-2,3) 64. Center (-1,-3), vertex (-7,-3), focus (-4,-3) 65. Foci (8,2) and (-2,2), major axis length 12 66. Foci (-1,5) and (-1,-3), major axis length 14 67. Vertices (3,4) and (3,-6), c= 2 68. Vertices (2,2) and (-4,2), c= 2 69. Center (1,3), focus (0,3), passes through point (1,5) 70. Center (-1,-2), focus (1,-2), passes through point (2,-2) 71. Focus (-15,-1), vertices (-19,-1) and (15,-1) 72. Focus (-3,2), vertices (-3,4) and (-3,-8) 596 Chapter 9 73. Whispering Gallery If an elliptical whispering gallery is 80 feet long and 25 feet wide, how far from the center of room should someone stand on the major axis of the ellipse to experience the whispering effect? Round to two decimal places. 74. Billiards Some billiards tables are elliptical and have the foci marked on the table. If such a one is 8 feet long and 6 feet wide, how far are the foci from the center of the ellipse? Round to two decimal places. 75. Planetary Orbits The orbits of planets around the sun are approximately elliptical with the sun as a focus. The aphelion is a planet’s greatest distance from the sun and the perihelion is its shortest. The length of the major axis is the sum of the aphelion and the perihelion. Earth’s aphelion is 94.51 million miles and its perihelion is 91.40 million miles. Write an equation for Earth’s orbit. 76. Satellite Orbits The orbit of a satellite around Earth is elliptical with Earth’s center as a focus. The satellite’s maximum height above the Earth is 170 miles and its minimum height above the Earth is 90 miles. Write an equation for the satellite’s orbit. Assume Earth is spherical and has a radius of 3960 miles. 77. Eccentricity e of an ellipse is the ratio a c where c is the distance of a focus from the center and a is the distance of a vertex from the center. Write an equation for an ellipse with eccentricity 0.8 and foci at (-4,0) and (4,0). 78. Confocal ellipses have the same foci. Show that, for k > 0, all ellipses of the form 1 6 2 2 = + + k y k x are confocal. 79. The latus rectum of an ellipse is a line segment with endpoints on the ellipse that passes through a focus and is perpendicular to the major axis. Show that a b2 2 is the length of the latus rectum of 1 2 2 2 2 = + b y a x where a > b. Section 9.2 Hyperbolas 597 Section 9.2 Hyperbolas In the last section, we learned that planets have approximately elliptical orbits around the sun. When an object like a comet is moving quickly, it is able to escape the gravitational pull of the sun and follows a path with the shape of a hyperbola. Hyperbolas are curves that can help us find the location of a ship, describe the shape of cooling towers, or calibrate seismological equipment. The hyperbola is another type of conic section created by intersecting a plane with a double cone, as shown below5. The word “hyperbola” derives from a Greek word meaning “excess.” The English word “hyperbole” means exaggeration. We can think of a hyperbola as an excessive or exaggerated ellipse, one turned inside out. We defined an ellipse as the set of all points where the sum of the distances from that point to two fixed points is a constant. A hyperbola is the set of all points where the absolute value of the difference of the distances from the point to two fixed points is a constant. 5 Pbroks13 ( “Conic sections with plane”, cropped to show only a hyperbola by L Michaels, CC BY 3.0 598 Chapter 9 Hyperbola Definition A hyperbola is the set of all points ( ) y x Q , for which the absolute value of the difference of the distances to two fixed points ( ) 1 1 1 , y x F and ( ) 2 2 2 , y x F called the foci (plural for focus) is a constant k: ( ) ( ) k F Q d F Q d = − 2 1 , , . The transverse axis is the line passing through the foci. Vertices are the points on the hyperbola which intersect the transverse axis. The transverse axis length is the length of the line segment between the vertices. The center is the midpoint between the vertices (or the midpoint between the foci). The other axis of symmetry through the center is the conjugate axis. The two disjoint pieces of the curve are called branches. A hyperbola has two asymptotes. Which axis is the transverse axis will depend on the orientation of the hyperbola. As a helpful tool for graphing hyperbolas, it is common to draw a central rectangle as a guide. This is a rectangle drawn around the center with sides parallel to the coordinate axes that pass through each vertex and co-vertex. The asymptotes will follow the diagonals of this rectangle. x y d(Q,F1) d(Q,F2) Q F1 F2 x y d(Q,F2) d(Q,F1) Q F2 F1 x y Vertex Focus Asymptote Center Co-vertex Transverse axis Conjugate axis Section 9.2 Hyperbolas 599 Hyperbolas Centered at the Origin From the definition above we can find an equation of a hyperbola. We will find it for a hyperbola centered at the origin ( ) 0 , 0 C opening horizontally with foci at ( ) 0 , 1 c F and ( ) 0 , 2 c F − where c > 0. Suppose ( ) y x Q , is a point on the hyperbola. The distances from Q to F1 and Q to F2 are: ( ) ( ) ( ) ( ) 0 , 2 2 2 2 1 y c x y c x F Q d + − = − + − = ( ) ( ) ( ) ( ) ( ) 0 , 2 2 2 2 2 y c x y c x F Q d + + = − + − − = . From the definition, the absolute value of the difference should be constant: ( ) ( ) ( ) ( ) k y c x y c x F Q d F Q d = + + − + − = − , , 2 2 2 2 2 1 Substituting in one of the vertices ( ) 0 , a , we can determine k in terms of a: ( ) ( ) k c a c a = + + − + − 0 0 2 2 2 2 k c a c a = + − − Since c > a, a c c a − = − k c a a c = + − − ) ( ) ( a a k 2 2 = − = Using a k 2 = and removing the absolute values, ( ) ( ) a y c x y c x 2 2 2 2 2 ± = + + − + − Move one radical ( ) ( ) 2 2 2 2 2 y c x a y c x + + + ± = + − Square both sides ( ) ( ) ( ) 2 2 2 2 2 2 2 4 4 y c x y c x a a y c x + + + + + ± = + − Expand ( ) 2 2 2 2 2 2 2 2 2 2 4 4 2 y c xc x y c x a a y c xc x + + + + + + ± = + + − Combining like terms leaves ( ) 4 4 4 2 2 2 y c x a a xc + + ± = − Divide by 4 ( ) 2 2 2 y c x a a xc + + ± = − Isolate the radical ( ) xc a y c x a − − = + + ± 2 2 2 Square both sides again ( ) ( ) 2 2 2 4 2 2 2 2 c x xc a a y c x a + + = + + Expand and distribute 2 2 2 4 2 2 2 2 2 2 2 2 2 c x xc a a y a c a xc a x a + + = + + + Combine like terms 2 2 2 2 4 2 2 2 2 x a c x a c a y a − = − + Factor common terms ( ) ( ) 2 2 2 2 2 2 2 2 x a c a c a y a − = − + 600 Chapter 9 Let 2 2 2 a c b − = . Since c > a, b > 0. Substituting 2 b for 2 2 a c − leaves 2 2 2 2 2 2 x b b a y a = + Divide both sides by 2 2b a 2 2 2 2 1 a x b y = + Rewrite 1 2 2 2 2 = −b y a x We can see from the graphs of the hyperbolas that the branches appear to approach asymptotes as x gets large in the negative or positive direction. The equations of the horizontal hyperbola asymptotes can be derived from its standard equation. 1 2 2 2 2 = −b y a x Solve for y         − = 1 2 2 2 2 a x b y Rewrite 1 as 2 2 2 2 x a a x         − = 2 2 2 2 2 2 2 2 x a a x a x b y Factor out 2 2 a x        − = 2 2 2 2 2 2 1 x a a x b y Take the square root 2 2 1 x a x a b y − ± = As x → ±∞ the quantity 2 2 x a → 0 and 2 2 1 x a − → 1, so the asymptotes are x a b y ± = . Similarly, for vertical hyperbolas the asymptotes are x b a y ± = . The standard form of an equation of a hyperbola centered at the origin C( ) 0 , 0 depends on whether it opens horizontally or vertically. The following table gives the standard equation, vertices, foci, asymptotes, construction rectangle vertices, and graph for each. Section 9.2 Hyperbolas 601 Equation of a Hyperbola Centered at the Origin in Standard Form Example 1 Put the equation of the hyperbola 4 4 2 2 = −x y in standard form. Find the vertices, length of the transverse axis, and the equations of the asymptotes. Sketch the graph. Check using a graphing utility. The equation can be put in standard form 1 1 4 2 2 = −x y by dividing by 4. Comparing to the general standard equation 1 2 2 2 2 = −b x a y we see that 2 4 = = a and 1 1 = = b . Since the x term is subtracted, the hyperbola opens vertically and the vertices lie on the y-axis at (0,±a) = (0, ±2). Opens Horizontally Vertically Standard Equation 1 2 2 2 2 = −b y a x 1 2 2 2 2 = −b x a y Vertices (-a, 0) and (a, 0) (0, -a) and (0, a) Foci (-c, 0) and (c, 0) where 2 2 2 a c b − = (0, -c) and (0, c) Where 2 2 2 a c b − = Asymptotes x a b y ± = x b a y ± = Construction Rectangle Vertices (a, b), (-a, b), ( a,-b), (-a, -b) (b, a), (-b, a), (b, -a), (-b, -a) Graph x y (0,b) (-c,0) (0,-b) (c,0) (a,0) (-a,0) x y (0,a) (0,c) (-b,0) (0,-a) (0,-c) (b,0) 602 Chapter 9 The length of the transverse axis is ( ) ( ) 4 2 2 2 = = a . Equations of the asymptotes are x b a y ± = or x 2 ± = . To sketch the graph we plot the vertices of the construction rectangle at (±b,±a) or (-1,-2), (-1,2), (1,-2), and (1,2). The asymptotes are drawn through the diagonals of the rectangle and the vertices plotted. Then we sketch in the hyperbola, rounded at the vertices and approaching the asymptotes. To check on a graphing utility, we must solve the equation for y. Isolating y2 gives us ( ) 2 2 1 4 x y + = . Taking the square root of both sides we find 2 1 2 x y + ± = . Under Y= enter the two halves of the hyperbola and the two asymptotes as 2 1 2 x y + = , 2 1 2 x y + − = , x y 2 = , and x y 2 − = . Set the window to a comparable scale to the sketch with xmin = -4, xmax = 4, ymin= -3, and ymax = 3. Sometimes we are given the equation. Sometimes we need to find the equation from a graph or other information. Section 9.2 Hyperbolas 603 Example 2 Find the standard form of the equation for a hyperbola with vertices at (-6,0) and (6,0) and asymptote x y 3 4 = . Since the vertices lie on the x-axis with a midpoint at the origin, the hyperbola is horizontal with an equation of the form 1 2 2 2 2 = −b y a x . The value of a is the distance from the center to a vertex. The distance from (6,0) to (0,0) is 6, so a = 6. The asymptotes follow the form x a b y ± = . From x y 4 3 = we see a b = 4 3 and substituting a = 6 give us 6 4 3 b = . Solving yields b = 8. The equation of the hyperbola in standard form is 1 8 6 2 2 2 2 = −y x or 1 64 36 2 2 = −y x . Try it Now 1. Find the standard form of the equation for a hyperbola with vertices at (0,-8) and (0,8) and asymptote x y 2 = Example 3 Find the standard form of the equation for a hyperbola with vertices at (0, 9) and (0,-9) and passing through the point (8,15). Since the vertices lie on the y-axis with a midpoint at the origin, the hyperbola is vertical with an equation of the form 1 2 2 2 2 = −b x a y . The value of a is the distance from the center to a vertex. The distance from (0,9) to (0,0) is 9, so a = 9. Substituting a = 9 and the point (8,15) gives 1 8 9 15 2 2 2 2 = −b . Solving for b yeilds ( ) 6 9 15 8 9 2 2 2 2 = − = b . The standard equation for the hyperbola is 1 6 9 2 2 2 2 = −x y or 1 36 81 2 2 = −x y . 604 Chapter 9 Hyperbolas Not Centered at the Origin Not all hyperbolas are centered at the origin. The standard equation for one centered at (h, k) is slightly different. Equation of a Hyperbola Centered at (h, k) in Standard Form The standard form of an equation of a hyperbola centered at C( ) k h, depends on whether it opens horizontally or vertically. The table below gives the standard equation, vertices, foci, asymptotes, construction rectangle vertices, and graph for each. Opens Horizontally Vertically Standard Equation ( ) ( ) 1 2 2 2 2 = − − − b k y a h x ( ) ( ) 1 2 2 2 2 = − − − b h x a k y Vertices ( h ± a, k ) (h, k ± a) Foci ( h ± c, k ) where b2 = c2 – a2 (h, k ± c) where b2 = c2 – a2 Asymptotes ( ) h x a b k y − ± = − ( ) h x b a k y − ± = − Construction Rectangle Vertices ( h ± a, k ± b ) ( h ± b, k ± a ) Graph x y (h,k+b) (h-c,k) (h+a,k) (h-a,k) (h,k) (h,k-b) (h+c,k) x y (h-b,k) (h,k-a) (h+b,k) (h,k) (h,k-c) (h,k+c) (h,k+a) Section 9.2 Hyperbolas 605 Example 4 Write an equation for the hyperbola in the graph shown. The center is at (2,3), where the asymptotes cross. It opens vertically, so the equation will look like ( ) ( ) 1 2 3 2 2 2 2 = − − − b x a y . The vertices are at (2,2) and (2,4). The distance from the center to a vertex is 1 3 4 = − = a . If we were to draw in the construction rectangle, it would extend from x = -1 to x = 5. The distance from the center to the right side of the rectangle gives 3 2 5 = − = b . The standard equation of this hyperbola is ( ) ( ) 1 3 2 1 3 2 2 2 2 = − − − x y , or ( ) ( ) 1 9 2 3 2 2 = − − − x y . Example 5 Put the equation of the hyperbola 43 16 4 18 9 2 2 = + − + y y x x in standard form. Find the center, vertices, length of the transverse axis, and the equations of the asymptotes. Sketch the graph, then check on a graphing utility. To rewrite the equation, we complete the square for both variables to get ( ) ( ) 16 9 43 4 4 4 1 2 9 2 2 − + = + − − + + y y x x ( ) ( ) 36 2 4 1 9 2 2 = − − + y x Dividing by 36 gives the standard form of the equation, ( ) ( ) 1 9 2 4 1 2 2 = − − + y x Comparing to the general standard equation ( ) ( ) 1 2 2 2 2 = − − − b k h a h x we see that 2 4 = = a and 3 9 = = b . Since the y term is subtracted, the hyperbola opens horizontally. The center is at (h, k) = (-1, 2). The vertices are at (h±a, k) or (-3, 2) and (1,2). The length of the transverse axis is ( ) ( ) 4 2 2 2 = = a . Equations of the asymptotes are ( ) h x a b k y − ± = − or ( ) 1 2 3 2 + ± = − x y . 606 Chapter 9 To sketch the graph we plot the corners of the construction rectangle at (h±a, k±b) or (1, 5), (1, -1), (-3,5), and (-3,-1). The asymptotes are drawn through the diagonals of the rectangle and the vertices plotted. Then we sketch in the hyperbola rounded at the vertices and approaching the asymptotes. To check on a graphing utility, we must solve the equation for y. ( )         − + ± = 1 4 1 9 2 2 x y . Under Y= enter the two halves of the hyperbola and the two asymptotes as ( )         − + + = 1 4 1 9 2 2 x y , ( )         − + − = 1 4 1 9 2 2 x y , ( ) 2 1 2 3 + + = x y , and ( ) 2 1 2 3 + + − = x y . Set the window to a comparable scale to the sketch, then graph. Note that the gaps you see on the calculator are not really there; they’re a limitation of the technology. Example 6 Find the standard form of the equation for a hyperbola with vertices at ) 5 , 2 ( − − and ) 7 , 2 (− , and asymptote 4 2 3 + = x y . Section 9.2 Hyperbolas 607 Since the vertices differ in the y -coordinates, the hyperbola opens vertically with an equation of the form ( ) ( ) 1 2 2 2 2 = − − − b h x a k y and asymptote equations of the form ( ) h x b a k y − ± = − . The center will be halfway between the vertices, at ) 1 , 2 ( 2 7 5 , 2 − =       + − − . The value of a is the distance from the center to a vertex. The distance from ) 1 , 2 (− to ) 5 , 2 ( − − is 6, so a = 6. While our asymptote is not given in the form ( ) h x b a k y − ± = − , notice this equation would have slope b a . We can compare that to the slope of the given asymptote equation to find b. Setting b a = 2 3 and substituting a = 6 gives us b = 4. The equation of the hyperbola in standard form is ( ) ( ) 1 4 2 6 1 2 2 2 2 = + − − x y or ( ) ( ) 1 16 2 36 1 2 2 = + − − x y . Try it Now 2. Find the center, vertices, length of the transverse axis, and equations of the asymptotes for the hyperbola ( ) ( ) 1 36 2 9 5 2 2 = − − + y x . Hyperbola Foci The location of the foci can play a key role in hyperbola application problems. To find them, we need to find the length from the center to the foci, c, using the equation 2 2 2 a c b − = . It looks similar to, but is not the same as, the Pythagorean Theorem. Compare this with the equation to find length c for ellipses, which is 2 2 2 c a b − = . If you remember that for the foci to be inside the ellipse they have to come before the vertices ) ( a c < , it’s clear why we would calculate 2 a minus 2 c . To be inside a hyperbola, the foci have to go beyond the vertices ) ( a c > , so we can see for hyperbolas we need 2 c minus 2 a , the opposite. 608 Chapter 9 Example 7 Find the foci of the hyperbola ( ) ( ) 1 5 3 4 1 2 2 = − − + x y . The hyperbola is vertical with an equation of the form ( ) ( ) 1 2 2 2 2 = − − − b h x a k y . The center is at (h, k) = (3, -1). The foci are at (h, k ± c). To find length c we use 2 2 2 a c b − = . Substituting gives 4 5 2 − = c or 3 9 = = c . The hyperbola has foci (3, -4) and (3, 2). Example 8 Find the standard form of the equation for a hyperbola with foci (5, -8) and (-3, -8) and vertices (4, -8) and (-2, -8). Since the vertices differ in the x -coordinates, the hyperbola opens horizontally with an equation of the form ( ) ( ) 1 2 2 2 2 = − − − b k y a h x . The center is at the midpoint of the vertices ( ) ( ) ( ) 8 , 1 2 8 8 , 2 2 4 2 , 2 2 1 2 1 − =       − + − − + =       + + y y x x . The value of a is the horizontal length from the center to a vertex, or 3 1 4 = − = a . The value of c is the horizontal length from the center to a focus, or 4 1 5 = − = . To find length b we use 2 2 2 a c b − = . Substituting gives 7 9 16 2 = − = b . The equation of the hyperbola in standard form is ( ) ( ) ( ) 1 7 8 3 1 2 2 2 = − − − − y x or ( ) ( ) 1 7 8 9 1 2 2 = + − − y x . Try it Now 3. Find the standard form of the equation for a hyperbola with focus (1,9), vertex (1,8), center (1,4). Section 9.2 Hyperbolas 609 LORAN Before GPS, the Long Range Navigation (LORAN) system was used to determine a ship’s location. Two radio stations A and B simultaneously sent out a signal to a ship. The difference in time it took to receive the signal was computed as a distance locating the ship on the hyperbola with the A and B radio stations as the foci. A second pair of radio stations C and D sent simultaneous signals to the ship and computed its location on the hyperbola with C and D as the foci. The point P where the two hyperbolas intersected gave the location of the ship. Example 9 Stations A and B are 150 kilometers apart and send a simultaneous radio signal to the ship. The signal from B arrives 0.0003 seconds before the signal from A. If the signal travels 300,000 kilometers per second, find the equation of the hyperbola on which the ship is positioned. Stations A and B are at the foci, so the distance from the center to one focus is half the distance between them, giving 75 ) 150 ( 2 1 = = c km. By letting the center of the hyperbola be at (0,0) and placing the foci at (±75,0), the equation 1 2 2 2 2 = −b y a x for a hyperbola centered at the origin can be used. The difference of the distances of the ship from the two stations is km 90 ) s 0003 . 0 ( s km 000 , 300 = ⋅ = k . From our derivation of the hyperbola equation we determined k = 2a, so 45 ) 90 ( 2 1 = = a . Substituting a and c into 2 2 2 a c b − = yields 3600 45 75 2 2 2 = − = b . The equation of the hyperbola in standard form is 1 3600 45 2 2 2 = − y x or 1 3600 2025 2 2 = − y x . To determine the position of a ship using LORAN, we would need an equation for the second hyperbola and would solve for the intersection. We will explore how to do that in the next section. A B C D P 610 Chapter 9 Important Topics of This Section Hyperbola Definition Hyperbola Equations in Standard Form Hyperbola Foci Applications of Hyperbolas Intersections of Hyperbolas and Other Curves Try it Now Answers 1. The vertices are on the y axis so this is a vertical hyperbola. The center is at the origin. a = 8 Using the asymptote slope, 2 8 = b , so b = 4. 1 16 64 2 2 = −x y 2. Center (-5, 2). This is a horizontal hyperbola. a = 3. b = 6. transverse axis length 6, Vertices will be at (-5±3,2) = (-2,2) and (-8,2), Asymptote slope will be 2 3 6 = . Asymptotes: ( ) 5 2 2 + ± = − x y 3. Focus, vertex, and center have the same x value so this is a vertical hyperbola. Using the vertex and center, a = 9 – 4 = 5 Using the focus and center, c = 8 – 4 = 4 2 2 2 4 5 − = b . b = 3. ( ) ( ) 1 9 1 16 4 2 2 = − − − x y Section 9.2 Hyperbolas 611 Section 9.2 Exercises In problems 1–4, match each graph to equations A–D. A. 1 9 4 2 2 = −y x B. 1 4 9 2 2 = −y x C. 1 9 2 2 = −x y D. 1 9 2 2 = −x y 1. 2. 3. 4. In problems 5–14, find the vertices, length of the transverse axis, and equations of the asymptotes. Sketch the graph. Check using a graphing utility. 5. 1 25 4 2 2 = −y x 6. 1 9 16 2 2 = −x y 7. 1 4 2 2 = −x y 8. 1 25 2 2 = −y x 9. 9 9 2 2 = −y x 10. 4 4 2 2 = −x y 11. 144 16 9 2 2 = − x y 12. 400 25 16 2 2 = − y x 13. 18 9 2 2 = −y x 14. 12 4 2 2 = −x y In problems 15–16, write an equation for the graph. 15. 16. 612 Chapter 9 In problems 17–22, find the standard form of the equation for a hyperbola satisfying the given conditions. 17. Vertices at (0,4) and (0, -4); asymptote x y 2 1 = 18. Vertices at (-6,0) and (6,0); asymptote x y 3 = 19. Vertices at (-3,0) and (3,0); passes through (5,8) 20. Vertices at (0, 4) and (0, -4); passes through (6, 5) 21. Asymptote y = x; passes through (5, 3) 22. Asymptote y = x; passes through (12, 13) In problems 23–30, match each graph to equations A–H. A. ( ) ( ) 1 4 2 9 1 2 2 = − − − y x E. ( ) ( ) 1 9 1 4 2 2 2 = − − − x y B. ( ) ( ) 1 4 2 9 1 2 2 = + − + y x F. ( ) ( ) 1 9 1 4 2 2 2 = + − + x y C. ( ) ( ) 1 16 2 9 1 2 2 = + − + y x G. ( ) ( ) 1 16 1 4 2 2 2 = + − + x y D. ( ) ( ) 1 16 2 9 1 2 2 = − − − y x H. ( ) ( ) 1 16 1 4 2 2 2 = − − − x y 23. 24. 25. 26. 27. 28. 29. 30. Section 9.2 Hyperbolas 613 In problems 31–40, find the center, vertices, length of the transverse axis, and equations of the asymptotes. Sketch the graph. Check using a graphing utility. 31. ( ) ( ) 1 4 2 25 1 2 2 = + − − y x 32. ( ) ( ) 1 36 5 16 3 2 2 = + − − x y 33. ( ) ( ) 1 2 9 1 2 2 = + − − x y 34. ( ) ( ) 1 6 25 1 2 2 = − − − y x 35. 12 8 4 2 2 = − − y x x 36. 20 9 16 4 2 2 = − + x y y 37. 1 2 16 4 2 2 = − − − x x y y 38. 29 6 16 4 2 2 = + − − y y x x 39. 4 8 4 36 9 2 2 = + − + y y x x 40. 36 96 16 36 9 2 2 − = − − + x x y y In problems 41–42, write an equation for the graph. 41. 42. In problems 43–44, find the standard form of the equation for a hyperbola satisfying the given conditions. 43. Vertices (-1,-2) and (-1,6); asymptote ( ) 1 2 2 + = − x y 44. Vertices (-3,-3) and (5,-3); asymptote ( ) 1 2 1 3 − = + x y In problems 45–48, find the center, vertices, length of the transverse axis, and equations of the asymptotes. Sketch the graph. Check using a graphing utility. 45. 1 9 4 2 − ± = x y 46. 1 9 4 1 2 + ± = x y 47. 10 18 9 2 1 1 2 + + ± = x x y 48. 8 18 9 2 1 2 + − ± − = x x 614 Chapter 9 In problems 49–54, find the foci. 49. 1 19 6 2 2 = −x y 50. 1 35 2 2 = −y x 51. ( ) ( ) 1 6 15 1 2 2 = − − − y x 52. ( ) ( ) 1 2 5 47 3 2 2 = + − − x y 53. 25 8 3 4 1 2 + + ± = x x y 54. 21 4 5 12 3 2 − − ± − = x x y In problems 55–66, find the standard form of the equation for a hyperbola satisfying the given conditions. 55. Foci (5,0) and (-5,0), vertices (4,0) and (4,0) 56. Foci (0,26) and (0, -26), vertices (0,10) and (0,-10) 57. Focus (0, 13), vertex (0,12), center (0,0) 58. Focus (15, 0), vertex (12, 0), center (0,0) 59. Focus (17, 0) and (-17,0), asymptotes x y 15 8 = and x y 15 8 − = 60. Focus (0, 25) and (0, 25), asymptotes x y 7 24 = and x y 7 24 − = 61. Focus (10, 0) and (-10, 0), transverse axis length 16 62. Focus (0, 34) and (0, -34), transverse axis length 32 63. Foci (1, 7) and (1, -3), vertices (1, 6) and (1,-2) 64. Foci (4, -2) and (-6, -2), vertices (2, -2) and (-4, -2) 65. Focus (12, 3), vertex (4, 3), center (-1, 3) 66. Focus (-3, 15), vertex (-3, 13), center (-3, -2) Section 9.2 Hyperbolas 615 67. LORAN Stations A and B are 100 kilometers apart and send a simultaneous radio signal to a ship. The signal from A arrives 0.0002 seconds before the signal from B. If the signal travels 300,000 kilometers per second, find an equation of the hyperbola on which the ship is positioned if the foci are located at A and B. 68. Thunder and Lightning Anita and Samir are standing 3050 feet apart when they see a bolt of light strike the ground. Anita hears the thunder 0.5 seconds before Samir does. Sound travels at 1100 feet per second. Find an equation of the hyperbola on which the lighting strike is positioned if Anita and Samir are located at the foci. 69. Cooling Tower The cooling tower for a power plant has sides in the shape of a hyperbola. The tower stands 179.6 meters tall. The diameter at the top is 72 meters. At their closest, the sides of the tower are 60 meters apart. Find an equation that models the sides of the cooling tower. 70. Calibration A seismologist positions two recording devices 340 feet apart at points A and B. To check the calibration, an explosive is detonated between the devices 90 feet from point A. The time the explosions register on the devices is noted and the difference calculated. A second explosion will be detonated east of point A. How far east should the second explosion be positioned so that the measured time difference is the same as for the first explosion? 71. Target Practice A gun at point A and a target at point B are 200 feet apart. A person at point C hears the gun fire and hit the target at exactly the same time. Find an equation of the hyperbola on which the person is standing if the foci are located at A and B. A fired bullet has a velocity of 2000 feet per second. The speed of sound is 1100 feet per second. 72. Comet Trajectories A comet passes through the solar system following a hyperbolic trajectory with the sun as a focus. The closest it gets to the sun is 3×108 miles. The figure shows the trajectory of the comet, whose path of entry is at a right angle to its path of departure. Find an equation for the comet’s trajectory. Round to two decimal places. 3×108 616 Chapter 9 73. The conjugate of the hyperbola 1 2 2 2 2 = −b y a x is 1 2 2 2 2 − = −b y a x . Show that 0 25 5 2 2 = + −x y is the conjugate of 0 25 5 2 2 = + −y x . 74. The eccentricity e of a hyperbola is the ratio a c , where c is the distance of a focus from the center and a is the distance of a vertex from the center. Find the eccentricity of 1 16 9 2 2 = −y x . 75. An equilateral hyperbola is one for which a = b. Find the eccentricity of an equilateral hyperbola. 76. The latus rectum of a hyperbola is a line segment with endpoints on the hyperbola that passes through a focus and is perpendicular to the transverse axis. Show that a b2 2 is the length of the latus rectum of 1 2 2 2 2 = −b y a x . 77. Confocal hyperbolas have the same foci. Show that, for 0 < k < 6, all hyperbolas of the form 1 6 2 2 = − − k y k x are confocal. Section 9.3 Parabolas and Non-Linear Systems 617 Section 9.3 Parabolas and Non-Linear Systems To listen for signals from space, a radio telescope uses a dish in the shape of a parabola to focus and collect the signals in the receiver. While we studied parabolas earlier when we explored quadratics, at the time we didn’t discuss them as a conic section. A parabola is the shape resulting from when a plane parallel to the side of the cone intersects the cone6. Parabola Definition and Vocabulary A parabola with vertex at the origin can be defined by placing a fixed point at ( ) p F , 0 called the focus, and drawing a line at p y − = , called the directrix. The parabola is the set of all points ( ) y x Q , that are an equal distance between the fixed point and the directrix. For general parabolas, The axis of symmetry is the line passing through the foci, perpendicular to the directrix. The vertex is the point where the parabola crosses the axis of symmetry. The distance from the vertex to the focus, p, is the focal length. 6 Pbroks13 ( “Conic sections with plane”, cropped to show only parabola, CC BY 3.0 x y (0,p) y=-p Q x y Focus Directrix Vertex Axis of symmetry 618 Chapter 9 Equations for Parabolas with Vertex at the Origin From the definition above we can find an equation of a parabola. We will find it for a parabola with vertex at the origin, ( ) 0 , 0 C , opening upward with focus at ( ) p F , 0 and directrix at p y − = . Suppose ( ) y x Q , is some point on the parabola. The distance from Q to the focus is ( ) ( ) ( ) ( ) 0 , 2 2 2 2 p y x p y x F Q d − + = − + − = The distance from the point Q to the directrix is the difference of the y-values: p y p y d + = − − = ) ( From the definition of the parabola, these distances should be equal: ( ) p y p y x + = − + 2 2 Square both sides ( ) ( ) 2 2 2 p y p y x + = − + Expand 2 2 2 2 2 2 2 p py y p py y x + + = + − + Combine like terms py x 4 2 = This is the standard conic form of a parabola that opens up or down (vertical axis of symmetry), centered at the origin. Note that if we divided by 4p, we would get a more familiar equation for the parabola, p x y 4 2 = . We can recognize this as a transformation of the parabola 2 x y = , vertically compressed or stretched by p 4 1 . Using a similar process, we could find an equation of a parabola with vertex at the origin opening left or right. The focus will be at (p,0) and the graph will have a horizontal axis of symmetry and a vertical directrix. The standard conic form of its equation will be px y 4 2 = , which we could also write as p y x 4 2 = . Example 1 Write the standard conic equation for a parabola with vertex at the origin and focus at (0, -2). With focus at (0, -2), the axis of symmetry is vertical, so the standard conic equation is py x 4 2 = . Since the focus is (0, -2), p = -2. The standard conic equation for the parabola is y x ) 2 ( 4 2 − = , or y x 8 2 − = Section 9.3 Parabolas and Non-Linear Systems 619 For parabolas with vertex not at the origin, we can shift these equations, leading to the equations summarized next. Equation of a Parabola with Vertex at (h, k) in Standard Conic Form The standard conic form of an equation of a parabola with vertex at the point ( ) k h, depends on whether the axis of symmetry is horizontal or vertical. The table below gives the standard equation, vertex, axis of symmetry, directrix, focus, and graph for each. Since you already studied quadratics in some depth earlier, we will primarily explore the new concepts associated with parabolas, particularly the focus. Horizontal Vertical Standard Equation ( ) ( ) h x p k y − = − 4 2 ( ) ( ) k y p h x − = − 4 2 Vertex (h, k) (h, k) Axis of symmetry y = k x = h Directrix x = h - p y = k - p Focus (h + p, k) (h, k + p) Graph An example with p < 0 An example with p > 0 x y (h+p,k) x=h-p (h,k) y=k x y (h,k+p) y=k-p (h,k) x=h 620 Chapter 9 Example 2 Put the equation of the parabola 2 ) 1 ( 8 2 + − = x y in standard conic form. Find the vertex, focus, and axis of symmetry. From your earlier work with quadratics, you may already be able to identify the vertex as (1,2), but we’ll go ahead and put the parabola in the standard conic form. To do so, we need to isolate the squared factor. 2 ) 1 ( 8 2 + − = x y Subtract 2 from both sides 2 ) 1 ( 8 2 − = − x y Divide by 8 ( ) 2 ) 1 ( 8 2 − = − x y This matches the general form for a vertical parabola, ( ) ( ) k y p h x − = − 4 2 , where 8 1 4 = p . Solving this tells us 32 1 = p . The standard conic form of the equation is ( ) ( ) 2 32 1 4 1 2 −       = − y x . The vertex is at (1,2). The axis of symmetry is at x = 1. The directrix is at 32 63 32 1 2 = − = y . The focus is at       =       + 32 65 , 1 32 1 2 , 1 . Example 3 A parabola has its vertex at (1,5) and focus at (3,5). Find an equation for the parabola. Since the vertex and focus lie on the line y = 5, that is our axis of symmetry. The vertex (1,5) tells us h = 1 and k = 5. Looking at the distance from the vertex to the focus, p = 3 – 1 = 2. Substituting these values into the standard conic form of an equation for a horizontal parabola gives the equation ( ) ( ) 1 ) 2 ( 4 5 2 − = − x y ( ) ( ) 1 8 5 2 − = − x y Note this could also be rewritten by solving for x, resulting in ( ) 1 5 8 1 2 + − = y x Section 9.3 Parabolas and Non-Linear Systems 621 Try it Now 1. A parabola has its vertex at (-2,3) and focus at (-2,2). Find an equation for this parabola. Applications of Parabolas In an earlier section, we learned that ellipses have a special property that a ray eminating from one focus will be reflected back to the other focus, the property that enables the whispering chamber to work. Parabolas also have a special property, that any ray eminating from the focus will be reflected parallel to the axis of symmetry. Reflectors in flashlights take advantage of this property to focus the light from the bulb into a collimated beam. The same property can be used in reverse, taking parallel rays of sunlight or radio signals and directing them all to the focus. Example 4 A solar cooker is a parabolic dish that reflects the sun’s rays to a central point allowing you to cook food. If a solar cooker has a parabolic dish 16 inches in diameter and 4 inches tall, where should the food be placed? We need to determine the location of the focus, since that’s where the food should be placed. Positioning the base of the dish at the origin, the shape from the side looks like: The standard conic form of an equation for the parabola would be py x 4 2 = . The parabola passes through (4, 8), so substituting that into the equation, we can solve for p: ) 4 )( ( 4 82 p = 4 16 82 = = p The focus is 4 inches above the vertex. This makes for a very convenient design, since then a grate could be placed on top of the dish to hold the food. Try it Now 2. A radio telescope is 100 meters in diameter and 20 meters deep. Where should the receiver be placed? x y 4 8 622 Chapter 9 Non-Linear Systems of Equations In many applications, it is necessary to solve for the intersection of two curves. Many of the techniques you may have used before to solve systems of linear equations will work for non-linear equations as well, particularly substitution. You have already solved some examples of non-linear systems when you found the intersection of a parabola and line while studying quadratics, and when you found the intersection of a circle and line while studying circles. Example 4 Find the points where the ellipse 1 25 4 2 2 = + y x intersects the circle 9 2 2 = + y x . To start, we might multiply the ellipse equation by 100 on both sides to clear the fractions, giving 100 4 25 2 2 = + y x . A common approach for finding intersections is substitution. With these equations, rather than solving for x or y, it might be easier to solve for 2 x or 2 y . Solving the circle equation for 2 x gives 2 2 9 y x − = . We can then substitute that expression for 2 x into the ellipse equation. 100 4 25 2 2 = + y x Substitute 2 2 9 y x − = ( ) 100 4 9 25 2 2 = + − y y Distribute 100 4 25 225 2 2 = + − y y Combine like terms 125 21 2 − = − y Divide by -21 21 125 2 = y Use the square root to solve 21 5 5 21 125 ± = ± = y We can substitute each of these y values back in to 2 2 9 y x − = to find x 21 64 21 125 21 189 21 125 9 21 125 9 2 2 = − = − =         − = x 21 8 21 64 ± = ± = x There are four points of intersection:         ± ± 21 5 5 , 21 8 . Section 9.3 Parabolas and Non-Linear Systems 623 It’s worth noting there is a second technique we could have used in the previous example, called elimination. If we multiplied the circle equation by -4 to get 36 4 4 2 2 − = − − y x , we can then add it to the ellipse equation, eliminating the variable y. 100 4 25 2 2 = + y x 36 4 4 2 2 − = − − y x Add the left sides, and add the right sides 64 21 2 = x Solve for x 21 8 21 64 ± = ± = x Example 5 Find the points where the hyperbola 1 9 4 2 2 = −x y intersects the parabola 2 2x y = . We can solve this system of equations by substituting 2 2x y = into the hyperbola equation. 1 9 4 ) 2 ( 2 2 2 = −x x Simplify 1 9 4 4 2 4 = −x x Simplify, and multiply by 9 9 9 2 4 = −x x Move the 9 to the left 0 9 9 2 4 = − −x x While this looks challenging to solve, we can think of it as a “quadratic in disguise,” since 2 2 4 ) (x x = . Letting 2 x u = , the equation becomes 0 9 9 2 2 = − −u u Solve using the quadratic formula 18 325 1 ) 9 ( 2 ) 9 )( 9 ( 4 ) 1 ( ) 1 ( 2 ± = − − − ± − − = u Solve for x 18 325 1 2 ± = x But 0 325 1 < − , so 18 325 1+ ± = x This leads to two real solutions x ≈ 1.028, -1.028 Substituting these into 2 2x y = , we can find the corresponding y values. The curves intersect at the points (1.028, 2.114) and (-1.028, 2.114). 624 Chapter 9 Try it Now 3. Find the points where the line x y 4 = intersect the ellipse 1 16 4 2 2 = −x y Solving for the intersection of two hyperbolas allows us to utilize the LORAN navigation approach described in the last section. In our example, stations A and B are 150 kilometers apart and send a simultaneous radio signal to the ship. The signal from B arrives 0.0003 seconds before the signal from A. We found the equation of the hyperbola in standard form would be 1 3600 2025 2 2 = − y x Example 10 Continuing the situation from the last section, suppose stations C and D are located 200 km due south of stations A and B and 100 km apart. The signal from D arrives 0.0001 seconds before the signal from C, leading to the equation 1 2275 ) 200 ( 225 2 2 = + −y x . Find the position of the ship. To solve for the position of the boat, we need to find where the hyperbolas intersect. This means solving the system of equations. To do this, we could start by solving both equations for 2 x . With the first equation from the previous example, 1 3600 2025 2 2 = − y x Move the y term to the right 3600 1 2025 2 2 y x + = Multiply both sides by 2025 3600 2025 2025 2 2 y x + = Simplify 16 9 2025 2 2 y x + = With the second equation, we repeat the same process 1 2275 ) 200 ( 225 2 2 = + −y x Move the y term to the right and multiply by 225 2275 ) 200 ( 225 225 2 2 + + = y x Simplify A B C D P Section 9.3 Parabolas and Non-Linear Systems 625 91 ) 200 ( 9 225 2 2 + + = y x Now set these two expressions for 2 x equal to each other and solve. 91 ) 200 ( 9 225 16 9 2025 2 2 + + = + y y Subtract 225 from both sides 91 ) 200 ( 9 16 9 1800 2 2 + = + y y Divide by 9 91 ) 200 ( 16 200 2 2 + = + y y Multiply both sides by 1456 91 16 = ⋅ 2 2 ) 200 ( 16 91 291200 + = + y y Expand and distribute 640000 6400 16 91 291200 2 2 + + = + y y y Combine like terms on one side 0 348800 6400 75 2 = − − y y Solve using the quadratic formula ≈ − − − ± − − = ) 75 ( 2 ) 348800 )( 75 ( 4 ) 6400 ( ) 6400 ( 2 y 123.11 km or -37.78 km We can find the associated x values by substituting these y-values into either hyperbola equation. When y ≈ 123.11, 16 ) 11 . 123 ( 9 2025 2 2 + ≈ x 71 . 102 ± ≈ x When y ≈ -37.78km, 16 ) 78 . 37 ( 9 2025 2 2 − + ≈ x 18 . 53 ± ≈ x This provides 4 possible locations for the ship. Two can be immediately discarded, as they’re on land. Navigators would use other navigational techniques to decide between the two remaining locations. 626 Chapter 9 Important Topics of This Section Parabola Definition Parabola Equations in Standard Form Applications of Parabolas Solving Non-Linear Systems of Equations Try it Now Answers 1. Axis of symmetry is vertical, and the focus is below the vertex. p = 2 – 3 = -1. ( ) ( ) 3 ) 1 ( 4 ) 2 ( 2 − − = − − y x , or ( ) ( ) 3 4 2 2 − − = + y x . 2. The standard conic form of the equation is py x 4 2 = . Using (50,20), we can find that ) 20 ( 4 502 p = , so p = 31.25 meters. The receiver should be placed 31.25 meters above the vertex. 3. Substituting x y 4 = gives ( ) 1 16 4 4 2 2 = −x x . Simplify 1 16 4 16 2 2 = −x x . Multiply by 16 to get 16 64 2 2 = −x x 504 . 0 63 16 ± = ± = x Substituting those into x y 4 = gives the corresponding y values. The curves intersect at (0.504, 2.016) and (-0.504, -2.016). Section 9.3 Parabolas and Non-Linear Systems 627 Section 9.3 Exercises In problems 1–4, match each graph with one of the equations A–D. A. x y 4 2 = B. y x 4 2 = C. y x 8 2 = D. 0 4 2 = + x y 1. 2. 3. 4. In problems 5–14, find the vertex, axis of symmetry, directrix, and focus of the parabola. 5. x y 16 2 = 6. y x 12 2 = 7. 2 2x y = 8. 8 2 y x − = 9. 0 4 2 = + y x 10. 0 8 2 = + x y 11. ) 1 ( 8 ) 2 ( 2 + = − y x 12. ) 2 ( 4 ) 3 ( 2 − = + x y 13. 4 ) 1 ( 4 1 2 + + = x y 14. 1 ) 1 ( 12 1 2 + + − = y x In problems 15–16, write an equation for the graph. 15. 16. In problems 17-20, find the standard form of the equation for a parabola satisfying the given conditions. 17. Vertex at (2,3), opening to the right, focal length 3 18. Vertex at (-1,2), opening down, focal length 1 19. Vertex at (0,3), focus at (0,4) 20. Vertex at (1,3), focus at (0,3) 628 Chapter 9 21. The mirror in an automobile headlight has a parabolic cross-section with the light bulb at the focus. On a schematic, the equation of the parabola is given as 2 2 4y x = . At what coordinates should you place the light bulb? 22. If we want to construct the mirror from the previous exercise so that the focus is located at (0,0.25), what should the equation of the parabola be? 23. A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 12 feet across at its opening and 4 feet deep at its center, where should the receiver be placed? 24. Consider the satellite dish from the previous exercise. If the dish is 8 feet across at the opening and 2 feet deep, where should we place the receiver? 25. A searchlight is shaped like a paraboloid of revolution. A light source is located 1 foot from the base along the axis of symmetry. If the opening of the searchlight is 2 feet across, find the depth. 26. If the searchlight from the previous exercise has the light source located 6 inches from the base along the axis of symmetry and the opening is 4 feet wide, find the depth. In problems 27–34, solve each system of equations for the intersections of the two curves. 27. 1 2 2 2 = − = x y x y 28. 1 2 1 2 2 = + + = y x x y 29. 1 4 11 2 2 2 2 = − = + y x y x 30. 1 4 2 2 2 2 2 = − = + x y y x 31. 16 6 2 2 2 = − = x y x y 32. 1 9 4 2 2 2 = + = y x y x 33. 1 4 1 2 2 2 2 = − = − x y y x 34. ) 1 ( 8 ) 2 ( 4 2 2 + = − = y x y x Section 9.3 Parabolas and Non-Linear Systems 629 35. A LORAN system has transmitter stations A, B, C, and D at (-125,0), (125,0), (0, 250), and (0,-250), respectively. A ship in quadrant two computes the difference of its distances from A and B as 100 miles and the difference of its distances from C and D as 180 miles. Find the x- and y-coordinates of the ship’s location. Round to two decimal places. 36. A LORAN system has transmitter stations A, B, C, and D at (-100,0), (100,0), (-100, -300), and (100,-300), respectively. A ship in quadrant one computes the difference of its distances from A and B as 80 miles and the difference of its distances from C and D as 120 miles. Find the x- and y-coordinates of the ship’s location. Round to two decimal places. 630 Chapter 9 Section 9.4 Conics in Polar Coordinates In the preceding sections, we defined each conic in a different way, but each involved the distance between a point on the curve and the focus. In the previous section, the parabola was defined using the focus and a line called the directrix. It turns out that all conic sections (circles, ellipses, hyperbolas, and parabolas) can be defined using a single relationship. Conic Sections General Definition A conic section can be defined by placing a fixed point at the origin, ( ) 0 , 0 F , called the focus, and drawing a line L called the directrix at p x ± = or p y ± = . The conic section is the set of all points ( ) y x Q , for which the ratio of the distance from Q to F to the distance from Q to the directrix is some positive constant e, called the eccentricity. In other words, ( ) e L Q d F Q d = ) , ( , . Warning: the eccentricity, e, is not the Euler constant e ≈ 2.71828 we studied with exponentials The Polar Form of a Conic To create a general equation for a conic section using the definition above, we will use polar coordinates. Represent ( ) y x Q , in polar coordinates so ( ) ( ) ) sin( ), cos( , θ θ r r y x = . For now, we’ll focus on the case of a horizontal directrix at p y − = , as in the picture above on the left. The distance from the focus to the point Q in polar is just r. The distance from the point Q to the directrix p y − = is ) sin( ) ( ) sin( θ θ r p p r + = − − The ratio of these should be the constant eccentricity e, so ( ) e L Q d F Q d = ) , ( , Substituting in the expressions for the distances, e r p r = + ) sin(θ x y F L: y = −p Q x y F L: x=p Q Section 9.4 Conics in Polar Coordinates 631 To have a standard polar equation, we need to solve for r. Start by clearing the fraction. ( ) ) sin(θ r p e r + = Distribute ) sin(θ er ep r + = Move terms with r to the left ep er r = − ) sin(θ Factor the r ( ) ep e r = − ) sin( 1 θ Divide ) sin( 1 θ e ep r − = We could repeat the same approach for a directrix at p y = and for vertical directrices to obtain the polar equations below. Polar Equation for a Conic Section A conic section with a focus at the origin, eccentricity e, and directrix at p x ± = or p y ± = will have polar equation: ) sin( 1 θ e ep r ± = when the directrix is p y ± = ) cos( 1 θ e ep r ± = when the directrix is p x ± = Example 1 Write the polar equation for a conic section with eccentricity 3 and directrix at 2 = x . We are given e = 3 and p = 2. Since the directrix is vertical and at a positive x value, we use the equation involving cos with the positive sign. ) cos( 3 1 6 ) cos( 3 1 ) 2 )( 3 ( θ θ + = + = r Graphing that using technology reveals it’s an equation for a hyperbola. Try it Now 1. Write a polar equation for a conic with eccentricity 1 and directrix at 3 − = y . 632 Chapter 9 Relating the Polar Equation to the Shape It was probably not obvious to you that the polar equation in the last example would give the graph of a hyperbola. To explore the relationship between the polar equation and the shape, we will try to convert the polar equation into a Cartesian one. For simplicity, we will consider the case where the directrix is 1 = x . ) cos( 1 θ e e r + = Multiply by the denominator ( ) e e r = + ) cos( 1 θ Rewrite r x = ) cos(θ e r x e r =      + 1 Distribute e ex r = + Isolate r ex e r − = Square both sides ( )2 2 ex e r − = Rewrite 2 2 2 y x r + = and expand 2 2 2 2 2 2 2 x e x e e y x + − = + Move variable terms to the left 2 2 2 2 2 2 2 e y x e x e x = + − + Combine like terms 2 2 2 2 2 2 ) 1 ( e y x e e x = + + − We could continue, by completing the square with the x terms, to eventually rewrite this in the standard form as 1 1 1 ) 1 ( 2 2 2 2 2 2 2 2 2 =        − +         − −         − y e e e e x e e , but happily there’s no need for us to do that. In the equation 2 2 2 2 2 2 ) 1 ( e y x e e x = + + − , we can see that: When e < 1, the coefficients of both 2 x and 2 y are positive, resulting in ellipse. When e > 1, the coefficient of 2 x is negative while the coefficient of 2 y is positive, resulting in a hyperbola. When e = 1, the 2 x will drop out of the equation, resulting in a parabola. Relation Between the Polar Equation of a Conic and its Shape For a conic section with a focus at the origin, eccentricity e, and directrix at p x ± = or p y ± = , when 0 < e < 1, the graph is an ellipse when e = 1, the graph is a parabola when e > 1, the graph is a hyperbola Section 9.4 Conics in Polar Coordinates 633 Taking a more intuitive approach, notice that if e < 1, the denominator ) cos( 1 θ e + will always be positive and so r will always be positive. This means that the radial distance r is defined and finite for every value of θ, including 2 π , with no breaks. The only conic with this characteristic is an ellipse. If e = 1, the denominator will be positive for all values of θ, except π − where the denominator is 0 and r is undefined. This fits with a parabola, which has a point at every angle except at the angle pointing along the axis of symmetry away from the vertex. If e > 1, then the denominator will be zero at two angles other than 2 π ± , and r will be negative for a set of θ values. This division of positive and negative radius values would result in two distinct branches of the graph, fitting with a hyperbola. Example 2 For each of the following conics with focus at the origin, identify the shape, the directrix, and the eccentricity. a. ) sin( 2 1 8 θ − = r b. ) cos( 2 3 6 θ − = r c. ) sin( 5 5 8 θ + = r a. This equation is already in standard form ) sin( 1 θ e ep r ± = for a conic with horizontal directrix at p y − = . The eccentricity is the coefficient of ) sin(θ , so e = 2. Since e = 2 > 1, the shape will be a hyperbola. Looking at the numerator, ep = 8, and substituting e = 2 gives p = 4. The directrix is 4 − = y . r < 0 r > 0 634 Chapter 9 b. This equation is not in standard form, since the constant in the denominator is not 1. To put it into standard form, we can multiply the numerator and denominator by 1/3. ( ) ) cos( 3 2 1 2 3 1 ) cos( 2 3 3 1 6 3 1 3 1 ) cos( 2 3 6 θ θ θ − =       −       = ⋅ − = r This is the standard form for a conic with vertical directrix p x − = . The eccentricity is the coefficient on ) cos(θ , so 3 2 = e . Since 0 < e < 1, the shape is an ellipse. Looking at the numerator, ep = 2, so 2 3 2 = p , giving p = 3. The directrix is 3 − = x . c. This equation is also not in standard form. Multiplying the numerator and denominator by 1/5 will put it in standard form. ( ) ) sin( 1 5 8 5 1 ) sin( 5 5 5 1 8 5 1 5 1 ) sin( 5 5 8 θ θ θ + =       +       = ⋅ + = r This is the standard form for a conic with horizontal directrix at p y = . The eccentricity is the coefficient on ) sin(θ , so e = 1. The shape will be a parabola. Looking at the numerator, 5 8 = ep . Since e = 1, 5 8 = p . The directrix is 5 8 = y . Notice that since the directrix is above the focus at the origin, the parabola will open downward. Try it Now 2. Identify the shape, the directrix, and the eccentricity of ) cos( 2 4 9 θ + = r Graphing Conics from the Polar Form Identifying additional features of a conic in polar form can be challenging, which makes graphing without technology likewise challenging. We can utilize our understanding of the conic shapes from earlier sections to aid us. Section 9.4 Conics in Polar Coordinates 635 Example 3 Sketch a graph of ) sin( 5 . 0 1 3 θ − = r and write its Cartesian equation. This is in standard form, and we can identify that 5 . 0 = e , so the shape is an ellipse. From the numerator, 3 = ep , so 3 5 . 0 = p , giving p = 6. The directrix is 6 − = y . To sketch a graph, we can start by evaluating the function at a few convenient θ values, and finding the corresponding Cartesian coordinates. 0 = θ 3 1 3 ) 0 sin( 5 . 0 1 3 = = − = r ) 0 , 3 ( ) , ( = y x 2 π θ = 6 5 . 0 1 3 2 sin 5 . 0 1 3 = − =       − = π r ) 6 , 0 ( ) , ( = y x π θ = 3 1 3 ) sin( 5 . 0 1 3 = = − = π r ) 0 , 3 ( ) , ( − = y x 2 3π θ = 2 5 . 0 1 3 2 3 sin 5 . 0 1 3 = + =       − = π r ) 2 , 0 ( ) , ( − = y x Plotting these points and remembering the origin is one of the foci gives an idea of the shape, which we could sketch in. To get a better understanding of the shape, we could use these features to find more. The vertices are at (0, -2) and (0, 6), so the center must be halfway between, at       + − 2 6 2 , 0 = (0, 2). Since the vertices are a distance a from the center, a = 6 – 2 = 4. One focus is at (0, 0), a distance of 2 from the center, so c = 2, and the other focus must be 2 above the center, at (0, 4). We can now solve for b: 2 2 2 c a b − = , so 10 2 4 2 2 2 = − = b , hence 10 ± = b . The minor axis endpoints would be at ( ) 2 , 10 − and ( ) 2 , 10 . We can now use the center, a, and b to write the Cartesian equation for this curve: 1 16 ) 2 ( 10 2 2 = − + y x 636 Chapter 9 Try it Now 3. Sketch a graph of ) cos( 2 1 6 θ + = r and identify the important features. Important Topics of This Section Polar equations for Conic Sections Eccentricity and Directrix Determining the shape of a polar conic section Try it Now Answers 1. ) sin( 1 ) 3 )( 1 ( θ − = r . ) sin( 1 3 θ − = r 2. We can convert to standard form by multiplying the top and bottom by 4 1 . ) cos( 2 1 1 4 9 θ + = r . Eccentricity = 2 1 , so the shape is an ellipse. The numerator is 4 9 2 1 = = p ep . The directrix is 2 9 = x . 3. The eccentricity is e = 2, so the graph of the equation is a hyperbola. The directrix is 3 = x . Since the directrix is a vertical line and the focus is at the origin, the hyperbola is horizontal. 0 = θ 2 2 1 6 ) 0 cos( 2 1 6 = + = + = r ) 0 , 2 ( ) , ( = y x 2 π θ = 6 1 6 2 cos 2 1 6 = =       + = π r ) 6 , 0 ( ) , ( = y x π θ = 6 2 1 6 ) cos( 2 1 6 − = − = + = π r ) 0 , 6 ( ) , ( = y x 2 3π θ = 6 1 6 2 3 cos 2 1 6 = =       + = π r ) 6 , 0 ( ) , ( − = y x Section 9.4 Conics in Polar Coordinates 637 Plotting those points, we can connect the three on the left with a smooth curve to form one branch of the hyperbola, and the other branch will be a mirror image passing through the last point. The vertices are at (2,0) and (6,0). The center of the hyperbola would be at the midpoint of the vertices, at (4,0). The vertices are a distance a = 2 from the center. The focus at the origin is a distance c = 4 from the center. Solving for b, 12 2 4 2 2 2 = − = b . 3 2 12 ± = ± = b . The asymptotes would be ( ) 4 3 − ± = x y . The Cartesian equation of the hyperbola would be: ( ) 1 12 4 4 2 2 = − − y x 638 Chapter 9 Section 9.4 Exercises In problems 1–8, find the eccentricity and directrix, then identify the shape of the conic. 1. ) cos( 3 1 12 θ + = r 2. ) sin( 1 4 θ − = r 3. ) sin( 3 4 2 θ − = r 4. ) cos( 2 7 θ − = r 5. ) cos( 5 5 1 θ − = r 6. ) cos( 8 3 6 θ + = r 7. ) cos( 2 7 4 θ + = r 8. ) sin( 3 4 16 θ + = r In problems 9–14, find a polar equation for a conic having a focus at the origin with the given characteristics. 9. Directrix x = −4, eccentricity e = 5. 10. Directrix y = −2, eccentricity e = 3. 11. Directrix y = 3, eccentricity e = 3 1 . 12. Directrix x = 5, eccentricity e = 4 3 . 13. Directrix y = −2, eccentricity e = 1. 14. Directrix x = −3, eccentricity e = 1. In problems 15–20, sketch a graph of the conic. Use the graph to help you find important features and write a Cartesian equation for the conic. 15. ) cos( 2 1 9 θ − = r 16. ) sin( 3 1 4 θ + = r 17. ) sin( 3 12 θ + = r 18. ) cos( 2 3 15 θ − = r 19. ) cos( 1 6 θ + = r 20. ) sin( 1 4 θ − = r Section 9.4 Conics in Polar Coordinates 639 21. At the beginning of the chapter, we defined an ellipse as the set of all points Q for which the sum of the distance from each focus to Q is constant. Mathematically, ( ) ( ) k F Q d F Q d = + 2 1 , , . It is not obvious that this definition and the one provided in this section involving the directrix are related. In this exercise, we will start with the definition from this section and attempt to derive the earlier formula from it. a. Draw an ellipse with foci at ( ) 0 , c and ( ) 0 , c − , vertices at ( ) 0 , a and ( ) 0 , a − , and directrixes at p x = and p x − = . Label the foci 1 F and 2 F . Label the directrixes 1 L and 2 L . Label some point ( ) y x, on the ellipse Q. b. Find formulas for ( ) 1 ,L Q d and ( ) 2 ,L Q D in terms of x and p. c. From the definition of a conic in this section, ( ) ( ) e L Q d F Q d = 1 1 , , . Likewise, ( ) ( ) e L Q d F Q d = 2 2 , , as well. Use these ratios, with your answers from part (b) above, to find formulas for ( ) 1 ,F Q d and ( ) 2 ,F Q D in terms of e, x, and p. d. Show that the sum, ( ) ( ) 2 1 , , F Q d F Q d + , is constant. This establishes that the definitions are connected. e. Let Q be a vertex. Find the distances ( ) 1 ,F Q d and ( ) 2 ,F Q D in terms of a and c. Then combine this with your result from part (d) to find a formula for p in terms of a and e. f. Let Q be a vertex. Find the distances ( ) 2 ,L Q D and ( ) 2 ,F Q D in terms of a, p, and c. Use the relationship ( ) ( ) e L Q d F Q d = 2 2 , , , along with your result from part (e), to find a formula for e in terms of a and c. 22. When we first looked at hyperbolas, we defined them as the set of all points Q for which the absolute value of the difference of the distances to two fixed points is constant. Mathematically, ( ) ( ) k F Q d F Q d = − 2 1 , , . Use a similar approach to the one in the last exercise to obtain this formula from the definition given in this section. Find a formula for e in terms of a and c. 640 Chapter 9
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A Quick Guide to Mole Conversions in Chemistry By John T. Moore Chris Hren Peter J. Mikulecky Updated 2016-03-26 07:50:42 From the book No items found. Share Chemistry All-in-One For Dummies (+ Chapter Quizzes Online) Explore Book Buy Now Subscribe on Perlego Chemistry All-in-One For Dummies (+ Chapter Quizzes Online) Explore Book Buy Now Subscribe on Perlego Moles are always challenging to deal with, and in a general chemistry class, you usually end up having to perform a lot of conversions involving moles (mol). Whether you’re converting from moles to grams, moles to volume, or moles to particles (atoms or molecules), use this quick guide to remind you of how to do each type of mole conversion: From mass (grams) to moles: Divide your initial mass by the molar mass of the compound as determined by the periodic table. From moles to mass (grams): Multiply your initial mole value by the molar mass of the compound as determined by the periodic table. From volume (liters) to moles: Divide your initial volume by the molar volume constant, 22.4 L. From moles to volume (liters): Multiply your mole value by the molar volume constant, 22.4L. From particles (atoms, molecules, or formula units) to moles: Divide your particle value by Avogadro’s number, 6.02 × 1023. Remember to use parentheses on your calculator! From moles to particles (atoms, molecules, or formula units): Multiply your mole value by Avogadro’s number, 6.02 × 1023. Mole-to-mole conversions: Use the coefficients from your balanced equation to determine your conversion factor. Be sure your units cancel out so you end up with the correct mole value. About This Article This article is from the book: No items found. About the book author: John T. Moore, EdD, is a chemistry professor at Stephen F. Austin State University. He’s the author of many chemistry titles, including all editions of Chemistry For Dummies. Christopher Hren is a high school chemistry teacher and former track and football coach. Peter J. Mikulecky, PhD, teaches biology and chemistry at Fusion Learning Center and Fusion Academy. This article can be found in the category: Chemistry Hot off the press Explore Related content ResourcesBooks Chemistry Cheat Sheet Chemistry All-in-One For Dummies Cheat Sheet View Cheat Sheet Cheat Sheet Button Text Chemistry Articles How to Make Unit Conversions When studying chemistry you often need to convert units, such as meters and inches. Here are some easy conversion methods to use. View Article Articles Button Text Chemistry Articles How to Convert Units Using Conversion Factors Learn how to quickly convert between measurement units using this handy conversion-factor chart and some simple algebra! 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https://nightsky.jpl.nasa.gov/club/attachments/Fun_Facts_About_the_Moon.pdf
Fun Facts About the Moon Size: With a diameter of 2,159 miles (3,475 kilometers), the Moon is just 1/4 the size of Earth. Distance from Earth: The Moon's average distance from Earth is 238,000 miles (383,500 km). Orbit around Earth: It takes the Moon 27.3 Earth days to revolve around our planet one time. Rotation: The Moon spins on its axis once every 27.3 Earth days. Surface: The Moon's surface is covered with craters, mountain ranges, rilles (long narrow channels), and lava plains. The vast, dark regions we see on the Moon's surface are called maria, or seas. They are actually very large, smooth lava beds. The bright, light areas on the Moon's surface are called highlands. The Moon is covered with a solid, rocky crust about 500 miles (800 km) thick. Underneath the crust, scientists think there is a partially molten zone that leads to a small core of iron. Craters on the Moon come in a wide variety of sizes. The largest crater measures 1,600 miles (2,575 km) across, while the smallest is the size of a pinprick. Atmosphere: The Moon has no long-lasting, significant atmosphere, so the footprints left by Apollo astronauts will last a long time. Temperature: The mean daytime temperature is 225°F (107°C), while the mean nighttime tempera-ture is –243° (–153° C). Escape velocity: To escape the Moon's gravity, you need to travel 5,200 miles (8,400 km) per hour, compared to 25,000 miles (40,200 km) per hour necessary to escape Earth's gravity. Earth's gravity is six times greater than the Moon's. Other information: Roughly 842 pounds (382 kilograms) of Moon rock and soil were brought back to Earth aboard the Apollo spacecraft. Brief Explanation of the Moon Phases The Sun always illuminates the half of the Moon facing the Sun (except during lunar eclipses, when the Moon passes through the Earth's shadow). When the Sun and Moon are on opposite sides of the Earth, the Moon appears "full" to us, a bright, round disk. When the Moon is between the Earth and the Sun, it appears dark, a "new" Moon. In between, the moon's illuminated surface appears to grow (wax) to full, then decreases (wanes) to the next new Moon. The edge of the shadow (the terminator) is always curved, being an oblique view of a circle, giving the Moon its familiar crescent shape. New Moon Waxing Crescent First Quarter Waxing Gibbous Full Moon Waning Gibbous Last Quarter Waning Crescent Have you seen these lunar features? Cool Lunar Facts NUMBER 1 Bye-bye Moon-As you read this, the Moon is moving away from us. Each year, the Moon steals some of Earth's rotational energy, and uses it to propel itself about 3.8 centimeters higher in its orbit. Re-searchers say that when it formed, the Moon was about 14,000 miles (22,530 kilometers) from Earth. It's now more than 280,000 miles, or 450,000 kilometers away. NUMBER 2 Ocean tug-Tides on Earth are caused mostly by the Moon, the Sun to a lesser degree. The Moon's gravity pulls on Earth's oceans. High tide aligns with the Moon as Earth spins underneath. Another high tide occurs on the opposite side of the planet because gravity pulls Earth toward the Moon more than it pulls the water. At full Moon and new Moon, the Sun, Earth and Moon are lined up, producing the higher than normal tides (called spring tides, for the way they spring up). When the Moon is at first or last quarter, smaller neap tides form. The Moon's 29.5-day orbit around Earth is not quite circular. When the Moon is closest to Earth (called its perigee), spring tides are even higher, and called perigean spring tides. All this tugging has another interesting effect: Some of Earth's rotational energy is stolen by the Moon, causing our planet to slow down by about 1.5 milliseconds every century. NUMBER 3 The Moon is a planet?-Our Moon is bigger than Pluto. And at roughly one-fourth the diameter of Earth, some scientists think the Moon is more like a planet. They refer to the Earth-Moon system as a "double planet." Pluto and its moon Charon are also called a dwarf double-planet system by some. NUMBER 4 Moonquakes-Apollo astronauts used seismometers during their visits to the Moon and discovered that the gray orb isn't a totally dead place, geologically speaking. Small moonquakes, originating several miles (kilometers) below the surface, are thought to be caused by the gravitational pull of Earth. Sometimes tiny frac-tures appear at the surface, and gas escapes. Scientists say they think the Moon probably has a core that is hot and perhaps partially molten, as is Earth's core. But data from NASA's Lunar Prospector spacecraft showed in 1999 that the Moon's core is small -- prob-ably between 2 percent and 4 percent of its mass. This is tiny compared with Earth, in which the iron core makes up about 30 percent of the planets mass. NUMBER 5 Egghead-The Moon is not round (or spherical). Instead, it's shaped like an egg. If you go outside and look up, one of the small ends is pointing right at you. And the Moon's center of mass is not at the geomet-ric center of the satellite; it's about 1.2 miles (2 kilometers) off-center. NUMBER 6 Sister moons-The Moon is Earth's only natural satellite. Right? Maybe not. In 1999, scientists found that a 3-mile- (5-kilometer-) wide asteroid may be caught in Earth's gravitational grip, thereby becoming a satellite of our planet. Cruithne, as it is called, takes 770 years to complete a horseshoe-shaped orbit around Earth, the scientists say, and it will remain in a suspended state around Earth for at least 5,000 years. NUMBER 7 Punching bag-The Moon's heavily cratered surface is the result of intense pummeling by space rocks between 4.1 billion and 3.8 billion years ago. The scars of this war, seen as craters, have not eroded much for two main reasons: The Moon is not geologi-cally very active, so earthquakes, volcanoes and mountain-building don't destroy the landscape as they do on Earth; and with virtually no atmosphere there is no wind or rain, so very little surface erosion occurs. NUMBER 8 Moon trees-More than 400 trees on Earth came from the Moon. Well, okay: They came from lunar orbit. Okay, the truth: In 1971, Apollo 14 astronaut Stuart Roosa took a bunch of seeds with him and, while Alan Shepard and Edgar Mitchell were busy sauntering around on the surface, Roosa guarded his seeds. Later, the seeds were germinated on Earth, planted at various sites around the country, and came to be called the Moon trees. Most of them are doing just fine. NUMBER 9 Locked in orbit-Perhaps the coolest thing about the Moon is that it always shows us the same face. Since both the Earth and Moon are rotating and orbiting, how can this be? Long ago, the Earth's gravitational effects slowed the Moon's rotation about its axis. Once the Moon's rotation slowed enough to match its orbital period (the time it takes the Moon to go around Earth) the effect stabilized. Many of the moons around other planets behave similarly. NUMBER 10 Making of the Moon-The Moon was created when a rock the size of Mars slammed into Earth, shortly after the solar system began forming about 4.5 billion years ago, according to the leading theory. NUMBER 11 Eclipses- A lunar eclipse happens when the Full Moon enters the shadow of the Earth. The Moon’s orbit is slightly inclined to the orbit of the Earth so this doesn’t happen every month. A solar eclipse happens when the Moon moves in front of the Sun, from the viewpoint of someone on Earth, during New Moon. The angular size of the Sun and the Moon are very nearly the same, ½ degree. The Moon is smaller, but much closer, so this spectacular phenomenon can occur. The next solar eclipse over the US is 8/21/2017.
8416
https://www.reddit.com/r/learnmath/comments/qn0lh3/whats_the_derivative_of_tanhx/
What's the derivative of tanh(x)? : r/learnmath Skip to main contentWhat's the derivative of tanh(x)? : r/learnmath Open menu Open navigationGo to Reddit Home r/learnmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to learnmath r/learnmath r/learnmath Post all of your math-learning resources here. Questions, no matter how basic, will be answered (to the best ability of the online subscribers). 403K Members Online •4 yr. ago QuantumC-137 What's the derivative of tanh(x)? Some sources say: sech(x)2 Others: 1-tanh(x)2 I tried finding the derivative of [senh(x)/cosh(x)] ,which is tanh(x), and got 1-tanh(x)2 { [senh(x)]' [cosh(x)] } - { [senh(x)] [cosh(x)]' } / [cosh(x)]2 Read more Share Related Answers Section Related Answers best explanation for derivative of tanh(x) derivative of cosh(x) and sinh(x) explained difference between tanh and tan functions Effective strategies for mastering algebra Best online tools for geometry practice New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of November 5, 2021 Reddit reReddit: Top posts of November 2021 Reddit reReddit: Top posts of 2021 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
8417
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems?srsltid=AfmBOopzZhsuaJCaQ8icWC-L-ekqB0ty_iz1zYeao-dJq9TLvyZBdMmC
Art of Problem Solving 1951 AHSME Problems - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 1951 AHSME Problems Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1951 AHSME Problems 1951 AHSC (Answer Key) Printable version: Wiki | AoPS Resources • PDF Instructions This is a 50-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. You will receive? points for each correct answer,? points for each problem left unanswered, and? points for each incorrect answer. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers. Figures are not necessarily drawn to scale. You will have ? minutes working time to complete the test. 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25•26•27•28•29•30•31•32•33•34•35•36•37•38•39•40•41•42•43•44•45•46•47•48•49•50 Contents [hide] 1 Problem 1 2 Problem 2 3 Problem 3 4 Problem 4 5 Problem 5 6 Problem 6 7 Problem 7 8 Problem 8 9 Problem 9 10 Problem 10 11 Problem 11 12 Problem 12 13 Problem 13 14 Problem 14 15 Problem 15 16 Problem 16 17 Problem 17 18 Problem 18 19 Problem 19 20 Problem 20 21 Problem 21 22 Problem 22 23 Problem 23 24 Problem 24 25 Problem 25 26 Problem 26 27 Problem 27 28 Problem 28 29 Problem 29 30 Problem 30 31 Problem 31 32 Problem 32 33 Problem 33 34 Problem 34 35 Problem 35 36 Problem 36 37 Problem 37 38 Problem 38 39 Problem 39 40 Problem 40 41 Problem 41 42 Problem 42 43 Problem 43 44 Problem 44 45 Problem 45 46 Problem 46 47 Problem 47 48 Problem 48 49 Problem 49 50 Problem 50 51 See also Problem 1 The percent that is greater than is: Solution Problem 2 A rectangular field is half as wide as it is long and is completely enclosed by yards of fencing. The area in terms of is: Solution Problem 3 If the length of a diagonal of a square is , then the area of the square is: Solution Problem 4 A barn with a flat roof is rectangular in shape, yd. wide, yd. long and yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is: Solution Problem 5 Mr. A owns a home worth dollars. He sells it to Mr. B at a profit based on the worth of the house. Mr. B sells the house back to Mr. A at a loss. Then: Solution Problem 6 The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to: Solution Problem 7 An error of is made in the measurement of a line long, while an error of only is made in a measurement of a line long. In comparison with the relative error of the first measurement, the relative error of the second measurement is: Solution Problem 8 The price of an article is cut To restore it to its former value, the new price must be increased by: Solution Problem 9 An equilateral triangle is drawn with a side length of A new equilateral triangle is formed by joining the midpoints of the sides of the first one. then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. the limit of the sum of the perimeters of all the triangles thus drawn is: Solution Problem 10 Of the following statements, the one that is incorrect is: Solution Problem 11 The limit of the sum of an infinite number of terms in a geometric progression is where denotes the first term and denotes the common ratio. The limit of the sum of their squares is: Solution Problem 12 At o'clock, the hour and minute hands of a clock form an angle of: Solution Problem 13 can do a piece of work in days. is more efficient than . The number of days it takes to do the same piece of work is: Solution Problem 14 In connection with proof in geometry, indicate which one of the following statements is incorrect: Solution Problem 15 The largest number by which the expression is divisible for all possible integer values of , is: Solution Problem 16 If in applying the quadratic formula to a quadratic equation it happens that , then the graph of will certainly: Solution Problem 17 Indicate in which one of the following equations is neither directly nor inversely proportional to : Solution Problem 18 The expression is to be factored into prime binomial factors and without a numerical monomial factor. This can be done if the value ascribed to is: Solution Problem 19 A six place number is formed by repeating a three place number; for example, or , etc. Any number of this form is always exactly divisible by: Solution Problem 20 When simplified and expressed with negative exponents, the expression is equal to: Solution Problem 21 Given: and . The inequality which is not always correct is: Solution Problem 22 The values of in the equation: are: Solution Problem 23 The radius of a cylindrical box is inches and the height is inches. The number of inches that may be added to either the radius or the height to give the same nonzero increase in volume is: Solution Problem 24 when simplified is: Solution Problem 25 The apothem of a square having its area numerically equal to its perimeter is compared with the apothem of an equilateral triangle having its area numerically equal to its perimeter. The first apothem will be: Solution Problem 26 In the equation the roots are equal when: Solution Problem 27 Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides forming six triangular sections. Then: Solution Problem 28 The pressure of wind on a sail varies jointly as the area of the sail and the square of the velocity of the wind. The pressure on a square foot is pound when the velocity is miles per hour. The velocity of the wind when the pressure on a square yard is pounds is: Solution Problem 29 Of the following sets of data the only one that does not determine the shape of a triangle is: Solution Problem 30 If two poles and high are apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is: Solution Problem 31 A total of handshakes were exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was: Solution Problem 32 If is inscribed in a semicircle whose diameter is , then must be Solution Problem 33 The roots of the equation can be obtained graphically by finding the abscissas of the points of intersection of each of the following pairs of equations except the pair: [Note: Abscissa means x-coordinate.] Solution Problem 34 The value of is: Solution Problem 35 If and , then Solution Problem 36 Which of the following methods of proving a geometric figure a locus is not correct? Solution Problem 37 A number which when divided by leaves a remainder of , when divided by leaves a remainder of , by leaves a remainder of , etc., down to where, when divided by , it leaves a remainder of , is: Solution Problem 38 A rise of feet is required to get a railroad line over a mountain. The grade can be kept down by lengthening the track and curving it around the mountain peak. The additional length of track required to reduce the grade from to is approximately: Solution Problem 39 A stone is dropped into a well and the report of the stone striking the bottom is heard seconds after it is dropped. Assume that the stone falls feet in t seconds and that the velocity of sound is feet per second. The depth of the well is: Solution Problem 40 equals: Solution Problem 41 The formula expressing the relationship between and in the table is: Solution Problem 42 If , then: Solution Problem 43 Of the following statements, the only one that is incorrect is: Solution Problem 44 If , where are other than zero, then equals: Solution Problem 45 If you are given and , then the only logarithm that cannot be found without the use of tables is: Solution Problem 46 is a fixed diameter of a circle whose center is . From , any point on the circle, a chord is drawn perpendicular to . Then, as moves over a semicircle, the bisector of angle cuts the circle in a point that always: Solution Problem 47 If and are the roots of the equation , the value of is: Solution Problem 48 The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as: Solution Problem 49 The medians of a right triangle which are drawn from the vertices of the acute angles are and . The value of the hypotenuse is: Solution Problem 50 Tom, Dick and Harry started out on a -mile journey. Tom and Harry went by automobile at the rate of mph, while Dick walked at the rate of mph. After a certain distance, Harry got off and walked on at mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was: Solution See also AMC 12 Problems and Solutions Mathematics competition resources 1951 AHSC (Problems • Answer Key • Resources) Preceded by 1950 AHSCFollowed by 1952 AHSC 1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25•26•27•28•29•30•31•32•33•34•35•36•37•38•39•40•41•42•43•44•45•46•47•48•49•50 All AHSME Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: AHSME Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
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https://www.youtube.com/watch?v=0V-NGwNzYRc
Step-by-step Approach to Graphing f(x) = x ln(x) Functions & Calculus by Professor Calculish 2080 subscribers 19 likes Description 915 views Posted: 30 Mar 2024 The purpose of the channel is to learn, familiarize, and review the necessary precalculus and trigonometry/geometry topics that form a basis for calculus (with a focus on functions). Calculus topics include limits, derivatives, integrals, formula derivations, derivations/proofs, area/volumes of curves and calculus applications. Utilize simple and easy techniques to get the material across to you. Disclaimer: I am not a professor by profession. Thank you! functions #graphingfunctions #limits #limitsandcalculus #naturallog #exponential #indeterminateforms #lhopital #derivatives #differentialcalculus #extrema #criticalpoint #globalminima #absoluteextreme #firstorderderivatetest #secondorderderivativetest #secondorderderivatives #roots #domain #inflectionpoint #increasingfunctions #decreasingfunction #extremevalues #lhospitalrule #productrule 12 comments Transcript: Introduction I welcome all of you to this video I am Mr is thank you for joining me we are using all the information all the tools we have available in our heads with regards to pre-calculus and calculus to make a respectable graph of what we see here the function X natural log X how can we graph it and not miss out on any of the key features which are involved in this function when you're looking at this it's made of two different functions one function is X natural log X you know it you see it's a product of two different functions what's the domain of X it's going to be minus infinity to Infinity however what's the domain of natural log X it's 0o to Infinity when you combine both of these into this product X natural log X this right here is what will constrain the domain it will determine the domain the domain of this right here will be 0 to Infinity but we need to verify this value for sure when we are talking about graphing it we must then start worrying about the roots of this graph when you're looking at this your equation is f ofx = to 0 you solve for x to determine your roots that's how you will do it I have two different entities x equal to 0 and natural log X is equal to0 and independently soft you can't do much over here so you're getting one root 0 but here you can you can push the natural log on the other side you have x equal to e ^ 0 which is giving you a one these are giving you your X values of roots you have a certain x value and that's what it is again I haven't forgotten about this there's a little bit of an issue involved here we have to examine it to very I it indeed is good when I put zero on the original function what do you get well you can get 0 natural log X there's a bit of a problem so I won't put anything over here what's the problem I'm doing 0 natural log of 0 natural log of 0 is technically not defined so we'll leave this out when I put one in here what do I have 1 natural log of 1 which is a zero it's a zero we're good we'll start developing our graph over here I know one root is 1A 0 and the other root could be right here maybe 0a 0 we'll find out but the part with regards to Roots is done we need to clarify this we need to clarify this and we will the clarification Clarification Procedure procedure can come with us understanding the limit limit as X approaches zero from right why do we have to do this clarification procedure because we have a natural log you know the natural log is a one-sided limit at zero you're always approaching Zero from the right for natural log if my function is X natural log X what happens when we evaluate this limit what happens to this function as you approach zero it will bring us Clarity with regards to this and with regards to this when I put Z in places of X what am I really getting zero and natural log of 0 when you think about the natural log graph it looks something like this 1 comma 0 and then here it is as your values of X approach Z your function is dipping towards minus infinity so you're looking here at this 0 comma minus infinity and indeterminate product type you know indeterminate product types is 0 comma Infinity but minus infinity equally applies and this is an indeterminate product type so we deal with it how do we deal with it we analyze it by means of the separation I have a f function and a g function and I'll convert this into coent forms I can do F / 1 or G or I can do G / 1/ f I would rather do this cuz it would be easy derivative with the numerator function which be just natural log X ID by derivative of the denominator function which will be the reciprocal I'm looking at this reciprocal of x 1 or X and I'll apply this zero coming from the right what's the Der of natural log X 1 or X X what's the derivative of 1x it's - 1x2 and then 0 from the right you know when you Shuffle these things around you have 1X - x^2 / 1 0 from the right this will cancel out with that the only thing which remains is a Min - x you putting Zero from the right and you're getting a zero this limit when you evaluated this function X natural log X indeed approaches zero so you can say your end result here is zero and it brings us Clarity with this we do have 0 comma Infinity the only problem is again when you put zero here in this function you have a natural log of Z which is a problem we do in open value so our graph will have two Roots one is a closed value one is an open value or you can say an empty value but it is still a value now what remains is the evaluation Extremes of extremes consider this as you're taking large values of X like 50 100 a th000 a million in Infinity this product here will become a large so you know you'll be increasing towards Infinity eventually limit as X approaches Infinity this function here the product of these two functions will also approach Infinity so we know it's going to exceed out of bounds but what about the extreme value if you do the first order derivative you can you have to do the paric rule you'll do x the D of natural log X which is 1x + natural log x the dtive of X which is just a one what do you get you have 1 plus natural log X that's your first order derivative but to determine a critical point you equal it to Z and you solve for x this is your formula you equal it to Z and you solve for x 1 + natural log x equal to 0 natural log x = -1 x here isal to e ^ -1 which is equal to 1 E what is this giving you your critical point x value my critical point has a certain x value which is 1 E I have to put this value right here and get my y value and I can I have 1 E natural log of 1 E how can you evaluate this think of your laws of exponents and properties of natural logs which will be 1 E natural log of e^ minus1 this can come right over here and hit with this I'll have -1 e natural log e natural log e is equal to 1 my y value is -1 e what does 1 e - 1 or E look like in terms of a calculator value you can just do exponential and do the reciprocal you're looking at 3678 comma minus 3678 critical point value we can estimate it and it'll be right over here 1 E comma minus 1 or E it's an extreme value a critical point but what type of extreme value is it this is where the second order dtive test will come into Second Order Dtive play We Know by means of the natural log there is no presence of this function here after and past the Y AIS everything is going to be located only on the right of the y- axis second order derivative what we have we just do the second order dtive of this or do the next order dtive of this it will will be 1/x and to classify you put your critical point into your second order derivative and see what you get by critical point x value is 1 or E I'm putting it right here 1 1 or E which is equal to e it's a positive that's all I want I'm looking if it's positive if it's a positive I'm dealing with a minimum a local minimum or relative minimum if it's a negative I'm looking at a local maximum or relative maximum it's a positive value so I know I'm looking at a local minimum or relative minimum however when I connect these I'm looking at a function which is coming like this why is it coming like this because I already alluded to at the very beginning as I put larger values of this as X becomes Infinity you have Infinity natural log Infinity you you will exceed out of bounds you are looking at an increasing function so that indeed is my graph and what is this is this a local minimum or relative minimum yes it is in the region of this area but in the overall domain from 0 to Infinity is my absolute or Global minimum it is certainly a local and relative minimum in the region of that critical point but in the overall domain it's your absolute or Global minimum do we have to worry about an inflection point no you don't because you only see a single type of concavity you see a concavity up you don't see a change from concavity up to down or concavity down to up so there will be no inflection point but hypothetically your inflection point would be determined as this and you solve for it X would here equal to zero I have my second order data function I have to make it equal to Z x would be equal to 1 or0 and you know this right here is problematic it's telling that this right here could be an inflection point if suppos there was some presence of the function on this side it would be an inflection point but it's not really a inflection point so we don't worry about inflection points we don't worry about concavities but what is it that we can worry about one more thing we have a decreasing function over here if you look at it it's decreasing where is it decreasing from 0o up to 1 over e notice I'm using a circular parenthesis here I'm not closing it out even though that value is part of the critical point I'm taking it up to but not including because when I look at the increasing aspect of this function then I'm keeping a circular parentheses one or E and it's going up to Infinity I'm not closing these out with a square parentheses on either side cuz I'm reserving the exact value here for the critical point then we have a decreasing function up to that point then we have an increasing function from that point and you can understand the difference but that right there is our graph in its entirety and it will be correct it has a root here at 1 comma 0 it has has what looks like a root at the origin but we know it to be an empty value cuz you cannot put zero in place of here natural log0 is not defined we have a critical point 1 E comma minus1 or E which is a local relative minimum in the neighborhood of that value but it is global and absolute minimum in the overall domain and our graph will be right you can verify it on your software it will be good but that's all we'll talk about for this function and it brings us to the end thank you for joining me have a good day
8419
https://www.scribd.com/document/706960893/Module-4-THE-THERMODYNAMICS-OF-ELECTROCHEMICAL-SYSTEMS-2023
Thermodynamics in Electrochemical Systems | PDF | Electrochemistry | Redox Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for days 0 ratings 0% found this document useful (0 votes) 209 views 35 pages Thermodynamics in Electrochemical Systems The document discusses electrochemical cells and thermodynamics. It begins by defining electrochemistry and describing oxidation/reduction reactions that can occur by direct contact or in an… Full description Uploaded by andreslloydralf AI-enhanced title and description Download Save Save Module 4 – THE THERMODYNAMICS OF ELECTROCHEMICAL S... For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Report 0 ratings 0% found this document useful (0 votes) 209 views 35 pages Thermodynamics in Electrochemical Systems The document discusses electrochemical cells and thermodynamics. It begins by defining electrochemistry and describing oxidation/reduction reactions that can occur by direct contact or in an electrochemical cell where reactants are separated. It then discusses the components of an electrochemical cell, including electrodes, electrolyte, and half-cell reactions. Next, it focuses on galvanic cells, where a spontaneous chemical reaction generates electricity. It describes the structure of galvanic cells, including separating redox reactions into half-reactions occurring at different electrodes, and using a salt bridge or porous disk to allow ion flow between solutions. Read more Download 0 ratings 0% found this document useful (0 votes) 209 views 35 pages Thermodynamics in Electrochemical Systems The document discusses electrochemical cells and thermodynamics. It begins by defining electrochemistry and describing oxidation/reduction reactions that can occur by direct contact or in an… Full description Uploaded by andreslloydralf AI-enhanced title and description Go to previous items Go to next items Download Save Save Module 4 – THE THERMODYNAMICS OF ELECTROCHEMICAL S... For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Report Download Save Module 4 – THE THERMODYNAMICS OF ELECTROCHEMICAL S... For Later You are on page 1/ 35 Search Fullscreen Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial Footer menu Back to top About About Scribd, Inc. Everand: Ebooks & Audiobooks Slideshare Join our team! Contact us Support Help / FAQ Accessibility Purchase help AdChoices Legal Terms Privacy Copyright Do not sell or share my personal information Social Instagram Instagram Facebook Facebook Pinterest Pinterest Get our free apps About About Scribd, Inc. Everand: Ebooks & Audiobooks Slideshare Join our team! Contact us Legal Terms Privacy Copyright Do not sell or share my personal information Support Help / FAQ Accessibility Purchase help AdChoices Social Instagram Instagram Facebook Facebook Pinterest Pinterest Get our free apps Documents Language: English Copyright © 2025 Scribd Inc. We take content rights seriously. Learn more in our FAQs or report infringement here. We take content rights seriously. Learn more in our FAQs or report infringement here. Language: English Copyright © 2025 Scribd Inc. 576648e32a3d8b82ca71961b7a986505
8420
https://basic-chemistry.eu5.org/Folder1/82-stp-ntp-satp.html
Understanding the Relations between STP, NTP, and SATP in Chemistry ⚫Intoduction: In the field of chemistry, the terms STP, NTP, and SATP are commonly used to denote specific standard conditions for temperature and pressure. These conditions are essential for standardizing measurements and comparing experimental results. This article aims to provide a detailed explanation of STP (Standard Temperature and Pressure), NTP (Normal Temperature and Pressure), and SATP (Standard Ambient Temperature and Pressure), highlighting their definitions, significance, and differences. ⚫Standard Temperature and Pressure (STP): STP refers to a set of standard conditions used for comparing gases. In the International Union of Pure and Applied Chemistry (IUPAC) convention, STP is defined as a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere (101.325 kilopascals). At STP, one mole of an ideal gas occupies approximately 22.4 liters of volume. STP is particularly useful in gas calculations, as it allows for easy conversions between moles and volumes, simplifying stoichiometry and other related calculations. ⚫Normal Temperature and Pressure (NTP): NTP is another set of standard conditions used in various fields, including engineering and metrology. However, it is important to note that NTP is not as universally defined or recognized as STP. The most commonly used definition for NTP is a temperature of 20 degrees Celsius (293.15 Kelvin) and a pressure of 1 atmosphere (101.325 kilopascals). NTP is often employed in practical applications, such as gas flow measurements, to provide a standardized reference point for comparison. However, it is crucial to confirm the specific definition of NTP being used, as it may vary depending on the context. ⚫Standard Ambient Temperature and Pressure (SATP): SATP is another set of standard conditions used in certain fields, such as environmental studies and gas analysis. SATP is defined as a temperature of 25 degrees Celsius (298.15 Kelvin) and a pressure of 1 atmosphere (101.325 kilopascals). These conditions are representative of typical environmental conditions. SATP is often used in experimental settings where ambient conditions are relevant, such as determining gas concentrations in air samples or studying gas behavior in specific environmental conditions. ⚫Differences and Significance: The main differences between STP, NTP, and SATP lie in their defined temperature values. STP uses 0 degrees Celsius as the reference temperature, while NTP uses 20 degrees Celsius and SATP uses 25 degrees Celsius. Additionally, the pressure values for all three conditions remain consistent at 1 atmosphere. These standard conditions are essential for standardizing measurements, ensuring consistency in experimental data, and allowing for comparisons between different experiments or research studies. They provide a common reference point for scientists and researchers to communicate and reproduce results accurately. ⚫Conclusion: STP, NTP, and SATP are standard conditions used in chemistry and various other fields to establish a common reference point for temperature and pressure measurements. While STP is universally defined as 0 degrees Celsius and 1 atmosphere, NTP and SATP may have slightly different definitions depending on the specific context. Understanding these standard conditions is crucial for accurate comparisons, calculations, and communication in scientific research and experimentation. By adhering to these standardized conditions, scientists can ensure consistent and reliable data across different studies and disciplines. ............... Click here to visit your Index Page to find more topics. ..... Connect us on : © Basic Chemistry 2023 . ® All Rights Reserved.
8421
https://www.mathworksheetscenter.com/mathtips/roundingnumbers.html
Math Worksheets Center Logo 5 Everyday Uses of Rounding Numbers Article Summary: A decimal number is composed of an integer for the whole number to the left of the decimal point and the fractional part to the right of the decimal point. Rounding numbers means shortening the number of digits to the right of the decimal point to make the number easier to work with. The numbers to the right of the decimal point usually show small amounts and sometimes it's okay to ignore very small amounts. Rounding numbers is a strange phrase, isn't it? The Arabic numerals like 0, 2, 3 and 9 already have a rounded shape so does rounding numbers mean changing the shape of 1, 4 and 7? No, we've just having some fun and joking a bit! Rounding numbers applies to decimal numbers and not to integers. A decimal number is composed of an integer for the whole number to the left of the decimal point and the fractional part to the right of the decimal point. Rounding numbers means shortening the number of digits to the right of the decimal point to make the number easier to work with. The numbers to the right of the decimal point usually show small amounts and sometimes it's okay to ignore very small amounts. When you round a decimal number, sometimes you change the last number that you show. If the decimal digit that you are truncating (removing) is 5 or more then you add 1 to the last digit that you keep. This is called rounding up. It sounds complicated but it isn't. Here are some examples. 12.3742 rounds to 12.37 because 4 is less than 5 18.4593 rounds up to 18.46 because 9 is greater than 5 Let's review how to convert fractions to decimal. You probably remember that 1/2 in fractions is the same as 0.5 in decimal. The 5 in 0.5 is the same as half of 10 or 10 divided by 2. What happens when you try to represent 1/3 in decimal? This time you are trying to divide 10 by 3. The answer in fractions is 3 1/3. But, wait a minute, 1/3 is what we trying to show in decimal in the first place. The fraction 1/3 becomes 0.3333333333 with as many three's at the end as you wish to show. There just isn't an exact match between the fraction 1/3 and how it's written in decimal. Use 1. Handling the mismatch between fractions and decimal One use of rounding numbers is shorten all the three's to the right of the decimal point in converting 1/3 to decimal. Most of the time, you will use the rounded numbers 0.33 or 0.333 when you need to work with 1/3 in decimal. In fact, you usually work with just two or three digits to the right of the decimal point when there is no exact equivalent to the fraction in decimal. How would you show 1/6 in decimal? Remember to round up! Use 2. Changing multiplied results What is the answer if you multiply 25 by 75? You get 1875. Now multiply 0.25 by 0.75. You get 0.1875. You started with 2 digits to the right of the decimal point and ended up with 4. Many times you will just round up the result to 0.19. Use 3. Taxes When you shop, do you pay a state sales tax? If the sales tax is 3% and you by something for $56.30 how much sales tax do you pay? $56.30 x 0.03 = $1.689 Tax is $1.69 Use 4. Do calculations in your head Let's pretend that you and your friends are going to a fast food restaurant and you want to be sure to have enough money to pay the bill. You can round the cost of each meal to the nearest dollar and add the amounts quickly and easily. Use 5. Getting an estimate Sometimes you want to round integers instead of decimal numbers. Usually you are interested in rounding to the nearest multiple of 10, 100, 1,000 or million. For example, in 2006 the census department determined that the population of New York City was 8,214,426. That number is hard to remember and if we say the population of New York City is 8 million it is a good estimate because it doesn't make any real difference what the exact number is.
8422
https://www.physicsclassroom.com/class/waves/Lesson-0/Motion-of-a-Mass-on-a-Spring
Physics Tutorial: Motion of a Mass on a Spring My Account × Custom Search Sort by: Relevance Relevance Date TPC and eLearning What's NEW at TPC? 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Classroom » Physics Tutorial » Vibrations and Waves » Motion of a Mass on a Spring Vibrations and Waves - Lesson 0 - Vibrations Motion of a Mass on a Spring Vibrational Motion Properties of Periodic Motion Pendulum Motion Motion of a Mass on a Spring Getting your Trinity Audio player ready... Hold down the T key for 3 seconds to activate the audio accessibility mode, at which point you can click the K key to pause and resume audio. Useful for the Check Your Understanding and See Answers. Report This Ad In a previous part of this lesson, the motion of a mass attached to a spring was described as an example of a vibrating system. The mass on a spring motion was discussed in more detail as we sought to understand the mathematical properties of objects that are in periodic motion. Now we will investigate the motion of a mass on a spring in even greater detail as we focus on how a variety of quantities change over the course of time. Such quantities will include forces, position, velocity and energy - both kinetic and potential energy. Hooke's Law We will begin our discussion with an investigation of the forces exerted by a spring on a hanging mass. Consider the system shown at the right with a spring attached to a support. The spring hangs in a relaxed, unstretched position. If you were to hold the bottom of the spring and pull downward, the spring would stretch. If you were to pull with just a little force, the spring would stretch just a little bit. And if you were to pull with a much greater force, the spring would stretch a much greater extent. Exactly what is the quantitative relationship between the amount of pulling force and the amount of stretch? To determine this quantitative relationship between the amount of force and the amount of stretch, objects of known mass could be attached to the spring. For each object which is added, the amount of stretch could be measured. The force which is applied in each instance would be the weight of the object. A regression analysis of the force-stretch data could be performed in order to determine the quantitative relationship between the force and the amount of stretch. The data table below shows some representative data for such an experiment. Mass (kg)Force on Spring (N)Amount of Stretch (m) 0.000 0.000 0.0000 0.050 0.490 0.0021 0.100 0.980 0.0040 0.150 1.470 0.0063 0.200 1.960 0.0081 0.250 2.450 0.0099 0.300 2.940 0.0123 0.400 3.920 0.0160 0.500 4.900 0.0199 By plotting the force-stretch data and performing a linear regression analysis, the quantitative relationship or equation can be determined. The plot is shown below. A linear regression analysis yields the following statistics: slope = 0.00406 m/N y-intercept = 3.43 x10-5 (pert near close to 0.000) regression constant = 0.999 The equation for this line is Stretch = 0.00406•Force + 3.43x10-5 The fact that the regression constant is very close to 1.000 indicates that there is a strong fit between the equation and the data points. This strong fit lends credibility to the results of the experiment. This relationship between the force applied to a spring and the amount of stretch was first discovered in 1678 by English scientist Robert Hooke. As Hooke put it: Ut tensio, sic vis. Translated from Latin, this means "As the extension, so the force." In other words, the amount that the spring extends is proportional to the amount of force with which it pulls. If we had completed this study about 350 years ago (and if we knew some Latin), we would be famous! Today this quantitative relationship between force and stretch is referred to as Hooke's law and is often reported in textbooks as F spring = -k•x where Fspring is the force exerted upon the spring, x is the amount that the spring stretches relative to its relaxed position, and k is the proportionality constant, often referred to as the spring constant. The spring constant is a positive constant whose value is dependent upon the spring which is being studied. A stiff spring would have a high spring constant. This is to say that it would take a relatively large amount of force to cause a little displacement. The units on the spring constant are Newton/meter (N/m). The negative sign in the above equation is an indication that the direction that the spring stretches is opposite the direction of the force which the spring exerts. For instance, when the spring was stretched below its relaxed position, x is downward. The spring responds to this stretching by exerting an upward force. The x and the F are in opposite directions. A final comment regarding this equation is that it works for a spring which is stretched vertically and for a spring is stretched horizontally (such as the one to be discussed below). Force Analysis of a Mass on a Spring Earlier in this lesson we learned that an object that is vibrating is acted upon by a restoring force. The restoring force causes the vibrating object to slow down as it moves away from the equilibrium position and to speed up as it approaches the equilibrium position. It is this restoring force which is responsible for the vibration. So what is the restoring force for a mass on a spring? We will begin our discussion of this question by considering the system in the diagram below. The diagram shows an air track and a glider. The glider is attached by a spring to a vertical support. There is a negligible amount of friction between the glider and the air track. As such, there are three dominant forces acting upon the glider. These three forces are shown in the free-body diagram at the right. The force of gravity (Fgrav) is a rather predictable force - both in terms of its magnitude and its direction. The force of gravity always acts downward; its magnitude can be found as the product of mass and the acceleration of gravity (m•9.8 N/kg). The support force (Fsupport) balances the force of gravity. It is supplied by the air from the air track, causing the glider to levitate about the track's surface. The final force is the spring force (Fspring). As discussed above, the spring force varies in magnitude and in direction. Its magnitude can be found using Hooke's law. Its direction is always opposite the direction of stretch and towards the equilibrium position. As the air track glider does the back and forth, the spring force (Fspring) acts as the restoring force. It acts leftward on the glider when it is positioned to the right of the equilibrium position; and it acts rightward on the glider when it is positioned to the left of the equilibrium position. Let's suppose that the glider is pulled to the right of the equilibrium position and released from rest. The diagram below shows the direction of the spring force at five different positions over the course of the glider's path. As the glider moves from position A (the release point) to position B and then to position C, the spring force acts leftward upon the leftward moving glider. As the glider approaches position C, the amount of stretch of the spring decreases and the spring force decreases, consistent with Hooke's Law. Despite this decrease in the spring force, there is still an acceleration caused by the restoring force for the entire span from position A to position C. At position C, the glider has reached its maximum speed. Once the glider passes to the left of position C, the spring force acts rightward. During this phase of the glider's cycle, the spring is being compressed. The further past position C that the glider moves, the greater the amount of compression and the greater the spring force. This spring force acts as a restoring force, slowing the glider down as it moves from position C to position D to position E. By the time the glider has reached position E, it has slowed down to a momentary rest position before changing its direction and heading back towards the equilibrium position. During the glider's motion from position E to position C, the amount that the spring is compressed decreases and the spring force decreases. There is still an acceleration for the entire distance from position E to position C. At position C, the glider has reached its maximum speed. Now the glider begins to move to the right of point C. As it does, the spring force acts leftward upon the rightward moving glider. This restoring force causes the glider to slow down during the entire path from position C to position D to position E. Sinusoidal Nature of the Motion of a Mass on a Spring Previously in this lesson, the variations in the position of a mass on a spring with respect to time were discussed. At that time, it was shown that the position of a mass on a spring varies with the sine of the time. The discussion pertained to a mass that was vibrating up and down while suspended from the spring. The discussion would be just as applicable to our glider moving along the air track. If a motion detector were placed at the right end of the air track to collect data for a position vs. time plot, the plot would look like the plot below. Position A is the right-most position on the air track when the glider is closest to the detector. The labeled positions in the diagram above are the same positions used in the discussion of restoring force above. You might recall from that discussion that positions A and E were positions at which the mass had a zero velocity. Position C was the equilibrium position and was the position of maximum speed. If the same motion detector that collected position-time data were used to collect velocity-time data, then the plotted data would look like the graph below. Observe that the velocity-time plot for the mass on a spring is also a sinusoidal shaped plot. The only difference between the position-time and the velocity-time plots is that one is shifted one-fourth of a vibrational cycle away from the other. Also observe in the plots that the absolute value of the velocity is greatest at position C (corresponding to the equilibrium position). The velocity of any moving object, whether vibrating or not, is the speed with a direction. The magnitude of the velocity is the speed. The direction is often expressed as a positive or a negative sign. In some instances, the velocity has a negative direction (the glider is moving leftward) and its velocity is plotted below the time axis. In other cases, the velocity has a positive direction (the glider is moving rightward) and its velocity is plotted above the time axis. You will also notice that the velocity is zero whenever the position is at an extreme. This occurs at positions A and E when the glider is beginning to change direction. So just as in the case of pendulum motion, the speed is greatest when the displacement of the mass relative to its equilibrium position is the least. And the speed is least when the displacement of the mass relative to its equilibrium position is the greatest. Energy Analysis of a Mass on a Spring On the previous page, an energy analysis for the vibration of a pendulum was discussed. Here we will conduct a similar analysis for the motion of a mass on a spring. In our discussion, we will refer to the motion of the frictionless glider on the air track that was introduced above. The glider will be pulled to the right of its equilibrium position and be released from rest (position A). As mentioned, the glider then accelerates towards position C (the equilibrium position). Once the glider passes the equilibrium position, it begins to slow down as the spring force pulls it backwards against its motion. By the time it has reached position E, the glider has slowed down to a momentary pause before changing directions and accelerating back towards position C. Once again, after the glider passes position C, it begins to slow down as it approaches position A. Once at position A, the cycle begins all over again ... and again ... and again. The kinetic energy possessed by an object is the energy it possesses due to its motion. It is a quantity that depends upon both mass and speed. The equation that relates kinetic energy (KE) to mass (m) and speed (v) is KE = ½•m•v 2 The faster an object moves, the more kinetic energy that it will possess. We can combine this concept with the discussion above about how speed changes during the course of motion. This blending of the concepts would lead us to conclude that the kinetic energy of the mass on the spring increases as it approaches the equilibrium position; and it decreases as it moves away from the equilibrium position. This information is summarized in the table below: Stage of Cycle Change in Speed Change in Kinetic Energy A to B to C Increasing Increasing C to D to E Decreasing Decreasing E to D to C Increasing Increasing C to B to A Decreasing Decreasing Kinetic energy is only one form of mechanical energy. The other form is potential energy. Potential energy is the stored energy of position possessed by an object. The potential energy could be gravitational potential energy, in which case the position refers to the height above the ground. Or the potential energy could be elastic potential energy, in which case the position refers to the position of the mass on the spring relative to the equilibrium position. For our vibrating air track glider, there is no change in height. So the gravitational potential energy does not change. This form of potential energy is not of much interest in our analysis of the energy changes. There is however a change in the position of the mass relative to its equilibrium position. Every time the spring is compressed or stretched relative to its relaxed position, there is an increase in the elastic potential energy. The amount of elastic potential energy depends on the amount of stretch or compression of the spring. The equation that relates the amount of elastic potential energy (PEspring) to the amount of compression or stretch (x) is PE spring = ½ • k•x 2 where k is the spring constant (in N/m) and x is the distance that the spring is stretched or compressed relative to the relaxed, unstretched position. When the air track glider is at its equilibrium position (position C), it is moving it's fastest (as discussed above). At this position, the value of x is 0 meter. So the amount of elastic potential energy (PEspring) is 0 Joules. This is the position where the potential energy is the least. When the glider is at position A, the spring is stretched the greatest distance and the elastic potential energy is a maximum. A similar statement can be made for position E. At position E, the spring is compressed the most and the elastic potential energy at this location is also a maximum. Since the spring stretches as much as compresses, the elastic potential energy at position A (the stretched position) is the same as at position E (the compressed position). At these two positions - A and E - the velocity is 0 m/s and the kinetic energy is 0 J. So just like the case of a vibrating pendulum, a vibrating mass on a spring has the greatest potential energy when it has the smallest kinetic energy. And it also has the smallest potential energy (position C) when it has the greatest kinetic energy. These principles are shown in the animation below. When conducting an energy analysis, a common representation is an energy bar chart. An energy bar chart uses a bar graph to represent the relative amount and form of energy possessed by an object as it is moving. It is a useful conceptual tool for showing what form of energy is present and how it changes over the course of time. The diagram below is an energy bar chart for the air track glider and spring system. The bar chart reveals that as the mass on the spring moves from A to B to C, the kinetic energy increases and the elastic potential energy decreases. Yet the total amount of these two forms of mechanical energy remains constant. Mechanical energy is being transformed from potential form to kinetic form; yet the total amount is being conserved. A similar conservation of energy phenomenon occurs as the mass moves from C to D to E. As the spring becomes compressed and the mass slows down, its kinetic energy is transformed into elastic potential energy. As this transformation occurs, the total amount of mechanical energy is conserved. This very principle of energy conservation was explained in a previous chapter - the Energy chapter - of The Physics Classroom Tutorial. Period of a Mass on a Spring As is likely obvious, not all springs are created equal. And not all spring-mass systems are created equal. One measurable quantity that can be used to distinguish one spring-mass system from another is the period. As discussed earlier in this lesson, the period is the time for a vibrating object to make one complete cycle of vibration. The variables that effect the period of a spring-mass system are the mass and the spring constant. The equation that relates these variables resembles the equation for the period of a pendulum. The equation is T = 2•Π•(m/k).5 where T is the period, m is the mass of the object attached to the spring, and k is the spring constant of the spring. The equation can be interpreted to mean that more massive objects will vibrate with a longer period. Their greater inertia means that it takes more time to complete a cycle. And springs with a greater spring constant (stiffer springs) have a smaller period; masses attached to these springs take less time to complete a cycle. Their greater spring constant means they exert stronger restoring forces upon the attached mass. This greater force reduces the length of time to complete one cycle of vibration. Looking Forward to Lesson 2 As we have seen in this lesson, vibrating objects are wiggling in place. They oscillate back and forth about a fixed position. A simple pendulum and a mass on a spring are classic examples of such vibrating motion. Though not evident by simple observation, the use of motion detectors reveals that the vibrations of these objects have a sinusoidal nature. There is a subtle wave-like behavior associated with the manner in which the position and the velocity vary with respect to time. In the next lesson, we will investigate waves. As we will soon find out, if a mass on a spring is a wiggle in time, then a wave is a collection of wigglers spread through space. As we begin our study of waves in Lesson 2, concepts of frequency, wavelength and amplitude will remain important. We Would Like to Suggest ... Why just read about it and when you could be interacting with it? Interact - that's exactly what you do when you use one of The Physics Classroom's Interactives. We would like to suggest that you combine the reading of this page with the use of ourMass on a Spring Interactive. You can find it in the Physics Interactives section of our website. TheMass on a Spring Interactiveprovides the learner with a simple environment for exploring the effect of mass, spring constant and duration of motion upon the period and amplitude of a vertically-vibrating mass. Visit: Mass on a Spring Interactive Check Your Understanding A force of 16 N is required to stretch a spring a distance of 40 cm from its rest position. What force (in Newtons) is required to stretch the same spring … a. … twice the distance? b. … three times the distance? c. … one-half the distance? See Answer Answer: Ut tensio, sic vis. As the extension, so the force. Force and stretch are proportional to one another such that if you double the force, the amount of stretch will double. If you triple the force, the amount of stretch will triple. If you half the force, the amount of stretch will halve. So the answers here are: a. 32 N (twice the stretch requires twice the force) b. 48 N (three times the stretch requires three times the force) c. 8 N (one-half the stretch requires one-half the force) Perpetually disturbed by the habit of the backyard squirrels to raid his bird feeders, Mr. H decides to use a little physics for better living. His current plot involves equipping his bird feeder with a spring system that stretches and oscillates when the mass of a squirrel lands on the feeder. He wishes to have the highest amplitude of vibration that is possible. Should he use a spring with a large spring constant or a small spring constant? See Answer Answer: Mr. H should use a spring with a low spring constant (k). The spring constant (k in the Hooke's law equation) is the ratio of the F/x. If this ratio is low, then there will be a relatively large displacement for any given F value. Being displaced furthest from the equilibrium position will set the spring into a relatively high amplitude vibrational motion. Referring to the previous question. If Mr. H wishes to have his bird feeder (and attached squirrel) vibrate with the highest possible frequency, should he use a spring with a large spring constant or a small spring constant? See Answer Answer: Mr. H should use a spring with a large spring constant (k). Using a large spring constant (k) will cause the period to be small. A small period corresponds to a high frequency. Get them squirrels, Mr. H! Use energy conservation to fill in the blanks in the following diagram. See Answer Answer: Energy is conserved. So the total amount of KE + PE must be the same for each location. Which of the following mass-spring systems will have the highest frequency of vibration? Case A: A spring with a k=300 N/m and a mass of 200 g suspended from it. Case B: A spring with a k=400 N/m and a mass of 200 g suspended from it. See Answer Answer: Case B has the highest frequency. Frequency and period are inversely related. The highest frequency will have the shortest (smallest) period. Both springs have the same mass; only the spring constant (k) is different. A spring with a higher spring constant will have a shorter period. This is consistent with the equation for period. Which of the following mass-spring systems will have the highest frequency of vibration? Case A: A spring with a k=300 N/m and a mass of 200 g suspended from it. Case B: A spring with a k=300 N/m and a mass of 100 g suspended from it. See Answer Answer: Case B has the highest frequency. Frequency and period are inversely related. The highest frequency will have the shortest (smallest) period. Both springs have the same spring constant; only the suspended mass (m) is different. A spring with a smaller suspended mass will have a shorter period. This is consistent with the equation for period. Jump To Next Lesson: Waves and Wavelike Motion Report This Ad Tired of Ads? Go ad-free for 1 year Privacy Manager privacy contact home about terms © 1996-2025 The Physics Classroom, All rights reserved. By using this website, you agree to our use of cookies. We use cookies to provide you with a great experience and to help our website run effectively. Freestar.comReport This Ad
8423
https://handwiki.org/wiki/Super-logarithm
Super-logarithm - HandWiki Anonymous Not logged in Create account Log in Hand W iki Search Super-logarithm From HandWiki bookmark [x] Namespaces Page Discussion More More Page actions Read View source History ZWI Export Short description: Arithmetic function In mathematics, the super-logarithm is one of the two inverse functions of tetration. Just as exponentiation has two inverse functions, roots and logarithms, tetration has two inverse functions, super-roots and super-logarithms. There are several ways of interpreting super-logarithms: As the Abel function of exponential functions, As the inverse function of tetration with respect to the height, As a generalization of Robert Munafo's large number class system, For positive integer values, the super-logarithm with base-e is equivalent to the number of times a logarithm must be iterated to get to 1 (the Iterated logarithm). However, this is not true for negative values and so cannot be considered a full definition. The precise definition of the super-logarithm depends on a precise definition of non-integer tetration (that is, y x for y not an integer). There is no clear consensus on the definition of non-integer tetration and so there is likewise no clear consensus on the super-logarithm for non-integer inputs. [x] Contents 1 Definitions 2 Approximations 2.1 Linear approximation 2.2 Quadratic approximation 3 Approaches to the Abel function 4 Properties 5 Super-logarithm as inverse of tetration 6 See also 7 References 8 External links Definitions The super-logarithm, written slog b(z), is defined implicitly by slog b(b z)=slog b(z)+1 and slog b(1)=0. This definition implies that the super-logarithm can only have integer outputs, and that it is only defined for inputs of the form b,b b,b b b, and so on. In order to extend the domain of the super-logarithm from this sparse set to the real numbers, several approaches have been pursued. These usually include a third requirement in addition to those listed above, which vary from author to author. These approaches are as follows: The linear approximation approach by Rubstov and Romerio, The quadratic approximation approach by Andrew Robbins, The regular Abel function approach by George Szekeres, The iterative functional approach by Peter Walker, and The natural matrix approach by Peter Walker, and later generalized by Andrew Robbins. Approximations Usually, the special functions are defined not only for the real values of argument(s), but to complex plane, and differential and/or integral representation, as well as expansions in convergent and asymptotic series. Yet, no such representations are available for the slog function. Nevertheless, the simple approximations below are suggested. Linear approximation The linear approximation to the super-logarithm is: slog b(z)≈{slog b(b z)−1 if z≤0−1+z if 0<z≤1 slog b(log b⁡(z))+1 if 1<z which is a piecewise-defined function with a linear "critical piece". This function has the property that it is continuous for all real z (C 0 continuous). The first authors to recognize this approximation were Rubstov and Romerio, although it is not in their paper, it can be found in their algorithm that is used in their software prototype. The linear approximation to tetration, on the other hand, had been known before, for example by Ioannis Galidakis. This is a natural inverse of the linear approximation to tetration. Authors like Holmes recognize that the super-logarithm would be a great use to the next evolution of computer floating-point arithmetic, but for this purpose, the function need not be infinitely differentiable. Thus, for the purpose of representing large numbers, the linear approximation approach provides enough continuity (C 0 continuity) to ensure that all real numbers can be represented on a super-logarithmic scale. Quadratic approximation The quadratic approximation to the super-logarithm is: slog b(z)≈{slog b(b z)−1 if z≤0−1+2 log⁡(b)1+log⁡(b)z+1−log⁡(b)1+log⁡(b)z 2 if 0<z≤1 slog b(log b⁡(z))+1 if 1<z which is a piecewise-defined function with a quadratic "critical piece". This function has the property that it is continuous and differentiable for all real z (C 1 continuous). The first author to publish this approximation was Andrew Robbins in this paper. This version of the super-logarithm allows for basic calculus operations to be performed on the super-logarithm, without requiring a large amount of solving beforehand. Using this method, basic investigation of the properties of the super-logarithm and tetration can be performed with a small amount of computational overhead. Approaches to the Abel function The Abel function is any function that satisfies Abel's functional equation: A f(f(x))=A f(x)+1 Given an Abel function A f(x) another solution can be obtained by adding any constant A f′(x)=A f(x)+c. Thus given that the super-logarithm is defined by slog b(1)=0 and the third special property that differs between approaches, the Abel function of the exponential function could be uniquely determined. Properties Other equations that the super-logarithm satisfies are: slog b(z)=slog b(log b⁡(z))+1 slog b(z)≥−2 for all real z Probably the first example of a mathematical problem where the solution is expressed in terms of super-logarithms, is the following: Consider oriented graphs with N nodes and such that oriented path from node i to node j exists if and only if i>j. If length of all such paths is at most k edges, then the minimum possible total number of edges is: Θ(N 2) for k=1 Θ(N log⁡N) for k=2 Θ(N log⁡log⁡N) for k=3 Θ(N slog N) for k=4 and k=5(M. I. Grinchuk, 1986; cases k>5 require super-super-logarithms, super-super-super-logarithms etc.) Super-logarithm as inverse of tetration f=s l o g e(z) in the complex z-plane. As tetration (or super-exponential) s e x p b(z):=z b is suspected to be an analytic function, at least for some values of b, the inverse function s l o g b=s e x p b−1 may also be analytic. Behavior of s l o g b(z), defined in such a way, the complex z plane is sketched in Figure 1 for the case b=e. Levels of integer values of real and integer values of imaginary parts of the slog functions are shown with thick lines. If the existence and uniqueness of the analytic extension of tetration is provided by the condition of its asymptotic approach to the fixed points L≈0.318+1.337 i and L∗≈0.318−1.337 i of L=ln⁡(L) in the upper and lower parts of the complex plane, then the inverse function should also be unique. Such a function is real at the real axis. It has two branch points at z=L and z=L∗. It approaches its limiting value −2 in vicinity of the negative part of the real axis (all the strip between the cuts shown with pink lines in the figure), and slowly grows up along the positive direction of the real axis. As the derivative at the real axis is positive, the imaginary part of slog remains positive just above the real axis and negative just below the real axis. The existence, uniqueness and generalizations are under discussion. See also Iterated logarithm Tetration References ↑М. И. Гринчук, О сложности реализации последовательности треугольных булевых матриц вентильными схемами различной глубины, in: Методы дискретного анализа в синтезе управляющих систем, 44 (1986), pp. 3—23. ↑Peter Walker (1991). "Infinitely Differentiable Generalized Logarithmic and Exponential Functions". Mathematics of Computation (American Mathematical Society) 57 (196): 723–733. doi:10.2307/2938713. ↑H.Kneser (1950). "Reelle analytische Losungen der Gleichung φ(φ(x))=e x und verwandter Funktionalgleichungen". Journal für die reine und angewandte Mathematik187: 56–67. doi:10.1515/crll.1950.187.56. ↑Tetration forum, Ioannis Galidakis, Mathematics, published online (accessed Nov 2007). W. Neville Holmes, Composite Arithmetic: Proposal for a New Standard, IEEE Computer Society Press, vol. 30, no. 3, pp.65–73, 1997. Robert Munafo, Large Numbers at MROB, published online (accessed Nov 2007). C. A. Rubtsov and G. F. Romerio, Ackermann's Function and New Arithmetical Operation, published online (accessed Nov 2007). Andrew Robbins, Solving for the Analytic Piecewise Extension of Tetration and the Super-logarithm, published online (accessed Nov 2007). George Szekeres, Abel's equation and regular growth: variations on a theme by Abel, Experiment. Math. Volume 7, Issue 2 (1998), 85–100. Peter Walker, Infinitely Differentiable Generalized Logarithmic and Exponential Functions, Mathematics of Computation, Vol. 57, No. 196 (Oct., 1991), pp.723–733. External links Rubstov and Romerio, Hyper-operations Thread 1[yes|permanent dead link|dead link}}] Rubstov and Romerio, Hyper-operations Thread 2[dead link] | Collapse v t e Hyperoperations | | Primary | Successor (0) Addition (1) Multiplication (2) Exponentiation (3) Tetration (4) Pentation (5) | | Inverse for left argument | Subtraction (1) Division (2) n th root (3) Super-root (4) | | Inverse for right argument | Subtraction (1) Division (2) Logarithm (3) Super-logarithm (4) | | Related articles | Ackermann function Bowers's operators Conway chained arrow notation Grzegorczyk hierarchy Knuth's up-arrow notation Steinhaus–Moser notation | Süper logaritma 0.00 (0 votes) Original source: Read more Retrieved from " Categories: All articles with dead external links Articles with dead external links from December 2023 Logarithms Hidden categories: Articles with invalid date parameter in template Articles with permanently dead external links Computing portal Encyclopedia of Knowledge Portals Main pageData analysisAstronomy & SpaceBiologyComputer conceptsChemistryMathematicsPhysicsEarth studiesUnsolved problems HistoryPhilosophySocial studiesReligionMedicine Engineering & TechSoftware programsFinance & Business BiographiesOrganizationsCompaniesPlaces BooksMonographsTutorialsManuals Navigation Navigation Add a new article Search in all topics Search in namespaces Search in categories Search using prefix Help About HandWiki FAQs How to edit Citation manager Formatting articles List of categories Recent pages Recent changes Random page Support & Donate Translate Select Language​▼ Wiki tools Wiki tools Special pages Cite this page Page tools Page tools User page tools More What links here Related changes Printable version Permanent link Page information Page logs Other projects In other languages Add links Categories Categories All articles with dead external links Articles with dead external links from December 2023 Logarithms Hidden categories Articles with invalid date parameter in template Articles with permanently dead external links This page was last edited on 8 February 2024, at 18:30. 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8424
https://www.cuemath.com/questions/a-coin-is-tossed-3-times-what-is-the-probability-of-getting-at-least-one-head/
A coin is tossed 3 times. What is the probability of getting at least one head? Solution: Given, a coin is tossed 3 times. We have to find the probability of getting at least one head. The possible outcomes are T H H H T H H H T T T H T H T H T T T T T H H H We observe that there is only one scenario in throwing all coins where there are no heads. The chances for one given coin to be heads is 1/2. The chance for all three to have the same result would be (1/2)3 = (1/2)(1/2)(1/2) = 1/8 So, the probability to have atleast one head is 1 - 1/8 = (8 - 1)/8 = 7/8 Therefore, the probability of getting at least one head is 7/8. A coin is tossed 3 times. What is the probability of getting at least one head? Summary: A coin is tossed 3 times. The probability of getting at least one head is 7/8. Math worksheets andvisual curriculum FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math ABOUT US Our Mission Our Journey Our Team QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad MATH TEST Math Kangaroo AMC 8 MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math FOLLOW CUEMATH Facebook Youtube Instagram Twitter LinkedIn Tiktok MATH PROGRAM Online math classes Online Math Courses online math tutoring Online Math Program After School Tutoring Private math tutor Summer Math Programs Math Tutors Near Me Math Tuition Homeschool Math Online Solve Math Online Curriculum NEW OFFERINGS Coding SAT Science English MATH CURRICULUM 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math MATH TEST CAASPP CogAT STAAR NJSLA SBAC Math Kangaroo AMC 8 ABOUT US Our Mission Our Journey Our Team MATH TOPICS Algebra 1 Algebra 2 Geometry Calculus math Pre-calculus math Math olympiad Numbers Measurement QUICK LINKS Maths Games Maths Puzzles Our Pricing Math Questions Blogs Events FAQs MATH ONLINE CLASSES 1st Grade Math 2nd Grade Math 3rd Grade Math 4th Grade Math 5th Grade Math 6th Grade Math 7th Grade Math 8th Grade Math Terms and ConditionsPrivacy Policy
8425
https://oec.xmu.edu.cn/virtual_attach_file.vsb?afc=tU4Tj2U4lbnR62owR-bnlLaM8nRUlQXRnNlYM8C8oR9aoRG0gihFp2hmCIa0nSybLSybnSyDnmNYM7l4M7lDnRlDozLanz-8nRnkM4L4nlrFnR9ZU8WkMzQFozNDol7Jv2nto4ODnmNYM7l4M7lDnRlDozLanz-8nRnkM4L4nlrFnR9ZU8WkMzQFozNDol7Jv2nto4xjqjXsqjTZ_ShXptQ0g47aMzC0MmUPLR7Jqd/nx&oid=1933934427&tid=1056&nid=64101&e=.pdf
２０１６年第３期 （总第８０期） 海外华文教育 ＯＶＥＲＳＥＡＳＣＨＩＮＥＳＥＥＤＵＣＡＴＩＯＮ Ｎｏ．３２０１６ ＧｅｎｅｒａｌＳｅｒｉａｌＮｏ．８０ 收稿日期：２０１５０３３１ 作者简介：刘吉力，男，北京师范大学文学院博士生，长沙师范学院初等教育系助教，研究方向为现代汉语共时状况及历时演 变。Ｅｍａｉｌ：２０１３３１０８００２５＠ｍａｉｌ．ｂｎｕ．ｅｄｕ．ｃｎ 海峡两岸量词“通”对比考察 刘吉力 （北京师范大学文学院，中国 北京 １００８７５） 摘　要：语法的差异与融合是海峡两岸现代汉语对比研究的重要内容。 对比考察两岸量词 “ 通” ，展现了它在语法分布倾向性、语法形式和语义色彩等三个方面存在较大差异。当前两岸量 词“ 通” 的融合主要是大陆与电话搭配的“ 通” 向台湾靠拢。文章以时间为线索，采用数据统计分 析的方法探讨两岸“ 通” 的融合，不仅关注数据变化以及语法形式本身的变化，还在重点考察书面 语的同时兼及口语，以期对两岸“ 通” 的融合作出较好的说明。 关键词：海峡两岸；量词；“ 通” 中图分类号：Ｈ１９３．５ 文献标志码：Ａ 文章编号：２２２１９０５６（２０１６）０３０３２３１０ ＤＯＩ编码：１０．１４０９５／ｊ．ｃｎｋｉ．ｏｃｅ．２０１６．０３．００４ 一、引　言 从“ 五四” 时期到上世纪５０年代，大陆的通用语被称作“ 国语” 。１９５０年代以后，国语在相互 隔绝的两个空间里发展，最终形成两个变体：大陆“ 国语” （普通话）和台湾“ 国语” （李行健、仇志 群，２０１４）。上世纪８０年代后期以来，随着海峡两岸逐步开放交流，两岸民族共同语（即大陆普通 话和台湾“ 国语” ）在差异当中已有不同程度的融合。 海峡两岸现代汉语对比属于两岸民族共同语内部的对比，它以两岸共同语之间的差异与融合 作为研究内容，涉及语音、文字、词汇、语法等诸多要素，研究目的主要在于揭示差异，描写融合，探 讨共同语发展变异的动因，并对其在两岸的发展趋势作一定预测，从而加深对共同语的研究（刁晏 斌，１９９７；邢福义、汪国胜，２０１２）。 语法的差异与融合是两岸现代汉语对比研究的重要内容，其中，量词在两岸共同语中存在不少 差异与融合。刁晏斌（２０００：１２１— １２５）较早指出两岸存有差异的量词，如“ 份” “ 间” “ 项” “ 通” 等 等。魏兆惠、华学诚（２００８）对量词“ 通” 的历时发展演变作了细致描写与分析，该文侧重“ 通” 在古 代汉语中的发展变化，对它在现代汉语中的发展情况讨论不多。 邹嘉彦、莫宇航（２０１３）主要比较 了量词“ 通” 在北京、上海、香港、台湾书面语中搭配对象上的差异，并尝试讨论了上海和香港“ 通” 的用法受到台湾影响的可能性，但该文对“ 通” 在上述四地的融合情况涉及较少。本文试图从现代 汉语的发展与演变出发，详细考察两岸量词“ 通” 的差异，并着重探讨两岸“ 通” 的融合。 · ３ ２ ３ · 海外华文教育 ２０１６年 二、两岸量词“ 通” 的差异 从两岸的语料来看，两岸量词“ 通” 在语法分布倾向性、语法形式和语义色彩等三个方面存在 较大差异，下面逐一加以论述。 （一）语法分布倾向性差异 邹嘉彦、莫宇航（２０１３）依据香港ＬＩＶＡＣ共时语料库，分析得出台湾书面语里量词“ 通” 主要的 搭配对象为电子通讯工具类名词，其中以“ 电话” 类名词最为突出。检索台湾“ 中研院” 现代汉语语 料库（以下简称“ 中研院语料库” ）约５００万词语料，我们得到下表数据： 表１　“ 中研院语料库” 量词“ 通” 搭配对象数据表 搭配对象 电话 ［１］ 言语、动作 其他 合计 数量 ５６ ２１ ２ ７９ 百分比 ７０．９％ ２６．６％ ２．５％ １００％ 表１显示与电话搭配的“ 通” 占比约７０％，而与言语、动作搭配的“ 通” 占比不足３０％，这印证了 邹嘉彦、莫宇航（２０１３）的上述观点，实例如下： （１）一通电话过去准有收获。 （２）教育厅连日来接获多通电话，指责省立美术馆的管理出现弊端。 （３）东尼摆出一派外交官姿态，天南地北，与人胡扯一通。 （４）有时候，两口子拌了嘴，就到我这儿来哭一通。（以上４例均出自“ 中研院语料库” ） 为了了解大陆量词“ 通” 的使用情况，我们检索北京大学 ＣＣＬ现代汉语语料库（约６亿字符）， 得到表２数据： 表２　ＣＣＬ现代汉语语料库量词“ 通” 搭配对象数据表 搭配对象 言语、动作 电报、文书、书信、石碑等 电话 合计 数量 １５０８ ２２１ ４２ １７７１［２］ 百分比 ８５．１％ １２．５％ ２．４％ １００％ 表２显示与言语或动作搭配的“ 通” 占比约８５％，与电报、文书、书信、石碑、电话等搭配的“ 通” 合共约占１５％，其中与电话搭配的“ 通” 不足２．５％，实例如下： （５）同样一通电话，小灵通和ＧＳＭ之间价差在１０倍左右。 （６）海藻今天被几通电话骚扰。 （７）这事传到扈尔干耳里，他把康古六喊去，狠狠训斥一通。 （８）日军１８３架飞机悄悄飞临，一通狂轰滥炸，眨眼之间珍珠港一片火海冲腾。 （９）信中说：有一件令我惊喜的事，就是我收到了宋美龄一通电报，为我的胜利致贺，并称我为 英雄。 （１０）山西省吉县日前发现一通记载有明代大地震情况的石碑。 （以上６例均出自ＣＣＬ现代汉语语料库） 统计表１和表２的数据，可以得到以下结论：台湾量词“ 通” 多数与电话搭配，与言语或动作搭 配的只占少数；大陆量词“ 通” 大多与言语或动作搭配，用于搭配电话的极少。 · ４ ２ ３ · 第３期 刘吉力：海峡两岸量词“ 通” 对比考察 两岸相关工具书对此已有所体现。台湾重编国语辞典修订本网络版（１９９４）对量词“ 通” 有如 下解释： 量词：（１）计算文书、电讯的单位。如：“ 一通电话” “ 三通电报” 。（２）计算敲击钟鼓次数的单 位。如“ 擂鼓三通” 。 这里举例说明“ 通” 计量电话，没有提到“ 通” 与言语搭配，与动作搭配也仅有“ 敲击钟鼓” 。 再看大陆。郭先珍《现代汉语量词用法词典》（２００２：１４６－１４７）把量词“ 通” 分为名量词和动量 词，名量词“ 通” 计量文书、告示、电报、书信等，动量词“ 通” 计量言语、动作。该词典没有提到“ 通” 计量电话的用法。 由此我们推断，两岸在选择与“ （接／打）电话” 组配的量词时也会存在差异。 基于语料统计与 分析，我们分别检索“ 中研院语料库” 语料和国家语委语料库在线现代汉语语料库（简称“ 语料库在 线” ）约２０００万字符语料，对两岸与“ （接／打）电话” 搭配的量词作了比较。请看下表： 表３　“ 中研院语料库” 与“ （接／打）电话” 搭配的量词数据表 量词 通 个 次 合计 数量 ５４ ４０ ８ １０２ 百分比 ５３％ ３９．２％ ７．８％ １００％ 表４　“ 语料库在线” 与“ （接／打）电话” 搭配的量词数据表 量词 个 次 合计 数量 ６２ ８ ７０ 百分比 ８８．６％ １１．４％ １００％ 表３表明台湾计量“ （接／打）电话” 的量词主要有“ 通” “ 个” “ 次” ，以“ 通” 和“ 个” 为主，尤其是 “ 通” ；表４说明大陆计量“ （接／打）电话” 的量词主要有“ 个” 和“ 次” ，以“ 个” 为主，未见“ 通” 。 考察两岸大规模共时语料库，我们得知两岸量词“ 通” 都可与电话、言语、动作等搭配，台湾如 例（１）－（４），大陆如例（５）－（８），这体现了两岸“ 通” 语法分布的相似性。但是，台湾“ 通” 倾向于与 电话组配，大陆则倾向于与言语、动作搭配，而且大陆“ 通” 的搭配范围除了言语、动作、电话外，还 有石碑等比较古旧的事物，如例（１０），较之台湾“ 通” ，大陆搭配范围要略广。 同样，两岸都选择与 “ （接／打）电话” 组配的量词有“ 个” 和“ 次” ，但台湾倾向于选择“ 通” ，而大陆倾向于使用“ 个” 。由 此看来，这种倾向性的差异在很大程度上可以看作是两岸量词“ 通” 的真正差异。储泽祥（２０１１）提 出在多样性基础上进行倾向性考察的语法比较思路。这对两岸语法对比研究思路的拓展具有方法 论上的创新意义。 （二）语法形式差异 与电话、言语、动作等搭配是两岸量词“ 通” 在搭配对象上的交集，这里的语法形式差异只就搭 配电话、言语、动作的“ 通” 及其组成结构而言。 １．搭配电话 两岸“ 通” 搭配电话时，其前多用数词“ 一” ，如上文例（１）（５），但不限于“ 一” ，如例（２）“ 多通 电话” 和例（６）“ 几通电话” ，这是它们语法形式一致的地方。不过，台湾“ 通” 与大陆相比，其语法 形式有不少差别，例如： （１１）张仁泽也表示，在这段期间，平均每天能接到十到二十通家长打来找家教的电话。（“ 中 研院语料库” ） · ５ ２ ３ · 海外华文教育 ２０１６年 （１２）台南市消防局表示，前晚到昨天清晨气温陡降，１１９不断接获求救电话，一通接一通，许多 来不及保暖的长者被紧急送往各大医院。（《中国时报》２０１３．１１．２９） （１３）陈菊拨、接给刘世芳及接收陈存永的简讯四通、秘书来电及拨打新闻局长丁允恭电话等 总共二十一通。（《自由时报》２０１４．９．９） 例（１１）“ 通” 前“ 十到二十” 为概数，“ 通” 与“ 电话” 之间隔着较长的修饰性成分；例（１２）“ 一通 接一通” 实际上是“ 一通电话接一通电话” ，这里“ 电话” 承前省略，“ 一通” 反复出现；例（１３）“ 来电” “ 电话” 提前，数量组合“ 二十二通” 后置。 由于“ 通” 在台湾主要搭配电话，“ 一通” 有时直接指代“ 一通电话” ，如： （１４）专线也曾接获学龄前小朋友反映功课问题，或希望在学校认识更多朋友，其中还有一通 是妈妈帮忙拨通后，让孩子向接线大哥哥大姊姊倾诉心声。（《国语日报》２００２．７．２５） （１５）代理台湾Ｓｋｙｐｅ的ＰＣｈｏｍｅ通讯应用服务部总监蔡文雄昨天宣布，将在年底前跟进降价， 可能降为一通二元或付月租费讲到饱；且年底前会解决Ｃａｌｌｉｎ问题，届时会员也可透过网路接收 来电。（《联合报》２００７．４．２０） 此外，“ 一通” 还可以重叠，与电话搭配表示电话数量之多，例如： （１６）上网，几乎成为现代人每天的例行公事，想要安排一趟旅行，能够在线上订位，就让人懒 得一通一通电话打去询问是否还有空位。（《经济日报》２００２．１．２７） （１７）一通通电话打进台南县政府要求办跨年晚会，县长苏焕智顺应民意，决定举办没有大卡 斯的公益跨年晚会，１０００个“ 无米乐” 福袋送给民众，还有捐发票摸彩活动。 （《联合报》２００８．１２． ３０） 总起来看，台湾搭配电话的“ 通” 及其组成结构在语法形式上比较丰富，上举各例灵活多样的 表达形式在大陆还比较少见。 ２．搭配言语、动作 两岸“ 通” 与言语、动作组配时，其前的数词一般都为“ 一” ，大陆“ 一通” 可位于动词前后作补 语或状语，如例（７）“ 训斥一通” 和例（８）“ 一通狂轰滥炸” ，但以作补语为主；而台湾“ 一通” 一般都 位于动词后作补语，多构成“ 胡／乱 Ｖ一通” 的固定语法框架，使用形式受限，如上文例（３）“ 胡扯一 通” 。再如： （１８）因为捕狗队根本搞不清楚是哪一只狗伤人，结果乱抓一通，问题不但没有解决，甚至让无 辜的狗遭到安乐死。（《台湾立报》２０１２．２．１２） （１９）万一搞不清楚状况，乱补一通，反而会造成反效果。（《自立晚报》２０１３．１．３） （２０）台东县农业处今天早上遭小偷光顾，办公室抽屉被宵小乱翻一通。（《自由时报》２０１４．８．８） 以上三例中“ 乱抓／补／翻一通” 都是“ 乱Ｖ一通” 形式。 据魏兆惠、华学诚（２００８），在汉语史上“ 通” 作动量词的使用频率并不高，而在现代汉语中“ 通” 的动量词用法已经比较常见 ［３］。考察现代汉语史第一阶段“ 通” 搭配言语、动作的使用情况 ［４］，在 我们搜集的３０个例句中，“ Ｖ一通” 形式共２９例 ［５］，如： （２１）两个浪子，打扮做小丑模样，大玩了一通回来了。（张爱玲《谈画》１９４４年） （２２）刘老九里里外外看了一通，见箩筐没有漏洞，麻索用力扯扯，也还可以挑挑，就颇为趁心 如意。（艾芜《丰饶的原野》１９４５年） （２３）然而后或续或改，非借尸还魂，即冥中另配，必令“ 生旦当场团圆” 才肯放手者，乃是自欺 欺人的瘾太大，所以看了小小骗局，还不甘心，定须闭眼胡说一通而后快。（鲁迅《坟》１９２５年） （２４）老孙头把坦白光荣这些新字眼，乱用一通，说得老万笑起来，把东屋萧队长笑醒了。（周 立波《暴风骤雨》１９４８年） · ６ ２ ３ · 第３期 刘吉力：海峡两岸量词“ 通” 对比考察 台湾“ 胡／乱Ｖ一通” 形式与例（２１）（２２）“ Ｖ一通” 基本一致，与例（２３）（２４）“ 胡说一通” “ 乱用 一通” 完全一致。 前面我们知道，大陆“ 通” 大多与言语、动作搭配，而台湾“ 通” 则主要与电话组配，这样大陆 “ 通” 与言语、动作搭配的使用频率比台湾要高出许多，这有助于它突破“ Ｖ一通” 形式的限制，从而 形成“ 一通Ｖ” 。台湾搭配电话的“ 通” ，其语法形式丰富也应与使用频率高存在一定关联。比较而 言，台湾“ 通” 与言语、动作搭配所组成的语法形式，更多的应该是对现代汉语史第一阶段的继承和 延续。 （三）语义色彩差异 一般来讲，搭配名词的“ 通” 没有语义色彩，这里的语义色彩专指搭配动词的“ 通” 。 从整个现代汉语史来看，动量词“ 通” 的语义色彩经过了由中性→贬义→趋向于中性的发展过 程（刁晏斌，２００６：１７１— １７２）。应该说，“ 由贬义趋向于中性” 的语义色彩主要是就现代汉语史第四 阶段的大陆普通话而言，例如： （２５）有一次在还没有改建的北京大学大讲堂里开了一个什么会，专门向同学们谈国学。当时 主席台上共坐着五位教授，每个人都讲上一通。（季羡林《病榻杂记》２００３年） （２６）专家们正要离开棚区结束讲座，贾宋镇郄村几个菜农丢下农活跑了过来，围着专家又是 一通提问… … （《人民日报》２００９．８．２） （２７）苦桷沟长满了郁郁葱葱的苦桷藤，小时候总爱到苦桷沟挖出苦桷根来，然后到盘龙河边 的滩边捶上一通，不一会儿就会有大片鱼儿翻着白肚子浮在水面上。（《人民日报》２０１４．６．２） 上面三例中“ 通” 的中性色彩都比较明显。 与大陆“ 通” 既用作贬义也用作中性不同，台湾 “ 通” 一般用作贬义，如上例（１８）（１９）“ 乱抓一通” “ 乱补一通” 。“ 中研院语料库” 搭配动词的“ 通” 共２１例，其中１８例带有贬义色彩，占８６％。再以台湾《自由时报》（２０１４．６．１０— ２０１４．９．１０）为例，与 动词搭配的“ 通” 共有２２例，全部用作贬义，且都是“ 乱 Ｖ一通” 的固定形式，如上例（２０）“ 乱翻一 通” 。台湾“ 通” 与带有贬义的“ 胡／乱” 组配，构成“ 胡／乱Ｖ一通” ，与之相适应的是，“ 通” 也带上了 贬义色彩。 三、两岸量词“ 通” 的融合 从上文来看，两岸量词“ 通” 在语法分布倾向性、语法形式和语义色彩上存在较大差异。 社会 分化造成语言分化，社会接触则造成语言不同程度的趋同（叶蜚声、徐通锵，２００９：１９６）。 周清海 （２００８）认为，中国改革开放之后各地华语与现代汉语标准语（即普通话）的相互冲击与交融的情 况，是前所未有的。两岸共同语本是同根同源，开放交流后，人员往来日益频繁，加之电话、电视、网 络等通讯、传播工具广泛使用，两岸量词“ 通” 已呈现融合态势。 刁晏斌（２０００：２）认为，海峡两岸语言的融合是一个趋同的过程，到目前为止，海峡两岸语言某 种程度的融合，主要是通过大陆语言的变化而实现的。据语料调查显示，当前两岸量词“ 通” 的融 合主要是大陆与电话搭配的“ 通” 向台湾靠拢。 这是一个引进、吸收进而融合的过程（刁晏斌， １９９７）。 在正式论述两岸“ 通” 的融合之前，我们首先讨论“ 通” 的搭配对象在台湾的发展变化，然后再 探讨大陆“ 通” 与台湾的融合。 （一）台湾“ 通” 搭配对象的发展变化 台湾“ 中研院语料库” 语料的收集始于１９９０年前后，该库于１９９７年１０月完成 ［６］，这说明在上 世纪９０年代前台湾量词“ 通” 即已主要搭配电话。我们以“ 通” 最常见的数量组合“ 一通” 为例，定 · ７ ２ ３ · 海外华文教育 ２０１６年 点调查了上世纪５０－９０年代初台湾《联合报》部分语料 ［７］，以每十年选取一个时间段，每段为该年 的前六个月，见下表： 表５　《联合报》“ 一通” 搭配对象数据表 搭配对象 时间段 电话 电报、文书、书信等 言语、动作 其他 合计 １９５２．１．１－６．３０ ０ １５／８８．２％ ２／１１．８％ ０ １７／１００％ １９６２．１．１－６．３０ ０ ８／４７％ ９／５３％ ０ １７／１００％ １９７２．１．１－６．３０ １１／４０．７％ １４／５１．９％ ２／７．４％ ０ ２７／１００％ １９８２．１．１－６．３０ ２７／５８．７％ ３／６．５％ １５／３２．６％ １／２．２％ ４６／１００％ １９９２．１．１－６．３０ ８３／７９．８１％ ２／１．９２％ １８／１７．３１％ １／０．９６％ １０４／１００％ 实例如下： （２８）她给读卖新闻的声明，系应该报之请，对六月九日拍给社会党党员松冈驹吉的一通由她 具名的电报加以说明，该电报中透露说：… … （《联合报》１９５２．６．１７） （２９）为什么装上一个原子炉就要骂一通祖宗？（《联合报》１９６２．１．１） （３０）下午六时，警察电台接到一通电话，一位叫做原勉的日籍观光客说… … （《联合报》 １９７２．５．２０） （３１）夏光莉虽然理由充足的推掉了柯俊雄的“ 美意” ，可是柯小生并不死心，仍然有事没事的 一通通电话打到深坑夏宅，操起他的台湾国语，发动他的影帝魅力，向夏小姐倾吐出一箩筐一箩筐 的甜言蜜语。（《联合报》１９８２．２．６） （３２）有关大专寒训通知，每年十二月份都由本课转发，去年就接到一通由某参加寒训学生的 家长打来询问的电话：… … （《联合报》１９８２．６．１５） 表５显示，在我们考察的５个时间段中，“ 一通” 搭配电话出现于上世纪７０年代初，在接下来的 １９８０、１９９０年代初，电话逐渐成为“ 一通” 最主要的搭配对象。 与“ 一通” 搭配电话不断增多相反， 在上世纪５０— ９０年代初，电报、文书、书信等由最初是“ 一通” 主要的搭配对象，进而逐步成为最次 要的搭配对象，到１９９０年代初，其所占比例不足２％；尤其是在１９７０— １９８０年代初，降幅十分明显。 而在１９５０— １９９０年代初，言语、动作一般都是“ 一通” 相对次要的搭配对象。 电报、文书、书信等均见于书面，我们可以将其概括为纸质通信类名词，上世纪五十年代以来， 它的使用范围逐渐缩小，而作为后起的电子通信类名词电话，它的使用频率不断增加，进而逐步取 代前者。由于计量对象本身的发展变化，“ 通” 由主要计量电报等逐渐发展为主要计量电话，这是 社会发展、科技进步在语言中的投射，同时也是“ 语言与社会共变” 之一例。 从调查结果来看，新世纪台湾“ 通” 依然主要搭配电话。 以《自由时报》（２０１４．６．１０— ２０１４．９． １０）为例，相关数据如下： 表６　《自由时报》量词“ 通” 搭配对象数据表 搭配对象 电话 言语、动作 其他 合计 数量 ９８ ２２ ５ １２５ 百分比 ７８．４％ １７．６％ ４％ １００％ · ８ ２ ３ · 第３期 刘吉力：海峡两岸量词“ 通” 对比考察 表６显示搭配电话的“ 通” 占比近８０％，这与表１数据基本吻合。 （二）大陆“ 通” 与台湾的融合 大陆ＣＣＬ现代汉语语料库语料多属于１９９６年之前的（邵敬敏，２００７），“ 语料库在线” 现代汉语 语料以２００２年前近２０年的语料为主 ［８］，它们反映的是普通话距今十多年前乃至更早时间里的状 况。为了动态地研究大陆量词“ 通” ，这里以规范程度很高，且能反映普通话历时变化的《人民日 报》为例，检索《人民日报》图文数据库（１９４６．５．１５— ２０１３．１２．３１）近６８年的数据，得到表７： 表７　《人民日报》量词“ 通” 搭配“ 电话” 数据表 年份 １９９２ ２００４ ２００９ ２０１２ ２０１３ 总数 次数 １ １ ２ ３ １１ １８ 表７显示“ 通” 搭配电话的用例首次出现是在１９９２年，即： （３３）去年，她很忙，我们只是偶尔在社交场合碰面，或是一通电话而已。（１９９２．１．４） 在往后的２１年里，“ 通” 搭配电话整体上呈递增趋势，而且首次出现时的“ 一通电话” 还有了新 的发展： （３４）一名在高雄市采访社会新闻３０多年的资深记者，有天傍晚接到一通自称绑架他女儿的 电话。（２００４．４．２１） 例（３４）“ 通” 与“ 电话” 之间隔着比较复杂的修饰性成分，这与台湾例（１１）类似，而且该例显示 出，大陆“ 通” 用于搭配电话，具有从台湾引进的痕迹。 接下来，“ 通” 前量词也不再限于“ 一” ，如： （３５）李风发挥了央视人干练的风格，几通电话后，问题及时解决… … （２００９．１２．１９） （３６）我一个星期要接到乡亲们打来的十几通电话，全天２４小时开机，有任何困难村民随时打 给我。（２０１２．１．１７） 语言接触不仅指不同民族语言的接触，而且指同一种语言的不同变体（包括地方变体、社会变 体、功能变体、地区变体和国际变体等）的接触（张兴权，２０１２：６）。《人民日报》量词“ 通” 与电话搭 配从无到有，由少到多，语法形式由简单到复杂，这应该与两岸共同语的不同地域变体（即大陆普 通话和台湾“ 国语” ）在过去二三十年里不断接触进而融合有着密切联系。 为了进一步探究大陆“ 通” 搭配“ 电话” 的使用情况，我们检索了“ 中青在线站内全文检索系 统” （简称“ 中青在线” ），与《人民日报》图文数据库语料相比，“ 中青在线” 规范度相对较低、时间跨 度比较短，但覆盖面更广，且能反映“ 通” 在最近十多年的发展变化，检索数据如下： 表８　“ 中青在线” 量词“ 通” 搭配“ 电话” 数据表 年份 ２００１ ２００２ ２００３ ２００５ ２００６ ２００７ ２００８ ２００９ ２０１０ ２０１１ ２０１２ ２０１３ 总数 次数 ２ ４ １ ２ １ ９ １０ ３６ ４２ ７０ ６０ ８８ ３２５ 表８显示“ 通” 搭配电话在整体上也呈递增趋势，而且在２００６年以后增幅明显。另外，在语法 形式上，不仅“ 通” 与“ 电话” 之间也隔着较长的修饰性成分，如： （３７）也许这是一通事关国家危亡的紧急电话。（中国日报网环球在线２００９．９．１０） “ 一通” 还有了新的变化： （３８）这一天，湖北山东商会会长周垂远也正忙着，一通又一通的电话，谈的只有一个主题——— 大武汉常福国际汽车乐园。（《人民政协报》２０１１．５．６） （３９）孩子们躲在空调房里给密友打着一通通电话，通讯越来越快，人与人之间的距离越来越 · ９ ２ ３ · 海外华文教育 ２０１６年 远。（《中国青年报》２０１１．８．１５） 例（３８）“ 一通又一通” 与台湾例（１６）“ 一通一通” 类似，例（３９）“ 一通通” 与台湾例（１７）相同。 以上这些都体现出量词“ 通” 及其相关语法形式趋于灵活，这与台湾是一致的。 通过对《人民日报》和“ 中青在线” 的全面考察，我们可以看到量词“ 通” 搭配电话在两岸不同 言语社区因为不断接触、渗透从而趋同的具体过程。 这里需要指出的是，量词“ 通” 搭配“ 电话” 在台湾书面语和口语中都不少见，而这在大陆都还 不常见，这说明当前“ 通” 计量电话在两岸的融合度还不太高。就口语来说，根据我们对台湾政治 大学“ 国语口语语料库” 部分会话的调查统计 ［９］，“ 通” 与电话搭配在其第２７段会话（共５７６０字）中 即出现３例 ［１０］，如： （４０）［然后］… … 结果呢… … 就是三通电话打出去… … 然后我们家人就说… … 腪… … （出自 “ 国语口语语料库” ） 再以“ 通” 搭配电话最常见的形式“ 一通电话” 为例，检索中国传媒大学“ 媒体语言语料库” 检 索系统 ［１１］，“ 一通电话” 在中央电视台新近六年（２００８．１．１— ２０１３．１２．３１）约１亿字符的口语语料中 出现１９例，其中９例，说话人具有大陆背景；其他１０例则出自节目《海峡两岸》、有关台湾的新闻 等，说话人都具有台湾背景。下面各举一例： （４１）在回家的路上他接到了一通电话，当时是陈水扁打给他的，陈水扁特别跟他谢谢说，我知 道你刚才拿了两百万，非常谢谢你那两百万。（台湾时事评论员刘宝杰语，出自节目《海峡两岸》 ２００８．８．２８） （４２）而且其实我们发现，在我们接到的每一通电话里，似乎这些保险业务员他们的表述几乎 都是相同的，当然他们的目的都是一致的，就是一定要说服你，让你能够买保险。（大陆主持人陈 伟鸿语，出自节目《今日观察》２０１０．１２．１５） 直接接触是接触双方在时间和空间上都不分离的语言接触，一般通过口语交际实现（贺阳， ２００４）。两岸口语的直接接触让我们进一步看到大陆量词“ 通” 受到了台湾的影响。 此外，从表４得知，以往大陆与“ （接／打）电话” 组配主要用“ 个” ，而表７和表８都表明近些年 来使用“ 通” 的例子逐渐增多。随着电话在大陆广泛普及，普通话对搭配“ （接／打）电话” 的量词的 需求也随之大增，吸收“ 通” 用于计量电话，这有利于丰富其表达。但是，能否据此预测大陆“ 通” 计 量电话的用例会越来越多，两岸“ 通” 的融合度会越来越高，这还需要未来语言事实的检验。 四、结　语 从整体来看，两岸共同语的共性是主流，个性是支流。以往两岸语法对比研究讨论了一些“ 此 有彼无” 或“ 此无彼有” 的比较显豁的差异点，但随着研究的逐步深入，两岸共同语之间的不断融 合，如今像这样的差异点已不太常见。两岸量词“ 通” 的差异说明，两岸语法差异的重要形式可能 体现在“ 同中有异，异中有同” ，而具体差异则大多体现在使用频率以及多样性与倾向性等方面。 两岸共同语的趋同融合是未来的发展趋势，这一趋势是不会改变的（储泽祥、张琪，２０１３）。本 文以时间为线索，采用数据统计分析的方法来探讨两岸“ 通” 的融合，不仅关注数据变化以及语法 形式本身的变化，还在重点分析书面语的同时兼及口语，以期对两岸“ 通” 的融合作出较好的说明。 至于两岸语言融合的动因、过程以及融合度高低的准确描写等问题，则是以后需要继续关注的课 题。 海峡两岸现代汉语对比研究是一个广阔的新兴领域，量词“ 通” 在两岸的差异与融合只是两岸 现代汉语对比研究的一个具体案例，类似的例子还有很多，值得进一步深入挖掘。 · ０ ３ ３ · 第３期 刘吉力：海峡两岸量词“ 通” 对比考察 注释： ［１］根据《现代汉语词典（第６版）》，“ 电话” 义项有三：一指通信方式；二指用电话装置传递的话；三指电话机。语 料调查显示，两岸“ 通” 只计量前两个义项的“ 电话” 。需要说明的是，“ 电话” 义项二也可以看作“ 言语” ，但“ 电 话” 在本文单列，原因有三：一是“ 电话” 义项一不属于“ 言语” ；二是“ 电话” 为名词，而本文“ 言语” “ 动作” 并 列，其一般为动词或动词性词组；三是两岸“ 通” 计量“ 电话” 的情况是全文研究重点。 此外，为了与指“ 电话 机” 的“ 电话” 相区别，文中部分地方用“ （接／打）电话” 表示前两个义项的“ 电话” 。 ［２］不包括港台６８个用例。 ［３］根据《现代汉语量词用法词典》（郭先珍，２００２：１４６! １４７），动量词“ 通” 计量言语、动作。 ［４］刁晏斌（２０００）把现代汉语史分为四个阶段：１９１９年至１９４９年为第一阶段，１９４９年至１９６６年为第二阶段，１９６６ 年至１９７６年为第三阶段，１９７８年至今为第四阶段。 ［５］这２９个例句取自鲁迅《阿Ｑ正传》（６例，以下用数字标明）、《朝花夕拾》（５）、《坟》（１）、《药》（１）、《致姚克》 （１）、《再谈香港》（１）、《二心集》（１），艾芜《丰饶的原野》（３）、《记我的一段文艺生活》（１）、《石青嫂子》（１）、 《花园中》（１），张爱玲《连环套》（１）、《谈画》（１），柔石《为奴隶的母亲》（１），叶紫《插田——— 乡居回忆之一》 （１），端木蕻良《记一二九》（１），周立波《暴风骤雨》（１），巴金《寒夜》（１）；另有１例“ 一通 Ｖ” 形式出自《药》中 “ 一通咳嗽” 。 ［６］根据《“ 中央研究院” 现代汉语标记语料库版简介》（ｈｔｔｐ：／／ａｐｐ．ｓｉｎｉｃａ．ｅｄｕ．ｔｗ）。另本文作者于２０１５年５月８日 就“ 中研院” 现代汉语语料库语料的时间范围问题请教过该库主要建设者之一黄居仁先生，他告诉笔者，“ 中研 院” 现代汉语语料库语料为上世纪９０年代前二三十年的语料。据此我们推断，该库语料主要反映了台湾“ 国 语” 上世纪６０— ９０年代的语言状况。 ［７］该语料由同学褚靓取自台湾联合知识库（ｈｔｔｐ：／／ｕｄｎｄａｔａ．ｃｏｍ／ｕｄｎ），在此表示感谢。 ［８］根据《现代汉语语料库数据及使用说明》（ｈｔｔｐ：／／ｗｗｗ．ｃｎｃｏｒｐｕｓ．ｏｒｇ／ＣｏｒｐｕｓＩｎｔｒｏ．ａｓｐｘ） ［９］关于“ 国语口语语料库” ，请参阅Ｃｈｕｉ＆Ｌａｉ（２００８）。 ［１０］具有台湾背景的黄居仁先生于２０１５年５月８日在讲座中举出“ 两通未接电话” 的例子，这也可以补充说明台 湾口语里量词“ 通” 计量“ 电话” 并不少见。 ［１１］ｈｔｔｐ：／／ｌｉｎｇ．ｃｕｃ．ｅｄｕ．ｃｎ／ＲａｗＰｕｂ／ 参考文献： 储泽祥、张　琪：《海峡两岸“ 透过” 用法的多样性与倾向性考察》，《语言文字应用》，２０１３年第４期。 储泽祥：《在多样性基础上进行倾向性考察的语法研究思路》，《华中师范大学学报》，２０１１年第２期。 刁晏斌：《差异与融合——— 海峡两岸语言应用对比》，南昌：江西教育出版社，２０００年。 刁晏斌：《论现代汉语史》，《辽宁师范大学学报》，２０００年第６期。 刁晏斌：《现代汉语史概论》，北京：北京大学出版社，２００６年。 刁晏斌：《新时期大陆汉语与海外汉语的融合及其原因》，《辽宁师范大学学报》，１９９７年第４期。 郭先珍：《现代汉语量词用法词典》，北京：语文出版社，２００２年。 贺　阳：《从现代汉语介词中的欧化现象看间接语言接触》，《语言文字应用》，２００４年第４期。 李行健、仇志群：《一语两话：现代汉语共同语的共时状态》，《云南师范大学学报》（哲学社会科学版），２０１４年第 ２ 期。 邵敬敏：《说“ Ｖ一把” 中Ｖ的泛化与“ 一把” 的词汇化》，《中国语文》，２００７年第１期。 魏兆惠、华学诚：《量词“ 通” 的历史发展》，《汉语学报》，２００８年第１期。 邢福义、汪国胜：《全球华语语法研究的基本构想》，《云南师范大学学报》（哲学社会科学版），２０１２年第６期。 叶蜚声、徐通锵著，王洪君、李　娟修订：《语言学纲要（修订本）》，北京：北京大学出版社，２００９年。 张兴权：《接触语言学》，北京：商务印书馆，２０１２年。 中国社会科学院语言研究所词典编辑室：《现代汉语词典（第６版）》，北京：商务印书馆，２０１２年。 周清海：《华语研究与华语教学》，《暨南大学华文学院学报》，２００８年第３期。 邹嘉彦、莫宇航：《港台书面语的历史与现状》，刊于冯胜利《汉语书面语的历史与现状》，北京：北京大学出版社， · １ ３ ３ · 海外华文教育 ２０１６年 ２０１３年，６９— ７３页。 Ｃｈｕｉ，Ｋ．＆Ｈ．Ｌ．Ｌａｉ．ＴｈｅＮＣＣＵｃｏｒｐｕｓｏｆｓｐｏｋｅｎＣｈｉｎｅｓｅ：Ｍａｎｄａｒｉｎ，Ｈａｋｋａ，ａｎｄＳｏｕｔｈｅｒｎＭｉｎ．ＴａｉｗａｎＪｏｕｒｎａｌｏｆＬｉｎ ｇｕｉｓｔｉｃｓ，２００８，６（２）． Ｔｈｅｃｏｍｐａｒａｔｉｖｅｓｔｕｄｙｏｆｑｕａｎｔｉｆｉｅｒ“ ｔｏｎｇ（通）” ｏｎｂｏｔｈｓｉｄｅｓｏｆｔｈｅＴａｉｗａｎＳｔｒａｉｔ ＬＩＵＪｉｌｉ （ＣｏｌｌｅｇｅｏｆＣｈｉｎｅｓｅＬａｎｇｕａｇｅａｎｄＬｉｔｅｒａｔｕｒｅ，ＢｅｉｊｉｎｇＮｏｒｍａｌＵｎｉｖｅｒｓｉｔｙ，Ｂｅｉｊｉｎｇ１００８７５Ｃｈｉｎａ） Ａｂｓｔｒａｃｔ：ＴｈｅｇｒａｍｍａｔｉｃａｌｄｉｆｆｅｒｅｎｃｅｓａｎｄｆｕｓｉｏｎｉｓａｎｉｍｐｏｒｔａｎｔｃｏｎｔｅｎｔｏｆｍｏｄｅｒｎＣｈｉｎｅｓｅｃｏｍｐａｒａｔｉｖｅ ｒｅｓｅａｒｃｈｏｎｂｏｔｈｓｉｄｅｓｏｆｔｈｅＴａｉｗａｎＳｔｒａｉｔ．Ｔｈｅｃｏｍｐａｒａｔｉｖｅｓｔｕｄｙｏｆｑｕａｎｔｉｆｉｅｒ“ ｔｏｎｇ（通）” ａｃｒｏｓｓｔｈｅ Ｓｔｒａｉｔ，ｉｔｓｈｏｗｓｔｈｅｄｉｆｆｅｒｅｎｃｅｓｉｎｔｅｎｄｅｎｃｙｏｆｇｒａｍｍａｔｉｃａｌｄｉｓｔｒｉｂｕｔｉｏｎ，ｇｒａｍｍａｔｉｃａｌｆｏｒｍａｎｄｓｅｍａｎｔｉｃ ｃｏｌｏｒ．Ｃｕｒｒｅｎｔｔｈｅｆｕｓｉｏｎｏｆｑｕａｎｔｉｆｉｅｒ“ ｔｏｎｇ（通）” ｏｎｂｏｔｈｓｉｄｅｓｏｆｔｈｅＳｔｒａｉｔｉｓｍａｉｎｌｙｔｈｅｍａｉｎｌａｎｄｔｏ Ｔａｉｗａｎｏｎｔｈｅ“ ｔｏｎｇ（通）” ｏｆｃｏｌｌｏｃａｔｉｎｇｐｈｏｎｅ．Ｔｈｉｓａｒｔｉｃｌｅｔａｋｅｓｔｈｅｔｉｍｅａｓｔｈｅｃｌｕｅ，ｂｙｕｓｉｎｇｔｈｅ ｍｅｔｈｏｄｏｆｄａｔａｓｔａｔｉｓｔｉｃｓａｎｄａｎａｌｙｓｉｓｔｏｄｉｓｃｕｓｓｔｈｅｆｕｓｉｏｎｏｆ“ ｔｏｎｇ（通）” ａｃｒｏｓｓｔｈｅＳｔｒａｉｔ，ｎｏｔｏｎｌｙｆｏ ｃｕｓｏｎｄａｔａｃｈａｎｇｅａｎｄｔｈｅｃｈａｎｇｅｏｆｇｒａｍｍａｔｉｃａｌｆｏｒｍｉｔｓｅｌｆ，ｂｕｔａｌｓｏｍａｉｎｌｙｐａｙａｔｔｅｎｔｉｏｎｔｏｗｒｉｔｔｅｎｌａｎ ｇｕａｇｅａｎｄｌｅｓｓｓｐｏｋｅｎｌａｎｇｕａｇｅ，ｉｎｏｒｄｅｒｔｏｍａｋｅａｂｅｔｔｅｒｅｘｐｌａｎａｔｉｏｎｏｆｔｈｅｆｕｓｉｏｎｏｆ“ ｔｏｎｇ（通）” ａｃｒｏｓｓｔｈｅＳｔｒａｉｔ． Ｋｅｙｗｏｒｄｓ：ｂｏｔｈｓｉｄｅｓｏｆｔｈｅＴａｉｗａｎＳｔｒａｉｔ；ｑｕａｎｔｉｆｉｅｒ；“ ｔｏｎｇ（通）” · ２ ３ ３ ·
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https://www.pewresearch.org/religion/2025/02/26/decline-of-christianity-in-the-us-has-slowed-may-have-leveled-off/
Numbers, Facts and Trends Shaping Your World Home Research Topics Religion Religious Landscape Study | X Facebook Threads LinkedIn WhatsApp Share Decline of Christianity in the U.S. Has Slowed, May Have Leveled Off Findings from the 2023-24 Religious Landscape Study By Gregory A. Smith, Alan Cooperman, Becka A. Alper, Besheer Mohamed, Chip Rotolo, Patricia Tevington, Justin Nortey, Asta Kallo, Jeff Diamant and Dalia Fahmy Table of Contents Table of Contents Decline of Christianity in the U.S. Has Slowed, May Have Leveled Off Trends in the size of U.S. religious groups and denominations Race in religious groups and congregations The narrowing gender gap in American religion Religion in childhood and adulthood Religion in U.S. families today Religious switching Signs of strengthening spirituality Attendance at religious services How religious beliefs change as people age What is happening to the ‘middle’ of American religion Trends in the religiousness of Christians Religion and political polarization Executive summary Executive summary I. Religious affiliation and religious switching 1. Religious identity 2. Religious switching 3. Identifying with a religion because of culture, ethnicity or family background II. Religion and family life 4. Religious intermarriage 5. Religious upbringing and childhood education 6. Religion, fertility and child-rearing III. Religious or spiritual beliefs and practices 7. Importance of religion and the Bible 8. Religious attendance and congregational involvement 9. Race and ethnicity in religious congregations 10. Prayer and other religious practices 11. Religious and spiritual beliefs 12. Spiritual experiences 13. Spiritual activities 14. Spiritual and religious self-descriptions IV. Social and political views 15. Religion, partisanship and ideology 16. Religion and views on LGBTQ issues and abortion 17. Religion and views on gender, parenting and workforce participation 18. Religion and views on immigration and diversity 19. Religion and views on the environment 20. Religion and views on the role of government V. Opinions on religion's place in society 21. Religion and views of right and wrong 22. Religion’s role in public life 23. Religion and views of science VI. Demographics of U.S. religious groups 24. Age, race, education and other demographic traits of U.S. religious groups Acknowledgments Acknowledgments Methodology Methodology Appendices Appendix A: Comparing results across Religious Landscape Studies Appendix B: Classification of Protestant denominations Appendix C: Comparing the Center’s religion trends with those of other major surveys How we did this Pew Research Center conducts the Religious Landscape Study (RLS) to provide authoritative estimates of the U.S. population’s religious composition, beliefs and practices. The new RLS was conducted in English and Spanish from July 17, 2023, to March 4, 2024, among a nationally representative sample of 36,908 respondents. Our two previous Religious Landscape Studies, conducted in 2007 and 2014, were administered entirely by telephone. Because response rates to telephone surveys have dropped sharply over the past decade, we changed the design of the RLS in 2023-24. To recruit people to take the survey, we used a method known as address-based sampling (ABS). This involved mailing invitation letters to randomly sampled addresses from the United States Postal Service’s Computerized Delivery Sequence File. This approach gave nearly all U.S. adults a chance of being selected to participate in the survey. People who received our invitation had the option of completing the survey online, on paper, or by calling a toll-free number and completing the survey by telephone with an interviewer. In total, 25,250 respondents participated online, 10,733 completed the survey on paper, and 925 participated by phone. The survey is designed to be representative of all 50 states and the District of Columbia. The study’s large size also makes it possible to describe the religious characteristics of 34 large metro areas The data is weighted (using benchmarks from the U.S. Census Bureau and other sources) to be representative of the adult population of each state (and D.C.) and the overall U.S. adult population by gender, age, race, ethnicity, education and other categories. The survey’s margin of error for results based on the full sample is plus or minus 0.8 percentage points. The response rate is 20%. For more details, refer to the Methodology. Previous research has shown that people may answer some questions differently when completing a survey online or on paper than they do when speaking to a live interviewer over the phone. This means that when we compare data from the three landscape studies to see how Americans’ religious beliefs and practices have changed over time, we must consider the possible effects of changing the primary mode of interviewing from telephone (in 2007 and 2014) to online and on paper (in 2023-24). These “mode effects” can be quite large on some survey questions but small or nonexistent on others. To help determine whether the results of particular questions in the new RLS are – or are not – directly comparable with our previous landscape studies, we conducted a nationwide telephone survey alongside the RLS. It included all the questions in the 2023-24 RLS that previously appeared in the 2007 and 2014 surveys. We used the findings of this experimental “bridge study” to inform the discussion of long-term trends throughout this report. For more details, go to Appendix A. This report was made possible by The Pew Charitable Trusts, which received support from the Lilly Endowment Inc., Templeton Religion Trust, The Arthur Vining Davis Foundations and the M.J. Murdock Charitable Trust. After many years of steady decline, the share of Americans who identify as Christians shows signs of leveling off – at least temporarily – at slightly above six-in-ten, according to a massive new Pew Research Center survey of 36,908 U.S. adults. The Religious Landscape Study (RLS) is the largest single survey the Center conducts, aiming to provide authoritative figures on the size of U.S. religious groups because the U.S. census does not collect that information. We have conducted three of these landscape surveys over the past 17 years, with more than 35,000 randomly sampled respondents each time. That’s enough to paint a statistical portrait of religion not only nationally, but also in all 50 states and the District of Columbia, as well as in 34 large metro areas. This introductory essay walks through the big-picture trends: evidence both of a long-term decline in American religion and of relative stability in the last few years, since 2020 or so. Jump to an executive summary of key findings. Search for data on religious groups. The first RLS, fielded in 2007, found that 78% of U.S. adults identified as Christians of one sort or another. That number ticked steadily downward in our smaller surveys each year and was pegged at 71% in the second RLS, conducted in 2014. The latest RLS, fielded over seven months in 2023-24, finds that 62% of U.S. adults identify as Christians. That is a decline of 9 percentage points since 2014 and a 16-point drop since 2007. But for the last five years, between 2019 and 2024, the Christian share of the adult population has been relatively stable, hovering between 60% and 64%. The 62% figure in the new Religious Landscape Study is smack in the middle of that recent range. The largest subgroups of Christians in the United States are Protestants – now 40% of U.S. adults – and Catholics, now 19%. People who identify with all other Christian groups (including the Greek and Russian Orthodox Churches, the Church of Jesus Christ of Latter-day Saints, Jehovah’s Witnesses and many others) total about 3% of U.S. adults. Both Protestant and Catholic numbers are down significantly since 2007, though the Protestant share of the population has remained fairly level since 2019 and the Catholic share has been stable since 2014, with only small fluctuations in our annual surveys. Meanwhile, the share of Americans who identify with a religion other than Christianity has been trending upward, though it is still in single digits. Today, 1.7% of U.S. adults say they are Jewish when asked about their religion, while 1.2% of respondents in the new survey are Muslim, 1.1% are Buddhist, and 0.9% are Hindu. Religiously unaffiliated adults – those who identify as atheists, agnostics or as “nothing in particular” when asked about their religion – account for 29% of the population in the new RLS. The size of the religiously unaffiliated population, which we sometimes call religious “nones,” has plateaued in recent years after a long period of sustained growth. Rates of prayer, attendance at religious services also relatively stable Other standard survey measures contribute to this emerging picture of stability: Though down significantly since 2007, the share of Americans who say they pray daily has consistently held between 44% and 46% since 2021. In the new RLS, 44% say they pray at least once a day. Similarly, since 2020, the percentage of U.S. adults who say they attend religious services monthly has hovered in the low 30s. In the new RLS, 33% say they go to religious services at least once a month. Spiritual beliefs are widespread Moreover, the survey shows that large majorities of Americans have a spiritual or supernatural outlook on the world. For example: 86% believe people have a soul or spirit in addition to their physical body. 83% believe in God or a universal spirit. 79% believe there is something spiritual beyond the natural world, even if we can’t see it. 70% believe in an afterlife (heaven, hell or both). But, despite these signs of recent stabilization and abiding spirituality, other indicators suggest we may see further declines in the American religious landscape in future years. Namely, younger Americans remain far less religious than older adults. For example, the youngest adults in the survey (ages 18 to 24) are less likely than today’s oldest adults (ages 74 and older) to: Identify as Christian (46% vs. 80%) Pray daily (27% vs. 58%) Say they attend religious services at least monthly (25% vs. 49%) And the youngest adults are more likely than the oldest Americans to be religiously unaffiliated (43% vs. 13%). Also, younger Americans are less likely than older adults to say they were raised in religious households.1 And, compared with older adults, fewer young people who were raised in religious households have remained religious after reaching adulthood. These are among the key findings of Pew Research Center’s 2023-24 Religious Landscape Study. Like the previous studies, the new survey offers a great deal of information on what Americans believe and how they practice a wide variety of religions. In this report, we cover in detail: Executive summary Executive summary I. Religious affiliation and religious switching 1. Religious identity 2. Religious switching 3. Identifying with a religion because of culture, ethnicity or family background II. Religion and family life 4. Religious intermarriage 5. Religious upbringing and childhood education 6. Religion, fertility and child-rearing III. Religious or spiritual beliefs and practices 7. Importance of religion and the Bible 8. Religious attendance and congregational involvement 9. Race and ethnicity in religious congregations 10. Prayer and other religious practices 11. Religious and spiritual beliefs 12. Spiritual experiences 13. Spiritual activities 14. Spiritual and religious self-descriptions IV. Social and political views 15. Religion, partisanship and ideology 16. Religion and views on LGBTQ issues and abortion 17. Religion and views on gender, parenting and workforce participation 18. Religion and views on immigration and diversity 19. Religion and views on the environment 20. Religion and views on the role of government V. Opinions on religion's place in society 21. Religion and views of right and wrong 22. Religion’s role in public life 23. Religion and views of science VI. Demographics of U.S. religious groups 24. Age, race, education and other demographic traits of U.S. religious groups Acknowledgments Acknowledgments Methodology Methodology Appendices Appendix A: Comparing results across Religious Landscape Studies Appendix B: Classification of Protestant denominations Appendix C: Comparing the Center’s religion trends with those of other major surveys Reasons for the long-term decline and recent stability The RLS and other recent Pew Research Center surveys suggest that two things are happening simultaneously in American religion: Over the long term (since 2007 in our data and going back further in other major surveys), there is a downward trend in several measures of religiousness, including affiliation with Christianity. In the short term (the last four or five years), these changes have slowed or perhaps even plateaued. Long-term decline One driver of the long-term trend is “generational replacement.” Older, highly religious, heavily Christian generations are passing away. The younger generations succeeding them are much less religious, with smaller percentages of Christians and more “nones.”2 In addition, the landscape surveys show that between 2007 and 2023-24, each birth cohort has become less religious, by several measures, as it has aged. What’s a birth cohort? Throughout this report, we use “birth cohort” to refer to a group of people who were born in the same years. We categorize U.S. adults into seven cohorts: people born in the 1940s (or earlier) and those born in the 1950s, the 1960s, the 1970s, the 1980s, the 1990s, and from 2000 through 2006. One advantage of analyzing these cohorts rather than conventional “generations” (like the Silent Generation, Baby Boomers, Gen X, Millennials and Gen Z) is that most of the cohorts are of equal length (a decade long). Another advantage is that the decade-long cohorts may not carry as many assumptions – or stereotypes – as the familiar generations do. For example, people within the oldest and youngest cohorts, as well as those in between, have become less likely to say they pray daily, less likely to identify with a religion (including Christianity), and less likely to believe in God or a universal spirit with absolute certainty. Recent stability Since 2020, however, our surveys indicate that the religiousness of most birth cohorts has remained relatively stable. For instance, people born in the 1950s are about as likely to report praying daily in the 2023-24 RLS (53%) and the 2024 National Public Opinion Reference Survey, or NPORS (57%), as they were in the 2020 NPORS (55%). Additionally, the new RLS finds that the youngest cohort of adults is no less religious than the second-youngest cohort in a variety of ways.3 Americans born in 2000 through 2006 (those ages 18 to 24 in the 2023-24 RLS) are just as likely as those born in the 1990s (now ages 24 to 34) to identify as Christians, to say religion is very important in their lives, and to report that they attend religious services at least monthly.4 Time will tell whether the recent stability in measures of religious commitment is the beginning of a lasting shift in America’s religious trajectory. But it is inevitable that older generations will decline in size as their members gradually die. We also know that the younger cohorts succeeding them are much less religious. This means that, for lasting stability to take hold in the U.S. religious landscape, something would need to change. For example, today’s young adults would have to become more religious as they age, or new generations of adults who are more religious than their parents would have to emerge. Religion and spirituality among young adults The dynamics of religion and spirituality among young people are key to understanding the country’s recent religious trajectory. The large size of the new survey makes it possible to paint a religious and spiritual profile of the nation’s youngest adults with unique precision. By a number of traditional measures, today’s young adults exhibit far lower levels of religiousness than older adults. For example, 27% of adults between the approximate ages of 18 and 24 in the new survey say they pray daily, as do 31% of those ages 24 to 34. But among adults ages 54 and older, half or more say they pray daily. Young adults also report attending religious services less often than older adults do, and they express lower levels of belief in God or a universal spirit. Compared with older adults, fewer young people identify as Christians, and more say they don’t identify with any religion. On the survey’s questions touching on matters of spirituality, however, the age gaps are smaller. For example, eight-in-ten or more adults in all age categories say they believe people have a soul or spirit in addition to their physical bodies. And about seven-in-ten adults ages 18 to 24 and 75% of those ages 24 to 34 believe there is something spiritual beyond the natural world, only modestly below the 81% of the oldest adults (ages 74 and older) who say this. In the rest of this Overview, we explore these and other key topics in detail: Trends in the size of religious groups and denominations Race in religious groups and congregations The narrowing gender gap in American religion Religion in childhood and adulthood Religion in U.S. families today Religious switching Signs of strengthening spirituality Attendance at religious services How religious beliefs change as people age What is happening to the ‘middle’ of American religion Trends in the religiousness of Christians Religion and political polarization Trends in the size of U.S. religious groups and denominations In the new Religious Landscape Study, 62% of U.S. adults describe themselves as Christians. The Christian share of the population is now 9 points lower than when the landscape study was last conducted in 2014, and 16 points lower than in 2007. The share of Americans who say they have no religion – identifying, instead, as atheist, agnostic or as “nothing in particular” – stands at 29% in the new RLS, up from 23% in 2014 and 16% in 2007. The long-term decline in the Christian share of the population and growth of religious “nones” is demographically broad-based. There are fewer Christians and more “nones” among men and women; people in every racial and ethnic category; college graduates and those with less education; and residents of all major regions of the country. But the changes are much more pronounced among ideological liberals than conservatives. Today, 37% of self-described liberals identify with Christianity, down from 62% in 2007, a 25-point decline. Meanwhile, 51% of liberals now say they have no religion, up from 27% in 2007, a 24-point increase. There are now more religious “nones” than Christians among liberals, a reversal since 2007.5 There also are fewer Christians and more “nones” among conservatives. But the changes in the religious composition of conservatives have been much less pronounced than among liberals, and a large majority of conservatives continue to identify with Christianity.6 These changes within ideological categories resemble long-term trends within political parties. A prior Pew Research Center analysis shows that both Republicans and Democrats include fewer Christians and more religious “nones” today than they did a decade or more ago. But the decline of Christianity and rise of religious “nones” has been much more pronounced among Democrats than Republicans.7 For a detailed discussion of trends in the religious composition of demographic groups, refer to Chapter 1. Trends within Christianity In the 2023-24 RLS, about four-in-ten U.S. adults identify as Protestants. That is 11 points lower than in the 2007 RLS. All three major Protestant traditions have seen their population shares tick downward at least slightly over that period: Evangelical Protestants now account for 23% of all U.S. adults, down from 26% in 2007. Mainline Protestants stand at 11%, down from 18% in 2007. Members of historically Black Protestant churches make up 5% of U.S. adults, down slightly from 7% in 2007. Like the overall Christian share of the population, however, the total Protestant share of the population has been quite stable in recent years, hovering between 40% and 42% since 2019. The Catholic share of the population has been steady over an even longer period. In 11 Pew Research Center surveys conducted since 2014, all but one have found between 19% and 21% of respondents identifying as Catholic. Members of the Church of Jesus Christ of Latter-day Saints (widely known as Mormons) account for 2% of respondents in the new RLS, which is virtually unchanged from both the 2007 and 2014 landscape surveys. Orthodox Christians make up 1% of the U.S. population. Jehovah’s Witnesses and other, smaller Christian groups round out the country’s Christian population. A closer look at Protestant denominational families Many Protestant denominational families are slightly smaller today than they were in 2014 – at least nominally, even if some of the changes are too small to be statistically significant. One exception to this pattern is nondenominational Protestantism. The share of Americans identifying with this group is slightly higher today than it was in 2014 (7.1% vs. 6.2%).8 Baptists continue to be the single largest Protestant denominational family in America. Today, 12% of U.S. adults are Baptists, down from 15% in 2014 and 17% in 2007. Members of Pentecostal churches account for 4% of the U.S. adult population. Methodists and Lutherans each make up 3% of the population, while 2% of U.S. adults identify as Presbyterians. Detailed information about Protestant denominations is not available from Pew Research Center’s annual NPORS, making it difficult to know whether the size of denominations has stabilized in recent years. Categorizing Protestants into one of three traditions based on their denomination One unique feature of the Religious Landscape Study (RLS) is its detailed questions about religious affiliation. All respondents are asked an initial question about their religious identity (“Are you Protestant, Catholic … Jewish, Muslim, etc.?”). Then, Protestants are asked a second question to get more information about what type of church they identify with (“Are you Baptist, Methodist, Lutheran, etc.?”). Finally, depending on how they answer that second question, Protestants are asked a third question to try to determine which denomination they identify with. (Complete details on how these branching questions work and the exact wording of these questions are available in the questionnaires. Refer to the English paper questionnaire; Spanish paper questionnaire; and combined English/Spanish questionnaire for the web/phone administration.) Pew Research Center researchers then take the most specific information respondents provide about their denominational identity and use it to group Protestants into one of three major traditions – the evangelical Protestant tradition, mainline Protestant tradition, or the historically Black Protestant tradition. These divisions within Protestantism are important, because each tradition has its own, distinctive set of beliefs, practices and histories. For example, churches in the evangelical tradition tend to share the conviction that personal acceptance of Jesus is the only way to salvation; to emphasize bringing other people to the faith; and to have originated in separatist movements against established religious institutions. Churches in the mainline tradition, by comparison, tend to take a less exclusive view of salvation and to place more emphasis on social reform. Churches in the historically Black Protestant tradition have been shaped uniquely by the experiences of slavery and segregation, which put their religious beliefs and practices in a special context. It’s important to note that not all Protestant respondents identify with a specific denomination. Many Protestants identify as “just Baptist” or “just Methodist,” say they don’t identify with any particular denomination, or describe themselves as “just Christian.” In these cases, respondents are categorized into one of the three Protestant traditions based on their answers to a question asking whether they think of themselves as born-again or evangelical Christians, and/or a question asking about their race. Protestants who do identify with a specific denomination are grouped into one of the three Protestant traditions based exclusively on their denominational affiliation. For example, all Southern Baptists are coded as evangelical Protestants regardless of their race, and all United Methodists are coded as mainline Protestants regardless of their answer to the question about whether they think of themselves as a born-again or evangelical Christian. Complete details on the categorization of Protestantism are available in Appendix B. The Southern Baptist Convention (SBC) – which Pew Research Center categorizes as part of the evangelical Protestant tradition – remains the nation’s single largest Protestant denomination. Today, 4.4% of U.S. adults say they identify with the SBC, down from 5.3% in 2014 and 6.7% in 2007. Some of the other large evangelical denominational groupings include: The Assemblies of God, which we classify as a Pentecostal church (1.1% of U.S. adults) The Lutheran Church-Missouri Synod (1.1%) Churches of Christ (1.1%)9 The Presbyterian Church in America (0.5%) The United Methodist Church – a mainline Protestant denomination in the Center’s coding – makes up 2.7% of the U.S. population, compared with 3.6% in 2014 and 5.1% in 2007. The United Methodist Church has splintered in recent years, and many of its former churches have “joined the more conservative Global Methodist Church,” according to The Associated Press.10 Other large mainline denominations include: The Evangelical Lutheran Church in America, or ELCA (1.4% of U.S. adults) The American Baptist Churches USA (1.0%) The Episcopal Church (0.9%) The Presbyterian Church (USA), at 0.8% The National Baptist Convention, USA – a denomination in the historically Black Protestant tradition – was named by 1.0% of survey respondents. Other large historically Black Protestant denominations in the survey include: The Church of God in Christ (COGIC), a Pentecostal denomination mentioned by 0.7% of respondents The National Baptist Convention of America (0.4%) The Progressive National Baptist Convention (mentioned by fewer than 0.3% of respondents) The African Methodist Episcopal Church (also mentioned by fewer than 0.3% of respondents) Identifying with religions other than Christianity The share of Americans who identify with a religion other than Christianity has been trending upward, from 4.7% in 2007 to 7.1% today. Overall, 1.7% of adults identify as Jewish when asked about their religion – on par with results from the previous landscape studies and Pew Research Center’s 2020 survey of Jewish Americans. Muslims, Buddhists and Hindus each account for roughly 1% of the U.S. adult population. All three of these groups are larger today than they were in 2007. An additional 0.3% of respondents identify with other world religions (including Bahai’ism, Daoism, Rastafarianism, Sikhism and traditional African religions). And 1.9% of U.S. adults identify religiously as something else, including 1.1% of respondents who identify with Unitarianism or other liberal faiths, and 0.7% who identify with New Age groups.11 Religious affiliation of U.S. immigrants About 14% of U.S. adults who were born outside the country identify with religions other than Christianity, including 4% of U.S. immigrants who are Muslim, 4% who are Hindu, and 3% who are Buddhists.12 Most immigrants to the U.S. who were born in other parts of the Americas are Christian (72%), including 45% who are Catholic. Among immigrants from Europe, 57% are Christian, 8% identify with other religions, and 34% are religiously unaffiliated. Immigrants born in the Asia-Pacific region are divided about evenly between Christians, adherents of non-Christian religions (including 14% who are Hindu, 11% who are Buddhist and 7% who are Muslim), and the religiously unaffiliated.13 The survey did not include enough respondents born in the Middle East-North Africa or sub-Saharan Africa regions to be able to report on them separately. Among respondents who were born in the U.S. but had at least one parent born outside the U.S. (i.e., second-generation Americans), 10% identify with religions other than Christianity. By comparison, among people born in the U.S. to parents both of whom also were born in the U.S. (i.e., at least third-generation Americans), 5% identify with non-Christian religions. More broadly, the survey finds that the long-term decline in Christianity and growth of the religiously unaffiliated population is evident among immigrants, second-generation respondents (people who were born in the U.S. but had at least one parent born elsewhere), and people whose families have been in the U.S. for three generations or more. For example, among respondents born outside the U.S., the Christian share of the population declined from 75% to 58% between 2007 and 2023-24. Over the same period, the religiously unaffiliated share of this group (immigrants to the U.S.) grew from 16% to 26%, and the share identifying with religions other than Christianity grew from 8% to 14%. For more details on trends in the religious composition of U.S. adults, refer to Chapter 1. Race in religious groups and congregations Like the U.S. public as a whole, both Christians and religious “nones” have experienced a decline in the shares who are White. Among Christians in the new survey, 61% are White (and non-Hispanic), 18% are Hispanic, 13% are non-Hispanic Black, 4% describe themselves as multiracial or in another way, and 3% are non-Hispanic Asian. In 2007, by comparison, 70% of Christians said they were White, 13% were Hispanic, 12% were Black, 3% identified as multiracial or in another way, and 1% were Asian. In the new survey, 92% of respondents in the historically Black Protestant tradition are Black themselves, while 4% are Hispanic and 3% are White. In most other Christian traditions large enough to be analyzed in the survey, seven-in-ten or more adherents are White. Catholics are an exception: 54% in the new survey are White, while 36% are Hispanic, 4% are Asian, and 2% are Black. Nine-in-ten Jewish respondents in the new survey are White. Among Hindu respondents, 84% are Asian, as are 56% of Buddhists in the survey. Muslim respondents in the new study include 30% who are White, 30% who are Asian, 20% who are Black and 11% who are Hispanic. Roughly three-quarters of atheists (75%) and agnostics (74%) in the new survey are White; fewer people who describe their religion as “nothing in particular” are White (57%). (For a detailed discussion of the racial and ethnic composition of religious groups, refer to Chapter 24.) The survey asked people who attend religious services at least a few times a year about the racial and ethnic makeup of the congregation they attend. And it asked a similar question about the racial and ethnic composition of the congregation that respondents attended as children. Among people who attend religious services, roughly one-third currently attend a congregation where they, themselves, are not part of a racial or ethnic majority. This includes 11% who say that most of their fellow worshippers have a different race or ethnicity than they do and 21% who say that no one racial or ethnic group makes up a majority. Fewer people (roughly one-fifth of respondents) say they grew up attending a congregation where they, themselves, weren’t in a racial or ethnic majority. This includes 10% who grew up going to religious services at least a few times a year at a house of worship where most people belonged to a different race or ethnicity than they did, and an additional 10% who say that no single racial or ethnic group predominated at those services. For more on race in religious congregations, refer to Chapter 9. The narrowing gender gap in American religion Recent news accounts suggest that among the youngest Americans, men are more religious than women. This would be a major reversal from the past. Historically, U.S. women consistently have exhibited higher levels of religiousness (on average) than men.14 In the new RLS, women continue to report higher levels of religious affiliation, belief and practice than men do. At the same time, there are signs that the gender gap in religion is narrowing, as it is smaller among younger people than among older Americans. For example, in 2007, the share of women who said they pray every day exceeded the share of men who did so by 17 percentage points. In the new survey, women still report praying at higher rates than men. But the difference is slightly narrower, at 13 points. Among the oldest adults in the new survey (ages 74 and older), the share of women who say they pray every day is 20 points higher than among men. By contrast, among the youngest adults (ages 18 to 24), the share of women who say they pray daily (30%) is similar to the share of men who say they do the same (26%); the 4-point gap is not statistically significant. While the gender gap in American religion appears to be narrowing, there are still no birth cohorts in which men are significantly more religious than women. In every age group, women are at least as religious as men, and in many birth cohorts, women are significantly more religious than men. Religion in childhood and adulthood The survey shows that Americans’ current religious identities, beliefs and practices are strongly linked with their upbringing. People who say they were raised in religious homes are much more likely to be religious as adults. More than half of people who say religion was very important in their families while they were growing up also say religion is very important to them today. By contrast, among people who say religion was not too important or not at all important to their families during childhood, just 17% say religion is very important to them today. The survey finds a similar pattern on questions about religious attendance. People who grew up attending religious services regularly (at least once a month) are more than twice as likely as those who didn’t grow up attending services regularly to say they now attend religious services at least monthly. And among all respondents who were raised in a religion (i.e., among those raised Protestant, Catholic, Jewish, Muslim, or as adherents of another religious tradition), those who were raised in highly religious homes are much more likely to have retained their childhood religious identity. Indeed, 74% of people who were raised in a religion and grew up attending weekly religious services in a family in which religion was very important still identify with their childhood religion today; 15% of respondents who grew up in this kind of environment now say they have no religion, and 10% identify with a religion different from the one in which they were raised. (Refer to the section on religious switching for details about retaining or leaving one’s childhood religion in adulthood.) By contrast, among people who were raised in a religion but grew up seldom or never attending religious services – and in a family in which religion was not too important or not at all important – fewer than half still identify with their childhood religion. Instead, most now say they have no religion (40%) or identify with a religion different from the one in which they were raised (16%). In short, for many people, a religious upbringing leads to a religious adulthood.15 But the survey also indicates that raising children in a religious environment is no guarantee that those children will grow up to be religious as adults. Indeed, 40% of U.S. adults say they attend religious services less often today than they did as children. Just 5% say they attend religious services more often today than they did as kids. And 32% of U.S. adults say religion is less important to them today than it was to their families when they were growing up. By contrast, 18% say religion is more important to them today than it was to their families when they were children. Compared with their elders, today’s youngest adults are less likely, by a variety of measures, to say they had a religious upbringing. For example, more than nine-in-ten adults ages 74 and older say they were raised in a religion, including 89% who were raised Christian. Among U.S. adults who were roughly between 18 and 24 when the survey was conducted, 75% were raised in a religion, including 67% who were raised Christian. Two-thirds of the oldest Americans say they grew up going to religious services at least once a week. Only about half of the youngest adults say the same. And while half of the oldest adults say that as children they received a lot of formal religious education (i.e., seven or more years), just 19% of today’s youngest adults say the same. People in the youngest age group are about twice as likely as those in the oldest age group to say they received no formal religious education at all (42% vs. 20%). There is one exception to this pattern: Young adults are not less likely than older Americans to say religion was very important to their families when they were children. Among adults who were roughly between the ages of 18 to 24 when the new survey was conducted, 47% say religion was very important to their families when they were growing up. Among adults ages 74 and older, 44% say the same. Moreover, the persistence of a high level of religiousness from childhood into adulthood – the “stickiness” of a religious upbringing – appears to be declining, while the stickiness of a nonreligious upbringing seems to be increasing. In the oldest cohort of U.S. adults (ages 74 and older), 51% of those who say they grew up attending religious services weekly in families for whom religion was very important are still highly religious in these ways (i.e., they still go to services weekly and still say religion is very important in their lives). And among people in this oldest cohort, 50% of those who say they grew up seldom or never attending religious services in families in which religion was not too important or not at all important still describe themselves as nonreligious in these ways (i.e., they still rarely or never go to services, and still say religion is not important in their lives). In other words, a highly religious upbringing has proved to be just as “sticky” as a nonreligious upbringing over the lifetimes of the oldest Americans (now 74 and older). By contrast, among the youngest U.S. adults in the survey (now ages 18 to 24), just 28% of those raised in highly religious homes are, today, highly religious themselves. Meanwhile, 76% of young adults who grew up rarely or never attending services, in families in which religion was unimportant, still say they don’t attend religious services and that religion is not important to them. In other words, a highly religious upbringing has been much less persistent (or “sticky”) than a nonreligious upbringing so far in the lifetimes of the youngest U.S. adults. Of course, it’s possible that the effect of a highly religious upbringing just needs time to develop. Maybe the people in the youngest cohort will grow more religious as they age, and if they are surveyed again in 20 or 30 years, a highly religious upbringing will appear stickier than it does now. However, there is no evidence in the three Religious Landscape Studies conducted since 2007 that any birth cohorts have grown more religious over the long term, as discussed below in the section on how religious beliefs change as people age. For more information on Americans’ religious upbringing, refer to Chapter 4. Religion in U.S. families today Raising children Among parents of minor children – i.e., people currently raising children under 18 in their homes – 40% say they send their child or children to some kind of religious education or private religious school. By comparison, 51% of parents say they, themselves, received a fair amount or a lot of religious education as children, while 16% received a little, and 33% had none. However, these questions about the respondent’s own religious education and the religious education of their kids are not exactly parallel. The survey asked respondents to report how many years of religious education they received as children, whereas parents were asked whether their minor children are currently enrolled in a religious private school or other religious education programs. Some parents may not have children enrolled right now but may enroll them later or may have enrolled them in the past. Among parents living with minor children in their homes, 42% say they read scripture or pray with their children. The survey did not ask respondents whether (or how often) they used to read scripture with their parents or prayed with their parents when they were children themselves. Overall, 26% of people who are currently raising children say they go to religious services at least weekly, and an additional 9% say they go to religious services once or twice a month. The survey did not ask current parents whether they take their children with them when they go to religious services. But, even if all 35% do take their children with them to services, this still would be a much smaller percentage than the roughly two-thirds of parents who say they went to services at least monthly when they were growing up.16 For more on how parents in different religious traditions are raising their children, refer to Chapter 6. Religious intermarriage In the new survey, 26% of married adults say their spouse has a religious identity different from their own. That’s virtually identical to what we found in the 2014 RLS (25%). One-quarter of married Catholics in the new survey say they are married to a non-Catholic spouse, including 14% who are married to a spouse from another Christian tradition and 9% who are married to a spouse with no religion. About one-in-five married Protestants say they have a non-Protestant spouse, including 8% who have a spouse who identifies with another Christian faith and 10% whose spouse has no religion. Roughly one-third of married “nones” say their spouse identifies with some religion, in most cases a branch of Christianity. Protestants, Catholics and Jews who are in religiously mixed marriages are far less religiously active, on average, than those with spouses who share their religion. For example, 68% of Protestants married to other Protestants say religion is very important in their lives, compared with 38% of Protestants married to non-Protestants who say this. And 49% of Catholics married to fellow Catholics say they go to church at least once or twice a month, compared with 28% of Catholics married to non-Catholics who say they attend this often. These differences carry over into child-rearing. Among parents of minor children, Protestants married to fellow Protestants are significantly more likely than Protestants married to non-Protestants to say they pray or read scripture with their kids and send their children to religious education. For the most part, a similar pattern holds for Catholics. These patterns raise an interesting question: Does being in a religiously mixed marriage make people less religious, or are nonreligious people more likely to enter religiously mixed marriages? The survey cannot answer this question; it’s possible that both things are at play. Meanwhile, religiously unaffiliated people married to spouses who identify with a religion tend to be a little more religiously active, on average, than religiously unaffiliated people married to spouses who are also unaffiliated. For instance, 12% of married “nones” whose spouses identify with a religion say they pray daily, compared with 9% of married “nones” who have a fellow “none” as a spouse. And 13% of religious “nones” who are currently parenting a minor child and who are married to a religiously affiliated spouse say they send at least one of their children to religious education, compared with 5% of religious “nones” who are married to a religiously unaffiliated spouse.17 The differences are quite small, though, as are the overall shares of “nones” who say they are religiously active in these ways – regardless of whether they are married to a fellow “none,” or not. For more on religious intermarriage, refer to Chapter 4. Religious switching Overall, 35% of U.S. adults have switched religions between childhood and adulthood. That is, they say they currently identify with a religion (or with no religion) that is different from the religion in which they were raised. This figure includes people who switched from one Christian religious tradition to another (e.g., those who say they were raised Protestant but now identify as Catholic or vice versa), as well as people who switched from one non-Christian religion to another (e.g., those who say they were raised Hindu and now identify as Buddhist or vice versa).18 It also includes people who switch from identifying with a religion to describing themselves as religiously unaffiliated (or vice versa). The share of people in the new survey who say they switched religions is on par with the share who said this in 2014. When we divide the data into just three categories – Christianity, other religions and no religion – it shows very clearly that Christianity loses far more people than it gains through religious switching. Fully 80% of U.S. adults say they were raised Christian, but upward of a quarter of them (22% of all U.S. adults) no longer identify as Christians. What is ‘religious switching’? We use the phrase “religious switching” rather than more familiar terms like “conversion” because changes in religious identity occur in all directions, including from having a religion to having no religion. Also, these changes can take place without any formal ritual or declaration. We measure religious switching with two survey questions. The first asks, “What is your present religion, if any?” The second asks, “Thinking about when you were a child, in what religion were you raised, if any?” By comparing each respondent’s answers to these questions, we can see whether they still have the same religion in which they were raised or whether they have switched. However, these questions may not capture all the switching that takes place. For instance, they would not reveal whether a person has switched multiple times, or whether a person left their childhood religion and later returned to it. Our analysis of religious switching focuses on the difference (if any) between a respondent’s religious identity at the time of the survey and during childhood, not the steps that may have occurred in between. By contrast, religious “nones” gain far more people than they lose through religious switching. Overall, 13% of U.S. adults say they were raised in no religion. But fully 20% of U.S. adults now say they are religiously unaffiliated after having been raised in a religion, including 19% who were raised Christian and 1% who were raised in other religions. Expressed as a ratio, these figures mean that there are six former Christians for every convert to Christianity in the United States. The balance is especially lopsided for Catholicism (which loses 8.4 people through religious switching for every convert to the religion). But Protestants also lose more people than they gain through switching, by a ratio of 1.8 to one. In stark contrast, the religiously unaffiliated gain nearly six people for every person they lose through religious switching. That is, there are about six times as many Americans who say they were raised in a religion and no longer identify with a religion than there are who say they were raised in no religion but now identify with one. In the aggregate, religions other than Christianity gain about as many people as they lose through religious switching. Overall, 3% of the public identifies with a religion other than Christianity after having been raised Christian or with no religion (or they did not answer the question about their childhood religion). And 2% of U.S. adults say they were raised in a religion other than Christianity but no longer identify with it. The survey shows big differences in religious switching patterns across age groups, which helps explain the long-term declines in religious affiliation, especially for Christianity. Vast majorities of older adults (89% of those born in the 1950s or earlier) say they were raised Christian, and most of them are still Christian today. Relatively few older adults say either that they were raised in no religion or that they have become religiously unaffiliated after having been raised as Christians. Smaller majorities of young adults were raised as Christians, and fewer of them have retained their Christian identity in adulthood. Young people are more likely than older adults to have been raised in no religion and, also, more likely to have become religiously unaffiliated after having been raised as Christians. Interestingly, among the youngest adults – those born between 2000 and 2006, who were ages 18 to 24 when the survey was conducted – the share who were raised Christian and are still Christian is about as high as among those born in the 1990s, who were 24 to 34 when the survey was conducted (41% vs. 42%). And the youngest cohort is slightly less likely than the second-youngest cohort to have left Christianity. Jump to Chapter 2 to see the retention rates of different religious groups (i.e., what percentage of all the people raised in each group still identify with that group today). Signs of strengthening spirituality When asked how their levels of spirituality may have changed over the course of their lifetimes, Americans who say they have become more spiritual outnumber those who say they have become less spiritual by a roughly four-to-one margin (43% vs. 11%). Moreover, the survey finds that Americans of all ages are more likely to say their spirituality has grown stronger than to say it has weakened.19 Large majorities of the U.S. public believe that people have a soul or spirit in addition to their physical bodies, that there is something spiritual beyond the natural world, and that there is an afterlife. And although younger Americans are less traditionally religious than older adults (as measured by rates of prayer, identifying with a religious group, attending religious services, and some other beliefs and practices), the age gaps are much smaller on several of the survey’s questions about spirituality.20 For instance, 82% of adults who were roughly between the ages of 18 and 24 when the survey was conducted say they believe people have a soul or spirit in addition to their physical body – only slightly lower than the share of today’s oldest adults who affirm the same belief. And 63% of today’s youngest adults say they feel a deep sense of wonder about the universe at least monthly, which is somewhat higher than the share of today’s oldest adults who say the same. The survey finds that Americans who say their spirituality has increased are more likely than other adults (especially those who say their spirituality has declined) to say they regularly feel awe at nature’s beauty, feel a sense of spiritual peace, and experience the presence of something from beyond this world. Americans who say their spirituality has grown also are more inclined than others to say they believe in God or a universal spirit, and to say they pray daily. These results suggest that spirituality and religion are not necessarily in tension with each other. We know from previous research that many people view spirituality and religion as complementary; some see no difference between them. Indeed, the new survey finds that people who identify with a religion are more likely than religiously unaffiliated Americans to say they have grown increasingly spiritual during their lives. Religiously affiliated people also are more inclined than religious “nones” to believe in God or a universal spirit, to pray daily, and to report that they experience a variety of spiritual sensations, such as “the presence of something from beyond this world.” Among the “nones,” those who say they’ve grown more spiritual during their lives are more likely than other religiously unaffiliated people to say that people have souls or spirits in addition to their physical bodies, that there is something spiritual beyond the natural world, and that they regularly feel a sense of awe at the beauty of nature. Attendance at religious services The 2023-24 RLS finds a substantial age gap in attendance at religious services. Most young adults say they go to religious services no more than a few times a year. Indeed, among people born since the 1980s, about half or more say they seldom or never attend religious services. By comparison, older adults report attending religious services at far higher rates. Furthermore, 57% of people in the oldest cohort say they are members of a religious congregation, compared with 27% of the youngest adults. Does this mean that attendance at religious services is declining? Not necessarily. The age gap, by itself, is not proof that fewer people are attending services, because it is theoretically possible that people participate in religious services at higher rates as they get older. Perhaps today’s young adults will go to a church, synagogue, mosque or other house of worship more often when they reach middle age or retirement age than they do today. Moreover, we do not have a clear, long-term trend line for attendance at religious services because the new survey’s findings about rates of religious attendance cannot be directly compared with the previous landscape studies. The earlier studies in 2007 and 2014 were conducted by telephone, while the new survey was conducted mainly online or on paper. Previous research shows that these different ways (or “modes”) of conducting surveys produce significantly different results on the question “How often do you attend religious services?” This makes it hard to determine, over the long term, how much religious attendance has changed. However, the Center’s telephone surveys were showing a decline in religious attendance in the years before we switched to online/paper surveys. The share of Americans who reported attending religious services at least monthly dropped from 54% in 2007 to 50% in the 2014 Religious Landscape Study and had fallen to 45% by 2018-19 (which is when the Center last regularly conducted telephone surveys that asked about religious attendance). But the short-term trend line for the share of Americans who attend religious services at least monthly is pretty flat. In Pew Research Center’s 2020 NPORS, 33% of U.S. adults reported attending religious services at least once or twice a month. That’s identical to what we found in the 2023-24 RLS and very similar to the 32% measured in the 2024 NPORS. How religious beliefs change as people age When asked to describe how their own religiousness has changed over their lifetimes, the most common response people give is that there has been no clear change in either direction – up or down – over the course of their lives (44% of U.S. adults say this). This includes people who say they have sometimes grown more religious and other times less religious; people who say their level of religiousness has not changed very much; and people who don’t answer the question. This is the most common kind of response among respondents in every age category. Among people who say their religiousness has changed over the course of their lives, about equal shares say they have become less religious (29%) and more religious (28%). The new RLS also finds that more young adults say their religiousness has decreased than say it has increased. By contrast, older U.S. adults are more likely to say their religiousness has increased than that it has decreased.21 However, on the questions in the new survey that can be directly compared with the previous studies, there is no evidence that people in older birth cohorts have grown more religious between 2007 and today. In the three Religious Landscape Studies conducted over the past 17 years, there are no birth cohorts – neither the oldest Americans, nor the youngest, nor any cohort in between – that have grown more prayerful, more certain in their belief in God, more likely to believe in an afterlife (heaven, hell or both), or more likely to identify with a religion, including Christianity. This may seem at odds with the retrospective answers that survey respondents give to questions about how their own religiousness and spirituality have changed. As previously noted, about four times as many U.S. adults say they have become more spiritual (43%) as say they have become less spiritual (11%) over their lifetimes, while nearly equal shares say they have become more religious (28%) and less religious (29%). But substantial numbers of Americans say they haven’t really changed spiritually (46%) or religiously (44%). And the answers people give to these questions may be shaped by many factors, such as how desirable they feel it is to be spiritual and/or religious, and how they understand the meaning of those terms. Another limitation in understanding how religiousness tends to change as people age is that Pew Research Center’s data on many questions about religion goes back no further than the first Religious Landscape Study in 2007. That 17-year span may not be long enough to capture all the change that has occurred in older birth cohorts. For a longer perspective, we turned to the General Social Survey (GSS), a national survey that has been conducted every year or two since 1972. The GSS shows that people do tend to become more prayerful as they get older. For example, when people born in the 1960s were just entering adulthood in the early 1980s, 36% said in the GSS that they prayed on a daily basis. By the late 2010s, when people in this birth cohort were in their late 40s and 50s, 64% said they prayed daily. The GSS also reports that people born in the 1950s and 1970s exhibited similar long-term increases in rates of daily prayer. But the GSS suggests that people do not become more likely to identify with a religion as they get older. There is also little evidence in the GSS that as Americans get older, they tend to become more likely to believe in God or to attend religious services regularly. For more analysis of GSS data on how the religiousness of people in different birth cohorts has changed or stayed the same over time, refer to Appendix C.22 What is happening to the ‘middle’ of American religion As the Christian share of the U.S. adult population has declined in recent decades and the public has come to look less religious on a variety of measures, the question arises whether these changes are evidence of religious decline across the board or, instead, a hollowing out of the “religious middle.” That is, are religious beliefs and practices declining all across the religious spectrum? Or is there a decline only in the share of the population with medium levels of religiousness, accompanied by stability or even growth at both ends of the spectrum of religiousness – resulting in more highly secular people, more highly religious people, and fewer people who are just moderately religious? The new RLS indicates that the long-term changes in American religion are broad-based and not a hollowing out of the religious middle. Perhaps the easiest way to illustrate this is by looking at trends in prayer frequency over time. This is the new study’s best indicator of religiousness that can safely and directly be compared with previous RLS results.23 The share of Americans who say they pray daily is clearly down. And the share of Americans who say they seldom or never pray is clearly up. Meanwhile, the share of Americans who say they occasionally pray (on a weekly or monthly basis) is relatively stable. There is no indication of a hollowing out of the “religious middle” on this key indicator. The survey finds a similar pattern in questions about belief in God. The share of Americans who say they believe in God or a universal spirit with absolute certainty has clearly declined, while nonbelief is trending upward. And the share of Americans occupying the middle ground on this question – saying they believe in God but with less than absolute certainty – shows no signs of hollowing out, but rather is rising. The share of adults in the new survey who believe in God but with less than absolute certainty is higher than it has ever been in an RLS survey.24 Another way to assess the evidence for a hollowing out of the religious middle is to look at differences across age groups within the new survey. This allows us to examine additional indicators of religiousness, because looking just at the results of the new survey sidesteps the difficulties posed by the change of survey modes between the previous two Religious Landscape Studies (conducted by telephone) and the new one (conducted mainly online and on paper). For this analysis, we used four key questions: prayer frequency, belief in God and/or a universal spirit, religion’s importance, and religious service attendance. Responses to each question were assigned numbers from 0 (low) to 2 (high) as follows: Prayer frequency: Coded as 0 for those who seldom or never pray, 2 for those who pray daily, and 1 for everyone else. Belief in God or a universal spirit: Coded as 0 for those who do not believe in God or a universal spirit, 2 for those who believe with absolute certainty, and 1 for everyone else. Religion’s importance: Coded as 0 for those who say religion is not too important or not at all important in their lives, 2 for those who say religion is very important in their lives, and 1 for everyone else. Religious attendance: Coded as 0 for those who say they seldom or never attend religious services, 2 for those who attend religious services at least once a month, and 1 for everyone else. We then added these indicators together to form a scale ranging from 0 (for people who scored 0 on all four measures) to 8 (for respondents who scored 2 on all four measures).25 And we subdivided the scale roughly into quartiles, as follows: Low religiousness: Scores of 0 to 1 Medium-low religiousness: Scores of 2 to 4 Medium-high religiousness: Scores of 5 to 6 High religiousness: Scores of 7 to 8 Here again, there is no evidence that religious changes underway in America reflect a shrinking of the “religious middle.” The religious middle is not smaller among the younger cohorts than among the older cohorts, which is what one might expect if a hollowing out of the religious middle were occurring. Rather, compared with older adults, younger Americans include far more people at the low end of the religiousness spectrum and far fewer people at the high end. The size of the middle categories of religiousness is in the same ballpark across most age groups. This is exactly the pattern that would be expected if broad-based religious decline (which some social scientists call a process of secularization) were underway. Trends in the religiousness of Christians We have already seen that over the long term (since 2007), the Religious Landscape Studies show a substantial rise in the percentage of U.S. adults who are religiously unaffiliated (sometimes called “nones”) and a substantial decline in the percentage who describe themselves as Christians – including declines in all three major Protestant traditions (evangelical, mainline and historically Black Protestant churches). This raises the question: As the Christian share of the population has shrunk, have Christians become more religious, on average? One might expect that as some Americans leave Christianity and join the ranks of the unaffiliated, those who stay behind and remain Christian would become a smaller but more committed group, with rising average levels of religious belief and practice. Has this happened? The new RLS offers a mixed answer, with a couple of indicators of religiousness holding steady or rising slightly among U.S. Christians, while two other indicators clearly have declined. The four questions in the 2023-24 RLS about religious practice and belief that can most readily be compared with the prior studies ask about prayer frequency, belief in God or a universal spirit, belief in heaven and, separately, belief in hell.26 The new survey shows that among Christians, belief in heaven and hell is on par with or slightly higher than in 2007. Belief in heaven now stands at 85% among Christians, compared with 83% in 2007. Meanwhile, 72% of Christians now say they believe in hell, up slightly from 68% in 2007. But there is no indication that Christians are any more prayerful today than they were in 2007. Indeed, the share of Christians who say they pray every day is markedly lower now (44%) than it was in 2007 (58%). The share of Christians who say they believe in God with absolute certainty also is lower today than it was in 2007. In the new survey, 73% of Christians say they are absolutely certain God or a universal spirit exists, down from 80% in 2007. Religion and political polarization The U.S. is politically polarized, and religion is closely associated with the country’s political divisions. But the new Religious Landscape Study – which was conducted mostly in 2023 and the early part of 2024, before the 2024 presidential election – also finds that the connections between religion and political partisanship vary a lot by race and ethnicity, as described below. In general, highly religious Americans tend to identify with or lean toward the Republican Party and express conservative views on a variety of social, political and economic questions at much higher rates than do the least religious Americans. Meanwhile, Americans with lower levels of religious engagement tend to identify with or lean toward the Democratic Party and express liberal views on the same gamut of social, political and economic issues.27 For example, a majority of people in the most highly religious quartile of the U.S. adult population (61%) say they identify with or lean toward the Republican Party. By contrast, in the least religious quartile of the population, just 27% say they identify with or lean toward the GOP.28 Highly religious U.S. adults also are far more likely than the least religious U.S. adults to say that abortion should be illegal, that homosexuality should be discouraged, and that children are better off if their mother doesn’t work and stays home to raise them instead. There also are religious differences on questions that aren’t directly about sexuality or gender roles. Highly religious people are more inclined than the least religious people to say that environmental regulations cost too many jobs and hurt the economy, as well as to say that too much openness to people from other countries is a threat to America’s identity. None of the foregoing discussion is meant to suggest that religiousness (or lack thereof) is the primary factor driving Americans’ political opinions. Numerous other factors can help shape political views, including age, gender, education, geography and socioeconomic status. And political partisanship is itself an important factor in people’s views about a host of issues.29 Moreover, the new RLS shows that the links between religion and politics vary quite a bit across racial and ethnic categories. The political differences between Americans with different levels of religiousness are especially large among White respondents. For example, compared with the least religious White Americans, the most highly religious White Americans are: 64 percentage points more likely to say abortion should be against the law in most or all cases (72% vs. 8%) 53 points more likely to say homosexuality should be discouraged by society (61% vs. 8%) 49 points more likely to identify with or lean toward the Republican Party (77% vs. 28%) 35 points more likely to say environmental regulations cost too many jobs and hurt the economy (56% vs. 21%) In other racial and ethnic groups, there also is a relationship between religiousness and political views, with more highly religious people generally expressing more conservative views. However, the differences in political views between people with different levels of religiousness tend to be smaller in other racial and ethnic groups, and they are not seen on every issue. For example, Black respondents who are highly religious are not more likely than the least religious Black respondents to identify with or lean toward the Republican Party; if anything, highly religious Black adults are somewhat less Republican than Black adults with the lowest levels of religious engagement. Highly religious Hispanic respondents are, on average, more likely than the least religious Hispanics surveyed to identify with the Republican Party. But the gap in Republicanism among highly religious Hispanic adults and Hispanic adults in the lowest religiousness category is 22 points, much smaller than the 49-point gap among White adults. Next: Executive summary ← Prev Page 1 2 3 4 5 6 … 31 Next Page → RECOMMENDED CITATION: Smith, Gregory, Alan Cooperman, Besheer Mohamed, Chip Rotolo, Patricia Tevington, Justin Nortey, Asta Kallo, Jeff Diamant, and Dalia Fahmy. 2025. “Decline of Christianity in the U.S. Has Slowed, May Have Leveled Off.” Pew Research Center. doi: 10.58094/4kqq-3112. Compared with older people, younger adults are less likely to say they were raised in a religion, less likely to say they grew up going to religious services at least monthly, and less likely to have received a lot of formal religious education as children. Young adults also are somewhat less likely than older people to say religion was “very important” to them personally when they were children, and more likely to say religion was “not too important” or “not at all important” to them personally when they were children. An exception to this pattern is that there is little difference across age categories in the share of people who say religion was “very important” to their families when they were kids.↩ For more on generational changes in American religion, read Hout, Michael, and Claude S. Fischer. 2014. “Explaining Why More Americans Have No Religious Preference: Political Backlash and Generational Succession, 1987-2012.” Sociological Science. Also refer to Voas, David, and Mark Chaves. 2016. “Is the United States a Counterexample to the Secularization Thesis?” American Journal of Sociology.↩ This is not the first time surveys have found young people entering adulthood with levels of religiousness roughly equal to those of the preceding birth cohort. The 2014 Religious Landscape Study (RLS) also found that the youngest adults (at the time) were about as likely as their immediate predecessors to identify as Christians and to report attending religious services weekly. One possible explanation could be that the youngest cohort of new adults includes many who still live at home with parents or guardians. As they gradually form independent households, their levels of religious engagement may change.↩ On some other questions, however, adults born in 2000 or later exhibit lower levels of religiousness than those born in the 1990s. For instance, they are somewhat less likely to say they pray every day, to believe in God with absolute certainty, and to have received a lot of religious education as children.↩ Some social scientists have long argued that younger, more liberal Americans are leaving religion in a backlash against the entanglement of religious institutions and leaders with conservative politics. Read Hout, Michael, and Claude S. Fischer. 2002. “Why More Americans Have No Religious Preference: Politics and Generations.” American Sociological Review. Also refer to Putnam, Robert D. and David E. Campbell. 2010. “American Grace: How Religion Divides and Unites Us.”↩ Of course, this correlation doesn’t mean that changes in one attribute (political ideology) cause changes in the other (religious identity). It’s also possible that people in various religious categories have changed their ideological outlook over time. For more on the complex relationship between people’s religious characteristics and political attributes, refer to Campbell, David E., Geoffrey C. Layman and John C. Green. 2020. “Secular Surge: A New Fault Line in American Politics.” Also refer to Margolis, Michele. 2018. “From Politics to the Pews: How Partisanship and the Political Environment Shape Religious Identity.” Also refer to Margolis, Michele. 2022. “Reversing the Causal Arrow: Politics’ Influence on Religious Choices.” Advances in Political Psychology.↩ Tracking partisanship over time is complicated by a “mode effect.” Compared with surveys conducted by telephone (like the 2007 and 2014 landscape studies), surveys conducted mainly online and on paper (like the 2023-24 RLS) produce far lower estimates of the shares of adults who don’t identify with or lean toward either major party, as well as higher estimates of the shares who identify with or lean toward one or the other of the two major U.S. political parties. The RLS does not adjust for this effect, and therefore, we are not able to analyze trends in partisanship across the three landscape studies. However, the Center’s April 2024 report “Changing Partisan Coalitions in a Politically Divided Nation” does make this adjustment, and it provides estimates of how the religious composition of the partisan coalitions has changed over time. The mode switch from phone to web/paper also complicates comparing results on a question about political ideology (conservative/moderate/liberal) in the new survey with previous surveys. But the impact of the mode switch is less pronounced on the question about ideology than on the questions about partisanship. For more details about the consequences of the mode switch for drawing comparisons across the three Religious Landscape Studies, refer to Appendix A.↩ In the Religious Landscape Studies, respondents are categorized as nondenominational Protestants only if they identify themselves explicitly as a “nondenominational Protestant” or a “nondenominational Christian.” The “Nondenominational Protestant” category does not include respondents who identify themselves as “just Christian” or “just a Protestant.” Nor does it include people who identify with Protestantism but not with any particular denomination; these latter types of responses are included in the “Nonspecific Protestant” category.↩ Churches of Christ describes itself as “undenominational,” and the organization has “no central headquarters” and “no organization superior to the elders of each local congregation.” Visit the organization’s website for additional details.↩ In the new RLS, fewer than 0.3% of respondents identify with the Global Methodist Church, which we categorize as an evangelical denomination.↩ The “Unitarians and other liberal faiths” category includes U.S. adults who describe themselves as Unitarians, “spiritual but not religious,” deists, humanists and others. The “New Age” category includes those who identify as Pagans, Wiccans, druids, Satanists and others.↩ Overall, 18% of respondents in the new survey were born outside the United States.↩ Refer also to the Center’s report “Religion Among Asian Americans.”↩ For more information on gender differences in religious identities, beliefs and practices both in the U.S. and around the world, refer to Pew Research Center’s report “The Gender Gap in Religion Around the World.”↩ The reverse also is true: Religious adults are more likely to give their children a religious upbringing, as described in Chapter 6.↩ The survey finds that, in general, parents who are raising minor children attend religious services at higher rates than adults of a similar age who are not parents. For instance, among parents between the ages of 25 and 34, 34% say they attend religious services at least monthly, compared with 20% of people in the same age group who are not parents. Among parents between the ages of 35 and 44, 34% say they attend religious services at least monthly, compared with 18% of people in the same age group who are not currently parenting a minor child. One exception to this pattern is the youngest age group: Among Americans ages 18 to 24, those who are parents do not report attending services at higher rates than those who are not parents. But the survey includes a relatively small number of interviews with 18- to 24-year-olds who are parents, and the results for this group have a correspondingly large margin of error (plus or minus 8.8 percentage points).↩ Relatively few religious “nones” married to a religiously affiliated spouse say their spouse is very religious. Overall, 17% of religious “nones” married to a religiously affiliated spouse say religion is very important in their spouse’s life, 36% say religion is somewhat important to their spouse, and 47% say religion is not too important or not at all important to their spouse.↩ This figure does not include switching within Protestantism (e.g., from Baptist to Methodist, or from Lutheran to Episcopalian), nor does it include switching within the religiously unaffiliated category (e.g., from atheist to agnostic, or from agnostic to “nothing in particular”). If these kinds of switching within Protestantism and among the religiously unaffiliated are added to the equation, then the share of U.S. adults who have switched religions between childhood and adulthood rises to 49%.↩ Young adults and older Americans may have very different reference points in mind when answering this question. Young adults have only their childhood and early adulthood as points of comparison. Older adults may be considering a much longer time horizon.↩ There is an active debate among scholars over how to define religion and spirituality. Previous Pew Research Center surveys have shown that many Americans see no real difference between the two terms. However, some people describe themselves as spiritual but not religious, and our surveys find that many of these people hold supernatural beliefs (such as in a higher power or spiritual force in the universe) but rarely participate in organized religious activities, such as attending religious services. With that finding in mind, we use “spirituality” loosely in this report to refer to beliefs or practices that don’t necessarily imply participation in, or identification with, any organized religion. At the same time, we do not wish to suggest that there is a clear line between religion and spirituality.↩ Of course, young adults and older Americans may have very different reference points when asked this question about the course of their lives. Young adults have only their childhood and young adulthood as points of comparison, whereas older adults have a much longer time horizon.↩ For more on religious change over the course of adulthood, refer to Bengtson, Vern L., et al. 2015. “Does Religiousness Increase with Age? Age Changes and Generational Differences Over 35 Years.” Journal for the Scientific Study of Religion. Also refer to Silverstein, Merril, and Vern L. Bengtson. 2017. “Return to Religion? Predictors of Religious Change among Baby-Boomers in their Transition to Later Life.” Journal of Population Ageing. Also refer to Bleidorn, Wiebke, et al. 2022. “Secularization Trends Obscure Developmental Changes in Religiosity.” Social Psychological and Personality Science.↩ The switch from telephone interviewing in the earlier landscape studies to mostly online/paper administration of the new RLS means that other key questions, such as frequency of religious attendance and the importance of religion, are not directly comparable across the different studies. Refer to Appendix A for details.↩ The 2023-24 RLS was conducted mainly online and on paper, whereas the previous landscape studies were conducted by telephone. This “mode switch” complicates comparisons between results from the new survey and the previous studies. However, our analysis finds that the new survey’s question about prayer frequency can be safely compared with the prior studies. The questions about belief in God can be cautiously compared with the previous results. Analysis of a companion “bridge study” we conducted by telephone (for testing purposes only) finds that the share who say they believe in God or a universal spirit with absolute certainty is similar in the main survey and in the bridge study, as is the share who say they believe in God or a universal spirit but are not absolutely certain. These categories can be safely compared with the 2007 and 2014 results. However, the share who say they do not believe in God/a universal spirit is higher in the main survey than the bridge study (by 4 percentage points), while the share expressing no opinion is slightly lower in the main survey than in the bridge study. Both the main survey and the bridge study find that absolutely certain belief in God or a universal spirit is trending downward. Both the main survey and the bridge study also suggest that the share of people who do not believe in God or a universal spirit is trending upward, though it is difficult to say by exactly how much; some of the apparent increase in nonbelief suggested by the main survey may be attributable to the switch in survey modes. Neither the main survey nor the bridge study suggests that the middle category – those who believe in God or a universal spirit, but not with absolute certainty – is shrinking. For additional details, refer to Appendix A.↩ Cronbach’s Alpha=.865.↩ The 2023-24 RLS was conducted mainly online and on paper, whereas the previous landscape studies were conducted by telephone. This “mode switch” complicates comparisons between results from the new survey and the previous studies. However, our analysis finds that these estimates from 2023-24 can be safely compared with the prior studies. For additional details, refer to Appendix A.↩ For a discussion of political differences between Americans who are affirmatively secular and those who are instead simply nonreligious, refer to Campbell, David E., Geoffrey C. Layman and John C. Green. 2020. “Secular Surge: A New Fault Line in American Politics.”↩ The scale of religiousness is based on four questions: prayer frequency, belief in God and/or a universal spirit, religion’s importance, and religious service attendance. Each question was coded from 0 (low) to 2 (high). Prayer frequency is coded as 0 for those who seldom or never pray, 2 for those who pray daily, and 1 for everyone else. Belief in God/a universal spirit is coded as 0 for those who do not believe in God or a universal spirit, 2 for those who believe with absolute certainty, and 1 for everyone else. Religion’s importance is coded as 0 for those who say religion is “not too important” or “not at all important” in their lives, 2 for those who say religion is “very important” in their lives, and 1 for everyone else. Religious attendance is coded as 0 for those who say they seldom or never attend religious services, 2 for those who attend religious services at least once a month, and 1 for everyone else. These indicators were added together to form a scale ranging from 0 to 8, and then subdivided roughly into quartiles (scores of 0 to 1 in the lowest quartile, scores of 2 to 4 in the medium-low quartile, scores of 5 to 6 in the medium-high quartile, and scores of 7 to 8 in the highest quartile).↩ For a discussion of demographics and political polarization, refer to Pew Research Center’s report “Changing Partisan Coalitions in a Politically Divided Nation.” For a discussion of how people in the two major party coalitions see each other, refer to the Center’s report “As Partisan Hostility Grows, Signs of Frustration With the Two-Party System.” For a discussion of issue attitudes, political values and polarization, refer to the Center’s report “The Political Values of Harris and Trump Supporters.”↩ Report Materials Report PDF Topline Bridge study topline English paper questionnaire English/Spanish web/phone questionnaire Spanish paper questionnaire Interactive: Religious Landscape Study Video: Religion and Spirituality in America 2023-24 Religious Landscape Study (RLS) Dataset Sign up for our weekly newsletter Fresh data delivered Saturday mornings Sign up for The Briefing Weekly updates on the world of news & information Sign Up Topics Atheism & Agnosticism Beliefs & Practices Buddhism Catholicism Christianity Evangelicalism Hinduism Historically Black Protestantism Islam Judaism Latter-day Saint (Mormon) Mainline Protestantism Muslim Americans Non-Religion & Secularism Orthodox Christianity Other Religions Pentecostalism Protestantism Religion Religion & Abortion Religion & Government Religion & LGBTQ Acceptance Religion & Politics Religion & Race Religion & Science Share This Link: X Facebook Threads LinkedIn WhatsApp Share Related Religious Landscape Study: Database Religious Landscape Study Spirituality Among Americans How Religion Intersects With Americans’ Views on the Environment Online Religious Services Appeal to Many Americans, but Going in Person Remains More Popular Most Popular Report Materials Report PDF Topline Bridge study topline English paper questionnaire English/Spanish web/phone questionnaire Spanish paper questionnaire Interactive: Religious Landscape Study Video: Religion and Spirituality in America 2023-24 Religious Landscape Study (RLS) Dataset Table of Contents Table of Contents Decline of Christianity in the U.S. Has Slowed, May Have Leveled Off Trends in the size of U.S. religious groups and denominations Race in religious groups and congregations The narrowing gender gap in American religion Religion in childhood and adulthood Religion in U.S. families today Religious switching Signs of strengthening spirituality Attendance at religious services How religious beliefs change as people age What is happening to the ‘middle’ of American religion Trends in the religiousness of Christians Religion and political polarization Executive summary Executive summary I. Religious affiliation and religious switching 1. Religious identity 2. Religious switching 3. Identifying with a religion because of culture, ethnicity or family background II. Religion and family life 4. Religious intermarriage 5. Religious upbringing and childhood education 6. Religion, fertility and child-rearing III. Religious or spiritual beliefs and practices 7. Importance of religion and the Bible 8. Religious attendance and congregational involvement 9. Race and ethnicity in religious congregations 10. Prayer and other religious practices 11. Religious and spiritual beliefs 12. Spiritual experiences 13. Spiritual activities 14. Spiritual and religious self-descriptions IV. Social and political views 15. Religion, partisanship and ideology 16. Religion and views on LGBTQ issues and abortion 17. Religion and views on gender, parenting and workforce participation 18. Religion and views on immigration and diversity 19. Religion and views on the environment 20. Religion and views on the role of government V. Opinions on religion's place in society 21. Religion and views of right and wrong 22. Religion’s role in public life 23. Religion and views of science VI. Demographics of U.S. religious groups 24. Age, race, education and other demographic traits of U.S. religious groups Acknowledgments Acknowledgments Methodology Methodology Appendices Appendix A: Comparing results across Religious Landscape Studies Appendix B: Classification of Protestant denominations Appendix C: Comparing the Center’s religion trends with those of other major surveys Based on your impressions of our research, how trustworthy is the information on this page? What is your highest level of education? Is there anything else you'd like us to know? Thank you for your feedback!
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https://nightbeforetheexam.com/real-estate-wiki/plain-language-definitions-real-estate-exam-prep/?intID=1324
Metes and Bounds Definition - Real Estate Exam Wiki NightBeforeTheExam.com Home Select State Exam Prep Features Overview Bonus Features Compare Testimonials FAQ Guarantee Free Practice Questions Real Estate Wiki Log In Buy Now log inBuy Now Home Select State The Best Real Estate Exam PrepSee All States Alabama Alaska Arizona Arkansas California Colorado Connecticut District of Columbia Delaware Florida Georgia Hawaii Idaho Illinois Indiana Iowa Kansas Kentucky Louisiana Maine Maryland Massachusetts Michigan Minnesota Mississippi Missouri Montana Nebraska Nevada New Hampshire New Jersey New Mexico New York North Carolina North Dakota Ohio Oklahoma Oregon Pennsylvania Rhode Island South Carolina South Dakota Tennessee Texas Utah Vermont Virginia Washington West Virginia Wisconsin Wyoming Exam Prep Features Nothing Like it Anywhere Overview Bonus Features Compare Testimonials FAQ Guarantee Free Practice Questions Real Estate Wiki Home Select State The Best Real Estate Exam PrepSee All States Alabama Alaska Arizona Arkansas California Colorado Connecticut District of Columbia Delaware Florida Georgia Hawaii Idaho Illinois Indiana Iowa Kansas Kentucky Louisiana Maine Maryland Massachusetts Michigan Minnesota Mississippi Missouri Montana Nebraska Nevada New Hampshire New Jersey New Mexico New York North Carolina North Dakota Ohio Oklahoma Oregon Pennsylvania Rhode Island South Carolina South Dakota Tennessee Texas Utah Vermont Virginia Washington West Virginia Wisconsin Wyoming Exam Prep Features Overview Bonus Features Compare Testimonials FAQ Guarantee Free Practice Questions Real Estate Wiki log inBuy Now Common Joe 'n Jane Real Estate Wiki Real estate exam prep made easy! Dive into our wiki for key concepts and study materials tailored for success in your exams. ABCDEFGHIJKLMNOPQRSTUVWXYZ <--Back to Wiki Home Define Metes and Bounds in Real Estate Metes and Bounds: Metes and Bounds is a method used to describe the boundaries of a piece of land. It involves using distances (metes) and directions (bounds) from a starting point, following a series of lines to create a shape that outlines the property. This helps people understand the size and shape of a specific piece of land. Example: Let's say you're describing a rectangular piece of land. You might start at a specific point, let's say the southwest corner, and describe the length of the west boundary (200 feet), the length of the north boundary (100 feet), the length of the east boundary (200 feet), and the length of the south boundary (100 feet). This would give someone a complete understanding of the size and shape of the piece of land. "Wit & Whimsy with the Dumb Ox: Unlocking Knowledge with Rhyme:" Metes and bounds, oh what a sound! It's a way to tell, where land can be found. You start at a point, like the tip of your nose, Then measure in feet, which way the land goes. North, south, east, or west, Each direction's important, to define the rest. It's like a treasure map, but for land, Showing exactly where it does stand. So if you're ever lost, or can't find your way, Remember metes and bounds, and you'll save the day! Invest in Your Future. Buy Access Now! Buy Now Wiki - Disclaimer THE BEST REAL ESTATE EXAM PREP Home Select your State Real Estate Exam Prep Bonus Features Compare Exam Preps Testimonials Exam Prep Articles REAL ESTATE EXAM HELP Login Buy it now! Hardship Program FAQ Guarantee Free Practice Questions Real Estate Wiki Notable EXAM PREP NEWS/ UPDATES New videos added with matching quizzes for practice! Additional State Specific content added for New York Additional State Specific content added for Missouri Massive upgrades to Real Estate Exam Terms Encyclopedia New site features, numerous upgrades QUESTIONS? Contact © 2025 -NightBeforeTheExam.com | Terms of Use
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https://www.nature.com/articles/s41559-018-0528-0
Skip to main content Article Published: Supraorbital morphology and social dynamics in human evolution Ricardo Miguel Godinho ORCID: orcid.org/0000-0003-0107-95771,2, Penny Spikins ORCID: orcid.org/0000-0002-9174-51683 & Paul O’Higgins ORCID: orcid.org/0000-0002-9797-08091 Nature Ecology & Evolution volume 2, pages 956–961 (2018)Cite this article 28k Accesses 43 Citations 723 Altmetric Metrics details Abstract Uniquely, with respect to Middle Pleistocene hominins, anatomically modern humans do not possess marked browridges, and have a more vertical forehead with mobile eyebrows that play a key role in social signalling and communication. The presence and variability of browridges in archaic Homo species and their absence in ourselves have led to debate concerning their morphogenesis and function, with two main hypotheses being put forward: that browridge morphology is the result of the spatial relationship between the orbits and the brain case; and that browridge morphology is significantly impacted by biting mechanics. Here, we virtually manipulate the browridge morphology of an archaic hominin (Kabwe 1), showing that it is much larger than the minimum required to fulfil spatial demands and that browridge size has little impact on mechanical performance during biting. As browridge morphology in this fossil is not driven by spatial and mechanical requirements alone, the role of the supraorbital region in social communication is a potentially significant factor. We propose that conversion of the large browridges of our immediate ancestors to a more vertical frontal bone in modern humans allowed highly mobile eyebrows to display subtle affiliative emotions. Access through your institution Buy or subscribe This is a preview of subscription content, access via your institution Access options Access through your institution Access Nature and 54 other Nature Portfolio journals Get Nature+, our best-value online-access subscription $32.99 / 30 days cancel any time Learn more Subscribe to this journal Receive 12 digital issues and online access to articles $119.00 per year only $9.92 per issue Learn more Buy this article Purchase on SpringerLink Instant access to full article PDF Buy now Prices may be subject to local taxes which are calculated during checkout Similar content being viewed by others Drimolen cranium DNH 155 documents microevolution in an early hominin species Article 09 November 2020 Predicting the eyebrow from the orbit using three-dimensional CT imaging in the application of forensic facial reconstruction and identification Article Open access 10 March 2023 The relevance of late MSA mandibles on the emergence of modern morphology in Northern Africa Article Open access 25 May 2022 References Moss, M. 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Article PubMed PubMed Central Google Scholar 74. Zelditch, M. L., Swiderski, D. L., Sheets, H. D. & Fink, W. L. Geometric Morphometrics for Biologists: A Primer (Elsevier, New York, 2004). Download references Acknowledgements R.M.G. is funded by the Portuguese Foundation for Science and Technology (PhD funding reference: SFRH/BD/76375/2011). We are grateful to L. C. Fitton and S. Cobb at Hull York Medical School, C. Stringer at the Natural History Museum and B. Waller at the University of Portsmouth for discussion about this work. We thank R. Kruszynski at the Natural History Museum for facilitating access to the computed tomography scans and the original fossil of Kabwe 1. We also thank W. Sellers at the University of Manchester for access to software (Geomagic) in his laboratory. We are also grateful to the reviewers for helpful comments and suggestions. Author information Authors and Affiliations Department of Archaeology and Hull York Medical School, University of York, York, UK Ricardo Miguel Godinho & Paul O’Higgins 2. Interdisciplinary Center for Archaeology and Evolution of Human Behaviour, Faculdade das Ciências Humanas e Sociais, Universidade do Algarve, Faro, Portugal Ricardo Miguel Godinho 3. Department of Archaeology, University of York, York, UK Penny Spikins Authors Ricardo Miguel Godinho View author publications Search author on:PubMed Google Scholar 2. Penny Spikins View author publications Search author on:PubMed Google Scholar 3. Paul O’Higgins View author publications Search author on:PubMed Google Scholar Contributions R.M.G., P.S. and P.O. designed the experiment. R.M.G. performed the simulations. R.M.G., P.S. and P.O. wrote the manuscript. Corresponding author Correspondence to Ricardo Miguel Godinho. Ethics declarations Competing interests The authors declare no competing interests. Additional information Publisher’s note: Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Supplementary information Supplementary Information Supplementary tables and figures Reporting Summary Rights and permissions Reprints and permissions About this article Cite this article Godinho, R.M., Spikins, P. & O’Higgins, P. Supraorbital morphology and social dynamics in human evolution. Nat Ecol Evol 2, 956–961 (2018). Download citation Received: Accepted: Published: Issue Date: DOI: Share this article Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Provided by the Springer Nature SharedIt content-sharing initiative Subjects Archaeology Biological anthropology This article is cited by Mandibular morphology and the Mesolithic–Neolithic transition in Westernmost Iberia Ricardo Miguel Godinho Cláudia Umbelino Patricia Smith Scientific Reports (2023) ### Geometric morphometric investigation of craniofacial morphological change in domesticated silver foxes Timothy M. Kistner Katherine D. Zink Daniel E. Lieberman Scientific Reports (2021) ### The sociability spectrum: evidence from reciprocal genetic copy number variations Alejandro López-Tobón Sebastiano Trattaro Giuseppe Testa Molecular Autism (2020) ### Assessing the Big Five personality traits using real-life static facial images Alexander Kachur Evgeny Osin Alexey Novokshonov Scientific Reports (2020) ### The evolutionary history of the human face Rodrigo S. Lacruz Chris B. Stringer Juan-Luis Arsuaga Nature Ecology & Evolution (2019)
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https://www.youtube.com/watch?v=nOWC_47T7M4
How to Convert Percent to Decimal Wrath of Math 290000 subscribers 493 likes Description 51685 views Posted: 2 Oct 2017 In this short math lesson we will go over how to convert a percent to a decimal. Converting percentages to decimals is a shot process and can be done without any calculator. All you need to do is divide your percentage by 100, so if you have 23% and you want it as a decimal, just divide by 100. In that case, you would have 0.23. Dividing by 100 will have the effect of moving the decimal place over two place values to the left. Percentage means out of 100, so it is easy to see why 23% should be 0.23, which is 23 hundredths. I hope you find this video helpful, and be sure to ask any questions down in the comments! Check out my music channel! Check out my blog! 92 comments Transcript: hello everyone welcome to wrath of math I'm your host Sean Ian's in today's video we are going over how to convert a percent to a decimal so percent basically means out of 100 so to convert it to a decimal all you have to do is divide your percentage by 100 so for example 100 percent as a decimal is just equal to 1 what you have to do is divide by 100 and the effect that has is moving the decimal point over two places to the left so here there is sort of an invisible decimal point there marking the end of this whole number when you divide it by 100 the decimal point moves over two spots and that's really all there is to it so let's just look at a few examples and hopefully you'll be well on your way for easily converting percentages to decimals so let's say we have 23 percent to make that a decimal divided by 100 you take your decimal point right there move it over two spots you're going to be left with 0.23 as a decimal that's 23 one hundredths so now let's say you had 4.5% sorry I didn't like what I was writing there now we'll divide this by 100 to turn it into a decimal you take our decimal point that's jumping right out at us move it over to places as a decimal that's going to be zero point we have a zero there and then a four and then we have our five and that's what it looks like as a decimal let's say you had 0.12 percent again we divide by 100 just move the decimal place over two spots so there's one and there's two and just like before there's another zero there that we don't see when I write this number because it's not necessary but when we put it in decimal form it is necessary so we have one zero two zeros one two and that is what this looks like in decimal form and that's really all there is to it if you have a percentage you divide by 100 which results in your decimal point moving over two places as as you have seen in these examples so I hope this video helped you understand how to convert from a percentage to a decimal let me know in the comments you have any questions need anything clarified or of any of the video requests thank you very much for watching I'll see you next time and be sure to subscribe Franky's math videos on the internet all the way up here dear would you please come to blue you live it appear dear there's a light where I found that erases it makes
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https://artsandculture.google.com/story/%E2%80%9Cwe-the-people%E2%80%9D-printings-of-the-us-constitution-from-the-gilder-lehrman-collection-the-gilder-lehrman-institute-of-american-history/OwWxYaRWjqV_IA?hl=en
“We the People”: Printings of the US Constitution from the Gilder Lehrman Collection — Google Arts & Culture Home Explore Play Nearby Profile Achievements Collections Themes Experiments Artists Mediums Art movements Historical events Historical figures Places About Settings View activity Send feedback Privacy&Terms • Generative AI Terms HomeExplorePlayNearbyFavorites Sign in Loading… + – “We the People”: Printings of the US Constitution from the Gilder Lehrman Collection By The Gilder Lehrman Institute of American History In 1787, fifty-five men met in secret to write a constitution for "a more perfect Union." This exhibition of five early printings of the US Constitution opens a window into the process by which the draft evolved into the Constitution we live by today. The Gilder Lehrman Institute of American History is the only institution to hold all of the first five printings of the US Constitution. The State House in Philadelphia (11/12/1904)The Gilder Lehrman Institute of American History The Constitutional Convention On May 25, 1787, the fifty-five delegates to the Constitutional Convention opened their first session in Philadelphia’s State House. They posted sentries at the doors to keep their secrets from flying out. Barring the press and the public, the delegates took a vow not to reveal to anyone the words spoken there. First printed draft of the US Constitution First printed draft of the US Constitution (1787-08-06) by Constitutional ConventionThe Gilder Lehrman Institute of American History US Constitution, First Draft This draft of the Constitution was printed secretly for the delegates on August 6, 1787, by Dunlap and Claypoole, the official printer to the Constitutional Convention. In order to make it easier for the delegates to take notes, it was printed with wide margins. “We the People of the States” In the August 6 preamble, delegates describe themselves as representatives of “the States of New-Hampshire, Massachusetts, Rhode-Island,” etc. They viewed the United States as a confederation of separate states that worked together in limited situations. This view would soon change. Pierce Butler, 11/21/1904, From the collection of: The Gilder Lehrman Institute of American History Show less Read more Pierce Butler, Delegate Pierce Butler, a delegate from South Carolina, ignored the directive to destroy all his notes from the Convention. His notes, shown on the next slide, trace the debates and compromises that led to the final Constitution. First printed draft of the US Constitution Page 4The Gilder Lehrman Institute of American History Congress and the Slave Trade As a slaveholder, Pierce Butler was particularly invested in parts of the Constitution that addressed slavery. In the margin of his personal copy, he noted down a clause that would prevent Congress from ending the slave trade until at least 1808. First printing of the second draft of the US Constitution First printing of the second draft of the US Constitution (September 12, 1787) by Constitutional ConventionThe Gilder Lehrman Institute of American History US Constitution, Second Draft This is the second draft of the US Constitution, printed for further discussion on September 12, 1787. The handwritten notes by Pierce Butler on this version show that many changes were still being considered less than a week before the official printing. George Washington (1852) by Peale, Rembrandt (1778-1860)The Gilder Lehrman Institute of American History George Washington, President of the Convention George Washington reluctantly accepted the position of president of the Constitutional Convention. He rarely spoke or took part in the deliberations, but his presence gave legitimacy to the proceedings. First printing of the second draft of the US Constitution page 2, From the collection of: The Gilder Lehrman Institute of American History Show less Read more On September 17, as the delegates were preparing to sign the document, a delegate suggested changing the number of Representatives in the House from at least one per 40,000 to one per 30,000. Washington believed the change would increase the chances for ratification, so the Convention voted unanimously to adopt the one per 30,000 rule. Benjamin Franklin's member's copy of the United States Constitution Benjamin Franklin's member's copy of the United States Constitution (ca. September 17, 1787) by Constitutional ConventionThe Gilder Lehrman Institute of American History US Constitution, Member’s Copy Late in the day on September 17, 1787, Dunlap and Claypoole received the final revisions to the US Constitution from the Convention. They worked through the night to print this copy in time to distribute it to the delegates on the morning of September 18. Benjamin Franklin (ca. 1800) by Martin, David (1737-1797)The Gilder Lehrman Institute of American History Benjamin Franklin, Oldest Delegate At 81 years of age, Franklin was the oldest delegate to the Convention. Benjamin Franklin owned the member’s copy of the Constitution included in this exhibition. He presented it to his nephew Jonathan Williams. The document was discovered in an old trunk by a descendant of the Williams family in 1972. Benjamin Franklin's member's copy of the United States Constitution page 2, From the collection of: The Gilder Lehrman Institute of American History Show less Read more In his copy of the final Constitution, Franklin underlined “uniform laws on the subject of bankruptcies.” Franklin believed that uniform laws throughout the country would lead to equality under the law for all citizens. Convention at Philadelphia (2/24/1905) by Kappes, Alfred (1850-1894)The Gilder Lehrman Institute of American History Delegates the Sign US Constitution Although fifty-five delegates attended the Convention over the course of four months, only thirty-nine signed the final document. Broadsheet printing of the Constitution Broadsheet printing of the Constitution (ca. September 17-18, 1787) by Constitutional ConventionThe Gilder Lehrman Institute of American History US Constitution, First Public Printing This broadside is the rarest printing of the US Constitution, with only two copies in existence. It is the first edition printed specifically for mass public circulation. “We, the People of the United States” The preamble in this printing reads “We, the People of the United States.” The preamble to the first printed draft of August 6, 1787, reads “We the People of the States of New-Hampshire, Massachusetts, Rhode-Island,” etc., listing each of the thirteen states. Between the writing of the draft and the final version, the idea of a single, unified nation had been born. The Pennsylvania Packet newspaper printing of the Constitution The Pennsylvania Packet newspaper printing of the Constitution (September 19, 1787) by Constitutional ConventionThe Gilder Lehrman Institute of American History US Constitution, First Newspaper Printing On September 19, the full text of the US Constitution was published in the Pennsylvania Packet, and Daily Advertiser, a newspaper owned by Dunlap and Claypoole. This printing has a unique a place in American history, as it made the Constitution available to a wide readership through the popular press. Freeman's Journal, October 24, 1787, From the collection of: The Gilder Lehrman Institute of American History Show less Read more Before radio, television, and the Internet, it was through newspapers that Americans read and debated the provisions of the Constitution. In newspapers such as the Freeman’s Journal, Federalists and anti-Federalists picked apart each article and section. Their fierce debates culminated in the ratification of a strong but flexible Constitution that has served as the cornerstone of our republic for 228 years. Credits: Story Developed by the Gilder Lehrman Institute of American History. Credits: All media The story featured may in some cases have been created by an independent third party and may not always represent the views of the institutions, listed below, who have supplied the content. The Gilder Lehrman Institute of American History Stories from The Gilder Lehrman Institute of American History Online Exhibit ### The American Revolution through the Eyes of Hamilton #### The Gilder Lehrman Institute of American History Online Exhibit ### Alexander Hamilton and the Ratification of the Constitution #### The Gilder Lehrman Institute of American History Online Exhibit ### Alexander Hamilton Establishes the US Economy #### The Gilder Lehrman Institute of American History Online Exhibit ### Alexander Hamilton and Washington's Presidency #### The Gilder Lehrman Institute of American History Online Exhibit ### The Private Life of Alexander Hamilton #### The Gilder Lehrman Institute of American History Online Exhibit ### The Hamilton-Burr Duel #### The Gilder Lehrman Institute of American History Online Exhibit ### Hamilton's New York: Lower Manhattan Walking Tour #### The Gilder Lehrman Institute of American History Online Exhibit ### Alexander Hamilton: A Chronology #### The Gilder Lehrman Institute of American History Online Exhibit ### ALEXANDER HAMILTON: THE MAN WHO MADE MODERN AMERICA #### The Gilder Lehrman Institute of American History Online Exhibit ### The USA Votes! #### The Gilder Lehrman Institute of American History Interested in Fashion? 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https://artofproblemsolving.com/wiki/index.php/Divisibility_rules?srsltid=AfmBOop5Adbkg-9mILqyCg1AUukOumyOwb27eLGs_imU2R7nWoipwKS-
Art of Problem Solving Divisibility rules - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Divisibility rules Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Divisibility rules These divisibility rules help determine when positive integers are divisible by particular other integers. All of these rules apply for base-10only -- other bases have their own, different versions of these rules. Contents [hide] 1 Divisibility Videos 2 Basics 2.1 Divisibility Rule for 2 and Powers of 2 2.2 Divisibility Rule for 3 and 9 2.3 Divisibility Rule for 5 and Powers of 5 2.4 Divisibility Rule for 7 2.5 Divisibility Rule for 10 and Powers of 10 2.6 Divisibility Rule for 11 2.7 General Rule for Composites 2.7.1 Example 3 Advanced 3.1 General Rule for Primes 3.2 Divisibility Rule for 13 3.3 Divisibility Rule for 17 3.4 Divisibility Rule for 19 3.5 Divisibility Rule for 29 3.6 Divisibility Rule for 49 4 Special 4.1 Mod-preserving tests 4.1.1 Mod-preserving for 7 4.1.2 Mod-preserving for 13 4.2 Block tests 4.2.1 Small blocks -- 101 and 1001 4.2.2 Bigger blocks -- 10001 and 10000001 4.2.3 Type 2 blocks -- 111 and 11111 5 Problems 6 Resources 6.1 Books 6.2 Classes 7 See also Divisibility Videos Basics Divisibility Rule for 2 and Powers of 2 A number is divisible by if and only if the last digits of the number are divisible by . Thus, in particular, a number is divisible by 2 if and only if its units digit is divisible by 2, i.e. if the number ends in 0, 2, 4, 6 or 8. Proof Divisibility Rule for 3 and 9 A number is divisible by 3 or 9 if and only if the sum of its digits is divisible by 3 or 9, respectively. Note that this does not work for higher powers of 3. For instance, the sum of the digits of 1899 is divisible by 27, but 1899 is not itself divisible by 27. Proof Divisibility Rule for 5 and Powers of 5 A number is divisible by if and only if the last digits are divisible by that power of 5. Proof Divisibility Rule for 7 Rule 1: Partition into 3 digit numbers from the right (). The alternating sum () is divisible by 7 if and only if is divisible by 7. Proof Rule 2: Truncate the last digit of , double that digit, and subtract it from the rest of the number (or vice-versa). is divisible by 7 if and only if the result is divisible by 7. Proof Rule 3: "Tail-End divisibility." Note. This only tells you if it is divisible and NOT the remainder. Take a number say 12345. Look at the last digit and add or subtract a multiple of 7 to make it zero. In this case we get 12380 or 12310 (both are acceptable; I am using the former). Lop off the ending 0's and repeat. 1238 - 28 ==> 1210 ==> 121 - 21 ==> 100 ==> 1 NOPE. Works in general with numbers that are relatively prime to the base (and works GREAT in binary). Here's one that works. 12348 - 28 ==> 12320 ==> 1232 +28 ==> 1260 ==> 126 + 14 ==> 14 YAY! Tiny extension to tell you the remainder: Count how many zeroes you lop off and mod 6. Multiply mod 7 with the corresponding number Zeroes (mod 6) Number to multiply by 0 1 1 3 2 2 3 6 4 4 5 5 And that's the remainder. Divisibility Rule for 10 and Powers of 10 If a number is power of 10, define it as a power of 10. The exponent is the number of zeros that should be at the end of a number for it to be divisible by that power of 10. Example: A number needs to have 6 zeroes at the end of it to be divisible by 1,000,000 because . Divisibility Rule for 11 A number is divisible by 11 if the alternating sum of the digits is divisible by 11. Proof General Rule for Composites A number is divisible by , where the prime factorization of is , if the number is divisible by each of . Example For the example, we will check if 55682168544 is divisible by 36. The prime factorization of 36 to be . Thus we must check for divisibility by 4 and 9 to see if it's divisible by 36. Since the last two digits, 44, of the number is divisible by 4, so is the entire number. To check for divisibility by 9, we look to see if the sum of the digits is divisible by 9. The sum of the digits is 54 which is divisible by 9. Thus, the number is divisible by both 4 and 9 and must be divisible by 36. Advanced General Rule for Primes For every prime number other than 2 and 5, there exists a rule similar to rule 2 for divisibility by 7. For a general prime , there exists some number such that an integer is divisible by if and only if truncating the last digit, multiplying it by and subtracting it from the remaining number gives us a result divisible by . Divisibility rule 2 for 7 says that for , . The divisibility rule for 11 is equivalent to choosing . The divisibility rule for 3 is equivalent to choosing . These rules can also be found under the appropriate conditions in number bases other than 10. Also note that these rules exist in two forms: if is replaced by then subtraction may be replaced with addition. We see one instance of this in the divisibility rule for 13: we could multiply by 9 and subtract rather than multiplying by 4 and adding. is any number so that Divisibility Rule for 13 Rule 1: Truncate the last digit, multiply it by 4 and add it to the rest of the number. The result is divisible by 13 if and only if the original number was divisble by 13. This process can be repeated for large numbers, as with the second divisibility rule for 7. Proof Rule 2: Partition into 3 digit numbers from the right (). The alternating sum () is divisible by 13 if and only if is divisible by 13. Proof Rule 3: Works for . Let . If is odd add 39 to . Round up to the nearest multiple of 80, call the result . Find . Check: Is . Proof Divisibility Rule for 17 Truncate the last digit, multiply it by 5 and subtract from the remaining leading number. The number is divisible if and only if the result is divisible. The process can be repeated for any number. Proof Divisibility Rule for 19 Truncate the last digit, multiply it by 2 and add to the remaining leading number. The number is divisible if and only if the result is divisible. This can also be repeated for large numbers. Proof Divisibility Rule for 29 Truncate the last digit, multiply it by 3 and add to the remaining leading number. The number is divisible if and only if the result is divisible. This can also be repeated for large numbers. Proof Divisibility Rule for 49 Why 49? For taking pesky out of a root. Useful below 4900. Round up to a multiple of 50, call it , and subtract the original number, call this . If , it is divisible by 49. Examples: Round up: . Difference: . ? Yes! Round up: . Difference: . ? No! Round up: . Difference: . ? Yes! Extension to work for all numbers. Floor divide by 4950, multiply by 50, and add to before calculating Proof Special Mod-preserving tests These tests allow you take the modulo operation easily. Mod-preserving for 7 Multiply the first digit by 3 and add it to the rest. Mod-preserving for 13 Multiply the first digit by 3 and subtract it from the rest Block tests As a bonus, these are also mod-preserving Small blocks -- 101 and 1001 The divisibility for 101 test is simple: Alternate adding and subtracting blocks of two digits starting from the end two, which are added. Ex. 1102314 by 101 01 + 10 - 23 + 14 ← last block is always two digits and positive =0 so 1102314 is divisible by 101 The divisibility for 1001 is the same, but with blocks of three. (Starting with the end three, this time) The 1001 test also works for all it's divisors. The most useful are 7, 11, and 13. Bigger blocks -- 10001 and 10000001 10001 has block size length 4, and factors nicely into 73137. 1000001 has block size 6, and factors into 175882353. 5882353 isn't much use, but 17 is, when we're testing a large number. Type 2 blocks -- 111 and 11111 A different type of test can be yielded from adding all the blocks, but again starting with the end. 111 has a block length of three, and factors into 37 and 3. 11111 has a length of five, and factors to 41 and 271. 1111111, with a length of seven, can provide a test for 239 and 4649, if you ever need it. Problems Practice Problems on Alcumus Divisibility (Prealgebra) 2000 AMC 8 Problems/Problem 11 2006 AMC 10B Problems/Problem 25 Resources Books The AoPS Introduction to Number Theory by Mathew Crawford. The Art of Problem Solving by Sandor Lehoczky and Richard Rusczyk. Classes AoPS Introduction to Number Theory Course See also Number theory Modular arithmetic Math books Mathematics competitions Retrieved from " Category: Divisibility Rules Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
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I am a student I am a teacher Try Study.com, risk-free As a member, you'll also get unlimited access to over 88,000 lessons in math, English, science, history, and more. Plus, get practice tests, quizzes, and personalized coaching to help you succeed. Get unlimited access to over 88,000 lessons. Try it risk-free It only takes a few minutes to setup and you can cancel any time. It only takes a few minutes. Cancel any time. Already registered? Log in here for access Back What teachers are saying about Study.com Try it risk-free for 30 days Already registered? Log in here for access 00:04 How to count protons &… 03:58 How to count protons &… Jump to a specific example Speed Normal 0.5x Normal 1.25x 1.5x 1.75x 2x Speed Jiwon Park, Kirsten Wordeman Instructors Jiwon Park Jiwon has a B.S. degree in the mathematics/ science field and over 4 years of tutoring experience. She fell in love with math when she discovered geometry proofs and that calculus can help her describe the world around her like never before. View bio Kirsten Wordeman Kirsten has taught high school biology, chemistry, physics, and genetics/biotechnology for three years. She has a Bachelor's in Biochemistry from The University of Mount Union and a Master's in Biochemistry from The Ohio State University. She holds teaching certificates in biology and chemistry. View bio Example SolutionsPractice Questions How to Count Protons and Electrons in Atomic Ions Step 1: Find the number of protons in an atomic ion. The number of protons is equal to the atomic number of the element. To find the number of protons, find the atomic number of the element in the periodic table. Forming an ion does not change the number of protons in an atom. The number of protons determines which kind of element an atom is. Therefore, if the number of protons is given, we can identify the element by looking up the element by the atomic number (= the number of protons) in the periodic table. Step 2: Find the number of electrons in an atomic ion In a neutral atom, the number of electrons is equal to the number of protons. For an ion, charged positively or negatively, the number of electrons is the atomic number (the number of protons) minus the charge of the ion. Step 3: Determine if an ion is a cation or an anion. Compare the number of electrons and the number of protons. An ion with more electrons than protons is an anion. An ion with fewer electrons than protons is a cation. To determine if an ion is a cation or an anion by looking at the symbol, check the charge of the ion written on the top right of the element. If the number is positive, the ion is a cation. If the number is negative, the ion is an anion. How to Count Protons and Electrons in Atomic Ions Vocabulary Protons: Positively charged components of an atom. Electrons: Negatively charged components of an atom. Atomic number in the periodic table: Let's practice counting protons and electrons in atomic ions with the next two examples. How to Count Protons and Electrons in Atomic Ions: Example 1 Complete the following table by filling in the blanks as necessary for the symbol, type of ion, number of protons, or number of electrons for an atomic ion. Symbol Type of Ion Number of protons Number of electrons K 18 Step 1: Find the number of protons in an atomic ion. The number of protons is equal to the atomic number of the element. The atomic number of potassium (K) is 19. Therefore, the number of protons is 19. Step 2: Find the number of electrons in an atomic ion The number of electrons is given, and it is 18. Step 3: Determine if an ion is a cation or an anion. Compare the number of electrons and the number of protons. An ion with more electrons than protons is an anion. An ion with fewer electrons than protons is a cation. There are 19 protons and 18 electrons. Since there are more protons than electrons, this ion is positively charged, therefore a cation. The complete table is as below: Symbol Type of Ion Number of protons Number of electrons KCation1918 How to Count Protons and Electrons in Atomic Ions: Example 2 Complete the following table by filling in the blanks as necessary for the symbol, type of ion, number of protons, or number of electrons for an atomic ion. Symbol Type of Ion Number of protons Number of electrons 55 54 Step 1: Find the number of protons in an atomic ion. The number of protons is given and it is 55. If the number of protons is given, we can identify the element by looking up the element by the atomic number (= the number of protons) in the periodic table). The element with the atomic number 55 is cesium (Cs). Therefore, the ion in the problem is an ion of a cesium atom. Step 2: Find the number of electrons in an atomic ion. The number of electrons is given and it is 54. Step 3: Determine if an ion is a cation or an anion. Compare the number of electrons and the number of protons. An ion with more electrons than protons is an anion. An ion with fewer electrons than protons is a cation. There are 55 protons and 54 electrons. Since there are more protons than electrons, this ion is positively charged, therefore a cation. Symbol Type of Ion Number of protons Number of electrons CsCation55 54 Get access to thousands of practice questions and explanations! 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https://bpac.org.nz/2023/vulval-cancer.aspx
Vulval cancer – early detection and referral - bpacnz Home Articles Browse By Category CPD informationNEW Activities BPJ Log In CPD information Prescribing Reports Audits CME Quizzes Peer Group Discussions My Bpac account Adverse drug reactions and interactionsAllergies and immunologyAntibiotic resistance and stewardshipBiochemistryCancer careCardiovascular systemChild healthComplementary and alternative medicinesCorrespondenceDebatesDermatologyDiabetesEar, nose and throatEndocrinology (excl diabetes)Feature lettersGastroenterologyGeneticsGenitourinary system (male)GuidelinesGynaecology and urinary tract disorders (female)HaematologyHealth informaticsHepatologyImmunisationInfectionsIntegrated Performance and Incentive FrameworkMāori healthMedicine indicationsMedicine subsidyMedicines managementMedico-legal issuesMental healthMusculoskeletalNephrologyNeurologyNewsNutritionOccupational medicineOlder person’s healthOncologyOphthalmologyOral healthPacific peoples healthPain managementPalliative carePeer group discussionsPharmacologyPHO Performance ProgrammePregnancy and reproductive healthProfessional practice and developmentPublic healthReportsResearch updatesRespiratory conditionsRheumatologySexual healthSmoking, alcohol, and drug misuseTrauma and surgical proceduresVirology Search Mybpac - Login Articles View Articles bpac nz Bulletin View Bulletins bpac nz Update Series View Update Series CPD information View CPD information Prescribing Reports View reports Audits View all audits CME Quizzes Do Quizzes Peer Group Discussions View all discussions About us FAQ Contact us Terms of use Hello there! please login to access my bpac Username Password [x] Remember me Dont have an account? Sign up here Forgot your login? Reset password Cancer careGynaecology and urinary tract disorders (female)Oncology Vulval cancer – early detection and referral Vulval cancer is a rare gynaecological cancer in New Zealand, with an average of 52 females newly diagnosed each year (from 2015 – 2020). Most cases are squamous cell carcinoma and are usually related to high-risk HPV infection or vulval inflammatory disorders such as lichen sclerosus. The physical appearance of vulval cancer and associated precursor lesions is variable, so a biopsy is necessary for diagnosis. Please login to save this article. 0comments save share feedback Log in download / print version Vulval cancer is an uncommon type of gynaecological cancer Types of vulval cancer Diagnosing vulval cancer Manage according to vulval biopsy results Other resources B-QuiCK summary Gynaecological Cancer Series Peer group discussion Gynaecological cancer quiz Published: 28 April 2023 | Updated: 11 March 2024 | What's changed? 11 March 2024 Cervical screening recommendations after treatment for cervical HSIL updated to align with the HPV Primary Screening programme Key practice points: The vulva is affected by the same types of cancer that affect skin elsewhere, including squamous cell carcinoma (SCC), basal cell carcinoma and melanoma SCC accounts for approximately 90% of all vulval cancers. It is usually preceded by vulval intraepithelial neoplasia (VIN), either differentiated VIN or high-grade squamous intraepithelial lesions (HSIL). Vulval SCC develops through two pathways: Chronic vulval inflammation, e.g. due to lichen sclerosus or erosive lichen planus, with vulval SCC usually preceded by differentiated VIN; referred to as the human papillomavirus (HPV)-independent pathway Infection with high-risk HPV types, with vulval SCC usually preceded by HSIL; referred to as the HPV-associated pathway. HSIL are more common than differentiated VIN, but are less likely to progress to invasive vulval SCC. Historically, most vulval SCC arose from the HPV-independent pathway, but the epidemiology has now changed in New Zealand and the proportions of vulval SCC from both pathways are almost equal Additional risk factors for vulval cancer include increasing age, cigarette smoking and immune deficiency The risk of developing VIN (HSIL or differentiated VIN) and therefore, squamous cell vulval cancer can be reduced through prophylactic HPV vaccination with Gardasil 9 (currently recommended vaccine; prevents new infection with HPV) and good control of inflammatory conditions such as lichen sclerosus In most cases, vulval cancer is diagnosed in females who are post-menopausal, although HPV-associated vulval cancers generally occur 10 – 15 years earlier than HPV-independent vulval SCC Symptoms and signs of vulval cancer may include vulval pruritus, pain, bleeding, discharge or a vulval lesion or mass. Patients can also have a lesion but be asymptomatic. There are no screening programmes for the early detection of vulval cancer, but some people are diagnosed incidentally after an examination for other reasons, e.g. cervical screening. Most people are diagnosed from targeted investigations after symptoms are reported. The diagnostic workup of a patient with suspected vulval cancer includes a focused history, considering relevant risk factors and a vulval examination with particular attention to the labia majora as this is the most common site of vulval cancer Vulval cancers can vary widely in appearance and may be multifocal or involve a single lesion. Features of malignancy include: irregular epithelial surfaces, swelling, red, white or other pigmented areas, visible lesion/mass, ulceration, bleeding. Patients with vulval symptoms but no visible lesion are unlikely to have vulval cancer A biopsy is necessary for the diagnosis of VIN and vulval cancer. Refer patients with abnormal examination findings, e.g. suspicious vulval lesion, to a gynaecologist for further assessment. N.B. The term “female” is used in this article to describe the biological sex of the patient population at risk for vulval cancer. However, we acknowledge that this may not reflect the identity of all patients, which will include transgender boys or men, intersex and non-binary individuals. For information on the follow-up and surveillance of a patient after curative-intent treatment for vulval cancer, see: Vulval cancer is an uncommon type of gynaecological cancer Vulval cancer accounts for approximately 5% of all gynaecological cancers in New Zealand with an average of 52 new diagnoses (1.4 per 100,000 females; from 2015 – 2020) and 16 deaths (0.4 per 100,000 females; from 2015 – 2018) each year;† fewer than all other gynaecological cancers except vaginal cancer.1 Squamous cell carcinoma (SCC) is the most common type of vulval cancer (approximately 90% of cases), and is usually associated with lichen sclerosus or high-risk human papillomavirus (HPV) infection.2 Mortality data are available for 2019, but are preliminary so have not been included. Mortality data for 2020 are not yet available. †There are differences in published data. For consistency across the gynaecological cancer series, incidence and mortality data has primarily been obtained from the publication – Cancer: Historical summary 1948 – 2020. There is no screening test for vulval cancer, so diagnosis relies on investigating reported vulval lesions or symptoms, opportunistic detection of vulval abnormalities, e.g. during cervical screening, and following up patients who are at increased risk, e.g. those with lichen sclerosus.3 The risk of vulval cancer increases with age and it primarily affects females who are post-menopausal with most diagnoses occurring in those aged > 70 years.2 Vulval cancer can, however, still occur in younger females; internationally the incidence in those aged < 60 years has increased since the early 1990s, likely due to an increase in the prevalence of HPV infection in younger age groups.2 In New Zealand, the incidence of vulval cancer associated with high-risk HPV infection is increasing in females aged > 50 years.4 The prognosis of females with vulval cancer is usually good if diagnosed early. Overall five-year survival for people with vulval SCC is 71%;5 this can be above 80% in those with no lymph node involvement.3 However, five-year survival is below 50% for people with spread to the inguinal lymph nodes and 10 – 15% if the iliac or other pelvic nodes are involved.3 People with HPV-independent vulval SCC have a significantly poorer prognosis (25% worse) than those with HPV-associated vulval SCC.6, 7 This is independent of age at diagnosis and cancer stage, with poorer outcomes and risk of recurrence even at an early stage.6, 7 Risk factors for vulval cancer The most significant risk factors for vulval SCC are chronic inflammatory conditions of the vulva, e.g. lichen sclerosus, erosive lichen planus, and infection with high-risk HPV, particularly type 16 (see: “Prophylactic HPV vaccination can reduce the risk of vulval cancer”).2 Estimates of lifetime risk of vulval cancer in people with lichen sclerosus range from 2.6 – 6.6%.2 For information on the treatment and follow-up of a patient with lichen sclerosus, see: “Lichen sclerosus”. Additional risk factors include Paget disease of the vulva, smoking and increasing age; factors that increase the likelihood of HPV infection (e.g. high lifetime number of sexual partners, young age at sexual activity onset, immune deficiency) also therefore increase the risk of vulval cancer.5, 8, 9 ### Prophylactic HPV vaccination can reduce the risk of vulval cancer Persistent infection with high-risk HPV types (particularly type 16) is associated with the development of high-grade squamous intraepithelial lesions (HSIL) and vulval cancer.2, 10 In New Zealand, HSIL makes up 95% of vulval cancer precursor lesions.11 Prophylactic HPV vaccination prevents new infection with HPV, including high-risk type 16, and therefore protects against the development of HPV-associated pre-cancerous vulval lesions.2, 10 It does not reduce the progression of established vulval lesions or cancer.10 HPV vaccination is most effective when administered prior to HPV exposure, however, it is still effective after exposure to HPV (see: “Expert tip”), and can prevent the development of up to 100% of pre-cancerous vulval lesions and vulval cancers.12, 13 There is also emerging evidence that HPV vaccination may be beneficial in reducing recurrence risk after treatment, however, further studies are required.14 Gardasil 9 is the currently recommended vaccine in New Zealand and has been used since 2017. It protects against nine types of HPV (6, 11, 16, 18, 31, 33, 45, 52, 58); seven of which cause HPV-related cancers and two cause genital warts (6, 11).10 HPV types 16, 18, 31, 33, 39 and 45 are responsible for the majority of HPV-associated vulval squamous cell carcinomas.11 HPV vaccination is recommended for all females (and males) ideally before the onset of sexual activity, and is funded for eligible people aged 9 – 26 years inclusive.10 School immunisation programmes and general practices generally offer HPV vaccination to students in Year Eight (around age 12 years). Expert tip.Vaccinating people who have already commenced sexual activity is still recommended as even if they have been infected with one or several HPV types, there are still other types of HPV that are associated with malignancy, and so it is unlikely that someone will have been infected with all of them. N.B. Gardasil 9 is registered for use in females aged 9 – 45 years and in males aged 9 – 26 years. However, there are no theoretical concerns that the efficacy or safety of the vaccine in males aged up to 45 years will differ significantly from females of the same age or younger males.10 The vaccine may have efficacy in people aged > 45 years, however, there is a lack of evidence of this. If the course is started prior to the patients 27th birthday, the rest of the course is funded. For further information on funded indications, see: www.health.govt.nz/publication/immunisation-handbook-2020 For further information on HPV vaccination, see: “Cervical cancer – early detection and referral” Types of vulval cancer Approximately 90% of all vulval cancers are of SCC histology, and are generally preceded by precursor lesions, either differentiated VIN or HSIL (see Table 1 for explanation of terminology).2, 8, 15 The remaining 10% of vulval cancers predominantly include melanoma, Bartholin gland carcinoma, basal cell carcinoma and Paget disease of the vulva (see: “Rare types of vulval cancer”).2 Table 1. Terminology used to describe vulval intraepithelial neoplasia (VIN). | | Current terminology | Previous terminology | Association with vulval cancer (also see Figure 1) | --- --- | | Vulval intraepithelial neoplasia (VIN) | | High-grade squamous intraepithelial lesions (HSIL) | Usual-type VIN (uVIN) VIN 2 or 3 Bowen disease of the vulva | HPV-associated; may progress to vulval SCC | | Differentiated vulval intraepithelial neoplasia (differentiated or dVIN) | No change | HPV-independent; more likely to progress to vulval SCC | | Low-grade squamous intraepithelial lesions (LSIL) | Flat condyloma HPV effect VIN 1 | HPV-associated; does not require treatment | Introduced in 2015 by the International Society for the Study of Vulvovaginal Disease: Bornstein J, Bogliatto F, Haefner HK, et al. The 2015 International Society for the Study of Vulvovaginal Disease (ISSVD) Terminology of Vulvar Squamous Intraepithelial Lesions. Obstetrics & Gynecology 2016;127:264–8. doi:10.1097/AOG.0000000000001285. Vulval SCC Vulval SCC and associated precursor lesions develop through two pathways (Figure 1) and have different clinical features depending on the pathway (Table 2).2 Traditionally most vulval SCC arose from the HPV-independent pathway, but the epidemiology has now changed in New Zealand and the proportions of vulval SCC from both pathways are almost equal.4 Figure 1. An overview of the HPV-independent and HPV-associated pathways for the development of vulval squamous cell carcinoma.2, 5, 9, 16 N.B. The larger arrows denote a higher likelihood of progression. HPV = human papillomavirus. The HPV-independent pathway is usually associated with lichen sclerosus (or other chronic inflammation), differentiated VIN† and older age.2, 5, 9 In contrast, the HPV-associated pathway is associated with high-risk HPV infection, high-grade squamous intraepithelial lesions (HSIL) and a younger age at diagnosis (peak prevalence at ages 40 – 44 and > 50 years).2, 9, 17 The prevalence of HPV-associated vulval SCC is increasing in females aged > 50 years in New Zealand.4 Other factors may also contribute to the development of differentiated VIN, e.g. mutations in tumour suppressor gene, p53 2, 5 †Differentiated VIN is considered the main precursor lesion associated with the HPV-independent pathway; other precursor lesions may include differentiated exophytic vulvar intraepithelial lesions (DEVIL) and vulvar acanthosis with alternative differentiation (VAAD)16 The majority of vulval cancer precursor lesions (approximately 95%)11 are HSIL (previously referred to as usual-type VIN, VIN 2/3 or Bowen disease of the vulva), but these are associated with a low risk and rate of progression to invasive vulval cancer.5, 15, 16 Data vary across the literature; some studies report that < 5% of HSIL progress to invasive vulval cancer, while others report a risk of progression of approximately 10% (within ten years in females with treated vulval HSIL; the risk is likely to be higher in females with untreated vulval HSIL).5, 15, 16 For information on the treatment and follow-up of a patient with HSIL, see: “Vulval HSIL”. In comparison, differentiated VIN, although less common than HSIL has a high risk and rapid rate of progression to invasive vulval cancer. 16–19 Estimates vary but approximately 50% of females with differentiated VIN develop vulval SCC within ten years with a median progression time of one to two years.15 A small local study found that the median progression time between biopsy showing differentiated VIN and invasive cancer was 43.5 months.20 For information on the treatment and follow-up of a patient with differentiated VIN, see: “Differentiated VIN”. The low prevalence of differentiated VIN may be due to the rapid rate of progression to invasive vulval cancer and underdiagnosis of the lesion 5 Summary of vulval squamous cell carcinoma Most vulval cancers are squamous cell carcinoma (SCC) Vulval SCC usually develop via two pathways: HPV-independent (typically related to lichen sclerosus or other chronic inflammation of the vulva and associated with differentiated VIN) and HPV-associated (related to HPV infection and associated with HSIL) The majority of precursor lesions are HSIL from the HPV-associated pathway However, the precursor lesions associated with the HPV-independent pathway (differentiated VIN) are more likely to progress to invasive squamous cell vulval cancer Therefore the HPV-independent and HPV-associated pathways are responsible for almost equal numbers of squamous cell vulval cancers Rare types of vulval cancer The remaining 10% of vulval cancers predominantly include (see Table 2 for details on how to visually identify these types): Primary vulval melanoma. Vulval melanoma is not related to exposure to sunlight or UV radiation; it can occur at any age but is most common among Europeans aged 40 – 60 years.21 Patients with vulval melanoma are typically diagnosed at a later stage and at an older age than those with other melanomas.5 Bartholin gland carcinoma. A mass in the location of a Bartholin’s gland is more likely to be malignant in patients who are aged ≥ 40 years compared to patients who are younger, particularly if it is fixed, firm or irregularly shaped.22 Therefore, conservative management of Bartholin cysts with Word catheters (a flexible tube with a small balloon that is inserted into a cyst or abscess for drainage) in older females is not recommended. Basal cell carcinoma (BCC). Usually diagnosed in females who are post-menopausal at an average age of 70 years.9, 21 Unlike BCC at other sites, vulval BCC is not associated with exposure to sunlight or UV radiation, although risk may be increased by immunosuppression, chronic irritation, pelvic radiation or trauma.21 The risk of metastases or recurrence is low and long-term follow-up is not usually required.9 Paget disease of the vulva. Paget disease of the vulva typically affects females who are post-menopausal aged in their 60s of European ethnicity.2, 5 It is usually a pre-cancerous intraepithelial lesion (adenocarcinoma in situ), but can progress to invasive adenocarcinoma.2, 5, 16 Invasive Paget disease of the vulva accounts for 1 – 2% of vulval cancers.2, 5 In fewer than 10% of cases, Paget disease of the vulva can be associated with other cancers, and further investigations, e.g. mammogram, urine cytology/cystoscopy, are often recommended. Diagnosing vulval cancer The diagnosis of vulval cancer involves the recognition of suspicious vulvovaginal symptoms from a focused patient history, followed by investigation with a pelvic examination to detect any abnormal changes to vulval appearance.3, 8 Referral to a gynaecologist is required for patients with suspicion for vulval cancer for consideration of a vulvoscopy or biopsy (diagnosis is made via biopsy). Most people with vulval cancer are symptomatic While a small number of people with vulval cancer (or precursor lesions) are asymptomatic, most experience some degree of symptoms such as vulval pruritus, pain, irritation, bleeding or discharge.3, 5, 16 A vulval lesion (i.e. mass or ulcer) is generally present.5, 9, 16 A study in the United Kingdom found that the risk of vulval cancer was 12.8% in patients with a suspicious vulval lesion and this risk was higher if the lesion was painful or bleeding.2 In some cases, the patient or their partner may have noticed a lesion themselves, or it may have been detected by a clinician during a pelvic examination for another reason, e.g. cervical screening or follow-up of patients with lichen sclerosus.3 Take a history and perform a vulval examination for patients with suspicion of vulval cancer Begin by taking a focused history with particular attention given to vulvovaginal symptoms and any risk factors, e.g. lichen sclerosus. Perform a pelvic examination, including speculum or bimanual examination, as indicated.3, 8, 9Palpate the groin to assess for any enlarged lymph nodes.9 Offer cervical screening to patients who are due. Swabs for sexually transmitted infections, e.g. genital herpes, are not usually indicated if a tumour is present, but may be useful if there is ulceration, bleeding or discharge, to exclude other potential causes. Examining the vulval region When examining the vulval region, pay close attention to the labia majora as this is the most common site of vulval cancer.8, 9 Changes to vulval appearance due to precursor lesions or vulval cancer are highly variable (see Table 2 for specific features). In general, look for:2, 3, 5 Changes to the surface of the skin, i.e. irregular epithelial surfaces Swelling Changes in colour, e.g. red, white or other pigmented areas A visible lesion, e.g. flat, raised, lump (or apparent genital warts in a post-menopausal patient) Ulceration or bleeding Table 2. Typical features of vulval cancer types and precursor lesions. HSIL/HPV-associated vulval SCCLesions have a variable appearance, but may be multifocal with red, white or other pigmented areas 18 Differentiated VIN/HPV-independent vulval SCCUnifocal, unicentric, rough, warty plaques or papular lesions with poorly defined edges.16, 18 Plaques are usually pink or grey-white and hyperkeratotic.16 Ulceration may be present.18 Can be difficult to distinguish from underlying lichen sclerosus.16, 18 Vulval melanomaAsymmetric macule, papule or nodule with irregular margins, with or without ulceration or pigmentation (i.e. amelanotic melanoma).2, 5 Often on the labia majora, labia minora or clitoris.5 The ABCDEFG checklist may help to differentiate vulval melanoma from benign naevi and melanosis of the vulva. For further information on melanoma, including use of dermatoscopy, see: bpac.org.nz/2021/melanoma-detection.aspx Bartholin gland carcinomaPainful mass in the area of a Bartholin’s gland that is firm, fixed or nodular on palpation.2 Age > 40 years should raise suspicion for malignancy.2 A regularly shaped mass that is soft, mobile and often tender to touch is more likely to be a Bartholin’s gland cyst or abscess than carcinoma.22 Vulval BCCDiscrete solitary nodule, papule or plaque, or raised, rolled-edge ulcer, usually on the labia majora 2, 9 Paget disease of the vulva (adenocarcinoma in situ and adenocarcinoma)Painful and erythematous (or pink) rash or eczematous lesion/plaque, typically asymmetric or unilateral on the labia majora.2, 5 Can be difficult to distinguish from other inflammatory dermatoses. Refer patients with a lesion (or lesions) suspected to be VIN or vulval cancer to a gynaecologist for further assessment which is likely to include vulvoscopy and/or biopsy.3 A fungating mass or palpable inguinal nodes indicate more advanced cancer and should prompt urgent referral.2 Patients with generalised vulval irritation but no visible lesion on examination are unlikely to have vulval cancer.2 Consider other causes of the patient’s symptoms, including candidiasis, an inflammatory dermatosis, e.g. lichen sclerosus or planus, or a sexually transmitted infection, e.g. genital herpes.18 Manage according to vulval biopsy results The biopsy results may indicate the presence of a vulval condition such as lichen sclerosus or Paget disease of the vulva, or it may show a precursor or malignant lesion. If the biopsy results indicate vulval cancer, the patient will be managed in a Gynaecological Oncology centre in Auckland, Wellington or Christchurch. Patients whose biopsy results show pre-cancerous lesions or inflammatory conditions of the vulva (see below) may require treatment to reduce the risk of progression to invasive vulval cancer and ongoing monitoring due to the risk of recurrence.18 Consider setting a recall for periodic follow-up of these patients. Lichen sclerosus Effective control of lichen sclerosus with topical corticosteroids reduces the risk of progression to differentiated VIN and invasive vulval cancer.23, 24 Patients should be reviewed at least annually, or more often if they notice changes to their symptoms or to the feel or appearance of their lesion(s).16, 25 Referral for an additional biopsy should be considered in patients with lichen sclerosus who do not respond to treatment.5, 25 Long-term or indefinite follow-up for a patient with lichen sclerosus is required; those with stable well controlled lichen sclerosus may be followed up in primary care (with periodic specialist review).25, 26 For further information on lichen sclerosus, see: bpac.org.nz/bpj/2014/september/vulvovaginal.aspx Paget disease of the vulva Consider investigating for co-existing malignancies, as a diagnosis of Paget disease of the vulva may represent spread of an adenocarcinoma from another site, e.g. rectum, bladder, urethra, cervix.2 There are multiple treatment options available for patients diagnosed with Paget disease of the vulva with the overall aim being to reduce the risk of progression to invasive adenocarcinoma. Surgery is usually the standard treatment, but imiquimod cream (unapproved indication) is an effective non-surgical option for some patients.2, 16 As the rate of recurrence after treatment is high (up to 70%), long-term follow-up is recommended.2 For further information on Paget disease of the vulva, see: dermnetnz.org/topics/extramammary-paget-disease Low-grade squamous intraepithelial lesion Low-grade squamous intraepithelial lesions (LSIL) indicate infection with HPV (in most cases with low risk types, 6 or 11), and do not require treatment.16 The infection is usually transient and self-resolves within one to two years.27 LSIL are classified as low risk but progression to HSIL is possible (uncommon); consider setting a recall for periodic follow-up until the lesion resolves.27 High-grade VIN Surgical resection is often the first-line treatment for patients with differentiated VIN or vulval HSIL as these lesions (especially differentiated VIN) are associated with an increased risk of invasive vulval cancer.16, 18 Recurrence is common after treatment for high-grade VIN, so patients may benefit from increased surveillance.16, 18 Differentiated VIN Differentiated VIN need to be excised. After excision, treatment of any underlying lichen sclerosus with ultrapotent topical corticosteroids is recommended to reduce the risk of recurrence and subsequent progression to vulval SCC.16, 19 European guidelines recommend ongoing follow-up at least every six months, depending on the severity of any associated lichen sclerosus.18 Vulval HSIL First-line treatment of vulval HSIL may include surgical excision or the topical application of imiquimod cream (unapproved indication).2, 16 Exclusion of malignancy by mapping biopsies may be performed prior to treatment.16 Recurrence of vulval HSIL after treatment is high at 20 – 40%.17 Vulval HSIL frequently co-exists with other HPV-associated cancers or pre-cancerous lesions. A history of cervical cancer or abnormal cervical cytology is present in 50% of females diagnosed with vulval HSIL and up to half of those diagnosed with vulval HSIL develop another HPV-associated genital cancer or pre-cancerous lesion, e.g. cervical, vaginal or anal.27 As the incidence of anal cancer is higher in people with a history of pre-cancerous vulval lesions or vulval cancer, consider investigations for anal cancer, e.g. digital anorectal examination, anal cytology, as indicated.2, 16 For patients with treated cervical HSIL, New Zealand Cervical Screening Guidelines recommend HPV and cytology testing (test of cure) before returning to their regular cervical screening interval (if both tests are negative/normal at 6 and 18 months post-treatment);28 there are no specific cervical screening recommendations for people treated for vulval HSIL. International guidelines recommend that all patients with a history of vulval HSIL have annual cervical screening (it is unclear whether this will change when HPV testing is introduced).18 In practice, decisions around cervical screening for those with a history of vulval HSIL will be made on a case-by-case basis in conjunction with a gynaecologist. For information on the follow-up and surveillance of a patient after curative-intent treatment for vulval cancer, see: There is a B-QuiCK summary available for this topic Acknowledgement Thank you to Dr Anju Basu, Consultant Gynaecologist, Wellington, and Clinical Senior Lecturer, Department of Obstetrics, Gynaecology & Women’s Health, University of Otago, Wellington, and Dr Lois Eva, Clinical Director Gynaecological Oncology, National Women’s Hospital, Te Toka Tumai, for expert review of this article. N.B. Expert reviewers do not write the articles and are not responsible for the final content. bpac nz retains editorial oversight of all content. This resource is the subject of copyright which is owned by bpac nz. You may access it, but you may not reproduce it or any part of it except in the limited situations described in the terms of use on our website. Article supported by Te Aho o Te Kahu, Cancer Control Agency. #### References Ministry of Health NZ. Cancer: historical summary 1948–2020. 2022. Available from: (Accessed Apr, 2023). Morrison J, Baldwin P, Buckley L, et al. British Gynaecological Cancer Society (BGCS) vulval cancer guidelines: recommendations for practice. Eur J Obstet Gynecol Reprod Biol 2020;252:502–25. Royal College of Obstetricians and Gynaecologists. Guidelines for the diagnosis and management of vulval carcinoma. 2014. Available from: Eva LJ, Sadler L, Fong KL, et al. Trends in HPV-dependent and HPV-independent vulvar cancers: The changing face of vulvar squamous cell carcinoma. Gynecologic Oncology 2020;157:450–5. Weinberg D, Gomez-Martinez RA. Vulvar cancer. Obstetrics and Gynecology Clinics of North America 2019;46:125–35. Eva LJ, Ganesan R, Chan KK, et al. Vulval squamous cell carcinoma occurring on a background of differentiated vulval intraepithelial neoplasia is more likely to recur: a review of 154 cases. J Reprod Med 2008;53:397–401. Eva L, Sadler L, Thompson JM, et al. HPV-independent and HPV-associated vulvar squamous cell carcinoma: two different cancers. Int J Gynecol Cancer 2022;:ijgc–2022–003616. Koh W-J, Greer BE, Abu-Rustum NR, et al. Vulvar cancer, Version 1.2017, NCCN Clinical Practice Guidelines in Oncology. Journal of the National Comprehensive Cancer Network 2017;15:92–120. Verma V, Haldar K. Invasive vulval cancer. Obstetrics, Gynaecology & Reproductive Medicine 2020;30:213–8. Ministry of Health. Immunisation handbook 2020. 2020. Available from: (Accessed Apr, 2023). Bigby SM, Eva LJ, Tous S, et al. Prevaccine human papillomavirus status in invasive and intraepithelial lesions of the vulva in New Zealand women. J Low Genit Tract Dis 2022;26:323–7. Staples JN, Duska LR. Cancer screening and prevention highlights in gynecologic cancer. Obstetrics and Gynecology Clinics of North America 2019;46:19–36. Huh WK, Joura EA, Giuliano AR, et al. Final efficacy, immunogenicity, and safety analyses of a nine-valent human papillomavirus vaccine in women aged 16-26 years: a randomised, double-blind trial. Lancet 2017;390:2143–59. Garland SM, Paavonen J, Jaisamrarn U, et al. Prior human papillomavirus‐16/18 AS04‐adjuvanted vaccination prevents recurrent high grade cervical intraepithelial neoplasia after definitive surgical therapy: Post‐hoc analysis from a randomized controlled trial. Int J Cancer 2016;139:2812–26. Thuijs NB, van Beurden M, Bruggink AH, et al. Vulvar intraepithelial neoplasia: incidence and long‐term risk of vulvar squamous cell carcinoma. Int J Cancer 2021;148:90–8. Preti M, Joura E, Vieira-Baptista P, et al. The European Society of Gynaecological Oncology (ESGO), the International Society for the Study of Vulvovaginal Disease (ISSVD), the European College for the Study of Vulval Disease (ECSVD) and the European Federation for Colposcopy (EFC) consensus statements on pre-invasive vulvar lesions. International Journal of Gynecologic Cancer 2022;32. Preti M, Igidbashian S, Costa S, et al. VIN usual type—from the past to the future. Ecancermedicalscience 2015;9:531. van der Meijden W i., Boffa M j, ter Harmsel B, et al. 2021 European guideline for the management of vulval conditions. J Eur Acad Dermatol Venereol 2022;36:952–72. Voss FO, Thuijs NB, Vermeulen RFM, et al. The vulvar cancer risk in differentiated vulvar intraepithelial neoplasia: a systematic review. Cancers 2021;13:6170. Bigby SM, Eva LJ, Fong KL, et al. The natural history of vulvar intraepithelial neoplasia, differentiated type: evidence for progression and diagnostic challenges. International Journal of Gynecological Pathology 2016;35:574–84. DermNet NZ. Vulval cancer. Available from: (Accessed Apr, 2023). Omole F, Kelsey RC, Phillips K, et al. Bartholin duct cyst and gland abscess: office management. Am Fam Physician 2019;99:760–6. Chin S, Scurry J, Bradford J, et al. Association of Topical Corticosteroids With Reduced Vulvar Squamous Cell Carcinoma Recurrence in Patients With Vulvar Lichen Sclerosus. JAMA Dermatol 2020;156:813. Lee A, Bradford J, Fischer G. Long-term management of adult vulvar lichen sclerosus: a prospective cohort study of 507 women. JAMA Dermatol 2015;151:1061. Yeon J, Oakley A, Olsson A, et al. Vulval lichen sclerosus: An Australasian management consensus. Aust J Dermatology 2021;62:292–9. Lewis FM, Tatnall FM, Velangi SS, et al. British Association of Dermatologists guidelines for the management of lichen sclerosus, 2018. British Journal of Dermatology 2018;178:839–53. DermNet NZ. Vulval intraepithelial neoplasia. Available from: (Accessed Apr, 2023). Health New Zealand, Te Whatu Ora. Clinical practice guidelines for cervical screening in Aotearoa New Zealand. 2023. Available from: (Accessed Mar, 2024). ##### Comment features You can add replies to a comment. Simply click the "Reply to comment" button and complete the form. Your reply, once approved, will appear below the comment to which you replied (if there are multiple replies to a comment, they will appear in order of submission). You can add a new comment by scrolling to the bottom of the discussion and clicking the "Add a comment" button. If someone adds a reply to one of your comments (or replies) you will receive an email notifying you of this. You can opt out of (or into if currently out) all comment notification emails by clicking the button below. Discussion There are currently no comments for this article. Please login to make a comment. bpac nz advocates for best practice in healthcare treatments and investigations across a wide range of health service delivery areas, and we are recognised nationally and internationally for our expertise and innovation. Post P.O. Box 6032 DUNEDIN 9059 Articles Mybpac About FAQ Contact Terms of use Subscribe to our emails Follow us on facebook Follow us on twitter © Copyright BPAC NZ Limited (bpac nz) 2025. All rights reserved. Made with by the bpac nz team Partner links
8434
https://dictionary.cambridge.org/us/example/english/meandering-river
meandering river meanings of meandering and river Your browser doesn't support HTML5 audio Your browser doesn't support HTML5 audio Your browser doesn't support HTML5 audio Your browser doesn't support HTML5 audio (Definition of meandering and river from the Cambridge English Dictionary © Cambridge University Press) Examples of meandering river Word of the Day Victoria sponge Your browser doesn't support HTML5 audio Your browser doesn't support HTML5 audio a soft cake made with eggs, sugar, flour, and a type of fat such as butter. It is made in two layers with jam or cream, or both, between them Blog Calm and collected (The language of staying calm in a crisis) New Words vibe coding © Cambridge University Press & Assessment 2025 © Cambridge University Press & Assessment 2025 Learn more with +Plus Learn more with +Plus {{message}} {{message}} There was a problem sending your report. {{message}} {{message}} There was a problem sending your report.
8435
https://www.openaccessjournals.com/articles/a-short-note-on-benzene-and-its-properties.pdf
241 Perspective Journal of Medicinal and Organic Chemistry J. Med. Org. Chem. (2024) 07(5), 241-242 Goran Ungar Department of Organic Chemistry, University of Patras, Patras, Greece Author for correspondence: g.ungar@shefield.ac.uk Received: 02-Jul-2024, Manuscript No. JMOC-24-140532; Editor assigned: 05-Jul-2024, PreQC No. JMOC-24-140532 (PQ); Reviewed: 19-Jul-2024, QC No. JMOC-24-140532; Revised: 01-Oct-2024, Manuscript No. JMOC-24-140532 (R); Published: 29-Oct-2024, DOI: 10.37532/ jmoc.2024.7(5).241-242 A Short Note on Benzene and its Properties Introduction Benzene, a simple yet remarkable molecule, is perhaps one of the most studied and celebrated organic compounds in chemistry. Its unique structure and properties have fascinated scientists for over a century since its discovery in the 19th century. This aromatic hydrocarbon, with the molecular formula C6H6, consists of a ring of six carbon atoms, each bonded to one hydrogen atom. The alternating double bonds between carbon atoms create a planar, hexagonal structure with delocalized π electrons, giving rise to its aromaticity and exceptional stability. Benzene is a fundamental organic compound with the molecular formula C6H6. It is a colorless and highly flammable liquid with a characteristic sweet odor. Benzene is one of the simplest aromatic hydrocarbons, consisting of a six-carbon ring with alternating double bonds. The discovery of benzene dates back to the early 19th century, when Michael Faraday isolated it in 1825 from the oily residue derived from whale oil. However, it was not until 1834 that Eilhard Mitscherlich determined its empirical formula and coined the name “benzene” in 1836. The structure of benzene posed a significant challenge to chemists until it was finally resolved by Friedrich August Kekule in 1865, who proposed the hexagonal structure with alternating single and double bonds. Description One of the most distinctive features of benzene is its aromatic nature. This term “aromatic” was originally used to describe compounds with fragrant odors, but in chemistry, it refers to compounds that possess a specific stabilization due to a conjugated ring of π-electrons. Benzene’s stability is significantly higher than what would be expected based on traditional models of double-bonded and single bonded carbon structures. This stability arises from the delocalization of its π-electrons over the entire ring, rather than being localized between specific carbon atoms. Benzene’s structure and aromaticity have profound implications for its chemical behavior and properties. It is a colorless, highly flammable liquid with a sweet odor. It has a boiling point of 80.1°C and a melting point of 5.5°C, making it volatile and easily vaporizable. Due to its aromatic stability, benzene undergoes substitution reactions rather than addition reactions typical of alkenes. This substitution pattern, known as electrophilic aromatic substitution, is a cornerstone of organic chemistry. The molecular structure of benzene is often depicted using resonance structures to illustrate the delocalization of π-electrons. These resonance structures show how the six carbon atoms and six hydrogen atoms are arranged in a planar hexagonal ring, with alternating single and double bonds between the carbon atoms. The actual structure of benzene, however, is best described as a hybrid of these resonance structures, where the electrons are not fixed between specific carbon atoms but rather delocalized around the entire ring. From a physical standpoint, benzene exhibits characteristics that reflect its aromaticity and molecular structure. It has a density of 0.879 g/cm³, which is lower than that of water and it, is insoluble in water but soluble in organic solvents like ether, alcohol and chloroform. 242 Perspective J. Med. Org. Chem. 2024 07(5) Ungar G. Benzene’s solubility in non-polar solvents and its non-polarity itself contribute to its versatility in chemical reactions and industrial applications. In terms of chemical reactivity, benzene’s stability makes it less reactive under normal conditions compared to alkenes and other unsaturated hydrocarbons. However, its aromatic ring structure makes it susceptible to electrophilic substitution reactions. These reactions involve the replacement of one of the hydrogen atoms on the benzene ring with an electrophilic species, such as a halogen or a nitro group. The electrophilic aromatic substitution mechanism proceeds through the formation of a resonance stabilized carbocation intermediate, followed by deprotonation to regenerate the aromaticity of the benzene ring. Benzene and its derivatives play crucial roles in both industrial and laboratory settings. Industrially, benzene is a precursor to a wide range of chemicals, including plastics, synthetic fibers, dyes and pharmaceuticals. It serves as a starting material in the production of styrene, which is used to make polystyrene and aniline, which is used in the manufacture of dyes. In the laboratory, benzene is used as a solvent for organic reactions and as a standard in NMR spectroscopy due to its distinctive chemical shifts in the NMR spectrum. Conclusion In conclusion, benzene is a fundamental molecule in organic chemistry, renowned for its aromatic stability, unique structure and diverse chemical reactivity. Its discovery and subsequent study have contributed significantly to our understanding of chemical bonding and aromaticity, paving the way for advancements in synthetic chemistry, materials science and pharmaceutical research. While its industrial applications are vast and essential, precautions must be taken to mitigate the risks associated with its toxicity, underscoring the dual nature of this iconic compound in science and society.
8436
https://rmccares.org/2025/05/08/shedding-light-on-the-risk-of-maternal-fetal-transmission-of-lyme-disease/
Shedding Light on the Risk of Maternal-Fetal Transmission During Lyme Disease Awareness Month Every May, Lyme Disease Awareness Month brings attention to the United States’ most common vector-borne disease (a disease spread by blood-feeding bugs like mosquitoes, ticks, and fleas). This year, we’re educating our readers on the possibility and effect of maternal-fetal transmission and what expecting mothers or women trying to get pregnant need to know about the most common disease that comes from a tick bite. Keep reading to learn more! Key Takeaways What Is Lyme Disease? While maternal-fetal transmission of Lyme disease is considered rare, over the last few decades, more research has emerged showing it does occur, especially if the mother does not receive treatment. Proving that Lyme disease, caused by the Borrelia burgdorferi bacterium and spread by tick bites, can be passed from the mother to the baby during pregnancy has been a challenge for multiple reasons. The top among them is that Lyme disease is difficult to diagnose in adults, let alone the unborn. As a disease with many common symptoms that are also symptoms of other common illnesses and diseases, Lyme disease is often misdiagnosed and underdiagnosed. The first stage of symptoms of Lyme disease includes headache, fatigue, flu-like symptoms, joint pain, and a rash that can appear in different shapes (not just the well-known bullseye rash). If the disease is not quickly diagnosed and treated, it can lead to long-term effects such as facial paralysis, neurological problems, cognitive impairment, hearing and vision problems, inflammation of the heart and brain, and more. It’s also worth noting that testing for Lyme disease can be somewhat unreliable, as test accuracy depends on how long you’ve been infected. This is because it’s an antibody test looking for signs that your immune system has been trying to fight off the infection, which can take a while (around 4–6 weeks). Additionally, it’s not easy to determine if a patient has been cured after completing their antibiotic treatment, as their blood will likely continue to test positive for Lyme bacteria antibodies for months or years after the bacteria are gone. Can Lyme Disease Cross the Placenta? In 2020, the CDC acknowledged that Lyme disease can infect the mother’s placenta during pregnancy and, in rare cases, the baby. This is known as congenital Lyme or referred to as maternal-fetal transmission or vertical transmission. The possibility of transmission increases if the mother doesn’t receive treatment with antibiotics. However, the Lyme disease community has long suspected this to be the case. A 2018 review published in PLOS One examined 45 studies and found that while it’s rare due to limited research, there is evidence of maternal-fetal transmission of Lyme disease bacteria. It goes on to suggest that adverse outcomes were more likely in cases where the disease was left untreated during the pregnancy. The review goes on to suggest that serious complications, including miscarriage or congenital abnormalities, were more likely when the mother didn’t receive immediate treatment. In other instances, doctors such as Rosalie Greenberg, M.D., Child and Adolescent Psychiatrist, have documented their experience diagnosing congenital Lyme in children of mothers with Lyme disease. She notes how, without more research for doctors to draw upon, more children will suffer as Lyme disease is underdiagnosed or misdiagnosed in children of mothers who have or had Lyme disease. Research Revelations and Shortcomings As we mentioned earlier, one of the challenges in studying congenital Lyme disease is that it is difficult to prove, as the symptoms in infants often mimic those of many other common conditions. These symptoms can range from developmental delays to neurological or cardiac abnormalities. Still, some physicians and researchers like Dr. Greenberg have been able to document cases that strongly suggest vertical transmission happens more often than is reported. Sue Faber, R.N., provides a literature review spanning 32 years, offering a comprehensive overview of repeated patterns across case studies. These include adverse pregnancy outcomes, the presence of spirochetes in fetal tissue, and infants born with Lyme-consistent symptoms, some of whom even test positive themselves. Although the CDC states that there’s no evidence of Lyme being passed through breastfeeding, the research on this aspect of transmission still appears to be inconclusive. More research is still needed to fully understand the scope and mechanics of congenital Lyme. Still, the existing evidence is enough to warrant caution and proactive care by expectant mothers and their OB-GYNs. Why Early Diagnosis Matters During Lyme Disease Awareness Month The good news is that timely antibiotic treatment is highly effective for mothers and babies. The CDC emphasizes that when Lyme disease is diagnosed early on in pregnant women, it is safe to treat with antibiotics like oral amoxicillin or cefuroxime; however, doxycycline can have adverse effects on fetuses and should be avoided during pregnancy. Early treatment dramatically decreases the risk of unborn babies contracting the disease and usually results in a full, quick recovery for mom. RMC Supports Expecting Mothers Year-Round As we recognize Lyme Disease Awareness Month, we want to ensure that all of our readers understand that RMC is here to take all of their health concerns seriously, especially those of women who are expecting. This joyful time also comes with concerns that can be overwhelming. If a tick has bitten you or you are experiencing multiple worrisome symptoms, our maternal medical care team is ready to ensure you and your baby are safe. If you’re pregnant and think you may have been exposed to a tick, don’t wait. Contact RMC today to talk with a provider about your symptoms and get the care that you need! Post navigation Post Categories RMC Anniston 400 East 10th Street Anniston, AL 36207 Stringfellow Campus of Regional Medical Center 301 East 18th Street Anniston, AL 36207 Helpful Links © 2025 The Health Care Authority of the City of Anniston. All rights reserved. Privacy Notice | Smoking and Tobacco Policy | Healthcare Civil Rights Web Development By Infomedia
8437
https://ejnmmipharmchem.springeropen.com/articles/10.1186/s41181-019-0075-2
Typesetting math: 100% Skip to main content EJNMMI Radiopharmacy and Chemistry Download PDF Review Open access Published: Auger electrons for cancer therapy – a review Anthony Ku1na1, Valerie J. Facca1na1, Zhongli Cai1 & … Raymond M. Reilly1,2,3,4 EJNMMI Radiopharmacy and Chemistry volume 4, Article number: 27 (2019) Cite this article 38k Accesses 308 Citations 3 Altmetric Metrics details Abstract Background Auger electrons (AEs) are very low energy electrons that are emitted by radionuclides that decay by electron capture (e.g. 111In, 67Ga, 99mTc, 195mPt, 125I and 123I). This energy is deposited over nanometre-micrometre distances, resulting in high linear energy transfer (LET) that is potent for causing lethal damage in cancer cells. Thus, AE-emitting radiotherapeutic agents have great potential for treatment of cancer. In this review, we describe the radiobiological properties of AEs, their radiation dosimetry, radiolabelling methods, and preclinical and clinical studies that have been performed to investigate AEs for cancer treatment. Results AEs are most lethal to cancer cells when emitted near the cell nucleus and especially when incorporated into DNA (e.g. 125I-IUdR). AEs cause DNA damage both directly and indirectly via water radiolysis. AEs can also kill targeted cancer cells by damaging the cell membrane, and kill non-targeted cells through a cross-dose or bystander effect. The radiation dosimetry of AEs considers both organ doses and cellular doses. The Medical Internal Radiation Dose (MIRD) schema may be applied. Radiolabelling methods for complexing AE-emitters to biomolecules (antibodies and peptides) and nanoparticles include radioiodination (125I and 123I) or radiometal chelation (111In, 67Ga, 99mTc). Cancer cells exposed in vitro to AE-emitting radiotherapeutic agents exhibit decreased clonogenic survival correlated at least in part with unrepaired DNA double-strand breaks (DSBs) detected by immunofluorescence for γH2AX, and chromosomal aberrations. Preclinical studies of AE-emitting radiotherapeutic agents have shown strong tumour growth inhibition in vivo in tumour xenograft mouse models. Minimal normal tissue toxicity was found due to the restricted toxicity of AEs mostly on tumour cells targeted by the radiotherapeutic agents. Clinical studies of AEs for cancer treatment have been limited but some encouraging results were obtained in early studies using 111In-DTPA-octreotide and 125I-IUdR, in which tumour remissions were achieved in several patients at administered amounts that caused low normal tissue toxicity, as well as promising improvements in the survival of glioblastoma patients with 125I-mAb 425, with minimal normal tissue toxicity. Conclusions Proof-of-principle for AE radiotherapy of cancer has been shown preclinically, and clinically in a limited number of studies. The recent introduction of many biologically-targeted therapies for cancer creates new opportunities to design novel AE-emitting agents for cancer treatment. Pierre Auger did not conceive of the application of AEs for targeted cancer treatment, but this is a tremendously exciting future that we and many other scientists in this field envision. Introduction Many radionuclides commonly used for imaging in nuclear medicine (e.g. 99mTc, 123I, 111In, 67Ga; Table 1; Fig. 1) decay by electron capture (EC) and/or internal conversion (IC). As a result of these decay processes, these high atomic number elements eject a series of low energy electrons in what is referred to as the Auger effect. Although Auger electrons originating from K-shell transitions can have energy higher than 25 keV, and up to 80 keV, their yield per decay is lower than 0.1. The majority of Auger electrons (AEs) have low energy (< 25 keV), which is deposited over short nanometre-micrometre distances in tissues. This extremely short range yields high linear energy transfer (LET), which makes AEs attractive for radiation treatment of cancer, especially if they are emitted in close proximity to cell sensitive targets such as DNA and the cell membrane. These high LET electrons were first described in independent work by Lise Meitner in 1922 (Meitner 1922) and Pierre Auger in 1923 (Auger 1923). Pierre Auger later published experiments to detect electron cascades from excited atoms in his published thesis (Auger 1926). Ionisation events were visualised in these early experiments by the use of cloud chambers, where ions produced in a dense water vapour were evidenced by the condensation of a water droplet upon liberation of an electron (Wilson 1923). Pierre Auger induced what is now known as the Auger effect in noble gases excited by incident X-rays, which resulted in a primary electron ejection event and multiple electron tracks (Auger 1975). The Auger effect describes the process in which a vacancy in an inner electron orbital (i.e. K-shell) is filled by the decay of an electron from a higher shell (i.e. L-shell) with lower binding energy. The energy difference of this transition is emitted as a characteristic X-ray (Fig. 2a), or transferred to another electron which is subsequently ejected (Fig. 2b). When the ejected electron is from the same principle energy level (i.e. L-shell), it is referred to as an Auger electron, this results in two electron vacancies within the shell, for example resulting in 2s and 2p holes. When the ejected electron is from a higher shell (i.e. M-shell), resulting in a vacancy in two different principle energy levels (i.e. holes in 2s and 3s) it is referred to as a Coster-Kronig electron, and if the resulting outer shell electron hole (i.e. 3s) is filled by an electron of the same primary shell and also ejects an electron from that same shell (i.e. holes in 3p and 3d), this is referred to as a Super Coster-Kronig electron (Howell 1992; McGuire 1975). Auger, Coster-Kronig and Super Coster-Kronig electrons are collectively referred to as Auger electrons and result from the propagation of the Auger process as a cascade of electron vacancies and ejections. The primary vacancy can be induced by an incident X-ray (as shown by Auger), or occur during the decay of unstable nuclei (radionuclides) through electron capture (EC) or internal conversion (IC) processes. EC is the process in which an inner, K-shell electron is absorbed by a proton-rich nucleus, converting the proton and absorbed electron to a neutron, maintaining the atomic mass but resulting in an excited state and a primary orbital vacancy (Intemann and Pollock 1967). IC occurs when a nucleus in an unstable excited state releases excess energy that is transferred to an inner orbital electron, overcoming its binding energy and ejecting the electron (Fig. 2c) with high kinetic energy (Choppin et al. 2002). In this review, we describe the radiobiological properties of AEs that make them attractive for cancer therapy, discuss their radiation dosimetry and describe methods for labelling biomolecules with AE-emitting radionuclides for targeted radiotherapy of cancer. We further review the preclinical studies of Auger electrons for cancer treatment and the limited but promising clinical studies that have been performed. Radiobiological properties Particles employed in systemic radiation therapy Key considerations in the selection of a radionuclide for systemic radiation therapy, i.e. radiation therapy using targeted biomolecules, are the energy of the emitted particles (e.g. α-particles, β-particles or AE or IC electrons), and the range of these particles in tissues. AEs have a very short range of energy deposition, offering the most precise irradiation for cancer treatment. These low energy electrons may travel up to several micrometres from the site of decay, but the majority of the electrons have sufficient energy to travel only nanometre distances (Reilly and Kassis 2010). AEs emitted from the lower electron shells, particularly propagating from K-shells, have the largest energies and travel the greatest distances, but these are produced in lower abundance than electrons ejected after transitions from the outer electron shells (M- through O-shell) which are ejected with reduced energies and travel only a few nanometres (Howell 1992). Due to the short range of AEs emitted in cascade, most result in a high LET between 1 and 23 keV/μm, which is potentially very potent for producing clustered damage in macromolecular targets of a cancer cell, particularly DNA and the cell membrane. This is different from much more energetic (MeV) β− particles, which deposit most of their energy along a longer (millimetre) track length, but at the track end, some of these particles yield high LET within a nanometre range causing repairable individual DNA lesions. Additionally, most AE-emitting radionuclides emit a small number of IC electrons, with higher energies and ranges up to several millimetres, providing a longer-range effect (Eckerman and Endo 2008; Howell 1992). In comparison, α-particles deposit extremely high energy (several MeV) over 5–10 cell diameters (50–100 μm), resulting in very high LET of 50–230 keV/μm. Historically, β-particles have attracted interest for treatment of cancer due to their average long range of energy deposition of 2–10 mm in tissue, which depends on the β-particle energy. This long range causes a cross-fire effect, which irradiates not only the targeted cells but additional neighbouring cells that are within the range of the β-particles. This aids in homogenisation of the radiation dose across millimetre sized solid tumours, but can also lead to greater haematological toxicity (Carr 2004; Salem et al. 2005; Stabin et al. 2001). For example, the cross-fire effect of the β-particles emitted by 177Lu or 90Y (maximum range of 1.7 or 11 mm in tissue, respectively) (Lai et al. 2017), radionuclides that have been complexed to monoclonal antibodies for radioimmunotherapy (RIT) of cancer, contributes to dose-limiting non-targeted hematopoietic toxicity (Vallabhajosula et al. 2005) through irradiation of bone marrow cells adjacent to skeletal regions of nonspecific uptake, or skeletal metastases, as well as circulating radioactivity perfusing the bone marrow. Targets for cytotoxicity from AEs Radiation induced cell death may proceed via several different mechanisms (Fig. 3). Direct DNA damage may be caused by traversal of the DNA duplex by high LET particles such as AEs or α-particles. Alternatively, indirect damage to DNA may be inflicted by reactive hydroxyl radicals (ROS) that are generated by interaction of AE, α-particles, β-particles, γ-photons or X-rays with water molecules and consequent radiolysis. The cascade of electrons and resultant local generation of free radicals may result in the formation of a concentrated region of macromolecular damage, which in the case of a DNA target can result in multiple and complex DNA breaks. Unrepaired damage can then result in cell death by several pathways, e.g. apoptosis or mitotic catastrophe (Haefliger et al. 2005; Kriehuber et al. 2004a; Urashima et al. 2006). DNA is considered the main target for causing radiation-induced cell death and indeed, the greater the unrepaired DNA damage the higher the incidence of lethality (Chan et al. 1976; Schneider and Whitmore 1963). Radiation-induced DNA double strand breaks (DSBs) are recognised by the protein kinase ataxia telangiectasia mutated (ATM), and through ATM/p53 pathways result in cell cycle arrest or apoptosis (Ismail et al. 2005). However, tumour cells of epithelial origin commonly have deficient pro-apoptotic pathways and are mainly killed through mitotic catastrophe as a consequence of attempting to enter mitosis with damaged DNA (Eriksson and Stigbrand 2010). Phenotypic features of cells entering pre-mature mitosis are aberrant chromosomes, multiple nuclei and micronuclei (Bhattathiri et al. 1998). Since DNA has historically been considered the principal target for radiation treatment of cancer (Burdak-Rothkamm and Prise 2009), the emission of AEs in or near the cell nucleus, and especially in close proximity to the DNA greatly amplifies the lethality of these electrons, particularly taking into account the subcellular range of these electrons (Faraggi et al. 1994; Fasih et al. 2012; Hoang et al. 2012). Indeed when decaying in close proximity to DNA, AEs emitted by 125I exhibited superior cell killing properties than the longer-range β-particles emitted by 131I (Chan et al. 1976). Kassis and Adelstein showed that the AEs emitted by 125I were most lethal to Chinese hamster V79 lung fibroblasts when emitted by 125I-iododeoxyuridine (125I-IUdR), a nucleoside analogue that is incorporated into DNA, than with 125I-iododihydrorhodamine which is localised in the cytoplasm (Kassis et al. 1987). Relative biological effectiveness (RBE) describes the ratio of the effectiveness of one form of radiation compared to another, with X-rays or γ-radiation often used for comparison. An RBE = 1 indicates that the radiation has equivalent biological effects (e.g. cytotoxicity) as X-rays or γ-radiation. For deterministic effects, AE have exhibited RBE in the range of 2–5 (Freudenberg et al. 2014; Kriehuber et al. 2004b; Regulla et al. 2002; Yasui et al. 2001). The high lethality of AEs emitted near the nucleus is apparent in a comparison of RBE considering whole cell or nuclear targeting. Cellular dose calculations for the AE-radionuclides 99mTc and 123I in rat thyroid PC Cl3 cells, assuming a whole cell or nuclear target for dose calculations, resulted in an increased RBE from 0.75 to 2.18, for 99mTc and from 1.87 to 3.43 for 123I (Freudenberg et al. 2014). The enhanced efficacy of AE in proximity to DNA is illustrated in one report that showed that the RBE for 125I incorporated into DNA was almost 8, and that these low energy AEs were equivalently cytotoxic as the 5.3 MeV α-particles emitted by 210Po (Rao et al. 1989). However, if AEs were emitted in the cytoplasm, the RBE was ~ 1, and similar to low LET radiation such as X-rays (Rao et al. 1990). A modelling study of dose deposition in the DNA duplex revealed higher energy deposition by the AEs emitted by 125I than 5 MeV α-particles, and that even at a distance in DNA as short as 5 base pairs from the AE emission, the energy deposition was reduced by a factor of 10 (Charlton 1986). Some insightful studies employed triplex-forming oligonucleotides or minor groove-binding ligands that hybridised to DNA labelled with the AE-emitters 125I or 111In, to illustrate the distant-dependent effects of AEs on causing DNA double-strand breaks (Karamychev et al. 2000; Lobachevsky et al. 2008; Panyutin and Neumann 1997). Panyutin and Neumann (1997) demonstrated using 125I-labelled DNA triplex forming oligonucleotides that the probability of DNA strand breaks was strongly correlated with distance from the position of radionuclide decay, with 90% of DNA breaks found within 10 base pairs (Panyutin and Neumann 1997). Free radical scavenging studies that employed dimethylsulfoxide (DMSO) to eliminate indirect DNA damage, illustrated that beyond a critical distance, the DNA DSBs induced by AEs were mediated mainly by indirect effects caused by oxygen free radicals (Balagurumoorthy et al. 2012; Lobachevsky et al. 2008; Panyutin and Neumann 1997). Detection of DNA damage Interaction of AEs or their induced reactive oxygen species (ROS) with DNA can cause complex and multiple DSB (Fig. 3). The accumulation of unrepaired DNA DSB is negatively correlated with cell survival (Cai et al. 2008) and the misrepair of these breaks from damaging AE can result in chromosomal aberrations, similarly associated with reduced cell survival (Beckmann et al. 1993; Woo et al. 1989). Assays for DSBs induced in cancer cells by AEs are thus useful for studying their cytotoxicity. Common assays used for radiation-induced DNA DSBs include γH2AX, comet assay, and pulsed-field gel electrophoresis (PFGE). In response to DSBs caused by ionizing radiation, the histone H2AX is rapidly phosphorylated (γH2AX) and accumulates at these sites of DNA damage (Mah et al. 2011). The phosphorylation of H2AX can be detected within one minute after exposure to ionizing radiation and reaches maximal levels within 10–30 min (Rogakou et al. 1998). γH2AX can be imaged by confocal immunofluorescence microscopy of the nucleus of cells and appears as discrete foci, which are a measure of radiation induced DNA DSB (Cai et al. 2009). Immunofluorescent imaging of γH2AX foci has proven useful in the study of AE-mediated DNA damage in cancer cells (Cai et al. 2008; Mah et al. 2011; Piron et al. 2014; Sedelnikova et al. 2002; Yasui et al. 2007). A direct correlation was found between the number of radionuclide decays and the number of foci observed for human glioblastoma SF-268 and fibrosarcoma HT-1080 cells treated with 125I-IUdR, demonstrating a similar response in two different cancerous cell lines (Sedelnikova et al. 2002). The single cell gel electrophoresis assay (SCGE) or comet assay, allows assessment of DNA damage at the single cell level. Cells are embedded in agarose and lysed with detergents at high concentrations of salt. Application of an electric field resolves fragmented DNA as a “tail” beyond the large, supercoiled, undamaged DNA “head” (Olive and Banath 1993). The comet assay has been useful to assess DNA damage caused by AEs (Haines et al. 2001; Hoyes et al. 2001; Olive and Banath 1993; Pedraza-López et al. 2000; Piron et al. 2014; Reske et al. 2007). For example, the comet assay was used to show increased DNA damage caused by 99mTc-hexamethylpropylenamineoxime (HMPAO) over 99mTc gentisic acid when internalised into the cytoplasm or remaining bound to the cell membrane of lymphocytes, respectively (Pedraza-López et al. 2000). It was also used to demonstrate extensive DNA damage caused by 125I-IUdR or 123I-4′-thio-2′-deoxyuridine (123I-ItdU) incorporated into DNA (Olive and Banath 1993; Reske et al. 2007). Pulsed field gel electrophoresis (PFGE) allows for improved separation of large DNA fragments beyond the capabilities of traditional gel electrophoresis by employing an alternating electric field. PFGE is useful for analysis of larger samples of DNA from a high number of cells, contrasting it from single cell gel electrophoresis. PFGE has been used to examine DSB caused by AEs from 125I-IUdR in hamster V79-379A cells (Elmroth and Stenerlöw 2005) and Chinese Hamster Ovary (CHO) cells (Iliakis et al. 1991). Cross-dose, bystander effects and cell membrane damage The cross-fire effect, most commonly attributed to β-particles, describes the irradiation of cells distant from those harbouring the radionuclide due to the long (several millimetres) range of these particles. Despite the short nanometre-micrometre range of AEs, there is a local cross-dose effect of AE-emitting radionuclides that deposits dose in tumour cells that are immediately adjacent to cells in which the radionuclide decays (Cai et al. 2010) (Fig. 3). This is mediated by the several micrometre range of some higher energy AEs and IC electrons. In addition, tumour cells with lethal DNA damage caused by radiation may release mediators that cause the death of distant non-irradiated cells through the bystander effect (Mothersill et al. 2018) (Fig. 3). The bystander effect has been observed for AEs in vitro in media transfer experiments in which growth medium from donor cells exposed to 123I-metaiodobenzylguanidine (123I-MIBG) was transferred to non-irradiated recipient cells causing decreased clonogenic survival of these cells (Boyd et al. 2006; Paillas et al. 2016). Diminished clonogenic survival and increased numbers of γH2AX foci in HCT116 colon cancer cells were observed by media transfer experiments following exposure of donor cells to 125I-labelled anti-epidermal growth factor receptor (EGFR) monoclonal antibodies (mAb) (Paillas et al. 2016). Other studies have shown greater inhibition of tumour growth in mice inoculated with a mixture of non-irradiated cells and pre-irradiated cells compared to non-irradiated cells alone, demonstrating an AE-mediated bystander effect in vivo (Xue et al. 2002). Due to the short range of most AEs, considerable attention has been focused on delivery of AE-emitting radionuclides to the nucleus or DNA (historically considered the primary cellular target of radiation damage) of tumour cells to maximise their cytotoxic effects. However, it has been shown that internalisation into cancer cells and delivery to the cell nucleus is not obligatory for cell killing, and that the lethal effects of AEs may be induced indirectly by free radical-mediated pathways (Goddu et al. 1996; Narra et al. 1995). Targeting the cell membrane has been proven to be an effective strategy for killing cancer cells with AEs (Paillas et al. 2016; Pouget et al. 2008; Santoro et al. 2009) (Fig. 3). In in vitro experiments, non-internalising 125I-anti-carcinoembryonic (CEA) mAbs bound to the surface of HCT116 colon cancer cells generated ROS that caused re-organisation of lipid rafts and activated receptor-mediated cell signalling pathways (ERK1/2, AKT, p38/JNK) and several phosphorylated protein mediators of Ca2+ levels (phospholipase C-γ and proline-rich tyrosine kinase 2 and paxillin) (Paillas et al. 2016). Cell membrane damage further induced γH2AX foci in the nucleus of donor cells exposed to 125I-anti-CEA mAbs and in recipient, non-exposed cells through a bystander effect. This study further revealed that DNA damage was quite homogeneous in CEA-positive A431 tumours in mice administered 125I-anti-CEA mAbs, despite radioactivity being localised mainly at the periphery of the tumour, suggesting a local bystander effect on non-targeted cells that could be mediated by damage to the cell membrane of targeted tumour cells (Paillas et al. 2016). 125I-labelled anti-CEA 35A7 was also found to be effective in vivo for treatment of small peritoneal tumours in mice, illustrating that internalisation and nuclear importation are not always required for the use of AEs for cancer therapy (Santoro et al. 2009). These findings are promising since they extend the targets for AE radiotherapy of cancer to non-internalising cell surface antigens overexpressed on tumour cells that are recognised by mAbs or other ligands. Dosimetric properties Organ and cellular dosimetry of AEs Radionuclides that emit AEs also release γ-rays and X-rays and IC electrons. AEs have energies from a few eV to tens of keV, and ranges in soft tissue from a few nm up to 100 μm, while IC electrons have higher energies (tens to hundreds keV) and ranges (tens of μm to several mm). In contrast, γ-rays and X-rays are penetrating and travel much farther, mostly much longer than cm distances, except that for ultra-soft X-rays, the range can be as low as a few μm (Eckerman and Endo 2008; Berger et al. 2005r ."); Hubbell and Seltzer 2004. Natl Inst Stand Technol. 2004; .")). Due to the diverse radiations and energy deposition distances and the dimension of critical targets, which range from single cells and subcellular compartments, to tumour masses and normal organs, in order to understand the whole picture of dosimetry for AE emitting radionuclides, both organ and cell doses should be considered (Bolch et al. 2009:477–84."); Goddu et al. 1997; Loevinger et al. 1988; Roeske et al. 2008; Stabin 2006:R187–202."); Thierens et al. 2001:187–93."); Vaziri et al. 2014:1557–64.")). Organ dosimetry estimates the absorbed dose at the whole organ level, to which photons are the main contributor when the target organ is different than the source organ. Electrons play the prominent role when the source and target organs are the same, i.e. radioactivity localised in a source organ deposits energy in that same organ. However, when we zoom in one million times to examine the absorbed dose at the subcellular level, we need cellular dosimetry. This is especially the case for short range radiation such as AEs. Cellular dosimetry is the study of energy deposition at the cellular level, which considers both the cellular targets and sources (e.g. cell, cell membrane, cytoplasm and nucleus) as a quantitative means of understanding the biophysical interactions of radiation with matter (Goddu et al. 1997; Roeske et al. 2008; Vaziri et al. 2014:1557–64.")). The Medical Internal Radiation Dose (MIRD) schema is the gold standard to estimate the internal absorbed dose from radiopharmaceuticals. It can be described by eq. 1: where D (rT,TD) is the mean absorbed dose to the target region (rT) over a dose-integration period (TD), from a radionuclide distributed uniformly within a source region (rS). à (rS, TD) is the time-integrated radioactivity in rS over TD, while S (rT ← rS) is the absorbed dose in rT per radioactive decay in rS. The MIRD schema is applicable to both organ and cellular dosimetry (Dewaraja et al. 2012; Goddu et al. 1997; Siegel et al. 1999; Vaziri et al. 2014). To calculate à (rS, TD), it is necessary to measure the source radioactivity at different time points to produce the time-radioactivity curve, followed by integration of this curve over TD. For clinical studies using AE emitting radionuclides such as 111In (Fisher et al. 2009; Vallis et al. 2014), 99mTc (Ocampo-Garcia et al. 2017) and 123I (Chin et al. 2014), planar imaging, SPECT/CT, non-imaging whole body radioactivity monitoring, or tissue sampling (e.g. blood and urine) may be used to assay the source radioactivity (Siegel et al. 1999). For animal studies, besides these imaging approaches used in the clinic, the source radioactivity is most commonly quantified by harvesting samples of source organs after euthanising animals at selected time points, then measuring the radioactivity in these tissue samples and the weight of the tissue sample to calculate the radioactivity per gram of tissue (Cai et al. 2016; Razumienko et al. 2016). Since organ dosimetry is based on à (rS, TD) in the entire organ, these tissue concentrations are multiplied by the organ weights to obtain the radioactivity per organ. For cellular dosimetry, cells are incubated with the radiopharmaceutical for different times, then cell fractionation is performed to separate the radioactivity bound to the cell membrane, internalised into the cytoplasm or imported into the nucleus, followed by measuring the radioactivity in each of these fractions by γ-counting (Ngo Ndjock Mbong et al. 2015). The selection of time points depends on both the biological (Tb) and physical (Tp) half-lives of the radiopharmaceuticals, which can be summarised by the effective half-life (Te): At least 3 time points (e.g. ~ 1/3, 1 and 3 × Te) should be selected for each exponential term of the curve (Siegel et al. 1999). For example, 5 time points at 0, 4–6 h; 1, 3, 6 days post-injection (p.i.) of 111In-ibritumomab tiuxetan (Zevalin) mAbs were selected for quantitative SPECT to estimate the organ doses in a study of 10 lymphoma patients (Fisher et al. 2009). A shorter time period was used for estimating the doses from 111In-DTPA-human epidermal growth factor (111In-DTPA-hEGF) in 15 patients with EGFR positive breast cancer using whole body planar images acquired at 1, 4–6, 24 and 72 h p.i., since radiopeptides are eliminated more rapidly than mAbs (Vallis et al. 2014). Whole-body images were obtained at 20 min, 2, 6, 24 h post injection of 99mTc-EDDA/HYNIC-Tyr (3)-octreotide to estimate the radiation doses in 4 healthy individuals (Ocampo-Garcia et al. 2017). The choice of collimators and scan time, as well as attenuation and scatter correction used in image reconstruction all have effects on the accuracy of image quantification (Dewaraja et al. 2012; Siegel et al. 1999). Sub-organ regional dosimetry for AE-emitting radionuclides is desirable for critical organs such as the kidneys and liver, as well as the tumour due to the spatial non-uniformities of radioactivity distribution (Dewaraja et al. 2012). This is particularly true for the kidneys, since it has been found that the sub-organ distribution greatly influences the nephrotoxicity of radiopeptides, due to differences in the range of the β-particles emitted by 90Y or 177Lu and the AEs emitted by 111In and the distance between radioactivity localised in the renal tubules and the more radiosensitive glomeruli (Konijnenberg et al. 2004). For patients, it might be challenging to acquire SPECT/CT with sufficient resolution to accurately quantify radioactivity at different regions in critical normal organs or a tumour. The spatial resolution of clinical SPECT is 5–10 mm (Dewaraja et al. 2012). Recently resolutions of 2.5 and 1.6 mm were reported in images of a phantom using a multi-pinhole clinical (Chen et al. 2018) or preclinical (Massari et al. 2019) SPECT system, respectively. However, for animal studies, it is possible to perform autoradiography in combination with immunohistochemical staining of excised tissues ex vivo to understand the heterogeneous radioactivity distribution in different regions of tumours and normal organs (Lee et al. 2010). In one study, to understand the effect of heterogeneous radioactivity distribution in the tumour on the treatment response, tumour cell spheroids were cultured and incubated with 111In-DTPA-hEGF or 111In-DTPA-trastuzumab, then fixed, frozen, and cryo-sectioned (Falzone et al. 2018, 2019). One section was used for micro-autoradiography to visualise the radioactivity distribution and the adjacent section was stained for γH2AX to assess DNA damage. à (rS, TD) can be calculated from the area under the curve (AUC) via numerical methods such as the trapezoidal rule, or analytical methods by fitting the time-radioactivity curve to a sum of exponentials, followed by integration from time 0 to TD. If the radioactivity of rS cannot be directly measured, its time-radioactivity curve may be derived via compartmental modelling of the measured data of other regions with which rS is physiologically interacted (Siegel et al. 1999). Recently a robust biodistribution reporting and publication standard were proposed (Kesner and Bodei 2018) and an International Atomic Energy Agency (IAEA) Radiotracer Biodistribution Template (IAEA RaBiT) allowing detailed and standardised input of biodistribution data (Kesner et al. 2017) for dosimetry estimates is free to download (IAEA Radiotracer Biodistribution Template (RaBiT) n.d./ .")). Figure 4 outlines the steps of acquiring à in vivo and in vitro for dosimetry estimates. S-values depend on the physical properties of the radionuclides, and the size and geometry of source and target regions. They can be calculated either via Monte Carlo simulation or analytical methods in the case of cellular dosimetry. Figure 5 shows representative models of reference man (A), the kidney (B), or cells (C) used in Monte Carlo Simulation of S-values (Stabin and Siegel 2018; Bouchet et al. 2003; Cai et al. 2017). Monte Carlo code used to calculate S-values includes MCNP6 (Di Maria et al. 2018), PENELOPE (Falzone et al. 2017; López-Coello et al. 2019), Geant4 (López-Coello et al. 2019) and GATE (López-Coello et al. 2019). They all allow event-by-event simulation of electron and photon transport. Though Monte Carlo code is mostly used in condensed history mode for organ dosimetry, event-by-event simulation is more accurate for cellular dosimetry and is required for electrons with energies lower than 1 keV (Tajik-Mansoury et al. 2017). Monte Carlo simulation is more accurate but also more time consuming and computationally demanding. However, S-values for organs of several human phantoms, mice and rat tissues, cells and cell compartments are available in MIRD Pamphlets (Bolch et al. 1999; Bouchet et al. 1999, 2003; Goddu et al. 1997; Vaziri et al. 2014), ICRP reports (Mattsson et al. 2015) and numerous other publications. IDAC-Dose2.1 (Abuqbeitah et al. 2018; Andersson et al. 2017), an internal dosimetry computer program for diagnostic nuclear medicine based on the ICRP adult reference voxel phantoms (both male and female), is free software (IDAC-Dose2.1 n.d./ .")) that allows calculation of the absorbed doses based on measured Ã. When human dosimetry is extrapolated from animal experiments, the radioactivity in normal organs of a human could be estimated from that in the respective organs of animals using the %kg/g method (Kirschner et al. 1975): (%ID/organ)human = (%ID/organ)animal × (animal body weight/animal organ weight) × (human organ weight/human body weight). Another free software package, MIRDcell V2.1 (MIRDcell, a Multicellular Dosimetry Tool [n.d.. Available at: .")), a Multicellular Dosimetry tool, calculates cellular doses where the source is defined as the whole cell, cell surface, cytoplasm and nucleus and the target as the whole cell, nucleus or cytoplasm with cell and nucleus radius, as well as distance between cells adjustable in 1 μm increments. The cell and nucleus shape can only be defined as concentric spheres. The multicellular geometry can be either single cell, 1-D (dimensional) cell pair, 2-D colony or a 3-D cluster with sphere, rod, cone or ellipsoid shape. Cell labelling can be selected as uniform, log-normal or normal distribution (Vaziri et al. 2014:1557–64.")). S values calculated with MIRDcell correlated well with those by Monte Carlo simulation, however, the relative error can be as high as 68%, depending on the source and target sizes as well as particle ranges and simulation codes. Comparable scoring size and particle range leads to the biggest differences between different Monte Carlo simulations (Tajik-Mansoury et al. 2017:N90–N106.")). Olinda/EXM (Stabin et al. 2005:1023–7.")) is more widely used than IDAC-Dose2.1, but is not free software. Olinda/EXM 1.0 includes 10 phantom models; adult male, adult female, 1, 5, 10 and 15-year old, newborn, 3-, 6-, 9-month pregnant woman as well as models of the prostate gland, peritoneal cavity, spheres (used for tumours), head and kidneys. The anthropomorphic phantoms define the body and its organs with geometrical shape. Olinda/EXM 2.0 replaces the above phantoms with realistic, non-uniform rational B-splines (NURBS) voxel-based models and uses updated decay data. It also includes phantoms for the mouse, rat and dog. IDAC-Dose2.1, MIRDcell V2.1 and Olinda/EXM all use radiation spectra obtained via the (MIRD) RADTABS program. Nuclear Decay Data in the MIRD Format () also produces continuingly evaluated radiation spectra of radionuclides including Auger electron emitters. However, it should be noted that the (MIRD) RADTABS program provides detailed Auger and Coster-Kronig electron spectra while the National Nuclear Data Centre only lists average energies and yields of Auger-K and Auger-L electrons. Recently it was reported that the yield of Auger, CK electrons from (MIRD) RADTABS was consistently higher than from the newly developed stochastic atomic relaxation model BrIccEmis (Falzone et al. 2017:2239.")). The difference in spectra had a significant impact on the S-values only at the nanoscale. Due to the subcellular and cellular range of AEs, for a single cell, SN ← N > > SN ← Cy > SN ← CS. For a monolayer of cells or a cell cluster, this trend remains the same, but the difference becomes smaller due to the contribution of cross-dose (Cai et al. 2010). Although nuclear DNA is the most sensitive target for radiation, recent studies show that radiation damage to mitochondria and cell membrane leads to apoptosis or cell death (Bavelaar et al. 2018; Kirkby and Ghasroddashti 2015; Pouget et al. 2018). A Monte Carlo mitochondrial dosimetry study on the β- and γ-emitter, 131I showed 2 order higher S-values for the mitochondrial source to mitochondrial target compartment compared to the nucleus to nucleus (Carrillo-Cázares and Torres-García 2012). It would be interesting to calculate S-values for the cell membrane source to cell membrane target and mitochondrial source to mitochondrial target for AE-emitting radionuclides, for correlation with the cytotoxic effects. The dosimetry of AEs is important for understanding their effects on radiation sensitive targets in cells, and for predicting their effectiveness for cancer treatment as well as any potential toxicities to critical organs (e.g. red marrow, kidneys, liver and spleen). Preclinically, cellular dosimetry may be estimated based on subcellular distribution of radioactivity measured by cell fractionation studies, and organ dosimetry may be estimated in mouse tumour xenograft models by organ biodistribution studies which can be combined with ex vivo cell fractionation to measure subcellular distribution (Costantini et al. 2007). Clinically, however, there are challenges in estimating subcellular doses from AEs, since information is not available on the subcellular radioactivity distribution. Nonetheless, tumour and normal organ dosimetry is feasible. It should nevertheless be realised that there are limitations to tumour dosimetry in predicting the therapeutic response from targeted radiotherapeutic agents in patients, due to heterogeneous tumour uptake, inaccuracies in dose estimates based on the imaging data, and radiobiological factors including the sensitivity of tumours to exponentially decreasing low dose rate radiation emitted by AEs and other forms of radiation used (in contrast to external radiation treatment which is much higher dose rate radiation). Organ dosimetry has nonetheless been found to be quite accurate in predicting normal tissue toxicity to the bone marrow and the kidneys in patients receiving targeted radiotherapeutic agents (Chalkia et al. 2015; Sarnelli et al. 2017; Strigari et al. 2014). Absorbed doses are only the first step to understand the dose-effect relationships. The microenvironments and radiosensitivity of the tumour and normal tissues, radiation chemistry reactions and radiobiological effects such as the bystander effect induced by physical dose absorption all contribute to the outcomes (Pouget et al. 2018; Ma et al. 2019; Dong et al. 2019). Radiolabelling approaches To achieve selective delivery of AE-emitting radionuclides to tumours for cancer treatment, these radionuclides must be attached to targeting ligands such as monoclonal antibodies (mAbs), or peptides that recognise cell surface receptors displayed on cancer cells (Fig. 6). In some cases, nanoparticles that incorporate polymers that complex AE-emitters and are modified on their surface with targeting ligands have also been used (Cai et al. 2016; Fonge et al. 2009; Hoang et al. 2012, 2013). Radiolabelling of mAbs and peptides with AE-emitters may be performed by radioiodination (125I or 123I), or by modification with bifunctional metal chelators (e.g. DTPA) that complex radiometals (111In, 67Ga, 64Cu; Table 1). Radioiodination requires oxidation of 125I or 123I from a valence state of − 1 as sodium iodide to + 1 for electrophilic substitution onto tyrosine amino acids in mAbs or peptides (Reilly 2007). Iodogen® (1,3,4,6-tetrachloro-3α,6α-diphenyl-glycoluril; Fraker and Speck Jr 1978) is the most commonly used oxidising reagent. However, since radioiodination efficiency is typically low (40–60%), radioiodinated mAbs and peptides must be purified post-labelling to remove free 125I or 123I. This does not allow kit formulation for preparing the radiopharmaceutical. Another disadvantage is that radioiodinated proteins are deiodinated in vivo by deiodinases, which are ubiquitous in the body and function to deiodinate thyroid hormones (Reilly 2007). In contrast, mAbs and peptides labelled with radiometals via chelation are relatively stable in vivo, and complexation of radiometals by chelators is very efficient (> 90–95%) enabling kit formulation. We previously reported the design of a kit for labelling DTPA-hEGF with 111In for AE radiotherapy of EGFR-positive breast cancer (Reilly et al. 2004). The most commonly used chelators for complexing AE-emitting radiometals are DTPA (diethylenetriaminepentaacetic acid), DOTA (1,4,7,10-tetraazacyclododecane-1,4,7,10-tetraacetic acid) and NOTA (1,4,7-triazacyclononane-1,4,7-triacetic acid). DOTA and NOTA are most suitable for labelling mAbs and peptides with 67Ga or 64Cu, but DOTA may also be used for 111In-labelling (Reilly 2007). DOTA and NOTA are macrocyclic chelators that provide a more stable radiometal complex than DTPA, but this causes slower labelling kinetics which requires mild heating of the DOTA-conjugated mAbs to 42 °C and longer incubation times (Razumienko et al. 2016). Peptides are more heat-stable than mAbs and thus DOTA-peptide conjugates may be heated up to 95 °C to achieve rapid and efficient complexation of radiometals (Brom et al. 2012). Several bifunctional chelators have been synthesised with a chemically-reactive group positioned on a side-chain for conjugation to mAbs to preserve the most stable octadentate radiometal-chelator complex. These include p-isothiocyanatobenzyl DTPA (p-SCN-Bn-DTPA), p-SCN-Bn-DOTA, and p-SCN-Bn-NOTA (Macrocyclics, Inc., Plano, TX, USA). The p-isothiocyanate group forms a thiourea linkage with ε-amino groups on lysines on the mAbs. Peptides may be assembled by solid-phase peptide synthesis which allows a chelator to be positioned at a specific location in the sequence to complex radiometals (Lever et al. 2019). An important consideration for cancer treatment with AEs is the specific activity (SA) of the radiopharmaceuticals, since this defines the amount of radioactivity delivered to a single cancer cell per receptor-recognition event. At the SA usually achieved for labelling mAbs with 111In (< 1 MBq/μg), there may be as few as 1 in 50 antibody molecules that are radiolabelled, which limits their effectiveness for killing cancer cells, especially cells that have low-moderate target receptor expression (Ngo Ndjock Mbong et al. 2015). A strategy which we have explored to increase the SA of trastuzumab labelled with 111In is conjugation of the antibodies to G4 polyamidoamine (PAMAM) dendrimers derivatised with multiple DTPA (Chan et al. 2013). This approach increased the SA by 100-fold compared to conjugation directly with DTPA and increased the cytotoxicity in vitro of these radioimmunoconjugates (RICs) on HER2-positive breast cancer cells by up to 9-fold. Another approach is conjugation of the mAbs with metal-chelating polymers (MCPs) that present multiple DOTA chelators for 111In (Aghevlian et al. 2018; Ngo Ndjock Mbong et al. 2015). Conjugation of trastuzumab to MCPs presenting 24–29 DTPA chelators increased the SA of trastuzumab labelled with 111In by 90-fold compared to trastuzumab conjugated to DTPA and increased the cytotoxicity of these RICs in vitro on HER2-positive breast cancer (BC) cells, including cells with low-moderate HER2 expression (Ngo Ndjock Mbong et al. 2015). Panitumumab modified with an MCP that presented 13 DOTA chelators increased the SA for labelling with 111In to > 70 MBq/μg (Aghevlian et al. 2018). Nanoparticles may be labelled with AE emitters by incorporation of polymers that present DTPA or DOTA for complexing 111In. For example, 111In-labelled block copolymer micelles (BCMs) modified with EGF to target EGFR-positive breast cancer cells were assembled by incorporating a DTPA-polyethylene glycol (PEG)-b-polycaprolactone (PCL) block polymer into these micelles (Fonge et al. 2009). 111In-labelled BCMs modified with trastuzumab to target HER2-positive breast cancer cells were similarly constructed (Hoang et al. 2012). We labelled gold nanoparticles (AuNPs) modified with trastuzumab by attaching 111In-DTPA-PEG polymers to the surface of the AuNPs via a gold-thiol linkage (Cai et al. 2016). Preclinical studies DNA-binding Radiotherapeutic agents Since AEs are most damaging to DNA and lethal when emitted in close proximity to DNA, the pioneering work on AEs for cancer treatment was performed by Kassis and Adelstein using 125I- or 123I-5-iodo-2-deoxyuridine (IUdR) (Kassis et al. 1987; Makrigiorgos et al. 1989). 125I/123I-IUdR is a nucleoside analogue that is incorporated directly into DNA. Emission of AEs causes DNA DSBs resulting in chromosomal aberrations in DNA, causing profound cytotoxicity (Chan et al. 1976; Kassis et al., 1987). Chan et al. compared the clonogenic survival of V79 Chinese hamster (CHO) cells exposed in vitro to 125I-IUdR with 3H-deoxyuridine and 131I-iododeoxyuridine (3H-TdR; 131I-IUdR). 125I-IUdR yielded a steeply declining clonal survival curve with a small shoulder, typical of high-LET radiation, and 125I-IUdR was more cytotoxic than 3H-TdR or 131I-IUdR, which emit low energy or intermediate energy β-particles, respectively. Incubation of the cells for one hour with only 0.0037 Bq of 125I-IUdR generated 3 DNA DSBs per cell, but a 10-fold greater amount of 131I-IUdR and 17-fold higher amount of 3H-TdR were required for the same DNA damaging effect. The radiotoxicity of DNA-incorporating 125I-IUdR has also been shown in preclinical in vivo studies. Sahu et al. demonstrated that 125I-IUdR treatment prevented paralysis caused by leptomeningeal gliosarcoma metastases in rats (Sahu et al. 1997). A single intrathecal administration of 18.5 MBq of 125IUdR, given over 5 daily fractions or by continuous 5-day infusion prolonged time-to-paralysis to 11, 12 and 15 days respectively, from 9 days for control saline-treated mice. These early demonstrations of DNA-targeted AE radiotherapy in mammalian cell lines and aggressive preclinical animal tumour models illustrate the high cytotoxicity of AEs when emitted close to DNA. More recently, a novel chemoradiotherapy strategy has been explored using AE-emitters. The chemotherapy drug cisplatin [cis-diaminedichloroplatinum (II)] forms DNA-platinum adducts in cells. This DNA crosslinking agent has been combined with radiotherapy, termed chemoradiotherapy, in the treatment of several solid cancers (Shrivastava et al. 2018; Psyrri et al. 2004; Zatloukal et al. 2004). Cisplatin and ionising radiation show synergism in their cell killing effects (Gorodetsky et al. 1998). Thus, combining cisplatin with radiotherapy using radionuclides of platinum that emit AEs (e.g. 195mPt and 191Pt) is a rational approach. 195mPt decay results in a larger number of AEs and a high energy per decay compared to other commonly used AE emitters such as 125I and 111In (Table 1). 195mPt produced from irradiation of 197Au in a high current linear accelerator [197Au(γ,np)195mPt] was extracted and used to synthesise 195mPt-cisplatin with a high specific activity of up to 3.7 TBq/mg by Bodnar and colleagues (Dykiy et al. 2007; Bodnar et al. 2015). Treatment of Ehrlich adenocarcinoma cells in vitro with 0.017 pg/mL 195mPt-cisplatin reduced cell viability to 3% within 6 h, an approximate 8-fold reduction compared to 7.5 μg/mL treatment with nonradioactive cisplatin, which decreased cell viability to only 25% (Bodnar et al. 2015). In vivo, male mice with subcutaneous Ehrlich adenocarcinoma xenografts in the right thigh received 5 intraperitoneal (i.p.) injections of normal saline (control) or 0.7 mg nonradioactive cisplatin on alternating days, or received a single injection of 0.017 pg 195mPt (Bodnar et al. 2015). Mice were monitored for body weight and tumour size for 21 days. The single treatment with 195mPt-cisplatin resulted in a 65% tumour growth inhibition, compared to 35% tumour growth inhibition with 5 treatments of nonradioactive cisplatin. While both nonradioactive and 195mPt-cisplatin treatments resulted in significant decreases in body weight compared to the normal saline treated control mice, there were no significant differences in the body weights between radioactive and nonradioactive cisplatin treatments. These promising results suggest that combining an AE emitter with cisplatin chemotherapy greatly enhances the therapeutic effect of cisplatin on tumours while not increasing normal tissue toxicity. Another platinum isotope, 191Pt has been investigated for chemoradiotherapy. 191Pt decays with a half-life of 2.8 days by EC, producing 17.8 KeV of AEs per decay (and 273 KeV/decay in γ-photons) (Table 1, Eckerman and Endo 2008). Areberg et al. produced 191Pt by proton irradiation (75–65 MeV) of gold foil [197Au(p,2p,5n)191Pt, 197Au(p,p6n)191Au➔191Pt, 197Au(p,7n)191Hg➔191Au➔191Pt] and synthesised 191Pt-cisplatin (Areberg et al. 1999). The in vitro cytotoxicity of 191Pt-cisplatin was assessed at increasing specific activities of 0–167 MBq/mg (0–20 μg/mL) on ME180 human cervical carcinoma cells to determine the IC50 values (Areberg et al. 2000) which is the concentration required to inhibit cell growth by 50%. The IC50 was reduced from 3.2 μg/mL for nonradioactive cisplatin, to 2.8 and 0.8 μg/mL for the lowest (48 MBq/mg) and highest (167 MBq/mg) specific activities, respectively. This resulted in enhancement ratios of 1.2 to 4.3 for 191Pt-cisplatin compared to nonradioactive cisplatin. The enhanced potency of 191Pt-cisplatin was subsequently shown in tumour-bearing BALB/c nude mice (Areberg et al. 2001). A cisplatin-sensitive patient-derived squamous cell carcinoma of the nasal cavity was subcutaneously inoculated into male and female mice. Treatment groups consisted of mice receiving i.p. normal saline, 5 mg/kg nonradioactive cisplatin administered i.p., or a single i.p. injection of 80 MBq/mg or 160 MBq/mg of 191Pt-cisplatin. Specific tumour growth delay (SGD, the normalised difference in tumour doubling time compared to normal saline treated control mice) was 2.1 for nonradioactive cisplatin, 3.0 for 80 MBq/mg 191Pt-cisplatin, and 3.9 for 160 MBq/mg 191Pt-cisplatin. It was calculated that for the same tumour growth delay effect obtained with a 5 mg/kg dose of 80 or 160 MBq/mg 191Pt-cisplatin, the dose of nonradioactive cisplatin would need to be 9 and 10 mg/kg, respectively. Other DNA-binding radiotherapeutic agents that emit AEs have been studied. Anthracyclines (e.g. doxorubicin and daunorubicin) are planar aromatic molecules that intercalate between base-pairs in DNA. Radiolabelled anthracyclines could cause DNA damage by AE emission in addition to the chemotoxicity of these drugs. Ickenstein et al. demonstrated in vitro the potent radiotoxicity of radioiodinated daunorubicin (Ickenstein et al. 2006). Exposure of SK-BR-3 human breast cancer cells to 125I-labelled daunorubicin (50 kBq/mL; 0.5 ng/mL) was more than 4-orders of magnitude more cytotoxic than unlabelled daunorubicin or doxorubicin, or stable 127I-iodinated daunorubicin, demonstrating the increased potency imparted on these anthracycline drugs by labelling with 125I. Imstepf et al. studied a doxorubicin derivative conjugated to a dipicolylamine chelator for radiolabelling with 99mTc tricarbonyl complex (Imstepf et al. 2015). The 99mTc-labelled doxorubicin produced a strong cytotoxic effect on radiosensitive murine melanoma B16F1 cells. After 36 h of incubation at the maximum concentration of 99mTc studied (10 MBq/mL), about 80% of B16F1 cells were killed, but only 30% and 25% of A431 human squamous carcinoma cells and HeLa cells were killed, respectively. The reduced cytotoxicity of 99mTc-doxorubicin in these cells may reflect differences in radiosensitivity to AEs. Nonetheless, incubation with 99mTcO4− which was not accumulated in the nucleus and did not bind to DNA, only killed about 10% of the cells for all three cell types. 99mTc is not ideal as an AE-emitter for cancer radiotherapy because it emits 4.4 AEs per decay but there is a low total AE energy released per decay (0.9 keV) and a low average energy per AE (0.2 keV) (Table 1). However, there is a higher energy IC electron with average energy of 13.8 keV and total energy released per decay of 15.2 keV. Acridine and acridone derivatives are an additional class of DNA intercalators that have been studied as platforms for delivery of AE-emitting radionuclides. Desbois et al. screened a panel of acridine and acridone derivatives for cytotoxicity in several human and murine melanoma cell lines and identified 7-iodo-acridone, and 5-iodo-acridine as two compounds that demonstrated cytotoxicity in vitro on M4Beu human melanoma cells (Desbois et al. 2008). The biodistribution of these compounds was studied in mice bearing B16F0 murine melanomas. Both 125I-labelled compounds accumulated in B16F0 tumours at 72 h post-injection, but 7-125I-iodo-acridone exhibited 2-fold greater uptake (17.5 vs. 6.9%ID/g). Gardette et al. investigated 125I-labelled acridine (125I-ICF01040) and acridone (125I-ICF01035) derivatives in both melanotic and amelanotic melanoma cells (Gardette et al. 2014). 125I-ICF01035 was concentrated mainly in melanosomes of B16F0 murine melanoma cells, but in A375 amelanotic human melanoma cells, 125I-ICF01035 was taken up into the nucleus. Despite these differences in subcellular distribution, 125I-ICF01035 was similarly cytotoxic to both cell types (50% growth inhibitory activity, A50 = 10–12 kBq/mL). 125I-ICF01040 which mainly accumulated in the cytoplasm, had significantly greater cytotoxicity in all cell lines and was the most cytotoxic in B16F0 cells (A50 = 2 kBq/mL). This study demonstrated that although nuclear localisation resulted in the greatest cytotoxicity, cytoplasmic distribution also resulted in killing of melanoma cells in vitro. Targeted Radiotherapeutic agents The epidermal growth factor receptor (EGFR) is the first member the EGFR family which also includes HER2, HER3 and HER4 (Olayioye et al. 2000). EGFR overexpression is a hallmark of many epithelial-derived cancers (Salomon et al. 1995). Michel et al. reported killing of EGFR-positive vulvar squamous cell carcinoma A431 cells in vitro by 111In- or 125I-labelled anti-EGFR antibodies (Michel et al. 2004). At the highest concentration studied (1.5 MBq/mL), 93–100% of these cells were killed. 131I-anti-EGFR antibodies also killed A431 cells, but these effects were not targeted, and mostly resulted from β-particle emissions in the growth medium (i.e. cross-fire effect). As discussed earlier, nuclear importation is an important property that maximises the lethal DNA damaging effects of AEs. Our group modified 111In-labelled anti-EGFR mAb nimotuzumab with 13-mer peptides CGYGPKKKRKVGG that harbour the SV-40 large T-antigen-derived nuclear localisation sequence (NLS; italicised) and studied the cytotoxicity of these radioimmunoconjugates towards EGFR-positive breast cancer cells (Fasih et al. 2012). The NLS peptides are recognised by importins α/β that function to shuttle cytoplasmic proteins across the nuclear pore complex (Costantini et al. 2008b). NLS peptide modification enhanced nuclear uptake 2-fold in EGFR-positive MDA-MB-468 human breast cancer cells compared to 111In-nimotuzumab without NLS. However, 111In-nimotuzumab without NLS still accumulated in the nucleus, probably mediated by an endogenous NLS in the transmembrane domain of the EGFR (Wang and Hung 2012). Nonetheless, the incorporation of NLS rendered 111In-NLS-nimotuzumab 7-fold more potent in reducing the CS of MDA-MB-468 cells and caused 2-fold more DNA DSBs detected by immunofluorescence for γH2AX foci in the nucleus (Fasih et al. 2012). 111In-DTPA-hEGF is an AE-emitting radiotherapeutic agent for EGFR-positive breast cancer previously studied by our group (Reilly et al. 2000).111In-DTPA-hEGF exhibited rapid receptor-mediated binding and internalisation into MDA-MB-468 cells and importation into the cell nucleus within 24 h, likely mediated by the endogenous NLS in the EGFR. MDA-MB-468 cells exposed to 111In-DTPA-hEGF exhibited decreased CS which was directly correlated with increased density of DNA DSBs in the cell nucleus assessed by immunofluorescence for γH2AX (Cai et al. 2008). 111In-DTPA-hEGF administered i.v. in fractionated doses over 5 weeks (total 28–92 MBq; 5–17 μg) significantly inhibited the growth of MDA-MB-468 tumours in mice, and was particularly effective for treatment of small tumour xenografts (Chen et al. 2003). Some decreases in white blood cell (WBC) and platelet counts were observed, but these remained within the normal range. There was no hepatic or renal toxicity. Subsequent preclinical translational bridge studies revealed that administered amounts of 111In-DTPA-hEGF up to 44 MBq (3–30 μg) in mice or 85.1 MBq (58 μg) in rabbits, corresponding to 44-times and 1-times the maximum planned human administered amount of radioactivity for a Phase 1 clinical trial were safe (Reilly et al. 2004). The projected whole-body effective dose in humans was 0.19 mSv/MBq and the projected doses to the liver and kidneys were within safe limits for the Phase 1 trial (see Clinical Studies section). Despite these encouraging results, bioactive peptides such as hEGF may cause adverse effects in humans when used for radiotherapy. In the subsequent Phase 1 trial of 111In-DTPA-hEGF, patients experienced flushing, hypotension and nausea, which was attributed to the hEGF moiety (see Clinical Studies section) (Vallis et al. 2014). We studied an 111In-labelled truncated form of hEGF (111In-EGFt) that exhibits decreased biological activity (Panosa et al. 2015). 111In-DTPA-EGFt bound with lower affinity to the EGFR than 111In-DTPA-hEGF (Kd = 6.0 × 10− 8 M vs. 1.3 × 10− 9 M, respectively) but nevertheless was internalised and imported into the nucleus of MDA-MB-468 cells. However, 111In-DTPA-EGFt was 2–8 fold less cytotoxic to MDA-MB-468 cells in vitro than 111In-DTPA-hEGF and exhibited 2.2-fold lower tumour uptake in vivo in mice with MDA-MB-468 xenografts. HER2 is the second member of the EGFR family. HER2 is overexpressed on 15–25% of breast cancers and is the target for treatment with trastuzumab (Herceptin; Roche), pertuzumab (Perjeta; Roche) and trastuzumab-emtansine (Kadcycla; Roche) (Larionov 2018). We modified trastuzumab with NLS peptides and derivatised these immunoconjugates with DTPA to complex 111In for AE RIT of HER2-positive breast cancer (Costantini et al. 2007) (Fig. 7). 111In-DTPA-NLS-trastuzumab was bound and internalised by HER2-positive breast cancer cells and transported to the nucleus. Exposure of SK-BR-3 human BC cells in vitro to 111In-DTPA-NLS-trastuzumab (70 μg/mL; 240 MBq/mg) caused extensive DNA DSBs, reducing their clonogenic survival by > 90% (Costantini et al. 2007). In contrast, unlabelled trastuzumab (70 μg/mL) decreased the survival of SK-BR-3 cells by only 35%. 111In-DTPA-NLS-trastuzumab exhibited high tumour uptake (12.1%ID/g) at 72 h p.i. in mice with HER2-overexpressing MDA-MB-361 human breast cancer xenografts (Costantini et al. 2007). For RIT studies, an administered radioactivity amount level of 111In-DTPA-NLS-trastuzumab that caused no observable adverse effects (NOAEL) on hematopoietic, liver or kidney function and on body weight was first determined (Costantini et al. 2010). A single injection at the NOAEL (9.25 MBq; 4 mg/kg) administered to mice with subcutaneous MDA-MB-361 xenografts significantly inhibited tumour growth by 4-fold (Costantini et al. 2010) (Fig. 8a). An equivalent single mass dose (4 mg/kg) of unlabelled trastuzumab was 3-fold less effective than 111In-DTPA-NLS-trastuzumab. Treatment of mice with MDA-MB-361 tumours with two injections of 111In-DTPA-NLS-trastuzumab (9.25 MBq; 4 mg/kg each) separated by two weeks significantly prolonged survival compared to mice treated with unlabelled trastuzumab or untreated mice (Fig. 8b). Our group extended this approach to RIT of acute myeloid leukemia (AML) using 111In-labelled anti-CD33 murine M195 and humanised HuM195 mAbs (Chen et al. 2006) similarly modified with NLS peptides to promote nuclear importation. 111In-NLS-HuM195 exhibited 2-fold greater potency than 111In-HuM195 without NLS for killing HL-60 human myeloid leukemia cells in vitro, and also decreased the clonogenic survival of CD33-positive primary patient AML specimens (Chen et al. 2006). Moreover, 111In-NLS-HuM195 retained its cytotoxicity in vitro against HL-60-MX-1 cells, which are a mitoxantrone-resistant subclone of HL-60 cells as well as on primary AML specimens that expressed multidrug resistance (MDR) transporters (Kersemans et al. 2008). These results suggest that RIT with AEs could be a promising approach to treatment of AML and may be able to overcome chemotherapy resistance, which is a challenge in treating this disease. Our subsequent studies focused on anti-CD123 (interleukin-3 receptor, IL-3, α-subchain) murine 7G3 or humanised CSL360 mAbs (CSL Ltd., Parkville, Australia) modified with NLS and labelled with 111In (Leyton et al. 2011). Expression of CD123 in the absence of CD131 (IL-3R β-subchain) is a phenotype that is expressed by leukemic stem cells (LSCs). LSCs are believed to cause AML and their survival after treatment is implicated in recurrence (Jordan et al. 2000). 111In-NLS-7G3 and 111In-NLS-CSL360 killed AML-5 human myeloid leukemia cells expressing the CD123+/CD131− phenotype of LSCs (Gao et al. 2016; Leyton et al. 2011). Combining 111In-NLS-CSL360 with inhibitors of DNA repair increased their cytotoxicity in vitro (Zereshkian et al. 2014). These RICs localised in the bone marrow (BM) and spleen and at other extramedullary sites of leukemia in NOD/SCID mice engrafted with AML-5 cells or primary AML specimens, detected by microSPECT/CT (Leyton et al. 2011, 2014). However, we encountered two challenges in studying RIT of AML with 111In-NLS-CSL360 preclinically in AML-engrafted mice. The mouse model which was used to identify LSCs and is also used to study new treatments for leukemia is the NOD/SCID mouse engrafted into the BM with primary AML cells (Bonnet et al. 1999). However, NOD/SCID mice harbour a germ-line mutation in DNA repair, which renders these mice unusually sensitive to radiation, including to RIT, and limits the amount of radioactivity which may be safely administered. Moreover, NOD/SCID mice require pre-treatment with 200 cGy of X-radiation to enable AML engraftment. We found that paradoxically, RIT with 111In-NLS-CSL360 encouraged AML engraftment rather than decreased engraftment, due to a boost in radiation to the BM caused by the γ-photons from 111In which further primed the BM niche for leukemic cell engraftment (Bergstrom et al. 2016). NRG (NOD-Rag1null IL2rgnull, NOD rag gamma) mice that do not harbour this DNA repair defect may be used to study RIT, but these mice are more immunocompromised than NOD/SCID mice and AML engraftment efficiency is often very high, resulting in a high tumour burden that is difficult to eradicate. Further refinements to the mouse model of AML are required to study AE-mediated RIT of AML. Other researchers in the AE field have also previously studied 111In- or 67Ga-labelled mAbs recognising MHC-II antigens or CD20 for RIT of B-cell lymphomas with promising results in vitro (Michel et al. 2003) as well as in vivo (Michel et al. 2005). Somatostatin receptor subtype-2 (sst-2) which is overexpressed on neuroendocrine malignancies is another attractive target for radiotherapy with AEs. Octreotide is a stable octapeptide analog of somatostatin that preferentially binds with high affinity to sst-2. Capello et al. studied 111In-DTPA-octreotide for peptide receptor radionuclide therapy (PRRT) exploiting the AE emissions of 111In (Capello et al. 2003). Exposure of sst-2 positive CA20948 rat pancreatic tumour cells in vitro to 111In-DTPA-octreotide reduced their survival to 0%, while non-targeted 111In-DTPA had no inhibitory effects and unlabelled octreotide (1 μM) reduced survival by only 40%. In a subsequent in vivo study, Capello et al. showed that treatment of rats with CA20948 tumours with a single injection of 370 MBq or 3 injections of 111 MBq each of 111In-DTPA-octreotide yielded complete responses in small tumours and partial responses in larger tumours (Capello et al. 2005). 111In-DTPA-Octreotide has also been examined clinically for AE radiotherapy in patients with neuroendocrine malignancies (see Clinical Studies section). Prostate-specific membrane antigen (PSMA) is a cell surface glycoprotein overexpressed on prostate cancer, but only found at low levels in other tissues such as the salivary gland, proximal small intestine and kidneys (Silver et al. 1997; Ghosh and Heston 2004). PSMA is therefore an attractive target for AE radiotherapy of prostate cancer. 161Tb is a radionuclide with a half-life of 6.89 days that emits a total of 5.1 keV of AE energy per decay with an average AE energy of 5.7 keV (Table 1). 161Tb also emits γ-photons at 49 keV and 75 keV suitable for SPECT imaging (Uusijärvi et al. 2006) and thus may have theranostic application for imaging and treatment of cancer. 161Tb-PSMA-617 was more potent for killing PSMA-positive PC-3 PIP tumour cells than PSMA-617 labelled with the β-emitter, 177Lu (177Lu-PSMA-617) when these cells were exposed in vitro to 0.05–10 MBq/mL (Müller et al. 2019). There was no effect on the viability of PSMA-negative PC-3 flu cells up to concentrations of 10 MBq/mL. In PC-3 PIP tumour-bearing athymic mice, 161Tb-PSMA-617 treatment achieved tumour growth delay, which was dependent on the amount administered. An administered amount of 10.0 MBq prolonged the time for tumour size to reach the end-point to 42 days vs. 28 days for 5.0 MBq and prolonged survival (65 days vs. 36 days, respectively). Tumour-inhibitory effects were more pronounced in mice treated with 161Tb-PSMA-617 than 177Lu-PSMA-617 when administered at 2.5 MBq, 5.0 MBq and 10.0 MBq. Overcoming receptor heterogeneity in AE radiotherapy A strength of AEs for cancer therapy is that their cytotoxicity is restricted mostly to cells that bind and internalise the targeting agent. There is no cross-fire effect but only a local cross-dose effect, which provides a high specificity for killing cancer cells. However, this poses a challenge since heterogeneity in target expression may allow some cells that are target-negative to escape the lethal effects of the AEs. In contrast longer range β-emitters, such as 177Lu and 90Y, have a cross-fire effect that is able to kill non-targeted cells within the range of the β-particles. The bystander effect of AEs may help to overcome the obstacle of receptor heterogeneity as AE-emitters from targeted irradiated cancer cells may still exert cytotoxic effects on non-targeted cells. The bystander effect describes the biological effects of radiation on cells that have not been directly irradiated (Marín et al. 2015). One of the first demonstrations of the bystander effect from AEs was reported by Xue et al. in nude mice inoculated with a mixture of untreated LS174T human colon cancer cells and cells loaded with lethal amounts of 125I-IUdR (Xue et al. 2002). Tumour growth was reduced significantly compared to inoculation of only untreated cells, yet the range of the AEs from 125I was subcellular (< 0.5 μm) and the absorbed dose deposited in the untreated cells from 125I-IUdR incorporated into the DNA of treated cells was < 10 cGy. The effects on tumour growth caused by the 125I-IUdR was therefore interpreted as a bystander effect. In another study, UVW human glioma cells and EJ138 human bladder carcinoma cells transfected with the noradrenaline transporter (NAT) gene that were exposed to medium from cells that accumulated 123I-MIBG showed a decrease in survival, indicating that mediators of the bystander effect were released into the medium from irradiated cells (Boyd et al. 2006). Another strategy to address receptor heterogeneity is to design radioimmunoconjugates that recognise more than one receptor. We synthesised bispecific radioimmunoconjugates (bsRICs) that bind HER2 and EGFR (Razumienko et al. 2013, 2016) since upregulation of the EGFR and co-expression of EGFR and HER2 is a frequent mechanism of resistance to HER2-targeted therapies (Gallardo et al. 2012). These bsRICs were constructed by conjugating trastuzumab Fab fragments which bind HER2 through a flexible 24-mer PEG (PEG24) spacer to human EGF (hEGF). The immunoconjugates were modified with DTPA for labelling with 111In or DOTA for complex 177Lu (Razumienko et al. 2016). We also designed analogous bsRICs labelled with 64Cu for PET of tumours that co-express HER2 and EGFR (Kwon et al. 2017). In clonogenic assays, monospecific 177Lu- and 111In-trastuzumab Fab or EGF only killed tumour cells that expressed HER2 or EGFR, respectively, while the bsRICs were able to kill cells that expressed HER2 or EGFR or both receptors. These bsRICs were also more potent than the monospecific agents. 111In-labelled bsRICs were less effective than 177Lu-labelled bsRICs for treatment of MDA-MB-231/H2N tumours in mice that co-express HER2 and EGFR, but in this study, the same amount of radioactivity was administered for both bsRICs (11.1 MBq; 10 μg). This administered amount caused no observable normal tissue toxicity (NOAEL), which suggests that higher amounts of the 111In-labelled bsRICs could be safely administered to increase the therapeutic effects. It has been previously shown that RIT with AE-emitters yields more potent tumour growth-inhibitory effects in mice than β-emitters when administered at equitoxic but not necessarily equal amounts of radioactivity (Behr et al. 2000). Both 111In- and 177Lu-labelled bsRICs were also effective for treatment of TrR1 human breast cancer xenografts that are HER2 and EGFR-positive, but have acquired resistance to trastuzumab, but these tumours were less sensitive than MDA-MB-231/H2N tumour xenografts (Razumienko et al. 2016). Modular nanotransporters (MNTs) represent an interesting and unique platform for targeting and internalising AE-emitters into cancer cells and delivering these radionuclides to the cell nucleus (Sobolev 2018). MNTs are recombinant multifunctional polypeptides that combine modules for receptor binding, internalisation, endosomal escape and nuclear importation. Slastnikova et al. reported that 125I-labelled EGFR-targeted MNTs (125I-N-succinimidyl-4-guanidinomethyl-3-[125I] iodobenzoate (SGMIB)-MNT) efficiently transported over 60% of the internalised radioactivity into the nucleus of EGFR-positive human epidermoid carcinoma A431 cells and human glioblastoma D247 MG cells after 1 h of incubation in vitro (Slastnikova et al. 2012). The cytotoxicity of the 125I-MNTs was dependent on EGFR density with greater killing observed for EGFR-overexpressing A431 cells than D247 cells with low EGFR expression. Requirement for nuclear localisation of AE-emitters Nuclear translocation may not be an absolute requirement for the cytotoxic effects of AE-emitters. Using 125I-labelled non-internalizing anti-CEA murine IgG1K 35A7 mAb, internalizing anti-HER2 trastuzumab mAb or anti-EGFR hybridoma murine m225 mAb and Tat cell penetrating peptide targeting the cell membrane, cytoplasm or the cell nucleus, respectively, Pouget et al. demonstrated that decreases in clonogenic survival of A-431 human vulvar squamous carcinoma and SKOV-3 human ovarian cancer cells were significantly higher for cell membrane than for cytoplasmic localisation (Pouget et al. 2008). Therefore, non-internalizing cell surface biomarkers such as carcinoembryonic antigen (CEA) may represent feasible targets for AE radiotherapies. Santoro et al. showed that 125I-35A7 anti-CEA mAbs significantly increased survival in nude mice bearing intraperitoneal EGFR-positive vulvar squamous cell carcinoma xenografts transfected to express CEA compared to treatment with unlabelled mAbs (59 days vs. 24 days) (Santoro et al. 2009). The effect was drastically reduced for 125I-labelled internalizing anti-EGFR mAbs and the corresponding unlabelled mAbs (77 days vs 76 days, respectively), however, this lower effectiveness could be explained by the catabolism of the internalised 125I-labelled mAbs with export of released 125I from the tumour cells (Santoro et al. 2009). Interestingly, Piron et al. showed the accumulation of unrepaired DNA DSBs over time after exposure to either 125I-labelled non-internalizing anti-CEA mAbs or internalizing anti-EGFR mAbs in human colorectal cancer HCT-116 cells (Piron et al. 2014). There was a lack of linear dose-effect relationship between the biological response observed and the absorbed dose deposited by the anti-CEA and anti-EGFR 125I-labelled mAbs, suggesting 125I may cause bystander effects on the cell-membrane (Piron et al. 2014). Radiation nanomedicines Nanoparticles labelled with AE-emitting radionuclides have also been studied for cancer treatment. Fonge et al. reported the construction of 111In-labelled BCMs modified with hEGF to target EGFR-positive breast cancer cells (Fonge et al. 2009). These 111In-hEGF-BCMs showed EGFR density-dependent cellular uptake and nuclear importation in a panel of human breast cancer cell lines (MDA-MB-468, MDA-MB-231 and MCF-7). Exposure of MDA-MB-468 cells with high EGFR expression to 111In-hEGF-BCMs reduced the clonogenic survival of these cells to 2.6%, but treatment of these cells with 111In-DTPA-hEGF was more effective, reducing the survival to 0.4%. Non-targeted 111In-BCMs were not effective for killing MDA-MB-468 cells, and 111In-hEGF-BCMs did not kill MDA-MB-231 or MCF-7 cells with intermediate or low EGFR expression, respectively. Hoang et al. reported the synthesis of BCMs incorporating polymers with DTPA to complex 111In, Fab fragments to bind HER2-positive breast cancer cells and NLS peptides to enable nuclear localisation (Hoang et al. 2012). The BCMs also incorporated methotrexate as a radiosensitiser, since this was previously shown to enhance the cytotoxicity of 111In-NLS-trastuzumab (Costantini et al. 2008a). Uptake of 111In-trastuzumab-Fab-BCMs in human breast cancer cells was dependent on the level of HER2 expression, and incorporation of NLS peptides enhanced the nuclear uptake of 111In. The clonogenic survival of SK-BR-3 cells, MDA-MB-361 and MDA-MB-231 cells with high, intermediate or low HER2 expression were reduced to approximately 23%, 45% and 77%, respectively, after exposure to 111In-NLS-trastuzumab-BCMs for 24 h in vitro. Gold nanoparticles (AuNPs) have also been studied for AE radiotherapy. Song et al. reported the synthesis of 111In-labelled EGF-conjugated AuNPs (111In-EGF-AuNPs) (Song et al. 2016). Binding and internalisation of 111In-EGF-AuNPs by breast cancer cells was EGFR-dependent and was 12-fold higher for MDA-MB-468 than MCF-7 human breast cancer cells with high or low EGFR expression, respectively. Exposure of MDA-MB-468 cells for 4 h to 111In-EGF-AuNPs reduced the clonogenic survival to 17%, while less than 10% decreased survival was found for MCF-7 cells. A major limitation to systemic (i.v.) administration of radiolabelled AuNPs is high liver and spleen uptake mediated by interactions with the mononuclear phagocyte system (MPS), resulting in low tumour uptake (1–2%ID/g) in mouse tumour xenograft models (Chattopadhyay et al. 2012). Surface coating of nanoparticles with polyethylene glycol (PEG) can reduce MPS recognition. Song et al. constructed 111In-labelled PEGylated AuNPs modified with EGF to target EGFR for AE radiotherapy (Song et al. 2017). In mice with EGFR-positive MDA-MB-468 xenografts, liver uptake of i.v. injected of 111In-EGF-PEG-AuNPs was 3-fold lower than non-PEGylated 111In-EGF-AuNPs. Co-administration of an excess (15 μg) of EGF to block uptake by EGFR on hepatocytes improved tumour uptake from 2.8%ID/g to 3.9%ID/g. An alternative to systemic administration of radiolabelled AuNPs for cancer treatment is intratumoural (i.t.) injection, since this greatly minimises liver and spleen uptake due to retention of the AuNPs at the local injection site. We synthesised trastuzumab-modified AuNPs labelled with 111In for local treatment of HER2-positive breast cancer (Cai et al. 2016) (Fig. 9a). These 111In-trastuzumab-AuNPs were bound by HER2-positive SK-BR-3 or MDA-MB-361 human breast cancer cells and were internalised to a peri-nuclear location, likely mediated by a nuclear translocation sequence (NLS) in HER2 (Chen et al. 2005) (Fig. 9b). The emission of AEs by 111In caused lethal DNA DSBs in SK-BR-3 cells (Fig. 9c) reducing their CS by 3-fold. A single i.t. injection of 111In-AuNPs (10 MBq; 2.6 × 1012 AuNPs) arrested the growth of s.c. MDA-MB-361 tumours in mice without normal tissue toxicity (Fig. 9d). These results are promising for local injection of AuNPs labelled with AE-emitters for tumours that are accessible. Clinical studies There have been only a few clinical studies of AEs for cancer therapy and some of these trials were conducted 20 years ago. Nonetheless, it is helpful to review the results of these trials to appreciate the potential clinical feasibility of AEs for cancer treatment. Macapinlac et al. conducted a pilot clinical trial of co-administration of 125I-IUdR (185 MBq) or 131I-IUdR (370 MBq) to evaluate dosimetry and safety in 4 patients with colorectal cancer metastatic to the liver using hepatic artery infusion (Macapinlac et al. 1996). No tumour responses were noted or expected at these low amounts of administered radioactivity. Images revealed no retention in the BM or in other normal tissues. No side effects or hematologic toxicity were observed. Tumour DNA samples showed higher incorporation of radioactivity compared to normal hepatocyte DNA. This study suggests that hepatic artery infusion may be a feasible route of delivery of AE radiopharmaceuticals for treatment of hepatic metastases, although studies at therapeutic doses would be required to assess the effectiveness of this approach. Rebischung et al. performed intrathecal injection of 125I-IUdR in a patient with advanced pancreatic cancer with resistant neoplastic meningitis (Rebischung et al. 2008). The patient was given 4 doses of methotrexate prior to and after administration of 1850 MBq of 125I-IUdR. The treatment yielded clinical improvement correlated with a dramatic decrease in cerebrospinal carbohydrate antigen 19.9 (CA19.9) from 202 U/mL on Day 0 to 9 U/mL on Day 26. No central nervous system toxicity was found. Unfortunately, the disease recurred and the patient ultimately died. Krenning et al. reported a Phase 1 clinical trial of 111In-DTPA-octreotide in 30 patients with advanced sst-2 positive neuroendocrine malignancies (Krenning et al. 1999). Up to 14 doses of 111In-DTPA-octreotide at 6 to 7 GBq each were administered with at least a two week interval and a maximum cumulative radioactivity amount of 74 GBq. Among the 21 patients who received a total of > 20 GBq, 8 patients showed stable disease (SD) and 6 patients demonstrated a reduction in tumour size. No major side effects were noted up to 2 years post-treatment except for a transient drop in platelets and lymphocytes in some patients. Valkeman et al. reported a Phase 1 trial of 111In-DTPA-octreotide administered at intervals of 2 weeks to several months in 12 or more doses at 2 to 11 GBq to a total of 20–160 GBq in 50 patients with sst-2-positive tumours (Valkema et al. 2002). Among 40 evaluable patients, therapeutic benefit was achieved in 21 patients (52.5%) with stable disease in 14 patients, minor remissions in 6 patients and a partial remission in 1 patient. Mild hematopoietic toxicity was found in most patients, but myelodysplastic syndrome or leukemia developed in 3/6 patients who received a total dose > 100 GBq. Impairment in spermatogenesis was indicated in male patients who showed a decrease in serum inhibin B and concomitant increase of serum FSH. The radiation dose deposited in the kidneys from 111In-DTPA-octreotide was 0.45 mGy/MBq, corresponding to a total of 45 Gy for a total administered radioactivity of 100 GBq, which is twice the accepted limit for external beam radiation. However, none of these patients developed the sequelae expected for renal toxicity such as hypertension, proteinuria or any changes in serum creatinine or creatinine clearance, suggesting that this organ dosimetry did not predict renal toxicity from 111In-DTPA-octreotide. As mentioned above (see Radiation Dosimetry section) the regional distribution of 111In-DTPA-octreotide in the kidneys mainly in the renal tubules combined with the very short range of the AEs emitted by 111In, may protect the more radiation sensitive glomeruli (Konijnenberg et al. 2004). Limouris et al. infused 111In-DTPA-octreotide (average administered amount 6.3 GBq) via the hepatic artery in 17 patients with inoperable sst-2-positive liver metastases (Limouris et al. 2008). One patient achieved a complete response, while 8 patients exhibited a partial remission and 3 patients had stable disease. The median survival among the 12 responding patients was 32 months. Mild (grade 1) erythrocytopenia, leukocytopenia and thrombocytopenia were found in 3 patients. Based on promising preclinical studies, our group conducted a Phase 1 clinical trial of 111In-DTPA-hEGF in 16 patients with metastatic EGFR-positive breast cancer administered 370–2220 MBq (0.25 mg) (Vallis et al. 2014) (Fig. 10). SPECT was used to assess the tumour and normal tissue uptake of 111In-DTPA-hEGF and to estimate radiation doses to normal organs. Toxicity was also evaluated. At these administered amounts, there were no hematopoietic, renal or hepatic toxicities. The estimated radiation dose to the whole body was 0.06 mGy/MBq, corresponding to 0.133 Gy at the maximum amount administered (2220 MBq). Following administration of 2200 MBq of 111In-DTPA-hEGF, the radiation dose to the kidneys (1.64 Gy) and liver (1.9 Gy) were within the radiation toxicity limit for these organs of 23 and 30 Gy, respectively, based on external beam radiotherapy (EBRT). However, some adverse effects were found that were related to the hEGF moiety. These included flushing, chills, nausea, and vomiting. One patient experienced grade 3 thrombocytopenia, but this was attributed to cancer metastasis to the BM rather than an adverse effect of 111In-DTPA-hEGF. No other patients experienced a serious adverse reaction. SPECT showed accumulation of 111In-DTPA-hEGF at known sites of breast cancer in 7/15 evaluable patients. Further dose-escalation of 111In-DTPA-hEGF is required to achieve a therapeutic effect, but the adverse effects associated with the hEGF moiety may require a higher SA than employed in this trial, in order to minimise the mass of hEGF injected. 111In-labelled anti-EGFR mAbs may be a promising alternative for AE-radiotherapy of EGFR-overexpressing breast cancer that would not cause these adverse effects (as discussed in the Preclinical Studies section). Our group is planning a Phase 1 clinical trial of 111In-NLS-trastuzumab to study its uptake in HER2-positive breast cancer by SPECT. Based on the results of this trial, we aim to conduct future therapeutic studies aimed at treatment of HER2-positive breast cancer with 111In-NLS-trastuzumab. Li et al. reported a Phase 2 clinical trial of adjuvant RIT with 125I-labelled murine anti-EGFR mAb 425 in 192 patients with glioblastoma multiforme (GBM) (Li et al. 2010). Up to 3 weekly intravenous injections of 125I-mAb 425 (1.8 GBq each) were administered with a maximum cumulative radioactivity amount of 5.4 GBq. Among these 192 patients, 132 patients received RIT alone, and 60 patients received RIT and temozolomide (TMZ) chemotherapy. The median overall survival of 97 patients who received RIT alone was 14.5 months (range 12.1–16.7 months), and for 51 patients who received RIT and TMZ, the overall median survival was 20.4 months (range 14.9–25.8 months). Both treatment arms resulted in significantly improved overall survival compared to a historical control group of 39 patients receiving standard-of-care treatment, who had a median survival of 10.2 months (range 8.4–12.0 months). No grade 3 or 4 toxicities were observed for either of the RIT treatment groups, with only 7 of the 192 patients experiencing acute side effects (transient flushing, grade 1 nausea, hypotension and skin irritation). Four patients developed human anti-mouse antibodies (HAMA) preventing further administration of 125I-mAb 425, which was a murine mAb. The development of humanised or chimeric (e.g. cetuximab) and fully human anti-EGFR mAbs (e.g. panitumumab) in recent years should obviate this immunogenicity issue. Conclusions AEs have very attractive properties for cancer therapy since their nanometre-micrometre range results in high LET that is potent for causing lethal damage in cancer cells. Biomolecules (mAbs and peptides), nucleosides and nanoparticles have been labelled with AE-emitting radionuclides (e.g. 111In, 67Ga, 99mTc, 123I or 125I) and studied for cancer treatment. Several concepts have emerged. Firstly, AEs are especially lethal to cancer cells when emitted in close proximity to the cell nucleus, and particularly if the AE-emitting radiotherapeutic agent is incorporated directly into DNA (e.g. 125I-IUdR). Strategies that promote the delivery of AE-emitters to the nucleus by conjugating internalising mAbs to NLS peptides (e.g. 111In-NLS-trastuzumab), or targeting a receptor that harbours an endogenous NLS to enable nuclear uptake (e.g. EGFR or HER2) amplifies the lethal DNA-damaging effects of AEs. Nonetheless, nuclear localisation is not an absolute requirement and AEs can also kill targeted cancer cells by damaging the cell membrane, or non-targeted cells by a local cross-dose effect or a longer range bystander effect. Numerous studies have shown that AEs can kill cancer cells in vitro in clonogenic assays by inflicting lethal DNA damage (e.g. DSBs) detected by immunofluorescence for γH2AX foci in the nucleus or by damaging the cell membrane. Preclinical studies of AE-emitting radiotherapeutic agents in vivo in mouse tumour xenograft models have further demonstrated that cancer treatment using AEs is feasible. Strong tumour growth inhibition has been achieved with minimal toxicity to normal tissues at the amounts administered due to the mostly restricted cytotoxicity of AEs towards cancer cells that bind these radiotherapeutic agents. A limited number of clinical studies of AEs for cancer therapy have been performed, and these mainly evaluated the tumour and normal tissue localisation of the radiotherapeutic agents, estimated normal organ dosimetry and assessed normal tissue toxicity at relatively low administered amounts. Nonetheless, some studies have shown promising results for treatment of cancer with AE-emitting radiotherapeutic agents (e.g. 125I-IUdR, 111In-DTPA-octreotide or 125I-mAb 425), achieving tumour remissions or improved survival in patients. The recent introduction of a large armamentarium of biologically-targeted therapies for cancer, especially humanised and fully human mAbs creates many new opportunities to design novel AE-emitting radiotherapeutic agents for cancer treatment. In only a few years, we will celebrate the 100th anniversary of the publication by Pierre Auger (Auger 1923) on the discovery of these electrons that bear his name. Pierre Auger did not conceive of the application of AEs for targeted cancer treatment, but this is a tremendously exciting future that we and many other scientists in this field envision. Availability of data and materials Not applicable. Abbreviations 123I-ITdU: : 123I-4′-thio-2′-deoxyuridine 123I-MIBG: : 123I-metaiodobenzylguanidine 125I-IUdR: : 125I-iododeoxyuridine 131I-IUdR: : 131I-iododeoxyuridine 3H-TdR: : 3H-deoxyuridine AEs: : Auger electrons AML: : Acute myeloid leukemia ATM: : Ataxia telangiectasia mutated AUC: : Area under the curve AuNPs: : Gold nanoparticles BC: : Breast cancer BCMs: : Block copolymer micelles BM: : Bone marrow bsRICs: : Bispecific radioimmunoconjugates CEA: : Carcinoembryonic CHO: : Chinese Hamster Ovary CI: : Confidence interval CS: : Clonogenic survival DMSO: : Dimethylsulfoxide DOTA: : 1,4,7,10-tetraazacyclododecane-1,4,7,10-tetraacetic acid DSBs: : Double strand breaks DTPA: : Diethylenetriaminepentaacetic acid EBRT: : External beam radiotherapy EC: : Electron capture EDDA: : Ethylene diamine N,N′-diacetic acid EGFR: : Epidermal growth factor receptor GBM: : Glioblastoma multiforme HAMA: : Human anti-mouse antibodies hEGF: : Human epidermal growth factor HER2: : Human epidermal growth factor receptor 2 HMPAO: : Hexamethylpropylenamineoxime HYNIC: : Hydrazinonicotinamide i.p.: : Intraperitoneal i.t.: : Intratumoural IAEA: : International Atomic Energy Agency IC: : Internal conversion IUdR: : 125I- or 123I-5-iodo-2-deoxyuridine LET: : Linear energy transfer LSCs: : Leukemic stem cells mAb: : Monoclonal antibody MCPs: : Metal-chelating polymers MDR: : Multidrug resistance MIRD: : Medical Internal Radiation Dose MNTs: : Modular nanotransporters MPS: : Mononuclear phagocyte system NLS: : Nuclear localisation sequence NOAEL: : No observable adverse effects NOTA: : 1,4,7-triazacyclononane-1,4,7-triacetic acid NRG: : NOD-Rag1null IL2rgnull, NOD rag gamma NURBS: : Non-uniform rational B-splines PAMAM: : Polyamidoamine PCL: : Polycaprolactone PEG: : Polyethylene glycol PEG24 : : 24 mer PEG PFGE: : Pulsed-field gel electrophoresis PRRT: : Peptide receptor radionuclide therapy p-SCN-Bn: : p-isothiocyanatobenzyl RaBiT: : Radiotracer Biodistribution Template RBE: : Relative biological effectiveness RICs: : Radioimmunoconjugates RIT: : Radioimmunotherapy ROS: : Reactive oxygen species rS : : Source region rT : : Target region SA: : Specific activity SCGE: : Single cell gel electrophoresis assay SD: : Stable disease SPECT: : Single-photon emission computed tomography sst-2: : Somatostatin receptor subtype-2 Tb : : Biological half life TD : : Dose-integration period Te : : Effective half life TMZ: : Temozolomide Tp : : Physical half-life WBC: : White blood cell References 65-Terbium-161. 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The human polynucleotide kinase/phosphatase (hPNKP) inhibitor A12B4C3 radiosensitizes human myeloid leukemia cells to Auger electron-emitting anti-CD123 111In-NLS-7G3 radioimmunoconjugates. Nucl Med Biol. 2014;41(5):377–83. Article CAS PubMed Google Scholar Acknowledgments Not applicable. Funding The authors acknowledge research grant funding from the Canadian Breast Cancer Foundation, Canadian Cancer Society, and the Canadian Institutes of Health Research. AK received a scholarship from the Centre for Pharmaceutical Oncology at the University of Toronto and a Queen Elizabeth II Graduate Scholarship in Science and Technology (QEII-GSST). VF was supported by the Polymer Nanoparticles for Drug Delivery (POND) CREATE training program sponsored by the Natural Sciences and Engineering Research Council of Canada and the MDS Nordion Graduate Scholarship in Radiopharmaceutical Sciences (OTSS) award. The publication of this article was supported by funds of the European Association of Nuclear Medicine (EANM). Author information Author notes Anthony Ku and Valerie J. Facca contributed equally to this work. Authors and Affiliations Department of Pharmaceutical Sciences, University of Toronto, Toronto, ON, Canada Anthony Ku, Valerie J. Facca, Zhongli Cai & Raymond M. Reilly 2. Department of Medical Imaging, University of Toronto, Toronto, ON, Canada Raymond M. Reilly 3. Joint Department of Medical Imaging and Toronto General Research Institute, University Health Network, Toronto, ON, Canada Raymond M. Reilly 4. Leslie Dan Faculty of Pharmacy, University of Toronto, 144 College St., Toronto, ON, M5S 3M2, Canada Raymond M. Reilly Contributions AK and VF contributed equally to this article. RMR conceived of the article design and edited the final manuscript. AK, VF and ZC researched the literature and drafted the manuscript. All authors reviewed and approved the final manuscript. Corresponding author Correspondence to Raymond M. Reilly. Ethics declarations Ethics approval and consent to participate Not applicable. Consent for publication Not applicable. Competing interests The authors declare that they have no competing interests. Additional information Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. Rights and permissions Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License ( which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Reprints and permissions About this article Cite this article Ku, A., Facca, V.J., Cai, Z. et al. Auger electrons for cancer therapy – a review. EJNMMI radiopharm. chem. 4, 27 (2019). Received: Accepted: Published: DOI: Share this article Anyone you share the following link with will be able to read this content: Provided by the Springer Nature SharedIt content-sharing initiative Keywords Auger electrons 111In Monoclonal antibodies Nanoparticles Peptides Dosimetry Radiolabelling Cancer treatment Preclinical studies Clinical studies
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https://brainly.com/question/44329493
[FREE] How do you solve the equation \tan(x) = 1 on the interval [- \pi/2, \pi/2]? - brainly.com 2 Search Learning Mode Cancel Log in / Join for free Browser ExtensionTest PrepBrainly App Brainly TutorFor StudentsFor TeachersFor ParentsHonor CodeTextbook Solutions Log in Join for free Tutoring Session +73k Smart guidance, rooted in what you’re studying Get Guidance Test Prep +20,6k Ace exams faster, with practice that adapts to you Practice Worksheets +8,3k Guided help for every grade, topic or textbook Complete See more / Mathematics Textbook & Expert-Verified Textbook & Expert-Verified How do you solve the equation tan(x)=1 on the interval [−π/2,π/2]? 1 See answer Explain with Learning Companion NEW Asked by woodkkk227 • 12/11/2023 0:00 / 0:15 Read More Community by Students Brainly by Experts ChatGPT by OpenAI Gemini Google AI Community Answer This answer helped 6608739 people 6M 0.0 0 Upload your school material for a more relevant answer To solve the equation tan(x)=1 within the interval [-π/2, π/2], we identify x as π/4 radians (or 45 degrees), since this is the angle where the tangent has a value of 1. Explanation To solve the equation tan(x)=1 on the interval [-π/2, π/2], we need to determine the angle x for which the tangent function has a value of 1. The tangent of an angle in a right-angled triangle is the ratio of the side opposite to the angle to the side adjacent to the angle. Since we know that tan(x) = 1, we are looking for an angle where the opposite and adjacent sides are equal, which is the case for a 45-degree angle, or π/4 radians. However, because the tangent function is periodic, we must consider the interval we are restricted to. On the interval [-π/2, π/2], which is the same as [-90 degrees, 90 degrees], the function tan(x) has a period of π, so it will repeat every π radians. But within this specified interval, there is only one angle for which tan(x) equals 1, and that is x = π/4 radians (or 45 degrees). Thus, the solution to the equation within the given interval is x = π/4. Answered by rajeshgpandey4 •30.7K answers•6.6M people helped Thanks 0 0.0 (0 votes) Textbook &Expert-Verified⬈(opens in a new tab) This answer helped 6608739 people 6M 0.0 0 Physics for K-12 - Sunil Singh Mechanics - Benjamin Crowell Light and Matter - Benjamin Crowell Upload your school material for a more relevant answer The solution to the equation tan(x)=1 on the interval [−π/2,π/2] is x=4 π​. This solution arises because this is the angle where the tangent function equals 1. Since the tangent is continuous and increasing within this range, there are no other solutions. Explanation To solve the equation tan(x)=1 on the interval [−π/2,π/2], we will identify the angle x for which the tangent function equals 1. Understanding Tangent Function: The tangent of an angle is the ratio of the opposite side to the adjacent side in a right triangle. This means tan(x)=1 occurs when the lengths of the opposite and adjacent sides are equal. Identifying the Angle: From trigonometry, we know that tan(4 π​)=1 (or equivalently, tan(4 5∘)=1). Thus, one solution for x is 4 π​. Considering the Interval: The given interval is [−π/2,π/2], which means we only consider angles between -90 degrees and 90 degrees. In this range, the tangent function is continuous and monotonic (increasing), indicating it crosses the value of 1 only once. Conclusion: Therefore, the only solution to the equation tan(x)=1 within the interval [−π/2,π/2] is: x=4 π​ (or 45 degrees). Examples & Evidence For instance, if you visualize a right triangle with equal sides, the angle opposite to both sides is 45 degrees, confirming that tan(4 5∘)=1. We can reference the unit circle, where the coordinates of the point corresponding to 4 5∘ or 4 π​ radians are (2 2​​,2 2​​), resulting in tan(4 π​)=2 2​​2 2​​​=1. Thanks 0 0.0 (0 votes) Advertisement woodkkk227 has a question! Can you help? Add your answer See Expert-Verified Answer ### Free Mathematics solutions and answers Community Answer 4.6 12 Jonathan and his sister Jennifer have a combined age of 48. If Jonathan is twice as old as his sister, how old is Jennifer Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)? Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD Community Answer 4.1 17 How many positive integers between 100 and 999 inclusive are divisible by three or four? Community Answer 4.0 9 N a bike race: julie came in ahead of roger. julie finished after james. david beat james but finished after sarah. in what place did david finish? Community Answer 4.1 8 Carly, sandi, cyrus and pedro have multiple pets. carly and sandi have dogs, while the other two have cats. sandi and pedro have chickens. everyone except carly has a rabbit. who only has a cat and a rabbit? Community Answer 4.1 14 richard bought 3 slices of cheese pizza and 2 sodas for $8.75. Jordan bought 2 slices of cheese pizza and 4 sodas for $8.50. How much would an order of 1 slice of cheese pizza and 3 sodas cost? A. $3.25 B. $5.25 C. $7.75 D. $7.25 Community Answer 4.3 192 Which statements are true regarding undefinable terms in geometry? Select two options. A point's location on the coordinate plane is indicated by an ordered pair, (x, y). A point has one dimension, length. A line has length and width. A distance along a line must have no beginning or end. A plane consists of an infinite set of points. Community Answer 4 Click an Item in the list or group of pictures at the bottom of the problem and, holding the button down, drag it into the correct position in the answer box. Release your mouse button when the item is place. If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers. Express In simplified exponential notation. 18a^3b^2/ 2ab New questions in Mathematics Write the fraction 40 4​ in simplest form. When f(x)=−3, what is x? How many roots, real or complex, does the polynomial 7+5 x 4−3 x 2 have in all? Subtract. $\begin{array}{r} -21-43=\square \ 1-(-4)=\square \end{array}$ The table represents a function. | x | f(x) | :--- | | -6 | 8 | | 7 | 3 | | 4 | -5 | | 3 | -2 | | -5 | 12 | Which value is an output of the function? A. 7 B. -6 C. -2 D. 4 Previous questionNext question Learn Practice Test Open in Learning Companion Company Copyright Policy Privacy Policy Cookie Preferences Insights: The Brainly Blog Advertise with us Careers Homework Questions & Answers Help Terms of Use Help Center Safety Center Responsible Disclosure Agreement Connect with us (opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab)(opens in a new tab) Brainly.com Dismiss Materials from your teacher, like lecture notes or study guides, help Brainly adjust this answer to fit your needs. Dismiss
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https://www.thesaurus.com/browse/poverty
Advertisement Skip to noun (1) Advertisement poverty noun as in want; extreme need, often financial Strongest matches Strong matches Weak match Example Sentences "If Plaid want to be taken seriously, they need to confirm whether they're going to tax Welsh families into poverty or cut public services to the bone," Stevens said. "Last week, we reached our 3,000th person for this calendar year…. which included nearly 1,000 children living in poverty," Fr Stevenson said. Several MPs from Liverpool were among those who wrote to Sir Keir ahead of the conference insisting the cap "is one of the most significant drivers of child poverty in Britain today". About one-fifth of the city’s nearly 800 people live in poverty, according to Census figures. It comes as an annual report from the Director of Public Health revealed more than 27,000 children and young people in Cornwall and the Isles of Scilly were growing up in relative poverty. Advertisement Related Words Words related to poverty are not direct synonyms, but are associated with the word poverty. Browse related words to learn more about word associations. nounas in insufficiency, scarcity nounas in indigence nounas in hardship, adversity nounas in hardship, adversity nounas in humiliation, shame Viewing 5/23 related words From Roget's 21st Century Thesaurus, Third Edition Copyright © 2013 by the Philip Lief Group. Advertisement Advertisement Advertisement Browse Follow us Get the Word of the Day every day! By clicking "Sign Up", you are accepting Dictionary.com Terms & Conditions and Privacy Policies.
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https://www2.latech.edu/~sajones/June/Study%20Guides/ChII.pdf
II-1 II. FLUID STATICS From a force analysis on a triangular fluid element at rest, the following three concepts are easily developed: For a continuous, hydrostatic, shear free fluid: 1. Pressure is constant along a horizontal plane, 2. Pressure at a point is independent of orientation, 3. Pressure change in any direction is proportional to the fluid density, local g, and vertical change in depth. These concepts are key to the solution of problems in fluid statics and lead to the following: 1. Two points at the same depth in a static fluid have the same pressure. 2. The orientation of a surface has no bearing on the pressure at a point in a static fluid. 3. Vertical depth is a key dimension in determining pressure change in a static fluid. If we were to conduct a more general force analysis on a fluid in motion with constant density and viscosity, we would obtain the following: ∇P = ρ g −a { } + µ ∇ 2V Thus the pressure change in fluid in general depends on: effects of fluid statics (ρ g), Ch. II inertial effects (ρ a), Ch. III viscous effects ( µ∇ 2V ) Chs VI & VII Note: For problems involving the effects of both (1) fluid statics and (2) inertial effects, it is the net " g −" a acceleration vector that controls both the magnitude and direction of the pressure gradient. II-2 This equation can be simplified for a fluid at rest (ie., no inertial or viscous effects) to yield: ∇ p = ρ g ∂p ∂x = 0 ; ∂p ∂y = 0 ; ∂p ∂z = dp dz = −ρ g P2 −P 1 = −ρ gd Z 1 2 ∫ It is appropriate at this point to review the differences between absolute, gage, and vacuum pressure. These differences are illustrated in Fig. 2.3 (shown below) and are based on the following definitions: • absolute pressure - Pressure measured relative to absolute zero. • gage pressure - Pressure > Patm measured relative to Patm • vacuum pressure - Pressure < Patm measured relative to Patm • Patm - local absolute pressure due to the local atmosphere only. standard Patm at sea level = 1atm = 101.3 kPa = 2116 lbf/ft2 P (kPa) Patm Local atmospheric pressure e.g at 1000 m Patm≈ 90 kPa = 0 Pa gage = 0 Pa vacuum Pressure above atmospheric e.g P = 120 kPa absolute or P = 30 kPa gage P (relative to Patm) = Pgage = 30 kPa gage 0 Absolute zero, perfect vacuum P(abs) = 0, P(vac) = 90kPa, P(gage) = - 90 kPa Pressure below atmospheric e.g. P = 50 kPa abs P(relative to Patm) = P(vac) = 40 kPa P(abs) = 50 kPa Fig. 2.3 Illustration of absolute, gage, and vacuum pressure II-3 Hydrostatic Pressure in Liquids For liquids and incompressible fluids, the previous integral expression for P2 - P1 integrates to P1 – P2 = -ρg (Z2 – Z1) Note: Z2 – Z1 is positive for Z2 above Z1. but P2 – P1 is negative for Z2 above Z1. We can now define a new fluid parameter: γ = ρg ≡ specific weight of the fluid 2 P 1 P2 x y z Free surface Pressure = Pa h = Z - Z 2 1 Z Z 1 With this, the previous equation becomes (for an incompressible, static fluid) P2 – P1 = - γ (Z2 – Z1) The most common application of this result is that of manometry. Consider the U-tube, multi- fluid manometer shown on the right. If we first label all intermediate points between A & a, we can write for the overall pressure change II-4 PA - Pa = (PA- P1) + (P1 - P2) + (P2 - Pa ) This equation was obtained by adding and subtracting each intermediate pressure. The total pressure difference is now expressed in terms of a series of intermediate pressure differences. Substituting the previous result for static pressure difference, we obtain PA - PB = - ρ g(ZA- Z1) – ρ g (Z1 – Z2) – ρ g (Z2 - ZB ) Again note: Z positive up and ZA > Z1 , Z1 < Z2 , Z2 < Za . In general, follow the following steps when analyzing manometry problems: 1. On the manometer schematic, label points on each end of the manometer and at each intermediate point where there is a fluid-fluid interface, e.g. A - 1 - 2 - B. 2. Express the overall pressure difference in terms of appropriate intermediate pressure differences. PA - PB = (PA - P1) + (P1 - P2) + (P2 - PB) 3. Express each intermediate pressure difference in terms of an appropriate product of specific weight (or ρ g) elevations change ( and watch the signs). PA - PB = -γ (zA - z1) + -γ (z1 - z2) + -γ (z2 - zB) When developing a solution for manometer problems, take care to: 1. Include all pressure changes. 2. Use correct ∆Z and γ with each fluid. 3. Use correct signs with ∆ Z. If pressure difference is expressed as PA – P1, the elevation change should be written as ZA – Z1. 4. Watch units. II-5 Manometer Example: Given the indicated manometer, determine the gage pressure at A. Pa = 101.3 kPa. The fluid at A is Meriam red oil no. 3. ρgw = 9790 N/m3 ρg A = S.G.ρgw = 0.839790 N/m3 ρg A = 8126 N/m3 ρgair = 11.8 N/m3 1 2 a A 1 10 cm 18 cm S.G. = .83 H 0 2 With the indicated points labeled on the manometer, we can write PA - Pa = PA (gage) = (PA- P1) + (P1 – P2) + (P2 - Pa ) Substituting the manometer expression for a static fluid, we obtain PA (gage) = - ρgA(zA- z1) – ρgw(z1 – z2) – ρga(z2 - za ) Neglect the contribution due to the air column. Substituting values, we obtain PA (gage) = - 8126 N/m3 0.10 m – 9790 N/m3 -0.18 = 949.6 N/m2 Ans Note why: (zA- z1) = 0.10 m and (z1 – z2) = -0.18 m, & we did not use Pa Review the text examples for manometry. See Table 2.1 for values of specific weight, γ, in both B.G. and S.I. units. II-6 Hydrostatic Pressure in Gases Since gases are compressible, density is a non-constant variable in the previous expression for dP/dz. Assuming the gas is an ideal gas, we can write d P d z = −ρ g = −P RT g or d P P 1 2 ∫ = ln P 2 P 1 = −g R d z T 1 2 ∫ For an isothermal atmosphere with T = To, this integrates to P 2 = P 1exp −g z2 −z1 ( ) RT o       Up to an altitude of approximately 36,000ft (l l,000 m), the mean atmospheric temperature decreases nearly linearly and can be represented by T ≈ To - B z where B is the lapse rate The following values are assumed to apply for air from sea level to 36,000 ft: To = 518.69˚R = 288.16˚K (15 ˚C) B = 0.003566˚R/ft = 0.650˚K/m Substituting the linear temperature variation into the previous equation and integrating we obtain P = P a 1−Bz T o       g / RB ( ) where g RB = 5.26 for air ( ) and R = 287 m2/(s2 ˚K) Review example 2.2 in the text. II-7 Hydrostatic Forces on Plane Surfaces Consider a plane surface of arbitrary shape and orientation, submerged in a static fluid as shown: If P represents the local pressure at any point on the surface and h the depth of fluid above any point on the surface, from basic physics we can easily show that the net hydrostatic force on a plane surface is given by (see text for development): F = PdA A ∫ = Pcg A Thus, basic physics says that the hydrostatic force is a distributed load equal to the integral of the local pressure force over the area. This is equivalent to the following: The hydrostatic force on one side of a plane surface submerged in a static fluid equals the product of the fluid pressure at the centroid of the surface times the surface area in contact with the fluid. Also: Since pressure acts normal to a surface, the direction of the resultant force will always be normal to the surface. Note: In most cases, since it is the net hydrostatic force that is desired and the contribution of atmospheric pressure Pa will act on both sides of a surface, the result of atmospheric pressure Pa will cancel and the net force is obtained by F = ρ gh cgA F = PcgA Pcg is now the gage pressure at the centroid of the area in contact with the fluid. II-8 Therefore, to obtain the net hydrostatic force F on a plane surface: 1. Determine depth of centroid hcg for the area in contact with the fluid. 2. Determine the (gage) pressure at the centroid Pcg. 3. Calculate F = PcgA. The following page shows the centroid, and other geometric properties of several common areas. It is noted that care must be taken when dealing with layered fluids. The procedure essentially requires that the force on the part of the plane area in each individual layer of fluid must be determined separately for each layer using the steps listed above. We must now determine the effective point of application of F. This is commonly called the “center of pressure - cp” of the hydrostatic force. Note: This is not necessarily the same as the c.g. Define an x – y coordinate system whose origin is at the centroid, c.g, of the area. The location of the resultant force is determined by integrating the moment of the distributed fluid load on the surface about each axis and equating this to the moment of the resultant force about that axis. Therefore, for the moment about the x axis: F y cp = y P d A ∫ A Applying a procedure similar to that used previously to determine the resultant force, and using the definition (see text for detailed development), we obtain Ycp = −ρgsinθ I xx PcgA ≤ 0 where: Ixx is defined as the Moment of Inertia, or the ∫ 2nd moment of the area Therefore, the resultant force will always act at a distance ycp below the centroid of the surface ( except for the special case of sin θ = 0 ). II-9 a PROPERTIES OF PLANE SECTIONS Geometry Centroid Moment of Inertia I x x Product of Inertia I x y Area b 1 h b b L y x L 2R b L b y x Fluid Specific Weight Seawater Glycerin Mercury Carbon .0752 57.3 62.4 49.2 11.8 8,996 9,790 7,733 64.0 78.7 846. 99.1 10,050 12,360 133,100 15,570 0 0 0 h 0 y x y x x y R y x y x s R Air Oil Water Ethyl 3 1bf /ft N /m3 bL3 12 b L ⋅ b + b 1 ( ) h 2 a = 4 R 3π π 16 − 4 9π R 4    1 8 − 4 9π R 4    π R 2 4 a = L 3 bL 3 36 b b −2s ( )L 2 72 1 2 b ⋅L b 3 , L 3 bL 3 36 − b 2L 2 72 b ⋅L 2 π R 2 2 R 4 π 8 − 8 9π    0 ,a = 4R 3π π R 4 4 0, 0 π R 2 b 2 , L 2 a ) = h b + 2b 1 ( 3 b + b1 ( ) ) 3 b 2 + 4bb 1 + b 1 2 ( 36 b + b1 ( ) N /m3 3 1bf/ft II-10 Proceeding in a similar manner for the x location, and defining Ixy = product of inertia, we obtain X cp = − ρgsinθ I xy PcgA where Xcp can be either positive or negative since Ixy can be either positive or negative. Note: For areas with a vertical plane of symmetry through the centroid, i.e. the y - axis (e.g. squares, circles, isosceles triangles, etc.), the center of pressure is located directly below the centroid along the plane of symmetry, i.e., Xcp = 0. Key Points: The values Xcp and Ycp are both measured with respect to the centroid of the area in contact with the fluid. Xcp and Ycp are both measured in the inclined plane of the area; i.e., Ycp is not necessarily a vertical dimension, unless θ = 90o. Special Case: For most problems where (1) we have a single, homogeneous fluid (i.e. not applicable to layers of multiple fluids) and (2) the surface pressure is atmospheric, the fluid specific weight γ cancels in the equation for Ycp and Xcp and we have the following simplified expressions: F = ρ g h cgA Ycp = −I xx sinθ h cgA Xcp = − Ixy sinθ h cgA However, for problems where we have either (1) multiple fluid layers, or (2) a container with surface pressurization > Patm , these simplifications do not occur and the original, basic expressions for F , Ycp , and Xcp must be used; i.e., take II-11 care to use the approximate expressions only for cases where they apply. The basic equations always work. II-12 Summary: 1. The resultant force is determined from the product of the pressure at the centroid of the surface times the area in contact with the fluid. 2. The centroid is used to determine the magnitude of the force. This is not the location of the resultant force. 3. The location of the resultant force will be at the center of pressure which will be at a location Ycp below the centroid and Xcp as specified previously. 4. Xcp = 0 for areas with a vertical plane of symmetry through the c.g. Example 2.5 Given: Gate, 5 ft wide Hinged at B Holds seawater as shown Find: a. Net hydrostatic force on gate b. Horizontal force at wall - A c. Hinge reactions - B 8’ θ Seawater • c .g. hc.g. A B 64 lbf/ft3 15’ 6’ 9’ a. By geometry: θ = tan-1 (6/8) = 36.87o Neglect Patm Since the plate is rectangular, hcg = 9 ft + 3ft = 12 ft A = 10 x 5 = 50 ft2 Pcg = γ hcg = 64 lbf/ft3 12 ft = 768 lbf/ft2 ∴ Fp = Pcg A = 768 lbf/ft2 50 ft2 = 38,400 lbf II-13 II-14 b. Horizontal Reaction at A Must first find the location, c.p., for Fp ycp = −ρ gsinθ Ixx P cg A = −Ixx sinθ hcg A For a rectangular wall: Ixx = bh3/12 Ixx = 5 103/12 = 417 ft4 Note: The relevant area is a rectangle, not a triangle. θ • c.g. • c.p. Bx Bz P Fw 8 ft 6 ft y c.p. θ Note: Do not overlook the hinged reactions at B. ycp = −417 ft 4 ⋅0.6 12 ft ⋅50 ft2 = −0.417 ft below c.g. xcp = 0 due to symmetry M B ∑ = 0 5 −0.417 ( )⋅38,400 −6P = 0 P = 29,330 lbf ← ← ← ← θ • c.g. • c.p. Bx Bz P Fw 8 ft 6 ft y c.p. θ II-15 c. F x ∑ =0, B x + Fsinθ −P= 0 Bx + 38,4000.6 - 29,330 = 0 Bx = 6290 lbf → → → → F z ∑=0, B z −Fcosθ =0 Bz = 38,400 0.8 = 30,720 lbf ↑ ↑ ↑ ↑ Note: Show the direction of all forces in final answers. Summary: To find net hydrostatic force on a plane surface: 1. Find area in contact with fluid. 2. Locate centroid of that area. 3. Find hydrostatic pressure Pcg at centroid, typically = γ γ γ γ hcg (generally neglect Patm ). 4. Find force F = Pcg A. 5. The location will not be at the c.g., but at a distance ycp below the centroid. ycp is in the plane of the area. II-16 Plane Surfaces in Layered Fluids For plane surfaces in layered fluids, the part of the surface in each fluid layer must essentially be worked as a separate problem. That is, for each layer: 1.) Identify the area of the plate in contact with each layer, 2.) Locate the c.g. for the part of the plate in each layer and the pressure at the c.g., and 3. Calculate the force on each layered element using F1 = Pc.g1• A1 . Repeat for each layer. Use the usual procedure for finding the location of the force for each layer. Review all text examples for forces on plane surfaces. II-17 Forces on Curved Surfaces Since this class of surface is curved, the direction of the force is different at each location on the surface. Therefore, we will evaluate the x and y components of net hydrostatic force separately. Consider curved surface, a-b. Force balances in x & y directions yield Fh = FH Fv = Wair + W1 + W2 From this force balance, the basic rules for determining the horizontal and vertical component of forces on a curved surface in a static fluid can be summarized as follows: Horizontal Component, Fh The horizontal component of force on a curved surface equals the force on the plane area formed by the projection of the curved surface onto a vertical plane normal to the component. The horizontal force will act through the c.p. (not the centroid) of the projected area. b a cp hcg Fh ycp a’ b’ Projected vertical plane Curved surface II-18 Therefore, to determine the horizontal component of force on a curved surface in a hydrostatic fluid: 1. Project the curved surface into the appropriate vertical plane. 2. Perform all further calculations on the vertical plane. 3. Determine the location of the centroid - c.g. of the vertical plane. 4. Determine the depth of the centroid - hcg of the vertical plane. 5. Determine the pressure - Pcg = ρ g hcg at the centroid of the vertical plane. 6. Calculate Fh = Pcg A, where A is the area of the projection of the curved surface into the vertical plane, ie. the area of the vertical plane. 7. The location of Fh is through the center of pressure of the vertical plane, not the centroid. Get the picture? All elements of the analysis are performed with the vertical plane. The original curved surface is important only as it is used to define the projected vertical plane. Vertical Component - Fv The vertical component of force on a curved surface equals the weight of the effective column of fluid necessary to cause the pressure on the surface. The use of the words effective column of fluid is important in that there may not always actually be fluid directly above the surface. (See graphic that follows.) This effective column of fluid is specified by identifying the column of fluid that would be required to cause the pressure at each location on the surface. Thus, to identify the effective volume - Veff: 1. Identify the curved surface in contact with the fluid. 2. Identify the pressure at each point on the curved surface. 3. Identify the height of fluid required to develop the pressure. II-19 4. These collective heights combine to form Veff. b a V eff P P P Fluid above the surface a b V eff P P P fluid No fluid actually above surface These two examples show two typical cases where this concept is used to determine Veff. The vertical force acts vertically through the centroid (center of mass) of the effective column of fluid. The vertical direction will be the direction of the vertical components of the pressure forces. Therefore, to determine the vertical component of force on a curved surface in a hydrostatic fluid: 1. Identify the effective column of fluid necessary to cause the fluid pressure on the surface. 2. Determine the volume of the effective column of fluid. 3. Calculate the weight of the effective column of fluid - Fv = ρgVeff. 4. The location of Fv is through the centroid of Veff. Finding the Location of the Centroid A second problem associated with the topic of curved surfaces is that of finding the location of the centroid of Veff. Recall: II-20 Centroid = the location where a point area, volume, or mass can be place to yield the same first moment of the distributed area, volume, or mass, e.g. xcgV 1 = xdV V 1 ∫ This principle can also be used to determine the location of the centroid of complex geometries. For example: If Veff = V1 + V2 then xcgVeff = x1V1 + x2V2 or for the second geometry VT = V1 + Veff xTVT = x1V1 + xcgVeff b a 2 V V 1 a b V1 V eff fluid Note: In the figures shown above, each of the x values would be specified relative to a vertical axis through b since the cg of the quarter circle is most easily specified relative to this axis. II-21 Example: Gate AB holds back 15 ft of water. Neglecting the weight of the gate, determine the magnitude (per unit width) and location of the hydrostatic forces on the gate and the resisting moment about B. • • 15 ft A B Water H F F V Width - W a. Horizontal component γ = ρg = 62.4 lbf/ft3 Rule: Project the curved surface into the vertical plane. Locate the centroid of the projected area. Find the pressure at the centroid of the vertical projection. F = Pcg Ap Note: All calculations are done with the projected area. The curved surface is not used at all in the analysis. • • A B a b h cg Pcg The curved surface projects onto plane a - b and results in a rectangle, (not a quarter circle) 15 ft x W. For this rectangle: hcg = 7.5, Pcg = γhcg = 62.4 lbf/ft3 7.5 ft = 468 lbf/ft2 Fh = Pcg A = 468 lbf/ft2 15 ftW= 7020 W lbf per ft of width Location: Ixx = bh3/12 = W 153 /12 = 281.25 W ft4 ycp = −Ixxsinθ hcg A = −281.25W ft4sin90o 7.5 ft15W ft2 = −2.5 ft The location is 2.5 ft below the c.g. or 10 ft below the surface, 5 ft above the bottom. II-22 b. Vertical force: Rule: Fv equals the weight of the effective column of fluid above the curved surface (shown by the dashed diagonal lines). A • c.g. C B • b F v • Q: What is the effective volume of fluid above the surface? What volume of fluid would result in the actual pressure distribution on the curved surface? Vol = Vol A - B - C Vrec = Vqc + VABC, VABC = Vrec - Vqc VABC = Veff = 152 W - π 152/4W = 48.29 W ft3 Fv = ρg Veff = 62.4 lbf/ft3 48.29 ft3 = 3013 lbf per ft of width Note: Fv is directed upward even though the effective volume is above the surface. c. What is the location? Rule: Fv will act through the centroid of the effective volume causing the force. A • c.g. C B • b F v • We need the centroid of volume A-B-C. How do we obtain this centroid? Use the concept which is the basis of the centroid, the “first moment of an area.” Since: Arec = Aqc + AABC Mrec = Mqc + MABC MABC = Mrec - Mqc II-23 Note: We are taking moments about the left side of the figure, ie., point b. WHY? (The c.g. of the quarter circle is known to be at 4 R/ 3 π w.r.t. b.) xcg A = xrec Arec - xqc Aqc xcg {152 - π152/4} = 7.5152 - {415/3/π} π152/4 xcg = 11.65 ft { distance to rt. of b to the centroid } Q: Do we need a y location? Why? d. Calculate the moment about B needed for equilibrium. M B ∑ = 0 clockwise positive. MB +5F h + 15−xv ( )F v = 0 A • c.g. C B • b• F H F v MB M B + 5⋅7020W + 15 −11.65 ( )⋅3013W = 0 MB +35,100W +10,093.6W = 0 MB = −45,194W ft −lbf per unit width Why is the answer negative? (What did we assume for an initial direction of MB?) The hydrostatic forces will tend to roll the surface clockwise relative to B, thus a resisting moment that is counterclockwise is needed for static equilibrium. II-24 Always review your answer (all aspects: magnitude, direction, units, etc.) to determine if it makes sense relative to physically what you understand about the problem. Begin to think like an engineer. II-25 Buoyancy An important extension of the procedure for vertical forces on curved surfaces is that of the concept of buoyancy. The basic principle was discovered by Archimedes. It can be easily shown that (see text for detailed development) the buoyant force Fb is given by: Fb = ρ g Vb where Vb is the volume of the fluid displaced by the submerged body and ρ g is the specific weight of the fluid displaced. Patm V b Fb • c.g. Thus, the buoyant force equals the weight of the fluid displaced, which is equal to the product of the specific weight times the volume of fluid displaced. The location of the buoyant force is through a vertical line of action, directed upward, which acts through the centroid of the volume of fluid displaced. Review all text examples and material on buoyancy. II-26 Pressure Distribution in Rigid Body Motion All of the problems considered to this point were for static fluids. We will now consider an extension of our static fluid analysis to the case of rigid body motion, where the entire fluid mass moves and accelerates uniformly (as a rigid body). The container of fluid shown below is accelerated uniformly up and to the right as shown. From a previous analysis, the general equation governing fluid motion is ∇ P = ρ( g −a ) + µ ∇ 2 V For rigid body motion, there is no velocity gradient in the fluid, therefore µ∇ 2V = 0 The simplified equation can now be written as ∇ P = ρ( g −a ) = ρG where G = g −a ≡ the net acceleration vector acting on the fluid. II-27 This result is similar to the equation for the variation of pressure in a hydrostatic fluid. However, in the case of rigid body motion: ∇ P = f {fluid density & the net acceleration vector- G = g −a } ∇ P acts in the vector direction of G = g −a . Lines of constant pressure are perpendicular to G . The new orientation of the free surface will also be perpendicular to G . The equations governing the analysis for this class of problems are most easily developed from an acceleration diagram. Acceleration diagram: For the indicated geometry: θ = tan −1 ax g + az       dP ds = ρG where G = a x 2 + (g + a z ) 2 { } 1 2 and P2 −P 1 = ρG(s 2 −s1 ) Note: P2 −P 1 ≠ρ g z2 −z1 ( ) and s2 – s1 is not a vertical dimension a -a g G θ θ Free surface P 2 P 1 s ax az Note: s is the depth to a given point perpendicular to the free surface or its extension. s is aligned with G . II-28 In analyzing typical problems with rigid body motion: 1. Draw the acceleration diagram taking care to correctly indicate –a, g, and θ, the inclination angle of the free surface. 2. Using the previously developed equations, solve for G and θ. 3. If required, use geometry to determine s2 – s1 (the perpendicular distance from the free surface to a given point) and then the pressure at that point relative to the surface using P2 – P1 = ρ G (s2 – s1) . Key Point: Do not use ρg to calculate P2 – P1, use ρ G. Example 2.12 Given: A coffee mug, 6 cm x 6 cm square, 10 cm deep, contains 7 cm of coffee. The mug is accelerated to the right with ax = 7 m/s2 . Assuming rigid body motion and ρc = 1010 kg/m3, Determine: a. Will the coffee spill? b. Pg at “a & b”. c. Fnet on left wall. a. First draw schematic showing the original orientation and final orientation of the free surface. a b 7 cm 10 cm ∆ z ax θ 6 cm ρc = 1010 kg/m3 ax = 7m/s2 az = 0 g = 9.8907 m/s2 We now have a new free surface at an angle θ where θ = tan −1 ax g + az       θ = tan−1 7 9.807 =35.5° ∆z = 3 tan 35.5 = 2.14 cm a -a g G θ II-29 hmax = 7 + 2.14 = 9.14 cm < 10 cm ∴ Coffee will not spill. b. Pressure at “ a & b.” Pa = ρ G ∆ sa G = {a2 x + g2}1/2 = { 72 + 9.8072}1/2 G = 12.05 m/s2 ∆ sa = {7 + z} cos θ ∆ sa = 9.14 cm cos 35.5 = 7.44 cm Pa = 1010 kg/m312.05m/s20.0744 m Pa = 906 (kg m/s2)/m2 = 906 Pa Note: a P g y G g ρ ρ ρ ρ ≠ ≠ ≠ ≠ ≠ ≠ ≠ ≠ a b 7 cm 10 cm ∆ z ax θ 6 cm θ ∆ sa Q: How would you find the pressure at b, Pb? c. What is the force on the left wall? We have a plane surface, what is the rule? Find cg, Pcg, F = Pcg. A Vertical depth to cg is: zcg = 9.14/2 = 4.57 cm ∆scg = 4.57 cos 35.5 = 3.72 cm Pcg = ρ G ∆scg Pcg = 1010 kg/m312.05 m/s2 0.0372 m Pcg = 452.7 N/m2 F = Pcg A = 452.7 N/m20.09140.06m2 F = 2.48 N ← a b 7 cm 10 cm ∆ z ax θ 6 cm θ • cg ∆ scg II-30 What is the direction? Horizontal, perpendicular to the wall; i.e. Pressure always acts normal to a surface. Q: How would you find the force on the right wall?
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Ésta se utiliza para propósitos estadísticos en la web. | Sesión | Almacenamiento Local HTML | | com.adobe.reactor.core.visitorTracking.pagesViewed | Adobe Inc. | Contiene datos de la navegación del usuario, interacción y tiempo pasado en la web y en sus subpáginas. Estos datos se utilizan para optimizar la relevancia de los anuncios y con propósitos estadísticos. | Persistente | Almacenamiento Local HTML | | com.adobe.reactor.core.visitorTracking.sessionCount | Adobe Inc. | Contiene datos de la navegación del usuario, interacción y tiempo pasado en la web y en sus subpáginas. Estos datos se utilizan para optimizar la relevancia de los anuncios y con propósitos estadísticos. | Persistente | Almacenamiento Local HTML | | com.adobe.reactor.core.visitorTracking.trafficSource | Adobe Inc. | Utilizada por los propietarios de la web para identificar cómo el visitante ha accedido a la web. Esto se utiliza para propósitos estadísticos. | Sesión | Almacenamiento Local HTML | | ecademy_course | Adobe Inc. | Pendiente | Sesión | Almacenamiento Local HTML | | ecademy_track | Adobe Inc. | Pendiente | Sesión | Almacenamiento Local HTML | | gpc_journey_data | Adobe Inc. | Pendiente | Sesión | Almacenamiento Local HTML | | tgt:2146778465:b | Adobe Inc. | Pendiente | Persistente | Almacenamiento Local HTML | | tgt:2146778465:h | Adobe Inc. | Pendiente | Persistente | Almacenamiento Local HTML | | user_journey_completed | Adobe Inc. | Pendiente | Sesión | Almacenamiento Local HTML | | __adroll_shared | Adroll Group | Recoge datos del visitante a través de las webs – Estos datos se usan para hacer la publicidad más relevante. | 13 meses | Cookie HTTP | | receive-cookie-deprecation[x3] | Adroll Group Adroll Google | Recoge información del comportamiento del visitante múltiples webs. Esta información se usa en la web para optimizar la relevancia de la publicidad. | 13 meses | Cookie HTTP | | collect/ | LinkedIn | Recoge datos sobre el comportamiento y la interacción de los visitantes - Esto se utiliza para optimizar la web y hacer más relevante la publicidad en la misma. | Sesión | Píxel de Seguimiento | | tt_appInfo | Adobe Inc. | Utilizada por la red social TikTok para rastrear el uso de servicios incrustados. | Sesión | Almacenamiento Local HTML | | tt_pixel_session_index | Adobe Inc. | Utilizada por la red social TikTok para rastrear el uso de servicios incrustados. | Sesión | Almacenamiento Local HTML | | tt_sessionId | Adobe Inc. | Utilizada por la red social TikTok para rastrear el uso de servicios incrustados. | Sesión | Almacenamiento Local HTML | | sync | Bidswitch | Recoge datos sobre el comportamiento y la interacción de los visitantes - Esto se utiliza para optimizar la web y hacer más relevante la publicidad en la misma. | Sesión | Píxel de Seguimiento | | cee | capig.stape.org | Registra una identificación única que identifica el dispositivo del usuario durante las siguientes visitas. Utilizada para el seguimiento de la conversión y para medir la eficacia de los anuncios online. | 3 meses | Cookie HTTP | | CMID | Casale Media | Recopila datos relacionados con las visitas del usuario al sitio web, como el número de visitas, el tiempo medio pasado en el sitio web y qué páginas han sido cargadas, con el propósito de mostrar anuncios específicos. | 1 año | Cookie HTTP | | CMPRO | Casale Media | Recoge información del comportamiento del usuario en diferentes webs para mostrar publicidad más relevante - También le permite a la web limitar el número de veces que el usuario está expuesto a un mismo anuncio. | 3 meses | Cookie HTTP | | CMPS | Casale Media | Recopila datos relacionados con las visitas del usuario al sitio web, como el número de visitas, el tiempo medio pasado en el sitio web y qué páginas han sido cargadas, con el propósito de mostrar anuncios específicos. | 3 meses | Cookie HTTP | | rum | Casale Media | Recopila datos relacionados con las visitas del usuario al sitio web, como el número de visitas, el tiempo medio pasado en el sitio web y qué páginas han sido cargadas, con el propósito de mostrar anuncios específicos. | Sesión | Píxel de Seguimiento | | lastExternalReferrer | Adobe Inc. | Detecta cómo el usuario encontró la página web al registrar su última dirección URL. | Persistente | Almacenamiento Local HTML | | lastExternalReferrerTime | Adobe Inc. | Detecta cómo el usuario encontró la página web al registrar su última dirección URL. | Persistente | Almacenamiento Local HTML | | log/error | Adobe Inc. | Empleado para detectar y registrar errores potenciales de seguimiento. | Sesión | Píxel de Seguimiento | | topicsLastReferenceTime | Adobe Inc. | Recoge datos del visitante a través de las webs – Estos datos se usan para hacer la publicidad más relevante. | Persistente | Almacenamiento Local HTML | | __adroll | Adroll | Registra una identificación única que identifica el dispositivo de un usuario que vuelve. La identificación se utiliza para los anuncios específicos. | 13 meses | Cookie HTTP | | cm/#/out | Adroll | Recoge información del comportamiento del visitante múltiples webs. Esta información se usa en la web para optimizar la relevancia de la publicidad. | Sesión | Píxel de Seguimiento | | cm/g/in | Adroll | Utilizada para presentar al visitante contenido y publicidad relevante - El servicio está provisto por grupos de proveedores de publicidad externos, que facilitan ofertas en tiempo real a los anunciantes. | Sesión | Píxel de Seguimiento | | cm/mk/MQIH2GT23FAYXPRQCYQCDZ/in | Adroll | Utilizada para presentar al visitante contenido y publicidad relevante - El servicio está provisto por grupos de proveedores de publicidad externos, que facilitan ofertas en tiempo real a los anunciantes. | Sesión | Píxel de Seguimiento | | dp | Adobe Inc. | El administrador de audiencias de un sitio web configura esta cookie para determinar si se puede configurar en el navegador del visitante alguna cookie de terceros adicional. Las cookies de terceros sirve para reunir información o rastrear el comportamiento del visitante en múltiples sitios web. Las cookies de terceros son configuradas por una empresa o sitio web de terceros. | Sesión | Cookie HTTP | | demdex | Adobe Inc. | A través de una identificación única que se utiliza para el análisis semántico del contenido, se registra la navegación del usuario en el sitio web y se vincula con datos sin conexión de encuestas y registros similares para mostrar anuncios específicos. | 180 días | Cookie HTTP | | dpm | Adobe Inc. | Establece un identificar único para el visitante que permite a anunciantes externos (terceras partes) dirigirse al visitante con publicidad relevante. Este servicio combinado está provisto por centros de publicidad, que facilitan ofertas en tiempo real a los anunciantes. | 180 días | Cookie HTTP | | activity;register_conversion=#;#=# | Google | Pendiente | Sesión | Píxel de Seguimiento | | IDE | Google | Utilizada por Google DoubleClick para registrar e informar sobre las acciones del usuario en el sitio web tras visualizar o hacer clic en uno de los anuncios del anunciante con el propósito de medir la eficacia de un anuncio y presentar anuncios específicos para el usuario. | 400 días | Cookie HTTP | | test_cookie | Google | Utilizada para comprobar si el navegador del usuario admite cookies. | 1 día | Cookie HTTP | | everest_g_v2 | Adobe Inc. | Utilizada para anuncios específicos para documentar la eficacia de cada anuncio individual. | 1 año | Cookie HTTP | | everest_session_v2 | Adobe Inc. | Utilizada para anuncios específicos para documentar la eficacia de cada anuncio individual. | Sesión | Cookie HTTP | | privacy_sandbox/topics/registration/ | Adobe Inc. | Pendiente | Sesión | Píxel de Seguimiento | | gmp\conversion# | Google | Pendiente | Sesión | Píxel de Seguimiento | | NID | Google | Pendiente | 6 meses | Cookie HTTP | | pagead/1p-conversion/#/ | Google | Rastrea el índice de conversión entre el usuario y los banners publicitarios de la web – Esto sirve para optimizar la relevancia de los anuncios de la web. | Sesión | Píxel de Seguimiento | | pagead/1p-user-list/# | Google | Utilizada para rastrear si el visitante ha mostrado un interés específico em productos o eventos a través de múltiples webs y detectar como el visitante navega entre webs - Esto se utiliza para la medida de los esfuerzos publicitarios y facilitar la tasa de emisión entre sitios. | Sesión | Píxel de Seguimiento | | __adroll_fpc | Adroll Group | Utilizada para identificar al visitante a través de visitas y dispositivos. Esto permite a la web presentar al visitante publicidad relevante - Este servicio está provisto por centros de proveedores de publicidad externos, que facilitan ofertas en tiempo real a los anunciantes. | 1 año | Cookie HTTP | | _fbp | Meta Platforms, Inc. | Utilizada por Facebook para proporcionar una serie de productos publicitarios como pujas en tiempo real de terceros anunciantes. | 3 meses | Cookie HTTP | | _gcl_au | Google | Utilizada por Google AdSense para experimentar con la eficiencia publicitaria a través de las webs usando sus servicios. | 3 meses | Cookie HTTP | | _ie_cpn | Grundfos | Contiene datos del comportamiento del visitante y su interacción con la web. Esto se utiliza en contexto con el servicio de email marketing Marketo.com, que permite a la web dirigirse a los visitantes vía email. | Sesión | Cookie HTTP | | _ie_eml | Grundfos | Utilizada en relación con la función de caché en la web - La función de caché facilita la transferencia de datos entre Adobe DTM y Adobe Launch. | Sesión | Cookie HTTP | | _mkto_trk | Marketo | Contiene datos del comportamiento del visitante y su interacción con la web. Esto se utiliza en contexto con el servicio de email marketing Marketo.com, que permite a la web dirigirse a los visitantes vía email. | 2 años | Cookie HTTP | | _ttp[x2] | TikTok | Utilizada por la red social TikTok para rastrear el uso de servicios incrustados. | 1 año | Cookie HTTP | | mbox | Adobe Inc. | Esta cookie se utiliza para recoger información no personal del comportamiento del visitante y estadísticas no personales del visitante, que pueden ser utilizadas por una agencia de publicidad de terceros. | 2 años | Cookie HTTP | | trwsb.cpv | Grundfos | Utilizada en relación con la función de caché en la web - La función de caché facilita la transferencia de datos entre Adobe DTM y Adobe Launch. | Sesión | Cookie HTTP | | trwsb.sid | Grundfos | Se dirige a anuncios según la realización de un perfil de comportamiento y la ubicación geográfica. | Sesión | Cookie HTTP | | trwsb.stu | Grundfos | Contiene datos del comportamiento del visitante y su interacción con la web. Esto se utiliza en contexto con el servicio de email marketing Marketo.com, que permite a la web dirigirse a los visitantes vía email. | Sesión | Cookie HTTP | | trwv.crd | Grundfos | Contiene datos del comportamiento del visitante y su interacción con la web. Esto se utiliza en contexto con el servicio de email marketing Marketo.com, que permite a la web dirigirse a los visitantes vía email. | Sesión | Cookie HTTP | | trwv.cvd | Grundfos | Se dirige a anuncios según la realización de un perfil de comportamiento y la ubicación geográfica. | Sesión | Cookie HTTP | | trwv.eml | Grundfos | Contiene datos del comportamiento del visitante y su interacción con la web. Esto se utiliza en contexto con el servicio de email marketing Marketo.com, que permite a la web dirigirse a los visitantes vía email. | Sesión | Cookie HTTP | | trwv.lvd | Grundfos | Contiene datos del comportamiento del visitante y su interacción con la web. Esto se utiliza en contexto con el servicio de email marketing Marketo.com, que permite a la web dirigirse a los visitantes vía email. | Sesión | Cookie HTTP | | trwv.uid | Grundfos | Recoge datos de las preferencias y el comportamiento del usuario en la web - Esta información se utiliza para producir contenidos y publicidad más relevantes para un usuario concreto. | 2 años | Cookie HTTP | | trwv.vc | Grundfos | Contiene datos del comportamiento del visitante y su interacción con la web. Esto se utiliza en contexto con el servicio de email marketing Marketo.com, que permite a la web dirigirse a los visitantes vía email. | Sesión | Cookie HTTP | | bcookie | LinkedIn | Utilizada por el servicio de networking social LinkedIn para rastrear el uso de servicios incrustados. | 1 año | Cookie HTTP | | lidc | LinkedIn | Utilizada por el servicio de networking social LinkedIn para rastrear el uso de servicios incrustados. | 1 día | Cookie HTTP | | w/1.0/sd | Openx | Registra datos de visitantes como su dirección IP, la localización geográfica, y la interacción publicitaria. Esta información se utiliza para optimizar la publicidad en las webs que usan los servicios de OpenX.net. | Sesión | Píxel de Seguimiento | | __adroll_consent_params[x2] | Adroll Group | Recoge información del comportamiento del visitante múltiples webs. Esta información se usa en la web para optimizar la relevancia de la publicidad. | Sesión | Cookie HTTP | | __ar_v4[x2] | Adroll Group | Optimiza la visualización de anuncios según el movimiento del usuario combinado y diversas campañas del anunciante para mostrar anuncios al usuario. | 1 año | Cookie HTTP | | _te [x2] | Adroll Group | Registra una identificación única que identifica el dispositivo de un usuario que vuelve. La identificación se utiliza para los anuncios específicos. | Sesión | Cookie HTTP | | getSessionStorage[x2] | Grundfos | Permite el seguimiento de los mismos datos de sesión en varias pestañas y sesiones del navegador. | Persistente | Almacenamiento Local HTML | | AdServer/Pug | PubMatic | Establece una marca temporal cuando el visitante entra en la web. Ésta se utiliza para propósitos estadísticos en la web. | Sesión | Píxel de Seguimiento | | KRTBCOOKIE_# | PubMatic | Registra una identificación única que identifica el dispositivo del usuario al volver a visitar sitios web que usan la misma red publicitaria. La identificación se utiliza para permitir anuncios específicos. | 3 meses | Cookie HTTP | | adroll_flgs | Adroll Group | Utilizada para presentar al visitante contenido y publicidad relevante - El servicio está provisto por grupos de proveedores de publicidad externos, que facilitan ofertas en tiempo real a los anunciantes. | Sesión | Almacenamiento Local HTML | | _boomr_akamaiXhrRetry | s.go-mpulse.net | Recoge información de las preferencias y/o interacción del usuario con contenido de campañas web – Esto se usa en la plataforma de campaña CRM, utilizada para propietarios de webs para promover eventos o productos. | Persistente | Almacenamiento Local HTML | | 1/i/adsct[x2] | Adobe Inc. | Recoge datos sobre el comportamiento y la interacción de los visitantes - Esto se utiliza para optimizar la web y hacer más relevante la publicidad en la misma. | Sesión | Píxel de Seguimiento | | muc_ads | Twitter Inc. | Recoge datos sobre el comportamiento y la interacción de los visitantes - Esto se utiliza para optimizar la web y hacer más relevante la publicidad en la misma. | 400 días | Cookie HTTP | | idsync/ex/receive/check | Tapad | El administrador de audiencias configura esta cookie para determinar la hora y frecuencia de sincronización de datos del visitante. La sincronización de la cookie de datos sirve para sincronizar y reunir datos del visitante de diversos sitios web. | Sesión | Píxel de Seguimiento | | TapAd_3WAY_SYNCS | Tapad | Sincronización de datos entre redes de anunciantes. | 2 meses | Cookie HTTP | | TapAd_DID | Tapad | Utilizada para determinar qué tipo de dispositivos (smartphones, tablets, ordenadores, TV, etc.) utiliza un usuario. | 2 meses | Cookie HTTP | | TapAd_TS | Tapad | Utilizada para determinar qué tipo de dispositivos (smartphones, tablets, ordenadores, TV, etc.) utiliza un usuario. | 2 meses | Cookie HTTP | | guest_id | Twitter Inc. | Recopila datos relacionados con las visitas del usuario al sitio web, como el número de visitas, el tiempo medio pasado en el sitio web y qué páginas han sido cargadas, con el propósito de personalizar y mejorar el servicio de Twitter. | 400 días | Cookie HTTP | | guest_id_ads | Twitter Inc. | Recoge información del comportamiento del visitante múltiples webs. Esta información se usa en la web para optimizar la relevancia de la publicidad. | 400 días | Cookie HTTP | | guest_id_marketing | Twitter Inc. | Recoge información del comportamiento del visitante múltiples webs. Esta información se usa en la web para optimizar la relevancia de la publicidad. | 400 días | Cookie HTTP | | _gcl_ls | Google | Rastrea el índice de conversión entre el usuario y los banners publicitarios de la web – Esto sirve para optimizar la relevancia de los anuncios de la web. | Persistente | Almacenamiento Local HTML | | #-# | YouTube | Se usa para rastrear la interacción del usuario con el contenido integrado. | Sesión | Almacenamiento Local HTML | | iU5q-!O9@$ | YouTube | Registra una identificación única para mantener estadísticas de qué vídeos de YouTube ha visto el usuario. | Sesión | Almacenamiento Local HTML | | LAST_RESULT_ENTRY_KEY | YouTube | Se usa para rastrear la interacción del usuario con el contenido integrado. | Sesión | Cookie HTTP | | nextId | YouTube | Se usa para rastrear la interacción del usuario con el contenido integrado. | Sesión | Cookie HTTP | | requests | YouTube | Se usa para rastrear la interacción del usuario con el contenido integrado. | Sesión | Cookie HTTP | | TESTCOOKIESENABLED | YouTube | Se usa para rastrear la interacción del usuario con el contenido integrado. | 1 día | Cookie HTTP | | yt.innertube::nextId | YouTube | Registra una identificación única para mantener estadísticas de qué vídeos de YouTube ha visto el usuario. | Persistente | Almacenamiento Local HTML | | yt.innertube::requests | YouTube | Registra una identificación única para mantener estadísticas de qué vídeos de YouTube ha visto el usuario. | Persistente | Almacenamiento Local HTML | | ytidb::LAST_RESULT_ENTRY_KEY | YouTube | Se usa para rastrear la interacción del usuario con el contenido integrado. | Persistente | Almacenamiento Local HTML | | YtIdbMeta#databases | YouTube | Se usa para rastrear la interacción del usuario con el contenido integrado. | Persistente | IndexedDB | | yt-remote-cast-available | YouTube | Registra las preferencias del reproductor de vídeo del usuario al ver vídeos incrustados de YouTube | Sesión | Almacenamiento Local HTML | | yt-remote-cast-installed | YouTube | Registra las preferencias del reproductor de vídeo del usuario al ver vídeos incrustados de YouTube | Sesión | Almacenamiento Local HTML | | yt-remote-connected-devices | YouTube | Registra las preferencias del reproductor de vídeo del usuario al ver vídeos incrustados de YouTube | Persistente | Almacenamiento Local HTML | | yt-remote-device-id | YouTube | Registra las preferencias del reproductor de vídeo del usuario al ver vídeos incrustados de YouTube | Persistente | Almacenamiento Local HTML | | yt-remote-fast-check-period | YouTube | Registra las preferencias del reproductor de vídeo del usuario al ver vídeos incrustados de YouTube | Sesión | Almacenamiento Local HTML | | yt-remote-session-app | YouTube | Registra las preferencias del reproductor de vídeo del usuario al ver vídeos incrustados de YouTube | Sesión | Almacenamiento Local HTML | | yt-remote-session-name | YouTube | Registra las preferencias del reproductor de vídeo del usuario al ver vídeos incrustados de YouTube | Sesión | Almacenamiento Local HTML | | __Secure-ROLLOUT_TOKEN | YouTube | Pendiente | 180 días | Cookie HTTP | | __Secure-YEC | YouTube | Registra las preferencias del reproductor de vídeo del usuario al ver vídeos incrustados de YouTube | Sesión | Cookie HTTP | | __Secure-YNID | YouTube | Pendiente | 180 días | Cookie HTTP | | VISITOR_INFO1_LIVE | YouTube | Intenta calcular el ancho de banda del usuario en páginas con vídeos de YouTube integrados. | 180 días | Cookie HTTP | | YSC | YouTube | Registra una identificación única para mantener estadísticas de qué vídeos de YouTube ha visto el usuario. | Sesión | Cookie HTTP | [x] Preferencias + Las cookies pertenecientes a esta categoría se usan para recordar las preferencias del visitante (como sus elecciones en cuanto a idioma o ubicación). Ningún tercero puede acceder a estas cookies. | Nombre | Proveedor | Propósito | Duración máxima de almacenamiento | Tipo | --- --- | companyCode | Adobe Inc. | La cookie determina el idioma preferido y la configuración de país del visitante - Esto permite a la web mostrar contenido más relevante para esa región e idioma. | 7 días | Cookie HTTP | | kndctr_#_AdobeOrg_cluster | Adobe Inc. | Asigna un identificador específico al visitante - Esto permite a la web contabilizar el número de visitas de un usuario específico para análisis y estadísticas. | 1 día | Cookie HTTP | | adroll#adroll[x2] | Grundfos | Almacena el estado de consentimiento de cookies del usuario para el dominio actual | Persistente | IndexedDB | | sizingHero[x2] | Grundfos | Esta cookie permite que el sitio web recuerde y muestre la talla de ropa correcta. | 30 días | Cookie HTTP | | userEvents | Azure | Pendiente | Persistente | Almacenamiento Local HTML | | AuthSess | sway.cloud.microsoft | Esta cookie permite al visitante dejar comentarios sobre su uso de la web. | Sesión | Cookie HTTP | [x] Estadística + Las cookies pertenecientes a esta categoría se usan para analizar el tráfico de nuestras webs y realizar mejoras (por ejemplo, optimizar la ruta de navegación de las páginas más vistas por nuestros visitantes o personalizar el contenido de nuestras webs de acuerdo con tus intereses). Ningún tercero puede acceder a estas cookies, a excepción de las cookies de Google, que son accesibles para Google. En cuanto a las cookies de Google, actuamos como responsables del tratamiento de los datos en conjunto con Google para la recogida y transmisión de los datos a Google. Fuera de la UE/el EEE, pueden participar otros colaboradores en tales operaciones. | Nombre | Proveedor | Propósito | Duración máxima de almacenamiento | Tipo | --- --- | tgt:tlm:# | Adobe Inc. | Pendiente | Persistente | Almacenamiento Local HTML | | tgt:tlm:lower | Adobe Inc. | Recopila datos relacionados con las visitas del usuario al sitio web, como el número de visitas, el tiempo medio pasado en el sitio web y qué páginas han sido cargadas, con el propósito de generar informes para optimizar el contenido del sitio web. | Persistente | Almacenamiento Local HTML | | tgt:tlm:upper | Adobe Inc. | Recopila estadísticas sobre las visitas del usuario al sitio web, como el número de visitas, el tiempo medio pasado en el sitio web y qué páginas se han leído. | Persistente | Almacenamiento Local HTML | | image.aspx | Adobe Inc. | Registra datos estadísticos del comportamiento del visitante en la web. Esto se utiliza para análisis internos por el operador de la web. | Sesión | Píxel de Seguimiento | | hjSession# | Hotjar | Recopila estadísticas sobre las visitas del usuario al sitio web, como el número de visitas, el tiempo medio pasado en el sitio web y qué páginas se han leído. | 1 día | Cookie HTTP | | hjSessionUser# | Hotjar | Recopila estadísticas sobre las visitas del usuario al sitio web, como el número de visitas, el tiempo medio pasado en el sitio web y qué páginas se han leído. | 1 año | Cookie HTTP | | hjTLDTest | Hotjar | Registra datos estadísticos del comportamiento del visitante en la web. Esto se utiliza para análisis internos por el operador de la web. | Sesión | Cookie HTTP | | _tt_enable_cookie | TikTok | Utilizada por la red social TikTok para rastrear el uso de servicios incrustados. | 1 año | Cookie HTTP | | AMCV# | Adobe Inc. | Identificador único de usuario que reconoce al usuario cuando vuelve a visitar la página | 2 años | Cookie HTTP | | AMCVS_#AdobeOrg | Adobe Inc. | Pendiente | Sesión | Cookie HTTP | | com.adobe.alloy.getTld | Adobe Inc. | Registra datos del comportamiento del visitante en la web. Esto se utiliza para el análisis interno y la optimización de la web. | Sesión | Cookie HTTP | | kndctr_#_AdobeOrg_identity | Adobe Inc. | Registra datos estadísticos del comportamiento del visitante en la web. Esto se utiliza para análisis internos por el operador de la web. | 13 meses | Cookie HTTP | | nmstat | Siteimprove | Esta cookie tiene una cadena de ID en la sesión en curso. Contiene datos no personales sobre las subpáginas a las que accede el visitante. Esta información sirve para mejorar la experiencia del visitante. | 1000 días | Cookie HTTP | | s_cc | Adobe Inc. | Utilizada para comprobar si el navegador del usuario admite cookies. | Sesión | Cookie HTTP | | trwsa.sid | Grundfos | Recoge información de la dirección de IP del visitante, localización geográfica y navegación. Esta información se utiliza para la optimización interna y estadísticas para el operador de la web. | 1 día | Cookie HTTP | | hjActiveViewportIds | Hotjar | Esta cookie tiene una cadena de ID en la sesión en curso. Contiene datos no personales sobre las subpáginas a las que accede el visitante. Esta información sirve para mejorar la experiencia del visitante. | Persistente | Almacenamiento Local HTML | | hjViewportId | Hotjar | Guarda el tamaño de pantalla del usuario para ajustar el tamaño de las imágenes del sitio web. | Sesión | Almacenamiento Local HTML | | QSI_HistorySession[x2] | Qualtrics | Se utiliza para detectar si el usuario ha respondido a las encuestas presentes en la web. Estas encuestas sirven al propósito de optimización de la web. | Sesión | Cookie HTTP | | PugT | PubMatic | Utilizada para determinar el número de veces que las cookies han sido actualizadas en el navegador del visitante. Se usa para optimizar la eficiencia del servidor de la web. | 30 días | Cookie HTTP | | Q_INTER | Qualtrics | Recopila estadísticas sobre las visitas del usuario al sitio web, como el número de visitas, el tiempo medio pasado en el sitio web y qué páginas se han leído. | Persistente | Almacenamiento Local HTML | | QSI_ActionSetHistory | Qualtrics | Asigna un identificador específico al visitante - Esto permite a la web contabilizar el número de visitas de un usuario específico para análisis y estadísticas. | Sesión | Almacenamiento Local HTML | | QSI_HistorySession | Qualtrics | Esta cookie se usa para determinar cuando el visitante ha visitado por última vez las diferentes subpáginas de la web. | Sesión | Almacenamiento Local HTML | | browser_support | Azure | Utilizada en relación con la función de vista 3D en la web - Colecciona datos de la interacción del usuario con el servicio para propósitos estadísticos. | Persistente | Almacenamiento Local HTML | | b/ss/#/1/#/s# | Grundfos | Registra datos del comportamiento del visitante en la web. 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Productos relacionados Encuentra los productos Grundfos relacionados con este tema. ### MAGNA1 Una opción más completa y sencilla para un trabajo satisfactorio. La MAGNA1 presenta una interfaz intuitivo que está especialmente focalizado para la sustitución de circuladoras antiguas. La mejor opción para las necesidades de funcionalidad básica. Diseño compacto y fácil instalación Índice EEI promedio < 0,23 Bajo nivel de ruido Vista ### MAGNA3 La mejor opción cuando el control, la supervisión y el confort del sistema son claves. El sencillo manejo permite una configuración intuitiva y adaptada a cada tipo de instalación. Fácil integración en sistemas BMS mediante módulos CIM opcionales. Pantalla a color con infografías en 3D Índice EEI promedio < 0,19 Entrada analógica configurables Vista ### NB Bombas con acoplamiento cerrado de aspiración final conforme a EN 733 Diseño de extracción trasera que facilita el mantenimiento de la bomba Sistema hidráulico optimizado en carcasa e impulsor para un caudal de líquido sin obstáculos Cierre de junta tórica entre carcasa de bomba y cubierta Vista ### NBE Bombas con acoplamiento cerrado y motor MGE de aspiración final conforme a EN 733 Diseño de extracción trasera que facilita el mantenimiento de la bomba Sistema hidráulico optimizado en carcasa e impulsor para un caudal de líquido sin obstáculos Cierre de junta tórica entre carcasa de bomba y cubierta Vista ### NBG Bombas de acoplamiento cerrado conforme a ISO 2858 Diseño de extracción trasera que facilita el mantenimiento de la bomba Sistema hidráulico optimizado en carcasa e impulsor para un caudal de líquido sin obstáculos Cierre de junta tórica entre carcasa de bomba y cubierta Vista ### NBGE Bombas de acoplamiento cerrado con motor MGE conforme a ISO 2858 Diseño de extracción trasera que facilita el mantenimiento de la bomba Sistema hidráulico optimizado en carcasa e impulsor para un caudal de líquido sin obstáculos Cierre de junta tórica entre carcasa de bomba y cubierta Vista ### TPE Serie 2000, TPE3 Bomba en línea simple de rotor seco, cierre mecánico según EN 12756, motor según estándar IEC y DIN con variador de frecuencia integrado con eficiencias de hasta IE5 y sensores de presión diferencial incorporados. Adaptación automática a los requisitos del sistema Sistema hidráulico optimizado para una alta eficiencia Diseño con parte superior extraíble para un desmontaje sencillo durante el mantenimiento Vista ### TPE Series 1000, TPE2 Bomba en línea simple de rotor seco, cierre mecánico según EN 12756, motor según estándar IEC y DIN con variador de frecuencia integrado y hasta clases de eficiencia IE5 Adaptación automática a los requisitos del sistema Sistema hidráulico optimizado para una alta eficiencia Diseño con parte superior extraíble para un desmontaje sencillo durante el mantenimiento Vista Aplicaciones relacionadas Encuentra las aplicaciones de Grundfos relacionadas con este tema. ### Aire acondicionado en edificación comercial ### Enfriamiento ### Refrigeración urbana Soluciones de bombeo AplicacionesSoluciones A-ZCategorías Soporte SoportePreguntas frecuentesGuías prácticasContactar con Service Formación Grundfos EcademyInvestigación y desarrolloWebinars Sobre nosotros Nuestra compañíaEmpleoNoticiasCasos Contacto Dónde comprar Enlaces rápidos Grundfos.comMyGrundfosGrundfos EcademyProduct CenterReciclaje de nuestros productos (en inglés)Centro de marketing Bombas GRUNDFOS España S.A. 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https://sparkedinnovations.net/Going%20Up%20(The%20Wild%20Blue%20Yonder%20Part%202).pdf
1 "You can never tell where the winds will blow you, what fantastic. good fortune they can lead you to. Long Live Balloons!" Design your balloon! Inspire students with a video showing a plethora of designs from rockets to elephants, monster trucks, and even the Old Lady Who Lived in a Shoe, such as this… If we’re setting off like Professor Sherman, we need our own hot air balloons! A hot air balloon is an incredibly personal representation of style, personality, image, performance and more. Deciding what your own balloon should become can involve limitless design possibilities - it can be as simple or complex as desired. Not sure where to start? Not surprising! But that’s all right, it all begins with the balloon. Going Up? The Wild Blue Yonder Image Credit: Dave Pimentel Find this image and more of his amazing art at All 2 Choosing the Shape To begin with, what is the preferred envelope style that you might want? Most hot air balloons are designed using a shape which minimizes the stress on the fabric. In the early days of upper atmosphere research using helium balloons it was common for the balloons to burst when they neared their intended float altitude. These failures were traced to the shape of the balloons loading the thin material until it 3 ruptured. The US military commissioned a company called General Mills to devise a shape which minimized the fabric stress. A design that is similar in appearance to one of the mainstream manufacturers' models might serve as a reasonable starting point - but students should feel free to customize to their heart's extent- even if it's very unique, a little oddball, or even close to full-blown crazy. Check out the 3-gore design in the chart, or these pictures of real-life balloons! You should consider the number of gores you'd probably like (8, 12, 16, 18 & 24 are common numbers of gores), the shape of the gore (bulbous, semi-smooth, or flat to any degree), and the orientation of the different panels comprising each gore (horizontal, vertical or diagonal). Quickly draft a 3D rendering of what the balloon will look like based on your design preferences. Tip: For help with how to draw a basic hot air balloon go to and watch: 4 12-gore Half-Gore construction 16-gore Flat Panel construction 20-gore Half-Gore construction 3-gore Extremely Bulbous! 8-gore Bulbous 16-gore Semi-Bulbous 24-gore Semi-Bulbous 18-gore Zig-Zag Bulbous 18-gore Zig-Zag Spiral Bulbous 5 Rising High! Choosing Your Colors Art and design influence so much of the world around us, and such creativity can lead to a career. Before that balloon was born, an artist had to make all those decisions about what it looked like. Any type of artwork or logos can be integrated into the design of your envelope, whether inlaid or overlaid (appliqué). Inlaid cuts of any angle, curvature or complexity can be added to change fabric colors within any panel at any location in the envelope. Painting directly on the fabric is sometimes useful for smaller accents, and is a technique often used to add detail to special shapes Each student must create three separate balloon designs and complete one original drawing of their final balloon concept, using Sharpies, colored pencils, crayons, and/or colored markers; and an autobiographical worksheet Discuss with students the difference between coloring a balloon and designing one. Have them use symbolism and imagery to tell something important about themselves, “There are a trillion things that make you you!” Be quick to praise a storytelling burst of creativity, and just as quick to have students push their ideas further, and not be satisfied with just a pretty color scheme. Show students examples of color scheme swatches from books or cards. The developed pattern dimensions for an inlaid parachute design Sewing the center panel of the parachute ...And the finished product! 6 How long does a balloon last? Depending on the care it is given, a balloon envelope may last 500 or more flying hours. Considering that most sport pilots fly from 35-50 hours a year, balloons can last a long time. The envelope wears out before the basket and burners. With proper maintenance, a basket will last indefinitely. Envelope Sizing & Intended Use How big was the Professor’s balloon? Why did he choose that size? (p. 40 in the novel). The most important consideration - apart from what your balloon will look like - is what you plan to do with it. It makes no sense to decide on a 56,000 cubic foot envelope and then later realize you wished you could be able to fly pilot plus two heavy passengers at all times of the year and temperature ranges. In the warmer months, that objective would be difficult to accomplish in a balloon that small without overheating. Have students put thought into at least the following considerations:  What is the maximum number of passengers I plan to carry at all times in addition to the pilot? Consider 200-225 pounds per occupant a reasonable average, which includes clothing and the normal personal cargo.  What temperature extremes will be expected in the typical area of operations? Of most importance is the maximum average temperature in the summer.  How high would I normally want to fly with a full payload?  How long do I hope to be able to fly for with a full payload? Simply adding more fuel tanks is not necessarily as beneficial to your flight duration as increased envelope size can be. A larger envelope flying at cooler temperatures will use less fuel than a smaller envelope with the same payload or weight.  How heavy will my envelope and basket be? The type of fabric and materials chosen can have a significant impact on the system weight. 7 Certainly, the larger an envelope is for a given payload (amount of weight), the cooler it will generally fly, the longer the flight duration will be, and the longer the fabric itself will last. It's always better to have a surplus of lift rather than risking overheating the envelope when it gets warm out. Why? An envelope flown at 225°F or less will tend to last longer than one flown constantly close to 250°F. 200-220°F is a good target range to stay within for the best combination of longevity and flight handling characteristics. Fuel economy increases at lower internal temperatures. Temperature and flight altitude are the two most important factors in determining payload for any given envelope volume. In general, a good rule of thumb to follow is that 1000 cubic feet of hot air will create 16 pounds of lift. In the real world, this figure will certainly differ due to ambient and envelope internal temperature as well as altitude above sea level, but it's a conservative rule of thumb to estimate the gross lift of any given envelope volume at average internal flying temperatures. This allows for a good margin of free lift available beyond the calculated gross weight, which may be important in situations where the balloon is flying hot, heavily loaded, and in situations where a rapid ascent may be required to clear obstacles. Most balloon manufacturers calculate the gross weight of their balloon envelopes based on a figure of 16 pounds per 1000 cubic feet, but some have chosen to certificate their balloons at a higher gross weight using a figure of 20 pounds per 1000 cubic feet. Here is a table of common envelope volumes for a given number of occupants: Capacity Volume Gross Lift Range Pilot Only 21-42,000 350-800 lbs. 2 People 42-65,000 650-1300 lbs. 3 People 65-77,000 1000-1500 lbs. 4 People 77-105,000 1200-2100 lbs. 4-6 People 105-120,000 1650-2400 lbs. 5-7 people 120-150,000 1900-3000 lbs. Types of Fabric 8 The amount of fabric in a balloon is obviously dependent on the size of the balloon a typical four passenger balloon has over two thousand square yards of fabric. Typical Balloon Pricing Like cars and boats, new balloons can vary in size and amenities. You can start with a smaller sport model for around $14,000. These balloons typically carry a pilot and maybe one additional person. The larger balloons that can carry two or ten passengers in addition to the pilot will range between $15,000 and $100,000. There are many good used balloons on the market that vary in price due to size, age and flight hours. The price for a custom hot air balloon will vary considerably depending on the design of the envelope, the many options involved, and the materials chosen. Because of this, all custom-designed balloons are quoted on a singular basis whether fully-constructed, in kit form, or as a set of easy-to-follow construction plans. The style of envelope design, number of gores, options such as turning vents, Nomex, skirt/scoop, inlaid or overlaid artwork, etc. factors greatly in the determination of the cost. 9 A rough estimate of a new envelope constructed of first-quality 1.3oz silicone coated ripstop nylon fabric would be in the ballpark of $9,000 complete. A 90 would be right around $10,000. For a precut "kit" build option, the price would be around $6,000, cut out and ready to sew with detailed construction plans included. Options such as fancier fabric or artwork would be extra and added "at cost" to the rough base prices above. Design your Wicker Basket Gondola. What was Professor’s reasoning for his basket? Who was his inspiration? (page 41 in the novel) What ideas was his basket built with? Remember Professor Sherman had two floors, an open air attic for storage and a porch wrapping around his basket. What was his basket made of? Why did he choose those materials? P. 41. Students don’t have to get this elaborate, though it is rather fun, but they need to make room for their passengers, supplies and equipment. Also keep in mind comfort, such as beds, chairs and entertainment. How will they handle laundry? Going to the bathroom? Taking a shower or a bath? (They could use wipes!) What will the furniture be made out of? What did the professor make his out of? Why? P. 43-44 in the novel. Everything was chosen for the idea of… Using graph paper provided by your instructor, create a front view of your gondola showing any rooms, stairs, furniture or other designs you decide to add. Now calculate the square footage of the walls, floor and ceiling (if any) of your basket. This figure will be needed for later, so be sure have students calculate this accurately. They will need to record in their journal what balloon they have decided to use in your travel with an explanation of why they chose this balloon versus the other balloon options. Also include the drawing of their basket gondola with the square footage calculations. What did the professor use for ballast? What will the students use for ballast? P. 43 in the novel Samples of Artist Renderings of imaginary Hot Air Balloon Designs follow. 10 Copyright 2013 Martin Bergquist. All Rights Reserved. Sample image is for non-commercial educational use only. 11 Copyright 2013 Martin Bergquist. All Rights Reserved. Sample image is for non-commercial educational use only. 12 Way Above Water Basket Weaving Tutorial found at: Craft Passion Copyright 2013. All Rights Reserved. Have students become apprentice basket weavers before they leave for Krakatoa! Before we can be trusted to construct large wicker baskets for hot air balloons we must make miniature baskets and become seasoned weavers! We may need those skills later to escape the volcanic island! This is a wonderful technique to make basket. With this method and concept, basically students can build any shape and size basket that they want. Here, it is shown in three shapes: triangle, round and square, but you can create a rectangle, pentagon, hexagon, heptagon & even octagon with just a little tweak in the design. The height can be modified to suit their preference. One thing they have to take note is that the sum of the “fingers” on the frame must be in odd number. Have students take a closer look at the template, especially the square basket, they should notice that one of the sides has 1 extra “finger” than the other 3 sides. To make faux wicker baskets, like hot air balloon baskets, use jute twine and they come out very sturdy and rigid! Note: Students can use any material to weave the frame of the basket, as long as they are “weave-able”, not too bulky and in long soft strips, eg: twine, yarn, fabric strip, ribbon, rope, glossy paper strip, plastic, raffia etc… Material: 1. Thick Cardboard or paperboard, non-corrugated (around 1/6″ thick). 2. Felt for bottom and base (optional) 3. Weaving material, example: twine Tools: 1. Sharp scissors (able to cut thick cardboard) 2. Hole puncher 3. Tapestry needle (#13) 4. Clothes pin 5. Craft / Tacky Glue 6. Pen or pencil 7. Ruler 13 Download the basket weaving template (templates for a triangle, round and square) in pdf format or use the following page. Print it and enlarge 200% to get the size in the example. Have students trace their chosen design on thick cardboard. After tracing the template on the thick cardboard, use a pair of strong and sharp scissors to cut the template out. Punch the holes with a paper hole-puncher. Trace base on felt (if desired), cut 2 pieces. Get the rest of the tools & materials ready. Option: Brush a layer of tacky glue on the bottom of the basket, press & stick the felt onto it. Repeat the same to adhere the felt to the base. Snip the edge of the felt follows the grooves between the fingers. 1. Being to weave over and under around the fingers of the frame. 2. Push the twine into the grooves of the felt that you have just snipped. 3. Keep on weaving over and under the fingers until the end of first round. 4. When begin the second round, you will notice that you are weaving in the reverse side of the first round. Both sides should be fully wrapped after you finish the second round. 14 5. Continue weaving and push the twine down to make them closer to each other. This will avoid gaps in between and will also hide the cardboard away. As you go higher, the frame will stand up vertically and form the side of the basket. 6. Continue until you just over the level of the holes. Allow a long length of twine for rim stitching later. 7. Put on clothes pin and allow some room for rim making. 15 8. To make the rim, insert another type of twine (ex. white cotton twine) through the space between the clothes pin by using a tapestry needle. Since the length is quite long, divide the twine in half and work both ends from middle. 9. Keep going round by round. Be careful on the corners and arrange them neatly. 10. Lay the twine until it fully covers the top portion of the basket. 11. Secure the rim by stitching the original twine through the holes. 16 12. Stitch again from the opposite direction to form crisscross pattern. 13. When reaching the corner, pierce through the gap and make another stitch there. 17 14. Hide all ends into the braid to neaten up the basket. The basket is ready to be used. 18 19 Up, Up, and Away? When thinking about balloons, and the Krakatoan escape vessel with all those balloons, we might think of the Pixar animation movie UP! So, what if we wanted to use helium balloons to travel? Could we really do it? What is the lifting power of helium? For the sake of time and a more detailed explanation, here’s a simple data table that will give you an idea of what it would take to lift (only) a person using a helium balloon. We’re assuming standard temperatures and pressures and no significant overpressure. For reference purposes, you should know that a large tank of helium – the kind you find at a grocery store or party shop, holds approximately 250 cubic feet of helium. Based on the calculations below, you can see that 250 cubic feet will lift roughly 17 pounds. In Steve Spangler’s test he found that 250 cubic feet lifted about 8 pounds of potatoes, plus the string and the weight of the 45 large balloons (2.3 pounds). All in total, they lifted just over 10 pounds. Why didn’t they lift the calculated weight of 17 pounds? As you can imagine, we’re working with lots of assumptions. The company supplying the helium told us that the larger cylinder holds about 250 cubic feet. Each red balloon held approximately 5 cubic feet of helium, and we were able to inflate just 45 balloons (that’s 225 cubic feet of gas). To top everything off, we lost 5 of our balloons during the test (a few pops and 3 into the sky). These are the results based on an impromptu science experiment with a single tank of helium. Do you get different results? Materials:  Potatoes  Bag  String  Helium  Balloons  Optional: Chair Try It! When the Professor got back to the United States, he had a hard time walking, what did the people do for him? They made him a floating chair using helium balloons. (p. 26-28) How many balloons would you need to make a chair (with the Professor sitting in it) float? What happened to the chair? (p. 30-31) 20 What about the little boy who tied the helium balloon around his waist in the story? What was the difference between his experience and his little brother’s? (pages 21-22) Can students make a sack of potatoes float, first? And then estimate how many balloons it would take to lift the Professor? Have students Record and calculate the results of testing their hypotheses. 21 Going UP? When 78-year-old retiree Carl Frederickson's house takes off into the air aided by the help of hundreds of helium balloons in Up!, viewers saw it is a heart-warming moment of pure fiction. But for some people, it became more than that. So, the conceit of the Disney/Pixar cartoon epic, Up, is that an old guy’s house gets attached to a bunch of helium balloons which lift it up out of the city and on a wonderful adventure. That got Wired Science thinking: Could that actually work? And if so, how many balloons would you need? They called Wolfe House Movers, which specializes in moving old structures and had Kendal Siegrist, a manager, take a look at the images from the movie to see how much the house might weigh. “A building like that, you’d figure right around 100,000 pounds,” Siegrist said. Then they did some calculations. Air weighs about 0.078 pounds per cubic foot; helium weighs just 0.011 pounds per cubic foot. A helium balloon experiences a buoyant upward force that is equal to the air it displaces minus its own weight, or 0.067 pounds per cubic foot of helium balloon. One more simple calculation — 100,000 pounds divided by 0.067 pounds per cubic foot — and you’ve got that it would take 1,492,537 cubic feet of helium to lift the house. Of course, you’d need some more balloons to keep getting it higher, but that’s the minimum. Now, let’s assume you’ve got a bunch of spherical balloons three feet in diameter. They’ve got a volume in 14.1 cubic feet, so you’d need 105,854 of them filled with helium to lift the house. Eyeballing the cluster of balloons above the house in Up, let’s say on average, it is 40 balloons across and deep and 70 balloons tall. Do the math and there could be 112,000 balloons in there. Cluster Ballooning fans actually do this sort of thing, but with people in harnesses, not enormous houses, and they generally use a lot less balloons. They tend to use bigger balloons, say, six feet in diameter. You’d only need 13,208 of those. But even if you could get the balloons and one very strong cable, could a house be pulled from the top like that? “If you go try picking it up, depending on what you’re doing, you can,” Siegrist said, “but for the most part, you want the house to bear the weight on its foundation.” The way real, professional house movers like to do it is to get into the house’s basement and lift from below. 22 And some engineers…went out and built it! For the show "How Hard Can It Be?" on the National Geographic Channel, engineers constructed a basic house structure and lifted it into the air for more than an hour by 300 weather balloons. It made headlines around the world a real-life UP! balloon house soaring more than 10,000 feet in the air. Now, see the full story behind this spectacular flight in How Hard Can It Be? Meet Vin, Paul and Eric, the three ultra-ambitious hosts who came up with the crazy idea to fly a balloon house. From concept to execution, we'll see how the guys mixed off-the-shelf technology and unconventional experiments to bring animation to life. Watch it happen at : or at Visit for pictures from the amazing recreation of the flying house from Disney/Pixar's "Up" animated film. If you’re wondering more about the science of helium… Helium is mined, or more exactly drilled for. In the Oklahoma and Texas panhandles are natural gas wells that contain up to 4% or more helium. This natural resource is very rare. The gas field must be encased in radioactive rock or no helium is produced. The alpha particle decay in the surrounding radioactive rock over millions and millions of years creates the helium. An alpha particle is just a helium nucleus. When it slows down and regains its two electrons, it becomes a helium molecule. Thus the radioactive rock makes helium, one molecule at a time, to accumulate in the same pocket as the natural gas. 23 In the 1930′s Germany asked the USA many times for helium for its Zepplins (large passenger carrying blimps). The US was concerned that helium had other military uses and horded it as a strategic material. For this reason, the Hindenburg was still lofted with explosive hydrogen gas on its last disastrous flight, instead of being converted to helium as Germany had been trying to do for years. Helium is a natural byproduct of the liquefaction of the natural gas for pipeline shipment from these special gas fields. Helium liquefies at a much lower temperature than natural gas, close to absolute zero (4 degrees Kelvin). The volume left over after liquefaction is mostly helium ready to be stripped off and sold to the US Government Bureau of Mines. Few people seem concerned that this is a non-renewable and expendable natural resource, tied to very few gas wells in the world. According to the National Public Radio (NPR) report by Ailsa Chang, there's a global shortage of refined helium, and it could get worse if the federal government doesn't stay in the business of selling helium. The Senate is considering legislation to prevent a global helium shortage from worsening in October of 2013. That's when one huge supply of helium in the U.S. is set to terminate. The House overwhelmingly passed its own bill last month to keep the Federal Helium Program going. That was a relief to industries that can't get along without helium. The gas is used in MRI machines, semiconductors, aerospace equipment, lasers and of course balloons. To understand how we got here, we need to go back to nearly a century ago to World War I. Germany started building huge inflatable aircraft, and to keep up, the U.S. started stockpiling helium. That federal helium reserve is located outside Amarillo, Texas. Sam Burton of the Bureau of Land Management helps manage the supply. Burton says "he lives and breathes helium," adding that he's a 24 "total helium geek." Burton says there are now 10 billion cubic feet of the gas stored in this federal reservoir — enough to fill about 50,000 Goodyear blimps. And it's all kept under a wide-open prairie dotted with coyotes and jack rabbits. "Imagine a layer cake being several thousand feet thick, layers of rock several thousand feet thick, you'd get an idea of how the gas has been stored in one particular layer," Burton explains. Over the decades, private companies learned how to extract helium too. But they weren't extracting that much of it, partly because the government was selling helium so cheaply it didn’t seem worth the effort. Then in 1996, Congress decided it was time to get the federal government out of the helium business so it wouldn't compete with private industry. Congress passed a law that would effectively end the helium program this October. The problem is, private companies haven't caught up with demand, and a big hole would be left in the market if Washington suddenly cut off supply as scheduled. Besides a Balloon what else do you need to go Flying?  A “chase vehicle is required to transport all but the tiniest balloons, Balloons travel with the wind and can cover considerable distance depending on wind speed and flight duration.  A gas powered fan is required to cold inflate the envelope.  Toy balloons and a helium tank are required unless you always fly with other pilots who have them. Pibals (pilot balloons) are critical for pre-flight studying the winds aloft.  A quality hand held compass to read Pibals is important. If you really want to be accurate a stopwatch, and an inclinometer, will allow you to determine the exact wind speed, and direction, at specific altitudes.  Helmets are a good Idea and are required equipment for some manufactures  Gloves to minimize the transfer of skin oils during envelope handling and to prevent rope burn from handling lines are a good idea. Pilot gloves that have long cuffs and offer protection from open flame and raw liquid propane frost burns are smarter still.  A Tie down line that has a simple reliable release under considerable tension, yet can restrain a balloon with the equivalent of over 5000 square feet of sail area, is a must.  It’s a good Idea to have an aviation Radio particularly if you fly in the vicinity of an airport. Some airspace is only accessible with a radio.  Ground communication between balloon and chase, including radios and cell phones are suggested both from a safety prospective and a “where are you” prospective.  Some Balloonist invest in tether systems, which allow a balloon to be flown up and down while tied to the ground. The maximum height and wind tolerance of these systems vary considerably. As does the associated price.  If a pilot desires to fly in the winter, special equipment (tank heaters or nitrogen fuel pressurization) is required to insure adequate fuel pressure. 25  Depending on where you fly a two wheel cart in the chase vehicle might be the difference between having willing crew, and crew who need to wash their hair when it’s flyable. At Sky Sail Balloons we employ electric four wheel cart, the ultimate crew assist device.  One other accessory that is nearly universally hated by crew but is sometimes a necessary evil is a tarp to lay out the envelope on. Dusty, damp, muddy or snowy landing fields can really take their toll on balloon fabric.  Crew preference also may dictate having an envelope “squeezer” to milk the air out of the envelope after flying. Like kitchen gadgets for cooks there are any number of other contrivances to help minimize labor and keep things orderly. In Case of Emergency… Challenges: What will we do in an emergency? What’s our plan? What things did the Professor do to address his in-case-of – emergency needs? P. 44 in the novel. What did he do during his real emergency? Pages 55-60 in the novel. What did he find out from Mr. F on page 72 of the novel? If the bird hadn’t pecked, what would Mr. F have done? Why would he have done that? Like presidential candidates and stock-car racers, adventure balloonists tend toward the optimistic. This is not a matter of predilection so much as necessity, since, as we’ve learned, the overwhelming majority of flights result in two outcomes: ditch or splosh. Even if everything goes right--which experienced pilots acknowledge almost never happens (look at what happened to the Professor in our story!)--the technical difficulty of searching out the right winds, staying at the proper altitude, and dealing with variances in temperature and weather make any flight, but especially a long distance flight an immensely problematic proposition. Many balloonists, not just imaginary ones, have run into some serious problems. Take Atlantic crossings, for instance: 14 sploshes (splash downs into water) and five deaths before Maxie Anderson, Ben Abruzzo, and Larry Newman's Double Eagle II finally made it across in 1978. Look at trans-American attempts: four ditches before John Shoecraft and Frederick Gorrell landed Super Chicken III on the Georgia coast in 1981. It's not simply that you crash. It's the way you crash. Accidents almost never happen suddenly. They unfold slowly, with precision and elephantine (lumbering) grace. It is significant that balloonist vocabulary lacks the term "crash landing." It's redundant, when you land, you crash. One of the dangers is called icing, for instance: You drift into super-cooled clouds, the envelope is covered by a thick rime of ice whose weight pulls the balloon earthward, toward warmer air in which the ice melts, drenching the occupants and dumping the weight, launching the balloon upward like a jack-in-the-box, only to be covered by more ice, only to sink again...boing, boing, ditch, splosh. 26 Or there’s the danger of the tiniest pinhole leak, which can bring a balloon down in three or four days. Or the trickery of swirling storm systems, which suck a balloon into a slow whirlpool, spinning irretrievably toward a storm. Or lightning, which can toss balloonists out of the sky wholesale--five killed in a single 1923 race. Or simply altitude: Drop too much ballast (weight) to fly high, and there's nothing left to slow your descent, as happened to the Soviet balloon Osoaviakhim in its 1934 return from a record altitude of 72,178 feet. As the three-man capsule dropped toward earth, buffeted by turbulence that eliminated any possibility of parachuting out, Soviet ground control received a final radio broadcast. "The bright sunlight... The gondola... Beautiful sky... The ground... This... The sky... The balloon... It..." The balloon finally ripped under the strain. All three crew members died. There are mechanical difficulties that we might have to face. Trying to avoid crossing the East German border in a 1983 race, Maxie Anderson and Don Ida were forced to ditch in the Bavarian Alps. Touching down, they flipped the switch that fired the explosive bolts to release the now-useless envelope from the gondola and safely release the basket from the balloon, and nothing happened. The switch malfunctioned, the wind gusted catching the balloon, and suddenly they were up in the air again, floating high above the ground. Suddenly the bolts fired (broke) and the basket dropped. Neither survived the fall. Then there is the more delicate wiring of international relations. 27 September 1995, after then end of the Cold War, while participating in the Gordon Bennett Cup long-distance race, American balloonists John Stuart-Jervis and Alan Fraenckel were shot down and killed when their drifting balloon came too close to a missile launch site in Belarus. The mild stir that followed ended with the usual diplomatic serve-and-volley: The United States expressed its sincere outrage and the Belarussians pledged a "full investigation." Translation: Sorry guys, but what do you expect, playing kick-the-can above the trigger-happy countries below? All of these problems are symptoms of the real hazard of ballooning: the wind. Balloons are blown into clouds of ice crystals, across unfriendly borders, and into storms. So, what’s our plan to keep safe and avoid trouble? What kind of emergency gear do we need to make sure we take? Riding the Wind But the wind is also the boon of ballooning. Balloons are perfectly enmeshed in the breeze, not holding steady against it, but moving along at exactly the same speed. In fact, because balloons travel at the speed of the wind, passengers don’t actually feel any breeze at all! Even if you’re zipping along, even roaring along at 200 mph in the jet stream the ride feels surprisingly safe and serene. Long-distance balloonists seek control of the wind through a constant feed of information from meteorological forecasts about the winds, which have different names depending on where you are. The Arabians have the nafhat, there is the beshabar of the Caucasus, and the Samiel blowing from Turkey; North Africa's solano, California's Santa Ana, and the sirocco of the Sahara, which can blow such quantities of red sand to Europe that rains of blood were reported in Portugal and Spain in 1901. Winds are part of legends and superstitions around the world. As late as the 1920s, sailors of 28 the Shetland Islands purchased benevolent winds from old women for a sixpence. The Payaguas of South America beat the air with their fists to frighten oncoming storms, and Herodotus tells of a Saharan army that attempted to halt the simoom, or poison wind, by donning its battle gear and marching directly into it. Though the aeronauts are able to seek out favorable winds by fine-tuning their altitude, and though they can master the Byzantine mathematics of trajectory plotting, their control over their destiny remains, to put it kindly, limited. Once aloft, other forces take control. They travel by the whim of the winds. On a long trip like ours, or the Professor’s, it really does come down to luck. But, If Pecos Bill could ride the winds , why not us? 29 Sources and Resources m
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https://www.scirp.org/html/11-3700342_38785.htm
Modeling a General Equation for Pool Boiling Heat Transfer Home Journals Books Conferences News About Us Jobs Advances in Chemical Engineering and Science Vol.3 No.4(2013), Article ID:38785,10 pages DOI:10.4236/aces.2013.34037 Modeling a General Equation for Pool Boiling Heat Transfer Mohammed Salah Hameed 1, Abdul Rahman Khan 2, A. A. Mahdi 3 ●Abstract ●Full-Text PDF ●Full-Text HTML ●Full-Text ePUB ●Linked References ●How to Cite this Article 1 Chemical Engineering Department, Higher Colleges of Technology, Abu Dhabi, United Arab Emirates 2 Department of Chemical Engineering, Kuwait University, Kuwait City, Kuwait 3 Mechanical Engineering Department, University of Technology, Baghdad, Iraq Email: mhameed@hct.ac.ae Copyright © 2013 Mohammed Salah Hameed et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Received July 26, 2013; revised August 26, 2013; accepted September 5, 2013 Keywords: Pool Boiling; Nucleate Boiling; Linear and Non-Linear Technique Methods; Heat Transfer ABSTRACT It is recognized that the nucleate pool boiling data available in literature are mainly related to four known correlations, each differs from the other by a varying magnitude of constant coefficients, depending on restrictive experimental conditions. The present work is concerned in developing an empirically generalized correlation, which covers the entire range of nucleate boiling with a minimum possible deviation from experimental data. The least squares multiple regression technique is used to evaluate the best coefficient value used in the correlations. An empirical correlation that fits a broader scope of available data has been developed by a non-linear solution technique leading to the following equation: where the coefficients R 1 and R 3 both represent the effect of surface-liquid combination. They are assessed independently for the used surface material and liquid. 1. Introduction Boiling is a complex process and an intensive work is needed for its understanding. Within the last decades, several nucleate boiling models were formulated and could be grouped in two main categories: a) Bubble Agitation Models and b) Macro/Micro Layer Evaporation Models. The bubble agitation models are based on the principle of agitating the liquid, but they carry away little heat. The heat transfer is considered within the turbulent forced convection. The obtained empirical pool boiling heat transfer models employ dimensionless groups based on both fluid and solid properties while the main constant in the model is found to depend on the geometry of the heater. The models found in literature are useful within the range of database used in developing their derivation. Bubble agitation mechanism together with Helmholtz-instability mechanism can be used either to explain the heat transfer at the low heat flux regime or to explain CHF (Critical Heat Flux). They cannot account for the continuity of the pool-boiling curve. On the other hand, the macro/micro layer evaporation reproduces the poolboiling curve from the nuclear boiling to transition boiling. The macro/micro layer models play an important role in high heat flux region. The liquid layer includes the micro layer underneath the bubble and the macro layer on the base of coalescence and dries out periodically . Heramura & Katto assume the liquid-vapor interface is stationary and the entire surface heat flux contributes to macro layer evaporation. Several numerical models were proposed based on the macro layer theory among that of Maruyama et al. . Zhao et al. put forward a model for transient pool boiling heat transfer. The model employed is too high heating rate to be realized in practical experiments for a horizontal surface. He et al. concluded that the macro layer model is more suitable for the high heat flux regime. Dhir confirmed that numerical simulations are not a substitute for detailed experiments. The experimental results are needed to validate the simulations. Numerical simulations provide additional insights into the boiling phenomena. Within the late decade, many researchers worked on viewing the pool boiling in microgravity (in the absent of buoyancy) to understand the lower limit of forced convection. Several workers are Lee , Herman , Wan & Zhao , and Kubota et al. . Ji et al. enhanced the pool boiling heat transfer in microgravity by using porous coating heating surface at atmospheric pressure and slightly moderate superheats. Other researchers [11,12] enhanced the pool boiling by using nanofluid (water mixed with extremely small amount of nanosized particles). They concluded the enhancement of the thermal conductivity and convection heat transfer capability of the suspended particles of nanometer in size for many volume fractions of nanofluids. The results of workers [6-12] can be used as a guidance in formulating proper equations that can be used in design. The aim of the present work is to use bubble agitation models to obtain a generalized empirical correlation that gives the best possible representation of collected data. Pool of data is collected from literature for various liquids effects with different plain test surfaces. For this purpose, linear and non-linear programming techniques were used in the evaluation of the proposed correlation. 2. Theoretical Analysis of Bubble Agitation Models The primary requirement for nucleation to occur or for a nucleus to subsist in a liquid is that the liquid should be superheated. There are two types of nuclei. One type is formed in a pure liquid; it can be either a high energy molecular group resulting from thermal fluctuations of liquid molecules, or a cavity resulting from a local pressure reduction such as that occurs in accelerated flow. The other type, formed on a foreign object can be either a cavity on the heating wall or suspended foreign material with a non-wetted surface. Rohsenow assumes that the movement of bubbles at the instant of breaking away from the heating surface is of prime importance and obtained Equation (1) for heat transfer in the region of nucleation pool boiling. (1) The recommended variation of r is within 0.8 to 2.0. Evaluation of C sf from experimental results of many workers prove to be a parameter which does not pick out only the nucleation ability of heating surface but contains the effect of physical properties of liquid. Rohsenow proposed the surface factor C sf to prescribe the condition of heating surface in nucleate boiling. Various investigators utilized this factor in their determination of empirical expressions. The surface factor is defined by Equation (1), generally known as Rohsenow empirical correlation. Forster and Zuber indicated that small bubbles grow rapidly and large ones slowly, but the degree of agitation in the surrounding liquid due to bubble growth remains the same. They derived the following empirical correlation: (2) Equation (2) predicts the same heat transfer coefficient for a liquid boiling on any hot surface (for all heterogeneous cases only) or boiling in bulk (for all homogeneous cases only). Rohsenow’s Equation (1) was developed and applied to the heterogeneous case only. Forster and Greif suggested a different approach by considering that the mechanism of high heat transfer rate, during nucleate boiling, is mainly due to the liquidvapor exchange. They obtained a dimensional empirical correlation, for the pool boiling heat flux q in water at 100 - 4763 kN/m 2, as shown in Equation (3). (3) This correlation is not as widely verified as that of Rohsenow. Gupta and Varshney obtained experimental data for boiling heat transfer, using distilled water, benzene and toluene as liquids over a heated horizontal cylinder made of stainless steel. Their data was correlated by the following dimensionless empirical correlation: (4) where Nu B and Pe B are the Nusselt and Peclet number of boiling respectively. Or it can be written as: (5) In order to derive a general correlation based on bubble agitation phenomena to be more versatile than the correlations existing in literature, a search was made through published work in literature and found that the following four well known empirical correlations referred to in most publications: a) Rohsenow correlation (Equation(1)) b) Forster and Zuber correlation (Equation (2)) c) Forster and Greif correlation (Equation (3)) d) Gupta and Varshney correlation (Equation (5)). For the sake of analysis, experimental data collected from many literatures [19-24] tabulated as heat flux (q), surface temperature (T s), heat transfer coefficient (h), and coefficient h (equal to h/q 0.7). Moreover, physical and thermodynamic properties collected at the reported experimental conditions from literature [25-29] to be used in the analysis. The properties include, liquid thermal conductivity (K L), liquid heat capacity (C L), density of vapor (r V), surface tension of liquid (s), saturation temperature (T sat), density of liquid (r L), latent heat of vaporization (H fg), and viscosity of liquid (m L). The above stated correlations are of dimensionless form with the exception of the Forster and Greif correlation (Equation (4)). These equations can be represented by general equation as shown in Appendix. Many modifications to linear correlations have been tried to minimize the sum of squares of errors and to conclude some general correlations. 3. Results and Discussion Boiling heat transfer studied earlier indicated that several variables are important in nucleate boiling such as pressure, fluid properties, surface condition, boiling temperature, kind and relative amount of impurities. The practical data showed that changes in magnitude of these properties and conditions could significantly affect pool boiling heat transfer. A graphical analyses for 56 sets of literature data was used in studying the effect of heat flux, (q), and operating pressure (P), on boiling heat transfer coefficient, (h). Figures 1-3 show the variation of heat transfer coefficient with heat transfer flux, (q). The lines in the figures are the best-fit lines of the reported data. Figure 1 plotted for various liquids at different operating pressure and test surface. Figure 2 corresponds to various liquid-surface combinations at constant atmospheric pressure. Figure 3 reflects the behavior of various metal surfaces and operating pressures for the same liquid. All the data can be represented by the empirical, Equation (6), with an average percentage error ranging from 0.012 to 11.8 (6) The proportionality constant h is proved to be a function of pressure and liquid-surface combination. Cichelli Figure 1. Heat transfer coefficient (h) versus heat flux (q) at various pressures. Figure 2. Heat transfer coefficient (h) versus heat flux (q) at atmospheric pressure. Figure 3. Heat transfer coefficient (h) versus heat flux (q) for various metal surfaces. and Bonilla confirmed that the coefficient of heat transfer increases with absolute operating pressure in the nucleate boiling zone. They reported the following correlation: (7) Figure 4 shows the variation of the proportionality constant (h) as function of pressure (P) for a definite liquid-surface combination. Equation (8) represents the relationship between h and P, that was obtained from best data fit of Figure 4. (8) The overall dependence of (h) on operating pressure (P) and heat flux (q) for different liquid-surface combinations is shown in Figure 5. 3.1. Linear Programming Analysis of Empirical Correlations Equation (A.4) in Appendix used to test the validity of the published correlations. It is used to formulate a correlation that shows the best fit of the experimental data. All the data cited in the literature from [19-24] classified as eight liquids (Water, Benzene, Methanol, Carbon Tetrachloride, n-Butanol, Isopropanol, n-Amyl Alcohol, and n-Heptane) and four surfaces (Brass, Copper, Nichrome, and Stainless Steel) at different operating pressures grouped in 56 data sets. The applicability of the four empirical correlations, Equations (1)-(3) and (5), in representing the data was test by using linear-programming; that by fixing some of Figure 4. Proportionality constant (h) versus operating pressure (P). Figure 5. Heat transfer coefficient (h) dependence on operating pressure and heat flux (q). the coefficients and evaluating the others and the average percentage error is used as test criteria for comparison purposes. 3.1.1. Rohsenow’s Correlation Equation (9) is a general expression for Rohsenow’s correlation while the exact expression stated as in Equation (1). (9) The coefficient C sf reported, in the literature, to vary with each liquid-surface combination and it is independent of pressure . The validity of Equation (9) was tested for the entire collected data by applying the leastsquares method. In the initial analysis of data, the pressure was assumed constant and the obtained results showed inconsistency in the calculated values of constants for various experimental conditions tested. The inconsistency in values of constants is most likely due to pressure effect, which was not considered as variable during the initial analysis of data. On the next try, a pressure parameter was introduced in an attempt to narrow the variation in the values of constants for different systems and to conclude general correlation. Four different expressions of pressure parameter cited from literature , and stated in Equation (10), was used and expected to have an effect on heat transfer in the region of pool boiling nucleation: (10) Each of these pressure expressions selected to replace the coefficient C sf in Equation (9) and the obtained results were compared and checked. The analysis found that the pressure parameter gives the lowest minimum percentage error between the others forms and selected to replace C sf and modify Equation (9) to the form showing in Equation (11). (11) To make the present work more general, the analysis was repeated for various liquids at certain test surface and different operating pressures, that by calculating all the coefficients (R 1, R 2, R 3, and R 4) of Equation (11) for each set of data. The values of coefficients were selected, by the help of Equation (A.4), on the basis of using Equation (11) with the lowest average percentage error. 3.1.2. Gupta and Varshney Correlation Equation (12) is a general expression for Gupta and Varshney empirical correlation, while Equation (5) represent the exact form. (12) The pressure term is not included in the correlation of Gupta and Varshney as was the case of Rohsenow’s empirical correlation, Equation (1). Equation (12) differs from Rohensow’s Equation (9) by including the density ratio term and classifying other dimensionless term in well-known groups. A similar data analysis used for Equation (12) as it was with the case of Rohsenow’s Correlation. By substituting the various pressure forms of Equation (10) in place of coefficient R 3 in Equation (12), it is found that pressure expression gave the minimum average percentage error. This term is then chosen as the best fit expression for pressure and used in Equation (12) to obtain Equation (13). (13) In order to make the present work more general, the analysis repeated for various liquids at specific test surface and different operating pressures then follow the same procedure as in case of Rohsenow’s correlation. The analysis concluded that the modified Equation (13) provides a better data fit than that of Gupta and Varshney Equation (5). 3.1.3. Forster and Zuber Correlation Equation (14) is a general expression for Forster and Zuber empirical correlation, while the exact form is given in Equation (2). (14) The pressure term is taken care of in Forster and Zubers correlation, Equation (2). By applying a similar procedure as in previous cases, it was found that the validity of the above equation is restricted to specific experimental data near critical temperature difference. By checking the above results for Forster and Zuber empirical correlation, Equation (14), it is found that the overall average percentage errors is very high in predicting the published data under consideration. 3.1.4. Forster and Greif Empirical Correlation Equation (15) is a general expression for Forster and Greif dimensional empirical correlation, while the exact form is given in Equation (3). (15) Using the same sets of data analyzed previously gave a higher average percentage error as compared with the dimensionless empirical correlations as shown in Table 1. The variation of some thermodynamic properties as function of pressure and temperature is not reported in literature and these properties were considered constant during the calculations, which may be the cause of the large average percentage error. Moreover, this empirical correlation dealt with fluid side effect of the problem and ignored the surface side effect on the nucleate boiling behavior. This might have added more error to the validity of this correlation. The linear programming analysis recommended the use of the modified correlations of Rohsenow and Gupta & Varshney as they give closer prediction to the experimental data than in case of using Forster & Zuber and Forster & Greif modified correlations as showing in Table 1. Hence, the last two correlations excluded from any further analysis. Table 1. Linear programming results for Equation (11) to Equation (15). Where APE = Average Percentage Errors; CC = Correlation Coefficient. 3.2. Non-Linear Programming Analysis of Empirical Correlations The data were re-analyzed by non-linear programming methods by using the modified correlations of Rohsenow, Equation (11), and Gupta & Varshney Equation (13) in an attempt to improve the correlations for lower average percentage error. By assume that: in Equation (4) is equal to X and in Equation (13) is equal to Y, then use either of the following equations: Binominal Expression: (16) or non-linear Expression: (17) to replace the in Equation (11) or in Equation (13) respectively. It is found from data fittings that a better representation can be obtained by using the binominal expression, Equation (16), with in place of in Equation (11) and using the binominal expression, Equation (16), with in place of in Equation (13). Various forms of expressions for the constant R 7 was tried for both Rohsenow and Gupta & Varshney modified correlations. It was found that keeping R 7 as a constant and independent of other parameters yielded a better data representation. It was also concluded that the average percentage error would not be improved by using the non linear equations instead of the linear equation as showing in Table 2. General Empirical Correlation By using the best data fit for Equation (11) for different surfaces (Nichrome, Copper, Brass, and Stainless Steel), the variation of the powers of pressure expression term and Prandtl number in the equation were found to be approximately equal to 0.08 and 1.0 respectively leading to the following generalized Equation (18). (18) The coefficients R 1 and R 3 represent the effect of surface-liquid combination. They are assessed independently for each surface by the least-squares linear regression method and the results are stated in Table 3. A similar analysis tried for Equation (13) and concluded that the powers of pressure term, Peclet number (Pe B), and density ratio term (r V/r L) were relatively independent of surface—liquid combination as compared with the coefficient R 3 and the power of Prandtl number, R 4. The best form of Equation (13) was tested for different data sets and concluded Equation (19). (19) A similar way was followed for Equation (19), to that of Equation (18), in finding the power R 4 and the coefficient R 3 and their best values are given in Table 3. The result of analysis of Equation (18) and Equation (19) listed in Table 3 suggested the use of Equation (19) in preference to Equation (18). The applicability of Equaiton (19) was examined for different surfaces as showing in Figures 6-9. The equation found to fit well for all the data with the exception of Brass. The deviation in the results for Brass is due to Table 2. Comparison of linear, binominal, and non linear expressions versions of Equations (11) and (13). Table 3. Linear programming results for Equations (18) and (19). Figure 6. Experimental data predictions using Equation (19) for Nichrome surface. Figure 7. Experimental data predictions using Equation (19) for Copper surface. Figure 8. Experimental data predictions using Equation (19) for Brass surface. Figure 9. Experimental data prediction using Equation (19) for Stainless Steel surface. limited available data at very low pressure. Equation (20) represents the dimensionless form of Equation (19). (20) The analysis concluded that Equation (20) is valid for the entire available data and represent a more generalized correlation than the correlations found in literature. 4. Conclusions A graphical analysis concluded that the empirical Equation (6) is showing the effect of heat flux (q) and operating pressure (P) on the boiling heat transfer coefficient (h). (6) where h is a function of pressure and for different liquid-surface combinations, it is found to vary with the pressure as follows: (8) 56 sets of literature data were tested on each of the four known correlations, Rohsenow, Forster & Zuber, Forster & Greif, and Gupta & Varshney, by using the linear and non-linear programming solution. The concluded results show that any of these correlations does not fit the entire data satisfactory. To improve their predictions, the correlations were modified, including additional parameters in an attempt to close up the deviation in the values of calculated parameters. The modified correlations of Rohsenow and Gupta & Varshney responded better to the applied modification than that of Foster & Zuber and Foster & Grief and they were considered for further analysis. The least squares multiple regression technique [30,31] is used to evaluate the best possible values of the constant coefficients in the correlation. The cumulative error squares were minimized by using an ordinary optimum seeking technique. Linear, binominal & non-linear correlations were tested in concluding the final correlation. Mohammed Salah Hameed, Abdul Rahman Khan, A. A. Mahdi (20) gives the best representation of the entire tested data. REFERENCES Y. Haramura and Y. Katto, “A New Hydrodynamic Model of Critical Heat Flux, Applicable Widely to Both Pool and Forced Convection Boiling on Submerged Bodies in Saturated Liquids,” International Journal of Heat and Mass Transfer, Vol. 26, No. 2, 1983, pp. 389-399. [Citation Time(s):2] S. Maruyama, M. Shoji and S. Shimizu, “A Numerical Simulation of Transition Boiling, Heat Transfer,” Proceedings of the 2nd JSME-KSME Thermal Engineering Conference, Kitakyushu, 19-21 October 1992, pp. 345- 348. [Citation Time(s):1] Y. H. Zhao, T. Masuoka and T. Tsuruta, “Theoretical Studies on Transient Pool Boiling Based on Microlayer Model (Mechanism of Transition from Nonboiling Regime to Film Boiling),” Trans. JSME (B), Vol. 63, No. 607, 1997, pp. 218-223. [Citation Time(s):1] Y. He’ M. Shoji and S. Maruyama; “Numerical Study of High Heat Flux Pool Boiling Heat Transfer,” International Journal of Heat and Mass Transfer, Vol. 44, No. 12, 2001, pp. 2357-2373. [Citation Time(s):1] V. K. Dhir; “Numerical Simulation of Pool-Boiling Heat Transfer,” AICHEJ, Vol. 47, No. 4, 2001, p. 813. [Citation Time(s):1] H. S. Lee, “Mechanisms of Steady-State Nucleate Pool Boiling in Microgravity,” Annals of the New York Academy of Sciences, Vol. 974, 2002, pp. 447-462. [Citation Time(s):1] H. Herman, “Some Parameter Boundaries Governing Microgravity Pool Boling Modes,” Annals of the New York Academy of Science, Vol. 644, 2006, pp. 629-649. [Citation Time(s):1] S. X. Wan and J. Zhao, “Pool Boiling in Microgravity: Recent Results and Perspectives for the Project DEPASJ10,” Microgravity Science and Technology, Vol. 20, No. 3-4, 2008, pp. 219-224. [Citation Time(s):1] C. Kubota, et al., “Experiment on nucleate Pool Boiling in Microgravity by using Transparent Heating Surface- Analysis of Surface Heat Transfer Coefficients,” Journal of Physics: Conference Series, Vol. 327, 2011, Article ID: 012040. [Citation Time(s):1] X. Ji, J. Xu, Z. Zhao and W. 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[Citation Time(s):1] Appendix Sum of Squares of Errors Equations (1) to (4) can be represented by a general equation: where (A.1) This equation represents a general form of those correlations and simplified by taking logarithms of both sides. or (A.2) For N number of data readings there will be N number of linear equations, while for the determination of k coefficients only k equations are required. A least squares multiple regression technique [30,31] was used to evaluate the best possible coefficient from raw data readings. The cumulative error squares minimized by an ordinary optimum seeking technique resulting into k number of equations to provide k number of coefficients for the entire data. These equations mathematically represented in the form: (A.3) The sum of squares of errors is expressed as: (A.4) Nomenclature C L—Heat capacity of liquid, J/kg×˚C C s f—Surface factor F m—Nucleation factor g—Acceleration of gravity, m/s 2 g c—Conversion ratio, kg m/kg·s 2 h—Heat transfer coefficient W/m 2·˚C h—Proportionality constant H fg—Latent heat of vaporation, J/kg K L—Thermal conductivity, W/m ˚C Nu B—Nusselt number for boiling = P—Operating pressure (kN/m 2) Pe B—Peolet number for boiling = Pr—Prandtl number q—Heat Flux, W/m 2 s —coefficients Re—Reynolds number Re b—Reynolds number for bubbles = Sr—Superheat ratio = T s—Test surface temperature, ˚C T sat—Saturation temperature, ˚C Greek Letters a L Thermal diffusivity, m 2/s, DP Pressure difference corresponds to (T s–T sat), kN/m 2 m L Viscosity of liquid, kg/m×s r V Density of vapor, kg/m 3 r L Density of liquid, kg/m 3 s Surface tension, kg/s 2 Subscripts b Refers to bubble property B Refers to boiling condition L Refers to liquid condition sf Refers to surface factor s Refers to surface condition sat Refers to saturation condition v Refers to vapor condition NOTES Corresponding author. Journal Menu >> ●Indexing ●View Papers ●Aims & Scope ●Editorial Board ●Guideline ●Article Processing Charges ●Paper Submission ● ACES Subscription ●Free Newsletter Subscription ●Most popular papers in ACES ●Publication Ethics Statement ●About ACES News ●Frequently Asked Questions ●Open Special Issues ●Published Special Issues ●Special Issues Guideline Top ABSTRACT Introduction Theoretical Analysis of Bubble Agitation Models Results and Discussion Conclusions REFERENCES Appendix NOTES Home | About SCIRP | Sitemap | Contact Us Copyright © 2006-2015 Scientific Research Publishing Inc. All rights reserved. E-Mail AlertJoin Peer-Review Program Submit your Manuscript Sign up
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What is the value of (X+Y)-Z where X=a+b, Y=b+c, Z=c+a? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Mathematics Simplifying Algebraic Exp... Variables and Constants Basic Algebra Mathematical Equations Variables (mathematics) Maths, Algebra & Variable... Equations and Expressions Algebraic Expression 5 What is the value of (X+Y)-Z where X=a+b, Y=b+c, Z=c+a? All related (40) Sort Recommended Komal A reflective space to understand emotions, values, strengths · Author has 376 answers and 123.8K answer views ·6y Upvote · 9 6 9 1 Sponsored by Grammarly Stuck on the blinking cursor? Move your great ideas to polished drafts without the guesswork. Try Grammarly today! Download 99 34 Related questions More answers below What is the value of X+Y-Z where X=2Y, Z=X, Y=a+b+c? Let X = {a,b, c}, Y= {a,c, b} and Z = {a,b, b,c, c,c}. What are the elements of X,Y and Z ? How are they related? What is Z, A, Y, B, X, C? What is X,Y value given a=2,b=4,X=a+(b++);Y=a+(++b)? X=7. Y=9 Z=++x + y-- - ++y - x--x-- + ++y, so what will be the value of z? Neha Aggarwal B.com graduate frm DU in Qualified UGC NET in First Attempt, Masters in Commerce (Graduated 2019) ·6y We have provided the values of X Y and Z . So will put the values in given sum (X+Y )- Z = ( a+b + b+c ) - (c+a ) = a+2b+c -c - a =2b ( ans) ( we are taking value of a in bracket as I is followed by a minus sign which will change sign of each variable) Then add the same variables like b+b becomes 2b After that we are left with third step .. here a - a becomes zero and same happens with c also So answer becomes 2b Upvote · 9 1 Mahesh Kumar Former Assistant Manager · Author has 300 answers and 339.7K answer views ·6y We have x = a+b, Y = b+ C, Z = C + a, Then ( X + Y ) - Z = ( a+ b + b + C ) - ( C - a) = ( a + 2b + c - c + a) = 2a + 2b or 2 ( a+b ) ANS Upvote · 9 3 9 1 Angeljohn Angel 6y (X+Y)-Z Given , X=a+b,Y=b+c,Z=c+a Sub x ,y,a value in equation (a+b+b+c)-c+a=0 a+b+b+c-c-a =0 We can cancel c ,a 2b=0 Therefore (X+Y)-z=2b Upvote · Related questions More answers below What is the value of k in k=x-z where x=6, z=a+b, a=1, b=3? Given that z=x +iy what are the values of x and y if 1/z + 2/z = 1+ I? What is the value of X(Y-Z) where X=30, Y=20, Z=10? If A=Z, B=Y, C=X, then what is A+B+C =? Let X = {a,b, c}, Y= {a,c, b} and Z = {a,b, b,c, c,c}. What are the elements of X,Y and Z ? How are x, y, and z related? Mohit Kyler 6y Here, X=a+b, Y=b+c, Z= c+a So, (X+Y)-Z = [ (a+b) + (b+c) ] - (c+a) = [ a+2b+c ] - (c+a) = a+2b+c-c-a (X+Y)-Z= 2b. Ans Upvote · Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 621 Laxman Kohar Studied at Sharda University · Author has 79 answers and 39.3K answer views ·6y here, X=a+b----1 Y=b+c-------2 Z=c+a-------3 now, first of all add x and y we get, x+y=a+b+b+c= a+2b+c ——-4 now, subtract equation 3from 4 we get, (X+Y) -z = a+2b+c-(c+a) =a+2b+c-c-a=2b which is our answer Upvote · Akhtar Khan Former Ex-Sales Manager (North) ·6y Value of (X+Y)-Z will be … (a+b+b+c)—(c+a) or a+2b+c-c-a or 2b. Answer. Thanks. Upvote · Sponsored by Reliex Looking to Improve Capacity Planning and Time Tracking in Jira? ActivityTimeline: your ultimate tool for resource planning and time tracking in Jira. Free 30-day trial! Learn More 99 24 Suruchi Bhasin SRM Analyst at British Telecommunications (BT) (2018–present) ·6y x=a+b y=b+c z=c+a as per the given equation (x+y)-z =(a+b+b+c)-(c+a) =(a+2b+c)-(c+a) =a+2b+c-c-a (opening the second bracket) =a-a+c-c+2b =2b Upvote · Virendra Kumar Yadav Works at Cognizant (company) ·6y Given value is X=a+b Y=b+c Z=c+a Than find (X+Y)-Z=(a+b+b+c)-(c+a). Here we are putting X,Y,Z value =a+2b+c-c-a =2b Your response is private Was this worth your time? This helps us sort answers on the page. Absolutely not Definitely yes Upvote · Sponsored by MailerLite Your free AI assistant completes time-consuming tasks in seconds. Once you connect your AI tool to MailerLite, you can start asking questions and giving it tasks. Learn More 9 1 Aditya Deshmukh Former Intern at Precision Transmission (2018–2018) · Author has 63 answers and 247.6K answer views ·6y Related What is the value of X+Y-Z where X=2Y, Z=X, Y=a+b+c? GIVEN X= 2Y Y= a+b+c Z = X PROBLEM X + Y - Z 2Y + (a + b + c) - X 2Y + (a+b+c) - 2Y . . . . . ( Since Z=X ) a+b+c = ANS That would be the value of equation on substitution. Ps - Please correct me if I solved it wrong. Or if I've missed something Upvote · 9 3 Rajiv K Author has 80 answers and 41.4K answer views ·6y 2b Substituting x, y and z in the above equation, = (a+b+b+c)-(c+a) = a+2b+c-c-a = 2b Upvote · Manisha Bharti User of WordPess ,Blogger · Author has 325 answers and 509.2K answer views ·6y Related What is the value of X+Y-Z where X=2Y, Z=X, Y=a+b+c? Putting the value of x,z,& y X+Y-Z 2y + a+b+c - x 2(a+b+c) + (a+b+c) - 2y 2 (a+b+c) +(a+b+c) -2 (a+b+c) Take a common (a+b+c) (a+b+c)(2+1–2) (a+b+c)(1) (a+b+c) Answer Upvote · 9 5 Rubin Roy Worked at Unemployed Currently ·6y Related What is the value of X+Y-Z where X=2Y, Z=X, Y=a+b+c? The value for X+Y-Z is a+b+c (X and Z will subtract each other and become zero since X=Z. This leaves us with Y where Y = a+b+c so the answer is a+b+c) Upvote · 9 1 Related questions What is the value of X+Y-Z where X=2Y, Z=X, Y=a+b+c? Let X = {a,b, c}, Y= {a,c, b} and Z = {a,b, b,c, c,c}. What are the elements of X,Y and Z ? How are they related? What is Z, A, Y, B, X, C? What is X,Y value given a=2,b=4,X=a+(b++);Y=a+(++b)? X=7. Y=9 Z=++x + y-- - ++y - x--x-- + ++y, so what will be the value of z? What is the value of k in k=x-z where x=6, z=a+b, a=1, b=3? Given that z=x +iy what are the values of x and y if 1/z + 2/z = 1+ I? What is the value of X(Y-Z) where X=30, Y=20, Z=10? If A=Z, B=Y, C=X, then what is A+B+C =? Let X = {a,b, c}, Y= {a,c, b} and Z = {a,b, b,c, c,c}. What are the elements of X,Y and Z ? How are x, y, and z related? What is the value of (X-Y) (X+Y) where X=a+b, Y=a-b? What is the value of Y+Z where Y=a+b-c, Z=a-b+c? What is the value of X+Y+Z where X=a-b-c, Y=a+b+c, Z=a? X Z Y + X Y Z = Y Z X. Find the three digits? Let a=x y+y z+z x,b=x z+y x+z y a=x y+y z+z x,b=x z+y x+z y and c=(x y+y z)(y z y+z x)(z x+x y).c=(x y+y z)(y z y+z x)(z x+x y). How do I find the value of |a b−c||a b−c|? Related questions What is the value of X+Y-Z where X=2Y, Z=X, Y=a+b+c? Let X = {a,b, c}, Y= {a,c, b} and Z = {a,b, b,c, c,c}. What are the elements of X,Y and Z ? How are they related? What is Z, A, Y, B, X, C? What is X,Y value given a=2,b=4,X=a+(b++);Y=a+(++b)? X=7. Y=9 Z=++x + y-- - ++y - x--x-- + ++y, so what will be the value of z? What is the value of k in k=x-z where x=6, z=a+b, a=1, b=3? 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The Y chromosome: a blueprint for men’s health? | European Journal of Human Genetics Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript. Advertisement View all journals Search Search Search articles by subject, keyword or author Show results from Search Advanced search Quick links Explore articles by subject Find a job Guide to authors Editorial policies Log in Explore content Explore content Research articles Reviews & Analysis News & Comment Current issue Collections Follow us on Twitter Sign up for alerts RSS feed About the journal About the journal Journal Information Open Access Fees and Funding About the Editors About the Partner Contact For Advertisers Subscribe Publish with us Publish with us For Authors & Referees Language editing services Submit manuscript Sign up for alerts RSS feed nature european journal of human genetics review article The Y chromosome: a blueprint for men’s health? Download PDF Download PDF Review Open access Published: 30 August 2017 The Y chromosome: a blueprint for men’s health? Akhlaq A Maan1, James Eales1, Artur Akbarov1, Joshua Rowland1, Xiaoguang Xu1, Mark A Jobling2, Fadi J Charchar3 & … Maciej Tomaszewski1,4 Show authorsEuropean Journal of Human Genetics volume 25, pages 1181–1188 (2017)Cite this article 42k Accesses 49 Altmetric Metrics details Abstract The Y chromosome has long been considered a ‘genetic wasteland’ on a trajectory to completely disappear from the human genome. The perception of its physiological function was restricted to sex determination and spermatogenesis. These views have been challenged in recent times with the identification of multiple ubiquitously expressed Y-chromosome genes and the discovery of several unexpected associations between the Y chromosome, immune system and complex polygenic traits. The collected evidence suggests that the Y chromosome influences immune and inflammatory responses in men, translating into genetically programmed susceptibility to diseases with a strong immune component. Phylogenetic studies reveal that carriers of a common European lineage of the Y chromosome (haplogroup I) possess increased risk of coronary artery disease. This occurs amidst upregulation of inflammation and suppression of adaptive immunity in this Y lineage, as well as inferior outcomes in human immunodeficiency virus infection. From structural analysis and experimental data, the UTY (Ubiquitously Transcribed Tetratricopeptide Repeat Containing, Y-Linked) gene is emerging as a promising candidate underlying the associations between Y-chromosome variants and the immunity-driven susceptibility to complex disease. This review synthesises the recent structural, experimental and clinical insights into the human Y chromosome in the context of men’s susceptibility to disease (with a particular emphasis on cardiovascular disease) and provides an overview of the paradigm shift in the perception of the Y chromosome. Similar content being viewed by others The effects of loss of Y chromosome on male health Article 02 January 2025 Genetics of female and male reproductive traits and their relationship with health, longevity and consequences for offspring Article 13 December 2024 Genome-wide analyses identify 25 infertility loci and relationships with reproductive traits across the allele frequency spectrum Article Open access 14 April 2025 A shift in thinking Views on the biological importance of the Y chromosome have peaked and troughed over the past 60 years.1, 2, 3 Although its perception as the key determinant of male sex has remained fundamentally unchanged, the potential association between the Y and human disease has been much more ambiguous. In the second half of the twentieth century, there was much interest in holding this unique chromosome to account for so-called ‘Y-linked’ or ‘holandric’ traits.4 At least 14 such traits generated interest because of their exclusive father-to-son transmission, including hypertrichosis pinnae auris (HPA) – having abnormally long hair on the outer ear (pinna). Despite problems of reporter bias and illegitimacy in the numerous pedigrees studied,1, 5, 6 the pattern of inheritance suggested that HPA was a Y-linked trait. This was until a 2004 study7 utilised contemporary Y-chromosome haplogroup classification to show that no single haplogroup predominated in the HPA cases, thus making it unlikely to be Y linked and more likely to be an autosomal trait with phenotypic expression limited to males. Several small case–control studies conducted in the 1970s found associations between the 47,XYY karyotype (males carrying two copies of the Y chromosome) and criminality8 and between the length of Y and physical activity levels.9 These studies exemplify the simmering interest at the time in linking the Y chromosome to psychological/physiological phenotypes. However, doubts about the validity of these Y-linkage studies2 led to an era dominated by the idea that the Y chromosome possessed little genetic content, and that its sole purpose was to trigger testis development in males.10 Such was the redundancy assigned to the Y chromosome that some commentators described it as a ‘genetic wasteland’ during this period.3 Subsequent work on the Y chromosome during the 1990s reinforced its role in the reproductive system with the localisation of the specific testis-determining factor to sex-determining region Y (SRY)11 and the definition of three distinct loci involved in spermatogenesis (azoospermia factor a, b and c), deletions of which are associated with varying degrees of spermatogenic failure in men.12 Recent advances in genetic technologies enabling mapping and sequencing of the Y chromosome have again altered scientific perspectives on the Y. Although the Human Genome Project confirmed that there is relatively sparse protein-coding material within the Y and a high degree of repetition, the finding that more than half of the active genes are expressed in non-gonadal tissue throughout the body3 has brought the Y to the forefront of research on men’s susceptibility to disease. Structure of the Y chromosome Comprehensive sequencing of the Y chromosome was first completed in 2003 by Skaletsky et al3 and their findings remain largely valid today. The male-specific region of the Y chromosome (MSY) makes up 95% of its length.3 Unlike autosomal chromosomes, the MSY does not undergo reciprocal recombination with a partner chromosome during meiosis.3 Only short regions at either tip of the chromosome undergo recombination with the X, and these are called the pseudoautosomal regions (PARs).3 Because of the lack of crossing over, the MSY is transmitted unaltered from father to son along the paternal line.3 Using a combination of genetic markers (usually single-nucleotide polymorphisms (SNPs)) it is possible to classify each individual Y chromosome into one of numerous haplogroups.13 In 2002, the Y Chromosome Consortium collated all phylogenetically informative SNPs discovered to date, constructing a robust maximum-parsimony tree, and assigning universal nomenclature to each recognised haplogroup;14 subsequently, this was updated through sporadic SNP discovery,15 and more recently thanks to large-scale resequencing projects that have yielded tens of thousands of SNPs.16, 17, 18 To simplify the task of selecting suitable SNPs for genotyping in medical and other studies, a stable minimal phylogeny containing 417 SNPs has also been described.19 The phylogenetic framework has been incorporated into studies of association between the Y chromosome and susceptibility to complex diseases. This important development has filled (at least to some extent) a void in genetic association discovery for the Y chromosome left by genome-wide association studies (GWASs) in which, due to its haploid nature, the Y chromosome was routinely ignored.20 Compared with all other nuclear chromosomes, the Y harbours the smallest number of genes at 568 and is considerably shorter than the X chromosome in length (~57 Mbp compared with ~156 Mbp).21 Based on Ensembl data (v86), only 71 of the Y-chromosome genes have protein-coding potential; however, several of the encoded proteins belong to the same protein families, leaving only 27 genes coding for distinct MSY proteins.21 These 27 protein-coding genes are displayed in Figure 1. The 109 genes produce long and short noncoding RNAs, all of which could have important effects on regulation of gene expression, but none of which have had their regulatory potential explored in greater detail.21 The remaining 388 genes are described as pseudogenes.21 It is important to note that many of the gene assignments are predictions and require biological validation. Figure 1 highlights key structural and functional characteristics of the 27 Y-chromosome genes that encode distinct MSY proteins and three noncoding genes. For full names of genes mentioned in the main text and figures, please see Table 1. Figure 1 Genes of the Y chromosome. Chromosome starts at the top of the circle and proceeds anticlockwise. Track 1=locations and lengths of Y genes: Ensembl v86 genes are shown as black tiles, with genes that either encode distinct MSY proteins or are known to produce biologically significant products labelled. Please note that the protein-coding gene AC009977.1 lies within the Y-chromosome coordinates for TXLNGY but is positioned on the reverse strand rather than the forward strand. For greater visual identification and separation of tiles for these two genes, the position of AC009977.1 has been shifted slightly proximally. Track 2=Y-chromosome regions: this track represents gross structural subdivisions of the Y. Green=short arm (Yp); light purple=ampliconic regions; grey=centromere; yellow=long arm (Yq). Track 3=Gene biotype: this track illustrates the current Ensembl biotype status for each of the labelled Y genes. Green=protein-coding; yellow=pseudogene; black=noncoding RNA. Track 4=Copy/isoform number: this track represents the number of copies or isoforms that each gene possesses on the Y chromosome. Single copy=light pink; multicopy=solid, dark red. Track 5=X paralogue and/or X–Y gene dosage sensitive: this track shows genes that have an X paralogue and/or have been classified as one of 12 X–Y dosage-sensitive gene pairs. Dark grey=gene has neither X paralogue nor is part of an X–Y gene dosage-sensitive pair; brown=gene has an X paralogue but is not part of an X-Y gene dosage-sensitive pair; blue=gene possesses an X paralogue and is part of an X–Y gene dosage-sensitive pair. Track 6=Biological functions: known or potential biological functions of the gene products. Brown=translation, red=transcription, pink=spermatogenesis, light yellow=cell adhesion, light green=biomineralisation, blue=T-cell activation, dark grey=unknown, light grey=brain development, orange=cell differentiation. Plot constructed using Circos software.66 Full size image Table 1 Gene abbreviations and acronyms used in text Full size table Of the 27 genes that encode distinct MSY protein, 9 are ubiquitously expressed; a further 14 are considered testis specific or show predominant expression in specific tissues such as the brain (eg, PCDH11Y) or the thyroid (eg, TBL1Y);22 the remaining 4 do not currently have validated tissue expression data available – Figure 2 shows the pattern of tissue expression for the 23 protein-coding genes whose tissue expression has been evaluated and validated. Many of the original genes described as testis-specific by Skaletsky et al3 have now been reclassified as ubiquitous using updated tissue expression data.22 All of the latter are X-degenerate genes – they have a paralogue on the X chromosome.3 Figure 2 Tissue expression of key Y-chromosome genes. This heatmap illustrates the relative tissue expression in a range of different tissues for 23 out of 27 Y-chromosome genes that encode distinct MSY proteins. The tissue expression profiles for protein-coding genes AC006386.1, AC009491.1, AC009977.1 and AC012005.2 have not been evaluated and validated at the time of this review and hence these genes have not been included in the heatmap. Data are based on RNA transcript values for each gene (Reads Per Kilobase of transcript per Million mapped reads (RPKM)) obtained from GTex Portal22 that have been transformed logarithmically. Lighter shades of blue represent lower log(RPKM) values and lower levels of expression in the particular tissue, whereas darker shades of blue represent higher log(RPKM) values and higher levels of expression in the particular tissue. Grey blocks represent no recorded expression of the gene in the tissue of interest. The nine genes that are ubiquitously expressed have been labelled with an arrow below the gene name. Full size image X–Y gene dosage Recent work by Bellott et al23 has identified 12 X–Y gene pairs that are collectively critical for survival (see track 5 in Figure 1). The X paralogue of each pair escapes X inactivation, implying a dose-sensitive relationship that requires both genes to be active.23 These genes are generally ubiquitously expressed and are understood to perform a variety of gene expression regulatory functions including chromatin modification, splicing and translation,23 and are thus potentially relevant to a wide range of physiological traits and susceptibility to disease. Hypertension Genetic crosses of spontaneously hypertensive rats (SHRs) and normotensive Wistar Kyoto (WKY) rats produced male offspring of SHR fathers with significantly higher blood pressure compared with the offspring of SHR mothers.24 Consomic techniques have been used to isolate and estimate approximate contributions of the Y chromosome and autosomes to blood pressure.25 Here, successive selective breeding of male offspring across several generations leads to male rats that possess the Y chromosome of interest on a known, defined genetic background of autosomes and X chromosome from the normal strain.25 This allows for isolated analysis of the phenotypic effects of the Y.25 To study Y-chromosome influence in the SHR phenotype, Turner et al25 developed two separate consomic strains: one strain possessing the Y chromosome from the SHR rat, X and autosomal chromosomes from the normotensive WKY rat and another with the opposite configuration.25 Such techniques revealed that the Y chromosome independently raised blood pressure by 34 mm Hg.25 A later review by Ely et al26 estimated the Y effect on blood pressure at a more modest 15–20 mm Hg. A linkage of Y to blood pressure has also been demonstrated in rat strains other than the Wistar Kyoto.27 Search for a potential locus mediating this effect in rats has focused on the SRY gene, already well established as the testis-determining factor. Whereas humans possess a single copy of SRY on the Y, normotensive experimental rats carry multiple highly similar copies.28 Sequencing techniques show the presence of an additional SRY3 copy in the SHR,29, 30 containing a proline-to-threonine amino acid substitution at position 76.30 Importantly, SRY is a transcription factor that, in synergy with androgen receptor and in a testosterone-dependent manner,30 regulates promoter regions for genes encoding angiotensinogen, renin, angiotensin-converting enzyme (ACE) and ACE229 – known for their key roles in blood pressure regulation. The threonine point mutation in SRY3 has been shown to reduce SRY3 promoter regulation,30 leading to an increase in transcription of angiotensinogen, renin and ACE, thus promoting formation of the vasoconstrictor angiotensin II (Ang II); in contrast, SRY3 has an inhibitory effect on ACE2 transcription, the enzyme important for formation of vasodilatory and blood pressure-lowering Ang-(1–7).29 Experimental delivery of SRY3 to normotensive rat kidneys raises blood pressure,30, 31 a rise that can be prevented by concomitant administration of olmesartan, a renin–angiotensin–aldosterone system (RAAS) inhibitor.30 The in silico analysis and transfection studies of Chinese Hamster Ovary cells show that the SRY X paralogue (SOX3) is also capable of influencing RAAS gene expression, although in vivo SOX3 is primarily transcribed in non-kidney tissues.32 This suggests that the SRY paralogue is unique in its pro-hypertensive effects in male rats. The translatability of SRY as a key blood pressure regulator to humans is uncertain. Rat SRY is significantly different to that of humans not only in terms of copy number but also in terms of the gross protein structure: human SRY lacks a polyglutamine (Q-)-rich motif present in rat SRY and the high-mobility group (HMG)-box region important for DNA binding is in a different location.28 Nevertheless, human SRY has been shown to influence expression of rat and human RAAS genes in vitro,33 suggesting potential to play a role in genetically acquired human hypertension. One of the earliest studies to suggest that blood pressure could be a Y-influenced phenotype in humans evaluated Japanese university students aged 17–21 years with and without hypertensive parents.34 Male students born to hypertensive fathers had significantly higher systolic and diastolic blood pressures than female students born to hypertensive fathers, suggesting a possible genetic susceptibility to higher blood pressure via paternal lineage and/or autosomal influence that was sex limited.34 However, the absence of significant difference in blood pressure between male students born to hypertensive mothers or fathers34 seemed to argue against Y linkage. Although these results were inconclusive, they prompted a series of studies to investigate associations between specific genetic variants of the Y chromosome and hypertension. One extensively studied variant is a _Hin_dIII restriction site polymorphism in the Y-chromosomal alphoid satellite DNA that divides Y chromosomes into two classes.35 The majority of these studies preceded the introduction of informative phylogenetic tree classification; however, the class showing absence of the restriction site is equivalent to the currently defined super-haplogroup P-M45.36 Whereas some studies found an association between _Hin_dIII variants and altered systolic/diastolic blood pressure with effect sizes ranging from 1.44 to 6.2 mm Hg,35, 37, 38 others failed to replicate this association.39, 40, 41, 42, 43 Phylogenetically based studies using haplogrouping strategies are considered a more efficient method of identifying an association between the Y chromosome and a phenotype compared with single, isolated variants. Such studies by our group44, 45 found no evidence of association between one of the most common European lineages of the Y chromosome (haplogroup I) and blood pressure. However, this does not rule out associations between other common haplogroups and blood pressure. Delineating such associations will benefit greatly from extensive databases such as the UK Biobank that includes phenotypic data for 230 000 men, genotyped using an array of markers with extensive coverage for the Y chromosome, thus facilitating haplogrouping.46 The evidence currently points to a strong Y-chromosome signal influencing blood pressure in rats with a relative paucity of evidence in human studies. The human genetic association studies had inherent limitations related to the availability of genotyping technology when they were conducted (only one to two polymorphisms were studied at a time). Currently, there are no convincing data that even if a blood pressure regulating gene exists on the human Y, it is the same gene that stimulates the blood pressure rise in rodents. Coronary artery disease Our earlier study established an association between the Y chromosome and coronary artery disease (CAD) in two separate British cohorts using reconstruction of the Y phylogeny.44 One of the most common European lineages, haplogroup I, was associated with a higher incidence of CAD compared with all others. This effect was present in both the cross-sectional British Heart Foundation Family Heart Study (BHF-FHS) and the prospective West of Scotland Coronary Prevention Study (WOSCOPS). Indeed, the magnitude of the effect was comparable across both studies, with the odds ratio for CAD with haplogroup I being 1.75 (95% CI 1.20–2.54, P\=0.004) in BHF-FHS and 1.45 (95% CI 1.08–1.95, P\=0.012) in WOSCOPS. In the prospective WOSCOPS, cardiovascular risk parameters were also available and haplogroup I was not associated with any traditional cardiovascular risk factors including hypertension, dyslipidaemia, high BMI, diabetes, elevated C-reactive protein (CRP), alcohol consumption or smoking. Importantly, the associations between haplogroup I and CAD were not affected by the adjustment for common autosomal variants linked to CAD identified in previous GWAS. In search of molecular mechanisms that may explain the association between haplogroup I and CAD, we explored monocyte and macrophage transcriptomes of men whose Y chromosomes were genetically characterised and haplogrouped.44 This transcriptome-wide analysis revealed differences in expression of 30 Kyoto Encyclopedia of Genes and Genomes (KEGG) pathways between men with haplogroup I and carriers of other haplogroups. Nineteen of these pathways belonged to inflammatory or immune signalling cascades. In general, there was a downregulation of genes in pathways involved in autoimmunity and adaptive immunity (such as antigen processing and presentation) combined with upregulation of genes in inflammatory pathways (such as transendothelial leukocyte migration). This is highly pertinent given the significant inflammatory component to atherosclerosis,47 exhibited by monocyte entry into the intimal layer of arteries and subsequent differentiation into macrophages that internalise lipids and stimulate intimal hyperplasia.48 The male-related phenotypes such as aggression and sex steroid levels (including androstenedione, testosterone) showed no differences between men who inherited haplogroup I from their fathers and those representing other paternal lineages.49 As such, these factors most intuitively linked to male sex are unlikely to explain the association between haplogroup I and increased susceptibility to CAD. Recent studies have looked at the role of Y haplogroup in determining risk of other cardiovascular diseases. A recent prospective study conducted in Cypriot men (in whom prevalence of haplogroup I is estimated at 2.4%) associated haplogroup K with a more than twofold increased risk of atherosclerotic plaque occurrence in the carotid and femoral artery bifurcations compared with all other haplogroups.50 Systolic blood pressure was also associated with haplogroup K in this analysis and was proposed as a potential intermediate phenotype of the identified association.50 In contrast, Haitjema et al51 studied histological vessel wall characteristics of Dutch patients who had undergone carotid endarterectomy or open aneurysmal repair, but found no significant differences in vessel wall characteristics including leukocyte infiltration, lipid, collagen and smooth muscle content between the major haplogroups present (including I with prevalence 24–28%). Caution should be exercised when interpreting the data from this analysis with the results of the previous study on CAD given the obvious differences between the mechanisms of CAD, carotid artery disease and abdominal aortic aneurysms.51 The association between haplogroup I and CAD presents a strong case for a Y-linked heritable component of CAD. The effects of haplogroup I are independent of traditional cardiovascular risk factors or the male-related phenotypes. It is anticipated that utilisation of larger cohorts (such as those derived from the UK Biobank) will increase the power to detect phenotypes that may mediate the link between haplogroup I and CAD. Immunity and inflammation The immune system and inflammation play key roles in atherosclerosis and the ensuing development of CAD. In this context, the emerging evidence for the role of the Y-chromosome genes in immunity and the inflammatory response strengthens the hypothesis that the association between Y-chromosome haplogroup and CAD is mediated by the immune system. Viral infections In a population of European Americans, CAD-predisposing haplogroup I was associated with faster progression of HIV to AIDS, a greater depletion of the CD4+ T-cell count and a higher mortality rate than other haplogroups more than 7 years after initial infection.52 Haplogroup I was also associated with a higher risk of malignancy, including the highly specific AIDS-defining malignancy Kaposi’s sarcoma.52 Moreover, individuals with haplogroup I were more resistant to highly active antiretroviral therapy, taking longer to achieve viral load suppression.52 This implies a prominent role for the Y chromosome in determining outcomes of HIV infection where systematic immune system targeting of virally infected CD4+ T cells is a key process underlying pathogenesis.53 Furthermore, in vitro studies show that DDX3X (the X paralogue of the Y gene DDX3Y) is a determinant of HIV-1 replication.54 These examples offer strong support for inherited differences in immune responses between men with haplogroup I and those from others.44 UTY Following the discovery of an association between haplogroup I and CAD, our group carried out gene expression analysis to compare the macrophage expression of X-degenerate genes of the MSY between men with CAD-predisposing haplogroup I and other haplogroups.45 Of the 14 X-degenerate genes with confirmed macrophage expression, 2 were associated with haplogroup I – men with this paternal lineage showed ∼0.61- and 0.64-fold lower expression of UTY (Ubiquitously Transcribed Tetratricopeptide Repeat Containing, Y-Linked) and PRKY (Protein Kinase, Y-Linked), respectively.45 Little is known about the biological functions of PRKY, although it is speculated to encode a ubiquitously expressed protein kinase that may have important signalling functions.21 The downregulation of UTY, on the other hand, is particular intriguing given the links of UTY and its X paralogue, UTX, with various aspects of inflammation and immunity. UTY encodes a minor histocompatibility antigen important for male stem cell allograft rejection55 – a process linked to one of the KEGG pathways associated with haplogroup I in transcriptome-wide analysis.44, 56 UTX is implicated in the proinflammatory response of macrophages.57 Structural analysis has identified a specific enzyme inhibitor of UTX and subsequent selective inhibition leads to a reduction in inflammatory cytokine release, including TNF-α, by human macrophages.57 Moreover, in vitro studies exemplify the importance of UTX for facilitation of T-follicular helper cell differentiation and indirectly the maturation of IgG-secreting plasma cells in the setting of chronic viral infections.58 UTX (also known as KDM6A) encodes an enzyme belonging to a family of lysine-specific histone demethylases (KDMs) that remove epigenetic marks at histone H3 Lysine 27 (H3K27).59 These KDMs regulate transcription and possess a Jumonji C (JmjC) domain that utilises iron as a cofactor.60 Protein sequence analysis suggests that UTY possesses particularly high sequence identity (≈96%) with UTX in two important domains: the tetratricopeptide repeat (TPR) regions and the JmjC catalytic domain that includes the principal iron-binding residues (Figure 3).61 UTY comparison with the autosomal-encoded, functional histone demethylase KDM6B (also known as JMJD3) suggests relatively high sequence homology in the pertinent JmjC domain but less so in the TPR regions (Figure 3) that have undetermined function.60 A high degree of conservation in the important JmjC domain region of UTY with functional KDMs suggests that UTY could be an active histone demethylase and, thus, implicated in similar inflammatory and immune processes to those associated with UTX. Indeed, although protein sequence analysis would suggest the high degree of JmjC sequence similarity between UTY and the functional KDMs (Figure 3), there is some conflict in the literature regarding whether UTY inherently possesses histone demethylase activity. UTX knockout experiments in mice embryos suggest that UTY exhibits redundancy for UTX activity.59 Follow-up in vivo analysis, however, found that both mouse and human UTY lacked inherent histone demethylase activity, suggesting that biological functions of UTY (and UTX) may be at least partly independent of the demethylase region.59 In contrast, a different group conducted in vitro analysis revealing conserved UTY histone demethylase activity, although reduced compared with UTX.60 Based on studies conducted to date, UTY appears capable of regulating gene expression, possibly (at least in part) via histone demethylation. It is tempting to speculate that altered UTY expression in carriers of haplogroup I may contribute to the observed changes in their macrophage expression of inflammatory and immune pathways. This will require further studies, in particular given the evidence for another Y gene (KDM5D, formerly known as SMCY) to exhibit histone H3 Lysine 4 (H3K4) demethylase activity21 and play a role in immunological complications of stem cell transplantation.62 Figure 3 Amino acid sequence homology of UTY and related histone demethylases. (a) Line graph representing sequence identity of UTY aligned with KDM6B; the colour of the line correlates with the degree of sequence identity (yellow→orange→red colours show increasing % identity). (b) Line graph representing sequence identity of UTY aligned with UTX, and (c) the structure of the UTY protein including relative positions of the Tetratricopeptide domains and the JmjC catalytic domains, together with important iron (Fe) and zinc (Zn) ion binding residues. Red triangles represent deletions within UTY compared with UTX protein sequence. UTY possesses particularly high sequence identity with UTX in the Tetratricopeptide repeat domains and the JmjC catalytic domain as compared with other areas of the protein. Although UTY does not display conservation of the Tetratricopeptide domains of KDM6B, there is strikingly high conservation of the JmjC domain. These protein sequence similarities in important domains imply the possibility of UTY possessing functional histone demethylase activity. Protein sequence and domain data were obtained from UniProt.61 UniProt accession numbers of sequences used: O14607 (UTY), 015550 (UTX/KDM6A) and 015054 (KDM6B). Full size image Autoimmunity Haplogroup I was also associated with downregulation of pathways involved in human autoimmunity.44 Further support for the Y chromosome as a potential autoimmunity locus comes from animal models of diseases with significant autoimmune components such as experimental allergic encephalomyelitis (EAE) and experimental myocarditis.63 Indeed, experiments on consomic strains of mice showed that the Y chromosome defining the strain strongly influenced the susceptibility to and severity of EAE and myocarditis.63 Copy numbers of mouse Y genes, SLY and RBMY, were correlated with disease severity and the strains with reduced susceptibility carried fewer copies of these genes.63 In addition, transcriptomic analysis showed 398 differentially expressed Y-chromosome transcripts in the macrophages and CD4+ T cells between the more and less susceptible strains.63 These observations have been mirrored (at least to some extent) in male patients with an early form of multiple sclerosis – clinically isolated syndrome (CIS) – a disease with a strong autoimmune component and the human correlate of EAE.63 Compared with healthy controls, CD4+ T cells from individuals with CIS showed differential expression of a large proportion of the same Y genes identified in the mouse autoimmune models, suggesting a common Y-determined genetic basis to autoimmunity in mice and humans.63 Future studies With the emerging availability of large data sets comprising clinical phenotypes and Y-chromosome genotypes, future research should fully utilise the power of phylogenetic analysis to explore the potential contribution of Y chromosome to complex polygenetic traits. In particular, the evidence suggests the Y chromosome can be a powerful determinant of male immunity, including autoimmunity. We therefore propose studies to determine associations between Y-chromosome haplogroup and autoimmune disorders (such as rheumatoid arthritis), many of which exhibit sexual dimorphisms.64 In addition, trans-ethnic mapping studies would benefit our understanding of the complex relationship between Y-chromosome haplogroup and CAD. In the original British cohorts, two haplogroups predominated – R1b1b2 (70.0–72.7%) and I (14.5–17.0%).44 Although the elevated CAD risk associated with haplogroup I was attributed to haplogroup I posing increased susceptibility, an alternative interpretation of the results could be that the main non-I haplogroup (R1b1b2) offered protection against CAD. The uncertainty regarding precise identification of the causal haplogroup is exacerbated by two haplogroups accounting for nearly 90% of the cohort. Analyses of populations of different ethnicities (such as East Asians) with greater haplogroup diversity and an absence of haplogroup I13 would enable greater understanding of the specific cause for the altered CAD risk. If an association was found between haplogroup R (the clade containing R1b1b2 as a subtype) and reduced CAD risk in such a population, this would suggest that rather than haplogroup I elevating CAD risk, the causal factor for the original association was haplogroup R1b1b2-mediated protection against CAD. In contrast, an association between a haplogroup completely absent in Europe (eg, O) and CAD in East Asians would imply a greater range of unique genetic variants underpinning Y-mediated CAD risk. By analysing the locations of these variants, particular Y genes likely to have altered expression (variants within promoter sites) or altered molecular function of encoded proteins (variants within exon sequences) could be identified. Further investigation of the inflammatory and immune systems as potential mediators of the link between Y lineage and CAD requires high-fidelity immune phenotyping studies. This should involve extensive RNA sequencing of a wide range of target immune cells and tissues involved in CAD pathogenesis (such as dendritic cells, T and B lymphocytes) to provide much needed insight into the expression of all MSY protein-coding genes and their influence on underlying inflammatory and immune responses. Furthermore, the functional roles of Y-chromosome long noncoding RNAs and pseudogenes, especially within control of genetic regulation, remain largely unexplored.65 It is anticipated that examination of Y non-protein-coding elements within the broader ENCODE project will identify potential functional pathways mediating the associations between the Y and disease. Conclusions Data from association studies have revealed a potential role for the genetic variation within the Y chromosome in determination of men’s health and susceptibility to disease. This contrasts with initial pessimistic views about the Y as a futile, redundant piece of DNA. One of the strongest pieces of evidence is the association between haplogroup I and increased CAD risk, in the context of inflammation and immunity. Future endeavours will need to concentrate on identifying specific MSY genes that directly influence inflammatory and adaptive immunity processes within atherosclerosis. The identification of 12 X–Y dosage-sensitive gene pairs has refined our focus for future studies. Three of these pairs warrant further attention given their prior associations with haplogroup I and/or immune processes: UTY/UTX, PRKY/PRKX and KDM5D/KDM5C, with UTY being the most promising functional candidate. References Tommasi C : Ipertricosi auricolare famigliare. 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Author information Authors and Affiliations Division of Cardiovascular Sciences, Faculty of Biology, Medicine and Health, University of Manchester, Manchester, UK Akhlaq A Maan, James Eales, Artur Akbarov, Joshua Rowland, Xiaoguang Xu & Maciej Tomaszewski Department of Genetics, University of Leicester, Leicester, UK Mark A Jobling School of Applied and Biomedical Sciences, Faculty of Science and Technology, Federation University, Mount Helen Campus, Ballarat, VIC, Australia Fadi J Charchar Division of Medicine, Central Manchester NHS Foundation Trust, Manchester Academic Health Science Centre, Manchester, UK Maciej Tomaszewski Authors Akhlaq A Maan View author publications You can also search for this author inPubMed Google Scholar James Eales View author publications You can also search for this author inPubMed Google Scholar Artur Akbarov View author publications You can also search for this author inPubMed Google Scholar Joshua Rowland View author publications You can also search for this author inPubMed Google Scholar Xiaoguang Xu View author publications You can also search for this author inPubMed Google Scholar Mark A Jobling View author publications You can also search for this author inPubMed Google Scholar Fadi J Charchar View author publications You can also search for this author inPubMed Google Scholar Maciej Tomaszewski View author publications You can also search for this author inPubMed Google Scholar Corresponding author Correspondence to Maciej Tomaszewski. 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Download citation Received: 28 February 2017 Revised: 16 June 2017 Accepted: 28 June 2017 Published: 30 August 2017 Issue Date: November 2017 DOI: Share this article Anyone you share the following link with will be able to read this content: Get shareable link Sorry, a shareable link is not currently available for this article. Copy to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Subjects Cardiovascular genetics Haplotypes Immunogenetics Ischaemia Phylogenomics This article is cited by Sex differences in disease: sex chromosome and immunity Zuxi Feng Minjing Liao Liansheng Zhang Journal of Translational Medicine (2024) Testosterone deficiency promotes arterial stiffening independent of sex chromosome complement Anil Sakamuri Bruna Visniauskas Benard O. 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Arch Psichaitr Neuropat Antropol Crim Med Leg 1907; 28: 60–67. Google Scholar Stern C : The problem of complete Y-linkage in man. Am J Hum Genet 1957; 9: 147–166. CAS PubMed PubMed Central Google Scholar Skaletsky H, Kuroda-Kawaguchi T, Minx PJ et al: The male-specific region of the human Y chromosome is a mosaic of discrete sequence classes. Nature 2003; 423: 825–837. Article CAS Google Scholar Enriques P : Hologynic heredity. Genetics 1922; 7: 583–589. CAS PubMed PubMed Central Google Scholar Gates RR : Y-chromosome inheritance of hairy ears. Science 1960; 132: 145. Article CAS Google Scholar Gates RR, Chakravartti MR, Mukherjee DR : Final pedigrees of Y chromosome inheritance. Am J Hum Genet 1962; 14: 363–375. CAS PubMed PubMed Central Google Scholar Lee AC, Kamalam A, Adams SM, Jobling MA : Molecular evidence for absence of Y-linkage of the Hairy Ears trait. Eur J Hum Genet 2004; 12: 1077–1079. Article CAS Google Scholar Jarvik LF, Klodin V, Matsuyama SS : Human aggression and the extra Y chromosome. Fact or fantasy? Am Psychol 1973; 28: 674–682. Article CAS Google Scholar Nielsen J, Nordland E : Length of Y chromosome and activity in boys. Clin Genet 1975; 8: 291–296. Article CAS Google Scholar Ohno S The basic difference in constitution between the mammalian X and the Drosophila X. In: Labhart A, Mann T, Samuels LT, Zander J (eds): Sex Chromosomes and Sex-Linked Genes 1st edn Berlin, Germany: Springer Berlin Heidelberg, 1967, pp 82–91. Book Google Scholar Berta P, Hawkins JB, Sinclair AH et al: Genetic evidence equating SRY and the testis-determining factor. Nature 1990; 348: 448–450. Article CAS Google Scholar Vogt PH, Edelmann A, Kirsch S et al: Human Y chromosome azoospermia factors (AZF) mapped to different subregions in Yq11. Hum Mol Genet 1996; 5: 933–943. Article CAS Google Scholar Jobling MA, Tyler-Smith C : The human Y chromosome: an evolutionary marker comes of age. Nat Rev Genet 2003; 4: 598–612. Article CAS Google Scholar The Y Chromosome Consortium: A nomenclature system for the tree of human Y-chromosomal binary haplogroups. Genome Res 2002; 12: 339–348. Article Google Scholar Karafet TM, Mendez FL, Meilerman MB, Underhill PA, Zegura SL, Hammer MF : New binary polymorphisms reshape and increase resolution of the human Y chromosomal haplogroup tree. Genome Res 2008; 18: 830–838. Article CAS Google Scholar Poznik GD, Henn BM, Yee M-C et al: Sequencing Y chromosomes resolves discrepancy in time to common ancestor of males versus females. Science 2013; 341: 562–565. Article CAS Google Scholar Scozzari R, Massaia A, Trombetta B et al: An unbiased resource of novel SNP markers provides a new chronology for the human Y chromosome and reveals a deep phylogenetic structure in Africa. Genome Res 2014; 24: 535–544. Article CAS Google Scholar Hallast P, Batini C, Zadik D et al: The Y-chromosome tree bursts into leaf: 13,000 high-confidence SNPs covering the majority of known clades. Mol Biol Evol 2015; 32: 661–673. Article CAS Google Scholar van Oven M, Van Geystelen A, Kayser M, Decorte R, Larmuseau MH : Seeing the wood for the trees: a minimal reference phylogeny for the human Y chromosome. Hum Mutat 2014; 35: 187–191. Article CAS Google Scholar Winham SJ, de Andrade M, Miller VM : Genetics of cardiovascular disease: importance of sex and ethnicity. Atherosclerosis 2015; 241: 219–228. Article CAS Google Scholar Aken BL, Ayling S, Barrell D et al: The Ensembl gene annotation system. Database 2016; 2016: baw093. Article Google Scholar The Broad Institute of MIT and Harvard. GTEx Portal, Version 6p [Internet]. [updated 16 May 2016 1 December cited 2016]. Available from Bellott DW, Hughes JF, Skaletsky H et al: Mammalian Y chromosomes retain widely expressed dosage-sensitive regulators. Nature 2014; 508: 494–499. Article CAS Google Scholar Ely DL, Turner ME : Hypertension in the spontaneously hypertensive rat is linked to the Y chromosome. Hypertension 1990; 16: 277–281. Article CAS Google Scholar Turner ME, Johnson ML, Ely DL : Separate sex-influenced and genetic components in spontaneously hypertensive rat hypertension. Hypertension 1991; 17: 1097–1103. Article CAS Google Scholar Ely D, Turner M, Milsted A : Review of the Y chromosome and hypertension. Brazilian J Med Biol Res 2000; 33: 679–691. Article CAS Google Scholar Kren V, Qi N, Krenova D et al: Y-chromosome transfer induces changes in blood pressure and blood lipids in SHR. Hypertension 2001; 37: 1147–1152. Article CAS Google Scholar Ely D, Underwood A, Dunphy G, Boehme S, Turner M, Milsted A : Review of the Y chromosome, Sry and hypertension. Steroids 2010; 75: 747–753. Article CAS Google Scholar Milsted A, Underwood AC, Dunmire J et al: Regulation of multiple renin-angiotensin system genes by Sry. 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Miller-Rabin primality test | Mathematical Cryptography Skip to primary content Skip to secondary content Mathematical Cryptography focusing on the math in public key cryptography Search Main menu Home About Tag Archives: Miller-Rabin primality test Post navigation ← Older posts Using a primality test to find square roots of 1 modulo a Carmichael number Posted on November 3, 2024 by Dan Ma 2 A square root of 1 is a solution to the congruence equation . We can always find two solutions to this equation, namely and (these two are the trivial solutions). Any solution to this equation that is not is a non-trivial solution. There exist no non-trivial solutions if the modulus is a prime. If the modulus is a product of distinct primes in general and a Carmichael number in particular, there would be non-trivial square roots of 1. We demonstrate how to find the non-trivial square roots of one modulo a Carmichael number using the Miller-Rabin test instead of using the Chinese Remainder Theorem (CRT). To use CRT, we would need to know the factoring of the Carmichael number in question. We use the Miller-Rabin primality test to show that the Carmichael number is not a strong pseudoprime to some bases. This process would identify non-trivial square roots of 1. We present an example on how this is done. See here for a discussion of the Miller-Rabin test. Background Information The Fermat Little Theorem states that if is a prime, then for any base that is relatively prime to . The condition can be used as a primality test. If this condition is satisfied, then we say is a probable prime to the base . Unfortunately, there are composite numbers such that mod for all that is relatively prime to (such composite numbers are called Carmichael numbers). The smallest Carmichael number is 561 = 17 x 33. Applying the Fermat Little Theorem on a Carmichael number , we might mistake it for a prime number. The Miller-Rabin test is a much better tool for identifying Carmichael numbers. We now describe how the Miller-Rabin test works. Miller-Rabin Test …. Let be a positive odd integer. Write where and is odd. Let be a positive integer that is relatively prime to . Perform the following series of modular exponentiation: …. ….(1)…. modulo …. The sequence (1) consists of numbers. The last term in the sequence is identical to modulo and the term before the last term is identical to modulo . After the first term is computed, each subsequent term is the square of the preceding term modulo . …. The Basis for the Test If is prime, then one of the following is true. …. ….(a) The first term in Sequence (1) is a 1. ….(b) The first 1 in Sequence (1) is preceded by -1. …. What the Test Tells Us If either (a) or (b) is true, then is said to be a strong probable prime to the base (i.e., passes the base- Mill-Rabin test). If is known to be composite and passes the base- Miller-Rabin test, then is said to be a strong pseudoprime to the base . …. ….If the modulus fails the Miller-Rabin test for one base , then we have conclusive ….proof that is composite. …. ….If the modulus passes the Miller-Rabin test for several randomly chosen bases , …. is likely a prime number. …. Comments The idea behind the test is the roots of unity modulo a prime, the idea that for prime moduli, there are only trivial square roots of 1 (see Lemma 1 here). With this idea, condition (a) or condition (b) is true if the modulus is prime (see Theorem 2 here). If a non-trivial square root of 1 is found, then the modulus must be composite. …. On the other hand, the Miller-Rabin test is more reliable. If the bases are randomly chosen and if enough bases are used, it is highly likely that a composite modulus will fail the Miller-Rabin test for some base. If is composite, then for at least 75% of the bases with , the Miller-Rabin test will fail (see Theorem 3 here). Thus, the Miller-Rabin test should be randomized. For example, there is a 337-digit number that is a strong pseudoprime to the prime number bases less than 200 (see here). If only small bases are used (such as 2, 3, 7, …), then this 337-digit number will pass the Miller-Rabin test. Therefore, in implementing the Miller-Rabin test, the bases should be randomly chosen. …. In contrast, the Fermat Little Theorem as a primality test is not 100% reliable. It will always flag Carmichael numbers as prime (i.e., Carmichael numbers are false positives). …. The modular exponentiation to produce Sequence (1) can be computed efficiently using the fast powering algorithm (described here). In the next section, we present an example to show how this test can help find non-trivial square roots of 1 modulo a Carmichael number. Finding Non-Trivial Square Roots Let be the number 24262,899886,6496173. This is an 18-digit Carmichael number with factoring 55339 x 151579 x 28924933. Since there are 3 factors, the total number of square roots of 1 is 8, with 2 of them being the trivial ones, and 6 of them being the non-trivial square roots of 1. In general, the total number of square roots would be 2 raised to the number of factors of the modulus. This number less 2 would be the total number of non-trivial square roots of 1. If the factoring of the modulus is not known, the process of finding square roots is less precise. Once several non-trivial square roots of 1 are found, we can use them to factor the Carmichael number (see below). The approach we take is to use prime bases starting with 2 until all six non-trivial square roots of 1 are found. Since the goal here is not primality testing, the bases need not be randomized. For each base we use, we calculate the sequence (1) described above. Here’s bases that are used: 2, 3, 5, 7, 11, 13, 17, 19, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113 The process stops with the base 113 (producing the sixth square root of 1). The bases highlighted in blue are the ones producing a new non-trivial square root of 1. The three bases highlighted in red are the bases at which the Miller-Rabin test passes. This means that the Carmichael number in question is a strong pseudoprime to both bases 7, 73 and 79. Write where is 2 and = 60657, 249716, 624043. Sequence (1) as described above has 3 terms, , and . We calculate the sequence for each prime base up to 113. The following displays the bases with a new square root of 1. ….Base 2……10700 697604 866079, .233393 985126 859505, .1 ….Base 3……9235 013739 636668, .1, .1 ….Base 11…..147345 506987 502923, .1, .1 ….Base 61…..86048 478139 356581, .1, .1 ….Base 71…..156580 520727 139592, .1, .1 ….Base 113….95283 491878 993250, .1, .1 We only display these 6 bases because the other bases produce the same square roots of 1. The number highlighted in blue are the non-trivial square roots of 1 modulo . Note that the blue numbers precede 1, indicating that the square of each blue number is 1. Of course, there are two more square roots of 1, , which are the trivial square roots of 1. Using Chinese Remainder Theorem Using the Chinese Remainder Theorem produces the following 8 systems of congruence equations where = 55339, = 151579 and = 28924933 (the 3 prime factors of the Carmichael number ). ….A…….. ….B…….. ….C…….. ….D…….. ….E…….. ….F…….. ….G…….. ….H…….. System A and system H give the trivial solutions of 1 and -1, respectively. The solutions of these systems of equations are: ….A…….. 1 ….B…….. 233393 985126 859505 ….C…….. 95283 491878 993250 ….D…….. 156580 520727 139592 ….E…….. 86048 478139 356581 ….F…….. 147345 506987 502923 ….G…….. 9235 013739 636668 ….H…….. -1 ……..all modulo 242628 998866 496173 The idea here is that we solve these three separate quadratic congruence equations: mod 55339, mod 151579, and mod 28924933. Each of the 3 moduli is prime. So each quadratic congruence equation has 2 solutions of 1 and -1 modulo the respective prime. These lead to the 8 systems of equations indicated above. Examples of Euler Pseudoprimes and Strong Pseudoprimes Because = 24262 899886 6496173 is a Carmichael number, it is a pseudoprime to all bases that are relative prime to it, i.e., mod for all that is relatively prime to . The calculation performed here shows that is not a strong pseudoprime to all prime bases up to 113, except for bases 7, 73 and 79. ….Base 7……..242628 998866 496172 -1, .1, .1 ….Base 73……242628 998866 496172 -1, .1, .1 ….Base 79……1, .1, .1 In each of these 3 sequences, either the first term is 1 or the term preceding the first 1 is -1. Thus, the Carmichael number passes the Miller-Rabin test to these 3 bases. This demonstrates that a Carmichael number can be a strong pseudoprime to some bases. However, for a Carmichael number (composite number in general), there are many more bases at which the number fail (at least 75% of the bases) than pass (at most 25% of the bases). This example tells us that, to use Miller-Rabin test as a primality test, we would need to randomize the bases. In this example if we only use bases 7, 73 and 79, we might mistake the Carmichael number as a prime. Of the bases with , the Carmichael number in question can pass the Miller-Rabin test for up to 25% of the bases. To have the probability working for us, we need to randomize the bases. We now give some examples of Euler pseudoprimes. First, evaluate the Jacobi symbol for the following bases and compare the Jacobi symbol with Sequence (1): …. ….Sequence (1):….9235 013739 636668, 1, 1 …. ….Sequence (1):….9235 013739 636668, 1, 1 …. ….Sequence (1):….9235 013739 636668, 1, 1 Because the Jacobi symbol agree with the term preceding the last term, is an Euler pseudoprime to each of the bases 3, 19 and 23. These are examples of Euler pseudoprimes but not strong pseudoprimes. Consider the following Jacobi symbols. …. ….Sequence (1):….10700 697604 866079, 233393 985126 859505, 1 …. ….Sequence (1):….1465 683865 229412, 233393 985126 859505, 1 …. ….Sequence (1):….158046 204592 369003, 233393 985126 859505, 1 For the last 3 bases, the Jacobi symbol does not agree with the term preceding the last term. This shows that is not an Euler pseudoprime to the bases 2, 5 and 23. However, is a pseudoprime to these bases because is a Carmichael numbers. Thus, this gives examples of pseudoprimes but not Euler pseudoprimes. More examples. Consider the Jacobi symbol for the following bases. …. ….Sequence (1):….86048 478139 356581, 1, 1 …. ….Sequence (1):….156580 520727 139592, 1, 1 …. ….Sequence (1):….95283 491878 993250, 1, 1 In each of the last 3 bases, the Jacobi symbol disagree with the term preceding the last term. The disagreement is of a more subtle kind – the Jacobi symbol is -1 and the term before the last term is 1. However, this also means that the Carmichael number in question is not an Euler pseudoprime to these 3 bases (for a more subtle reason). For more discussion on strong pseudoprimes and Euler pseudoprimes and their relationship, see here and here. Here’s the relationships among three classes of pseudoprimes. Partial Factoring of Carmichael Numbers In case a Carmichael number is large or its factorization is not known, we can obtain a factoring if enough non-trivial square roots of 1 are known. If is a square root of 1, we have mod . Then we have for some integer . Next, we would evaluate the GCD of and to further reduce. We show how this is done using the two square roots 9235 013739 636668 and 95283 491878 993250. From these 2 square roots, we can recover the three factors of the Carmichael number . Using the square root 9235 013739 636668 ….1. 9235 013739 636667 x 9235 013739 636669 = x ….2. = 351 595711 063851 (the left-hand-side divided by ) ….3. GCD(351 595711 063851, 9235 013739 636669) = 1 100949 ….4. 9235 013739 636667 x 8388 230281 = x 319 275199 (reduced by the GCD) ….5. GCD(319 275199, 9235 013739 636667) = 319 275199 ….6. 28 924933 x 8388 230281 = 242628 998866 496173 (reduced by the GCD) In the above factoring, 28 924933 is a prime factor. Instead of factoring 8388 230281, we perform the same calculation with another square root in order to find another prime factor. Using the square root 95283 491878 993250 ….1. 95283 491878 993249 x 95283 491878 993251 = x ….2. = 37419 038396 353263 (the left-hand-side divided by ) ….3. GCD(37419 038396 353263, 95283 491878 993251) = 628606 151769 ….4. 95283 491878 993249 x 151579 = x 59527 (reduced by the GCD) ….5. GCD(59527, 95283 491878 993249) = 59527 ….6. 1 600676 867287 x 151579 = 242628 998866 496173 (reduced by the GCD) ….7. 1 600676 867287 = 28 924993 x 55339 (the remaining prime factor) The two iterations give us two prime factors: 28 924993 and 151579. From these we derive the third factor. If there are more than 3 prime factors, we can keep using more square roots of 1 to find the prime factors. The factoring method just demonstrated is quite efficient. It involves a series of modular exponentiation to find the non-trivial square roots of 1. With the non-trivial square roots, we have instances of difference of squares , which gives us a way to factor the modulus in question. This is somewhat related to Farmet’s differences of squares method in factoring, except that in this case we obtain the difference of squares by finding non-trivial square roots of 1 in the process of searching for strong pseudoprimes. The calculation for this factoring method is quite efficient since it only involves computing modular exponentiation (to find the non-trivial square roots of 1) and computing GCDs. Of course, this is a limited factoring algorithm since it only works on a small subset of the integers. For this method to work, the modulus needs to be a pseudoprime to several bases and needs to be not a strong pseudoprime to those same bases. Such numbers would contain the Carmichael numbers. But such numbers are rare. Dan Ma Miller-Rabin Primality Test Daniel Ma Miller-Rabin Primality Test Dan Ma Strong Pseudoprimes Daniel Ma Strong Pseudoprimes Dan Ma Carmichael Number Daniel Ma Carmichael Number 2024 – Dan Ma Posted: November 3, 2024 Posted inPrimality Testing | TaggedCarmichael numbers, Euler Pseudoprime, Miller-Rabin primality test, Pseudoprime, Strong pseudoprime | 2 Replies Miller-Rabin Primality Test Posted on November 19, 2022 by Dan Ma 2 This post is about the Miller-Rabin primality test. Fermat Primality Test The calculation behind the Miller-Rabin primality test is closely related to that of the Fermat primality test. It is instructive to recall the calculation of the Fermat test. The Fermat test is based on the modular exponentiation modulo , where is the odd positive integer being tested and is a positive integer that is relatively prime to . According to the Fermat Little Theorem, if is prime, then modulo . This fact is the basis of the Fermat primality test. If modulo , this provides evidence that may be prime since this is a property that is shared by all prime numbers. Thus, any integer is called a probable prime to base if modulo . If is a probable prime to base and if is composite, then is said to be a pseudoprime to base . Furthermore, if is not a probable prime to base , i.e., modulo , then the base is said to be a Fermat witness for (the compositeness of) . This is how the Fermat test is run. Given a positive integer whose “prime or composite” status is not known, choose several bases that are relatively prime to at random. Compute modulo for the chosen bases . If for one base , modulo , we have conclusive proof that is composite. If modulo for all the chosen bases , we have very strong evidence for the primality of . The Fermat test is discussed in the two preceding posts (here and here). The Miller-Rabin Calculation The Miller-Rabin test rests on the following idea. Lemma 1 (Roots of Unity Modulo a Prime) Let be a prime number. Suppose that modulo . Then modulo or modulo . Clearly, both and modulo are solutions to the congruence equation modulo . These two solutions are the trivial solutions to the equation. We claim that there is no other solution (no non-trivial solution) when the modulus is prime. To this end, we show that if modulo , then or modulo . Let be such that modulo . Then divides . Since is prime, either divides or divides (due to Euclid’s Lemma). If divides , then modulo . If divides , then modulo . The calculation for the Miller-Rabin test is the same modular exponentiation but built up in several steps. Let be an odd positive integer. Express where and is odd. Let be a base relatively prime to . Instead of computing the exponentiation directly, we compute and then square times to obtain . In other words, we calculate the terms in the following sequence modulo . (1)……………. The first term in the sequence is modulo . Each subsequent term is the square of the preceding term. This is done until we obtain . For all odd primes , the above sequence follows a certain pattern, which is described in the following theorem. Theorem 2 Let be an odd prime number. Let be a positive integer that is relatively prime to . Then either one of the following holds. modulo .……………………………………………………………….(a) modulo for some with .………………….(b) The first bullet point says that the first term in the sequence (1) is congruent to 1 modulo . The second bullet point says that the term immediately preceding the first 1 in the sequence is congruent to -1 modulo . The proof of Theorem 2 relies on Lemma 1. The proof goes something like this. First, the last term in the sequence must be a 1 due to the Fermat Little Theorem. If modulo , then the entire sequence consists of 1’s. Then (a) is satisfied. Suppose the sequence is not entirely 1’s. Then the term that is the first 1 must have a preceding term. Then the square of the preceding term is congruent to 1 modulo . Since is prime, that preceding term can only be 1 or -1. However, the preceding term can only be a -1 since it precedes the term that is the first value of 1. Two Pieces of Information About Primality Theorem 2 describes a property possessed by all old primes. Whenever an odd positive integer satisfies this property with a certain base , i.e., the sequence (1) satisfied either (a) or (b), we consider that as evidence that may be a prime. Hence, we have this definition. Suppose is an odd positive integer with the “prime or composite” status not known. Let be a base that is relatively prime to . Compute the sequence (1). If either condition (a) or condition (b) holds, then is said to be a strong probable prime to base . The underlying calculation in the sequence (1) attempts to capture two pieces of information about primality. The first piece is based on the Fermat Little Theorem (whether is congruent to 1). The second piece of information is the roots of unity modulo a prime (Lemma 1). If the Miller-Rabin calculation indicates a probable prime, the evidence is stronger than that from the Fermat test. Two pieces of evidence is more corroborative than just one piece of evidence. The Millar-Rabin test will be more capable to differentiate composite numbers. This is because there are two ways to detect compositeness embedded in the calculation of the sequence (1) described above. The compositeness of a number can be revealed by the Fermat Little theorem (whether is congruent to 1 or not) and by whether there exists a non-trivial square root of 1 modulo . This is demonstrated in the examples in the next section. Examples of the Millar-Rabin Calculation In this section, we demonstrate how the Miller-Rabin calculation can detect the compositeness of Carmichael numbers, which the Fermat test fails to do. Example 1 Consider 29,341. This is a composite number with 29,341 = 13 x 37 x 61. This is a Carmichael number. Thus, modulo 29341 for any coprime base . As a result, the Fermat test will not be able to detect the compositeness of this and other Carmichael numbers. Note that . We first perform base 2 calculation, which produces the sequence . The resulting sequence is: 26424, 29340, 1 The last term is 1 (due to 29341 being a Carmichael number). The term preceding the last term is 29340, which is congruence to -1. Thus, condition (b) is satisfied. This means that 29341 is a strong probable prime to base 2. Let’s try base 3 calculation, which is the sequence . The resulting sequence is: 22569, 1, 1 The term preceding the first 1 is 22569, which is neither 1 nor -1. This violates conditions (a) and (b) described above. This gives a conclusive proof that 29341 is a composite number. The number 22569 is a non-trivial square root of 1 modulo 29341. The existence of the non-trivial square root shows that the modulus 29341 cannot be prime. Example 2 Consider 172,947,529. This is a Carmichael number with 172,947,529 =307 x 613 x 919. First, we have where 21,618,441. Let’s try the base 2 calculation, producing the sequence , resulting in: 40063806, 2257065, 1, 1 Note that the term preceding the first 1 is 2,257,065, which is neither 1 nor -1. This means that the conclusion of Theorem 2 is not satisfied. As a result, the Carmichael number 172,947,529 is shown to be composite. The two examples demonstrate that the Miller-Rabin works correctly where the Fermat test fails. Referring back to the point about the two pieces of evidence for primality, the first way of detecting compositeness does not work in both Example 1 and Example 2, i.e., the congruence is 1. However, the second piece of evidence enables the Miller-Rabin calculation to detect the compositeness. Two pieces of evidence is better than just one. Therefore, the Miller-Rabin test is superior than the Fermat test. For the Miller-Rabin calculation to work correctly, we need to find the right bases to use. In Example 1, using base 2 does not detect the compositeness. Luckily, trying base 3 works. How likely is it to find a base that will detect a composite number? If a number is composite, the calculation described in Theorem 2 will detect the compositeness of with at least 75% of the bases coprime to . See Theorem 3 below. The Implementation To implement the Miller-Rabin test, randomly select several bases that are relatively prime to , the number to be tested. For each chosen base , compute the sequence (1). If, for one chosen base , both conditions (a) and (b) in Theorem 2 do not hold, then is proven to be composite. If, for all chosen bases , one of the conditions (a) and (b) holds, then is a strong probable prime to all the chosen bases, which indicates that we have obtained strong evidence for primality and is likely a prime number. Strong Pseudoprime and Miller-Rabin Witness Additional terminology will help the discussion. For a given base , if neither condition (a) nor condition (b), as described in Theorem 2, holds, then is proven to be composite, in which case, the base is said to be a Miller-Rabin witness for (the compositeness of) . For simplicity, we use the term witness for Miller-Rabin witness in the remainder of this article. If is a strong probable prime to base and if is composite, then is said to be a strong pseudoprime to base . Between the Fermat test and the Miller-Rabin test, there are two notions of probable primes, two notions of pseudoprimes and two notions of witnesses. How do they relate? We go back to the above section on the two pieces of evidence for primality. A probable prime under the Miller-Rabin test must also be a probable prime under the Fermat since having two pieces of evidence is stronger than one piece of evidence. Therefore, we have the following implications. Strong probable prime to base Probable prime to base Strong pseudoprime to base Pseudoprime to base The above implications cannot be reversed. The number in Example 1 is a Carmichael number, hence is a pseudoprime to base 3. However, it is not a strong pseudoprime to base 3. In general, a number can be a weaker form of pseudoprime and not be a stronger form of pseudoprime. Strong pseudoprime to base Pseudoprime to base How do the two notions of witnesses relate? We have the following implications. Fermat witnessMiller-Rabin witness Fermat witnessMiller-Rabin witness A Fermat witness violates one piece of evidence of primality, namely that from the Fermat Little Theorem. It means that it would be a witness from the point of view of the Miller-Rabin test. On the other hand, a Miller-Rabin witness does not have to be a Fermat witness. It is possible that a Miller-Rabin witness is to violate the second piece of evidence by producing a non-trivial square root of 1 while satisfying the Fermat Little Theorem. In Example 1, 3 is a (Miller-Rabin) witness for the integer 29,341 and 3 is not a Fermat witness for 29,341. In Example 2, the base 2 is a witness for 172,947,529, while base 2 is not a Fermat witness. It is because in both examples, the Miller-Rabin calculation produces a non-trivial square root of unity. Performance The above discussion and examples indicate that the Miller-Rabin test is superior than the Fermat test in that it can detect the Carmichael numbers as composite. The advantage of the Miller-Rabin test is real. For any composite number , there is at least one base that is a witness for , even when is a Carmichael number (as shown in the above examples). In fact, if is composite, at least 75% of the possible choices for bases are (Miller-Rabin) witnesses for . Consider the following theorem. Theorem 3 Let be an odd composite number. Then at least 75% of the numbers with are (Miller-Rabin) witnesses for . In other words, the composite number is a strong pseudoprime to at most 25% of the possible bases . With Theorem 3, the Miller-Rabin test can be made a truly probabilistic test for primality. This means that the test can be implemented in such a way that the error probability can be made as small as we wish. When we randomly select a base out of all the bases coprime with the composite , there is at least a 0.75 probability that is a witness, which means that the probability of error is at most 0.25. If we choose 50 or so bases at random and if the test indicates prime, the error probability is practically zero. Thus, the Miller-Rabin test is an efficient and practical test for finding large numbers that are prime with high probability. Dan Ma Fermat primality test Dan Ma Miller-Rabin primality test Daniel Ma Fermat primality test Daniel Ma Miller-Rabin primality test Dan Ma Carmichael numbers Daniel Ma Carmichael numbers 2022 – Dan Ma Posted inPrimality Testing | TaggedFermat primality test, Fermat witness, Miller-Rabin primality test, Probable prime, Pseudoprime, Strong probable prime, Strong pseudoprime, Witness | 2 Replies Another version of the primality test using a theorem of Lucas Posted on December 3, 2014 by Dan Ma Reply In the previous post, we prove a theorem that is attributed to Lucas and give examples on how to use it to prove primality. The task of using this theorem to prove primality is essentially the task of finding a primitive root. So this primality test could be a little tricky to use since there is no simple formula for finding primitive root. In this post, we discuss a variation (but an equivalence) of Lucas’ theorem that does not require using primitive root but still proves primality. The theorems In the previous post, we discuss and prove the following theorem. Theorem 1 (Lucas) Let be a positive integer. Then is prime if and only if there exists some integer such that , for all prime factor of . The value of in the statement of the theorem is a primitive root modulo . Though the theorem is fairly easy to prove, the theorem is important and has far reaching consequences. The basic idea in this theorem is the foundation of many other deterministic primality tests. So it is a good idea to examine this theorem in great details. One way to do that is to examine variations of it. In this post, we examine the following theorem. Theorem 2 Let be a positive integer. Then is prime if and only if for each prime factor of , there exists some integer such that , . The existential quantifier and the universal quantifier are switched in the statements of the two theorems. In Theorem 1, one value of works for all prime factors of . In Theorem 2, each prime factor of has one value of that satisfies some conditions. More importantly, the value of in Theorem 2 cannot be expected to be a primitive root. Computationally speaking Theorem 2 should be a little easier to use. The proof of Theorem of 2 uses the following basic lemma. Lemma 3 Suppose such that for . Then if divides for each , then divides . Proof of Lemma 3 Suppose that divides for each . Assume that . By the division algorithm, we can write where and are integers and . We claim that . Once the claim is proved, the lemma will follow. Suppose that . it follows from that divides for each . It follows that for some integers , the number can be expressed as follows: Because the numbers are mutually relatively prime, each with is a factor of . Thus for some integer . We have , leading to . This is a contradiction since is an integer. In the above argument, we make the assumption that . Suppose . Using the division algorithm, we have for some integers and with . As before, all the divides . The same argument shows that for some positive integer . Now , contradicting . So the assumption of is valid. Proof of Theorem 2 The direction is clear (see the proof of the same direction in Theorem 1). Now suppose that the number has the following prime factorization where each is prime and each . Suppose that for each prime factor , there exists an such that , . . For each , let be the least positive integer such that . In other words, is the order of modulo . It follows that divides . On the other hand, does not divide . If it did, we would have . For each , let for some integer . Note that cannot divide . Otherwise would divide . Since none of the prime factors of can be , it follows that must divide . On the other hand, divides . So divides . By Lemma 3, divides , giving . It is always that case that . We have shown that . By Lemma 2 in this previous post, is prime. The primality test from the theorems The above theorems are equivalent. So both theorems give essentially the same primality test. It is just that the implementation based on Theorem 2 is somewhat easier and in some cases much faster. No matter which theorem is used, it is always a good idea to apply the strong probable prime test (the Miller-Rabin test) first, just in case the compositeness can be easily detected. Of course, the limitation for the primality discussed here is that the factorization of must be known if is the number being tested. If Lucas’ theorem is not applicable and if probabilistic primality testing is not 100% satisfactory, other primality proving theorems may be explored. See . Examples In the previous post, we show how to apply Theorem 1. We now work some examples illustrating the use of Theorem 2. Example 1 Consider the number 38709183810571. The following is the prime factorization of . The last factor can easily be verified as prime by checking all prime numbers below . Use as a starting point. Note that the number works for all prime factors of except for 3. Now use . In conclusion, for all prime factors of except 3, the value satisfies the conditions in Theorem 2. For the prime factor 3 of , use . By Theorem 2, the number 38709183810571 is proved to be prime. Example 2 Use Theorem 2 to perform primality testing on the following number 3825123056546413183. This is Example 2 in this previous post. The factorization of is . The last factor 16157348744821 is a prime number (see the previous post). We start with . With , the above calculation works for all prime factors of , except 2. Now try . By Theorem 2, the number 3825123056546413183 is proved prime. Exercises Prove the compositeness or primality of each of the following numbers. In proving primality, use Lucas’ theorem according to Theorem 2. 204482919124364689 3825123056546413093 3825123056546413133 3825123056546413211 3825123056546413213 Reference Brillhart J., Lehmer D. H., Selfridge J.L., New primality criteria and factorizations of , Math. Comp., 29, no. 130, 620-647, 1975. Lehmer D. H., Tests for primality by the converse of Fermat’s theorem, Bull. Amer. Math. Soc., 33, 327-340 1927. Posted inPrimality Testing, Prime Numbers | TaggedMiller-Rabin primality test, Primality testing, Primitive root, Probable prime, Pseudoprime, Strong probable prime, Strong probable prime test, Strong pseudoprime | Leave a reply A primality proving algorithm using a theorem of Lucas Posted on December 1, 2014 by Dan Ma 2 In this post we discuss a theorem that can be used as a primality test that actually proves primality rather than just giving strong evidence for primality. The theorem is originally due to Lucas in 1891. The version we discuss here is proved by Lehmer . The primality test using the theorem is a limited one since the test requires that must be completely factored if is the number being tested. Though a limited test, this is a useful test in situations where has small prime factors or the factorization of is known in advance (e.g. factorial primes and primorial primes). The theorem Theorem 1 (Lucas) Let be a positive integer. Then is prime if and only if there exists some integer such that , for all prime factor of . The function is Euler’s phi function. If is a prime number, . The following lemma shows that the converse is true. The lemma is used in proving Theorem 1. Lemma 2 If , then is prime. Proof of Lemma 2 By definition is the number of integers where such that , i.e., and are relatively prime. First of all, . Suppose is composite. Then there is some where such that is a divisor of . Clearly . It follows that . Therefore if If , then must be prime. Proof of Theorem 1 It is well known that for any prime number , there exists a primitive root modulo (see here). A primitive root modulo is a number such that but for all . In other words, is the least exponent such that . Let be a prime number. Then . By the point that is just stated, there exists an that is a primitive root modulo . This primitive root would satisfy and for any prime divisor of . Now suppose that there exists an such that and for any prime divisor of . Let be the least positive integer such that . The number is said to be the order of modulo . We claim that . From the assumption that , we have . Thus . Suppose that . Then is an integer that is greater than 1. Let be a prime factor of . Set for some integer . Then , contradicting the assumption stated above. Thus . On the other hand, always holds (see here). Since is least, we have . It is always the case that . Thus we have established that . By Lemma 2, is prime. Remark The above proof shows that is prime from the existence of such that and for any prime divisor of . The proof also shows that is the least number such that . That is, the number is a primitive root modulo . Whenever is proved prime by finding such an , keep in mind that the task of primality proving using this theorem is essentially the task of finding a primitive root. The primality test from the theorem Theorem 1 can be fashioned into a primality test, or rather a primality proving algorithm. The key requirement to proving is prime is that must be completely factored. Because of this obstacle, the primality test is a limited one. The following are some of the cases for which Lucas’s theorem is suitable as a proof for the primality of . All prime factors of are small. For each prime factor of , either it is small or its primality can be established using Lucas’s theorem. The second case points to the fact that sometimes Lucas’ theorem can be used recursively to establish the primality of a number. Specifically, in factoring , we may come across a factor that is a probable prime. We then can use Lucas’ theorem to prove its primality. In doing so, we may come across another number that is a probable prime. We can then use Lucas’ theorem again and prove its primality and so on. As mentioned at the end of the previous section, the task of primality proving here is essentially the task of finding a primitive root. There is no easy formula for finding primitive roots. One can always start with and work the way up to a value of that works. But this may be a lengthy process. A faster option is to use randomly chosen values of . To use Theorem 1 to prove that is prime, it is a matter of finding a value of with that satisfies the two conditions in the theorem. If such an is not found after a reason number of iterations, then is probably composite. In this case, it makes sense to subject to a probabilistic primality test such as the Miller-Rabin test to check for compositeness. In light of the comment made in the preceding paragraph, before using Theorem 1, we should apply a probabilistic primality test on the number being tested. If the number is shown to be composite by the Miller-Rabin test, then there is no need to use Theorem 1. If the number is a strong probable prime to base 2 for example, then we can proceed to use Theorem 1. Even when Theorem 1 is applicable, there may still be a need to switch back to a probabilistic primality test if a suitable value of cannot be found in a reasonable number of iterations. Additional remark is given at the end of Example 1. Examples The examples demonstrated here are small enough to be realistically proved as prime by factoring (by computer). However, they are large enough to be excellent demonstrations of the method using the theorem of Lucas. Example 1 Consider the number 3825123056546413057. As a preliminary check, we find that it is a strong probable prime to base 2. The following is the prime factorization of . Choose a random number in the interval . Then calculate the following 8 congruences. If the first one is a 1 and the other 7 congruences are not 1, then we have a proof that 3825123056546413057 is prime. Using the random number 986534828637101811, we have the following results. The first congruence is a 1 and the remaining 7 congruences are not 1. Thus by Lucas’ theorem, the number 3825123056546413057 is prime. Remarks Upon choosing a value of (random or otherwise), if the first congruence is not a 1, then is composite by Fermat’s little theorem, and we are done. Suppose the first congruence is a 1. If the second congruence is not , then is composite. This is because if is prime, then the square root of 1 modulo must be . Since Lucas’ theorem requires that , the only legitimate value of the second congruence is -1. This is something we should look for. If the second congruence is not a -1, then choose another value of and start over. In fact, the second congruence should be the one to do first. If it is not a -1, then use another value of . Example 2 Using Lucas’ theorem, perform primality testing on the following number 3825123056546413183. As a preliminary check, it is a strong probable prime to base 2. Next, factor as far as possible. The last factor 16157348744821 of is an 14-digit number. It is a probable prime to base 2, meaning that . So we have a good reason to believe that might be a prime. To test whether it is prime, we can use the same method to test. So now we focus our attention on the number 16157348744821. Factor as far as possible. The last factor 58300313 is a prime number. The square root of is 7635.46. None of the prime numbers below 7635 is a factor of . So the above factorization is the prime factorization of . We calculation the following: We use as a starting point. We have the following results. We are lucky that the first we try works. Now we have proof that the number is a prime. As a result, the prime factorization of for the original number 3825123056546413183 is complete. To prove the primality of , we calculate the following. We try and then , both not working, with and . Next we try , with the following results. The above calculation for base shows that the number 3825123056546413183 is prime. Example 3 Consider the number 219944603708904241. The following is the factorization of . The last factor 1199465273 is a probable prime to base 2, meaning that . We have good evidence that is prime. To confirm, we apply Lucas’ theorem on this number. The following is the prime factorization of , where each factor is small and is easy to be determined as prime. To prove the primality of , we calculate the following. We try two random choices for and do not have the desired results. The next random choice is 526979896. The following calculation proves that is prime. With the prime factorization of being complete, we turn our attention to the following calculations. We generate the following 8 random choices of 013397886753078290 193080712858269996 012695419760523254 096180046746919966 134541150430885987 212892893489065625 209448807773524821 141141720485036352 and find that for all these 8 values of . This is highly unusual if the number is prime. We then realize that we forget to check for strong probable primality of . With where 13746537731806515, we calculate the following: The above calculation shows that the number 219944603708904241 is composite; it is a strong pseudoprime to base 2. This goes to show that before applying Lucas’ theorem, it makes sense to do some strong probable prime test to rule out numbers that happen to be composite. The number in Example 3 is a Carmichael number. It is the product of the prime factors , and where 55365. Exercises Prove the compositeness or primality of each of the following numbers. In proving primality, use Lucas’ theorem as shown in the above examples. 204482919124364689 3825123056546413093 3825123056546413133 3825123056546413211 3825123056546413213 Reference Lehmer D. H., Tests for primality by the converse of Fermat’s theorem, Bull. Amer. Math. Soc., 33, 327-340 1927. Posted inPrimality Testing, Prime Numbers | TaggedCarmichael numbers, Fermat primality test, Miller-Rabin primality test, Number theory, Primality testing, Probable prime, Strong probable prime, Strong pseudoprime | 2 Replies Large Carmichael numbers that are strong pseudoprimes to several bases Posted on November 29, 2014 by Dan Ma 1 There is a very strong theorem in the paper that has impactful theoretical results. For example, Theorem 1 in implies that for any given finite set of bases, there are infinitely many Carmichael numbers that are strong pseudoprimes to all the bases in the given finite set. Another implication of Theorem 1 of is that there are infinitely many Carmichael numbers that have no small prime factors. Given an odd number whose “prime versus composite” status is not known, set where and is odd. The number is said to be a probable prime to base if the conclusion of Fermat’s little theorem holds, i.e., . A probable prime to base could be prime or could be composite. If the latter, it is said to be a pseudoprime to base . The number is said to be a strong probable prime to base if or there exists some such that . In other words, the number is a strong probable prime to base if in the following sequence the first term is a 1 or there exists a -1 followed by a 1. A strong probable prime to base could be prime or could be composite. If the latter, it is said to be a strong pseudoprime to base . According to , there are only 4842 strong pseudoprimes to base 2 below . Only 13 of these 4842 numbers are strong pseudoprimes to the bases 2, 3 and 5. The paper has a listing of all strong pseudoprimes to the bases 2, 3 and 5 that are below . This listing has only 101 numbers. Only 9 these 101 numbers are strong pseudoprimes to bases 2, 3, 5 and 7. So there are indications that strong pseudoprimes are rare and that strong pseudoprimes to several bases are even rarer. However, there are infinitely many strong pseudoprimes to a given base. In particular, there are infinitely many strong pseudoprimes to base 2 (two proofs are given in this previous post and in this previous post). According to what is said at the beginning of the post, there are infinitely many Carmichael numbers that are strong pseudoprimes to all the first trillion prime bases! Furthermore, for these Carmichael numbers, any prime factors must be larger than the first trillion primes bases! It makes sense to examine this theorem so that we can further understand the implied results. The theorem is obviously a deep theoretical result, established by heavy duty machinery (theoretically speaking). Fermat’s little theorem implies that if happens to be a prime number, then is a probable prime to all bases that are relatively prime to . It is easy to see that if is a prime number, then is a strong probable prime to all bases that are relatively prime to . Equivalently, if is not a strong probable prime to some base , then is proved to be composite. Such a base is said to be a witness for the compositeness of . So a witness is a base, when used in the calculation for (1), will prove the compositeness of a number. Here’s one reason that the concept of strong pseudoprime is better than the concept of pseudoprime. If is composite, there is at least one base that is a witness for (see this previous post). A Carmichael number is a composite number such that it is a pseudoprime to all the bases relatively prime to . Thus the Fermat test will never detect the compositeness of Carmichael numbers. On the other hand, the strong probable prime test (the Miller-Rabin test) will likely to give the correct result on Carmichael numbers. Let be an odd positive composite number. Let denote the smallest witness for the compositeness of . Since any composite number has at least one witness, there is a least one. So is well defined. It is clear that if and only if is a strong pseudoprime to all bases . So any composite number that has a large least witness is a strong pseudoprime to many bases. The following is the statement of Theorem 1 of . Theorem 1 There are infinitely many Carmichael numbers such that Furthermore, whenever is sufficiently large, the number of such below is at least We show how Theorem 2 is derived from Theorem 1. Theorem 2 Given a finite set of bases, there are infinitely many Carmichael numbers that are strong pseudoprimes to all the bases in the finite set. Proof of Theorem 2 Consider the function defined in the right-hand side of the inequality (2) in the statement of Theorem 1. The key facts about this function are that: The function is an increasing function for sufficiently large . The function is increasing without bound, i.e., for each sufficiently large , there exists an such that . We use these two facts to derive the intended results. These two key facts will be discussed at the end. By Theorem 1, there exist Carmichael numbers such that (2) holds for each term in the sequence, i.e., for each . Based on the key facts, the following sequence is increasing without bound. Let be a finite set of positive integers . Choose some such that for each , is greater than all elements of . Let be the maximum of all the integers in the finite set . Now it is clear that for each , and that . It follows that for each of the following infinitely many Carmichael numbers the least witness is greater than . Thus all these Carmichael numbers are strong psuedoprimes to all bases , hence to all the bases in the finite set . To see that the function is increasing pass some sufficiently large , take the derivative of . The above derivative is positive if and only if the following is true: This last inequality always holds for sufficiently large . Thus for sufficiently large , the derivative of is positive, meaning that is increasing. To see that increases without bound, fix a sufficiently large . Choose an integer large enough such that and . Let where . We have the following. The above derivation shows that the function increases without bound. Theorem 2 implies the following theorem. For the infinitely many Carmichael numbers that are strong pseudoprimes to all the bases below a bound, no prime numbers within the bound can be a factor for these Carmichael numbers. Theorem 3 There are infinitely many Carmichael numbers that have no small prime factors. The Carmichael numbers that are claimed to exist by Theorem 3 will likely not be detected as composite by factoring. For example, consider a Carmichael number that is a strong pseudoprime to all the prime bases below . Its compositeness will likely not be detected by factoring based on the current computing technology and on the current factoring algorithms. This is another reason that the strong probable prime test (the Miller-Rabin test) is better than the Fermat test. The latter will declare the Carmichael number in question as prime while the former will likely be correct. Example Consider the following 397-digit number. 28871482380507712126714295971303939919776094592797 22700926516024197432303799152733116328983144639225 94197780311092934965557841894944174093380561511397 99994215424169339729054237110027510420801349667317 55152859226962916775325475044445856101949404200039 90443211677661994962953925045269871932907037356403 22737012784538991261203092448414947289768854060249 76768122077071687938121709811322297802059565867 The above is a 397-digit Carmichael number found in . It is a strong pseudoprime to all the prime bases up to 300. According to Theorem 2, there are infinitely many other Carmichael numbers that are strong pseudoprimes to all the prime bases below 300. The number is generated using the following formula. 29674495668685510550154174642905332730771991799853 04335099507553127683875317177019959423859642812118 8033664754218345562493168782883 Reference Alford W., Granville A., and Pomerance C., On the difficulty of finding reliable witnesses, L. Adleman and M.-D. Huang, editors, Algorithmic Number Theory: Proc. ANTS-I, Ithaca, NY, volume 877 of Lecture Notes in Computer Science, pages 1–16. Springer–Verlag, 1994. Alford W., Granville A., and Pomerance C., There are infinitely many Carmichael numbers, Ann. of Math. 139, 703-722, 1994. Arnault F., Constructing Carmichael numbers which are pseudoprimes to several bases, J. Symbolic Computation, 20, 151-161, 1995. Jaeschke G., On strong pseudoprimes to several bases, Math. Comp., Volume 61, No. 204, 915-926, 1993. Jiang Yupeng, Deng Yingpu, Strong pseudoprimes to the first 9 prime bases, arXiv:1207.0063v1 [math.NT], June 30, 2012. Pomerance C., Selfridge J. L., Wagstaff, S. S., The pseudoprimes to , Math. Comp., Volume 35, 1003-1026, 1980. Posted inPrimality Testing, Prime Numbers | TaggedCarmichael numbers, Miller-Rabin primality test, Number theory, Primality testing, Probable prime, Pseudoprime, Strong probable prime, Strong probable prime test, Strong pseudoprime, Witness | 1 Reply Large examples of strong pseudoprimes to several bases Posted on November 23, 2014 by Dan Ma 1 A strong pseudoprime to base is a composite number that passes the strong probable prime test (i.e. the Miller-Rabin test) in the base . There are indications that strong pseudoprimes are rare. According to , there are only 4842 numbers below that are strong pseudoprimes to base 2. Strong pseudoprimes to several bases are even rarer. According to , there are only 13 numbers below that are strong pseudoprimes to bases 2, 3 and 5. The paper tabulates the numbers below that are strong pseudoprimes to bases 2, 3 and 5. That listing contains only 101 numbers. These examples show that strong pseudoprimes to several bases may be rare but they do exist. In this post we discuss two rather striking examples of large numbers that are strong pseudoprimes to the first 11 prime bases for the first one and to the first 46 prime bases for the second one. One important lesson that can be drawn from these examples is that the implementation of the strong probable prime test must be randomized. The strong probable prime test Given an odd number whose “prime versus composite” status is not known, set where and is odd. Then calculate the following sequence of numbers: where is some integer relatively prime to . The first term can be efficiently calculated using the fast powering algorithm. Each subsequent term is the square of the preceding term. Each term is reduced modulo . If (i.e. the first term in sequence (1) is a 1) or for some (i.e. the term preceding the first 1 in the sequence is a -1), then is said to be a strong probable prime to the base . A strong probable prime to base could be a prime or could be composite. If the latter, it is said to be a strong pseudoprime to base . In fact, most strong probable primes to one base are prime. The strong probable prime test consists of checking whether is a strong probable prime to several bases. If is not a strong probable prime to one of the bases, then is composite for sure. If is a strong probable prime to all of the bases being used, then is likely a prime number in that the probability that it is composite is at most if is the number of bases. The last probability of is what makes the 337-digit number defined below striking. Here we have a number that is a strong probable prime to 46 bases. What could go wrong in declaring it a prime number? The problem is that using a pre-determined set of bases is not a randomized implementation of the strong probable prime test. Examples The following is a 46-digit found in that is a strong pseudoprime to the first 11 prime bases (the prime numbers from 2 to 31). 1195068768795265792518361315725116351898245581 The following is a 337-digit number found in that is a strong pseudoprime to all 46 prime bases up to 200 (the prime numbers from 2 to 199). 80383745745363949125707961434194210813883768828755 81458374889175222974273765333652186502336163960045 45791504202360320876656996676098728404396540823292 87387918508691668573282677617710293896977394701670 82304286871099974399765441448453411558724506334092 79022275296229414984230688168540432645753401832978 6111298960644845216191652872597534901 The following sets up the calculation for the Miller-Rabin test (strong probable prime test). where and are odd and 298767192198816448129590328931279087974561395 20095936436340987281426990358548552703470942207188 95364593722293805743568441333413046625584040990011 36447876050590080219164249169024682101099135205823 21846979627172917143320669404427573474244348675417 70576071717774993599941360362113352889681126583523 19755568824057353746057672042135108161438350458244 6527824740161211304047913218149383725 Random bases Though the second number is very striking, the author of has an even larger example in , a 397-digit Carmichael number that is a strong pseudoprime to all the 62 prime bases under 300! One lesson from these examples is that the implementation of the strong probable prime test should be randomized, or at least should include some randomly chosen bases in the testing. Any algorithm that implements the strong probable prime test in any “fixed” way (say, only checking the prime bases up to a certain limit) may incorrectly identify these rare numbers as prime. Let’s apply the strong probable prime test on the above numbers and using some random bases. Consider the following randomly chosen bases and where and . 932423153687800383671087185189848318498417236 23876349986768815408041169070899917334655923628776 58344592618224528502905948639172375368742187714892 08287654410018942444630244906406410549094447554821 42803219639200974486541191341820595453041950723891 75748815383568979859763861820607576949539863746244 98291058421101888215044176056586791235119485393994 789287924346696785645273545040760136 The calculation using random bases Here’s the calculation for the 46-digit number . where , and are: 1042866890841899880275968177787239559549445173 826876320842405260107866407865475914456579581 1195068768795265792518263537659322626328455738 Note that the last number is not a 1. So the number is not a strong probable prime to base . This means that it is composite. Here’s the calculation for the 337-digit number . where , and are: 43050290968654965761881145696359381339174664947842 93659429396009893693594328847691223585119425166890 43134041173054778367051375333950357876719375530986 40705386242996844394887879855798166233504226845778 76290707027478869178569806270616567220414388766208 75314254126730991658967210391794715621886266557484 525788655243561737981785859480518172 69330060289060873891683879069908453474713549543545 79381982871766126161928753307995063953670075390874 26152164526384520359363221849834543643026694082818 11219470237408138833421506246436132564652734265549 18153992550152009526926092009346342470917684117296 93655167027805943247124949639747970357553704408257 9853042920364675222045446207932076084 80383745745363949125707961434194210813883768828755 81458374889175222974273765333652186502336163960045 45791504202360320876656996676098728404396540823292 87387918508691668493091034289810372813316104284761 45244183107779151749589951324358651494383103941607 35777636976785468267798055151468799251924355070195 1772723904683953622590747688533861698 Interestingly, the last number agrees with the modulus from the first digit (the largest) to digit 167. It is clear that is clearly not a 1. So the 337-digit number is not a strong probable prime to base , meaning it is composite. Even though the number is a strong pseudoprime to all of the first 46 prime bases, it is easy to tell that it is composite by using just one random base. This calculation demonstrates that it is not likely to make the mistake of identifying large pseudoprimes as prime if randomized bases are used in the strong probable prime test. Some verification We also verify that the 46-digit is a strong pseudoprime to the first 11 prime bases. The asterisks in the above table mean that those cells have numbers that are modulo . Clearly the table shows that the number passes the strong probable prime test for these 11 bases. We also verify that is not a strong probable prime to base 37, the next prime base. where 977597583337476418144488003654858986215112009 368192447952860532410592685925434163011455842 1195068768795265792518263537659322626328455738 Note that the last number agrees with the modulus for the first 22 digits and they differ in the subsequent digits. It is clear that is not 1. So the strong pseudoprimality of stops at the base 37. We do not verify the number for all the 46 prime bases. We only show partial verification. The following calculation shows the pseudoprimality of to bases 197 and 199, and that the strong pseudoprimality of fails at 211, the next prime base. where 48799236892584399744277334997653638759429800759183 35229821626086043683022014304157526246067279138485 99367014952239377476103536546259094903793421522217 99291447356172114871135567262925519534270746139465 22832800077663455346130103616087329184090071367607 57478119260722231575606816699461642864577323331271 2974554583992816678859279700007174348 80191643327899921083661290416909370601037633208226 50175490124094760064341402392485432446383194439467 16432633017071633393829046762783433857505596089159 3600905184063673203 Note that is clearly not congruent to -1. Thus the number is not a strong pseudoprime to base 211 (though it is a pseudoprime to base 211). Clearly, the above calculation indicates that the number is a strong pseudoprime to bases 197 and 199. Remark Because strong pseudoprimes are rare (especially those that are strong pseudoprimes to several bases), they can be used as primality tests. One idea is to use the least pseudoprimes to the first prime bases . This method is discussed in this previous post. Another idea is to use the listing of strong pseudoprimes to several bases that are below a bound. Any number below the bound that is strong probable primes to these bases and that is not on the list must be a prime. For example, Table 1 of lists 101 strong pseudoprimes to bases 2, 3 and 5 that are below . This test is fast and easy to use; it requires only three modular exponentiations to determine the primality of numbers less than . The numbers and are not Carmichael numbers since and , making the random numbers and Fermat witnesses for these two numbers respectively. Another way to see that they are not Carmichael is that each of the two number and is also a product of two distinct primes according to . An even more striking result than the examples of and is that it follows from a theorem in that for any finite set of bases, there exist infinitely many Carmichael numbers which are strong pseudoprimes to all the bases in the given finite set. This is discussed in the next post. Reference Alford W., Granville A., and Pomerance C., _On the difficulty of finding reliable witnesses_, L. Adleman and M.-D. Huang, editors, Algorithmic Number Theory: Proc. ANTS-I, Ithaca, NY, volume 877 of Lecture Notes in Computer Science, pages 1–16. Springer–Verlag, 1994. 2. Arnault F., Constructing Carmichael numbers which are strong pseudoprimes to several bases, J. Symbolic Computation, 20, 151-161, 1995. 3. Arnault F., Rabin-Miller primality test: composite numbers that pass it, Math. Comp., Volume 64, No. 209, 355-361, 1995. 4. Jaeschke G., On strong pseudoprimes to several bases, Math. Comp., Volume 61, No. 204, 915-926, 1993. 5. Jiang Yupeng, Deng Yingpu, Strong pseudoprimes to the first 9 prime bases, arXiv:1207.0063v1 [math.NT], June 30, 2012. 6. Pomerance C., Selfridge J. L., Wagstaff, S. S., The pseudoprimes to , Math. Comp., Volume 35, No. 151, 1003-1026, 1980. Posted inPrimality Testing, Prime Numbers | TaggedCarmichael numbers, Fast powering algorithm, Fermat witness, Fermat's little theorem, Miller-Rabin primality test, Number theory, Primality testing, Probable prime, Pseudoprime, Strong probable prime, Strong probable prime test, Strong pseudoprime | 1 Reply A primality proving algorithm using least strong pseudoprimes Posted on November 22, 2014 by Dan Ma Reply This post is the continuation of this previous post. In this post, we discuss a deterministic primality proving algorithm that uses the least strong pseudoprimes to several prime bases. After describing the test, we present several examples. The previous post discusses the notion of witness for the strong probable prime test (the Miller-Rabin test). One important characteristic of the strong probable prime test is that for any composite number, there is always at least one witness (in fact lots of them). This means that the strong probable prime test is not going to be tripped up on a Carmichael number like it is for the Fermat test. When there is a guarantee that every composite number has a witness for its compositeness, it makes sense to talk about the least witness for a composite number . The statement that is equivalent to the statement that is a strong pseudoprime to all the bases less than or equal to . Strong pseudoprimes to base 2 are rare. Strong pseudoprimes to multiple bases are even rarer. According to , there are only 13 numbers below that are strong pseudoprimes to all of the bases 2, 3 and 5. Thus it is rare to have composite numbers whose . Because they are rare, knowing about strong pseudoprimes can help us find the primes. The test in question comes from . It had been improved and sharpened over the years. The paper seems to contain the best results to date regarding this test. To see how the method evolved and got improved, any interested reader can look at the references provided in . Let be the least strong pseudoprime to all of the first prime bases. The paper presents the following 11 least strong pseudoprimes. 2047 1373653 25326001 3215031751 2152302898747 3474749660383 341550071723321 3825123056546413051 To illustrate, the number 25326001 is the smallest strong pseudoprime to all the bases 2, 3 and 5. For any odd number less than 25326001, check whether is a strong probable to these 3 bases. If it is, then has to be prime. Otherwise, is a strong pseudoprime to bases 2, 3 and 5 that is less than 25326001! Of course, if happens to be not a strong probable prime to one of the 3 bases, then it is a composite number. The test using represents a primality test that actually proves primality rather than just giving strong evidence for primality. Using , the test only requires modular exponentiations. This test is a limited test since it only applies to numbers less than . However, it is interesting to note that the notions of strong probable primes and strong pseudoprimes give a deterministic primality test (though limited) that is fast and easy to use in addition to the usual Miller-Rabin probabilistic primality test. Examples Example 1 Consider the number 2795830049. This number is below . So we check for probable primality of to the bases 2, 3, 5, and 7. First of all, where 87369689. Here’s the calculation. Note that the first term that is a 1 in the above sequence is . The preceding term is a -1. Thus 2795830049 is a strong probable prime to base 2. Now the base 3 calculation. Note that the first term that is a 1 in the above sequence is the last term . The preceding term is a -1. Thus 2795830049 is a strong probable prime to base 3. Now the base 5 calculation. The base 7 calculation. Both the base 5 and base 7 calculations show that 2795830049 is a strong probable prime to both bases. The calculations for the 4 bases conclusively prove that 2795830049 is a prime number. Example 2 Consider the number 62834664835837. This number is below . So we check for probable primality to the bases 2, 3, 5, 7, 11, 13, and 17. First, where 15708666208959. The calculation for all bases shows that 62834664835837 is a strong probable primes to all 7 prime bases. This shows that 62834664835837 is prime. Example 3 Consider the number 21276028621. This is a 11-digit number and is less than . The algorithm is to check for the strong probable primality of to the first 5 prime bases – 2, 3, 5, 7, 11. First, where 5319007155. Things are going well for the first 3 prime bases. The number 21276028621 is a strong pseudoprime to the first 3 prime bases. However, it is not a strong probable prime to base 7. Thus the number is composite. In fact, 21276028621 is one of the 13 numbers below that are strong pseudoprimes to bases 2, 3 and 5. Exercises Use the least strong pseudoprime primality test that is described here to determine the primality or compositeness of the following numbers: 58300313 235993423 1777288949 40590868757 874191954161 8667694799429 1250195846428003 Reference Yupeng Jiang, Yingpu Deng, Strong pseudoprimes to the first 9 prime bases, arXiv:1207.0063v1 [math.NT], June 30, 2012. Pomerance C., Selfridge J. L., Wagstaff, S. S., The pseudoprimes to , Math. Comp., Volume 35, 1003-1026, 1980. Posted inPrimality Testing, Prime Numbers | TaggedMiller-Rabin primality test, Number theory, Primality testing, Probable prime, Pseudoprime, Strong probable prime, Strong pseudoprime, Witness | Leave a reply Looking for a good witness Posted on November 21, 2014 by Dan Ma Reply For some primality tests, the approach of proving a number as composite is to produce an auxiliary number that shows that a property of prime numbers is violated. Such an auxiliary number is called a “witness” for compositeness. When using such a primality test on a composite number, for the test to be correct, at least one “witness” must be found. In this post, we compare the Fermat primality test and the Miller-Rabin test from the standpoint of producing “witnesses”. In this regard, the Fermat test works correctly most of the time. But there is a rare but infinite class of composite numbers for which no Fermat test witness can be found. On the other hand, using the Miller-Rabin test on a composite number, at least one witness can be found (in fact, there are lots of them). Fermat witnesses According to Fermat’s little theorem, if is a prime number, then the following congruence holds for all numbers that are coprime to . If there exists a number such that , then is proved to be composite, in which case the number is said to be a Fermat witness for the compositeness of . If the congruence (1) holds for several values of , then is a likely a prime. Suppose is a large odd number whose “prime versus composite” status is not known. If the congruence (1) holds for some positive integer , then is said to be a probable prime to base . A probable prime to the base could be prime or could be composite. If the latter, the number is said to be a pseudoprime to base . This is how the Fermat test (or probable prime test) works. Given that is a large odd number whose “prime versus composite” status is not known, if is not a probable prime to one base , then is proven to be composite and is a Fermat witness for . On the other hand, if is a probable prime to several randomly selected bases, then is a probable prime. Suppose that you apply the Fermat test on the number and find that is a probable prime to several bases. There are two possibilities. Either is prime or is a pseudoprime to the several bases that are being used. Pseudoprimess to one base are rare. Pseudoprimes to several bases at once are even rarer. For example, according to , there are only 21853 pseudoprimes to the base 2 that are less than . The number of primes below is roughly . So most probable primes to base 2 are primes. Of the 21853 pseudoprimes to the base 2 that are below , 4709 of them are pseudoprimes to the bases 2 and 3, 2522 of them are pseudoprimes to the bases 2, 3 and 5, and 1770 of them are pseudoprimes to the bases 2, 3, 5 and 7. Thus there is indication that the numbers that are pseudoprimes to several bases are very rare. Therefore, the number being a simultaneous pseudoprime to several bases is very strong evidence that is prime. But this is not the end of the story for the Fermat test. The Fermat test can detect the compositeness of most composite numbers. These are the numbers that are probable prime to one base but are not probable prime to another base. For example, 341 is a probable prime to base 2 since . On the other hand, . Thus 341 is not a probable prime to base 3 (i.e. 3 is a Fermat witness for the compositeness of 341). For integers like 341, they may be probable primes to one base, but they are not probable primes to another base. If there exists at least one Fermat witness for a composite number, there is a good chance that the Fermat test will detect the compositeness. A odd composite number is said to be a Carmichael number if the congruence (1) holds for all numbers coprime to . The smallest Carmichael number is 561. There is no danger of mistaking a small Carmichael number such as 561 as prime because factoring is possible. Carmichael numbers are rare but there are infinitely of them. So there are large Carmichael numbers (e.g. with hundreds of digits or thousands of digits). The Fermat test fails completely with these large Carmichael numbers. In other words, no matter how many bases are used, it is impossible to find a number that will witness the compositeness of a Carmichael number. Unlike the number 341, the Fermat test will fail to detect the compositeness of a Carmichael number. Looking for a better type of witness Let be an odd number with where and is odd. Then compute the following sequence of numbers modulo : where is some number coprime to . The first term in the sequence can be calculated by using the fast powering algorithm. Each subsequent term is the square of the preceding one. Of course, the last term is the same as , the calculation for (1). Then if this calculation is done on a prime number , the last term in (2) is always a 1. But more can be said about the sequence (2). The better witnesses and primality test we discuss here are based on this theorem about prime numbers. If is a prime number, then either one of the following conditions holds: for some for each number that is coprime to . The theorem just mentioned is the basis of the Miller-Rabin test. If there exists some such that both (2a) and (2b) are not true, then the number is proved to be composite, in which case the number is said to be a witness for the compositeness of . So a witness from the Fermat test standpoint is called Fermat witness. The term witness here refers to the calculation of (2). If the conditions (2a) or (2b) holds for several values of , then is likely a prime number. Suppose is a large odd number whose “prime versus composite” status is not known. If the conditions (2a) or (2b) holds for some positive integer , then is said to be a strong probable prime to base . A strong probable prime to the base could be prime or could be composite. If the latter, the number is said to be a strong pseudoprime to base . This is how the Miller-Rabin test (or strong probable prime test) works. Given that is a large odd number whose “prime versus composite” status is not known, if is not a strong probable prime to one base , then is proved to be composite and the number is a witness for . If is a strong probable prime to several randomly chosen bases, then the probability that is composite is smaller than (if bases are used). In other words, the error probability can be made arbitrarily small by using more random bases in the calculation for sequence (2). The last point about the small error probability is the most important advantage of the Miller-Rabin test over the Fermat test. Let’s look at a small example first. Recall that 561 is the smallest Carmichael number. Note that . The following is the calculation for (2). The number 561 is a probable prime to base 2 (from a Fermat perspective). Notice that the condition (2a) is not met. The condition (2b) is also not met since the term preceding the first 1 is 67, which is not congruent to -1. So the number 2 is a witness for the compositeness of 561. The Fermat test will have virtually no chance of detecting the compositeness of 561. Yet the strong probable prime test takes only one calculation for 561. In this case at least, it is pretty easy to find a witness. At least one witness can be found One characteristic of the strong probable prime test is that there are lots of witnesses when testing composite numbers. In fact, if is a composite number, over 75% of the possible choices of bases are witnesses for . We prove a simpler theorem that if is composite, at least one witness for the compositeness of can be found. Even this much weaker theorem is a big improvement over the Fermat test. Theorem 1 Let be a composite positive odd number. Then there a number that is coprime to such that is not a strong pseudoprime to base . Proof of Theorem 1 Let such that are odd and are coprime. Consider the following two congruence equations. By the Chinese remainder theorem (CRT), there is an such that We can assume that . If not, reduce modulo . Then . If , then for some . This would mean that , contradicting . Similarly . So we have . On the other hand, we have By CRT, . Let where and odd. Let for some integer . The following calculates and . The first term in (2) that is a 1 is . The preceding term is not a -1. This shows that the composite number is not a strong pseudoprime to base . Thus for any composite number , there always exists a witness for the compositeness of . The proof of Theorem 1 is essentially that if is composite, there is a square root other than . This nontrivial square root is a witness to the compositeness of . The least witness If is a composite number, there exists a witness for according to Theorem 1. Of all the witnesses for a given composite number , there must be the smallest one. Let be the least witness for the compositeness for the composite number . The number is an interesting one. In general, the larger is, the harder to find the number . This stems from the fact that strong pseudoprimes are very rare. Most composite numbers are not strong pseudoprimes to base 2. For these composite numbers, . For the strong pseudoprimes to base 2, . For , the composite numbers must be strong pseudoprimes to base 2 and base 3. For , the composite numbers must be strong pseudoprimes to the prime bases 2, 3 and 5. There are only 13 such numbers below . Thus it is not easy to find composite numbers that have large least witnesses. Two more comments to make. One is that knowing the least integer such that is greater than the first prime numbers can be as a deterministic primality test . This is discussed in the next post. The other is that even though a large means that the composite numbers are rare (under a specific bound). But there are infinitely many of them (over the entire number line). It follows from a theorem in that for any finite set of integers, there are infinitely many Carmichael numbers whose is not found in the finite set. Reference Alford W., Granville A., and Pomerance C., _On the difficulty of finding reliable witnesses_, L. Adleman and M.-D. Huang, editors, Algorithmic Number Theory: Proc. ANTS-I, Ithaca, NY, volume 877 of Lecture Notes in Computer Science, pages 1–16. Springer–Verlag, 1994. 2. Jiang Yupeng, Deng Yingpu, Strong pseudoprimes to the first 9 prime bases, arXiv:1207.0063v1 [math.NT], June 30, 2012. 3. Pomerance C., Selfridge J. L., Wagstaff, S. S., The pseudoprimes to , Math. Comp., Volume 35, 1003-1026, 1980. Posted inPrimality Testing, Prime Numbers | TaggedCarmichael numbers, Fast powering algorithm, Fermat primality test, Fermat witness, Fermat's little theorem, Miller-Rabin primality test, Number theory, Primality testing, Probable prime, Pseudoprime, Strong probable prime, Strong pseudoprime, Witness | Leave a reply There are infinitely many strong pseudoprimes Posted on November 19, 2014 by Dan Ma Reply Pseudoprimes are rare. Strong pseudoprimes are rarer still. According to , there are 21853 pseudoprimes to base 2 and 4842 strong pseudoprimes to base 2 below . According to the prime number theorem, there are over 1 billion prime numbers in the same range. When testing a random number, knowing that it is a strong probable prime to just one base is strong evidence for primality. Even though most of the strong probable primes are prime, for a given base, there exist infinitely many strong pseudoprimes. This fact is captured in the following theorem. Theorem 1 For a given base , there are infinitely many strong pseudoprimes to base . For a proof, see Theorem 1 in . We give a simpler proof that there exist infinitely many strong pseudoprimes to base 2. Theorem 1a There are infinitely many strong pseudoprimes to base 2. Proof of Theorem 1a We make the following claim. Claim Let be a pseudoprime to base 2. Then is a strong pseudoprime to base 2. In a previous post on probable primes and pseudoprimes, we prove that there exist infinitely pseudoprimes to any base . Once the above claim is established, we have a proof that there are infinitely many strong pseudoprimes to base 2. First of all, if is composite, the number is also composite. This follows from the following equalities. Thus is composite. Note that . Let , which is an odd integer. Because is a pseudoprime to base 2, . Equivalently, for some integer . Furthermore, it is clear that . It follows that . This means that is a strong pseudoprime to base 2. In the previous post probable primes and pseudoprimes, it is established that there are infinitely many pseudoprimes to any base . In particular there are infinitely many pseudoprimes to base 2. It follows that the formula gives infinitely many strong pseudoprimes to base 2. Example Theorem 1a can be considered a formula for generating strong pseudoprimes to base 2. The input is a pseudoprime to base 2. Unfortunately the generated numbers get large very quickly and misses many strong pseudoprimes to base 2. The smallest pseudoprime to base 2 is 341. The following is the 103-digit . 44794894843556084211148845611368885562432909944692 99069799978201927583742360321890761754986543214231551 Even though is a strong pseudoprime to base 2, it is not strong pseudoprime to bases 3 and 5. In fact, it is rare to find a strong pseudoprime to multiple bases. To determine the strong pseudoprimality of for other bases, note that where is the following 103-digit number. 22397447421778042105574422805684442781216454972346 49534899989100963791871180160945380877493271607115775 Calculate and modulo . Look for the pattern and or the pattern pattern and . If either pattern appears, then is a strong pseudoprime to base . See the sequence labeled (1) in the previous post on strong pseudoprimes. Exercise Verify that is not a strong pseudoprime to both bases 3 and 5. Reference Pomerance C., Selfridge J. L., Wagstaff, S. S., The pseudoprimes to , Math. Comp., Volume 35, 1003-1026, 1980. Revised July 4, 2015 Posted inPrimality Testing, Prime Numbers | TaggedMiller-Rabin primality test, Number theory, Primality testing, Probable prime, Pseudoprime, Strong probable prime, Strong pseudoprime | Leave a reply Strong probable primes and strong pseudoprimes Posted on November 18, 2014 by Dan Ma 1 This post is the first in a series of posts to discuss the Miller-Rabin primality test. In this post, we discuss how to perform the calculation (by tweaking Fermat’s little theorem). The Miller-Rabin test is fast and efficient and is in many ways superior to the Fermat test. Fermat primality test is based on the notions of probable primes and pseudoprimes. One problem with the Fermat test is that it fails to detect the compositeness of a class of composite numbers called Carmichael numbers. It is possible to tweak the Fermat test to by pass this problem. The resulting primality test is called the Miller-Rabin test. Central to the working of the Miller-Rabin test are the notions of strong probable primes and strong pseudoprimes. Fermat’s little theorem, the basis of the Fermat primality test, states that if is a prime number, then for all numbers that are relatively prime to the modulus . When testing a prime number, the Fermat test always gives the correct answer. What is the success rate of the Fermat test when it is applied on a composite number? The Fermat test is correct on most composite numbers. Unfortunately the Fermat test fails to detect the compositeness of Carmichael numbers. A Carmichael number is any composite integer such that () is true for any that is relatively prime to . Fortunately we can tweak the calculation in () to get a better primality test. Recall that a positive odd integer is a probable prime to base if the condition () holds. A probable prime could be prime or could be composite. If the latter, then is said to be a pseudoprime to base . Setting up the calculation Let be an odd positive integer. Instead of calculating , we set where is an odd number and . Then compute the following sequence of numbers: Each term in (1) is reduced modulo . The first term can be computed using the fast powering (also called fast exponentiation) algorithm. Each subsequent term is the square of the preceding term. Of course, the last term is . It follows from Fermat’s little theorem that the last term in the sequence (1) is always a 1 as long as is prime and the number is relatively prime to . The numbers used in the calculation of (1) are called bases. Suppose we have a large positive odd integer whose “prime or composite” status is not known. Choose a base . Then compute the numbers in the sequence (1). If is prime, we will see one of the following two patterns: In (1a), the entire sequence consists of 1. In (1b), an asterisk means that the number is congruent to neither 1 nor -1 modulo . In (1b), the sequence ends in a 1, and the term preceding the first 1 is a -1. These two patterns capture a property of prime numbers. We have the following theorem. The theorem behind the Miller-Rabin test Theorem 1 Let be an odd prime number such that where is an odd number and . Let be a positive integer not divisible by . Then the sequence (1) resembles (1a) or (1b), i.e., either one of the following two conditions holds: The first term in the sequence (1) is congruent to 1 modulo . The term preceding the first 1 is congruent to -1 modulo . The proof of Theorem 1 is not complicated. It uses Fermat’s little theorem and the fact that if is an odd prime, the only solutions to the congruence equation are . The proof goes like this. By Fermat’s little theorem, the last term in sequence (1) is a 1, assuming that is an odd prime and is relatively prime to . If the first term in (1) is a 1, then we are done. Otherwise, look at the first term in (1) that is a 1. The term preceding the first 1 must be a -1 based on the fact that the equation can have only the trivial solutions . It is an amazing fact that Theorem 1 is easily proved and yet is the basis of a powerful and efficient and practical primality test. Next we define the notions of strong probable primes and strong pseudoprimes. Strong probable primes and strong pseudoprimes Suppose we have a large positive odd integer whose “prime or composite” status is not known. We calculate sequence (1) for one base . If the last term of the sequence (1) is not a 1, then is composite by Fermat’s little theorem. If the last term is a 1 but the sequence (1) does not match the patterns (1a) or (1b), then is composite by Theorem 1. So to test for compositeness for , we look for a base such that the sequence (1) does not fit the patterns (1a) or (1b). Such a base is said to be a Miller-Rabin witness for the compositeness of . Many authors refer to a Miller-Rabin witness as a witness. When we calculate the sequence (1) on the odd number for base , if we get either (1a) or (1b), then is said to be a strong probable prime to the base . A strong probable prime could be prime or could be composite. When a strong probable prime to the base is composite, it is said to be a strong pseudoprime to the base . To test for primality of , the Miller-Rabin test consists of checking for strong probable primality for several bases where that are randomly chosen. For an example of a primality testing exercise using the Miller-Rabin test, see the post The first prime number after the 8th Fermat number. Small examples of strong pseudoprimes Some small examples to illustrate the definitions. Because , the number 341 is a probable prime to the base 2. Because 341 is composite with factors 11 and 31, the number 341 is a pseudoprime to the base 2. In fact, 341 is the least pseudoprime to base 2. Now the strong probable prime calculation. Note that . The calculated numbers in sequence (1) are 32, 1, 1, calculated as follows: Because the sequence 32, 1, 1 does not fit pattern (1a) or (1b) (the term before the first 1 is not a -1), the number 341 is not a strong pseudoprime prime to base 2. How far do we have to go up from 341 to reach the first strong pseudoprime to base 2. The least strong pseudoprime to base 2 is 2047. Note that . Note that the congruences and . The sequence (1) is 1, 1, which is the pattern (1a). Thus 2047 is a strong pseudoprime to base 2. Note that 2047 is composite with factors 23 and 89. It can be shown (at least by calculation) that all odd integers less than 2047 are not strong pseudoprime to base 2. In other words, if a positive odd integer is less than 2047 and if it is a strong probable prime to base 2, then must be a prime number. Consider a slightly larger example. Let 65281. Set . The following is the calculation for the sequence (1) using base 2. The pattern is , , , -1, 1, 1, 1, 1, 1, which is (1b) (the term preceding the first 1 is a -1). So 65281 is strong probable prime to base 2. The following computation using base 3 will show that 65281 is a composite number, thus is a strong pseudoprime to base 2. Looking at the last term in the base 3 calculation, we see that the number 65281 is composite by Fermat’s little theorem. Because the pattern is , , , , , , , , , 65281 is not a strong pseudoprime to base 3. How does pseudoprimality and strong pseudoprimality relate? There are two notions of “pseudoprime” discussed here and in previous posts. One is based on Fermat’s little theorem (pseudoprime) and one is based on Theorem 1 above (strong pseudoprime). It is clear from the definition that any strong pseudoprime to base is a pseudoprime to base . The converse is not true. Let’s start with the number 341. It is a pseudoprime to base 2. This means that the Fermat test cannot detect its compositeness using base 2. Yet the strong pseudoprimality calculation as described above can detect the compositeness of 341 using base 2. The 341 is not a strong pseudoprime to base 2 since the least strong pseudoprime to base 2 is 2047. Let’s look at a slightly larger example. Take the number 25761. It is a pseudoprime to base 2 since and its factors are 3, 31 and 277. Let refine the calculation according to sequence (1) as indicated above. Note that . The pattern of sequence (1) is , , 1, 1, 1, 1. The term preceding the first 1 is not a -1. Thus the strong pseudomality method does detect the compositeness of 25761 using base 2. In general, strong pseudoprimality implies pseudoprimality (to the same base). The above two small examples show that the converse is not true since they are pseudoprimes to base 2 but not strong pseudoprimes to base 2. Why look at pseudoprimes and strong pseudoprimes? The most important reason for studying these notions is that pseudoprimality and strong pseudoprimality are the basis of two primality tests. In general, pseudoprimality informs primality. In a previous post on probable primes and pseudoprimes, we point out that most probable primes are primes. The same thing can be said for the strong version. According to , there are only 4842 strong pseudoprimes to base 2 below . Using the prime number theorem, it can be shown that there are approximately many prime numbers below . Thus most strong probable primes are primes. For a randomly chosen , showing that is a strong probable prime to one base can be quite strong evidence that is prime. Because strong pseudoprimality is so rare, knowing what they are actually help in detecting primality. For example, according to , there are only 13 numbers below that are strong pseudoprimes to all of the bases 2, 3 and 5. These 13 strong pseudoprimes are: Strong pseudoprimes to all of the bases 2, 3 and 5 below 25 billion 25326001, 161304001, 960946321, 1157839381, 3215031751, 3697278427, 5764643587, 6770862367, 14386156093, 15579919981, 18459366157, 19887974881, 21276028621 These 13 strong pseudoprimes represent a deterministic primality test on integers less than . Any odd positive integer less than that is a strong probable prime to all 3 bases 2, 3 and 5 must be a prime number if it is not one of the 13 numbers on the list. See Example 1 below for an illustration. This primality is fast since it only requires 3 exponentiations. Best of all, it gives a proof of primality. However, this is a fairly limited primality test since it only works on numbers less than . Even though this is a limited example, it is an excellent illustration that strong pseudoprimality can inform primality. Example 1 Consider the odd integer 1777288949, which is less than . Set . The proof of primality of requires only the calculation for 3 bases 2, 3 and 5. Base 2 Base 3 Base 5 The patterns for the 3 calculations fit either (1a) or (1b). So 1777288949 is a strong probable prime to all 3 bases 2, 3 and 5. Clearly 1777288949 is not on the list of 13 strong pseudoprimes listed above. Thus 1777288949 cannot be a composite number. Exercise Use the strong pseudoprime test to show that the following numbers are composite. 3277 43273 60433 60787 838861 1373653 Use the 13 strong pseudoprimes to the bases 2, 3 and 5 (used in Example 1) to show that the following numbers are prime numbers. 58300313 99249929 235993423 2795830049 Reference Pomerance C., Selfridge J. L., Wagstaff, S. S., The pseudoprimes to , Math. Comp., Volume 35, 1003-1026, 1980. Posted inPrimality Testing, Prime Numbers | TaggedCarmichael numbers, Fermat primality test, Fermat's little theorem, Miller-Rabin primality test, Number theory, Primality testing, Probable prime, Pseudoprime, Strong probable prime, Strong pseudoprime | 1 Reply Post navigation ← Older posts Dan Ma Archives November 2024 October 2024 February 2023 January 2023 November 2022 October 2022 October 2016 December 2014 November 2014 October 2014 Recent Posts Factoring a 397-digit Carmichael number Using a primality test to find square roots of 1 modulo a Carmichael number Lucas-Lehmer test, a test for Mersenne primes A strong pseudoprime is an Euler pseudoprime Three classes of pseudoprimes Pollard’s p-1 factorization algorithm The Goldwasser-Micali Cryptosystem Quadratic residuosity problem Diffie-Hellman Key Exchange The discrete logarithm problem Categories Factorization Algorithm Primality Testing Prime Numbers Public Key Cryptography Tags Carmichael numbers Cryptography Diffie-Hellman Key Exchange Discrete Logarithm Discrete Logarithm Problem Elite Prime Number Euler Pseudoprime Extended Euclidean algorithm Fast powering algorithm Fermat's little theorem Fermat number Fermat primality test Fermat witness Goldwasser-Micali Jacobi Symbol Legendre Symbol Lucas-Lehmer Test Mersenne prime Miller-Rabin primality test Number theory Pollard's p-1 Primality testing Primitive root Probabilistic Encryption Probable prime Pseudoprime Public key cryptography Quadratic Residue Quadratic Residuosity Problem RSA RSA algorithm Square and multiply algorithm Strong probable prime Strong probable prime test Strong pseudoprime Weakly prime numbers Witness Create a free website or blog at WordPress.com. 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8448
https://users.math.utoronto.ca/preparing-for-calculus/3_logic/we_3_negation.html
Loading Web-Font TeX/Main/Regular Logic and Mathematical Statements Worked Examples Negation Sometimes in mathematics it's important to determine what the opposite of a given mathematical statement is. This is usually referred to as "negating" a statement. One thing to keep in mind is that if a statement is true, then its negation is false (and if a statement is false, then its negation is true). Let's take a look at some of the most common negations. Negation of "A or B". Before giving the answer, let's try to do this for an example. Consider the statement "You are either rich or happy." For this statement to be false, you can't be rich and you can't been happy. In other words, the opposite is to be not rich and not happy. Or if we rewrite it in terms of the original statement we get "You are not rich and not happy." If we let A be the statement "You are rich" and B be the statement "You are happy", then the negation of "A or B" becomes "Not A and Not B." In general, we have the same statement: The negation of "A or B" is the statement "Not A and Not B." Negation of "A and B". Again, let's analyze an example first. Consider the statement "I am both rich and happy." For this statement to be false I could be either not rich or not happy. If we let A be the statement "I am rich" and B be the statement "I am happy", then the negation of "A and B" becomes "I am not rich or I am not happy" or "Not A or Not B". Negation of "If A, then B". To negate a statement of the form "If A, then B" we should replace it with the statement "A and Not B". This might seem confusing at first, so let's take a look at a simple example to help understand why this is the right thing to do. Consider the statement "If I am rich, then I am happy." For this statement to be false, I would need to be rich and not happy. If A is the statement "I am rich" and B is the statement "I am happy,", then the negation of "A B" is "I am rich" = A, and "I am not happy" = not B. So the negation of "if A, then B" becomes "A and not B". Example. Now let's consider a statement involving some mathematics. Take the statement "If n is even, then \frac{n}{2} is an integer." For this statement to be false, we would need to find an even integer n for which \frac{n}{2} was not an integer. So the opposite of this statement is the statement that "n is even and \frac{n}{2} is not an integer." Negation of "For every ...", "For all ...", "There exists ..." Sometimes we encounter phrases such as "for every," "for any," "for all" and "there exists" in mathematical statements. Example. Consider the statement "For all integers n, either n is even or n is odd". Although the phrasing is a bit different, this is a statement of the form "If A, then B." We can reword this sentence as follows: "If n is any integer, then either n is even or n is odd." How would we negate this statement? For this statement to be false, all we would need is to find a single integer which is not even and not odd. In other words, the negation is the statement "There exists an integer n, so that n is not even and n is not odd." In general, when negating a statement involving "for all," "for every", the phrase "for all" gets replaced with "there exists." Similarly, when negating a statement involving "there exists", the phrase "there exists" gets replaced with "for every" or "for all." Example. Negate the statement "If all rich people are happy, then all poor people are sad." First, this statement has the form "If A, then B", where A is the statement "All rich people are happy" and B is the statement "All poor people are sad." So the negation has the form "A and not B." So we will need to negate B. The negation of the statement B is "There exists a poor person who is not sad." Putting this together gives: "All rich people are happy, but there exists a poor person who is not sad" as the negation of "If all rich people are happy, then all poor people are sad." Summary. | | | --- | | Statement | Negation | | "A or B" | "not A and not B" | | "A and B" | "not A or not B" | | "if A, then B" | "A and not B" | | "For all x, A(x)" | "There exist x such that not A(x)" | | "There exists x such that A(x)" | "For every x, not A(x)" | MathJax Main 
8449
https://www.khanacademy.org/math/cc-fifth-grade-math/divide-fractions/imp-dividing-fractions-and-whole-numbers-word-problems/v/dividing-fractions-and-whole-number-word-problems
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https://www.quora.com/How-do-you-round-numbers-to-four-decimal-places
Something went wrong. Wait a moment and try again. Accurate Decimal Places Rounded Numbers Lining Up Decimals Decimal Notation Floating Numbers Decimal Degrees 5 How do you round numbers to four decimal places? Aaron Briseno B.S in Mathematics & Teaching, University of California, Los Angeles (Graduated 2010) · Author has 1.3K answers and 3.1M answer views · 4y Take your exact answer (or calculator response). Look at the fifth and sixth digit, if these two digits form a number greater than 50 then you round the fourth digit up one unit. Let x=1.453456 Round the value of x to 4 decimals. x=1.453456 Since 56 is over half of 100 (greater than 50) we’d round the fourth digit up 1 x≈1.4535 Example 2: x=1.3455503 Since 50 is half of 100 we expand our bold portion to see if we end up with >500 x=1.3455503 Since 503 is > half of 1000, we round the fourth digit up one unit x≈1.3456 Now if the bolded sections are less than half of the 100 or 1000 (resp Take your exact answer (or calculator response). Look at the fifth and sixth digit, if these two digits form a number greater than 50 then you round the fourth digit up one unit. Example Let x=1.453456 Round the value of x to 4 decimals. x=1.453456 Since 56 is over half of 100 (greater than 50) we’d round the fourth digit up 1 x≈1.4535 Example 2: x=1.3455503 Since 50 is half of 100 we expand our bold portion to see if we end up with >500 x=1.3455503 Since 503 is > half of 1000, we round the fourth digit up one unit x≈1.3456 Now if the bolded sections are less than half of the 100 or 1000 (respectively) then you’d just chop the decimal portion after the 4th decimal “off” example 3 x=1.556322 22 is not more than half of 100, so we round “down” to x≈1.5563 Apply good rounding rules depending on your situation too! Like if you are buying supplies to craft a certain number of items, make sure you have enough of your “limiting reagent” so you can make “enough” (whatever that number is) Promoted by Coverage.com Johnny M Master's Degree from Harvard University (Graduated 2011) · Updated Sep 9 Does switching car insurance really save you money, or is that just marketing hype? This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. This is one of those things that I didn’t expect to be worthwhile, but it was. You actually can save a solid chunk of money—if you use the right tool like this one. I ended up saving over $1,500/year, but I also insure four cars. I tested several comparison tools and while some of them ended up spamming me with junk, there were a couple like Coverage.com and these alternatives that I now recommend to my friend. Most insurance companies quietly raise your rate year after year. Nothing major, just enough that you don’t notice. They’re banking on you not shopping around—and to be honest, I didn’t. It always sounded like a hassle. Dozens of tabs, endless forms, phone calls I didn’t want to take. But recently I decided to check so I used this quote tool, which compares everything in one place. It took maybe 2 minutes, tops. I just answered a few questions and it pulled up offers from multiple big-name providers, side by side. Prices, coverage details, even customer reviews—all laid out in a way that made the choice pretty obvious. They claimed I could save over $1,000 per year. I ended up exceeding that number and I cut my monthly premium by over $100. That’s over $1200 a year. For the exact same coverage. No phone tag. No junk emails. Just a better deal in less time than it takes to make coffee. Here’s the link to two comparison sites - the one I used and an alternative that I also tested. If it’s been a while since you’ve checked your rate, do it. You might be surprised at how much you’re overpaying. Related questions How do I round 0.3895672 to four decimal places? How do you round .3999676 to 4 decimal places? What does it mean when they say round to four decimal places? Is 3.1416 π rounded to 4 decimal places? How do you round a decimal to 2 decimal places? Mason Flannery 7y All rounding is the same, once you know how to do it it's really easy. First, let's start basic. 3.6 so if you wanted to round to the whole number, you look at the number in front of it. The whole number is 3, and the number in front of it is 6. Since its at least 5, it rounds the 3 to a 4. If that didn't make sense let's try this: 1.15 if you wanted to round to the first place, or the tenths place, look at that number, which is 1. The number in front of it is 5. Since it is at least 5 it goes up to 1.2 If it is 5 or above it is rounded up, and if it's below 5 it isn't changed, so: 1.24 is rounded t All rounding is the same, once you know how to do it it's really easy. First, let's start basic. 3.6 so if you wanted to round to the whole number, you look at the number in front of it. The whole number is 3, and the number in front of it is 6. Since its at least 5, it rounds the 3 to a 4. If that didn't make sense let's try this: 1.15 if you wanted to round to the first place, or the tenths place, look at that number, which is 1. The number in front of it is 5. Since it is at least 5 it goes up to 1.2 If it is 5 or above it is rounded up, and if it's below 5 it isn't changed, so: 1.24 is rounded to 1.2, make sense? Now a step forward, or your question. Start with this: 1.12345 Go to the 4th decimal place, or the ten thousenths place. Look at the number in front of that. Is it 5 or above or less than 5? Take a guess at the answer. If you got it right it should be 1.1235 Hope this helps. Good luck! :) Curt Weinstein HS in Physics & Mathematics, The Bronx High School of Science (Graduated 1969) · Author has 42K answers and 27.4M answer views · 2y Look at the 5th decimal place. One way (there are more than one): If 5,6,7,8,9 → add one to the 4th decimal place. Drop all places greater than 4. If 0,1,2,3,4 → keep the 4th decimal place (drop all places greater than 4) There was a “statistical” method where 0 was dropped out of consideration, and the fourth place was made to be even when the 5th place was “5”. I never liked that method. Assistant Bot · Sep 6 Rounding to four decimal places means keeping four digits after the decimal point and adjusting the last kept digit based on the digit immediately to its right. Procedures below work for manual calculation, mental rounding, spreadsheets, and programming. Quick rule (step-by-step) Identify the fourth decimal place (the thousand-thousandth place). Example: in 3.1415926 the fourth decimal digit is 5 (3.1415). Look at the fifth decimal digit (the digit immediately to the right of the fourth). In the example the fifth digit is 9 (3.14159). If the fifth digit is 0–4, leave the fourth digit unchanged and Rounding to four decimal places means keeping four digits after the decimal point and adjusting the last kept digit based on the digit immediately to its right. Procedures below work for manual calculation, mental rounding, spreadsheets, and programming. Quick rule (step-by-step) Identify the fourth decimal place (the thousand-thousandth place). Example: in 3.1415926 the fourth decimal digit is 5 (3.1415). Look at the fifth decimal digit (the digit immediately to the right of the fourth). In the example the fifth digit is 9 (3.14159). If the fifth digit is 0–4, leave the fourth digit unchanged and drop all digits to the right. If it is 5–9, add 1 to the fourth digit and drop the remaining digits. If adding 1 produces a carry (e.g., 0.12995 → fourth digit 9 + 1 = 10), propagate the carry left through the decimal digits and into the integer part if necessary. Ensure the result shows exactly four decimal places (pad with trailing zeros if needed): for example, 2 → 2.0000 and 1.5 → 1.5000. Examples 2.345678 → fourth digit = 6, fifth = 7 → round up → 2.3457 0.12344 → fourth digit = 4, fifth = 4 → round down → 0.1234 0.99996 → fourth digit = 9, fifth = 6 → round up with carry → 1.0000 5 → written as 5.0000 Spreadsheet and calculator notes Excel/Google Sheets: use =ROUND(number, 4). Most scientific calculators have a rounding function or display setting; otherwise compute manually and format to 4 decimals. Programming examples Python: round(x, 4) — beware of binary floating-point representation when formatting for display; use format(x, ".4f") or f"{x:.4f}" for correct string output. JavaScript: Number(x.toFixed(4)) or x.toFixed(4) (returns string). Java/C#: use formatting functions (String.format("%.4f", x) / x.ToString("F4")). Common pitfalls Binary floating-point can produce results like 2.3457000000000002; format to 4 decimals for presentation. Bankers’ rounding vs. round-half-up: default rounding behavior differs by language/library (Python’s round uses ties-to-even). If you require always rounding 5 up, use a specific routine (e.g., Decimal in Python with ROUND_HALF_UP). Use these steps and the appropriate language/utility formatting to produce consistent four-decimal-place results. Related questions How do I round off 1.52646494 to five decimal places? How many decimal places does 6.050 have? What is 75,604 carried to four decimal places? How do you round two decimal places? What is the number of decimal places in 8.02456? Eric Barnes Author has 4.3K answers and 2M answer views · 6y Originally Answered: How do I round 3.65266 to the fourth decimal place? · When rounding to a specific place, take a look at the digit immediately to the right of the target, in this case, the fifth decimal place. If that digit is 4 or less, zero everything to the right of the target out. If that digit is 5 through 9, increase the target by one and zero everything to the right of it out. If the target is a 9, increasing it will turn it to zero and increase the digit to the left of the target by one. 3.65266 rounded to the fourth decimal place is 3.6527 Promoted by The Penny Hoarder Lisa Dawson Finance Writer at The Penny Hoarder · Updated Sep 16 What's some brutally honest advice that everyone should know? Here’s the thing: I wish I had known these money secrets sooner. They’ve helped so many people save hundreds, secure their family’s future, and grow their bank accounts—myself included. And honestly? Putting them to use was way easier than I expected. I bet you can knock out at least three or four of these right now—yes, even from your phone. Don’t wait like I did. Cancel Your Car Insurance You might not even realize it, but your car insurance company is probably overcharging you. In fact, they’re kind of counting on you not noticing. Luckily, this problem is easy to fix. Don’t waste your time Here’s the thing: I wish I had known these money secrets sooner. 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Charles Holmes Studied Financial Markets & Mathematics (Graduated 1990) · Author has 16.6K answers and 13.6M answer views · Updated 3y Originally Answered: How do you round .3999676 to 4 decimal places? · y=0.4000 (4000/10,000 or 4 thousand ten-thousandths) PREMISES y=decimal number 0.3999676 written to 4 decimal places CALCULATIONS To round a decimal number to 4 decimal places start by examining the 5th decimal place. If the 5th decimal place is 5 or more, then add 1 to the 4th decimal number and carry the addition over to the 3rd and 2nd and 1st decimal places as necessary. For example, 5.99998 rounded to 4 decimal places becomes 5.99998+0.00002=6.0000. y=0.3999676 satisfies y=0.3999676+0.00004 y= 0.4000 (4000/10,000 or 4 thousand ten-thousandths) C.H. Angela Lin 6y Originally Answered: How do you round .3999676 to 4 decimal places? · You look at the fifth number which in this case is 6. When the number is 5 or above we round up. In this case it’s the fourth number 9. Nine rounding up makes the other two nine round up too. Which the 3 rounds to a 4. Altogether it would be 0.4000 Sponsored by Grammarly Is your writing working as hard as your ideas? Grammarly’s AI brings research, clarity, and structure—so your writing gets sharper with every step. Jake Taraj Former Chief at Deep Sea Diver (US Navy) (1998–2018) · 7y Originally Answered: How do you round .3999676 to 4 decimal places? · Since the fifth place is 5 or greater you round up the forth place to 10. This carries the third and second place to 10 and the first to 4. The answer is .4000, ensure you keep the three zeros as they show accuracy and in chemistry called significant figures. Lukas Schmidinger I have graduate CS and my studies included math courses. · Author has 27.7K answers and 14.9M answer views · 7y Originally Answered: How do you round .3999676 to 4 decimal places? · You will get 0.4. First let’s write down the number: 0.3999676 Mark the decimal place you want round to: 0.3999676 Notice the digit next to the decimal place 0.3999¯¯¯676 It is 6 and 6≥5 so we round up so we increase the digit by 1 but this would get us 1 so it has an effect on the next number place and this repeats until we end up with 0.4000 or 0.4 for short. Promoted by Spokeo Spokeo - People Search | Dating Safety Tool Dating Safety and Cheater Buster Tool · Apr 16 Is there a way to check if someone has a dating profile? Originally Answered: Is there a way to check if someone has a dating profile? Please be reliable and detailed. · Yes, there is a way. If you're wondering whether someone has a dating profile, it's actually pretty easy to find out. Just type in their name and click here 👉 UNCOVER DATING PROFILE. This tool checks a bunch of dating apps and websites to see if that person has a profile—either now or in the past. You don’t need to be tech-savvy or know anything complicated. 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Name and phone number searches usually give the best and most accurate results The site scans billions of public records in just a few seconds. It also scans over 120 dating and social media websites to see if the person has a profile It will ask you a few quick questions to narrow down the results (like location) Once the search is done, you’ll see blurred preview with: Their full name Dating profiles & social media accounts All known phone numbers Current and past addresses A list of family members Any available court or criminal records And more useful background info ⚠️ KEY CALL OUTS ⚠️ Its not free. You will need to pay to see everything, but its pretty cheap. If nothing shows up, it doesn’t always mean they’re in the clear — some people use fake names or burner emails. So it’s worth digging a little deeper just to be sure. 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Kurt Behnke PhD in Mathematics & Theoretical Physics, University of Hamburg (Graduated 1981) · Author has 8.3K answers and 14.2M answer views · 7y You copy the first 3 decimal places, and take a look at the 4th and 5th position. If the 5th decimal is 5 or above, the 4th decimal is rounded UP, otherwise the 4th decimal stays what it is. Look at your example: 3 digits copied: Now look at 5 and 6 in 4th and 5th decimal position. 6 is greater than or equal to 5, so the 4th position gets rounded up (from 5 to 6). So your result is Mark Christian Mathematics Educator at Department of Defense Education Activity (DoDEA) · 7y Originally Answered: How do you round .3999676 to 4 decimal places? · The answer is 0.4000. To really appreciate this, think of the question, “Is 0.3999676 closer to 0.3999 or 0.4000?”, because these are the nearest numbers that are written to 4 decimal places. Charles Holmes Studied Financial Markets & Mathematics (Graduated 1990) · Author has 16.6K answers and 13.6M answer views · 5y Originally Answered: How do I round 0.3895672 to four decimal places? · y=0.3896 rounded up to 4d. p. y=0.3895 rounded down to 4d. p. PREMISES y=decimal number 0.3895672 rounded up/down to 4d. p CALCULATIONS y= 0.3896 rounded “up” to 4d. p If the 4d.p is 5 or “less” y= 0.3895 rounded “down” to 4d. p C.H. Duncan Watson BSc (Hons) in Physics, University of Natal, Durban (Graduated 1968) · Author has 1K answers and 855.6K answer views · 7y Originally Answered: How do you round .3999676 to 4 decimal places? · the zeros show that it is accurate to four decimal places Related questions How do I round 0.3895672 to four decimal places? How do you round .3999676 to 4 decimal places? What does it mean when they say round to four decimal places? Is 3.1416 π rounded to 4 decimal places? How do you round a decimal to 2 decimal places? How do I round off 1.52646494 to five decimal places? How many decimal places does 6.050 have? What is 75,604 carried to four decimal places? How do you round two decimal places? What is the number of decimal places in 8.02456? How do you roundoff 3.377801 to 4 decimal places? How do you write 8 decimal places in numbers? How do you round to 2 decimal places in Excel? What are the names of decimal places? What is the process for rounding a decimal number to three decimal places? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
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Percentage Word Problem Worksheet | TPT Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn Winter Spring Summer Back to school End of year ELA ELA by grade PreK ELA Kindergarten ELA 1st grade 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Character education Classroom community School counseling School psychology Social emotional learning Specialty Specialty Career and technical education Child care Coaching Cooking Health Life skills Occupational therapy Physical education Physical therapy Professional development Service learning Vocational education Other (specialty) Percentage Word Problem Worksheet 82+results Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Search Grade Subject Supports Under $5 Format All filters (2) Filters Under $5 Centers Clear all Grade Elementary 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Higher education Adult education More Subject Math Algebra Applied math Arithmetic Basic operations Decimals Financial literacy Fractions Graphing Math test 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administrators More Language En español More Programs & methods Programs GATE / Gifted and Talented More Supports ESL, EFL, and ELL Special education Life skills Specialty Other (specialty) More Percentages Word Problems Part Whole Percent of a Number Worksheets Activity Created by Games 4 Gains Make solving word problems with percentages fun with this Percents Word Problems Color-by-Number Worksheet and Task Cards Activity! This part, whole, and percent of a number activity includes 24 task cards that require students to solve real-world percent problems and a color-by-number worksheet. The cards contain both one-step and multi-step problems and increase gradually in difficulty. The coloring aspect of this percents activity motivates even the most reluctant learners to work through the 6 th - 7 th Math Test Prep, Other (Math) CCSS 6.RP.A.3 , 6.RP.A.3c $3.75 Original Price $3.75 Rated 4.88 out of 5, based on 60 reviews 4.9(60) Add to cart Wish List Football Percentages Worksheets | Discounts | Word Problem Percentages 6th Grade Created by Teaching Mini People In this Football Percentages Worksheet packet, students practice solving percentages in word problems. Included is both the DIGITAL and PRINT versions. They have to answer word problems, fill in tables, and fill in guest checks in restaurants. These problems practice with sales tax, % more than, % less than, tips, commission, and discounts. These 6th grade percent worksheets are great for review, small groups, homework, or extra credit. Students will need to be able to find the “part”, “whol 6 th - 7 th Applied Math, Math, Other (Math) CCSS 7.RP.A.3 $3.97 Original Price $3.97 Rated 4.67 out of 5, based on 6 reviews 4.7(6) Add to cart Wish List Percentages Word Problems 6th Grade Unit Review Worksheets and Google Slides Created by Think in Numbers In this 6th Grade math percentages resource, students will review all the essential concepts within the percent unit using word problems. A total of 40 questions are provided in the resource. These percentages worksheets and google slides review the following concepts: Percent of a numberFind the percent given the part and whole.Converting fractions, decimals and percents.Comparing and ordering percentages.Find the whole given the part and percent.Percentages using visual models.FORMATThere are 6 th - 7 th Algebra, Math Test Prep, Other (Math) Also included in:6th Grade Percentages Unit Mastery Activities Practice and Review EDITABLE $3.50 Original Price $3.50 Add to cart Wish List Percentages Word Problems Worksheets, Percent Of A Number Halloween Math Mystery Created by Master Xuan Looking for NO PREP percentages word problems worksheets that allow students to have some FUN while doing some rigorous word problems practices? Then word problems with percentage + mystery puzzles will be perfect for you! Especially after you teach your students the basic percentage and now it is time to do percentage word problem worksheet. But if you keep giving them worksheets for percentages, they will be bored and unmotivated. Therefore, turn them into Math Detective with this resourc 6 th - 7 th Math CCSS 6.RP.A.3c Also included in:Percent Word Problems Worksheets, Proportions Percentages Craft Activity Bundle $3.99 Original Price $3.99 Add to cart Wish List Super Bowl 2025 Calculating Averages &Percentages Word Problems Worksheets Created by First Tries and Sunny Skies Bring the excitement of the Super Bowl to your classroom with this engaging set of 8 Super Bowl-Themed Word Problems Worksheets! Designed specifically for 5th and 6th grade students, this resource combines essential math skills like calculating averages and percentages with fascinating football history to create a fun and educational experience. Product Highlights:Real Super Bowl Stats: Each word problem features actual statistics from legendary Super Bowl games, providing students with real-w 5 th - 6 th Math, Other (Math) $3.00 Original Price $3.00 $1.00 Price $1.00 Add to cart Wish List Fractions, Decimals, Percentages& Discounts Word Problems (Vol. 2) | 5th Grade Created by Math World by Zara1989 Build real-world problem-solving skills with this Grade 5–6 pack of fraction, decimal, and percentage word problems. You’ll get 20 topics across 25 student practice pages—from everyday shopping and tax/tips to surveys, sports stats, measurement, and mixed multi-step challenges. Each page blends core practice with reasoning prompts so students develop accuracy and explain their thinking. "I designed these word problems to help students see that fractions, decimals, and percentages are simply 5 th - 6 th Basic Operations, Decimals, Fractions CCSS 5.NF.A.1 , 5.NF.B.3 , 5.NF.B.7 +3 $4.50 Original Price $4.50 Add to cart Wish List August Math Word Problem Worksheets, 6th,7th,8th Grade Created by FUN LEARNING WITH JANET August Math Word Problem Worksheets Grades 6–8Start of the school year with 22 pages of engaging August-themed math word problems designed specifically for 6th, 7th, and 8th grade students! This no-prep resource includes 40 real-world math word problems that build confidence, boost problem-solving skills, and review essential concepts – all while connecting to the fun and seasonal spirit of August. Each worksheet features a mix of multi-step problems, critical thinking, and grade-appropriate mat 6 th - 8 th Basic Operations, Fractions, Mental Math $5.00 Original Price $5.00 $4.75 Price $4.75 Add to cart Wish List Equivalent Fractions and Percentages #classroomprepsale Created by LittleStreams UPDATED April 2024: To include three differentiated student booklets, making it easier to print and go. This pack is filled with engaging and fun worksheets designed to introduce students to equivalent fractions converting between fractions, decimals and percentages. There are 10 worksheets, each differentiated 3 ways. There are two files. One with the worksheets and one with the solutions and teachers notes. This pack contains the following: Worksheet 1 Bar Equivalence Worksheet 2 Equi 3 rd - 6 th, Adult Education Decimals, Fractions, Numbers CCSS 3.NF.A.3 , 3.NF.A.3b , 4.NF.A.1 +4 Also included in:Introducing Fractions, Decimals & Percentages Bundle Fun Printables/worksheets $5.00 Original Price $5.00 $1.00 Price $1.00 Rated 4.89 out of 5, based on 18 reviews 4.9(18) Add to cart Wish List Real World Money Math : Coupon & Discount Math Worksheets : Percentage Off Math Created by The Millennial Teacher Store Bring real-life math into your classroom with this Percentage Off Coupon Worksheet Pack, perfect for middle school students or functional math groups! This resource gives students practical experience using percent discounts and money math—skills they'll use beyond the classroom. What's Included: ✔️ Worksheet 1 – Visual Discount Practice 9 real-world problems with product images (like shoes, backpacks, and tech accessories) Students calculate the sale price after applying a percentag 6 th - 8 th Applied Math, Decimals, Financial Literacy CCSS 7.EE.B.3 , 6.RP.A.3c , 7.RP.A.3 $1.99 Original Price $1.99 Add to cart Wish List 5th Grade Ratio & Proportion Worksheet Pack | Math Activities Created by Hannah Murphy This math pack is composed of 5th grade ratio and proportion worksheets that will support your fifth graders in their math lessons. This packet includes 8 engaging math worksheets that will help your students practice negative numbers, place value to 1,000,000, rounding, and much more! This pack is perfect for whole-class activities, math centers, review and homework. Save OVER 30% with the bundle!5th Grade Math Worksheets MEGA BundleWorksheets included:Comparing PercentagesPercentages of Amount 4 th - 6 th Decimals, Fractions, Math Also included in:5th Grade Math Worksheets MEGA Bundle $3.00 Original Price $3.00 Add to cart Wish List Percentage& Profit-Loss Word Problems | Final Review Grade 5-6 Created by Math World by Zara1989 Help your students master percentage skills with this comprehensive 10-page worksheet set, complete with 11 pages of detailed answer keys! This resource covers everything from basic percentage conversions to real-life applications involving profit, loss, selling price, and cost price. Whether you’re teaching in the classroom or assigning independent practice, this pack is perfect for: ✅ Classwork & homework ✅ Small group intervention ✅ Test prep & revision ✅ Math centers & fast finisher acti 5 th - 6 th Basic Operations, Decimals, Fractions CCSS 5.NF.B.3 , 6.RP.A.3b , 6.RP.A.3c $2.00 Original Price $2.00 Add to cart Wish List Money Word Problems Printable Worksheets for 5th and 6th Grade (2.MD.8) Created by Learning Reading Fun Money Word Problems for 5th and 6th Grade (2.MD.8)This collection of money word problems is designed for 5th and 6th graders to build financial literacy and problem-solving skills. The problems cover addition, subtraction, multiplication, division, percentages, tax, discounts, budgeting, savings, interest, and profit calculations in real-world scenarios. Students will practice: Shopping calculations (discounts, sales tax, total cost)Budgeting and saving (weekly savings, goal setting)Earning a 5 th - 6 th Math, Numbers, Other (Math) Also included in:Money Word Problems Printable Worksheets (2.MD.8) Bundle $3.00 Original Price $3.00 Add to cart Wish List Percent Word Problems | NO PREP Math Printable Worksheets Created by Rach's Resources Looking for engaging, real-life math practice for your middle school students? These Percent Word Problems Worksheets are perfect for reinforcing key concepts like discounts, tax, tip, percent increase/decrease, commission, percent error, and interest. With over 35 real-world word problems, your students will develop confidence applying percent skills in everyday situations! Perfect for:Grade 7 & 8 MathPercent Unit PracticeBellwork / Warm-UpsHomeworkMath CentersSub PlansTest Prep Skills Cove 6 th - 10 th Basic Operations, Financial Literacy, Numbers CCSS 6.RP.A.3c , 7.RP.A.3 , 8.F.B.4 +3 Also included in:Middle School Math Worksheets BUNDLE | ALL YEAR NO PREP Printables $2.00 Original Price $2.00 Add to cart Wish List Valentine's Day Percents of a Number Word Problems Color by Code Worksheet Created by That One Cheerful Classroom Are your students struggling to stay engaged when practicing percents of a number word problems? Look no further, this percentages color by code worksheet is sure to keep your students focused with a fun Valentine's Day theme. Quick and easy setup, plus clear student directions make this activity perfect for centers or substitute days, too! Lead your students to practice their finding the percent of a number skills with this engaging Valentine's Day color by code. They will solve the percentages 6 th - 7 th Basic Operations, Math CCSS 6.RP.A.3c , 7.RP.A.3 $1.75 Original Price $1.75 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Percentage Practice: Detective Math Mystery | Worksheet [8th Grade] Created by Mettry Teaches Calling all math detectives! The Pufferfish Potion Pack has vanished from the Coral Reef Research Center, and it's up to your students to crack the case using their percentage skills! What's Included?Case File Cover Page: Introduces the mystery and sets the scene for the Pufferfish Potion Pack investigation.Answer Key.4 Mission Sheets: Each sheet focuses on a different aspect keeping your detectives engaged: Mission: What?Mission: Who?Mission: Where? Mission: Why? Mission: When? - Calculating 7 th - 9 th Fractions, Math $3.75 Original Price $3.75 Rated 5 out of 5, based on 1 reviews 5.0(1) Add to cart Wish List Easter Math Worksheet Digital Activity Distance Learning Created by Tiarra's Teaching Techniques This Easter Color by Code activity will be a great way to add a spiral review to your classroom while being productive learning about Ratios and Percentages. These ten Easter Word Problems give students some statistical facts requiring them to break down the information. Many of the questions use information gained from the previous questions making it a bit of a puzzle too! Perfect for 6th and 7th graders or remedial 8th. The problems include real-world information taken about the Easter Hol 6 th - 8 th Math, Other (Math) CCSS 7.EE.B.3 , 6.RP.A.3b , 6.RP.A.3c $2.50 Original Price $2.50 Rated 4.69 out of 5, based on 8 reviews 4.7(8) Add to cart Wish List Percent, Decimal, Fraction Math Stations or Worksheets: Test Prep 5th 6th 7th Created by Grace Under Pressure This ready-to-go set of five middle school math stations will supplement a unit about fractions, decimals, and percent in grades 5, 6, and 7. You can use the pages as individual worksheets or set them up as stations around your room that students can rotate through. There are word problems, real world math activities, games, and a journal prompt. These worksheets could be a great test prep activity for your class! Station 1: Students solve four word problems using percentages. (optional: set 5 th - 7 th Decimals, Math Test Prep, Other (Math) CCSS 5.NBT.A.1 , 5.NBT.A.3 , 5.NBT.B.7 +7 Also included in:Math Stations or Worksheets BUNDLE 50 Pages & Rubric: Math Test Prep 5th 6th 7th $3.00 Original Price $3.00 Add to cart Wish List Multiplication and Division Worksheet Centers | equal groups | word problems Created by Kiddos and Caffeine Need a multiplication and division center for your students? This set of four worksheets focuses on equal groups, arrays, word problems, and equations. Super Six math includes a daily multi-part challenge problem designed to test students' conceptual understanding of multiplication and division. The daily Super Six problem has three to four components that students are required to answer. Students keep track of their mastery progress using the included student tracker. I like to set a percentage 3 rd Basic Operations, Math, Other (Math) CCSS 3.OA.A.1 , 3.OA.A.2 , 3.OA.B.5 Also included in:Multiplication and Division Worksheet Center Bundle | word problems | arrays $2.00 Original Price $2.00 Add to cart Wish List Multiplication and Division Worksheet Centers | equal groups | word problems Created by Kiddos and Caffeine Need a multiplication and division center for your students? This set of four worksheets focuses on equal groups, arrays, word problems, and equations. Super Six math includes a daily multi-part challenge problem designed to test students' conceptual understanding of multiplication and division. The daily Super Six problem has three to four components that students are required to answer. Students keep track of their mastery progress using the included student tracker. I like to set a percentage 3 rd Basic Operations, Math, Other (Math) CCSS 3.OA.A.1 , 3.OA.A.2 , 3.OA.A.3 +1 Also included in:Multiplication and Division Worksheet Center Bundle | word problems | arrays $2.00 Original Price $2.00 Add to cart Wish List Percentages of an Amounts One Step Word Problems Colouring Activity Created by Ms Alami's Classroom This is a resource that focuses on one-step word problems involving percentages, offering 20 questions to strengthen students’ problem solving skills. As a teacher, you’ll find this activity easy to integrate into your lessons, providing a valuable tool to reinforce percentage concepts in a practical and enjoyable way. Once students have finished answering the 20 questions, they find the selected answer in the colouring sheet and colour the number in the required colour. Students will learn h 5 th - 6 th Basic Operations, Math, Numbers Also included in:Percentages of an Amounts Differentiated Word Problems Colouring Activity BUNDLE $3.00 Original Price $3.00 Add to cart Wish List Percentages of an Amounts Multi Step Word Problems Colouring Activity Created by Ms Alami's Classroom This is a resource that focuses on Multistep word problems involving percentages, offering 20 questions to strengthen students’ problem-solving skills. As a teacher, you’ll find this activity easy to integrate into your lessons, providing a valuable tool to reinforce percentage concepts in a practical and enjoyable way. Once students have finished answering the 20 questions, they find the selected answer in the colouring sheet and colour the number in the required colour. Students will learn 5 th - 6 th Basic Operations, Math, Numbers Also included in:Percentages of an Amounts Differentiated Word Problems Colouring Activity BUNDLE $3.00 Original Price $3.00 Add to cart Wish List Percentages of an Amounts Two Step Word Problems Colouring Activity Created by Ms Alami's Classroom This is a resource that focuses on two-step word problems involving percentages, offering 20 questions to strengthen students’ problem solving skills. As a teacher, you’ll find this activity easy to integrate into your lessons, providing a valuable tool to reinforce percentage concepts in a practical and enjoyable way. Once students have finished answering the 20 questions, they find the selected answer in the colouring sheet and colour the number in the required colour. Students will learn h 5 th - 6 th Basic Operations, Math, Numbers Also included in:Percentages of an Amounts Differentiated Word Problems Colouring Activity BUNDLE $3.00 Original Price $3.00 Add to cart Wish List Multiplication and Division Worksheet Centers | equal groups | word problems Created by Kiddos and Caffeine Need a multiplication and division center for your students? This set of four worksheets focuses on equal groups, arrays, word problems, and equations. Super Six math includes a daily multi-part challenge problem designed to test students' conceptual understanding of multiplication and division. The daily Super Six problem has three to four components that students are required to answer. Students keep track of their mastery progress using the included student tracker. I like to set a percentage 3 rd Basic Operations, Math, Other (Math) CCSS 3.OA.A.1 , 3.OA.A.2 , 3.OA.A.3 +1 Also included in:Multiplication and Division Worksheet Center Bundle | word problems | arrays $2.00 Original Price $2.00 Add to cart Wish List A Trip to the Museum - Money, Rates and Percentage Word Problem Created by Recycled Systems Analyst A quick quiz on rates and percentages. This is a series of questions (word problems presented in stages) about a calculating the cost of trip to the Museum. It could also be used as a worksheet. 6 th - 12 th Applied Math, Arithmetic, Decimals $2.00 Original Price $2.00 Add to cart Wish List 1 2 3 4 Showing 1-24 of 82+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. 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https://www.youtube.com/watch?v=tK2y0GV2BY4
Solving a Nice Problem Using Vieta's Formulas | Sum and Product of Roots Anon Math 1700 subscribers 33 likes Description 656 views Posted: 13 Oct 2021 ⭕️ Subscribe for more: ⭕️ 🔴 Solving a Nice Problem Using Vieta's Formulas | Sum, and Product of Roots | How to use Vieta formulas Hey there. In this video, we want to deal with a problem relating to the roots of a quadratic. we are given that p and q are the roots of 2x^2-4x+1 and we want to use that information to evaluate p^2+q^2 p^3 + q^3 If we solve the quadratic to evaluate p and q, we will see that it will not be easy to use the value of p and q to evaluate p^2+q^2 and p^3 + q^3. Instead, we will use Viteta's formulas, which are some powerful tools that connect the roots of a quadratic to the coefficints. 🔴I hope you enjoy watching this video.🔴 Don't forget to: ✅ Leave a comment ✅ Subscribe ✅ Hit the like button ✅ Ring the bell topics covered in this video: how to use Vieta formulas the roots of a quadratic solving a quadratic problem sum and product of roots vietafomulas #math 7 comments Transcript: hello everyone this is anan from nn math channel in this video we're going to take a look at a very nice problem related to vieta's formulas the problem is if p and q are roots of this expression 2x squared minus 4x plus 1 then we need to compute these four expressions to do that as i said we're going to use vedas formulas so if you're interested just watch the rest of this video but before we get to vieta's formulas let's just take a look at a different idea for solving this problem the idea is if p and q are roots of this expression why wouldn't we take this expression and set it equal to zero and then solve it for x and find p and q because it kind of feels like if we have the value of p and q it wouldn't be super hard to find p squared plus q squared or the rest of the expressions we're supposed to find but i gotta tell you actually it's going to be super hard to do that and i'm going to show you why let's say we take this expression 2x squared minus 4x plus 1 and we set it equal to 0 and then we solve it for x using the quadratic formula so we should get x equals 4 plus or minus square root of 16 minus 8 over 2 times 2 which is 4 which simplifies to 4 plus or minus 2 root 2 divided by 4 which reduces to 2 plus or minus root 2 divided by 2. and that means that p can be 2 plus root 2 over 2 and q can be 2 minus root 2 divided by 2 which means if we want to find p squared plus q squared we would have to square them and then add the result if you want to find p q plus q cubed we would have to cube them and then add the results but the thing is since these numbers are rational numbers with radicals it's not easy to do that so probably it doesn't make any sense to solve this problem using this method i mean if you had something like x squared minus 5 x plus 6 which can be factored as x minus 2 times x minus 3 then in that case it would make sense to solve this problem in this method because the roots are 2 and 3 so we can easily square them cube them and etc but about our situation since the roots are irrational numbers with radicals it's not easy to do that so in that case it makes sense to solve using this method but in our case it doesn't especially when we know about vieta's formulas now what are veera's formulas beta's formulas state a very beautiful relation between roots of a polynomial and its coefficient for a quadratic expression such as ax squared plus bx plus c it expresses that p plus q the sum of the roots equals negative b over a and the product of the roots p times q equals c over a so in our case p plus q equals negative negative 4 over 2 which equals 2 and p times q equals 1 over 2. so what we are going to do is to use these two values to find the expressions we're supposed to find so as i said based on vieta's formulas the sum of the roots equals two and the product of the roots equals one over two and we want to use these two to find p squared plus q squared how can we do that well we can rewrite p squared plus q squared as p squared plus q squared plus 2pq minus 2pq basically we add and subtract 2pq now the reason we do that is that if we put these three together we get a perfect square p plus q quantity squared so basically we completed this square and we get p plus q quantity squared minus 2pq so the idea is to rewrite these expressions in a form that we would have p plus q and p times q in order to evaluate these expressions we would have to rewrite them in terms of p plus q and p times q and since we know p plus q is two and p times q is one over two we can evaluate this and this equals four minus one which equals three so this one equals three now the second one is p cubed plus q cubed to evaluate this one we can use two approaches first we can factor this as a sum of two cubes we can factor it as p plus q times p squared minus pq plus q squared which is the same thing as p plus q times p squared plus q squared minus pq and we know p plus q is two p squared plus q squared is three and p times q is one over two so this equals two times three minus one over two which equals six minus one which equals five so the second one is five and notice that we also could say since p plus q equals two then p plus q quantity cubed equals eight and if you expand the left-hand side using the binomial theorem we should get p cubed plus three times p squared times q plus three p times q squared plus q cubed equals eight and if we isolate p cubed plus q cubed on the left hand side we should get p cubed plus q cubed equals uh eight minus three p q times p plus q and then this equals eight minus three times one over two times two which equals five so again we found out p cubed plus q cubed equals five now in the third one we want to find p over q plus q over p to do that i'm going to get a common denominator and to get a common denominator we would have to multiply both the top and the bottom of the first fraction by p and the second one by q so we get p squared over pq plus q squared over pq which equals p squared plus q squared over p times q and as we already found out p squared plus q squared equals three and p times q is one over two so this equals three over one over two which is the same thing as three times two which equals six so this one equals 6. and the last one is square root of p plus square root of q to evaluate this first i'm going to set it equal to a and then i'm going to square both sides if i do that i get p plus q plus 2 times square root of p times q which equals a squared and then we know that p plus q equals two and p times q is one over two so this gives us a squared equals two plus two times square root of one over two which is the same thing as two plus 2 times root 2 over 2 which is the same thing as 2 plus root 2. so a squared equals 2 plus root 2 and we want to find a so we need to take the square root of both sides and if we do that we get a equals positive or negative square root of 2 plus root 2 but notice that since this is positive and this is positive and sum of two positive numbers is going to be a positive value so a is a positive number so that means that a must be positive square root of two plus root two so the last one equals square root of two plus root two and this brings us to the end of this video thank you for watching this video as always if you enjoyed watching this video click like click subscribe for more content and don't forget to leave a comment and click on the notification bell so you wouldn't miss any of my future videos thank you for watching this video again goodbye and i will see you later as always in another video [Music] [Music] you
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Search How do integrate with dy/dx? Ghost3k New member SAMUELK New member Ghost3k New member renegade05 Full Member Ghost3k New member renegade05 Full Member
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https://www.phy.pku.edu.cn/xzli/group-theory-part1.pdf
《群论一》讲义 李新征 北京大学物理学院 ii 课程导言 很高兴这个学期继续为大家担任 《群论一》 课程的主讲教师。 今年选课的学 生比较多，尤其是本科生。研究生这边因为我们这两年教学改革，我建议不要把 这门课列为一些专业的必选课， 而是按核心课对待。 因此人数稍微比往年少一些。 但总选课人数还是两百多。 我不知道本科生中有没有一部分人是因为我去年在课 上说今年最后一次讲而选的。如果有，先声明一下应该不会是最后一次。因此， 大家不要急着选。按我一会儿在课上讲的学习规律来。 之后， 我们这堂课按三部分展开。 先说一下 《群论一》 与 《群论二》 的关系。 再说一下这学期的课程内容与前几个学期的不同。最后，按惯例，对《群论》课 程的起源、特性、参考书这些东西进行一个详细的介绍。目的，是在大家进入很 细节的学习之前，对课程有个整体的认识。 先说 《群论一》 是啥？它与 《群论二》 的关系。 以前我们这门课就叫 《群论》 。 这两年我们在进行教学改革。 大伙的一致共识是不应该叫 《群论》 ， 而应该叫 《群 论一》 ，因为物理学科用到的群论一个学期我们肯定cover 不住，还有李群李代 数。我们北大实际上一直是按两个学期开的，但不按两个学期叫。其它高校，多 数就讲一个学期。我自己的感觉，是不可能有很好的教学效果。因为《群论一》 讲群论基础，重点是有限群。当然，还cover 转动群、双群这些东西。教学重点 是群表示论及其应用。学完之后，大家在凝聚态物理、原子分子物理、光学、理 论化学、材料科学这些研究中能够把课程内容用起来。 《群论二》 ，重点是李群这 种连续群的代数结构。我前一段时间看到一句话，有限群讲表示、李群讲代数， 感觉挺准确的。这两年，我们的《群论一》正常情况下还是我讲， 《群论二》由 王一男老师负责。大家学完《群论一》之后，可根据自身情况，选择是否学习李 群李代数。 第二点是这学期的课程与前几个学期的不同。 前面这些年， 根据课程讲义， 我们总结了一个教材。2019 年出版，大家可能知道。出版后，因为讲授的方式相 对简单，得到了一些读者的认可。在读者的反馈中，不少人提到希望加入一些关 于李群李代数的介绍。于是，我这边从2021 年底开始着手对教材进行第二版修 订，特意增加第七章来覆盖这部分内容。添加完了之后，我自己觉得在我们这个 学期的课程上，还是应该把这部分内容给加进去。目的当然不是为了把《李群李 代数》彻底讲清楚，我也没这个能力。但是在准备的过程中，我觉得还是发现李 群李代数里面一些基本的概念，这些概念很抽象。但是，它们的产生背后其实有 一个很简单的关于人类对自然界的认识， 以及对为了描述这种认识所发展的数学 的认识， 由简单到复杂演化的基本的逻辑。 这个逻辑如果我们参考讲历史与逻辑 相统一的讲授方式1， 是很容易说起清楚的。 把这个说清楚以后， 我们的学生在进 一步学习 《群论二》 的时候， 一定会容易很多。 基于这个考虑， 我们在 《群论一》 里面增加第七章。目的，将这些理论在历史上的发展进行一个简单的总结，专注 于对多数读者而言比较容易吸收的内容，为学生继续选《群论II》的课程进行一 些铺垫。 1 这个词是我最近在武汉大学哲学系赵林老师在喜马拉雅（一个音频分享平台）上的一个系列讲 座 《古希腊文明的兴衰》 中听到的。 据赵老师介绍， 最早来自黑格尔 （Georg Wilhelm Friedrich Hegel， 1770-1831）。黑格尔谈哲学时，最大的特点是系统、丰富、完整。其中，逻辑在先是他最大的利 器。他的论述多会基于简单的逻辑展开，然后使其逐渐复杂化，进而完整地体现学科的进展。当 然，在历史学家看来，这种方式具有一定的形而上的特质。但不得不说，这种方式对于人们系统 的理解各种知识，并很好地掌握继续发展它的技能，是很有效的。实际上，我在过去很多年地教 学中， 不自觉地就是采用这个方式来进行的。 这里， 冒昧地直接采用这个并不属于自己专业的名 词，特此说明一下。如有不妥，也请谅解！ iv 讲完这两点， 我们回到第三点， 传统的课程导言， 讲我们这门课的基本情况。 我们按下面五句展开： 1） 群论课程性质与特点； 2） 教材情况与需要的基础知识； 3） 教学内容； 4） 什么是群论； 5） 群论的历史以及在近代物理学、化学研究中的应用。 这部分是这门课最轻松的地方，因为有历史、不枯燥。但《群论》从本质上 是一门数学，群、环、域这些概念本身就是近世代数里面的基本语言，群论本身 更是在十九世纪末彻底发展起来的一个近世代数的分支 （人们在解一元多次方程 的时候，发现其解蕴藏着一个数学结构，进而发展出一个代数的分支） 。二十世 纪初，随着Emmy Noether（诺特，女，1882-1935，德国犹太人）就一个物理系 统的对称性与它的守恒量之间关系的认识（1915 年的工作，发表于1918 年，原 始文献） 、量子力学的发展、以及Eugene Paul Wigner（魏格纳，1902-1995， 匈牙利人，后期加入美国籍）和Hermann Klaus Hugo Weyl（外尔，1885-1955， 德国人） 在建立量子力学数学基础的过程中对对称性原理的使用[2-4]， 人们逐渐 认识到群论作为一门数学在物理学研究中的作用。 再后来物理学发展中人们所使 用的规范场 （gauge） 的方法， 是这些研究的进一步延伸 （后面我们会稍微展开讨 论） 。因此，不夸张地说，群论学习是我们物理学专业学生在从事具体研究工作 前所受基础教育中必不可少的环节。 这是群论课程特点在数学和物理方面的体现。 在我们的兄弟学科化学上， 在 量子力学建立后，敏锐的理论化学家们，以Linus Carl Pauling（鲍林）为代表， 已经认识到化学分子的存在形式以及化学反应的发生本质上是由量子力学与统 计物理基本原理支配的。既然对称性在量子力学中具备上述重要性，与之相应， 在描述由量子力学基本原理所决定的反应物、过渡态、生成物特性（比如电子能 级、振动谱等）描述中，对称性原理的数学语言（即群论）必然也会发挥重要的 作用。因此，在近代化学（特别是物理化学）的研究中，人们也认识到由对称性 决定的内在规律对人们理解这些物性与过程的本质至关重要。 换句话， 要想真正 地在分子设计的层面理解化学2，对称性的知识同样必不可少。 因为这些原因， 群论应该说是为数不多的这样一门课： 在好一些的大学的数 学系、物理系、化学系的课程设置中都有涉及。当然，不同的系会有不同的侧重 点。数学系会侧重这门课的数学属性，高一个层面，是我们在物理和化学中展开 应用的基础。而物理和化学系的同学，如果想理解这些应用，必须首先理解这门 课的数学基础部分（说白了就是掌握语言） ，再进行实例分析。物理系的同学， 就专业不同，所需掌握内容也会不同。以凝聚态、光学专业的同学为例，需要掌 握的内容绝大部分集中于有限群理论部分， 当然也需要转动群与双群的知识， 这 些在《群论一》课程中均有涉及。如想进一步理解规范场理论（连续变换下的对 称性与某守恒量的关系） ，李群也应适度学习。而对于理论物理专业的同学特别 是粒子物理专业的同学，李群部分的内容掌握也是必需。化学学院中理论化学、 物理化学专业同学所需掌握内容与物理学院中凝聚态物理、 光学专业类似， 以有 限群部分内容为主。 2 化学的本质是分子设计，这个可以说是目前多数人对化学的理解。此理解最早的提出者应该 也是Linus Pauling 教授。北京大学化学学院的全称是化学与分子工程学院，其内在涵义也在这 个地方。我想这个和唐有祺先生早期是Linus Pauling 的博士应该有一定关系。此观点与化学学 院的部分老师进行过交流，放在这里供大家参考。 vi 不管哪个具体专业，要想理解群论在其关心的具体问题中的应用， “掌握这 些应用的数学基础” （具体而言就是“群基础理论”与“群表示理论”两部分内 容）都是第一步。因此，我们这门课的前1/3 部分本质上就是数学性质的讲解。 就课程特点来说这部分是比较枯燥的。 如果没有学进去， 到了后半部分我们讨论 应用的时候，你就是在听一门没完全学过的外语。因此，必须说明：如果想学这 门课的话，前面两章必须啃下，否则别学！ 与此同时，在学习之前，笔者还需说明：在后面的实例说明中，我们会讲到 一些现在物理学研究中用到的例子。 理解这些例子， 对于这门课的学习是和掌握 理论基础同样重要的目标！ 因为没有这些例子， 你不可能理解到学习这些东西有 什么用？要掌握这部分内容， 我们需要的课程储备是 《量子力学》 与 《固体物理》 ， 没有选过这两门课， 也千万别看这本书和选这门课， 这个是由课程的特点决定的， 需要尊重！ 关于教材， 前三年我都是基于其他老师的教材手写自己的讲义， 每年重复并 更新。第四年把讲义的电子版整理出来，之后每年改进，但整体还比较肤浅。具 体的、深入的讨论大家可以参考： 1. 韩其智、孙洪洲 《群论》 北京大学出版社 2. 王宏利 《群论讲义》 （未出版，网上可以找到） 3. 徐婉棠、 喀兴林 《群论及其在固体物理中的应用》 高等教育出版社 4. M. S. Dresselhaus, G. Dresselhaus, A. Jorio, 《Group Theory: Applications to the Physics of Condensed Matter》. Springer 5. Zhongqi Ma, 《Group Theory for Physicists》, World Scientific 这五本是课程主要参考如果想扩展阅读，也可参考： 6. 马中骐 《物理学中的群论》 科学出版社 （上面那本书的中文版） 7. Anthony Zee, 《Group Theory in a Nutshell for Physicists》, Princeton University Press. （徐一鸿， 科普读物中经常称为阿热， 题目大意为： 物理学家眼 中的群论简言，in a nutshell 本意是：一言以蔽之，简约的） 8. Wu-Ki Tung (董无极) 《Group Theory in Physics》世图有影印版 9．F. Albert Cotton, 《Chemical Applications of Group Theory》, John Willey & Sons. Inc（化学家写的群论中的经典，对读者很友善） 10. 陶瑞宝 《物理学中的群论》 高等教育出版社（这本书很全，包含了 很多群论在物理中的应用，理论性也很强） 11. 张端明、李小刚、何敏华 《应用群论》 科学出版社 12. 俞文海 《晶体结构的对称群》 科大出版社 其中阿热先生的书是在此讲义基本定型后才有幸研读， 看完诸多体会。 如果 能早日看到，讲义本身质量应该会有很大提升。 前言的第三部分是课程的内容。主体是七章：1) 群的基础知识、2) 群表示 理论、3) 点群和空间群、4) 群论与量子力学、5) 转动群、6) 置换群、7）李群 李代数初步。 其中前两章是基础， 提供我们在进行后面的讨论的时候必须用到的 “语言” ，是我们的基本交流工具。在这两章学完之后，下面两个章节是3) 点群 与空间群，4) 群论与量子力学。其中点群、空间群是我们在分子、团簇、凝聚态 体系中遇到的群， 关于它们的性质自然是我们学习的重点。 群论与量子力学这一 章， 在现行教课书中并没有一个统一的路子。 但笔者认为是我们这门课里最重要、 最有用的部分！ 大家学完这门课之后， 有时间的话一定要不断地阅读和这部分相 关的教科书（特别是Dresselhaus 那本） ，这是加深我们对这门课理解的关键！此 viii 部分内容有点像金庸小说中常提到的任督二脉， 掌握好了， 能在科研中合理运用 群论， 课程学习才成功， 科研也会做得更好 （Dresselhaus 本人就是一个最好的例 子） 。 剩下的三章， 转动群不说大家也能感受到它的重要， 早期的原子体系和很多 现在还在用的中心力场理想体系都具备这样的对称性。 此讲义主要关注的主体是 有限群， 转动群本质上是一个连续群， 但它的一些最基本的属性我们在不学习 《群 论二》的情况下也能理解。如果你以后做和电子自旋相关的研究，背后的物理基 本也在这部分内容中。 置换群是一种有限群， 也是在全同粒子体系普遍存在的一 种对称群。我们的课程内容会覆盖到从置换群的基本特性、到其分类（杨图） 、 在到其不等价不可约表示分类（杨盘定理） 、以及简单的如何求置换群的表示这 些内容。对于不学理论物理的同学，一般我们用不上。对学理论并且要选《群论 二》的同学，这些基本的理论储备应该也够，深入的讲解你们下个学期会接触。 李群李代数初步，更是为下学期进行的准备。 上面说的课程内容都可以直接由章节的题目反映出来。 大家如果看其它教材 的话，其实还会注意到两个东西，我们目前还没有提及：一个叫投影算符、一个 叫幂等元，这两者有些联系。在我们的讲义中，分别会在第四章和第六章用到之 前作介绍，不单独把它们作为一章来讲。 导言的第四部分我们想说的是一个具体的问题：什么是群论？ 要明白这个问题的话我们可以先想一下什么是“群论”中的“群” 。这个对 应的英语的词源是group theory 中的group，汉语的翻译很贴切，就是“群”这个 字。汉字拆分，可以把它分为两个部分，一个“君” 、一个“羊” ，背后隐藏的一 个逻辑就是一个君管理了一群羊。在这里羊是一个集合，而君不单指一个人，更 代表一个管理者。他/她和羊在一起，大家可以理解为一个“具有一定结构特征 的集合” ，因为“君”这个管理者就是要给你这个集合建立一个结构特征，并且 要利用这个结构特征去实施管理的。 而群论呢？很自然的就是： 研究这个集合的 结构特征及其生成的规律的一门学科。 根据这个理解，我们回到前面提到的课程内容，很自然，我们就可以简单理 解一下刚才讲到的各章都是干什么的？ 1. 群的基础知识：集合总体的结构特征及其规律； 2. 群表示理论：对这些规律进行数学描述要用到的数学语言（基础是线性 代数） ； 3. 点群、 空间群： 人们面对分子、 晶体系统的时候， 系统具有的对称性操作 的集合。它们是我们在掌握前两章（群论的理论基础）后面对的第一类 具体的群； 4. 群论和量子力学：群论在近代的物理、化学等学科研究中的应用； 5. 转动群：是中心力场系统的对称群（物理体系中的一类对称群） ； 6. 置换群：是全同粒子系统的对称群（物理体系中的一类对称群） ； 7. 李群李代数初步：有限群向李群的过渡。 根据这个理解， 我们同时还很容易明白群论从本质上而言是研究数的结构及 其生成规律的，是数学，不是物理。我们物理研究的是物质运动的内在规律，一 般先强调“物” ，针对“物”来理解“理” 。而群论这门学科发展的初期，是人们 对一些“理”的认识，这些“理”是“数理” ，不是“物理” 。人们基于对这些“数 理”的认识，建立起了一套理论。后来人们又逐渐意识到它在物理、化学上有很 大的用途，才开始要求物理、化学这些专业背景的人来学习，以期对本学科中的 x 问题有更深入的认识。 就教学而言， 物理上教 《群论》 的老师分两拨。 一拨是做得比较理论的老师。 相应教材的特点是严格、抽象、深入。另一拨是做物质科学相关研究的，相应教 材比较直观、便于理解，但内容不包括《群论二》的部分。笔者的背景是后者， 此讲义只希望将《群论一》讲清楚。 至此， 《群论》是什么样的一门课大家应该有些概念了。但在学之前，出于 好奇，可能我们还是想知道一下作为一门学科《群论》是如何发展起来的？它现 在处在一个什么样的位置？这个就把我引到了我在引言中想解释的第五句话： 群 论的历史以及在物理和化学中的应用。 前面提到， 群论是近世代数的一个重要的分支， 它是在19 世纪发展起来的。 在发展的初期，数学上的另外三个分支是基础。这三个分支分别是： 1) 几何学，从19 世纪开始，有个德国数学家，叫August Ferdinand Möbius （莫比乌斯，1790-1865，德国人） 。他在研究一些非欧几何的问题的时候，就开 始使用了一些对称操作的概念。 和莫比乌斯相关的另外一个我们现在用的比较多 的概念是莫比乌斯环， 就是把一个纸条连成环的过程中翻一下， 这样的一个环和 正常的环比起来就不再有A、B 面了。这个概念在拓扑上比较有用； 2) 数论，这个是在18 世纪下半叶，欧拉在研究数论中的模算术的时候，用 到过一些群论中尚处在雏形阶段的概念； 3) 第三个基础是代数方程理论。应该说是它直接导致了群论作为一门学科 的诞生。 更准确地说就是人们在求解一元高次方程根式解的时候， 引入了置换群 的概念，进而建立起了群论这个理论体系。 现在， 人们会认为由这三个方面研究所诱发出来的群论是近世代数 （抽象代 数）中很重要的部分，并把它作为十九世纪最伟大的数学成就来看待。 而关于这个学科诞生的细节很数学。 想真正理解的话， 需要对抽象代数这门 课有深刻的认识（笔者自己曾经尝试着去看了一些，花了很大精力，但最后发现 这个确实超出能力范围） 。这里跟大家分享的，只是一些hand-waving（没有坚实 的理论基础， 试图显得有效， 但并没有触及实质内容） 的图像层面的认识。 网上， 大家也能找到一些资源，不如“妈咪说”就有一个分成五节的视频，我看完觉得 说得非常好！强烈推荐大家看！他真的很用心，并且写稿的水平很高。中间还用 了一些趣味数学的内容，比如1870s 由Samuel Loyd 提出的一个叫15 puzzle 的东 西，来说明置换的一些基本属性。大家把我们第一章内容学完，在结合一点第六 章第一节的内容， 应该就可以完全把他那五节视频看懂。 由于这个资源已经在那 儿，而我们的课程内容又已经足够多了，因此我们还是按自己的节奏来，不重复 他那里的内容。 刚才提到， 最直接的导致群论诞生的诱因是代数方程理论的发展。 代数方程， 大家都知道，一元一次的是ax + b=0，一元二次的是ax2+bx+c=0。它们的解析根 式解我们在中学的时候就学过。 一元三次方程和一元四次方程有没有和它们类似的根式解？ 答案：有。 对一元三次和四次方程， 早期人们是可以利用配方和换元的方法把它们变成 低次方程来求解的。 比方说ax3+bx2+cx+d=0 这样一个式子，a≠0。人们怎么做呢？ 先换元，取𝑦= 𝑥+ b 3a，把它代入上式，企图把𝑥的一般的一元三次方程变成 𝑦的一元三次方程。 而这个𝑦的一元三次方程， 不再是一个一般的一元三次方程， xii 而是具有特殊形式的一元三次方程。过程如下： a (𝑦−b 3a) 3 + b (𝑦−b 3a) 2 + c (𝑦−b 3a) + d = 0 a (𝑦3 −b a 𝑦2 + b2 3a2 𝑦−b3 27a3) + b (𝑦2 −2b 3a 𝑦+ b2 9a2) + c (𝑦−b 3a) + d = 0 a𝑦3 + (c −b2 3a) 𝑦+ (d + 2b2 27a2 −bc 3a) = 0 二次项不见了，一元三次方程变成了y3+py+q=0，这里p、q 都是由a、b、 c、d 确定的常数。而这样的一个特殊形式的一元三次方程，是有根式解的。这个 里面有个故事，时间是16 世纪，地点是意大利。当时在欧洲的数学界，去寻求 一元三次方程的解是一个时尚， 就像我们现在物理学界对高温超导机制的研究一 样3。因为当时的历史背景是文艺复兴（Renaissance） ，所以在学术上最活跃的地 区很自然的就是意大利（大家可以去想，哥白尼、布鲁诺、伽利略这三个现代科 学的鼻祖里，两个意大利人，一个哥白尼是波兰人，但基本在意大利生活） 。代 表人物有两个，Niccolo Fontana（冯塔纳，1499-1557）和Girolamo Cardano（卡 尔达诺， 也叫卡丹， 1501-1576） 。 传说第一个想出这个特殊方程根式解是冯塔纳， 但此君比较喜欢通过故弄玄虚来显示自己的聪明，不把话说明。因此，虽然当时 有很多人相信他会解这个方程， 但没有任何文献记录 （当时的出版业并没有现在 这么发达， 不然一个arXiv 就解决问题了） 。 而卡丹呢， 比较低调务实， 传说中他 跟冯塔纳讨教过， 这个冯塔纳用很隐晦的语言进行了提示， 但他认为以卡丹的悟 性根本理解不了。但结果是人家愣把它想明白了，并且在他的著作《大术》 （Ars Magna，1545）中给了一些详细的解释。因为这个，现在我们在讨论一元三次方 程的根式解的时候， 想到的第一个人物往往是卡丹， 只是在很少的文献中才会对 3当时人们经常针对类似问题进行数学比武。 当时冯塔纳的工作有所提及。 上面那个特殊一元三次方程的解， 人们也习惯于叫 卡丹公式： 𝑦1 = √−𝑞 2 + √(𝑞 2) 2 + (𝑝 3) 3 3 + √−𝑞 2 −√(𝑞 2) 2 + (𝑝 3) 3 3 𝑦2 = 𝜔∙√−𝑞 2 + √(𝑞 2) 2 + (𝑝 3) 3 3 + 𝜔2 ∙√−𝑞 2 −√(𝑞 2) 2 + (𝑝 3) 3 3 𝑦3 = 𝜔2 ∙√−𝑞 2 + √(𝑞 2) 2 + (𝑝 3) 3 3 + 𝜔∙√−𝑞 2 −√(𝑞 2) 2 + (𝑝 3) 3 3 这个里面𝜔= −1 + √3i。 这个是关于一元三次方程根式解的故事。 过程其实已经很麻烦了， 不然不会 让冯塔纳犯那个错误。对一元四次，之后人们又用相似方法做了努力，由卡丹的 学生Lodovico Ferrari（费拉里，1522-1565，意大利人）给出了根式解，这个结果 也是在卡丹的那本1545 年的《Ars Magna》里面发表的。那么五次、六次及其以 上又是什么情况呢？同样，在1545 年以后也继续成为欧洲数学界的时尚。但两 百年过去了，却始终没有任何进展。 图0.1 一元高次方程根式解 当这个问题有下一步进展的时候，也就到了我们《群论》作为一门学科出现 的时候了。这个前后发展的时间有一百多年，从1770 年代开始，到19 世纪末结 束。其中的代表人物包括Joseph-Louis Lagrange（拉格朗日，1736-1813，意大利 xiv 人，绝大部分时间工作在德国与法国） 、Paolo Ruffini（鲁菲尼，1765-1822，意大 利人） 、Évariste Galois（迦罗瓦，1811-1832，法国人） 、Niels Henrik Abel（阿贝 尔，1802-1829，挪威人） 、Arthur Cayley（凯莱，1821-1895，英国人） 、Ferdinand Georg Fröbenius（费罗贝尼乌斯，1849-1917，德国人） 、William Burnside（勃恩 赛德， 1852-1927， 英国人） 、 Friedrich Heinrich Schur （舒尔， 1856-1932， 德国人） 、 Marius Sophus Lie（李，1842-1899，挪威人）这些数学家。其中前面这些人（到 阿贝尔） ，他们工作的初衷是求一元五次方程的解，但结果是建立了群论。而后 面这些人，从凯莱开始，他们的主要工作，就是完善这个由前人提出的理论了。 这些名字以及与他们相关的定理，在后面的教学中我们会慢慢接触到。 怎么把解一元五次方程和《群论》这门学科联系起来，背后的道理其实很简 单。前面我们提到了，在费拉里之后，两百多年，欧洲各位顶级的数学家都尝试 着利用配方、换元这些数学手段去求四次以上方程的根式解，但都没成功。这种 情况下，按科学规律而言，一般传统思维肯定就不行了。这个就像我们把自己关 在一个屋子里找东西，你的前辈科学家，各个聪明绝顶，他们把这个屋子的每个 角落都进行了仔细的搜寻， 都没有找到。 这个时候你应该去意识到是不是这个屋 子有另外一个维度你并不知道？你需要打破传统思维去找到这个维度？物理史 上我们都知道的一个例子就是人们黑体辐射， 在19 世纪末20 世纪初， 传统经典 的思想是怎么都不可能在长波和短波区域同时给出合理解释的。 这个时候， 人们 就需要去拓展自己的思维了， 而把思维扩展开来的这些人， 就是我们眼中的天才 了，比如普朗克（当然普朗克常数的产生更多的是数学上的处理，而不是思想深 处的理解或信念） 、比如爱因斯坦（大家一定不要受一些科普读物的误导，他实 际上是量子力学发展最大的一个推动者之一， 从思想层面。 光电效应是他解释的， 德布罗意的波粒二象性也是他最早支持的，这些都是突破思维定式的典型例子。 只是在后期， 他从一个数学物理学家的视角， 不喜欢玻尔这些人对量子力学的一 些实用性解释。除了量子力学，狭义与广义相对论也是更典型的例子） 。 在五次及其以上一元方程根式解的问题上， 认识到这一点的最早的人物是拉 格朗日。他实际上是看到了一种新的数学结构。从他开始，到鲁菲尼，到伽罗瓦 与阿贝尔， 他们做的事情是开始从群的结构， 也就是常说的群论的角度去考虑这 些问题了。具体而言，如果把一个一元n 次方程的根式解作为变换对象，那么它 就会对应一个n 次置换群。其中拉格朗日、鲁菲尼干的事情是利用置换的概念， 去理解了三次和四次方程为什么有解。 之后就是伽罗瓦和阿贝尔了， 他们干的事 情是彻底地在代数方程的可解性与其对应的置换群之间建立了联系， 指出了n 次 方程有解的充要条件， 以及说明一般的五次方程没有根式解。 一个n 阶置换群是 可解群的条件， 是它的不变子群形成的不变子群列具有一个特殊的性质。 就是前 一个不变子群对后一个不变子群的商群，都是Abel 群。对于置换群而言，前四 个都有这个性质。第五个及其以上，没有。因此，一元五次及其以上方程没有根 式解。这个就是数学的魅力！它与哲学一样，对应的都是一些最基本的智慧。由 哲学中的自然哲学发展出的诸多自然科学的分支中，物理学也具备这样的特质。 这也是我们常说的物理学最能代表自然哲学的根本原因。在拉格朗日、鲁菲尼、 伽罗瓦、阿贝尔这四个人里面，前面两个是奠定基础的，迦罗瓦与阿贝尔是真正 利用群这个概念去解决这个问题的。我们现在会把后两个当作是《群论》这门学 科的奠基人4。 4建议到wikipedia去看一下这两个少年天才的生平。 笔者在课上尽量介绍每个科学家的生平， 就是希望我们在学习科学的同时， 不要脱离科学家本身所处的时代背景。 科学上重大进步的 产生，都是以由科学家本身的时代背景、学科背景综合起来诱发的。学生时代应尽量了解这 xvi 在数学家建立了《群论》的概念体系之后，我们物理学家做了什么呢？我想 比较有代表性的是下面三个方面的工作： 1. 几何晶体学的发展，晶体点阵、点群、空间群这些概念的诞生以及他们 在晶体学中的应用。这个的主要发展时间是19 世纪末、20 世纪初，代 表人物是Arthur Moritz Schöneflies（熊夫利，1853-1928，德国犹太人） 、 Carl Hermann （赫尔曼， 1998-1961， 德国人） 、 Charles Victor Mauguin （毛 古因， 1878-1958， 法国人） 。 后面讲点群空间群的时候我们会讲到他们。 2. 对称性与守恒量之间的关系，这个代表人物是诺特，她是个典型的数学 物理学家。她没得诺奖，不过这个不影响她本身的伟大。物以类聚，套 用现在的语言就是如果用微信的话她朋友圈是爱因斯坦、希尔伯特这种 人，她也被这些人称为数学史上最伟大的女性。诺特定理的基本内容是 “any differentiable symmetry of the action of a physical system has a corresponding conservation law” ，也可以说是任何一个保持拉格朗日量不 变的微分算符， 都对应一个守恒的物理量。 下面这张图是我从 《赛先生》 上面2015 年6 月20 号发的一篇文章上摘下来的（作者是UT Austin 的 些， 这样你才会对你的学科发展的规律产生一定的理解。 只有理解了每个发现背后那些让人 热血沸腾的故事与逻辑， 你才能真正理解教科书上那些冷冰冰的文字背后的内涵。 伽罗瓦被 认为是浪漫主义天才的代表。传说他投稿三次，第一次的审稿人是柯西，第二次的审稿人是 傅立叶，两次都没有发表，柯西让他把论文写的好懂一些，他没有听，傅立叶接到稿件没几 天自己都挂了，第三次投稿的时候，迦罗瓦本人已经因为决斗牺牲了，他的朋友帮他投的， 这次的审稿人是雅可比和高斯， 但这些大佬其实没有时间仔细看。 后来这个稿件又沉睡多年， 在得到了刘维尔的肯定后最终发表。从这些审稿人，我们应该可以感受到19 世纪法国数学 的强大。阿贝尔生平最大的标签，除了天才，就是贫穷。他是挪威人，挪威在当时欧洲科学 的版图中可以说是彻底的边缘。 他自己本身很优秀， 但找教职一直不顺。 27 岁死于贫困与疾 病，死后收到了柏林大学的聘书，令人唏嘘。 ） 张天蓉博士） ，很形象得描述了这个规律： 图0.2 诺特定理，对称性与守恒量关系示意图 比如空间平移对称性对应动量守恒、时间平移对称性对应能量守恒、旋 转对称性对应角动量守恒，等等。这些规律我们现在其实是都把它们当 常识了。它们究竟怎么来的？我们一会儿会用平移不变性对应动量守恒 作为一个例子（经典力学范畴内的一个问题） ，来个推导。 同时需要说明：我们目前都知道的Gauge Theory（规范场论） ，应该说是 沿着这条路继续的、 更加深入的发展。 它的基本思想是系统的Lagrangian （拉格朗日量）在一个连续的局域变换（规范变换）下保持不变。规范 这个词本意是scale（伸缩因子） ，但后来人们发现它的真实物理对应其 实是相位。现代物理研究中，它通指拉格朗日量多余的自由度。不同规 范间的变换 （也就是我们常说的规范变换） ， 形成了一个可以解析表达的、 具有微分流形性质的连续群，就是李群。在《群论二》 ，我们会学到每个 李群都有自己的群生成元。每个群生成元，会产生一个矢量场，也对应 一个代数结构。这个矢量场，就是规范场。这些对经典理论和量子理论 都是成立的。在量子理论中，这个规范场的量子被称为规范玻色子。以 我们最熟悉的电磁场为例， 量子电动力学理论就是个阿贝尔的规范理论， 它的阿贝尔的对称群是U(1)群， 它的规范场就是由电势𝜙与磁势𝐴 ⃗形成的 xviii 四分量矢量场(𝜙，𝐴 ⃗)，它的规范玻色子就是光子。近年来应该说用规范 场的理论去统一很多模型，比如量子力学、电动力学、量子色动力学， 是物理学最大的挑战。群论在中间发挥着重要的作用。 3. 对称性在量子力学中的应用，这个代表人物是维格纳[2-3]。他也因为这 方面的研究得了1963 年的诺贝尔物理奖（1/2，另外两个人分那1/2） ， 他获奖原因， 原话是 “for his contributions to the theory of the atomic nucleus and the elementary particles, particularly through the discovery and application of fundamental symmetry principles” 。 例1．平移不变性与动量守恒 考虑一个封闭的力学系统， 无外力， 那么这个系统的运动方程是由其作用量 （Action）决定的。这个Action 是 I = ∫L[q(t), q̇(t)] t2 t1 dt 设Q(t)这个函数是粒子的实际轨道， 而δq(t)是对这个实际轨道的偏移。 那么， 由 最小作用量原理，我们知道对实际轨道有： δ𝐼 δ𝑞(𝑡) = ∫ { 𝜕𝐿 𝜕𝑞(𝑡) −d d𝑡 ∂𝐿 𝜕𝑞̇(𝑡)}| 𝑞(𝑡)=𝑄(𝑡) d𝑡= 0 𝑡2 𝑡1 对任意t1、t2 成立。 既然它对任意t1、t2 成立，自然就会有： { 𝜕𝐿 𝜕𝑞(𝑡) −𝑑 𝑑𝑡 𝜕𝐿 𝜕𝑞̇(𝑡)}| 𝑞(𝑡)=𝑄(𝑡) = 0 这样一个式子。这个式子就是牛顿方程（我们理解这个问题的第一步） 。 现在引入平移不变性（第二步） ，对任意平移𝑎，有 𝐼[𝑄(𝑡2) + 𝑎, 𝑄(𝑡1) + 𝑎] = 𝐼[𝑄(𝑡2), 𝑄(𝑡1)] 这个式子左边为： 𝐼[𝑄(𝑡2) + 𝑎, 𝑄(𝑡1) + 𝑎] = ∫𝐿[𝑄(𝑡) + 𝑎, 𝑄̇(𝑡)]d𝑡 𝑡2 𝑡1 （平移不改变微分项） ，继续等于： ∫𝐿[𝑄(𝑡), 𝑄̇(𝑡)]d𝑡 𝑡2 𝑡1 + ∫ 𝜕𝐿 𝜕𝑄(𝑡) 𝑎 d𝑡 𝑡2 𝑡1 + ∆(a2) 而右边为： 𝐼[𝑄(𝑡2), 𝑄(𝑡1)] = ∫𝐿[𝑄(𝑡), 𝑄̇(𝑡)]d𝑡 𝑡2 𝑡1 等式对任意𝑎、任意𝑡1、𝑡2都成立，所以有： 𝜕𝐿 𝜕𝑄(𝑡) = 0 这个式子，代入运动方程，就有： d d𝑡 𝜕𝐿 𝜕𝑄̇(𝑡) = 0 而𝜕𝐿/𝜕𝑄̇(𝑡)对应动量，故由平移不变性可推出动量守恒。 现在数学说完了、物理说完了，前面我们提到， 《群论》这门课是在一个正 常的大学里面数学、物理、化学三个系都会开的。化学家在我们这门学科的发展 过程中起了什么样的作用呢？应该说和我们物理学家同等重要。 具体而言就是他 们在将这个理论应用到具体物性研究中扮演了重要的角色。 最具代表性的领域是 理论化学，很关键的一个人物是鲍林（Linus Pauling） 。这个人非常了不起，如果 说他是最具影响力的几个化学家之一与最具影响力的理论化学家（没有之一） ， 应该不为过。他是第一个将量子力学基本原理、分子轨道、分子设计这些概念引 入到化学研究中的人。也是我们现在公认的量子化学、分子生物学的开创人。 这里为什么要这样推崇鲍林呢？原因很简单，我们科学，往广义的说，就是 用理性的观点去认知客观世界， 这个理性的基本工具是数学。 在我们认知的过程 xx 中，由于侧重点的不同，科学会分化出很多学科，比如物理、比如化学、比如生 物。 我们物理关注的是物质的存在形式与运动规律， 化学关注的是不同物质放在 一起的反应，而生物关注的是生命的行为。我们相互之间是不应该排斥的。以物 理和化学为例，笔者在早期受教育的时候，始终认为它们是两个东西。直到做科 研， 才意识到现在的凝聚态物理的研究中其实是非常需要化学知识的； 而同时量 子化学， 说白了， 就是将量子力学基本原理应用到具体分子与凝聚态体系的行为 描述中去。 应该说是物理和化学两个大的学科的交融， 才使得两者都发展到了目 前的这个相当成熟的状态，而最早去推动这种交融的人，鲍林就是代表。这个人 本身是个化学家，美国人，笔者认为对他科研影响最大的一段经历，应该是他在 1926-1927 年在欧洲游学的这个时候。在这里他接触到了Arnold Sommerfeld（索 末菲，1869-1951，德国人） 、Niels Bohr（玻尔，1885-1962，丹麦人） 、Erwin Schrödinger（薛定谔，1887-1961，奥地利人）等人。他在这里接受了量子力学的 训练， 之后他敏锐得意识到这个东西在化学中的应用， 并且开始用这些原理去研 究化学中的现象，比如分子轨道、分子振动谱，等等。这些都是我们目前的科学 研究中运用群论的最为直接的例子， 在后面我们会详细讲。 在之前推荐的参考书 中，Albert Cotton 的那本《Chemical applications of group theory》就是一个典型的 化学家写的群论教材。相比于我们物理学家写的教材，会更实际、易读。 最后总结一下， 我们这个学期要学习的群论， 确实是人类文明在过去两百多 年间发展出来的一个精华，是我们认识我们所处在的这个世界本质的重要工具， 在我们日常的科学研究中，起着非常重要的作用。像杨先生总结20 世纪物理学 关键词：量子化、相位、对称性。其中后两个均与本课程相关。当然，此讲义内 容为基础部分。但虽说来简单，要想学明白，也需要花费很大的功夫。因此，谨 慎选课、认真对待！ 目录 前言 ............................................................................................................ 错误!未定义书签。 课程导言 .................................................................................................................................... ii 目录 ........................................................................................................................................... 1 第一章 群的基本概念 ............................................................................................................. 4 1.1 群 ..................................................................................................................................... 4 1.2 子群与陪集 ..................................................................................................................... 8 1.3 类与不变子群 ............................................................................................................... 12 1.4 同构与同态 ................................................................................................................... 18 1.5 变换群 ........................................................................................................................... 26 1.6 直积与半直积 ............................................................................................................... 30 1.7 习题与思考 ................................................................................................................... 38 第二章 群表示理论 ............................................................................................................... 41 2.1 群表示 .......................................................................................................................... 41 2.2 等价表示、不可约表示、酉表示................................................................................ 51 2.3 群代数与正则表示 ....................................................................................................... 62 2.4 有限群表示理论 ........................................................................................................... 69 2.5 特征标理论 ................................................................................................................... 88 2 2.6 新表示的构成 ............................................................................................................... 96 2.7 习题与思考 ................................................................................................................. 115 第三章 点群与空间群 ......................................................................................................... 117 3.1 点群基础 .................................................................................................................... 117 3.2 第一类点群 ................................................................................................................. 134 3.3 第二类点群 ................................................................................................................. 150 3.4 晶体点群与空间群 ..................................................................................................... 162 3.5 晶体点群的不可约表示 ............................................................................................. 190 3.6 习题与思考 ................................................................................................................. 200 第四章 群论与量子力学 ..................................................................................................... 202 4.1 哈密顿算符群与相关定理 ........................................................................................ 203 4.2 微扰引起的能级分裂 ................................................................................................. 214 4.3 投影算符与久期行列式的对角化.............................................................................. 218 4.4 矩阵元定理与选择定则、电偶极跃迁 ...................................................................... 236 4.5 红外、拉曼谱、和频光谱 ......................................................................................... 241 4.6 平移不变性与Bloch 定理 .......................................................................................... 250 4.7 布里渊区与晶格对称性 ............................................................................................. 255 4.8 时间反演对称性 ......................................................................................................... 258 4.9 习题与思考 ................................................................................................................. 262 第五章 转动群 ..................................................................................................................... 264 5.1 SO(3)群与二维特殊酉群SU(2) .................................................................................. 264 5.2SO(3)群与SU(2)群的不可约表示 ............................................................................... 274 5.3 双群与自旋半奇数粒子的旋量波函数 ...................................................................... 281 5.4 Clebsch-Gordan 系数 ................................................................................................... 294 第六章 置换群 ..................................................................................................................... 297 6.1 n 阶置换群 .................................................................................................................. 298 6.2 杨盘及其引理 ............................................................................................................. 305 6.3 多电子原子本征态波函数 ......................................................................................... 319 参考文献 ............................................................................................................................... 334 附录A 晶体点群的特征标表 ............................................................................................. 337 附录B 空间群情况说明 ...................................................................................................... 352 附录C 晶体点群的双群的特征标表 .................................................................................. 355 附录D 置换群部分相关定理与引理证明 ......................................................................... 368 4 第一章 群的基本概念 1.1 群 定义1.1 群：设𝐆是一些元素（操作）的集合，记为𝐆= {⋯，g，⋯}，在𝐆中定义 了乘运算，如果𝐆中元素对这种运算满足下面四个条件： 1) 封闭性：∀两个元素（操作）的乘积仍属于这类元素（操作）的集合5； 2) 结合律：对∀三个元素（操作）𝐟、𝐠、𝐡，有(𝐟𝐠)𝐡= 𝐟(𝐠𝐡)； 3) 有唯一单位元素e，使得对∀𝐟∈𝐆，有𝐞𝐟= 𝐟𝐞= 𝐟； 4) 对∀𝐟∈𝐆，存在且唯一存在𝐟−𝟏属于G，使𝐟−𝟏𝐟= 𝐟𝐟−𝟏= 𝐞； 这时我们称𝐆是一个群，其中元素是群元，𝐞为其单位元素，𝐟−𝟏为𝐟的逆。 去理解这个定义，我们先看一些例子。 例1.1. 一个集合有两个操作E和I，E作用三维欧式空间中任一向量r ⃗上，得到r ⃗本 身，I作用这个r ⃗上，得到−r ⃗。问{E，I}是否形成一个群？ 考虑这种问题的时候，就去想群的定义。两个元素，操作组合有4 种，E ∙ E、E ∙I、I ∙E、I ∙I，其中任何一个作用到r ⃗上，结果不是r ⃗就是−r ⃗，所以效 果与E或者I作用到r ⃗上一致，封闭性满足。 结合律，类似(E ∙I) ∙E = E ∙(I ∙E)的关系对于这三个位置怎么填都成立。 唯一单位元素E。 逆元素，E的逆是E，I的逆是I。 所以{E，I}形成一个群，称为空间反演群。 例2. 这样一系列操作的集合，它们中每一个操作，都把1、2、……、n 这n 个 数，一对一的对应到1、2、……、n 这n 个数上。比如 5 包含元素和其本身乘积。 𝑃= ( 1 2 … … m1 m2 … … n mn) 式2.1 就是把1、 2、 ……、 n 对应到m1、 m2、 ……、 mn上， 其中m1、 m2、 ……、 mn是1、2、……、n 的任意排列。 注意， 在这个标记中( 1 2 … … m1 m2 … … n mn)与( 2 1 … … m2 m1 … … n mn)是一样的， 因为它们的效果都是把1 变为m1、2 变为m2、…（以此类推） 现在我们来看这样的操作的集合是否形成群？ 1． 封闭性：不管怎么变，1、2、……、n 这n 个数都是变到这n 个数上， 封闭性满足； 2． 结合律：设P1 是把1 变为2，P2 是把2 变为3，P3 是把3 变为4，那 么 (P1P2)P3=(1→2)(2→3)=(1→3)(3→4)=(1→4) 式2.2 P1(P2P3)=(1→2)[(2→3)(3→4)]=(1→2)(2→4)=(1→4) 式2.3 两者相等，结合律成立 3． 单位元：有且唯一，什么都不变的那个操作 4． 逆：存在且唯一，你变过去我再变回来 所以这些元素的集合也构成一个群， 我们称为n 阶置换群， 它的群元的个 数是n!。在物理上，处理全同粒子体系的时候，会经常用到这一类群，我 们后面会专门介绍。 例3. 是个几何图形的对称性， 有三维欧式空间的一个正三角形， 顶点是A、 B、 C。 6 图1.1 D3 群示意图 对于这样一个三角形， 它有六个纯转动可以使自身回到自身的操作， 分别 是： 1. e：不动； 2. d：绕z 轴转2π/3； 3. f：绕z 轴转4π/3； 4. a：绕1 轴转π； 5. b：绕2 轴转π； 6. c：绕3 轴转π； 现在问：这六个操作是否形成群？ 这个答案肯定还是按我们上面的思路来走，看它是否满足那四个条件？ 但与此同时， 我们还可以借用一下前面讲的置换群的概念， 因为这些操作 无非是将（A、B、C）对应到（A、B、C）上去，而这六个几何操作又恰 恰和三阶置换群的六个变换一一对应。因此它们形成一个群。 既然形成一个群，现在我们看它们的乘法关系。d 操作干的事情是 (A B C B C A)，而a 操作干的事情是(A B C A C B)，因此： d ∙a = (A B C B C A) (A B C A C B) = ( (A B C A C B) (A C B B A C) ) = (A B C B A C) = c 式2.4 重复类似运算，可得完整乘法表： e d f a b c e e d f a b c d d f e c a b f f e d b c a a a b c e d f b b c a f e d c c a b d f e 表1.1 D3 群乘法表 这样一个群叫D3 群，它是图形中的三角形的纯转动群，当我们除了转动 还包含反射、 反演这些操作的时候， 群元就会再多一些， 群也不是这个D3 群了。这里的D3 群只包含转动操作。 例4. 定义群的乘法为数的加法， 则全体整数构成一个群， 0 是其中的单位元素， n 与-n 互逆。 同理，全体实数也在这个乘法规则下构成一个群，全体复数也是。 但如果我们把乘法定义为数乘， 那么它们就不再是群了， 因为这种情况下 单位元素只能是1，而0 是没有逆的。 现在我们知道群是定义了乘法且满足一定规律的元素的组合， 下面我们看一 下跟群相关的两个定义与一个定理。 定义1.2 有限群与无限群：群内元素个数称为群的阶，当群阶有限时，称为有限 群，当群阶无限时，称为无限群。 （这个学期我们主要讲有限群） 定义1.3 Abel 群：群的乘法一般不可交换（这个在群的定义里面没有体现，因此 在一般的群中也不需要遵守，比如D3 中ad 就不等于da） ，当群中元素乘法可以 任意互换时，这个群称为Abel 群。 （由这个定义我们很容易想象Abel 群的乘法 8 表都应该是相对于对角线对称的） 定理1.1 重排定理： 设𝐆= {⋯，𝐠𝛂，⋯}， 对∀𝐮∈𝐆， 当𝐠𝛂取遍𝐆中所有元素时， 𝐮𝐠𝛂给出且仅仅一次给出𝐆中所有元素。 证明： 两个方面，1)任何G中元素都可以由ugα给出，2)仅仅一次给出。 1. 给出。对任意gβ属于G，可取u−1gβ ∈G，使得：u(u−1gβ) = gβ 2. 仅仅一次给出。 反证：设有gα ≠gα′，使得ugα = ugα′，那么就会有：u−1ugα = u−1ugα′，进 而gα = gα′，与假设矛盾。 至此，第一节结束。四个内容，三个定义（群，有限、无限群、Abel 群） ， 一个定理（重排） 。这些讲的都是群本身的性质，不牵扯其内部结构。既然要理 解群这个元素集合的结构特性， 对其内部结构的认识不可避免。 下面的内容很自 然与内部结构有关，子群与陪集。 1.2 子群与陪集 定义1.4 子群：设H 是群G 的一个子集（部分元素的集合） ，若对群G 相同的 乘法运算，H 也构成一个群，则称H 为G 的子群。 这里需要注意的地方是和G 相同的乘法。 同时， 上面我们定义群的时候， 用 了四个条件。原则上，定义子群也需要这四个条件1) 封闭性、2) 结合律、3) 单 位元、4) 每个元素唯一逆。但因为H 属于G，又是相同的乘法，所以结合律自 然成立。同时，如果4)满足，则有f属于H 时，f −1也属于H，只要封闭性成立， e 自然属于H。因此，在证明子集为子群时，只要1)与4)成立就可以了。 显然{e}与G 本身都是G 的子群，由于太明显，所以称为显然子群，或平庸 子群。而群G 的非平庸子群称为固有子群。一般我们找群G 的子群的时候找的 是它的固有子群（非平庸子群） 。 例5. n 阶循环群，它的定义是an = e，由{a、a2、⋯、an−1、an = e}组成。这 样的群是Abel 群， 乘法可易。 以6 阶循环群为例， G = {a、a2、⋯、a5、 a6 = e}，其中{e}与G是显然子群。{a2、a4、e}与{a3、e}为固有子群。 例6. 在定义群的乘法为数的加法的时候， 整数全体形成的群是实数全体形成的 群的子群。 例7. 绕固定轴k ⃗⃗转动的元素形成的群{Ck ⃗ ⃗⃗(Ψ)}， 是绕轴上某一点转动 （过这点可 以有无数个轴）的群SO(3)群的子群。 定义1.5 群元的阶： 对任意一个有限群𝐆， 从中取一个元素𝐚， 从𝐚出发作幂操作， 总是可以构成𝐆的一个循环子群𝐙𝐤的，这个𝐙𝐤等于{𝐚、𝐚𝟐、⋯、𝐚𝐤−𝟏、𝐚𝐤= 𝐞}， 这时称𝐤（满足这个性质的最小的𝐤）为群元𝐚的阶。 这个概念很好理解， 但有个地方需要说明一下， 就是你凭什么说 “从a出发， 总能构成G的一个循环子群的” ？这是因为如果a = e， 则Zk等于{e}， 问题解决。 如果，a ≠e，则a2 ≠a（不然a = e） ，这时，如果a2 = e，则问题又解决了。如 a2 ≠e， 则它必为e与a之外的另一个元素， 我把a2、 a放到我的子集中， 继续做a3， 同样a3 ≠a2（不然a = e） 、也不等于a（不然a2 = e） ，如果a3 = e，问题解决，如 a3 ≠e，再把a3放到那个子集中。依次类推，因为G是有限群（阶为n） ，所以必然 存在一个k小于等于n， 使得ak = e， 来结束这个过程。 这时， {a、a2、⋯、ak−1、 ak = e}这个集合自然就形成了k阶循环子群了。 例8. 群元的阶的例子， 对D3 群， 六个元素e、d、f、a、b、c。 对d， 从它出发， d2 = f，d3 = e，所以由d形成的循环子群是{e、d、f}，d的阶是3。 10 对f，f 2 = d，f 3 = e，所以由f形成的循环子群也是{e、d、f}，f的阶也是 3。类似，a，a2 = e，由a出发形成的循环子群是{e、a}，a的阶是2；b， b2 = e，由b出发形成的循环子群是{e、b}，b的阶是2； c与a、b一样。 说完了子群与群元的阶，下一个概念是陪集。 定义1.6 陪集： 设𝐇是群𝐆的子群， 𝐇= {𝐡𝛂}， 由固定的𝐠∈𝐆， 可生成子群𝐇的左 陪集：𝐠𝐇= {𝐠𝐡𝛂|𝐡𝛂∈𝐇}，也可生成子群𝐇的右陪集：𝐇𝐠= {𝐡𝛂𝐠|𝐡𝛂∈𝐇}。 这个定义做两点说明。一是当H是有限子群时，陪集元素个数等于H的阶。 因为不可能存在hα ≠hα′但ghα = ghα′或hαg = hα′g的情况。 也就是说子群中元素 与陪集中元素一一对应。二是根据这个定义，陪集可以为子群本身。如果上面取 的g ∈H，就是。如果不属于，就不是。关于子群和陪集，除了这两点，还有一个 很重要的性质，就是陪集定理。 定理1.2 陪集定理：设群𝐇是群𝐆的子群，则𝐇的两个左（或右）陪集或者完全相 同，或者没有任何公共元素。 换句话说， 对于一个群G， 如果它按照其子群H来进行分割： G = {g0H、g1H、 g2H、⋯}，g0 = e，其中任意两个陪集giH与gjH的关系是它们要么完全相同，要 么根本没有任何公共元素。 证明： （以左陪集为例） 设uH、vH是不同陪集。 再假设uH与vH中间有一个公共元素uhα = vhβ，则有v−1uhα = hβ，进而v−1u = hβhα −1，v−1u ∈H。 这个时候，由于H本身是个群，所以由重排定理知道v−1uH = H，那么v(v−1uH) 自然等于vH。而同时v(v−1uH) = uH。也就是说uH=vH，与假设矛盾。 （证毕，右陪集证明类似） 有了这个性质， 我们在面对一个群的时候， 按照它的一个子群和这个子群的 陪集去进行分割，我们会得到什么？ 群是G，子群是H，它本身就是一个陪集。除了H，我们可以再取u1 ∈G，但 ∉H， 建一个H的陪集u1H， 这个u1H与H是没有任何公共元素的， 因为如果u1hα = hβ，则u1 = hβhα −1，与u1 ∉H矛盾。这时可以将G写为H、u1H、以及它们以外的 元素的集合。这个时候在继续取u2属于G，但它不属于H，也不属于u1H。做陪集 u2H，u2H中的u2不在eH中，也不在u1H中，所以u2H与eH、u1H完全不同。 依次类推， 设G 的阶是n， H 的阶是m。 则每个陪集给出的都是全新的m 个 群G 中的元素。重复这个过程，直到把G 中元素穷尽。那么这个H 的陪集的个 数就应该是n/m。这个n/m 必须是个整数，这也就意味着子群H 的阶m 必为群 G 的阶n 的因子。这个性质就是我们的第三个定理。 定理1.3 拉格朗日（Lagrange）定理：有限群子群的阶，必为群阶的因子。 由这个定理， 我们再去分析D3 群的子群， 我们说过它的子群有{e}、 G、 {e、 d、f}、{e、a}、{e、b}、{e、c}，这些子群的阶分别是1、6、3、2、2、2，都是 6 的因子。用{e、d、f}来分割群的话，G={{e、d、f}、a{e、d、f}}，其中a{e、 d、f}对照前面的乘法表，给出的恰恰是{a、b、c}。 至此，前两节结束。讲了六个定义：1) 群，2) 群的阶、有限群、无限群，3) Abel 群，4) 子群，5) 循环子群与群元的阶，6) 陪集。三个定理：1) 重排定理， 2) 陪集定理，3) 拉格朗日定理。在导言里面我们提到群是一个有结构的元素的 集合，这三个定理是最基本的结构性质，以后很多定理的证明都会用到。第二节 12 讲的子群和陪集是群中元素结构关系的一个方面， 下面一节要讲的类与不变子群 是其另一个方面。 1.3 类与不变子群 此节中，类与不变子群是我们要重点阐明的概念。理解它们需要一个基础， 这个基础是共轭。 定义1.7 共轭：所谓共轭，指的是群𝐆中两个元素𝐟、𝐡，如果在𝐆中存在一个𝐠， 使得𝐟、𝐡可以通过𝐠𝐟𝐠−𝟏= 𝐡联系起来，则称𝐟、𝐡共轭，记为𝐟~𝐡。 由此定义， 我们首先知道共轭是相互的。 因为如果gfg−1 = h， 则g−1hg = f， 也就是g−1h(g−1)−1 = f， 其中g−1 ∈G。 其次， 我们知道共轭有传递性， 也就是说 f1~f2，f2~f3，则f1~f3，为什么？因为由f1~f2，我们知道，一定存在g ∈G，使得 gf1g−1 = f2。 而由f2~f3， 我们又知道一定存在h ∈G， 使得hf2h−1 = f3。 把第一个 式子代入第二个，就会有hgf1g−1h−1 = f3，进而hgf1(hg)−1 = f3。而hg是属于G 的，所以f1~f3。 由此传递性，我们可以去定义类。 定义1.8 类：群G 中所有相互共轭的元素形成的集合，称为群G 的一个类。 由于共轭关系的传递性， 我们知道一个类是可以被其中任何一个元素所确定 的。这个有些像现实生活中的犯罪团伙，逮着一个，其他的也就差不多了。此比 喻不一定准确，但有相似的地方。当然，你也可以用朋友类比，因为朋友一定程 度上也有相互性、传递性、以及物以类聚这些特点。 而就由类中任何一个元素确定这个类的操作步骤而言， 很简单， 对一个类中 元素f，取任意g属于G，做操作gfg−1。当g走遍G 中所有元素的时候，那么f 的 所有同类元素，没跑，就一个个全出现了。 （地毯式搜捕） 由这个共轭关系和类的定义，我们还可以得出： 1) 一个群中的单位元素自成一类，因为对任意f 属于G，fef −1 = e； 2) Abel 群的所有元素都自成一类，因为对任意f 属于G，取任意h 属于G，做 hfh−1 = hh−1f = f； 3) 设群元素f 的阶为m，即f m = e，则与它同类的元素的阶也为m。这是因为 它的同类元素可以写为gfg−1， 对于这个元素， (gfg−1)k = gf kg−1。 当k 1 也就是说可约表示的特征标内积总是大于1 的。 同时关于群函数与这里讲的类函数， 还有一个关系以及由它引起的性质必须 说一下。 这个性质跟我们前面说的Burnside 定理有类似的地方， 是一个从维度引 出的群表示的性质。 这个性质解释的话最好还是从群函数和类函数的关系说起。 群函数我们之前 提到， 就是你给我一个群元， 我给你一个数。 群函数空间和群空间同维， 都是n。 而类函数，我们之前提到过，就是你给我一个类，我给你一个数。按照这个对应 关系，类函数空间的维度是多少？ 答案：类函数空间的维度就是这个群中类的个数。 由于类函数要求群中同类的群元给出的数相同， 这个可以说是对群函数加了 一个额外的限制，因此，我们也可以说类函数是一种特殊的群函数。 现在群函数和类函数的关系明确了， 我们想说的下一个性质与有限群不等价 不可约表示的特征标有关。 定理2.9 有限群的所有不等价不可约表示的特征标在类函数空间是完备的。 证明： 设群G 是我们关心的有限群，Ap是它的所有不等价不可约表示，其中p = 1， ⋯，q。q 是这个群的不等价不可约表示的个数。对有限群，由于每个表示都有 等价的酉表示，所以Ap这个系列有另外一个和它们等价的不等价不可约酉表示， 记为：A′p。 由之前讲的有限群不等价不可约酉表示矩阵元在群函数空间完备这个性质， 我们 知道对任意的一个群函数f(g𝑖)，它都可以写成： f(g𝑖) = ∑ aμ,ν p p,μ,ν A′ μ,ν p (g𝑖) 式2.10 而类函数，由于f(g𝑖) = f(g𝑗 −1g𝑖g𝑗)，可以写为： f(g𝑖) = 1 n ∑f(g𝑗 −1g𝑖g𝑗) n 𝑗=1 这样由式2.10 可知这个类函数可以进一步写成： f(g𝑖) = 1 n ∑∑aμ,ν p p,μ,ν A′ μ,ν p (g𝑗 −1g𝑖g𝑗) n j=1 = 1 n ∑∑∑aμ,ν p A′ μ,λ p (g𝑗 −1)A′ λ,σ p (g𝑖)A′ σ,ν p (g𝑗) λ,σ p,μ,ν n j=1 = ∑∑aμ,ν p A′ λ,σ p (g𝑖) (1 n ∑A′ μ,λ p (g𝑗 −1)A′ σ,ν p (g𝑗) n j=1 ) λ,σ p,μ,ν = ∑∑aμ,ν p A′ λ,σ p (g𝑖) (1 n ∑A′ λ,μ p∗(g𝑗)A′ σ,ν p (g𝑗) n j=1 ) λ,σ p,μ,ν = ∑∑aμ,ν p A′ λ,σ p (g𝑖) ( 1 Sp δλ,σδμ,ν) λ,σ p,μ,ν = ∑∑1 Sp aμ,μ p A′ λ,λ p (g𝑖) λ p,μ = ∑( 1 Sp ∑aμ,μ p μ ) (∑A′ λ,λ p (g𝑖) λ ) p = ∑apχ′p(g𝑖) p = ∑apχp(g𝑖) p 这也就是说任何一个类函数， 都可以用我这个有限群的不等价不可约表示的特征 标函数来展开。换句话说，类函数的空间维度，就应该等于我不等价不可约表示 的个数。 （证毕） 92 这个时候我们再结合上面分析类函数空间性质的时候得到的一个群群元可 以分多少个类，它的类函数空间就是多少维。很自然，我们就可以得到“一个群 的不等价不可约表示的个数就等于这个群中类的个数”这样一个性质了。 这个由特征标函数的完备性推出的性质， 和由不等价不可约酉表示表示矩阵 函数的完备性推出的Burnside 定理，合在一起，就是我们对一个有限群的不等 价不可约表示情况分析时的最为有力的工具了。 我们现在知道了特征标作为群函数的正交性、完备性。按照上一节，有限群 不等价不可约酉表示性质描述那一部分的路子， 大家可能会觉得我们要讲的应该 都结束了。 但实际情况是这一节我们还有两个关键的内容没讲。 这两个内容都和 上面那个性质“有限群不等价不可约表示数等于群中类的个数”有关。 在我们之前讲的特征标第一正交定理中， 我们的加和号是走遍群元， 或者说 走遍类的，具体形式是： (χp|χr) = 1 n ∑χp∗(g𝑖)χr(g𝑖) n 𝑖=1 = 1 n ∑n𝑖χp∗(K𝑖)χr(K𝑖) q′ 𝑖=1 = δpr 在讲上面这个类空间完备性定理的时候， 我们还不知道有限群的不等价不可约表 示数q 等于它类的个数q′。现在我们知道了q′ = q，那么这个加和号会出现另外 一种情况：加和到不等价不可约表示这个指标上，这种情况下，会不会出现正交 指标存在于类这个index 上面的情况呢？也就是： 1 n ∑n𝑖χp∗(K𝑖)χp(Kj) q p=1 = δij 这个答案是：会。而这种正交关系我们称为特征标第二正交定理。 定理2.10 𝟏 𝐧∑𝐧𝒊𝛘𝐩∗(𝐊𝒊)𝛘𝐩(𝐊𝐣) 𝐪 𝐩=𝟏 = 𝛅𝐢𝐣 证明： 我们可以设计一个矩阵F， q × q的，行指标走遍不等价不可约表示，列指标走遍 类。F 的形式为（第r 行，第i 列） ： Fri = √n𝑖 n χr(K𝑖) 这样一个矩阵的矩阵元很明显具备这样的性质： 𝐹 pi ∗= √n𝑖 n χp∗(K𝑖) = (F+)ip 之前我们讲过的特征标第一正交定理， 1 n ∑n𝑖χp∗(K𝑖)χr(K𝑖) q 𝑖=1 = δpr 按照这个矩阵形式，可表述为： ∑Fri(F+)ip q 𝑖=1 = (FF+)rp = δpr 这也就意味着FF+ = E，F+ = F−，进而：F+F = E。这个式子，写成矩阵元的形 式，就是： (F+F)𝑖j = ∑(F+)irFrj q r=1 = ∑[√n𝑖 n χr∗(K𝑖)√n𝑗 n χr(K𝑗)] q r=1 = δij 由于i，j 不等时两边都为零，所以也可以写为： 1 n ∑n𝑖χr∗(K𝑖)χr(Kj) q r=1 = δij （证毕） 由特征标的两个正交关系， 我们可以引入的一个很重要的东西就是有限群表 示的特征标表。它的具体定义我们可以这样理解：既然特征标是类函数，它又对 不同不等价不可约表示可以不同，那么原则上，在分析一个有限群的特征标时， 94 我们可以按两个轴来展开。 第一个轴是有限群的类， 第二个轴是群的不等价不可 约表示指标。 因为类数等于不等价不可约表示数， 所以由这两个轴做出的表的行 数和列数是相同的，都等于有限群的类的个数。在这个表的每一列，我们记为 n𝑖{K𝑖}， 其中{K𝑖}是类， n𝑖是其中元素个数。而每一行，我们记为Ap，是不等价不 可约表示的index。这个表的具体形式是： n1{K1} n2{K2} ⋯ n𝑞{K𝑞} A1 χ1(K1) χ1(K2) ⋯ χ1(Kq) A2 χ2(K1) χ2(K2) ⋯ χ2(Kq) ⋮ ⋮ ⋮ ⋱ ⋮ A𝑞 χq(K1) χ𝑞(K2) ⋯ χq(Kq) 表2.1 特征标表示意图 在这里，行列之间都有正交，对应的就是特征标的两个正交定理。 例10. n 阶循环群{a，a2，⋯，an = e}，它是 Abel 群，有n 个类，所以每个不 等价不可约表示都是一维的。A(a)是a 的表示，它是一个数，它要满足的 特性是： (A(a)) n = 1。 所以， A(a)可以是： exp[2π𝑖∗(p −1)/n]， 其中i 是 虚部因子，p 是不等价不可约表示的指标，从1 到n。 它所对应的特征标表就是： 1{e} 1{a} 1{a2} 1{a3} A1 1 1 1 1 A2 1 i −1 −i A3 1 −1 1 −1 A4 1 −i −1 I 表2.2 四阶循环群特征标表 这里行列的正交性大家都可以检验一下。 到这儿， 跟特征标函数相关的概念和定理我们都说完了。 在这一节开始的时 候，我们提过要确定两个表示等价，最笨的方法是找相似变换矩阵。因为这个太 麻烦了，我们才会引入特征标这个概念。现在我们把特征标的所有性质讲完了， 回到这个问题， 也就是这节课最后一句话： 两个表示等价的充要条件是它们的特 征标相等。 必要性不用解释了，等价特征标必相同，因为相似变换不改变特征标。充分 性怎么理解？这个要从类函数空间中不等价不可约表示特征标函数的正交性与 完备性出发。两个表示的特征标相同，说明它们对应的类函数相同。同一个类函 数， 在类函数空间， 是可以分解为相同的不等价不可约表示特征标函数的线性组 合的。也就是说表示是A 与B，它们都可能可约，但如果它们的特征标相同，那 就意味着当我把它们分解为不等价不可约表示的直和的时候，⊕mpAp的形式是 完全一样的。这也就意味着下面关系： 图2. 5 特征标相同则表示等价示意图 1 与2 通过相似变换称为具有相同结构的豆腐块3 与4，而3 与4 之间每个豆腐 块相互又是等价的，所以表示的特征标相同与表示等价互为充要条件。 96 到这个地方，我们这门课理论部分最重要的两节就结束了。还是那句话，这 是个坎儿，迈过了后面内容很自然你就能理解了。 这一章的最后一节是新表示的构成。 它说的是一个什么事情呢？在前面的学 习中，我们应该有这样一个感觉，对一个群对称性的分析，最重要的地方就是要 知道它的特征标表。在求这个特征标表的时候，我们说了，Burnside 定理、群元 的分类、 两个正交条件我们都可以利用。 当群比较小， 它的分类比较简单的时候， 我们利用这些条件把特征标表求出来。 但是当群比较大的时候， 这样就不太容易 了。这个时候，我们要做的，就是从这些比较复杂的群的因子，也就是比较简单 的群出发，去构造这些比较复杂的群的表示。那么新表示的构成这一节，讲的基 本就是这样一个事情。 它由四个部分构成：群表示的直积、直积群的表示、诱导表示、群表示在其 子群上的缩小。其中第四个概念最简单，第三个最复杂。第二、第三个最有用。 2.6 新表示的构成 先看群表示的直积，这个概念比较直接，但在讲它之前，我们还是先看一下 矩阵直积。 我们知道一个n × n的矩阵A=(aik)， 一个m × m的矩阵B=(bjl)， 这两个矩阵 的直积为C = A ⊗B，它的群元为cij,kl = (aikbjl)。这是一个(n × m) × (n × m)的 矩阵，行指标为ij，列指标为kl。用这个行列指标时，为方便起见，我们规定格 式为： C = ( a11B a12B a21B a22B … a1nB … a2nB ⋮ ⋮ an1B an2B ⋱ ⋮ … annB ) 这样规定的格式和其它方式规定的格式，差的是一个相似变换。 由这种方式定义的直积矩阵显然具备如下四个性质： 1. 两个单位矩阵的直积是单位矩阵； 2. 两个对角矩阵的直积是对角矩阵； 3. 两个酉矩阵的直积仍为酉矩阵； 4. 若A(1) 与A(2) 是阶相同的矩阵，B(1) 与B(2) 是阶相同的矩阵，则(A(1) ⊗ B(1))(A(2) ⊗B(2)) = (A(1)A(2)) ⊗(B(1)B(2))。 这四个性质里面1、2 很好理解。3、4 需要解释一下。先解释4，3 利用4 就 很好理解了。对4 中的式子，左边A(1)的行列指标为(i，k)，B(1)的行列指标为 (j，l)， A(1) ⊗B(1)的指标为(ij，kl)。 同样A(2)的行列指标为(i′，k′)， B(2)的行列 指标为(j′，l′)，A(2) ⊗B(2)的指标为(i′j′，k′l′)。当A(1) ⊗B(1)与A(2) ⊗B(2)作乘 法时，指标加和的要求是k = i′、l = j′。产生的矩阵指标为(ij，k′l′)。 而式子右边，A(1)A(2)的指标为(i，k′)，加和要求是k = i′；B(1)B(2)的指标为 (j，l′)，加和要求为l = j′。在A(1)A(2)与B(1)B(2)作直积时，没有加和的要求，只 有指标的组合，组合后结果为(ij，k′l′)。这样在进行右边的操作的时候，总结一 下， 就是加和要求为k = i′、 l = j′， 产生的矩阵指标为(ij，k′l′)。 和左边完全相同。 这样的话(A(1) ⊗B(1))(A(2) ⊗B(2)) = (A(1)A(2)) ⊗(B(1)B(2))成立。 由这个式子成立， 我们再去看第3 点性质。 这一点说的是A 是一个酉矩阵， 具备A+A = En， B 是一个酉矩阵， 也具备B+B = Em。 要说明A⊗B 也是酉矩阵， 满足(A ⊗B)+A ⊗B = En×m。 这个时候根据第4 点， 很容易有(A ⊗B)+A ⊗B = (A+ ⊗B+)A ⊗B = (A+A) ⊗B+B = En ⊗Em = En×m。 （补充：如果上述理解不直接，也可以换个角度，因为A 是酉矩阵，满足A+A = En，这个关系按照矩阵元的性质表示，就是： 98 ( a11 ∗ a21 ∗ a12 ∗ a22 ∗ … an1 ∗ … an2 ∗ ⋮ ⋮ a1n ∗ a2n ∗ ⋱ ⋮ … ann ∗ ) ( a11 a12 a21 a22 … a1n … a2n ⋮ ⋮ an1 an2 ⋱ ⋮ … ann ) = En 进而∑ aik ∗aik′ n i=1 = δkk′。同样，对B 也有：∑ bjl ∗bjl′ m j=1 = δll′。 当A 与B 作直积C 时，矩阵元为cij,kl = aikbjl，要看C 是否为酉矩阵，就是要看 C+C是否为单位矩阵， 体现在矩阵元上， 就是∑cij,kl ∗ cij,k′l′ ij 是否等于δkk′δll′？这个 时候，把C 的矩阵元代入，就有： ∑cij,kl ∗ cij,k′l′ ij = ∑(aik ∗bjl ∗)aik′bjl′ ij = (∑aik ∗aik′ n i=1 ) (∑bjl ∗bjl′ m j=1 ) = δkk′δll′ 补充完毕） 现在有了矩阵直积的定义与性质，我们来看群表示的直积。 定义2.18 群表示的直积：群G 有两个表示A 与B，作表示矩阵的直积𝐂(𝐠𝜶) = 𝐀(𝐠𝜶) ⊗𝐁(𝐠𝜶)， 这个直积矩阵群保持群的乘法规律不变， 因为对∀𝐠𝜶，𝐠𝜷∈𝐆， 有： 𝐂(𝐠𝜶)𝐂(𝐠𝜷) = (𝐀(𝐠𝜶) ⊗𝐁(𝐠𝜶)) (𝐀(𝐠𝜷) ⊗𝐁(𝐠𝜷)) = (𝐀(𝐠𝜶)𝐀(𝐠𝜷)) ⊗(𝐁(𝐠𝜶)𝐁(𝐠𝜷)) = 𝐀(𝐠𝜶𝐠𝜷) ⊗𝐁(𝐠𝜶𝐠𝜷) = 𝐂(𝐠𝜶𝐠𝜷) 这时，C 也是群G 的一个表示。这个表示称为A 表示与B 表示的直积表示。 因为矩阵直积的迹等于其因子迹的乘积， 所以直积表示的特征标等于其因子 的特征标的乘积： χC(g𝛼) = χA(g𝛼)χB(g𝛼) 如果A与B 都是群G的不可约表示， 那么(χA|χA) = (χB|χB) = 1， 而(χA|χB) = 0。这个时候直积表示的特征标内积为： (χC|χC) = 1 n ∑χA∗(gi)χB∗(gi)χA(gi)χB(gi) n i=1 这个加和是没法做分解的，一般情况下结果大于1，对应可约表示。这个也很好 理解，群空间你没有变，但你把表示矩阵变胖了很多。我这个群空间允许的不可 约表示都是那些由Burnside 定理和群的分类情况确定的瘦矩阵， 你来个胖子， 肯 定不属于我们这个瘦人集团。 我自己能想到两个不可约表示的直积仍为不可约表示的情况是：A 与B 中 间任何一个 （比如B） 是一维表示， 且满足χB∗(gi)χB(gi)这个数， 对任意gi都等于 1。 看个例子。 例11. 我们最常见的D3 群它的特征标表是： 1{e} 2{d} 3{a} A1 1 1 1 A2 1 1 -1 A3 2 -1 0 表2.3 D3 群特征标表 由这个特征标表，我们知道下面几个直积群的特征标分别为： A1 ⊗A3 2 -1 0 与A3等价，不可约； A2 ⊗A3 2 -1 0 与A3等价，不可约； A1 ⊗A2 1 1 -1 与A2等价，不可约； 100 A3 ⊗A3 4 1 0 可约了。我们把它记为C，由(χC|χA1) = 1、(χC|χA2) = 1、(χC|χA3) = 1， 我们知道它约化完的结果为A1 ⊕A2 ⊕A3。 以上就是群表示的直积部分，下面我们看直积群的表示。这里，G1 = {g1𝛼} 与G2 = {g2𝛽}是两个我们已知的群，G 是它们的直积群。由上一章我们讲的，这 个关系意味着任意一个G 中元素， 可唯一写成g1𝛼g2𝛽的形式， g1𝛼g2𝛽= g2𝛽g1𝛼， 且两个G 中元素g1𝛼g2𝛽与g1𝛼′g2𝛽′的乘积，可由g1𝛼g1𝛼′g2𝛽g2𝛽′得到，G1与G2中 元素乘法互易，它们都是G 的不变子群。 这是一个非常强的关系。 当这个关系成立的时候， 如果A 是G1的表示， B 是 G2的表示，那么对任意G 中元素g1𝛼g2𝛽，都有一个矩阵A(g1𝛼) ⊗B(g2𝛽)与之对 应。且这种对应关系会保持G 中元素的乘法规律不变，因为g1𝛼g2𝛽这个G 中元 素与g1𝛼′g2𝛽′这个G 中元素对应的矩阵乘法满足： C(g1𝛼g2𝛽)C(g1𝛼′g2𝛽′) = (A(g1𝛼) ⊗B(g2𝛽)) (A(g1𝛼′) ⊗B(g2𝛽′)) = (A(g1𝛼)A(g1𝛼′)) ⊗(B(g2𝛽)B(g2𝛽′)) = A(g1𝛼g1𝛼′) ⊗B(g2𝛽g2𝛽′) = C(g1𝛼g1𝛼′g2𝛽g2𝛽′) = C(g1𝛼g2𝛽g1𝛼′g2𝛽′) 左边是g1𝛼g2𝛽对应的矩阵与g1𝛼′g2𝛽′对应的矩阵的乘积， 右边是g1𝛼g2𝛽与g1𝛼′g2𝛽′ 乘积对应的矩阵，形成表示。这样的表示我们称为直积群的表示。 定义2.19 直积群的表示：群G 是另外两个群𝐆𝟏与𝐆𝟐的直积，𝐆𝟏与𝐆𝟐分别有表 示A 与B。这时，根据直积分解规则， 𝐂(𝐠𝟏𝜶𝐠𝟐𝜷) = 𝐀(𝐠𝟏𝜶) ⊗𝐁(𝐠𝟐𝜷)，构成群 G 的表示。这个表示称为𝐆𝟏与𝐆𝟐直积群的表示。 这个直积群的表示和我们前面讲的群表示的直积有什么区别呢？最重要的 一点点就是在群表示的直积那里， 我已经知道了它的两个表示， 我干的事情是用 这两个表示矩阵做直积，来形成新的表示。而在直积群的表示这里，我们本来是 不知道G 的任何表示的，我们知道的是它的两个不变子群的表示，我们同时知 道这个群可以写成这两个不变子群的直积。我们是基于这个关系构造的群G 的 表示。由于这个差别，在前面我们说群表示的直积形成的新的表示时，两个不可 约表示形成的新的表示可能可约。而在这里，如果A 与B 是G1与G2的不可约表 示，那么A(g1𝛼) ⊗B(g2𝛽)对应的G 的表示也不可约。因为设G1与G2的阶为n 和 m 时，有： (χC|χC) = 1 nm ∑∑χA∗(g1α)χB∗(g2β)χA(g1α)χB(g2β) m β=1 n α=1 = (1 n ∑χA∗(g1α)χA(g1α) n α=1 ) (1 m ∑χB∗(g2β)χB(g2β) m β=1 ) 当A 与B 不可约时，1 乘上1 还是1。换句话说，对G，如果它太复杂，我们可 以利用直积群的表示这个概念，由它因子的不可约表示来构造它的不可约表示。 下面我们看诱导表示。 这个概念的难度在我们这门课里面是数一数二的， 当 然，有用程度也很高。不然我们没必要自找麻烦。 它的主要作用是什么呢？前面我们讲过， 新表示的构成这一节的一个中心思 想是从已知表示构造新的表示。实际研究中，最经常面对的情况是群G 比较复 杂， 它的子群比较简单， 我们要从它子群的表示出发构造群G 的表示。 前面讲的 直积群的表示处理的实际就是这样一个情况。 但是我们必须要清楚， 在直积群的 表示那里，我们对群G 结构特征的要求是非常高的。这个群本身必须可以分解 为两个子群的直积，实际情况下这种条件很难成立。这里要讲的诱导表示，面对 的就是这样的情况。 对它而言， 我只要求群G 存在一个子群就行了， 我们从这个 102 子群的表示出发构造群G 的表示，我们不要求这个子群是不变子群，我们更不 要求这个群G 可以直积分解。 用图画出来，这里的结构关系是这样的： 图2.6 诱导表示示意图一 我们有群G，它很复杂，我们不知道它有什么表示。我们知道的是它有一个子群 H， 这个H 简单很多， 它有一个表示B 我们知道， 同时我们知道这个表示的表示 空间是W。我们想知道由这些信息，能否给出G 的一个表示？ 这里“诱导”也是这个意思了，就是有已知的H 的信息，得出未知的G 的 信息。把这个目的明确后，我们就去看怎么诱导了？这个说白了，就是要去定义 一个线性变换群，使得G 与它同态。而定义这个线性变换群，说白了三步： 1. 定义一个线性空间，使得线性变换可以作用到这个线性空间的向量上； 2. 定义这个线性变换； 3. 把这些线性变换放在一起，看它们是否形成群，是否与G 同态（也就是保持 乘法规律不变） 。 当这些都做完以后，我们就可以说我们从H 诱导出了G 的一个表示了。下 面我们的讨论也会按这个思路来进行。 一. 把诱导表示记为U，表示空间记为V，这个V 是什么？ 这个V 我们把它定义为一个函数空间，既然已知条件是H、B、W，那么这 个函数空间必然与这些已知条件有关。怎么个相关法？ 我们可以定义V 中向量是这样一个函数， 这个函数的特点是你给我一个G 中 的群元，我给你一个W 空间中的向量，我们把这个函数记为f。 这个相当于是把W 空间的信息利用起来了，把G 这个群利用起来了，但是 H、B 还没有利用。为了把这个B 是H 的表示这个信息再利用起来，我们规 定，对任意h 属于H，任意g 属于G，有：f(hg) = B(h)f(g)。 这是一个限制条件，也就是说f 并不是所有你给我一个G 中群元，我给你一 个W 空间中的向量，这样一个函数。我还要求f(hg) = B(h)f(g)对任意h、g 成立。在这个等式中，左边是W 空间中一个向量，右边f(g)也是W 空间中一 个向量， B(h)是W 空间上的线性变换， 作用到W 空间的向量f(g)上， 得到W 空间的另一个向量B(h)f(g)。这个等式要求这个向量与f 这个函数直接以hg 为输入得到的W 空间的向量相同。 二. 线性变换U(g)是什么？ 上面是对诱导表示的表示空间的定义，现在我们看表示的线性变换。我们对 它的要求是它作用到V中向量f上， 得到的V中向量U(g)f， 满足： U(g)f = f(g′′g)。 注意这里[U(g)f]是一个V 中向量，换句话说也是一个函数，对这个函数你给 它一个G 中群元，它给你一个W 空间向量。我们对这个V 中向量的要求是 如果你给它一个群元g′′，那么它给你的W 空间的向量，与函数f 以g′′g为输 入得到的W 空间中的向量相等。 这个是对线性变换U(g)的要求。 在第一点中， 我们说了V 中的向量f 要满足： 104 f(hg) = B(h)f(g)。 在这里， U(g)作用到f 上， 得到V 中的另一个向量[U(g)f]， 显然，[U(g)f]也是要满足要求：U(g)f = B(h)U(g)f的。这个要求 满不满足呢？答案是满足，因为对这个式子，有： U(g)f = f(hg′′g) = B(h)f(g′′g) 再根据U(g)f = f(g′′g)，继续有： U(g)f = B(h)f(g′′g) = B(h)U(g)f 这也就是说U(g)作用到V 中的一个向量上，得到的确实是V 中的另一个向 量。 三. 这个线性变换群保持抽象群G 的乘法规律不变 （两点： 形成群、 保持乘法） 。 现在我们是定义完了诱导表示的表示空间V，它是一个函数空间，其中函数 是f，f 的特征是你给我一个群G 中群元，我给你一个H 表示B 的表示空间 W 中的向量，且f 满足：f(hg) = B(h)f(g)，对∀h ∈H、∀g ∈G成立。 同时我们定义完了线性变换U(g)， 它满足： U(g)f = f(g′′g)对∀g、g′′ ∈G， ∀f ∈V成立。 现在我们看{U(g)}是否形成群？以及G 与{U(g)}是否同态？同态要满足的关 系是： U(g′)U(g)f = U(g′g)f 对∀g、g′、g′′ ∈G，∀f ∈V成立。我们先证这个关系成立，然后说明{U(g)}形 成群。两者合在一起，G 与{U(g)}同态自然成立。 证明这个关系成立比较简单： U(g′)U(g)f = U(g′)f（先对后一半作用，同时记f(g′′g)这个函数为 φ(g′′)，因为对这个式子而言，变量是g′′，线性变换作用到的，是一个以g′′为 变量的函数） 因此，它可以继续写为： U(g′)U(g)f = U(g′)f = U(g′)φ = φ(g′′g′) 这里U(g)f这个函数φ依然属于V，因此U(g′)作用到φ(g′′)上等于φ(g′′g′)。 这个时候把φ本来的样子代入，就有：φ(g′′g′) = f(g′′g′g)。进而： U(g′)U(g)f = φ(g′′g′) = f(g′′g′g) 而U(g′g)f = f(g′′g′g)。因此： U(g′)U(g)f = U(g′g)f 对∀g、g′、g′′ ∈G，∀f ∈V成立。 关系成立后，第二点：{U(g)}是否形成群呢？答案是肯定的。首先由于 U(g0)f = f(g′′)，U(g0)是单位元素；其次U(g−1)与U(g)互逆；第三是上 面说了U(g′)U(g) = U(g′g)，乘法关系保持（封闭） ；最后(U(g′′)U(g′))U(g) = U(g′′g′)U(g) = U(g′′g′g) = U(g′′)U(g′g) = U(g′′)(U(g′)U(g))，结合律也成立。 四点结合，{U(g)}形成群。 {U(g)}形成群与前面说的U(g′)U(g) = U(g′g)这两点再一结合，{U(g)}就是G 的一个表示。我们把这个表示称为诱导表示，记作：HUG B、HUB、或UB。 为了更清楚地把它的产生过程再说明一遍，我们把刚才那个图补完： 106 图2.7 诱导表示示意图二 （从1 到7 依次讲下来，加上f(hg) = B(h)f(g)、U(g)f = f(g′′g)两个限制条 件，我们就可以证明{U(g)}就是G 的表示了） 上面是搞清楚了诱导表示的基本原理。实际应用中，除了这个基本原理，可 能更重要的是我们要知道如何去产生诱导表示的矩阵？ 要搞清楚这个问题， 在前面求群表示的时候我们已经说过， 标准程序是什么？ 答案：如下三步。 1. 搞清楚表示空间的维度，在这里就是V 这个函数空间的维度； 2. 在这个函数空间中，取这个维度个线性无关的函数，作为基； 3. 将线性变换{U(g)}作用到这些基上， 得到新函数， 把每个新函数按旧基展开， 展开系数对应表示矩阵的列。 先看第一步，这个函数空间有多少维？ 1. 要确定一个函数空间的维度，说白了就是要看我们需要用几个数来确定一个 函数。在群代数以及它对应的群函数那里，我们需要群的阶n 个数来确定一 个群函数， 所以群函数空间的维度为n。 在类函数那里， 我们需要类的个数个 数来确定一个类函数，所以类函数的维度为类的个数。 在这里，我们的函数空间V 具备这样的特质，你给我一个群元，我给你一个 W 空间中的向量。群元有n 个选择（自由度为n） ，这个W 空间中的向量是 d 维的（自由度为d） ，两方面加在一起，要确定这样一个函数，我们有n × d 个自由度，这个函数空间是n × d维的。 但是在得出这个结论的时候，我们忽略了一个对V 中函数f 的限制，也就是 f(hg) = B(h)f(g)，对∀h ∈H、∀g ∈G成立。这个限制有什么意义呢？这个限 制其实意味着对f，我们不需要知道所有f(g)的值，只需要知道一部分，就可 以把所有f(g)的值通过f(hg) = B(h)f(g)得到。为什么呢？ 因为我们可以把群G， 按其子群H 进行陪集分解， G = {Hg1、Hg2、⋯、Hg𝑙}， 其中g1是G 中单位元，l=n/m，n 是G 的阶，m 是H 的阶。这样分割的话， 你给我一个任意群元g， 我都可以把它定位在某一个陪集中， 写成hgi， 这样， 它所对应的W 空间中的向量，就可以通过f(hgi) = B(h)f(gi)得到。这里B(h) 已知， 也就是说我只需要知道在上面的陪集分解中每个gi对应的f(gi)， 那么所 有的f(g)就都知道了。这样V 的维度就缩小为 n m d，以后记𝑙= n m。 2. 有了这样一个认识，我们就开始选基了。这个基应该有l×d 个，每个都是这 样一个函数，这个函数干的事情是你给我一个上面陪集分解时的gi，我给你 一个W 空间中的向量。 我们用两个指标来标记这一组基，r 与j，这组基的形式为：erj(gk) = δjke ⃗ ⃗r， 其中r 是W 空间的维度指标，走遍1 到d，e ⃗ ⃗r是W 里面d 个线性无关的向量 （H 表示B 的表示空间） 。j 是从1 到l 中间的一个数，这样rj 有l × d个组 合。 这里erj(gk)起的作用呢？是你给我一个G 中按H 的陪集分解得到的， g1、 g2、 到g𝑙中的一个元素，我给你一个W 空间中的矢量。当我这个函数的输入不是 g1、g2、到g𝑙中的一个元素，而是G 中任意一个元素时，我这个函数会根据 你这个元素在我G的陪集分解中的定位， 把你写成hgi的形式， 然后我erj(hgi)， 就等于B(h)erj(gi)。B(h)已知，erj(gi)已知，erj(hgi)自然就知道了。 也就是说，确定这组基组，最终会落定在r 与j 分别取遍从1 到d、1 到l 的 值时，erj(gk)分别对应什么样的函数？ 108 我们先看r=1，j=1 的情况，这个时候，对erj(gk)这个函数，你给我g1，我给 你W 中的e ⃗ ⃗1，你给我g2，我给你W 空间中的零向量，你给我g3到g𝑙，我都给 你W 空间中的零向量。 r=2，j=1 时，对erj(gk)这个函数，你给我g1，我给你W 中的e ⃗ ⃗2，你给我g2， 我给你W 空间中的零向量，你给我g3到g𝑙，我都给你W 空间中的零向量。 r=3，j=1 等情况依次类推。 r=d，j=1 时，对erj(gk)这个函数，你给我g1，我给你W 中的e ⃗ ⃗d，你给我g2， 我给你W 空间中的零向量，你给我g3到g𝑙，我都给你W 空间中的零向量。 r=1，j=2 时，对erj(gk)这个函数，你给我g1，我给你W 中的零向量，你给我 g2，我给你W 空间中的e ⃗ ⃗1，你给我g3到g𝑙，我都给你W 空间中的零向量。 r=2，j=2 时，对erj(gk)这个函数，你给我g1，我给你W 中的零向量，你给我 g2，我给你W 空间中的e ⃗ ⃗2，你给我g3到g𝑙，我都给你W 空间中的零向量。 r=3，j=2 等情况依次类推。 r=d，j=2 时，对erj(gk)这个函数，你给我g1，我给你W 中的零向量，你给我 g2，我给你W 空间中的e ⃗ ⃗d，你给我g3到g𝑙，我都给你W 空间中的零向量。 j=3 到l，对每个j，r 等于1 到d 的情况依次类推。当r、j 的组合走遍d× 𝑙个 维度。V 中的l × d个基就定了。 3. 定义完这样一个基以后，我们就要求诱导表示的表示矩阵了。这个说白了， 就是要看U(g)作用到erj(gk)上的效果，根据U(g)的定义，我们知道： U(g)erj(gk) = erj(gkg) 这里，erj(gkg)对应什么样的W 空间中的向量，我们可以根据G 按H 进行陪 集分解的具体形式， 确定gkg为hgi， 然后再根据erj(hgi) = B(h)erj(gi)来确定。 这个没有问题。但是，当我回到线性变换本身，要想把erj(gkg)写成以gk为变 量的函数，并且把它用erj(gk)来展开，就没那么容易了。 这个时候我们首先需要明确的是g 是一个确定的群元，我们想求的是它的线 性变换矩阵。 这个U(g)作用到erj(gk)上， 得到的是erj(gkg)。 这个erj(gkg)我们 不知道是什么，我们需要通过H 对G 的陪集分解，把它确定为某个hgi。g 定 了，gk定了，这个h 与gi也就定了。然后我们通过erj(gkg) = erj(hgi) = B(h)erj(gi)可以得到： U(g)erj(gk) = B(h)erj(gi) B(h)是一个我们已知的变换矩阵。 我们现在需要做的， 是把erj(gi)的变量变作 gk，这样的话我们就可以得出U(g)的矩阵表示了。 我们怎么做到这一点？我们用的性质是gi是g1、 g2、 到g𝑙中的一个群元， gk也 是g1、g2、到g𝑙中的一个群元。当我g1、g2、到g𝑙排好位置以后，gi与gk差的 是一个平移。假设i+t 等于k，那么erj(gi) = erj+t(gk)。把j+t 记为m，那么 上面那个变换就变成了： U(g)erj(gk) = B(h)erm(gk) erm(gk)是我们基矢组中的一个，展开系数就可以由右边确定了。 这个式子的右边要想不为零， 由erm(gk)的定义， 要求m=k。 同时， 由erm(gk) = erj(gi)这个条件，要求i=j。与此同时由gkg = hgi，知h=gkggi −1。结合m=k、 i=j，我们就知道：gmggj −1 = h ∈H。注意，这里m、j 是基erm(gk)、erj(gi)的 第二个指标， 这也就是说只有这个基的第二个指标m、 j 所对应的gm、 gj通过 gmggj −1的方式作用到群元g 上得到的群元gmggj −1 = h ∈H时，U(g)erj(gk)的 展开系数才不为零。 这时展开系数由B(h)， 即B(gmggj −1)决定。 其它情况下， 110 结果都是零。 基于这个理解，表示矩阵的具体形式就非常简单了。对∀g ∈G，U(g)等于： U(g) = ( B ̇ (g1gg1 −1) B ̇ (g1gg2 −1) B ̇ (g2gg1 −1) B ̇ (g2gg2 −1) … B ̇ (g1gg𝑙 −1) … B ̇ (g2gg𝑙 −1) ⋮ ⋮ B ̇ (g𝑙gg1 −1) B ̇ (g𝑙gg2 −1) ⋱ ⋮ … B ̇ (g𝑙gg𝑙 −1)) 式2.11 其中： B ̇ (gmggj −1) = {B(gmggj −1), if gmggj −1 ∈H 0, otherwise 在这个表述中，我们是把B ̇ (gmggj −1)这个d × d的矩阵作为基本单元，把U(g) 写成了𝑙× 𝑙块。 在这𝑙× 𝑙中， 对一个特定的m， 因为gmg只能分解到一个H 的 陪集中，所以只有一个B ̇ (gmggj −1)非零。同样，不同gmg与gm′g，必对应不同 陪集（因为不然由gmg = h1gj与gm′g = h2gj可得gm = h1h2 −1gm′，进而gm ∈ Hgm′，与Hgm、Hgm′是不同陪集矛盾） ，这样U(g)这个矩阵以B 这个d × d的 矩阵作为基本单元，写出来的样子就是： U(g) = ( 0 0 0 B … 0 … 0 ⋮ ⋮ B ⋮ 0 0 ⋮ 0 ⋱ ⋮ … 0 B 0) 这种类似于正则表示的样子了。 举个例子，还是用D3。 例12. 这个G={e、d、f、a、b、c}，H={e、d、f}。H 是个三阶循环群，有表示 B(e) = 1、B(d) = ε = exp [ 2πi 3 ]、B(f) = ε2 = exp[4πi/3]。求由它诱导出的 G 的表示？ 这里需要的信息是D3 群乘法表，也就是表1.1。 G={Hg1、Hg2}，g1 = e、g2 = a。 由上面给出的诱导表示的式子，很直接： UB(e) = (B ̇ (eee−1) B ̇ (eea−1) B ̇ (aee−1) B ̇ (aea−1)) = (B(e) 0 0 B(e)) = (1 0 0 1) UB(d) = (B ̇ (ede−1) B ̇ (eda−1) B ̇ (ade−1) B ̇ (ada−1)) = (B(d) 0 0 B(f)) = (ε 0 0 ε2) UB(f) = (B ̇ (efe−1) B ̇ (efa−1) B ̇ (afe−1) B ̇ (afa−1)) = (B(f) 0 0 B(d)) = (ε2 0 0 ε) UB(a) = (B ̇ (eae−1) B ̇ (eaa−1) B ̇ (aae−1) B ̇ (aaa−1)) = ( 0 B(e) B(e) 0 ) = (0 1 1 0) UB(b) = (B ̇ (ebe−1) B ̇ (eba−1) B ̇ (abe−1) B ̇ (aba−1)) = ( 0 B(f) B(d) 0 ) = (0 ε2 ε 0) UB(c) = (B ̇ (ece−1) B ̇ (eca−1) B ̇ (ace−1) B ̇ (aca−1)) = ( 0 B(d) B(f) 0 ) = (0 ε ε2 0) 这个诱导表示特征标为2、-1、0，是不可约表示。 由前面的式2.11，我们还知道诱导表示的特征标整体可写为： 𝜒U(g) = ∑ trB ̇ (gjggj −1) 𝑙 j=1 式2.12 在这个加和的时候，j 走遍从1 到l=n/m。这里gjggj −1 ∈H时B ̇ 是B，否则是零矩 阵。我们可以注意到这样一个性质，就是我们这个g1到g𝑙其实是在对G，依据H 进行陪集分解的时候用到的H 外的那些群元， 整个G 可以写成Hg1到Hg𝑙的并集。 对一个陪集，比如Hgj中的任何一个元素 hgj，把它作用到g 上，都会有这样的性 质： 就是如果gjggj −1 ∈H， 则hgjggj −1h−1 ∈H， 而如果gjggj −1 ∉H， 则hgjggj −1h−1 ∉ H。因为如果不然，就会反推出gjggj −1 ∈H。同时trB ̇ (gjggj −1) = trB ̇ (hgjggj −1h)。 这样的话，式2.12 中那个对gj的加和，就可以扩展到G 中所有元素上，唯一需 要注意的是对H 中元素遍历的时候，给了m 遍（H 的阶）相同的结果，所以要 把这个重复的部分扣除，最终有： 112 𝜒U(g) = 1 m ∑ trB ̇ (tgt−1) 𝑡∈G 式2.13 在诱导表示这个硬骨头啃下来之后，我们这一节就剩下两个简单很多的内容了： 1、群表示在子群上的缩小，2、Frobenius 定理。这两个都可以几句话说清。 先说群表示到其子群的缩小。 说的是这样一个事情， A 是群G 的一个表示， 对∀g ∈G， 都有一个线性变换A(g)与之对应， 且有A(g1g2) = A(g1)A(g2)。 那么这 个对应关系对G 的子群H 中的元素自然也成立，即对∀h ∈H ⊂G，都有一个线 性变换A(h)与之对应，且有A(h1h2) = A(h1)A(h2)。也就是说线性变换群{A(h)} 形成了G 的子群H 的一个表示。我们把这样的一个表示称为G 的表示A，到其 子群H 的缩小，记为A|H。 对于类似表示的缩小，如果A 是G 的不可约表示，那么A|H是H 的不可约 表示吗？ 答案是：不一定。比如对于D3 群，有二维不可约表示： A(e) = (1 0 0 1)、A(d) = (ε 0 0 ε2)、A(f) = (ε2 0 0 ε) A(a) = (0 1 1 0)、A(b) = (0 ε2 ε 0 )、A(c) = ( 0 ε ε2 0) 它在D3 的子群H={e、d、f}上的缩小，就不是这个三阶循环群的不可约表示。 这个就是群表示在子群上得缩小说的意思了。那么Frobenius 定理呢？说的 是群G 有个子群H，G 有不可约表示A，H 有不可约表示B，它们本来是两个道 上的车，看起来没啥关系。但是由我们前面的讨论我们知道由H 的不可约表示 B，可以诱导出G 的一个表示U，这个表示U 对G 来说可能是可约的。同时对 G 的不可约表示A，它在H 上有个缩小A|H，这个缩小对H 来说也可能是可约 的。 Frobenius 定理说的是：由于诱导表示和群表示的缩小这两个定义本身蕴藏 的结构关系， 我们可以知道G 的不可约表示A 在由H 的不可约表示B 所诱导出 来的G 的表示U 中的重复度，等于H 的不可约表示B 本身在G 的不可约表示 A 对H 的缩小（也就是H 的表示A|H）中的重复度。用图形的方式表达，就是： 图2.8 Frobenius 定理示意图 用数学式子写出来，就是： (χA|χU) = (χB|χ) 其中χA、χU是群G 的不可约表示A 与由H 的不可约表示B 所诱导出来的G 的 诱导表示U 的特征标；χB、χ是群H 的不可约表示B 与由G 的表示A 在H 上的 缩小构成的H 的表示A|H的特征标。 怎么证？我们可以这样看，式子的左边为： (χA|χU) = 1 n ∑χA∗(g)χU(g) g∈G = 1 n ∑χA∗(g) {1 m ∑trB ̇ (tgt−1) t∈G } g∈G = 1 nm ∑∑χ𝐴∗(g)trB ̇ (tgt−1) g∈G t∈G 这里t 与g 的加和都走遍G，记s = tgt−1，则g = t−1st，那么对t 与g 的加和， 是可以换作对s 与t 的加和的，也就是上式可继续等于 = 1 nm ∑∑χA∗(t−1st)trB ̇ (s) s∈G t∈G 114 由B ̇ 定义，它继续等于 = 1 m ∑{1 n ∑χA∗(t−1st) t∈G } s∈H χB(s) 因为对χB求和，要求s 属于H，所以加和范围自然缩小到H。同时，又由于在t 走遍G 时，χA∗(t−1st) = χA∗(s)都成立，所以，上式继续等于 = 1 m ∑χA∗(s) 𝑠∈H χB(s) = (χ|χB) 由于重复度为实数，它继续等于 = (χB|χ) 也就是 (χA|χU) = (χB|χ) Frobenius 定理成立。 再回到刚才D3 群的例子。 例13. 我们刚才从{e、 d、 f}的不可约表示B(e) = 1、 B(d) = ε = exp[2πi/3]、 B(f) = ε2 = exp[4πi/3]推出了A 的诱导： A(e) = (1 0 0 1)、A(d) = (ε 0 0 ε2)、A(f) = (ε2 0 0 ε)、 A(a) = (0 1 1 0)、A(b) = (0 ε2 ε 0 )、A(c) = ( 0 ε ε2 0) 这个诱导表示我们不保证是D3 的不可约表示，但这里凑巧它就是，A1、 A2、A3 在它上面的重复度由特征标表2.3 可以得出，分别为：0、0、1。 这个时候我们来验证Frobenius 定理。如果我取D3 的不可约表示为A1 或 A2，这个A1 或A2在H 上的缩小都是A(e)=A(d)=A(f)=1，H 的不可约表示 在它上面的重复度都是0，这个0 恰好与A1 或A2本身在这个诱导表示上 得重复度相同，这也就是Frobenius 定理说的内容。 如果我们把D3 的不可约表示换为A3，A3 在B 的诱导表示上的重复度为 1。而同时，A3 在H 上的缩小为： A(e) = (1 0 0 1)、A(d) = (ε 0 0 ε2)、A(f) = (ε2 0 0 ε) H 本身的不可约表示B(e) = 1 、B(d) = ε = exp[2πi/3] 、B(f) = ε2 = exp[4πi/3]在A|H上得重复度为： (χB|χ) = 1 3 [2 × 1 + (ε + ε2) × ε + (ε2 + ε) × ε2] = 1 3 [2 + (ε + ε2) × (ε2 + ε)] = 1 Frobenius 依然成立。 2.7 习题与思考 1. 设A(g)是群G = {g}的一个表示， 证明： 复共轭矩阵A∗(g)也是G的一个表示。 如 A(g)是不可约的或者幺正的，则A∗(g)也是不可约的或者幺正的。 2. 设A(g)是群G = {g}的一个表示， 证明： 转置逆矩阵[At(g)]−1、 厄密共轭逆矩阵 [A+(g)]−1也是G的一个表示。如A(g)是不可约的或者幺正的，则[A𝑡(g)]−1、 [A+(g)]−1也是不可约的或者幺正的。 3. 如A(g)是群G = {g}的一个表示，那At(g)、A+(g)是吗？为什么？ 4. 设A(g)是有限群G的一个不可约表示，C是G中一个共轭类，λ为常数，E是单位 矩阵。证明：∑ A(g) g∈C = λE。体会此证明中有限群这个条件的使用。 5. 证明有限群G中属于同一类的各元的表示矩阵之和，必与群G的一切元的表示 矩阵互易。 6. 求三阶群的所有不等价不可约表示。 7. 设A(g)是有限群G的一个不可约表示， B(g)是有限群G的一个一维非恒等表示， 证明：A(g)⨂B(g)也是群G的一个不可约表示。 8. 设V = {e、a、b、c}是满足如下乘法表的四阶群，求其所有不等价、不可约表 116 示。 9．求出D3群的所有不等价、不可约酉表示，并检验群表示的正交定理。 10. 求出D3群在二次齐次函数空间{𝛹1(𝑟 ⃗) = 𝑥2、 𝛹2(𝑟 ⃗)=𝑦2、 𝛹3(𝑟 ⃗) = 𝑧2、 𝛹4(𝑟 ⃗) = 𝑥𝑦、𝛹5(𝑟 ⃗) = 𝑦𝑧、𝛹6(𝑟 ⃗) = 𝑥𝑧}的群表示，并写出其包含的不可约表示。 11. 写出四阶循环群的左正则与右正则表示。 12. 设Ap(g)与Ar(g)是群G的两个不等价、 不可约表示， 直积表示Ap(g)⨂ Ar∗(g)包 含其恒等表示吗？Ap(g)⨂ Ap∗(g)呢？为什么？如包含，包含几次？ 13. 取子群H的表示为左正则表示， 由其诱导出的群G的表示是什么样子？为什么？ 14. 有限群G的非恒等的不可约表示的特征标之和，∑ χp(g) g∈G ，等于几？为什么？ 15. 以f1(r ⃗) = x2、f2(r ⃗) = y2、f3(r ⃗) = xy为基，写出D3群的三维表示，并约化。 16. D3群，基于其子群{e、a}的恒等表示，写出诱导表示，并约化。 17. 写出D3群所有不可约表示相互之间的直积，并约化。 第三章 点群与空间群 3.1 点群基础 前面我们说过很多次，第一、第二章是这门课的理论基础，讲它们是为我们 后面讲具体的群做准备的。第三章讲点群和空间群。历史上，物理学家（包括地 质学家） ，特别是其中研究晶体的那部分人，就对称性的认识，在早期也集中于 这个领域。 现在， 可以说点群与空间群的对称性描述是我们在讨论固体和分子的 结构、电子结构、振动谱等物理性质时不可或缺的科学语言。不懂这个，物质科 学类（Physical Sciences）杂志上的很多文章（比如Physical Review 系列）对我 们来说，会读起来很困难。 点群最基本的特征是我们在进行对称操作的时候， 操作对象至少有一个点保 持不动。这种对称性在晶体、分子、准晶材料中都会出现。而空间群，是基于点 群概念的。 它描述的是晶体的对称性。 因为这个原因， 我们这一章的基础是点群。 在点群中，我们不要求系统具有平移对称性。在讲完点群之后，我们会说，如果 我们再加一个限制，也就是我面对的系统需要有平移对称性，那么这个时候，就 不是所有的点群都可以在晶体中存在了。能在晶体中存在的只是其中的一部分。 比如在分子中是允许五次轴，也就是绕一个轴转2π/5 角的对称操作存在的，但 是在晶体中就不允许。这也就引出了晶体点群，它是点群的一部分。和晶体点群 的概念对应，我们还有晶体点阵的概念。 同时，我们可以注意一下，在晶体出现的时候，它对点群加的是一个限制， 效果是使得很多非周期性体系中的对称性在晶体中不能存在。 但是任何事物都有 两面性， 晶体的平移周期性结构对点群的对称操作加了这样一个限制， 作为补偿， 在其它的对称群描述中， 晶体周期性结构对其中群元的要求就会有所放松。 这个 118 在空间群这个概念中可以详细体现。 具体而言， 就是在晶体中进行一定角度的旋 转后，做个晶格长度分数倍的平移，系统还可以回到与之前不可分辨的状态。这 种操作对应某空间群元素， 其对称元素是我们常说的螺旋轴。 在不做这个旋转前， 晶格长度分数倍的平移是不被晶体中原子的周期性排布允许的。 同时， 晶体在对 一个平面进行反射后再做一个晶格长度分数倍的平移， 也可回到与之前不可分辨 的状态。 这个对称操作的对称元素是我们常说的滑移面。 这些操作因为不保持一 个固定的点不变了，所以它不是点群操作，但它能够保持晶体的不变。为了描述 这种转动与平移对称性的集合， 我们引入空间群这个概念。 由这个介绍大家可以 暂时理解它是基于点群， 但又跟点群不同。 具体的细节我们会在后面花几节课的 时间仔细讲。这里，先给大家做一个整体的介绍。 同时，如果你不满足于此讲义的内容，去看不同教材的话，你也应该会注意 到大体而言人们会通过两个途径来引入点群的概念： 一、从点群的母体， 三维实正交群O(3)出发。 这个O(3)群是由转动、 反演、 以及 它们的组合构成的，是一个无限群。这个后面我们会细讲。点群呢，是这个 O(3)群的有限子群； 二、从具体的多面体或分子出发， 去讨论它们的对称性， 总结归纳出一些共同点， 再引入点群的概念。 而比较有意思的是，如果你去看书的作者以及他/她们所采取的方式，你会 发现作者的背景和他/她们采取的写作方式是存在一定关联的。一般来说，学物 理出身的人倾向于使用前者，就是我把你的母体说明白了，再讲你，事情就很 trivial 了。 韩老师与孙老师、 马老师她/他们的教材采取的都是这个路子。 而化学 出身的，比如Frank Albert Cotton，还有我们一些物理化学教材里面讲对称性的 时候，就使用的是后者。当然，我只说有一定关联，也不排除例外。对我而言， 我觉得两个路子各有好处， 前者严格、 后者直观。 这里我们倾向于使用一个折中， 把三维实正交O(3)群与点群合在一起，作为“点群基础”来讲。 这里点群是一种群， 它对应的是一个实际系统在三维实空间中具有的对称性 的集合，这些操作有个特征，就是进行操作的时候，三维空间中有一个点不动。 这个实际系统可以有限大，比如分子、团簇，也可以无限大，比如晶体。 对点群的讨论可以从其中最重要的两个概念， 对称操作 （Symmetry Operation） 与对称元素（Symmetry Element） ，开始。先看几个例子： 例1. 立方体，有八个顶点，a 到h，中心是O，x、y、z 轴分别过中心： 图3.1 立方体对称性示意图 先看什么是对称操作？这里我们参考Albert Cotton 书上18 页给出的定义。 定义3.1：A symmetry operation is a movement of a body such that, after the movement has been carried out, every point of the body is coincident with an equivalent point of the body in its original orientation. In other words, if we note the position and orientation of a body before and after a movement is carried out, that movement is a symmetry operation if these points and orientations are indistinguishable. 由这个定义，我们知道对称操作是一个movement，是一个使物体到达与其 120 原状态不可分辨的另一个状态的movement。根据这个定义，我们看上面的立方 体有多少个对称操作？ 1． 沿立方体的四个对角线ag、fd、hb、ce 进行2π/3、4π/3 转动的操作，这样 的操作有4×2 = 8 个； 2． 沿x、y、z 轴转π/2、π、3π/2 的操作，这样的操作有3×3 = 9 个； 3． 立方体有12 个棱，两条相对的棱的中点连接起来，绕这个连线转动π 角 的操作，这样的操作有6 ×1 = 6 个； 4． 再加上根本就不动这个操作，一共8+9+6+1 = 24 个纯转动操作。 在这里转动是对称操作，与每个转动相应，有个旋转轴，这个旋转轴，就是 下面我们要讲得对称元素。同样，我们参考Albert Cotton 书上的定义。 定义3.2：A symmetry element is a geometrical entity such as a line, a plane, or a point, with respect to which one or more symmetry operations may be carried out. 由这个定义，我们知道对称元素是一个几何实体（geometrical entity） ，对称 操作是相对于这个几何实体进行的，在上面的例子中，它就是旋转轴。当然，实 际的对称元素不一定非得是旋转轴，也可以是反演中心、反射面、转动反演轴等 等，这些我们以后会讲。它们的共同特点就是它们都是几何实体，对称操作都是 相对它们进行的。 在上面讲的立方体的例子中，如果我们再加上反演操作I，也就是把三维实 空间中任意一个(x, y, z)的点变换到(−x, −y, −z)， 这个反演操作对刚才的立方体 实际上也是对称操作。这个I 与刚才24 个纯转动操作结合，就又可以给出24 个 新的对称操作。 这样加在一起就有48 个对称操作了。 这48 个对称操作的集合构 成一个群， 这个群就是这个立方体的对称群， 我们记为Oh， 它是一个点群 （因为 对这48 个操作，立方体中心那个点都不变） 。 现在我们可以去想， 如果我把这个立方体揉成一个球， 那么这48 个Oh 群的 对称操作是否保持这个球不变？答案：保持，是吧？同时，我们需要指出的是对 于这个球而言， 它的对称操作远远不止这些。 以后我们学到的所有点群的对称操 作，对这个球都成立。换句话说，如果我们把所有保持这个球不变的对称操作放 在一起， 它们是形成一个群的。 这个群是如此强大以至于任何的一个我们后面学 到的点群，都是这个群的子群。在这里，我们把它称为点群的母体，它的名字叫 三维实正交群，记为O(3)。下面我们会花些时间来介绍一下这个O(3)群。 但在讲这个O(3)群之前， 我们需要先介绍一下它里面的基本操作， 就是三维 实空间（R3）中的正交变换。 这个三维实空间的一个性质是在定了一组正交归一基i ⃗、 j ⃗、k ⃗⃗以后， 其中的任 意一个向量都可以写成下述列矩阵形式，比如： r ⃗= ( x1 x2 x3 )、r ⃗′ = ( x1 ′ x2 ′ x3 ′ ) 两个向量的内积为： (r ⃗∙r ⃗′) = ∑xixi ′ 3 i=1 其中向量r ⃗的长度为： |r ⃗| = (r ⃗∙r ⃗)1/2 = (∑xixi 3 i=1 ) 1/2 而向量r ⃗与r ⃗′的夹角φ满足： cos φ = (r ⃗∙r ⃗′)/(|r ⃗||r ⃗′|) （这些都是R3 中向量的性质） 而R3 中的正交变换，是保持R3 中任意两个向量内积不变的变换，也就是说 122 对∀r ⃗、r ⃗′ ∈R3，线性变换O 要满足： (Or ⃗∙Or ⃗′) = (r ⃗∙r ⃗′) 也就是： OtO = E 这里Ot是O 的转置（实矩阵） 。 这个时候， 如果我们对比之前讲的酉变换， 我们会发现酉变换的定义对应的 是任意内积空间中的保内积变换。而这里，正交变换，则对应的是三维欧式空间 这个实向量空间中的保内积变换，它是一种特殊的酉变换。 基于这样一个正交变换，我们可以引出三维实正交群O(3)。 定义3.3 由三维欧式空间中所有的正交变换构成的群， 称为三维实正交群， 记为 O(3)。 关于O(3)群其中元素最重要的性质，我们可以由这个式子得到： OtO = E 由这个式子，我们知： det(OtO) = (det(O)) 2 = 1 从而det(O) = ±1，也就是说正交变换为非奇异变换，其行列式为+1 或-1。 同时，由于 det(O1O2) = det(O1)det(O2) 我们还知道O(3)群中行列式为1 的正交变换形成它的一个子群，因为1 ×1=1。 并且由： det(O2 −1O1O2) = det(O2 −1)det(O1)det(O2) = det(O2 t )det(O1)det(O2) = det(O1) 可知O(3)群的所有行列式为1 的元素形成的子群还是一个不变子群。我们把它 记为SO(3)。具体定义如下。 定义3.4 O(3)群的所有行列式为1 的正交变换形成的不变子群称为SO(3)群，记 为： 𝐒𝐎(𝟑) = {𝐎∈𝐎(𝟑)|𝐝𝐞𝐭(𝐎) = 𝟏} 这个SO(3)群的一个特征是其中任意的一个元素，作用到三维欧式空间的一 组向量上，不光不改变它们之间的内积，还不改变这组向量的手性关系。比如我 们的三维空间以i ⃗、j ⃗、k ⃗⃗为基，在这组基下，我一个线性变换的矩阵形式是： O = ( a11 a12 a13 a21 a22 a23 a31 a32 a33 ) 我现在说这个线性变换是一个SO(3)群中的元素，那么有：det(O)=1。 我现在把它作用到三个向量i ⃗、 j ⃗、 k ⃗⃗， 它们相互关系是个右手系， 有i ⃗∙(j ⃗× k ⃗⃗) = 1。现在把O 作用到它们三个上面，有： Oi ⃗∙(Oj ⃗× Ok ⃗⃗) = ( a11 a21 a31 ) ∙[( a12 a22 a32 ) × ( a13 a23 a33 )] = | a11 a12 a13 a21 a22 a23 a31 a32 a33 | = 1 不改变手性关系。 同样，如果作用到j ⃗、i ⃗、k ⃗⃗，它们的相互关系是左手系，有j ⃗∙(i ⃗× k ⃗⃗) = −1。 现在把O 作用到它们上面，也有： Oj ⃗∙(Oi ⃗× Ok ⃗⃗) = ( a12 a22 a32 ) ∙[( a11 a21 a31 ) × ( a13 a23 a33 )] = | a12 a11 a13 a22 a21 a23 a32 a31 a33 | = −1 同样不改变手性关系。 而在O(3)群中，有一个空间反演操作，其矩阵形式为： I = ( −1 0 0 0 −1 0 0 0 −1 ) 这个操作行列式等于−1， 它也会改变三维实空间三个向量的手性关系。 同时， 这 个反演操作与恒等操作E 还可以构成一个空间反演群{E、 I}， 由于E、 I 都与O(3) 群中任意元素互易， 它也是O(3)群的不变子群。 这个不变子群与二阶循环群{1、 124 −1}同构。 这样的话，对O(3)群而言，我们可以把其中行列式为1的部分取出来，就是 SO(3)群，作为一部分；而属于O(3)，又不属于SO(3)的元素作为另一部分。这 样，就构成一个O(3)群到二阶循环群的映射： 图3.2 O(3)群与二阶循环群同态示意图 由于det(O1O2) = det(O1)det(O2)，这个映射是个同态映射。 同时我们可以取I 为行列式为−1元素的代表，把O(3)群进行一个SO(3) ∪I ∙ SO(3)的分解。 由于E、 I 与O(3)群中任意元素互易， 套用我们第一章讲的一个概 念，这个O(3)群就可以写成是SO(3)群与空间反演群{E、I}的直积。 到这个地方，O(3)群与SO(3)群的定义我们就讲完了。简单的说，O(3)群是 三维实空间中的保内积变换群，而SO(3)群，是三维实空间的保内积、保手性变 换群。下面我们介绍几个SO(3)群的性质。 定理3.1 对∀𝐠∈𝐒𝐎(𝟑)，可在R3 中找到一个向量𝐤 ⃗，使𝐠𝐤 ⃗= 𝐤 ⃗。 证明： （说白了就是要证对∀g ∈SO(3)其对应的(g −E)k ⃗⃗= 0有非零解。因为这个东西 有非零解，也就意味着存在向量k ⃗⃗，使gk ⃗⃗= k ⃗⃗。那么(g −E)k ⃗⃗= 0有非零解的充要 条件是什么？答：det(g −E) = 0。 这也就意味着现在要证的就是对∀g ∈SO(3)都有det(g −E) = 0。怎么证呢？很 简单，由g 的定义（正交变换） ，有： det(g −E) = det(g −E)t = det(gt −E) = det(g−1 −E) 式3.1 而 g−1 −E = g−1E(E −g) = g−1(−E)(g −E) 代入式3.1，会产生： det(g −E) = det(g−1)det(−E)det(g −E) = 1 ∙(−1) ∙det(g −E) 一个实数等于它的负，这个实数必为零。 （证毕） 由这个定理，我们还可以得到如下性质： 1. 在gk ⃗⃗= k ⃗⃗的解得到后，我们可以取它为k ⃗⃗，与之垂直的两个向量为i ⃗、j ⃗，在以 这样的i ⃗、j ⃗、k ⃗⃗为基时，这个转动g 表示为： g(Ψ) = ( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 ) = Ck ⃗ ⃗⃗(Ψ) 其中k ⃗⃗为转轴，Ψ为转角，取值范围是[0，π]，k ⃗⃗为向量，有个方位角θ、φ。θ 是它与z 轴的夹角， φ 是其xy 平面投影与x 轴的夹角 （群表示部分的内容） 。 2. 由于定理3.1 的存在，SO(3)又称为转动群。它的所有变换的行列式都是1， 两个SO(3)群中元素g1与g2的乘积g3由于行列式为1，也是一个纯转动，用 Ck ⃗ ⃗⃗(Ψ)的形式写出来就是： Ck ⃗ ⃗⃗1(Ψ 1)Ck ⃗ ⃗⃗2(Ψ2) = Ck ⃗ ⃗⃗3(Ψ3) Ck ⃗ ⃗⃗3(Ψ3)也是绕某一轴（k ⃗⃗3）的一个转动，k ⃗⃗3、Ψ3由k ⃗⃗1、Ψ1、k ⃗⃗2、Ψ2定，只 不过关系不是看起来很直接罢了。 126 上面说到在一组基(i ⃗、j ⃗、k ⃗⃗)下，转动Ck ⃗ ⃗⃗(Ψ)的表示为( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 )这 样一个简单的形式。 在任意一组基(i ⃗′、 j ⃗′、 k ⃗⃗′)下， Ck ⃗ ⃗⃗(Ψ)的表示矩阵又会是什么样 子？ 换句话来说，如果在三维欧式空间中，我已经选择了一组基(i ⃗′、j ⃗′、k ⃗⃗′)，你 现在在这个空间中任意找了一个轴k ⃗⃗，你绕它做了一个转动Ψ角的操作，然后你 问我这个Ck ⃗ ⃗⃗(Ψ)在(i ⃗′、j ⃗′、k ⃗⃗′ )下的表示矩阵是什么？这个问题的答案我们后面在 关于点群的讨论中也会经常用到，现在花时间细说一下。 这里我们可以取(i ⃗′、j ⃗′、k ⃗⃗′)为旧基，(i ⃗、j ⃗、k ⃗⃗)为新基，旧基到新基的变换矩阵 是Q，Q = ( a11 a12 a13 a21 a22 a23 a31 a32 a33 )，那么这个变换形式就是： (i ⃗、j ⃗、k ⃗⃗) = (i ⃗′、j ⃗′、k ⃗⃗′) ( a11 a12 a13 a21 a22 a23 a31 a32 a33 ) 这里Q 为实正交变换，Q−1 = Q𝑡。 空间有个矢量r ⃗，它在Ck ⃗ ⃗⃗(Ψ)作用下，变为r ⃗′。在坐标系(i ⃗、j ⃗、k ⃗⃗)下，这个变换 表现为： (i ⃗、j ⃗、k ⃗⃗) ( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 ) ( x1 x2 x3 ) 其中( x1 x2 x3 )为r ⃗在(i ⃗、j ⃗、k ⃗⃗)下的坐标。 在坐标系(i ⃗′、j ⃗′、k ⃗⃗′)下，这个变换表示为： (i ⃗′、j ⃗′、k ⃗⃗′)A ( x1 ′ x2 ′ x3 ′ ) 其中( x1 ′ x2 ′ x3 ′ )是r ⃗在(i ⃗′、j ⃗′、k ⃗⃗′)下的坐标，A 是我们想求的Ck ⃗ ⃗⃗(Ψ)在(i ⃗′、j ⃗′、k ⃗⃗′)下的 表示矩阵。 现在问题清楚了，就是要用已知条件，求A 的形式？ 这个时候，我们用的第一个条件就是r ⃗′这个矢量不管是在(i ⃗、j ⃗、k ⃗⃗ )下，还是 在(i ⃗′、j ⃗′、k ⃗⃗′)，它是同一个矢量，所以： (i ⃗、j ⃗、k ⃗⃗) ( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 ) ( x1 x2 x3 ) = (i ⃗′、j ⃗′、k ⃗⃗′)A ( x1 ′ x2 ′ x3 ′ ) 式3.2 同理，r ⃗也是同一个矢量，所以： (i ⃗、j ⃗、k ⃗⃗) ( x1 x2 x3 ) = (i ⃗′、j ⃗′、k ⃗⃗′) ( x1 ′ x2 ′ x3 ′ ) 式3.3 那么现在我们要做的事情就是从式3.2 与3.3 求A。这个时候，由： (i ⃗、j ⃗、k ⃗⃗) = (i ⃗′、j ⃗′、k ⃗⃗′)Q 代入3.3 我们知道，得： Q ( x1 x2 x3 ) = ( x1 ′ x2 ′ x3 ′ ) 这个式子，代入3.2，进一步有： (i ⃗、j ⃗、k ⃗⃗) ( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 ) ( x1 x2 x3 ) = (i ⃗′、j ⃗′、k ⃗⃗′)AQ ( x1 x2 x3 ) 由于这个式子对任意( x1 x2 x3 )成立，所以有： (i ⃗、j ⃗、k ⃗⃗) ( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 ) = (i ⃗′、j ⃗′、k ⃗⃗′)AQ 再把(i ⃗、j ⃗、k ⃗⃗) = (i ⃗′、j ⃗′、k ⃗⃗′)(Q)继续代入，有： (i ⃗′、j ⃗′、k ⃗⃗′)Q ( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 ) = (i ⃗′、j ⃗′、k ⃗⃗′)AQ 两个向量相等，它们的所有分量都要相等，所以： 128 Q ( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 ) = AQ 进而： A = Q ( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 ) Q−1 这个说明什么？这个说明在三维欧式空间中， 如果你事先选定了一组基的话， 那么绕空间任意一个轴转Ψ的操作，都可以写成： Q ( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 ) Q−1 的形式。 这里，( cos Ψ −sin Ψ 0 sin Ψ cos Ψ 0 0 0 1 )是这个转动在以k ⃗⃗为z 轴，以与它垂直的两个单 位向量i ⃗、j ⃗为x、 y 轴的坐标系下的表示矩阵。 Q 是这三个向量在原来选定的坐标 系下的展开，即：(i ⃗、j ⃗、k ⃗⃗) = (i ⃗′、j ⃗′、k ⃗⃗′)Q。 由于相似变换并不改变矩阵的迹，所以tr(QCk ⃗ ⃗⃗(Ψ)Q−1) = tr (Ck ⃗ ⃗⃗(Ψ)) = 1 + 2 cos Ψ。也就是说绕任何一个轴，转动Ψ角的转动，它的迹都是1 + 2 cos Ψ，转 动轴的选取，只影响转动的矩阵表达形式，不影响它的迹。 除了转动的迹只与转角有关， 下面两段话的分析对我们以后点群分类也特别 重要： 1. 这里我们直接用Ck ⃗ ⃗⃗(Ψ)代表绕k ⃗⃗轴的转动，因此，Ck ⃗ ⃗⃗(Ψ)满足Ck ⃗ ⃗⃗(Ψ)k ⃗⃗= k ⃗⃗。取 SO(3)群中任意元素g′ ，作用到这个等式上，有g′Ck ⃗ ⃗⃗(Ψ)k ⃗⃗= g′k ⃗⃗，进而 g′Ck ⃗ ⃗⃗(Ψ)g′−1g′k ⃗⃗= g′k ⃗⃗，也就是说g′C𝑘 ⃗⃗(Ψ)g′−1代表的是在(i ⃗、j ⃗、k ⃗⃗)下绕g′k ⃗⃗轴 转动Ψ角的操作。这里g′为SO(3)群中任意元素，因此g′k ⃗⃗可以是空间任意指 向，这也就是说SO(3)群中绕空间任意轴转动相同转角的操作都同类。 2. 把这个性质展开：对点群而言，转动相同转角的操作是否同类？答案是不一 定，因为上面的推导是针对SO(3)群的，要求g′属于SO(3)群。点群是O(3)群 有限子群。其中可能会有两个转动转角相同，但我这个点群中没有任何一个 操作，可以把这两个转动的转轴联系起来，这样，就不能从Ck ⃗ ⃗⃗(Ψ)k ⃗⃗= k ⃗⃗推到 g′Ck ⃗ ⃗⃗(Ψ)g′−1g′k ⃗⃗= g′k ⃗⃗了。因为没有一个点群中元素g′，使得这两个轴一个为 k ⃗⃗，一个为g′k ⃗⃗。 这些讨论总结一下就是： SO(3)群中所有转动相同转角的元素都同类， 但点 群中不一定，要看有没有一个点群中元素将他们的转动轴联系起来？ 这些讨论同时还告诉我们SO(3)群可以写成{Ck ⃗ ⃗⃗(Ψ)}， 其中k ⃗⃗取遍过原点O 的 所有轴，Ψ取遍[0，π]。前面我们还说过，O(3)可以写成SO(3)⊗{E、I}。因此， O(3)也可以写成{C𝑘 ⃗⃗(Ψ)、I C𝑘 ⃗⃗(Ψ)}。 对其中任意一个元素Ck ⃗ ⃗⃗(Ψ)，对其进行共轭操作，情况有两种： 1. Ck ⃗ ⃗⃗′(Ψ′)Ck ⃗ ⃗⃗(Ψ)Ck ⃗ ⃗⃗′(Ψ′)−1， 2. ICk ⃗ ⃗⃗′(Ψ′)Ck ⃗ ⃗⃗(Ψ) (ICk ⃗ ⃗⃗′(Ψ′)) −1 = Ck ⃗ ⃗⃗′(Ψ′)Ck ⃗ ⃗⃗(Ψ)Ck ⃗ ⃗⃗′(Ψ′)−1。 也就是不管怎么说，同类操作都是绕某一轴转动相同角度的纯转动操作。 而对其中任意一个转动反演操作ICk ⃗ ⃗⃗(Ψ)， 也具有同样地性质。 因此， 在O(3) 群中， 所有转动相同转角的纯转动操作为一类， 所有转动相同转角的转动反演操 作为一类。 现在我们是讲完了点群的母体， O(3)群的所有性质。 开始看点群， 先看概念。 定义3.5（点群）三点： 1. 点群是三维实正交群O(3)群的有限子群； 2. 如果点群只包含转动元素，它是SO(3)群的有限子群，称为第一类点群； 130 3. 如果点群还包含转动反演元素，称为第二类点群。 关于点群，有个重要的性质。 定理3.2，设群G 是绕固定轴𝐤 ⃗转动生成的n 阶群，则G 由元素𝐂𝐤 ⃗(𝟐𝛑/𝐧)生成。 证明： 由群G 是绕k ⃗⃗转动生成的n 阶群，知G 中只有n 个元素，且都是绕k ⃗⃗轴转动，我 们可以记为Ck ⃗ ⃗⃗(θi)，i 为从0 到n-1。 我们选θ0 = 0， 对应单位元素， θ1是非零转动中的最小转角。 其它θi可以写成θi = miθ1 + φi的形式。其中0 ≤φi < θ1，mi为正整数。 既然有这个关系，我们就知道： [Ck ⃗ ⃗⃗(θ1)] −miCk ⃗ ⃗⃗(θi) = Ck ⃗ ⃗⃗(φi) ∈G 前面说过，θ1是非零转动中的最小转角，所以φi必都为零。也就是说所有转角都 可以写成miθ1的样子了。在θ0对应单位操作，之后依次取n-1 个mi，这样就有n 个转动操作了。 第n+1 个与单位操作重复， 所以就有nθ1 = 2π， 进而θ1 = 2π/n， 这个转动群由Ck ⃗ ⃗⃗(2π/n)生成。 （证毕） 到这儿， 我们知道了一个点群其中元素必可写为{Ck ⃗ ⃗⃗(2π/n)、I Ck ⃗ ⃗⃗′(2π/n′)}的 形式。其中k ⃗⃗、n、k ⃗⃗′、n′有多种选择。对于Ck ⃗ ⃗⃗(2π/n)、I Ck ⃗ ⃗⃗′(2π/n′)这些操作，它 们的对称元素有几种情况，后面关于点群讨论的时候经常用到，这里说一下： 1. ICk ⃗ ⃗⃗(2π/n)中n取1，这时对应的是I，中心反演操作，对称元素为反演中心； 2. ICk ⃗ ⃗⃗(2π/n)中n取2， 对应操作为ICk ⃗ ⃗⃗(π)。 这个操作干的事情是绕k ⃗⃗轴转动π， 再 对原点作反演。 以(i ⃗、j ⃗、k ⃗⃗)为坐标轴的话， 这个操作干的事情就是先把( x y z )变 到( −x −y z )，再把( −x −y z ) 变到( x y −z )，最终使( x y z )与( x y −z )等价。这里，我们可以说 对称元素是过原点，与k ⃗⃗垂直的反射面； 3. Ck ⃗ ⃗⃗(2π/n)普遍情况，对称元素就是转动轴； 4. ICk ⃗ ⃗⃗(2π/n)普遍情况，对称元素就是转动反演轴。 需要说明的是对于有些教材， 他们可能不喜欢用转动反演轴来讲东西， 而喜 欢利用一个叫转动反射面的东西，记为Sk ⃗ ⃗⃗(2π/n)，指的是绕k ⃗⃗轴作个转动2π/n， 再对与k ⃗⃗轴垂直的镜面作反射，也就是σk ⃗ ⃗⃗Ck ⃗ ⃗⃗(2π/n)。它与转动反演的关系是： σk ⃗ ⃗⃗Ck ⃗ ⃗⃗(2π/n) = ICk ⃗ ⃗⃗(2π n + π) ，如下图所示。 图3.3 转动反演轴与转动反射面相互关系示意图 到这里， 我们是把点群的一些最基本的特性给说清楚了。 简单的说就是点群 是一个由{Ck ⃗ ⃗⃗(2π/n)、I Ck ⃗ ⃗⃗′(2π/n′)}形成的对称操作的集合，这里k ⃗⃗、n、k ⃗⃗′、n′有 有限个取值。我们之前说过，群本身是有很强的结构特征的，当我们说一个集合 形成群时，看似简单的几句话，其实都为它们之间的相互关系留下了伏笔。这个 大家在第一章应该有体会。同态，看似简单，但蕴藏着同态核定理。在这里其实 也是，我们说k ⃗⃗、n、k ⃗⃗′、n′取有限个取值的{Ck ⃗ ⃗⃗(2π/n)、I Ck ⃗ ⃗⃗′(2π/n′)}这样的一个 132 集合的时候，这个{Ck ⃗ ⃗⃗(2π/n)、I Ck ⃗ ⃗⃗′(2π/n′)}本身是不是也要符合什么规律呢？ 答案是肯定的。这个规律是我们这一章的定理3.3。 定理3.3 设G 是点群，K 是G 的纯转动部分，由于纯转动部分的乘积以及逆元 必属于这个纯转动部分， 所以K 也是G 的纯转动子群， 也就是说K = 𝐆∩𝐒𝐎(𝟑)。 这个时候，关于G 与K 的关系，存在三种可能： 1. G = K，这个时候G 是SO(3)的有限子群，这样的点群称为第一类点群，它只 包含纯转动操作； 2. 当G 不只包含纯转动操作时，如果G 中包含纯反演操作I，那么G 与K 的 关系必然是G = 𝐊∪𝐈𝐊； 3. 当G 不只包含纯转动操作， 且G 中不包含纯反演操作I 时， G 必与一个纯转 动群𝐆+同构， 这里𝐆+ = 𝐊∪𝐊+， 而𝐊+的定义是： 𝐊+ = {𝐈𝐠|𝐠∈𝐆，但𝐠∉𝐊}。 这三个性质，说出了所有点群的情况！大家一定要好好理解。其中第一个性 质不需要证明，很直白，就是有一类点群只包含纯转动操作。第二、第三个性质 是需要我们证明的。 证明之前我先问一下大家我们知道这个定理有什么用？一句话： 如果这个定 理成立，那么我们了解了所有第一类点群，也就很自然地了解所有点群了。 证明： G 不是纯转动元素集合时的情况，G = {{Ck ⃗ ⃗⃗(Ψ)}，{ICk ⃗ ⃗⃗′(Ψ′)}} = {K，IK+}。 这个时候，如果I 属于G，那么由于重排定理IG 应该也等于G，这样的话G 这 个集合就要写成{IK，K+}， 它要与{K，IK+}相同， 这就要求K+ = K。这也就是我 们说的第二种情况。 当I 不属于G 时，由上面G= {{Ck ⃗ ⃗⃗(Ψ)}，{ICk ⃗ ⃗⃗′(Ψ′)}}的形式，我们是可以构造一 个集合G+ = {{Ck ⃗ ⃗⃗(Ψ)}，{Ck ⃗ ⃗⃗′(Ψ′)}}的。 这个集合与G 存在一一对应的关系是显然 的。那么下面我们只需要证明G+是一个群，且保持乘法规则不变，就可以了。 先看G+ = {{Ck ⃗ ⃗⃗(Ψ)}，{Ck ⃗ ⃗⃗′(Ψ′)}}是不是一个群，由G = {{Ck ⃗ ⃗⃗(Ψ)}，{ICk ⃗ ⃗⃗′(Ψ′)}}是 一个群出发。 封闭性：由G= {{Ck ⃗ ⃗⃗(Ψ)}，{ICk ⃗ ⃗⃗′(Ψ′)}}是一个群，知两个{Ck ⃗ ⃗⃗(Ψ)}中元素相乘，结 果属于{Ck ⃗ ⃗⃗(Ψ)}；两个{ICk ⃗ ⃗⃗′(Ψ′)}中元素相乘，结果属于{Ck ⃗ ⃗⃗(Ψ)}；而一个{Ck ⃗ ⃗⃗(Ψ)} 中元素与一个{ICk ⃗ ⃗⃗′(Ψ′)}中元素相乘，结果属于{ICk ⃗ ⃗⃗′(Ψ′)}。 因此对应G+ = {{Ck ⃗ ⃗⃗(Ψ)}，{Ck ⃗ ⃗⃗′(Ψ′)}}中元素。 两个{Ck ⃗ ⃗⃗(Ψ)}中元素相乘， 结果属于 {Ck ⃗ ⃗⃗(Ψ)}； 两个{Ck ⃗ ⃗⃗′(Ψ′)}中元素相乘， 结果属于{Ck ⃗ ⃗⃗(Ψ)}； 而一个{Ck ⃗ ⃗⃗(Ψ)}中元素与 一个{Ck ⃗ ⃗⃗′(Ψ′)}中元素相乘，结果属于{Ck ⃗ ⃗⃗′(Ψ′)}。 故{{Ck ⃗ ⃗⃗(Ψ)}，{Ck ⃗ ⃗⃗′(Ψ′)}}封闭。 单位元素，在{Ck ⃗ ⃗⃗(Ψ)}中。 逆元， 由G 是群知对任意{Ck ⃗ ⃗⃗(Ψ)}中元素其逆属于这个集合， 任意{ICk ⃗ ⃗⃗′(Ψ′)}中元 素，其逆也属于这个集合。与此对应，G+中，任意{Ck ⃗ ⃗⃗(Ψ)}中元素，其逆属于这 个集合，任意{Ck ⃗ ⃗⃗′(Ψ′)}中元素，其逆也属于这个集合。这条也成立。 结合律，由G 是群且I 与任意其中元素互易得到。 因此G+是一个群且与G 一一对应。 而乘法规则呢？我们可以对比{{Ck ⃗ ⃗⃗(Ψ)}，{Ck ⃗ ⃗⃗′(Ψ′)}}与{{Ck ⃗ ⃗⃗(Ψ)}，{ICk ⃗ ⃗⃗′(Ψ′)}}， 由 I 与其中任意元素互易得到。 这个时候我们再提一句之前提到的讲这个定理的初衷， 也就是说明当我知道了所 有的第一类点群之后， 我们很自然的可以通过这个定理的第二、 第三条去构造所 有的第二类点群。 134 （证毕） 到这儿，我们点群基础这一节就讲完了，我们讲地比较细，内容看似很多， 不过总结起来，就是下面几句： 1. O(3)群是三维欧式空间中的实正交变换群， 包含转动群SO(3)与转动反演部分 I∙SO(3)； 2. SO(3)群中， 如果两个转轴可以由一个转动g 联系起来， 也就是说一个轴是k ⃗⃗， 一个轴是gk ⃗⃗， 则由Ck ⃗ ⃗⃗(Ψ)k ⃗⃗= k ⃗⃗可知gCk ⃗ ⃗⃗(Ψ)g−1gk ⃗⃗= gk ⃗⃗。 也就是说gCk ⃗ ⃗⃗(Ψ)g−1代 表的是绕gk ⃗⃗轴转动Ψ角的操作。 这也就意味着SO(3)群中所有转动相同转角的 操作，实际上是同类的； 3. 与之相似，I∙SO(3)中具有相同转角的转动反演操作也彼此同类； 4. 点群是O(3)群的有限子群， 可以用{{Ck ⃗ ⃗⃗(Ψ)}，{ICk ⃗ ⃗⃗′(Ψ′)}}的方式来分析， 其中 k ⃗⃗、Ψ、k ⃗⃗′、Ψ′具有有限个取值； 5. k ⃗⃗、k ⃗⃗′为整数阶轴； 6. 点群中对称元素包括：转动轴、转动反演轴、反演中心、反射面，其中后两 者是前两者的特殊情况； 7. 最后是定理3.3， 就是点群的三种情况。这三种情况意味着我们只要把第一类 点群分析清楚，依据这个定理，就可以把所有点群分析清楚了。 3.2 第一类点群 这一节，说白了，就是利用我们第一节给出的点群的基本知识，结合第一章 里面的群的基本理论， 对第一类点群进行一个系统地研究。 假设一个第一类点群 是G，它有转动轴Cn1、Cn2、⋯、Cni、⋯，其中ni为大于等于2 的整数，也就是 说Cni代表的是不同轴，ni是代表它几次。这个时候我们可以去想，对于Cni这个 轴来讲，它会对我这个纯转动群贡献几个非恒等的转动操作？答案：ni −1。这 个是我目前知道的关于群G 与其转动轴的信息。现在我们要从这个信息里面去 推出群G 有多少个元素。怎么推？ 这里一个很有用的工具就是球形图。它是以原点为中心，以任意一个正数r 为半径的球面，记为Sr。点群G 中元素的作用，就是把这个Sr，转到一个和它等 价的构型。 对G 中的ni阶轴Cni来说， 它与Sr有两个交点， 我们把这两个交点记为 ri ⃗ ⃗⃗与−ri ⃗ ⃗⃗。这两个矢量在Cni、Cni 2 、⋯、Cni ni−1这些非恒等操作下，是不变的。称为 这些操作的极点。 那么，对于r ⃗i与−r ⃗i，群{E、Cni、Cni 2 、⋯、Cni ni−1}是不是群G 对它们的迷向 子群？（第一章中的概念，对于一个变换群G，变换对象为X，其中有一点x， 如果群G 有个子群Gx作用到x 上，还是得到x，那么Gx是G 对x 的迷向子群） 除了迷向子群，第一章变换群那一节我们还有一个概念要用到，就是r ⃗i的G 轨道。说的是取任意g 属于G，让g 作用到r ⃗i上，当g 走遍G 中元素的时候，r ⃗i 就会走遍所有与它等价的点，这些点的集合就是G 对r ⃗i的G 轨道。 迷向子群与G 轨道这两个概念说了，它们之间有什么联系呢？我们在第一 章，有个定理1.9，说的是r ⃗i的G 轨道上点的个数可以由群G 的阶n 与迷向子群 的阶ni通过n/ni求出。 那么在这里， 这样一个信息能告诉我什么样的事情呢？那就是通过这个关系， 我们知道对任意一个ni阶轴Cni的极点r ⃗i而言，它会有n/ni个与它等价的ni阶轴的 极点。这些极点的集合我们称为一个极点的G 轨道。这个轨道上每个点能贡献 (ni −1)个非恒等操作，一共有n/ni个点，那么这个轨道可以贡献的群G 中非恒 等变换的个数就是： 136 n ni (ni −1) 这个要搞清楚。 然后呢， 我们把球面上可以由对称变换联系起来的所有的极点都归纳为一个 G 轨道，不能由对称变换联系起来的极点归为不同G 轨道，同一个G 轨道上的 极点所对应的轴的阶数是必须相等的 （因为一个轴不可能即使3 次轴， 又是4 次， 诸如此类） ，这样就可以把极点按照G 轨道分类了吧？我们假设一共有l 条G 轨 道。 那么这些极点的集合，能够贡献的群G 中非恒等变换的个数按照上面的逻 辑是不是就是：∑ n ni (ni −1) 𝑙 i=1 ？ 但在这个逻辑中， 需要指出的是我们还有一个漏洞， 就是r ⃗i与−r ⃗i给出的非恒 等操作是重复的，对吧？这种重复可能以两种方式出现： 1. 我们在算l 的时候，也就是极点的G 轨道个数的时候，如果r ⃗i与−r ⃗i不在一个 G 轨道上的时候，那么在上面的加和式子中，i 从1 到l，就会有两个不同的 i，给出的非对称操作其实是一样的，这是不是意味着在上面的加和式子中， 我们应该除个2？ 2. 第二种情况，当r ⃗i与−r ⃗i在同一个G 轨道上的时候，意味着什么？意味着并不 是所有这个G 轨道上的极点，给出的都是不同的非恒等变换。比如r ⃗i与−r ⃗i都 在这个G 轨道上，它们给出的非恒等变换就是重复的。同样gr ⃗i与−gr ⃗i也是， 它们给出的非恒等操作也重复了。这时，对于 n ni (ni −1)，我们是不是也应该 除上一个2？ 两者综合一下，那就是对 ∑n ni (ni −1) 𝑙 i=1 这个2 是必须除的。要么，是在i 从1 到l 的时候算重了，要么是 n ni (ni −1)这部 分本身就应该是 n 2ni (ni −1)。 综合起来， 我们得到的是群G 中非恒等操作的个数 就是： ∑n 2ni (ni −1) 𝑙 i=1 这个是由上面对极点G 轨道顶点的分析得到的G 中非恒等操作的个数，与 此同时，对这样一个纯转动群，它是n 阶的，那么它里面非恒等操作的个数本身 就应该是(n −1)。两者做一个结合，对这些有限次轴阶数ni以及极点G 轨道个 数l 的约束就可以通过下面这个式子反映出来了： ∑n 2ni (ni −1) 𝑙 i=1 = n −1 这个式子等价于： ∑(1 −1 ni ) 𝑙 i=1 = 2(1 −1 n) 式3.4 在这里，我们要注意的一个基本关系就是n 要大于等于ni，而ni要大于等于2。 也就是1/2 ≥1/ni ≥1/n。这个方程称为第一类点群的基本方程。它是我们这节 课的重点，下面，我们会从它出发，去分析一共会有哪些第一类点群出现？ 这里l 是正整数， 也就是1、 2、 3 等等。 但需要说明的是当l 等于1 的时候， 上面这个式子能不能成立？答案是不能，因为左边是：(1 − 1 ni)，小于1，右边是 2 − 2 n，大于等于1，两边不可能相等。 与此同时，当l 大于等于4 的时候，又会发生什么情况？这个时候，左边是 138 大于等于4 −∑ 1 ni 4 i=1 的， 而∑ 1 ni 4 i=1 要小于等于2， 所以左边是大于等于2。 而右边， 2 − 2 n肯定小于2。一个大于等于2 的数与一个小于2 的数是肯定不可能相等的。 所以l 也不可能大于等于4。 综合上面的讨论，l 只能是2，或者3。下面，我们会用穷举的方式把上面那 个基本方程的所有解都求出来。 1. l=2，我们把ni按从小到大的顺序排。这样的话上面那个基本方程就变成了： ∑(1 −1 ni ) 2 i=1 = 2(1 −1 n) 也就是： 2 −1 n1 −1 n2 = 2 −2 n 进而： 1 n1 + 1 n2 = 2 n 由于n1 ≤n2 ≤n，所以这种情况只能有一个解：n1 = n2 = n。 这时对应的实际情况，就是我有一个n 阶轴，它把我的球面戳了两个大洞。 由于没有其它的对称操作把这两个洞给联系起来， 所以它们表现为两个G 轨 道。实际上它们对应的是一个对称轴。 2. l =3，这样的话基本方程就变成了： 3 −1 n1 −1 n2 −1 n3 = 2 −2 n 进而： 1 n1 + 1 n2 + 1 n3 = 1 + 2 n 式3.5 由于n1 ≤n2 ≤n3 ≤n，所以 1 n1 ≥1 n2 ≥1 n3 ≥1 n 当n1 ≥3时，式3.5 的左边小于等于1，右边大于1，不成立。这也就是说n1 只能为2。这时，式3.5 可化为： 1 2 + 1 n2 + 1 n3 = 1 + 2 n 式3.6 而n2如果大于等于4，那么上式左边小于等于1，右边大于1，又不成立。所 以现在的情况是l = 3、n1 = 2、n2等于2 或者3。 当n2等于2 时，式3.6 变为： 1 n3 = 2 n 也就是说n = 2n3，这里n 等于4、6、8 等等都可以。这是第二个解。 3. l = 3、n1 = 2、n2 = 3，这时基本方程变成了： 1 2 + 1 3 + 1 n3 = 1 + 2 n 进而： 1 n3 = 1 6 + 2 n 式3.7 这时如果n3 = 3，则n=12。解为：l = 3、n1 = 2、n2 = 3、n3 = 3、n=12。 4. l = 3、n1 = 2、n2 = 3、n3 = 4，这时n=24，第4 个解。 5. l = 3、n1 = 2、n2 = 3、n3 = 5，这时n=60，第5 个解。 之后如果n3 ≥6，式3.7 的左边小于等于1/6，右边大于1/6，又不相等了。 所以只有以上五种解。 上面我们是从点群基本方程出发， 讨论了它存在的五种解 的可能，下面我们仔细看一下它们分别对应什么实际情况。 1. n1 = n2 = n。这个上面说过，就是一个n 阶轴，将球戳了两个洞。由于这两 140 个洞没法通过其它对称操作联系起来，所以它们是两个极点的G 轨道。这样 的群称为Cn 群。 例子： 图3.4 Cn 群例子 这样的群是Abel 群，每个群元是一类，一共有n 个类。 2. l = 3、n1 = 2、n2 = 2、n3 = m、n = 2m，m = 2、3、4、⋯ 全部极点分成三个轨道，第一个与第二个轨道上各有 2m 2 = m个极点。它们是 m 个二阶轴与球面的交点。第三个轨道上有 2m m = 2个极点，它们是一个m 阶 轴与球面的两个交点，这两个交点在一个G 轨道上。 由于这两个m 阶轴的极点在同一个G 轨道上， 说明必有一个二阶轴与这个m 阶轴垂直， 而反过来， 这个m 阶轴又可以将这个二阶轴转到m 个与之等价的 位置。 这样的群我们称为二面体群，比如我们总用到的D3 群，还有D4 群： 图3.5 D3 群、D4 群示意图 这个其实是非常好的两个例子，来说明我们在引入第一类点群基本方程时， 通过极点G 轨道上点的个数求群中非恒等操作群元个数的时候，为什么在 ∑n 2ni (ni −1) 𝑙 i=1 这个表示式中必须有1/2 这个因子？ 这里n1 = 2、n2 = 2，对D3 群来说，每个二阶轴与球面的两个交点不在一个 G 轨道上，所以在算极点G 轨道个数也就是l 的时候，是有double counting 的。也就是说在这里n1与n2对应的操作，本身就是同样的操作，我在对G 轨 道进行加和的时候， 把n1与n2都算了， 所以理所当然的要在最后除上一个2。 而对D4 群来说，我一个二阶轴与球面的两个交点r ⃗i与−r ⃗i在同一G 轨道上， 所以在对轨道个数进行加和的时候没有double counting。但是，我在算一个 轨道上的极点r ⃗i等价点的个数的时候， 是不是把−r ⃗i也算进来了？这时， double counting 还是存在的，它发生在我算一个G 轨道内等价极点的个数的时候。 两者结合起来，还是那两句话： 1． 当r ⃗i与−r ⃗i不在一个G 轨道的时候， double counting 是发生在我们对l 的统 计中； 2． 当r ⃗i与−r ⃗i在一个G 轨道的时候， double counting 是发生在计算r ⃗i的G 轨道 上的等价点对群本身非恒等操作的贡献上。 不管怎样，double counting 总是存在的。 现在我们知道了第一类点群基本方程的第二个解对应的是二面体群Dn 的情 况。下面我们来看它是怎么分类的。我们的基本思路还是：两个具有相同转 角的转动，如果其转轴可以通过群中另外一个元素联系起来，则它们同类。 142 由于这个原因，我们很容易知道对于上面D3 与D4，它们的分类情况就会不 一样。 对于D3，是不是它的所有二阶轴都可以通过绕3 次轴的转动联系起来，而对 于D4，它的二阶轴必须分为两类。与此同时，对于D3 这种奇数次的二面体 群，它绕n 阶轴的非恒等转动中，Cn k与Cn n−k同类，这里k 的取值有(n −1)/2 个。n=3 时为1，n=5 时为2。这样这种Dn 群的总的类的个数就是1+(n-1)/2+1=(n + 3)/2个。 这里第一个1 是恒等变换， 第二个(n −1)/2是绕n 次轴 的非恒等操作，最后一个1 对应所有2 次轴。 而对于D4 这种偶数次的二面体群， 它绕n 阶轴的非恒等转动中， Cn k与Cn n−k同 类，这里k 的取值有(n −2)/2个。同时恒等操作是一类，转π角的操作是一 类。同时2 次转动分为两类，所以总的类数是：1 + n−2 2 + 1 + 2 = n 2 + 3个。 除了Cn 与Dn，下面的三种情况是： 3. l = 3（有三个极点轨道） ，n=12（群里有12 个对称操作） n1 = 2、n2 = 3、n3 = 3 第一个极点轨道是二阶轴的，一共有6 个点； 第二个极点轨道是三阶轴的，一共有4 个点； 第三个极点轨道也是三阶轴的，一共也有4 个点。 怎么去理解这个群呢？我们可以从其中最高阶轴的极点出发（后面的讨论也 是同样的路子，因为高阶轴极点少，好在三维空间中构建图像） 。 假设r ⃗1、r ⃗2、r ⃗3、r ⃗4是第二个极点轨道上的四个极点，G 中的任意一个元素， 作用到这四个极点上，得到的集合还是这四个极点。 取r ⃗1对应的三次转动C3， 它作用到r ⃗1上得到的还是r ⃗1。 但它作用到其它三个极 点上，得到的C3r ⃗2、C3r ⃗3、C3r ⃗4是不是就要落到r ⃗2、r ⃗3、r ⃗4这个集合上？一定 不能产生新的点，不然这个G 轨道上就不止四个点了。 要想让这种情况成立的话，r ⃗1、r ⃗2、r ⃗3、r ⃗4的位置必须满足什么样的情况？答 案是：当我以r ⃗1作为北极的话，r ⃗2、r ⃗3、r ⃗4必须是在同一个纬度上，经度相差 120 度的三个点。 图3.6 T 群产生示意图一 同样，对于r ⃗2，取一个绕它的三次转动，我们也可以得到r ⃗1、r ⃗3、r ⃗4必处在以 它为极点的纬度线上等间距分布这样一个结论。 对r ⃗3、r ⃗4也可做类似处理。把这些结论和在一起，那么r ⃗1、r ⃗2、r ⃗3、r ⃗4的分布 就只能有一种情况：就是它们是球面上相互之间都等间距分布的四个点，构 成一个正四面体。 图3.7 T 群产生示意图二 对于这个正四面体，r ⃗1、r ⃗2、r ⃗3、r ⃗4对应的−r ⃗1、−r ⃗2、−r ⃗3、−r ⃗4，也构成一个 极点G 轨道，这个就是我们的解中n3 = 3对应的情况。 144 同时，二阶轴所对应的六个极点，也可以通过三次转动联系起来，比如： 图3.8 T 群示意图 这个图里面四面体中心到A、…、F 这些点的连线与球面的交点，对应的都 是二次轴极点，它们相互可由三次转动联系起来，所以只形成一个轨道，对 应解中n1 = 2的情况。 这样的一个正四面体群，我们称为T 群。在有些教材中在讲到正四面体群的 时候，会跟四阶置换群做一个类比。背后更大的背景是所有的有限群均同构 与置换群的子群。这里我们先用点群与置换群及其子群做个类比。 T 群的每一个对称操作， 说白了， 就是对正四面体的四个顶点进行一个置换， 比如： C3 = (1 2 3 4 1 3 4 2) 比如： C3 2 = (1 2 3 4 1 4 2 3) 对于一个四阶循环群来说， 它的阶数是4 的阶乘， 也就是24。 而T 群只有12 个元素，为什么？ 原因很简单，就是T 群是个第一类点群，它包含的是纯转动操作，在变换过 程中，1、2、3、4 四个点之间手性发生变化的变换是不包括的。因为这个原 因，在有些教材上，会说到T 群是S4 的偶置换子群。 在T 群中， 如果再加上不保手性的变换， 也就是我的对象还是这个正四面体， 但我允许它的顶点进行不保手性的变换，后面会说到这个群对应的是一个第 二类点群Td，这样的一个群跟S4 就同构了。 再展开来说，在点群中，随着顶点数的增加，几何限制会使得点群与置换群 的差别会越来越大。当只有三个顶点的时候，前面讲过D3 与S3 直接同构； 当我有四个顶点的时候，T 与S4 只有在加入非纯转动操作的时候才同构；对 于五个以上顶点的情况，即使假如转动反演操作，也不行，因为顶点之间的 任意置换是不可能被点群中的几何操作（转动或转动反演）所允许的。这些 点群只能与相应置换群的子群同构，这个越往后你们会越能体会。 关于T 群中群元的分类。 E 自成一类；二次转动C2、C2 ′ 、C2 ′′，由于转轴可通过三次转动联系起来，所 以成为一类；转2𝜋/3角的操作C3、C3 ′ 、C3 ′′、C3 ′′′，由于其转轴可以通过其它三 次转动联系起来，所以也成为一类；而C3与C3 2之间，由于其转轴没法通过T 中元素联系起来，所以不是一类，但C3 2、C3 ′2、C3 ′′2、C3 ′′′2之间是可以通过T 中 元素联系起来的，所以C3 2、C3 ′2、C3 ′′2、C3 ′′′2组成一个类。 综合起来， 就是T 中有四个类，分别是：{E}，{C2、C2 ′ 、 C2 ′′}， {C3、C3 ′ 、C3 ′′、 C3 ′′′}、{C3 2、C3 ′2、C3 ′′2、C3 ′′′2}。 这个分类情况再结合Burnside 定理，我们就会知道T 群的不等价不可约表示 维数情况是什么？ S1 2 + S2 2 + S3 2 + S4 2 = 12，一维恒等表示S1 = 1，其它维度只能是S2 = 1、S3 = 1、S4 = 3。也就是说T 群有三个一维不等价不可约表示，一个三维不等价不 可约表示。 4. Cn、Dn、T 之后的第四种情况。 146 𝑙= 3，三个极点G 轨道； n = 24，24 个对称操作； n1 = 2，第一个G 轨道对应的是2 次轴的极点，上面有12 个点； n2 = 3，第二个G 轨道对应的是3 次轴的极点，上面有8 个点； n3 = 4，第一个G 轨道对应的是4 次轴的极点，上面有6 个点； 分析的话还是和上面一个例子一样，从最高阶轴的极点出发。有r ⃗1、r ⃗2、r ⃗3、 r ⃗4、r ⃗5、r ⃗6六个极点。这个群本身的转动都是2、3、4 次转动。 我们把一个4 次转动作用到r ⃗1、r ⃗2、r ⃗3、r ⃗4、r ⃗5、r ⃗6上，得到的必是这六个点 的集合自身。 设这个C4本身对应的是r ⃗1轴，那么r ⃗2、r ⃗3、r ⃗4、r ⃗5、r ⃗6这六个点怎么配置呢？ 只能是其中的是个放在同一纬度线上等间距分布，随后一个放到r ⃗1正对的那 个极点，也就是这个样子： 图3.9 O 群产生示意图 与此同时，r ⃗1、r ⃗2、r ⃗3、r ⃗4、r ⃗5、r ⃗6完全等价，现在上面那个图是只考虑了绕 r ⃗1的四次转动，绕r ⃗2、r ⃗3、r ⃗4、r ⃗5、r ⃗6，其实我们也是有同样要求的。如果把 这些所有的要求都考虑在一起，那么很自然，我的顶点就只能是这种情况： 图3.10 O 群示意图 r ⃗1、r ⃗2、r ⃗3、r ⃗4、r ⃗5、r ⃗6是一个正八面体的顶点，因此这个群也就叫O 群 （Octahedron）了。这个正八面体，外面还可以接一个立方体。也就是说只考 虑转动的情况下，这个群描述的是立方体，或正八面体的对称性。 这个正八面体（里面虚线） ，相对正三角形的中心连线是个三阶轴（或者说事 外接立方体相对顶点连线） 。这样的三阶轴一共有4 条，对应8 个极点。这些 极点相互之间由于都可以用4 阶转动联系起来，所以G 轨道上是8 个点，对 应的就是我们的基本方程给出的n2 = 3的情况了。 同时，正八面体相对棱的中点连线，对应的是2 次轴（也是立方体相对棱的 中点连线） 。这样的2 次轴有6 条，对应12 个极点。这些极点之间可以由4 次转动联系起来， 所以G 轨道上是12 个点， 对应的就是基本方程中n1 = 2的 情况了。 最后，由于这些四次轴、三次轴、二次轴的极点都可以由群中其它元素联系 起来，所以G 轨道就三个。转2π/3与转4π/3的操作同类，转π/2与转3π/2的 操作同类。而转π的操作，存在两种情况，一种是绕2 次轴的转动，一种是绕 148 4 次轴的转动， 但转两次。 这两种转动的转动轴， 是不能通过群中其它元素联 系起来的。这样的话总共24 个操作的分类是： {E}、{C2 (1)、C2 (2)、C2 (3)、C2 (4)、C2 (5)、C2 (6)}、 {C3、C3 ′ 、C3 ′′、C3 ′′′、C3 2、C3 ′2、C3 ′′2、C3 ′′′2}、{C4、C4 ′ 、C4 ′′、C4 3、C4 ′3、C4 ′′3}、 {C4 2、C4 ′2、C4 ′′2} 其中二次轴与四次轴的夹角是45 度， 这样的话虽然绕四次轴转两次和绕二次 轴转一次转动角相同，但转轴不能通过群中其它元素联系起来，就不是一类 了。 同时再做前面说过的，D3 与S3 的类比，我们会发现这里T（四个顶点）是S4 （24 个元素）的一个子群，O（六个顶点）是S6 的（720 个元素）的一个子 群。只不过因为这6 个顶点相互位置不能乱换，O 群的阶比S6 群的阶小得很 多。 5. Cn、Dn、T、O 之后最后一种第一类点群： 𝑙= 3，三个极点G 轨道； n = 60，60 个对称操作； n1 = 2，第一个G 轨道对应的是2 次轴的极点，上面有30 个点； n2 = 3，第二个G 轨道对应的是3 次轴的极点，上面有20 个点； n3 = 5，第一个G 轨道对应的是5 次轴的极点，上面有12 个点； 还是从最高阶轴开始， 设r ⃗1、 r ⃗2、⋯⋯、r ⃗12对应这12 个极点， 他们的集合要 在五次转动下不变。 以绕r ⃗1的转动为例， 这些转动要把r ⃗2、⋯⋯、r ⃗12这11 个极点转到它们本身。 与前面的分析类似，这种情况，只能将这11 个点分成三组，前两组每组五个 点，等间距分布在以r ⃗1为极点的等纬度线上。最后一个点，与r ⃗1相对。 同时，这十二个点要相互等价，这样的话对每个极点，另11 个都要满足这样 的性质。综合起来，结果就只有一种可能： 图3.11 二十面体群 这是一个由20 个正三角形组成的二十面体， 因此这个群也称为正二十面体群 （Icosahedron） ，记为Y。这个正二十面体也可以内接一个正十二面体，每个 面试个正五边形。它们的对称性是完全一样的。 就二十面体而言， 每两个相对的极点的连线都是一个五次轴， 一共有12 个极 点，6 个轴，相对极点可由其它变换（二次转动）联系起来，所以这个轨道有 12 个极点。 同时每两个相对的正三角形中点的连线是个三阶轴， 三角形有20 个， 所以三 阶轴是10 个。相对极点同样可通过二次转动联系起来，这个轨道上有20 个 点。 每两个相对棱的中点连线为二次轴，一共有30 条棱，也就是15 个二阶轴。 同样，相对极点可由其它二阶转动联系起来，所以轨道上有30 个点。 在这个群中，由于它已经越来越像球了，对称性也越来越高，这个时候具有 相同转角的转动就越有可能被其它转动操作联系起来，从而形成一类。Y 群 150 的60 个元素的分类情况也就是： {E}、 {C2 (1)、C2 (2)、⋯、C2 (15)}（15 个） 、 {C3 (1)、⋯、C3 (10)、C3 (1)2、⋯、C3 (10)2}（20 个） 、 {C5 (1)、⋯、C5 (6)、C5 (1)4、⋯、C5 (6)4}（12 个） 、 {C5 (1)2、⋯、C5 (6)2、C5 (1)3、⋯、C5 (6)3}（12 个） 而这个正二十面体，你如果把每条棱都三等分，然后再把头去掉，那么你最 终得到的，就是一个足球。相关的诺奖是1996 年化学奖，C60。 图3.12 足球与二十面体群 3.3 第二类点群 前面我们是通过第一类点群的基本方程， 把第一类点群的所有情况作了一个 介绍。根据这个介绍，我们现在就可以再之前我们说的定理3.3 的第二与第三种 情况，把所有的第二类点群推出来了。 先看第二种情况：当G 不只包含纯转动操作时，如果G 中包含纯反演操作 I，那么G 与K 的关系必然是G= K ∪IK；这个时候，因为有五种第一类点群， 与之对应的第二类点群也就是五种。 下面我们用[II， 中]来表示这些第二类点群， II 是第二类点群的意思， ‘中’代表具有中心反演对称性。这里由于中心反演对 称操作与任意转动操作互易， 所以IK 这个部分不改变K 的分类， 只是把K 中所 有的类又在乘上I 之后重复了一遍。 1. Cn ∪ICn，这是一个2n 阶的Abel 群，Cn中的每个元素自成一类，ICn中的每 个元素也自成一类。 2. Dn ∪IDn，这里Dn是2n 阶的， Dn ∪IDn就是4n 阶的。 前面说了， n 为奇数时， Dn有(n+3)/2 个类，Dn ∪IDn就有n+3 个类。N 为偶数时，Dn有n/2+3 个类， Dn ∪IDn就有n+6 个。 3. T ∪IT阶为24，有8 个类，记为Th。 4. O ∪IO阶为48，有10 个类，记为Oh。 5. Y ∪IY阶为120，有10 个类，记为Yh。 （后面会解释为什么用这个h） 除了上面说的[I]、 [II， 中]， 第三种情况说的是： 当G 不只包含纯转动操作， 且G 中不包含纯反演操作I 时， G 必与一个纯转动群G+同构， 这里G+ = K ∪K+， 而K+的定义是：K+ = {Ig|g ∈G，但g ∉K}。 换句话说这里的第二类点群不包含纯反演操作， 同时它与一个第一类点群同 构。 这样的话我们只需要找到与之同构的第一类点群K ∪K+， 然后对它做K ∪IK+ 这样一个变换就可以了。 而对K ∪K+这样一个第一类点群，如果我们把K 当作一部分，K+当作一部 分，它与二阶循环群是同态的，为什么？ （因为K ∪IK+本身是一个群，IK+中的K+就是我们这个第一类点群的第二部分。 由于重排定理，对K ∪IK+，取K 中任何一个元素，它乘上K 这个集合，必给出 K 这个集合；而它乘上IK+这个集合，必给出IK+这个集合。同时IK+中任意元 素，乘上K 这个集合，必给出IK+这个集合；它乘上IK+这个结合，给出K。由 152 于I 是可以单独提出的，所以对K ∪K+这个第一类点群，必存在这样的性质：K 乘上K 等于K，K 乘上K+等于K+，K+乘上K 等于K+，K+乘上K+等于K。也就 是说{K，K+}与二阶循环群同态） 。 这样的话K 必须是{K， K+}的不变子群， 并且阶数是它的一半。 根据这样一 个规则， 现在我们就可以来通过我们知道的第一类点群， 构造这种第二类点群了。 在Cn、Dn、T、O、Y 里面符合这个条件的，只有C2n、Dn、D2n、O 这四种 情况，下面我们来分开介绍（符号在前面五种第二类点群的基础上继续） ： 6. C2n，对它来讲，{C2n 2 、C2n 4 、⋯、C2n 2n = E}形成不变子群。 {C2n、C2n 3 、⋯、C2n 2n−1}为其陪集，形成的第二类子群是： {{C2n 2 、C2n 4 、⋯、C2n 2n = E}、I{C2n、C2n 3 、⋯、C2n 2n−1}}。 7. 与Dn同构的第二类点群， 这里纯转动部分是： {Cn、Cn 2、⋯、E}， 另外一部分 是：I{C2 (1)、C2 (2)、⋯、C2 (n)}。 由于I 与任意元素互易，所以它的分类情况与Dn完全相同，n 为奇数时是 (n+3)/2，偶数时是n/2+3。 8. 对Dn，当n 为偶数时，也就是D2n这种情况，它除了C2n这个不变子群，还有 Dn也是它的不变子群。这个时候，如果根据Dn去做这个第二类点群，又能得 到一种情况。这个时候纯转动部分是： {C2n 2 、C2n 4 、⋯、C2n 2n = E、C2 (2)、C2 (4)、⋯、C2 (2n)} 转动反演部分是： I{C2n、C2n 3 、⋯、C2n 2n−1、C2 (1)、C2 (3)、⋯、C2 (2n−1)} 以D6为例， 这里是取它的不变子群D3， 这里C2 (1)、C2 (2)、⋯、C2 (2n)的分布就是： 图3.13 D6群对称性 由于I 与任意群元互易，所以它的分类情况与D2n 群完全一样，也是 2n/2+3=n+3。 9. 最后一种情况，基于的第一类点群是O 群，它的不变子群是T。 这个怎么理解呢？我们有一个立方体，它的纯转动对称群是O 群，现在针对 它做一个内接正四面体，这个四面体，纯转动操作相对于O 群减少了一半， 但是除了这些纯转动操作，它多了一些转动反演操作。这些转动反演操作加 上保留下来的纯转动操作， 形成的第二类点群与O 群同构， 也是24 个操作。 比如这个正四面体，如果我对z 轴转90 度，再做反演，它回到它本身。这个 操作就是这个正四面体具有的，它是一个转动反演操作，与立方体本身转90 度的纯转动操作对应。 154 图3.14 O 群对称性 这里K 等于： {C3、C3 ′ 、C3 ′′、C3 ′′′、C3 2、C3 ′2、C3 ′′2、C3 ′′′2、E、C4 2、C4 ′2、C4 ′′2} IK+等于： I{C2 (1)、C2 (2)、C2 (3)、C2 (4)、C2 (5)、C2 (6)、C4、C4 ′ 、C4 ′′、C4 3、C4 ′3、C4 ′′3} 同样，由于I 与任意元素互易，这个群与O 群分类相同，5 个类。 到这儿， 我们是讲完了所有的点群， 它们是如何分类的？都具有哪些对称操 作？但光有这些了解， 你们还没法看文献， 因为文献中一般都会用一些符号去标 识一个特定的点群， 比如Td、 Th、 Oh。 这些符号直接告诉了我们这个点群是什么， 而不用让我们去写比如：与O 同构的不含反演操作的第二类点群，或者T 与中 心反演操作结合形成的点群，这些东西。 这些符号叫熊夫利符号， 因为它是我们在工作中经常接触的东西， 我们在这 里详细说一下。 我们先说明的一点是熊夫利符号与点群的对称元素有着最直接的 联系。同时，在对称元素的描述中，熊夫利符号倾向于使用转动反射面，而不是 转动反演轴。在前面的讨论中，因为转动反演轴更便于我们讲点群分类，所以我 们一直采用它，后面要换一下。前面讲过，它与转动反射面的关系很简单，是： σk ⃗ ⃗⃗Ck ⃗ ⃗⃗( 2𝜋 n ) = ICk ⃗ ⃗⃗( 2𝜋 n + 𝜋)。 图3.15 转动反演与转动反射 在熊夫利的命名规则中，一个点群的最高阶转轴、与某反射面垂直的轴、或某个 转动反演轴会被当作是主轴。 命名的时候是基于第一类点群的符号， 结合反射面 或转动反射轴来进行的。 主轴的方向一般我们称为垂直方向。 水平的放射面记为 σh，而过主轴与水平面垂直的反射面记为σv或σd。这里v 与d 还有些区别，我们 一会儿再讲。 熊夫利命名规则，说白了，就是根据前面我们分析出来的各种群的特征，找 到一个可以标识它的对称元素的符号， 用这个符号来代表这个群。 或者说就是起 名，起名的根据是对称元素。这句话说起来简单，但做起来并不好做。下面我们 具体得来分析，根据这个图： 图3.16 点群分类图一 156 其中第一类点群名字已经有了。 对第二类点群， 我们最终是要把它分成的情况是： 图3.17 点群分类图二 下面我们按照兰色标记的十三个数字逐个讲，这13 个数字，其中在熊夫利 符号中重复的情况，比如10、11。它们在我们之前以定理3.3 来推各种点群的时 候，属于定理3.3 的不同情况，但这里它们由相似的对称元素标识。因此，这13 种可能性又可以在熊夫利的框架下缩小为9 种标识的情况。 下面， 我们先走遍这 13 个兰色数字，再对它们进行黑框中的归纳： 1. 先看 T ∪IT。 T 的操作是：{E、C2、C2 ′ 、C2 ′′、C3、C3 ′ 、C3 ′′、C3 ′′′、C3 2、C3 ′2、C3 ′′2、C3 ′′′2}，并上 IT，也就是I{E、C2、C2 ′ 、C2 ′′、C3、C3 ′ 、C3 ′′、C3 ′′′、C3 2、C3 ′2、C3 ′′2、C3 ′′′2}，这是 是它所有的操作。 这些操作中，有纯转动，有转动反射。在熊夫利的命名规则中，把IC2这个转 动反演操作所对应的反射面当作非纯转动部分的代表，并把这个反射面放到 水平的位置。这样我用T 可以把纯转动部分包括，用σh对T 做陪集就可以把 转动反演部分包括。子群与陪集结合产生所有这个群里面的元素，与之相应 这个群也就可以记为：Th。 2. 对O，一样，有IC4 2这个操作，我们可以把它对应的平面定为水平面，用σh与 O 来标识这个群，记为：Oh。 3. 对Y， 有IC2， 把它对应的平面定为水平面， 用σh与Y 来标识这个群， 记为： Yh。 4. 与Dn同构的第二类点群，对称操作为： {{Cn、Cn 2、⋯、E}、I{C2 (1)、C2 (2)、⋯、C2 (n)}}。 Cn这个高阶转动轴为主轴，形成子群{Cn、Cn 2、⋯、E}。陪集部分是I{C2 (1)、 C2 (2)、⋯、C2 (n)}。由于C2 (i)在水平面上，所以他们代表的是垂直水平面的反射 面σv，这个v 代表vertical。两者结合起来，我们用Cnv来标识这样的群。 5. 与O 同构的第二类点群，包含的操作是： {C3、C3 ′ 、C3 ′′、C3 ′′′、C3 2、C3 ′2、C3 ′′2、C3 ′′′2、E、C4 2、C4 ′2、C4 ′′2} ∪I{C2 (1)、C2 (2)、C2 (3)、C2 (4)、C2 (5)、C2 (6)、C4、C4 ′ 、C4 ′′、C4 3、C4 ′3、C4 ′′3} 纯转动部分是个Ｔ群，最高转动轴为三次轴。转动反演部分有三个４次转动 反演轴。 对于４次转动反演轴， 我们可以在实空间画一下它对应的等价的点： 图3.18 四次转动反演示意图 从这些点我们可以看出４次转动反演轴等价于4 次转动反射轴。我们以这个 158 转动反射轴为主轴。这个轴是本身O 群的4 次轴，也就是下图中的z 轴： 图3.19 O 群的四次轴 这个时候，我们对这个与O 群同构的第二类点群，纯转动部分是T，转动反 演部分只要找出一个操作，就可以通过陪集的方式把这个群确定下来了。在 上个图中， 我们知道O 群会有两个水平的二次轴， 对应C2 (i)中的两个， 它们乘 上I，给出的是垂直于水平面的反射面。这样的话我们就可以用T 与这个反 射面来标识这个群了。 但需要注意的是，前面我们说垂直于水平面的反射面，会有两种情况，σv与 σd。σv就是一个一般的垂直水平面的反射面，前面我们也用过。而σd，说的 是这个反射面除了垂直于水平面，还平分水平面上两个二次轴的夹角。d 是 diagonal 的意思。 好多教材没有解释这个， 大家有兴趣的话， 可以看一下朗道 《量子力学》第93 节那部分17，或者Dresselhaus 那本《Group Theory Applications o Physics of Condensed Matter》的第45 页。 这样的话对与O 同构的第二类点群，我们就可以用Td来标识了。 6. Cn ∪ICn，n 为奇数，可记为C2n+1 ∪IC2n+1，这个群包含的元素是： 17 我第一年上课的时候，一个叫杨康的学生帮我发现的，谢谢！ {C2n+1、C2n+1 2 、⋯、C2n+1 2n+1 = E} ∪I{C2n+1、C2n+1 2 、⋯、C2n+1 2n+1 = E} 这些元素的集合与S4n+2这个转动反射轴作为基本生成元， 形成的群是一样的， 所以在熊夫利符号中，记为S4n+2。 以S6为例，通过这个操作，我们可以知道一个点的等价点的集合为： 图3.20 六次转动反射示意图 从这个等价的点的集合，我们很容易想象它的对称群是： {C3、C3 2、E} ∪I{C3、C3 2、E} 7. 与C2n同构的第二类点群中的一种情况，当2n 为4 的整数倍，也就是4n，或 者直接说与C4n同构的第二类点群。这时这个群为C2n ∪I(C4n −C2n)，它包含 的群元有： {C4n 2 、C4n 4 、⋯、C4n 4n = E} ∪I{C4n、C4n 3 、⋯、C4n 4n−1} 这些元素的集合与S4n这个转动反射轴作为基本生成元，形成的群是一样的， 所以在熊夫利符号中，记为S4n。 以S4为例，通过这个操作，我们可以知道一个点的等价点的集合为： 160 图3.21 四次转动反射等价点 从这个等价的点的集合，我们很容易想象它的对称群是： {C4 2、E} ∪I{C4 1、C4 3} 8. Cn ∪ICn时，n 为偶数，也就是C2n ∪IC2n。这个群包含的元素是： {C2n、C2n 2 、⋯、C2n 2n = E} ∪I{C2n、C2n 2 、⋯、C2n 2n = E} 这里，既有C2n，又有转动反射IC2n n = 𝜎h，因此可记为C2nh。 9. 与C4n+2同构的第二类点群，群元： {C4n+2 2 、C4n+2 4 、⋯、C4n+2 4n+2 = E} ∪I{C4n+2、⋯、C4n+2 2n+1、⋯、C4n+2 4n+1} 有C2n+1这个第一类点群的所有操作，有IC4n+2 2n+1 = 𝜎h，因此记为C2n+1h。 10. D2n ∪ID2n，有D2n的全部对称操作，有IC2n n = 𝜎h，因此可记为D2nh。 11. 与D4n+2同构的第二类点群，群为D2n+1 ∪I(D4n+2 −D2n+1)。它有D2n+1的全 部对称性，同时还包括IC4n+2 2n+1 = 𝜎h，所以也记为D2n+1h。 12. D2n+1 ∪ID2n+1，有2n+1 个垂直反射面IC2 (1)、⋯、IC2 (2n+1)。 以D3 ∪ID3为例： 图3.22 D2n+1d中的𝜎d示意图 图中𝜎d (1) = IC2 (1)、 𝜎d (2) = IC2 (2)、 𝜎d (3) = IC2 (3)， 都是垂直于水平面的反射面， 我 们在这里用d 而不是用v，是因为它们刚好又在二次轴的平分线上。由于这 个群有D2n+1的全部对称性，又有σd (i)，它被记为D2n+1d。 13. 与D4n同构的第二类点群， D2n ∪I(D4n −D2n)， 以D2 ∪I(D4 −D2)为例， D4有 四个水平二阶轴， 现在两个保持 （记为C2 (1)、C2 (3)） ， 两个变为转动二阶轴 （记 为IC2 (2)、IC2 (4)） 。它们的关系见下图： 图3.23 D4nd中的𝜎d示意图 在这里𝜎d (2) = IC2 (2)，它在C2 (1)、C2 (3)平分线上，用d 标识。这个群又有D2n的 所有对称性，所以记为D2nd。 162 这13 种情况中，6（S4n+2） 、7（S4n）都对应S2n；8（C2nh） 、9（C2n+1h）都 对应Cnh；10（D2nh） 、11（D2n+1h）都对应Dnh；12（D2n+1d） 、13（D2nd）都对 应Dnd。 所以合在一起就是S2n、 Cnh、 Dnh、 Dnd、 Cnv、 Th、 Oh、 Yh、 Td九种情况。 3.4 晶体点群与空间群 现在我们是把点群基础、 第一类点群、 与第二类点群以及点群的熊夫利符号 讲完了。 利用这些知识， 我们可以具体的去研究一个在我们物理学近一百多年的 发展史中占据核心地位的系统――晶体。 我们会先研究一下晶体里面的点群， 然 后再把点群这个概念进行一个扩充，讲一下空间群。 所谓晶体，简单的说，就是这个系统必须要由全同的、由原子的集合构成的 结构单元组成，并且这个结构单元在三维空间中可以“无限地”重复。它要有转 动（含转动反演）与平移两种对称性。晶体点群，是忽略了平移对称性，能够将 晶体变到其本身的转动与转动反演对称性的集合。 在点群的所有对称操作中， 晶 体有一个点是不动的。 晶体点群，相对于普通点群（比如说分子中的点群） ，最重要的一个性质就 是晶体制约定理。 定理3.4 设G 是晶体点群，则G 中转动元素只能是𝐂𝟏（也就是E） 、𝐂𝟐、𝐂𝟑、 𝐂𝟒、𝐂𝟔，所有转动反演元素，只能是I、𝐈𝐂𝟐、I𝐂𝟑、𝐈𝐂𝟒、I𝐂𝟔。 （在证明这个定理的时候中要用到的一个概念是晶格， 它是由三个线性无关的向 量a ⃗ ⃗1、a ⃗ ⃗2、a ⃗ ⃗3的整数线性组合组成的。而晶体，可以看作是那些重复单元，坐在 晶格上形成的） 证明： 设晶格L 是G 不变的，L 的基本向量是a ⃗ ⃗1、a ⃗ ⃗2、a ⃗ ⃗3，则对∀g ∈G，有： ga ⃗ ⃗i = ∑cjia ⃗ ⃗j 3 j=1 对这个等式我们先看左边， ga ⃗ ⃗i这个向量本身必须在晶格上， 因此右边的cij必为整 数。这个是我们从左边向量性质推出的右边系数的性质。 而另一方面， G 是O(3)群的子群， 上式同时还代表着如果选a ⃗ ⃗1、 a ⃗ ⃗2、 a ⃗ ⃗3为基的话， 由系数cji形成的矩阵C(g)是群G 的表示矩阵。与之相应的{C(g)}就是G 的一个 表示。 这个表示和我们用三维欧式空间中正交的单位向量i ⃗、 j ⃗、 k ⃗⃗形成的表示差的就是一 个相似变换，从i ⃗、j ⃗、k ⃗⃗到a ⃗ ⃗1、a ⃗ ⃗2、a ⃗ ⃗3的相似变换，对吧？这个相似变换是不改变 表示的特征标的。 也就是说对{C(g)}这个表示， 它的特征标还等于±(1 + 2 cos Ψ)， 其中Ψ是转角。 这也就意味着tr(C(g))一方面等于∑ cii 3 i=1 ，另一方面等于±(1 + 2 cos Ψ)。 前面说过cii为整数， 那么±(1 + 2 cos Ψ)就必为整数， 这样的话这里的Ψ只能为0、 π/3、π/2、2π/3、π。纯转动操作为C1（也就是E） 、C2、C3、C4、C6，转动反 演操作为I、IC2、IC3、IC4、IC6。 （C5、C7、C8、⋯就不会存在） （证毕） 有了晶体制约定理，再回到我们之前讲的第一类、第二类点群，我们就很容 易知道： 一. 第一类点群中可以在晶体中出现的是Cn里面的C1、C2、C3、C4、C6，Dn里面 的D2、D3、D4、D6，O，T，共11 个。Y 不能存在因为它有五次轴。 二. 第二类点群里面可以在晶体中出现的包括： 1. S2n中的S2、S4、S6。因为S4n+2 = C2n+1 ∪IC2n+1，这个里面可以存在的是 S2和S6，S10及其以上都不能存在；S4n = C2n ∪I(C4n −C2n)，这里n 只能 164 为1，对应S4。 （3 个） 2. Cnv（与Dn同构的第二类点群）中的C2v、C3v、C4v、C6v。 （4 个） 3. Cnh中的C1h、 C2h、 C3h、 C4h、 C6h。 因为C2nh = C2n ∪IC2n， 它可以贡献C2h、 C4h、C6h；C2n+1h = C2n+1 ∪I(C4n+2 −C2n+1)，它可以贡献C1h、C3h。 （5 个） 4. Dnh中的D2h、D3h、D4h、D6h。因为D2n+1 h = D2n+1 ∪I(D4n+2 −D2n+1)， 它可以D3h；D2nh = D2n ∪ID2n，它可以贡献D2h、D4h、D6h。 （4 个） 5. Dnd中的D2d、 D3d。 其中D2nd = D2n ∪I(D4n −D2n)可以贡献D2d； D2n+1d = D2n+1 ∪ID2n+1可以贡献D3d。 （2 个） 6. Th = T ∪IT。 （1 个） 7. Td = T ∪I(O −T)。 （1 个） 8. Oh = O ∪IO。 （1 个） 共21 个第二类点群。 两者加在一起就是32 种晶体点群。任何晶体，它的转动与转动反演对称性 的集合，必属于这32 种点群。利用这32 种点群，我们是可以对晶体进行一个分 类的。具有相同点群的晶体，它们会有一些共同的特征，比如红外谱、拉曼谱、 能带，这个我们后面第四章会专门讲到。 除了按这32 种点群对晶体进行分类，在这32 种点群中，依据各点群间对称 操作的对称元素的相似性， 这32 种点群还可以进行一个划分。 以T、 O、 Td、 Th、 Oh这五种点群为例， 具有这些对称性的晶体， 它们都有4 个三次轴。 这4 个三次 轴，会让它们的宏观对称性呈现出一些相似性。我们把它们归为一类，这样的一 个类，称为一个晶系。有4 个三次轴的晶系是立方晶系，它们是晶体中对称性最 高的一个群体，宏观性质上，x、y、z 轴三者等价。如果沿z 轴方向做个拉伸， 使得x、y 等价，z 轴和它们不等价，造成的一个结果就是系统只在一个方向（z 轴） 有四阶转动轴或四阶转动反演轴。 对称操作的对称元素具备这个特征的点群 再形成一个群体，包括C4、D4、C4h、S4、D4h、C4v、D2d，我们称为四方系。 为什么要进行这样的分类呢？之前我们说过， 最早研究晶体结构的是晶体学 家，早于X 射线、早于电子衍射谱。因此，晶体的宏观特征是他们在描述晶体内 部结构时一定要抓住的特征。T、O、Td、Th、Oh这五种点群，它们都有4 个三 次轴，肉眼上可以看到的性质就是x、y、z 三个轴等价。晶体中类似的晶系有七 种，它们的划分依据，是共同的对称操作的对称元素，比如立方系的三次轴。在 描述这些点群之间对称操作的对称元素共同特征的时候， 我们可以借助于晶胞这 个反映晶体对称性的最小结构单元。 这些对称操作的对称元素的共同特征可以由 晶胞参数（a ⃗ ⃗、b ⃗⃗、c ⃗、α、β、γ，定义见下图）反映出来。 图3.24 晶胞参数关系示意图 下面我们就按照这七种晶系的定义（对称操作的对称元素的相似性） ，以及 它们晶胞参数的特性，依次把32 种晶体点群进行归纳： 1. 三斜晶系(Triclinic Crystal System) 这种晶系中，不同点群的对称元素的共性是只存在一重转动轴或一重转动反 演轴。 166 在上面的讨论中， 我们知道满足这个要求的点群包括： S2 = C1 ∪IC1、 C1两种。 其晶胞参数的限制就是a ⃗ ⃗、b ⃗⃗、c ⃗长度不同，夹角任意，都不为90 度。S2对应 的是有中心反演对称性的情况，它在此晶系中对称性最高。C1对应的是连I都 没有的情况。 2. 单斜晶系(Monoclinic Crystal System) 这种晶系中，不同点群的对称元素的共性是只在一个轴方向存在二重转动轴 或二重转动反演轴。 在上面的讨论中，满足这个要求的点群包括： 1) 第一类点群：C2 2) 第二类点群：C2 ∪IC2 = C2h、C1 ∪I(C2 −C1) = C1h 在这三种点群中，C2h对称性最高。 对其晶胞，要求是：a ≠b ≠c，α = β = 90°，γ ≠90°。 3. 正交晶系(Orthorhombic Crystal System) 这种晶系中，不同点群的对称元素的共性是有三个相互垂直的二重转动轴或 二重转动反演轴。 在上面的讨论中，满足这个要求的点群包括： 1) 第一类点群：D2 2) 第二类点群：D2 ∪ID2 = D2h、C2 ∪I(D2 −C2) = C2v 在这三种点群中，D2h对称性最高。 对其晶胞，要求是：a ≠b ≠c，α = β = γ = 90°。 4. 四方晶系(Tetragonal Crystal System) 这种晶系中，不同点群的对称元素的共性是在唯一高次轴方向有四重转动轴 或四重转动反演轴。 在上面的讨论中，满足这个要求的点群包括： 1) 第一类点群：C4、D4 2) 第二类点群：包括从C4出发的C4 ∪IC4 = C4h、C2 ∪I(C4 −C2) = S4；以及 由D4出发的D4 ∪ID4 = D4h、C4 ∪I(D4 −C4) = C4v、D2 ∪I(D4 −D2) = D2d。 在这些点群中，D4h对称性最高。 对其晶胞，要求是：a = b ≠c，α = β = γ = 90°。 5. 三方晶系(也叫三角晶系，Trigonal Crystal System) 这种晶系中，不同点群的对称元素的共性是在唯一高次轴方向有三重转动轴 或三重转动反演轴。 在上面的讨论中，满足这个要求的点群包括： 1) 第一类点群：C3、D3 2) 第二类点群：包括从C3出发的C3 ∪IC3 = S6；以及由D3出发的D3 ∪ID3 = D3d、C3 ∪I(D3 −C3) = C3v。 在这些点群中，D3d对称性最高。 晶胞方面， 有两种取法， 第一种是a = b ≠c， α = β = 90°，γ = 120°； 或者a = b = c，α = β = γ ≠90°。 6. 六角晶系(Hexagonal Crystal System) 这种晶系中，不同点群的对称元素的共性是在唯一高次轴方向有六重转动轴 或六重转动反演轴。 在上面的讨论中，满足这个要求的点群包括： 168 1) 第一类点群：C6、D6 2) 第二类点群：包括从C6出发的C6 ∪IC6 = C6h、C3 ∪I(C6 −C3) = C3h；以 及由D6出发的D6 ∪ID6 = D3h、C6 ∪I(D6 −C6) = C6v、D3 ∪I(D6 −D3) = D3h。 在这些点群中，D6h对称性最高。 晶胞方面， a = b ≠c，α = β = 90°，γ = 120°。 三方系与六角系，当三方系的晶胞按第一种取法取时，从晶胞参数上看不出 任何差别。 它们的本质的差别还是体现在主轴的对称性上， 对六角系是六次。 7. 立方晶系(Cubic Crystal System) 这种晶系中，不同点群的对称元素的共性是四个三次轴。 在上面的讨论中，满足这个要求的点群包括： 1) 第一类点群：T、O 2) 第二类点群：包括从T出发的T ∪IT = Th ；以及由O出发的O ∪IO = Oh、T ∪I(O −T) = Td。 在这些点群中，Oh对称性最高。 晶胞方面， a = b = c，α = β = γ = 90°。 有了这些知识， 你们就应该知道在以后事科研工作中， 如果老板问你研究的 晶体是什么晶系（Crystal System）？它的点群是什么？他/她指的是什么意思。 现在我们清楚了晶系与晶体点群两个概念， 下一步是往空间群过渡。 七种晶 系可容纳32 种点群。对于一个点群，它属于某晶系，晶胞特征可以由a ⃗ ⃗、b ⃗⃗、c ⃗之 间的关系描述。对于由a ⃗ ⃗、b ⃗⃗、c ⃗形成的六面体，我们也只是想象它的顶点有个东 西。以立方晶系中的Oh群为例。如果只是由a ⃗ ⃗、b ⃗⃗、c ⃗形成的六面体顶点有原子， 它可以形成一种晶体。 如果我们将这个六面体的体心加一个原子， 可以形成另一 种晶体， 这个晶体也具备Oh对称性。 把每个面心都加一个原子， 同样也可以在不 破坏点群对称性的基础上生成另一种晶体。 这些晶体无疑是不一样的 （空间群不 一样） ，但点群与晶系这两个概念不能描述这种差别。要描述这种差别，我们需 要引入另一个概念， 就是晶格 （布拉菲格子Bravais Lattice） ， 一共14 种。 说到底， 晶格是一个空间群的概念。 和前面讲点群一样，这14 个晶格也是分别属于前面我们提到的类似于晶系 的东西。 前面是7 种晶系 （Crystal System） ， 每种容纳几个点群， 加在一起容纳32 个点群。这里是7 种晶格系统（Lattice System） ，每种容纳几个布拉菲格子，加在 一起容纳14 个布拉菲格子。前面7 种晶系（Crystal System）和这里的7 种晶格系 统（Lattice System）大部分相同，但略有区别，这里必须解释一下。晶系，是基 于晶体点群对称性对晶体的分类， 它的基础是点群中对称操作的对称元素的共性。 晶格系统，是基于晶体的空间群的对称性对晶体进行的分类，它的基础是晶格。 要详细说明这个差别， 我们先做一个对比。 前面的7 种晶系 （Crystal System） 分别是：Triclinic（三斜） 、Monoclinic（单斜） 、Orthorhombic（正交） 、Tetrogonal （四角） 、Trigonal（三方） 、Hexagonal（六角） 、Cubic（立方） 。这里的7 种晶格 系统 （Lattice System） 分别是： Triclinic （三斜） 、 Monoclinic （单斜） 、 Orthorhombic （正交） 、Tetrogonal（四角） 、Rhombohedral（菱方） 、Hexagonal（六角） 、Cubic （立方） 。 其中， 这两者之间的Triclinic （三斜） 、 Monoclinic （单斜） 、 Orthorhombic （正交） 、Tetrogonal（四角） 、Cubic（立方）这五种是完全一样的。差别出现在 晶系 （Crystal System） 中的Trigonal （三方） 、 Hexagonal （六角） 和晶格系统 （Lattice System）中的Rhombohedral（菱方） 、Hexagonal（六角）上面，它们之间有交叉。 170 以后你们看文献，说到Rhombohedral（菱方） ，你们一定要知道说的是晶格系统 （Lattice System） ， 它指的是Bravais Lattice 的分类； 说到Trigonal （三方或三角） ， 你们一定要知道说的是Crystal System， 它指的是晶系的分类； 说到Hexagonal （六 角） ，即可以指晶系（Crystal System）也可以指晶格系（Lattice System） 。后面我们 会通过表格来详细解释这个差别。这里先通过一个图来说明：点群、晶系、晶格 系统、布拉菲格子之间的关系。通过这种关系，引出14 布拉菲格子。然后再详 细介绍它们之间以及它们与空间群之间的关系。 图3.25 晶系、晶格系统、点群、布拉菲格子、空间群关系示意图 在这个图中，大家注意，中间的地方是一个晶体。它可以根据点群对称性说 属于七种晶系中的一种， 也可以根据空间群对称性说属于其中晶格系统中的一种。 七种晶系可丁可卯得容纳32 种点群，七种晶格（Lattice System）可丁可卯得容纳 14 种晶格（布拉菲格子） 。点群与布拉菲格子之间可以相互组合，从而构造出空 间群。晶系（Crystal System）和晶格系统（Lattice System）处在中心的位置向两边 辐射，两边可交叉。 在产生14 种布拉菲格子的过程中，我们总是针对某个晶格系统里晶胞的六 面体，在体心、面心这些高对称点添加与六面体顶点相同的重复单元（原子或原 子的集合）使晶格发生变化。7 种晶格系统，通过添加这些修饰，我们可以产生 14 种布拉菲格子。对于这种修饰，我们有两个要求： 1. 修饰完了以后不改变晶胞的点群对称性（因此修饰都在体心、单面心、或全 面心这些高对称点进行） ； 2. 修饰完了以后，晶胞不能被进一步简化为更小的可反映点群对称性的晶胞。 下面，我们以三斜和单斜为例，讲一下这14 种布拉菲格子是怎么出来的： 1. 三斜晶系：特点是a ≠b ≠c，α ≠β ≠γ ≠90°。 在这样的晶胞中，如果我们在体心、面心加修饰的话，我们总能找到更小的 反应点群对称性的晶胞，比如下图： 图3.26 三斜格子中等价点示意图 结果就是对三斜晶系来说，修饰起不到任何作用，它所对应的布拉菲格子只 有一个。在以后的讨论中，我们会把只在顶点有东西的格子称为简单格子， 记为P 格子。 2. 单斜晶系：特点是a ≠b ≠c，α = β = 90°，γ ≠90°。 在这样的一个晶系中， 除了P 格子， 如果我们在上下表面的中心加一个修饰， 我们可以回到另一个P 格子，如图： 172 图3.27 单斜格子中等价点示意图一 如果我们在一个侧面的面心加修饰，那我们可以得到一个新的格子，记为A 面心： 图3.28 单斜格子中等价点示意图二 在另一个侧面的面心加修饰是同一个效果。 在体心加修饰的话，我可以把过体心的面当作一个面心，回到侧面面心修饰 的状态： 图3.29 单斜格子中等价点示意图三 如果在三个侧面的面心同时加修饰， 可以通过重新选择a、 b 轴回到体心的情 况： 图3.30 单斜格子中等价点示意图四 由体心回到一个侧面心，则按照前面讨论的方式来进行。 综合一下，对单斜晶体，布拉菲格子就是P 与A 两种。 3. 正交晶系：特点是a ≠b ≠c，α = β = γ = 90°。 P 格子单独存在； 随便在哪个面心加个修饰，也单独存在； 体心加修饰，单独存在； 三个面心同时加修饰，也单独存在； 所以对正交格子，有P、A、I、F 四种布拉菲格子。 依次重复下去，最终我们会得到这样一张图： 174 图3.31 布拉菲格子示意图 一共是14 种布拉菲格子。 其中，前面提到的晶系（Crystal System）与晶格系统（Lattice System）的差别 在这里就有体现。 除了三方晶系与六角晶系， 其它五种晶系Crystal System与Lattice System 两个概念完全重合， 1-9、 12-14 共12 种格子分别属于某一个晶系， 没有模 糊的地方。 对三方和六角晶系（Crystal System） ，在Lattice System 的概念中，它们的集 合对应菱方格子 （Rhombohedral Lattice） 与六角格子 （Hexagonal Lattice） 的集合。 但相互之间个体并不对应个体。在三方晶系（Crystal System）的晶体中，依据格 子的选取， 有些在晶格系统 （Lattice System） 中属于菱方格子 （Rhombohedral Lattice） 的晶体， 有些属于六角格子 （Hexagonal Lattice） 的晶体。 六角晶系 （Crystal System） 的晶体对应的都是晶格系统中属于六角格子（Hexagonal Lattice）的晶体。换句话 说，晶系中的三方晶系材料一部分具有菱方格子，一部分具有六角格子。而六角 晶系材料都是六角格子。用表格的方式表达，就是： 晶系 点群 布拉菲格子 晶格系统 三斜 2（S2、C1） 1 三斜 单斜 3（C2h、C2、C1h） 2 单斜 正交 3（D2h、D2、C2v） 4 正交 四方 7（D4h、C4、S4、D4、 C4v、C4h、D2d） 2 四方 三方 Trigonal 5（D3d、S6、C3、C3v、D3） 1 菱方Rhombohedral 1 六角Hexagonal 六角 Hexagonal 7（D6h、C6、C3h、C6h、 C6v、D6、D3h） 立方 5（Oh、T、O、Th、Td） 3 立方 共7 种 共32 种 共14 种 共7 种 表3.1 晶系与晶格系统相互关系 176 在这个表格中， 左边两列晶系、 点群， 是基于点群概念描述晶体的转动对称性的。 右边两列，布拉菲格子、晶格系统，是基于空间群概念描述晶体的全部对称性的 （包含转动与平移） 。这个差别是前面提到晶系中的三方与六角和晶格系统中的 菱方与六角概念间出现模糊的最本质的原因。 最后一部分我们要讨论的是Hermann-Mauguin 符号，也是目前点群空间群 描述中的国际符号。 目前人们是在对分子对称性描述的时候， 倾向于用熊夫利符 号； 对晶体结构描述的时候， 倾向于用Hermann-Mauguin 符号， 因为后者对平移 对称性的描述更方便。 这个国际符号的基本特征是用不等价的轴或平面来标记晶体的对称性18。这 个轴包含纯转动轴与转动反演轴，纯转动轴是1、2、3、4、6，转动反演轴是1 ̅、 2 ̅、3 ̅、4 ̅、6 ̅。在熊夫利符号中，我们用转动反射面S1、S2、S3、S4、S6，它们与 转动反演轴的关系是：S1 = 2 ̅、S2 = 1 ̅、S3 = 6 ̅、S4 = 4 ̅、S6 = 3 ̅（下去自己画图 理解） 。由这个关系，我们就知道32 种点群在国际符号下分别是： 1. S2、S4、S6：1 ̅、4 ̅、3 ̅； 2. C1、C2、C3、C4、C6：1、2、3、4、6； 3. D2：222； D3：32（有一组同类的2 次轴与主轴垂直） ； D4：422（两组二次轴不同类） ； D6：622（两组二次轴不同类） ； 4. T：23（主轴是2 次轴，3 次轴都同类） ； O：432（主轴4 次轴，3 次、2 次都同类） ； 18前面熊夫利符号的特点是根据不变子群以及陪集中元素的对称元素。 5. C1h：m； C2h：2/m（主轴为2 次轴，m 与之垂直） ； C3h（说白了是S3） ：6 ̅； C4h：4/m； C6h：6/m； 6. C2v：mm2（过主轴两类与水平面垂直的反射面） ； C3v：3m（三个反射面同类） ； C4v：4mm（反射面不同类） ； C6v：6mm（反射面不同类） ； 7. D2h = D2 ∪ID2：2/m 2/m 2/m（三个2 阶轴都有一个与之垂直的反射面，且 相互不同类） ； D4h = D4 ∪ID4：4/m m m（既有与主轴垂直，又有过主轴的发射面，且过主 轴的反射面分两类） ； D3h = D3 ∪I(D6 −D3)： 6 ̅ m 2 （6 ̅为主轴， m 为过主轴反射面， 2 为二阶轴） ； D6h = D6 ∪ID6： 6/m 2/m 2/m （6/m 代表6 阶轴以及与之垂直的反射面， 2/m 代表2 阶轴以及与之垂直的反射面，2/m 有两类） ； 8. D2d = D2 ∪I(D4 −D2)：4 ̅ 2 m； D3d = D3 ∪ID3：3 ̅ 2/m； 9. Th = T ∪IT：2/m 3 ̅； Td = T ∪I(O −T)：4 ̅ 3 m； 10. Oh = O ∪IO：4/m 3 ̅ 2/m； 现在我们把我们知道的晶系，点群的熊夫利、国际符号做个汇总，写成下面 178 这个表格： 晶系（全面对称群） 熊夫利符号 国际符号 三斜系（S2） S2 C1 1 ̅ 1 单斜系（C2h） C2h C2 C1h 2/m 2 M 正交系（D2h） D2h D2 C2v 2/m 2/m 2/m 222 mm2 四角系（D4h） D4h C4 S4 D4 C4v C4h D2d 4/m m m 4 4 ̅ 422 4mm 4/m 4 ̅ 2 m 三方系（D3d） D3d S6 C3 C3v D3 3 ̅ 2/m 3 ̅ 3 3m 32 六角系（D6h） D6h C6 C3h C6h C6v D6 D3h 6/m 2/m 2/m 6 6 ̅ 6/m 6mm 622 6 ̅ m 2 立方系（Oh） Oh T O Th Td 4/m 3 ̅ 2/m 23 432 2/m 3 ̅ 4 ̅ 3 m 表3.2 晶体点群按晶系分类情况 你把这个表看懂了， 在文献中跟晶体点群相关的东西你也就都明白了。 之后 在这个讲义中我想提一下的一个概念是极射赤面投影图。 它是对点群对称性的一 种图形表述。理解这种图，大家抓住6 点，就很容易理解： 1. 极射赤面投影图是什么样的图？ 它是用球形图Sr 的赤道面来描述点群对称性的图。 它的来源是球形图Sr，这个球的南北极连线S-N 是晶体点群的主轴。 极射赤面投影图显示的，是赤道面上的东西，这个需要把球面上的东西往赤 道面做投影，因此叫极射赤面投影图，也叫测地投影图。 2. 怎么描述？ 用两个赤面来描述。一般左边那个描述的是球面上的任意一点，在点群G 的 所有操作下形成的轨道的投影； 右边那个描述的是G 的所有对称操作的对称 元素，在赤道面的投影。 3. 左边那个投影图怎么画？ 它画的必须是Sr 上面的一个普通的点， 在G 中所有元素的作用下， 得到的G 轨道。 所谓普通，就是这个点在G 中的迷向子群必为{E}，这样它的G 轨道才能把 G 中所有对称操作的对称元素反映出来。 当这个轨道上的点P 在Sr 的北半球时， 我们做投影的时候是把这个点与南极 连起来，连线与赤面的交点用实心点表示。 180 图3.32 极射赤面投影示意图一 当这个轨道上的点Q 在Sr 的南半球时，我们做投影的时候是把它与北极连 起来，连线与赤面的交点用空心的圈表示。 当赤面上出现 时，说明这个轨道上在南北半球有可以被镜面反射联系起来 的点。 后面三点是关于右边这个反映对称操作对称元素的图怎么画的： 4. 对反射面，它与南北半球都有交线，我们在赤面上反映的只是它与北半球的 交线的投影，画成粗线。 作为一个结果：与水平面垂直的反射面在赤面上反映为直的粗线；水平的反 射面反映为绕赤道的粗圆周；斜的反射面反映为粗的曲线。 5. 转动轴与转动反演轴，类似，只取它们与北半球的交点做投影。轴水平时， 赤面圆周的相对的两端各出现一次。在这个过程中，转动轴、转动反演轴的 符号分别是： 图3.33 极射赤面投影示意图二 6. 右边图中应标出点群的所有对称操作的对称元素，这个与国际符号中只写出 能够确定点群的最小的对称元素不同。 看几个例子： 1. C1 群，没有非E 对称元素，只画左边一个图就可以。任意一点，轨道上也就 一点，可以画为： 图3.34 极射赤面投影示意图三 2. C2 群，有一个二次轴，Sr 上任意一点的轨道有两个点，转180 度，所以是： 图3.35 极射赤面投影示意图四 3. C1h = C1 ∪I(C2 −C1)，对应的图形是： 图3.36 极射赤面投影示意图五 4. C2h = C2 ∪IC2，对应的图形是： 182 图3.37 极射赤面投影示意图六 5. S4 = C2 ∪I(C4 −C2)，对应的图形是： 图3.38 极射赤面投影示意图七 6. C4h = C4 ∪IC4，对应的图形是： 图3.39 极射赤面投影示意图八 7. D6，对应的图形是： 图3.40 极射赤面投影示意图九 其他的投影图可总结为： 184 图3.41 极射赤面投影示意图汇总，内容摘自韩其智、孙洪洲老师教材，图的制 作由张小伟、张雪峰同学完成 最后讲一下空间群，这是一个什么概念呢？ ⚫ 点群，大家都知道是只考虑了晶体的转动对称性的群，对称操作为R； ⚫ 空间群呢，是在考虑了晶体转动的基础上，同时考虑它的平移不变性，从而 得到的所有对称操作形成的群。对称操作为{R|t ⃗}，晶体在{R|t ⃗}操作下回到自 身。 举个例子，前面我们讲晶体的时候，我们说了它根据所允许的点群，可分成 7 种晶系。 同时， 我们还可以利用7 种晶格系统来划分14 种布拉菲格子。 我们前 面提到，点群和格子之间可以相互组合。 以立方晶系为例， 它允许简单、 体心、 面心三种格子， 同时它允许T、 O、 Th、 Td、Oh五种点群。当这个晶体的点群为T 时，我们可以取简单、体心、面心三种 格子， 这样我们得到的三种{R|t ⃗}的组合一样吗？这里是R 相同， t ⃗不同， 见下图： 图3.42 空间群元素 结果是肯定不一样。同样，晶体点群是O 时，我们可以做同样的处理。以此类 186 推，对立方晶系，我们如果只考虑这种简单的组合，我们是可以得到3 × 5 = 15 种空间群。 现在我们把前面提到的晶系、布拉菲格子、点群的相互关系拿出来，做一个 简单的估计： 晶系 点群 布拉菲格子 晶格系统 简单空间群 三斜 2 1 三斜 2 单斜 3 2 单斜 6 正交 3 4 正交 12 四方 7 2 四方 14 三方 Trigonal 5 1 菱方 Rhombohedral 5 1 六角 Hexagonal 5 7 六角 Hexagonal 7 立方 5 3 立方 15 共7 种 共32 种 共14 种 共7 种 共66 种 表3.3 简单空间群的计算方式 这种最简单的组合可以给出66 种空间群，在这些空间群里面，平移矢量t ⃗是 平移周期性最小重复单元的整数倍。 对所有R， 取t ⃗=0， 也就是不做平移时， {R|0} 都是空间群中的元素。这样的空间群我们称为简单空间群。 不过需要说明的是上面表格里面我们对简单空间群的计算方式还是太简单 了。实际上简单空间群并不止我们上面算的66 种，而是73 种。这个原因是对有 些布拉菲格子，它与点群的结合方式不止一种。比如正交晶系的底心格子，我们 再算布拉菲格子的时候，把底心和侧面面心算成是一种格子，但当它与C2v = C2 ∪I(D2 −C2)结合的时候，如果取主轴为z 轴，这两种格子给出的空间群是不 一样的。对我们这7 种晶系，类似情况可以再多给出7 种简单空间群。所以总的 简单空间群个数是73 个。 除了这些简单空间群，在晶体中，{R|t ⃗}操作还存在另外一种情况，就是t ⃗并 不是空间平移对称性最小重复单元的整数倍， 而是它的分数倍。 比如下面两种情 况： 1. 螺旋轴（Screw axis） ，说的是存在这样的操作，我相对于某个轴转一定角度之 后，再沿着这个轴做一个平移，系统回到了与原来不可分辨的状态。比如： 图3.43 螺旋轴示意图 这样实线和虚线是用来区别z 轴不同的高度， 它们代表的结构单元是一样的。 右边是俯视图。纯平移的周期性是a，但转动60 度再做a/2 的平移也是系统 的对称操作。 它所对应的空间群元素{R|t ⃗}中的R 不可以脱离t ⃗单独存在。 与之 相应，拥有这样操作的空间群也不属于简单空间群。 188 2. 滑移平面 （Slide plane） ， 指的是对一个平面先做反射， 再平移一个t ⃗/m， 比如： 图3.44 滑移面示意图 这里沿x 轴方向平移a 是对称操作，平移a/2 不是。但这个a/2 与镜面反射结 合就是了。 加上这些对称操作后， 空间群的情况就会多很多。 这种群中对称元素平移部 分包含平移对称性分数倍情况的空间群，称为非简单空间群，它有157 种。两者 加在一起是230 种。弄明白这个事情，是19 世纪下半叶欧洲的一些晶体学家、 数学家的研究热点， 但大家可以想象这个其实是非常具有挑战性的。 这个里面最 早做出工作的是个德国数学家，叫Leonhard Sohncke，时间是1879 年，他推出了 66 种空间群。这个和我们前面讲的通过点群和格子的组合生成的66 种不完全一 样，因为当时人们还没有我们现在的这种理解。他的66 种空间群里面，还有一 些重复的情况。但这个工作之后引起了德国人熊夫利和俄国人Fedorov 的注意， 他们之后是在相互独立，但又有一些交流的情况下，分别于1891 年与1892 年发 表专著解释了空间群的230 种情况。这230 种空间群，根据前面提到的晶系、晶 格系统的划分情况是： 晶系 点群 空间群 布拉菲格子 晶格系统 三斜 2 2 1 三斜 单斜 3 13 2 单斜 正交 3 59 4 正交 四方 7 68 2 四方 三方Trigonal 5 7 1 菱方Rhombohedral 18 1 六角 Hexagonal 六角Hexagonal 7 27 立方 5 36 3 立方 共7 种 共32 种 共230 种 共14 种 共7 种 表3.4 空间群分类统计 现在人们需要用到空间群的时候， 一般就是先知道它的国际符号 （Hermann-Mauguin 符号） ， 然后查这个群中的对称操作{R|t ⃗}， 通过这些对称操作， 就可以把 晶体结构彻底搞清。 同时这些对称操作{R|t ⃗}对我们理解系统的电子态分布也非常 有用，下一章我们会做详细介绍。 空间群的完整的对称操作的表格，有一个很好的网站，叫Bilbao Crystallographic Server（在google 中输入Bilbao Crystal 就可以找到）。进入这 个网站后， 有个GENPOS 的选项， 点入后直接输入你的空间群序号， 比如122、 150，所有的对称操作就出来了。在这个网站中，你还可以看具有每个空间群的 晶体的布里渊区的样子，以及很多其它你们以后研究中可以用到的信息。最近， 这个网站还增加了关于后面我们要讲的双群以及这个讲义中并不触及但在磁性 材料物性研究中很有用的磁空间群的内容。 和这个网站差不多，英国也有一个网站19： 19 这个网站是某一年的一个学生（来自北京大学量子中心的陈玉琴同学）告诉我的。 190 里面有类似的功能，这两个网站都建议收藏。 有了这些对称操作后， 在读很多文献的时候， 他们给晶体结构只会给晶体中 不等价的原子的位置与晶体空间群。 你们可以做的事情是对每个原子， 用这些空 间群对称操作操作一遍，如果得到的原子已经出现了，就掠过，如果没有，就保 存，从而得到具体的晶体结构。现在，一些通行的材料模拟方面的软件（比如 Material Studio） 也会提供类似功能。 应该说对于需要应用此方面知识的从业人员， 辅助工具越来越强大了。 只要我们能够理解这些基本原理， 在应用这些知识的时 候，我们拥有前人无法想象的友善的软环境。 3.5 晶体点群的不可约表示 最后一部分是晶体点群的不可约表示。 在学完前两章之后， 我们后面讲任何 一种具体的群，基本都是这个路子。先讲这个群是什么，再分析它有多少种群， 每种群的分类情况， 最后看它们的不等价不可约表示是什么？为什么要看这些不 等价不可约表示，我们下一章《群论与量子力学》中会做详细说明。 找所有晶体点群的不等价不可约表示这个任务乍一看挺吓人的， 因为我们有 32 种晶体点群。但实际上，我们之前讲的一些定理可以在这里大大简化这个问 题。这些定理里面的第一个是第二章里面的：直积群G = G1 ⊗G2的所有不等价 不可约表示， 可以由G1与G2的不等价不可约表示给出。 第二个是我们这章的定理 3.3：点群三种情况，纯转动、纯转动与{E、I}的直积、与某个纯转动同构的不含 I 的第二类点群。这就意味着对点群，只要我们把所有的第一类点群的不等价不 可约表示搞明白了，我们就可以很自然的通过定理3.3 的第二点与第三点得到所 有的第二类点群的不等价不可约表示。 第一类点群有11 种：C1、C2、C3、C4、C6、D2、D3、D4、D6、T、O。下面 我们就一一地把这11 种情况过一遍。 1. 先看Cn 群，这些群都是循环群，我们前面说过，n 阶就有n 个类，也就是n 个不等价不可约表示，对生成元a，它的表示是： Ap(a) = exp [(p −1)2πi/n] 其中p 是不等价不可约表示的index，为1、2、⋯、n 中的一个数。 这样11 种第一类点群的前五种情况就包括了。 2. 第六种情况是D2 群，它有四个元素：e、绕z 轴转π角的操作a、绕x 轴转π 角的操作b、绕y 轴转π角的操作c。 {e、 a}是它的不变子群， 因为z 这个轴不能通过群中其它元素变到x 与y 轴。 同理{e、b}也是，且ab=ba，这样的话D2 ={e、a}⊗{e、b}。{e、a}与{e、b} 的特征标表为： e a 1 1 1 1 1 -1 表3.5 二阶循环群特征标表 那么D2的特征标表就是： e(ee) a(ae) b(eb) c(ab) 1 1 1 1 1 1 1 -1 1 -1 1 1 1 -1 -1 1 1 -1 -1 1 表3.6 D2群特征标表 3. D3群，有三个类，6 个元素，C3 是它的不变子群，含{e、d、f}，D3/C3 同构与 二阶循环群， 于是有了A2 这个表示。 再由正交定理， 得A3。 整体的特征标表 192 是： 1{e} 2{d} 3{a} A1 1 1 1 A2 1 1 -1 A3 2 -1 0 表3.7 D3群特征标表 4. D4 群，五个类，为： {E}、{C4、C4 3}、{C4 2}、{C2 (1)、C2 (3)}、{C2 (2)、C2 (4)} S1 2 + S2 2 + S3 2 + S4 2 + S5 2 = 8 解为： S1 = S2 = S3 = S4 = 1、S5 = 2 D4 群有不变子群：{E、C4、C4 2、C4 3}、{E、C4 2、C2 (1)、C2 (3)}、{E、C4 2、C2 (2)、 C2 (4)}，这样的话由每个不变子群产生的商群都与二阶循环群同构，可以得到 D4 群到二阶循环群的三个同态映射，其中恒等部分相同，非恒等部分给出三 个一维非恒等表示。 之后再由正交关系确认最后一个二维表示，得到特征标表为： 1{E} 1{C4 2} 2{C4} 2{C2 (1)} 2{C2 (2)} A1 1 1 1 1 1 A2 1 1 1 -1 -1 A3 1 1 -1 1 -1 A4 1 1 -1 -1 1 A5 2 -2 0 0 0 表3.8 D4群特征标表 5. D6，有12 个元素，为： {E、C6 2、C6 4、C2 (1)、C2 (3)、C2 (5)；C6 1、C6 3、C6 5、C2 (2)、C2 (4)、C2 (6)} 其中前六个是D3 群，它是D6 的不变子群，后面6 个是它的陪集。在这个陪 集里面C6 3自成一类（它与其它六次轴转动角度不同，与二次轴转动虽角度相 同但转轴无法通过群中元素联系起来） ，因此{E、C6 3 }也是一个不变子群， D6=D3⊗{E、C6 3}。 6. T 群，有{E}、{C2 (1)、C2 (2)、C2 (3)}、{C3 ′ 、C3 ′′、C3 ′′′、C3 ′′′′}、{C3 ′2、C3 ′′2、C3 ′′′2、 C3 ′′′′2}四个类。 S1 2 + S2 2 + S3 2 + S4 2 = 12 解为： S1 = S2 = S3 = 1、S4 = 3 T 的不变子群是：{E、C2 (1)、C2 (2)、C2 (3)}，把T 和它做商群，可以产生一个三 阶循环群，三阶循环群有三个一维表示，所以给出A1 到A3。最后一个三维 表示由正交关系得到： {E} 3{C2 (1)} 4{C3 ′ } 4{C3 ′2} A1 1 1 1 1 A2 1 1 ε ε2 A3 1 1 ε2 ε A4 3 -1 0 0 表3.9 T群特征标表 7. 最后一个，是这节最难的一个，O 群，有24 个元素，分五类： {E}、 {C2 (1)、C2 (2)、⋯、C2 (6)}、 {C3 ′ 、C3 ′′、C3 ′′′、C3 ′′′′、C3 ′2、C3 ′′2、C3 ′′′2、C3 ′′′′2}、 {C4 ′ 、C4 ′′、C4 ′′′、C4 ′3、C4 ′′3、C4 ′′′3}、{C4 ′2、C4 ′′2、C4 ′′′2} S1 2 + S2 2 + S3 2 + S4 2 + S5 2 = 24 解为： S1 = S2 = 1、S3 = 2、S4 = S5 = 3 194 O 有不变子群T，包含： {E、C3 ′ 、C3 ′′、C3 ′′′、C3 ′′′′、C3 ′2、C3 ′′2、C3 ′′′2、C3 ′′′′2、C4 ′2、C4 ′′2、C4 ′′′2} 因此通过O/T 商群与二阶循环群的同构，可以得到两个一维表示。 二维表示怎么办？我们可以利用前面讲的诱导表示的概念。 我们想用T 群，这个O 群的子群的非恒等的一维表示A2，得到O 群的一个 二维表示。我们要做的事情，就是把O 分成T 与它的陪集，O={g0T、g1T}， 这里g0 是单位元素，g1 有多种选择，我们取g1=C2 (1)。 T 群与O 群的关系是： 我们利用诱导表示的表示矩阵： UB(g) = (B ̇ (g0gg0 −1) B ̇ (g0gg1 −1) B ̇ (g1gg0 −1) B ̇ (g1gg1 −1)) 其中： B ̇ (giggj −1) = { A2(giggj −1)ifgiggj −1 ∈T 0 otherwise 而： A2(giggj −1) = { 1 if giggj −1 ∈{E、C4 ′2、C4 ′′2、C4 ′′′2} 𝜀 if giggj −1 ∈{C3 ′ 、C3 ′′、C3 ′′′、C3 ′′′′} 𝜀2 if giggj −1 ∈{C3 ′2、C3 ′′2、C3 ′′′2、C3 ′′′′2} 再利用公式（2.12） ： 𝜒U(g) = 1 𝑚∑trB ̇ (tgt−1) 𝑡∈O 这里m 是T 的阶，我们很容易得到： 𝜒𝑈(E) = 1 12 ∑trB ̇ (tEt−1) 𝑡∈O = 1 12 ∙24 ∙1 = 2 𝜒𝑈(C3 ′ ) = 1 12 ∑trB ̇ (tC3 ′ t−1) 𝑡∈O = 1 12 ∙12 ∙(𝜀+ 𝜀2) = −1 𝜒𝑈(C4 ′ ) = 1 12 ∑trB ̇ (tC4 ′ t−1) 𝑡∈O = 0 ⋯⋯tC4 ′ t−1 ∉T 𝜒𝑈(C4 ′2) = 1 12 ∑trB ̇ (tC4 ′2t−1) 𝑡∈O = 1 12 ∙24 ∙1 = 2 ⋯⋯tC4 ′2t−1 ∈{C4 ′2、C4 ′′2、C4 ′′′2} 𝜒𝑈(C2 (1)) = 1 12 ∑trB ̇ (tC2 (1)t−1) 𝑡∈O = 0 ⋯⋯tC2 (1)t−1 ∉T 这个诱导表示对O 群的五个类而言特征标分别是： 2、 0、 2、 -1、 0。 做内积， 满足归一条件，为不可约表示。 {E} 6{C4 ′ } 3{C4 ′2} 8{C3 ′ } 6{C2 (1)} A1 1 1 1 1 1 A2 1 -1 1 1 -1 A3 2 0 2 -1 0 A4 3 χ4(C4 ′ ) χ4(C4 ′2) χ4(C3 ′ ) χ4(C2 (1)) A5 3 χ5(C4 ′ ) χ5(C4 ′2) χ5(C3 ′ ) χ5(C2 (1)) 表3.10 O群特征标表 最后就剩下两个三维表示了。求法和上面一样，还是利用诱导表示。不过这 196 里用的是T 的三维表示，导出的O 的表示是六维的。它的特征标是： 𝜒𝑈(E) = 1 12 ∑trB ̇ (tEt−1) 𝑡∈O = 1 12 ∙24 ∙3 = 6 𝜒𝑈(C3 ′ ) = 1 12 ∑trB ̇ (tC3 ′ t−1) 𝑡∈O = 0 ⋯⋯T 的三维表示中C3 ′的特征标也是零 𝜒𝑈(C4 ′ ) = 1 12 ∑trB ̇ (tC4 ′ t−1) 𝑡∈O = 0 ⋯⋯tC4 ′ t−1 ∉T 𝜒𝑈(C4 ′2) = 1 12 ∑trB ̇ (tC4 ′2t−1) 𝑡∈O = 1 12 ∙24 ∙(−1) = −2 ⋯⋯tC4 ′2t−1 ∈{C4 ′2、C4 ′′2、C4 ′′′2} 𝜒𝑈(C2 (1)) = 1 12 ∑trB ̇ (tC2 (1)t−1) 𝑡∈O = 0 ⋯⋯tC2 (1)t−1 ∉T 这个六维表示肯定是可约的。如果它是两个A4 或两个A5 的直和，对应的A4 或A5 的特征标为：3、0、-1、0、0，做内积的话不等于1，不是不可约表示。 还有一种可能是它是A4与A5的直和，如果是这样的话，我们就有： χ4(C4 ′ ) + χ5(C4 ′ ) = 0 χ4(C4 ′2) + χ5(C4 ′2) = −2 χ4(C3 ′ ) + χ5(C3 ′ ) = 0 χ4(C2 (1)) + χ5(C2 (1)) = 0 对这样的条件，有一种解是： χ4(C4 ′ ) = 1、χ4(C4 ′2) = −1、χ4(C3 ′ ) = 0、χ4(C2 (1)) = −1 χ5(C4 ′ ) = −1、χ5(C4 ′2) = −1、χ5(C3 ′ ) = 0、χ5(C2 (1)) = 1 这个解是满足特征标归一条件的，正交性也满足，所以对应两个三维不可约 表示（填到上面那个表中） 。这样O 群的不等价不可约表示特征标表就也出 来了。 最后一个需要提一下的，你们在看一些文献上给出的点群特征标表的时候， 经常会看到这样的例子。比如Oh 特征标表，在有些文献中，不可约表示会被标 识为会Γ 1、 Γ 2、 Γ 12、 Γ 15、 Γ 25、 Γ 1 ′、 Γ 2 ′、Γ 12 ′ 、 Γ 15 ′ 、 Γ 25 ′ ，其中Γ 𝑖 ′有时也会被写成Γ 𝑖 +。 在另一些文献中，不可约表示又被标识为A1 +、A2 +、E+、T 1 +、T2 +、A1 −、A2 −、E−、 T 1 −、T2 −。为了给大家一个直观的印象，下面是两个截图。 图3.45 Oh 特征标表截图一 与 198 图3.46 Oh 特征标表截图二 第一种表述方法遵循的是Wigner 在1936 年和Bouckert、Smoluchowski 合写的 一篇文章中的习惯。 他们在这里的基本逻辑是对一些高对称的空间群的晶体， 如 果我把它的布里渊区画出来的话， 那么它的布里渊区里面不同的点具有不同的点 群对称性。比如一个晶体是简立方的晶格，那么它的布里渊区就是这个样子： 图3.47 立方系空间群布里渊区 （立方系，简单晶格对应的空间群的布里渊区，第221 个空间群） 这样的一个布里渊去里面，不同的点的点群对称性分别如下。因为这个原因，对 Oh这个点群，由于只有Γ 点具有它的对称性，所以在标识Oh的不可约表示的时 候，用了Γ。这个是做固体物理的人比较喜欢的习惯。 图3.48 立方系空间群布里渊区高对称k 点的点群对称性 另一拨人（比如原子分子物理、理论化学专业） ，可能又比较喜欢用A、B、E、 T 来标记点群不可约表示，A、B 是一维的，E 是二维的，T 是三维的。当然，做 固体物理的很多时候也会使用A、B、E、T 这种标识。因此，这个就是一个习惯 问题，背后有这样一个故事，不绝对。 同时，还有上标。第二个表中，上标是对应奇偶对称性的。对A、B、E、T 这种表述，上面的例子是用‘+’ 、 ‘-’这些上标给区分出来。在有些其它的文献 中，也会用Ag、Au 这类g 和u 的下标来区分。这里g 是gerade，德语even 的意 思，u 是ungerade，德语odd 的意思，对应的还是宇称。第一个表中，Γ 带不带 撇与第二个表中的正负号不严格对应。 最后，在basis function 那一栏，好多文献中会给出一些x、y、z 的齐次函数 的例子。 后面我们会讲一个量子力学系统的本征态往往对应一个不可约表示的基， 这样的话给出这些齐次函数的例子就很重要了。 这个后面我们会细讲， 这里大家 先记住是不可约表示的一些比较典型的基就可以了。 200 3.6 习题与思考 1. 设O是三维实正交群一个元素， Ck ⃗ ⃗⃗(θ)是其中的某纯转动操作，Sk ⃗ ⃗⃗(θ)是其中的 某转动反射操作。问OCk ⃗ ⃗⃗(θ)O−1是什么样的操作？OSk ⃗ ⃗⃗(θ)O−1是什么样的操作？ 2. 某点群有个垂直xy 水平面的二阶轴C2，和过C2的反射面σv，它是什么点群？ 以{𝑥𝑦, 𝑥𝑧, 𝑦𝑧}为基，它的表示是什么？它可约吗？如可约，请约化。 3. 以{2𝑥𝑦, 𝑥2 −𝑦2}、 {𝑅𝑥= 𝑦𝑝𝑧−𝑧𝑝𝑦, 𝑅𝑦= 𝑧𝑝𝑥−𝑥𝑝𝑧}为基， 求D3的表示矩阵。 4. 某点群有奇数阶转动轴S2𝑛+1， 证明： 必存在独立的转动轴C2𝑛+1及水平反射面。 5. 4n 阶转动反射轴S4𝑛为生成元，能否产生反演操作I？ 6. 求出二维实空间中所有点群。 7. 用熊夫利符号， 说出如下点群： 1） 基于C6群中， 增加空间反演操作； 2） 在C5h 群中，去掉所有转动反演操作；3）基于Td群中，增加空间反演操作。 8. 1)上Bilbao Crystallographic Server 的网站查基于D2点群一个空间群P222 的所 有对称操作，写出来。2)某篇关于晶体结构的文章在给结构的时候只给出非等价 的原子的位置。 如果给出的这个原子是C， 它在晶胞内的相对坐标是(0.125, 0.125, 0.125)， 写出晶胞内所有原子的位置。 3） 如果这个空间群是P2122， 情况又怎样？ 4）空间群是C222，情况又怎样？（本题写相对坐标就可以） 9. 一个立方体， 点群是什么？沿对角线方向拉伸， 点群变成什么？如果它是完美 单晶，这个晶体会由什么晶系变成什么晶系？ 10. 参考附录内容中三方晶系包含的空间群，体会本节描述晶系、点群、布拉菲 格子、晶格系统那个表格，指出哪些是基于菱方的晶格系统？ 11. 指出下列分子的点群： 12. 基于本节中已知第一类点群特征标表， 不要参考附录， 完整地写出C4h群的特 征标表（要求过程，同时写明每个类包含哪些元素） 。 13. 基于本节中已知第一类点群特征标表， 不要参考附录， 完整地写出D4d群的特 征标表（要求过程，同时写明每个类包含哪些元素） 。 202 第四章 群论与量子力学 现在我们把我们这门课的理论基础以及我们在以后面对实际材料的时候遇 到的最为重要的点群、 空间群讲完了。 我们 《群论一》 这门课还有两类群需要讲： 转动群与置换群。在讲之前，以我自己学这门课和讲这门课的经验来看，最大的 困难是我们在学了很多很难理解的概念和定理之后并不知道它有什么用？这个 坑必须现在先填上。 这一章的作用就是填坑。方法是用一些我知道的量子力学、固体物理、原子 分子物理中的例子来理解群论到底怎么去用？这里面量子力学、 固体物理、 原子 分子物理是物理，群论是数学，由于量子力学在物理这边起最基础的作用，这一 章的题目我们把它叫作《群论与量子力学》 。 具体而言，作为物理系、化学系的学生，面对微观世界，我们是需要用量子 力学的概念去理解其物理、化学性质的。我们观测到的任何东西，最终解释的时 候都会落到你所观测的系统的一些本征态上。 以分子或凝聚态体系为例， 这些本 征态可以是电子态， 可以是声子态， 也可以是它们之间的耦合形成的其它准粒子 态。 这些态是如何产生的？它与你所观测的系统的对称性存在什么样的关联？你 观测的系统在一个外界扰动下， 从一个本征态向另一个本征态跃迁会呈现什么样 的规律？这些都是我们在以后的科研中要遇到的实际问题。 在我们花费了巨大精 力学习完这门课后， 如果以后在科研中遇到类似问题， 不懂得如何用对称性原理 去理解，这门课就白学了。也正是因为这个原因，我们把《群论与量子力学》这 章当成我们这门课里重点的重点。 要讲的内容分为八节。 第一节哈密顿算符群与相关定理是总体的讲量子力学 与对称性之间的关系，其中的概念与定理是对我们这一章内容进行理解的基础。 第二节讲微扰引起的能级分裂， 是一个具体的系统对称性的变化导致物性变化的 例子。 这一节具体会说明如何用对称性的语言去描述和理解这种现象。 第三节讲 投影算符与久期行列式的对角化。 其中投影算符是数学基础部分， 它将教会我们 如何让一组基函数具备物理系统的对称性。 久期行列式的对角化是对称化的基函 数在解薛定谔方程的时候的应用， 它可以简化这个求解过程。 第四节讲一些矩阵 元定理与选择定则。它说的是本来系统有个哈密顿量，有一系列的本征波函数。 根据我们第一节讲的内容， 我们应该知道这些本征态基函数能反映系统的对称性。 现在我加了个微扰， 系统会在原来哈密顿量的本征态之间发生一些跃迁。 第四节 会告诉我们如何利用对称性的知识去理解哪些跃迁可以发生、 哪些跃迁不能发生？ 第五节我想讲一下你们以后做实验应该会经常接触到的红外谱、 拉曼谱、 和频光 谱。 第六节与第七节回到固体物理， 从平移与转动两个角度去讲解晶体中的这些 对称性对晶体能带的影响。 最后一节讲时间反演对称性。 这些内容只是根据我在 过去五年中的课程进度放入的内容， 使得讲义本身可以在一个学期讲完。 更多应 用建议大家仔细阅读Dresselhaus 那本书。 4.1 哈密顿算符群与相关定理 这部分讨论的基础是哈密顿量在系统的对称操作下的变换性质， 把它搞明白 了，所有的概念与定理就清楚了。而要明白哈密顿量的变换性质，首先就要清楚 哈密顿量是什么？ 我自己是一个做计算，不做理论的人。按我的简单理解，哈密顿量应该有这 样的性质： 1. 它是一个算符； 2. 在坐标表象下，依赖于坐标（这个坐标是个广义的坐标，暂记为x ⃗ ⃗） ； 204 3. 我们一般把它记为H ̂(x ⃗ ⃗)； 4. 它与一个物理系统对应。 因为它依赖坐标， 那么如果我们对它的变量空间做变换g 的话， 坐标会变成 gx ⃗ ⃗。这个时候，如果H ̂(x ⃗ ⃗) = H ̂(gx ⃗ ⃗)（注意，这个要求是整个哈密顿量不变，包含 动能项与势能项） ，我们称g 是保持哈密顿量不变的一个操作。换句话说，g 是 让系统回到与之前不可分辨状态的一个操作。 类似对称操作的存在会给系统带来什么样的性质呢？既然g 是作用到x ⃗ ⃗上的 一个变换，对于波函数所在希尔伯特空间的任意一个向量φ(x ⃗ ⃗)，那它肯定对应一 个线性变换算符P ̂ g。根据我们之前讲过的变换规则，P ̂ g作用到φ(x ⃗ ⃗)上后果为： P ̂ gφ(x ⃗ ⃗) = φ(g−1x ⃗ ⃗)。 如果我们定义g 这个作用到x ⃗ ⃗上的变换与f 这个作用到x ⃗ ⃗上的变换的乘积fg 所对应的线性变换算符P ̂ fg = P ̂ fP ̂ g，那么我们就会有： P ̂g−1P ̂ gφ(x ⃗ ⃗) = P ̂g−1 (P ̂ gφ(x ⃗ ⃗)) = P ̂g−1φ(g−1x ⃗ ⃗) = φ(x ⃗ ⃗) 这个等式对希尔伯特空间中的任意一个函数φ(x ⃗ ⃗)都成立，因此：P ̂g−1 = P ̂ g −1。 这样的话，对H ̂(x ⃗ ⃗)φ(x ⃗ ⃗)，就会有： H ̂(x ⃗ ⃗)φ(x ⃗ ⃗) = P ̂ gP ̂g−1H ̂(x ⃗ ⃗)φ(x ⃗ ⃗) = P ̂ gH ̂(gx ⃗ ⃗)φ(gx ⃗ ⃗) = P ̂ gH ̂(gx ⃗ ⃗)P ̂g−1φ(x ⃗ ⃗) = P ̂ gH ̂(gx ⃗ ⃗)P ̂ g −1φ(x ⃗ ⃗) 对任意一个希尔伯特空间中的函数φ(x ⃗ ⃗)成立。因为这个原因，此哈密顿量满足 H ̂(x ⃗ ⃗) = P ̂ gH ̂(gx ⃗ ⃗)P ̂ g −1这个条件。而我们之前说过g 是保持系统哈密顿量不变的一 个操作，满足：H ̂(x ⃗ ⃗) = H ̂(gx ⃗ ⃗)，因此我们可进一步得到：H ̂(x ⃗ ⃗) = P ̂ gH ̂(x ⃗ ⃗)P ̂g−1 = P ̂ gH ̂(x ⃗ ⃗)P ̂ g −1。这意味着： P ̂ gH ̂(x ⃗ ⃗) = H ̂(x ⃗ ⃗)P ̂ g 因此， 如果g 是保持哈密顿量𝐇 ̂(𝐱 ⃗ ⃗)不变的坐标空间的变换， 那么它所对应希尔伯 特空间的函数变换算符P ̂ g与𝐇 ̂(𝐱 ⃗ ⃗)互易。 这个性质是我们这章经常会用到的一个性质， 由它可以得到很多的定理。 在 讲这些定理之前，我们先看两个概念，分别是哈密顿算符的群（既系统对称群） 与哈密顿算符群（对称变换所对应的函数变换算符的群） 。 定义4.1 哈密顿算符的群：所有保持哈密顿量𝐇 ̂(𝐱 ⃗ ⃗)不变的变换g 的集合形成的 群，称为哈密顿算符的群，或薛定谔方程的群，或系统对称群，记为： 𝐆𝐇= {𝐠|𝐇 ̂(𝐠𝐱 ⃗ ⃗) = 𝐇 ̂(𝐱 ⃗ ⃗)} 定义4.2 哈密顿算符群（也叫薛定谔方程群） ：由哈密顿算符的群中群元对应的 函数变换算符形成的群，称为哈密顿算符群，或薛定谔方程群，记为： 𝐏𝐆𝐇= {𝐏 ̂𝐠|𝐠∈𝐆𝐇} 由于系统对称群的群元是使系统回到与之不可分辨状态的变换， 它反映的是 系统的全部对称性。 哈密顿算符群， 是系统对称群中群元所对应的函数变换算符 形成的群，它必须与某线性空间对应。它们的关系是：哈密顿算符群是系统对称 群的一个表示（这个表示依赖于哈密顿算符群的线性空间） 。 后面，为了讲起来方便， “哈密顿算符群”与“哈密顿算符群表示”这两个 名词我们会混着用，因为哈密顿算符群本身就是一个表示，系统对称群的表示。 我们讲哈密顿算符群的时候， 我们指的就是系统对称群的哈密顿算符群表示。 在 一组基下，它体现为一个矩阵群，这组基往往是一组本征态。 基于这两个定义， 我们去看群的表示理论与量子力学的联系。 其中第一个是 关于具有相同本征能量的本征函数的。 定理4.1 哈密顿量算符𝐇 ̂(𝐱 ⃗ ⃗)的具有相同本征能量的本征函数， 构成哈密顿算符群 206 表示的基函数。 （这个意思就是说H ̂(x ⃗ ⃗)不是有一系列本征能级吗？每个能级上， 都有一个或几个 本征态。 现在以能级为基本单元， 把它上面的本征态的线性组合为向量形成线性 空间。如果这个能级不简并，这个线性空间一维；如果n 重简并，这线性空间n 维。 这个定理说的是这些线性空间都是这个系统的哈密顿算符群表示的表示空间， 以其中本征态为基，它体现为一个矩阵群。系统对称群与之同态） 证明： 取一个本征能级En， 设它是l 重简并的。 这样的话就有l 个线性无关的本征函数， Ψ𝑖(x ⃗ ⃗)，其中i 从1 到l。它们形成的线性空间记为WH。 对∀P ̂ g ∈P GH，由于有P ̂ gH ̂(x ⃗ ⃗) = H ̂(x ⃗ ⃗)P ̂ g，因此P ̂ g作用到∀Ψ𝑖(x ⃗ ⃗)上，都有： H ̂(x ⃗ ⃗)P ̂ gΨ𝑖(x ⃗ ⃗) = P ̂ gH ̂(x ⃗ ⃗)Ψ𝑖(x ⃗ ⃗) = P ̂ gEnΨ𝑖(x ⃗ ⃗) = EnP ̂ gΨ𝑖(x ⃗ ⃗) 这也就意味着P ̂ gΨ𝑖(x ⃗ ⃗)仍然是H ̂(x ⃗ ⃗)本征值为En的本征函数。 之前我们说过En是l 重简并的，对应的线性空间为WH，那么P ̂ gΨ𝑖(x ⃗ ⃗)必为这个空 间中的向量，对应： P ̂ gΨ𝑖(x ⃗ ⃗) = ∑∆𝑖′𝑖 (n)(g)Ψ𝑖′(x ⃗ ⃗) 𝑙 𝑖′=1 {∆(n)(g)}这个矩阵群就是l 维的哈密顿算群表示以Ψ𝑖(x ⃗ ⃗)为基形成的矩阵群。 （证毕） 这里说明一下。 这个定理说的是一个能级上的本征态形成的线性空间可以承 载这个哈密顿算符群表示。这个表示，要么不可约，要么可以约化为一系列不可 约表示的直和。 但是对这个里面的每个不可约表示， 它们的基对应的本征态的本 征能量是一样的。 跟这个定理对应，还有一个定理，说的是不同能级上的本征态，如果通过线 性组合形成表示空间的话，承载的不可能是一个不可约表示。总结起来，是下面 一句话。 定理4.2 承载哈密顿算符群不可约表示的本征函数必属于同一能级。 证明： （反证） 设H ̂(x ⃗ ⃗)的l 个本征函数Ψ𝑖(x ⃗ ⃗)构成第α个哈密顿算符群不可约表示，而它们的能量 并不相同。取最简单的例子，前𝑙−1个属于能级E，最后一个属于能级E′。也就 是E𝑖= E，for 𝑖≤𝑙−1；E𝑖= E′，for 𝑖= 𝑙。 这样的话，由薛定谔方程，我们知： H ̂(x ⃗ ⃗)Ψ𝑙(x ⃗ ⃗) = E′Ψ𝑙(x ⃗ ⃗) 取∀P ̂ g ∈P GH，作用到这个方程两边， 左边= P ̂ gH ̂(x ⃗ ⃗)Ψ𝑙(x ⃗ ⃗) = H ̂(x ⃗ ⃗)P ̂ gΨ𝑙(x ⃗ ⃗) = H ̂(x ⃗ ⃗) ∑∆𝑖𝑙 (α)(g)Ψ𝑖(x ⃗ ⃗) 𝑙 𝑖=1 其中∆𝑖𝑙 (α)(g)是第α个不可约表示的第𝑖行、第l 列。它继续等于： ∑∆𝑖𝑙 (α)(g)E𝑖Ψ𝑖(x ⃗ ⃗) 𝑙 𝑖=1 对右边， 右边= E′P ̂ gΨ𝑙(x ⃗ ⃗) = E′ ∑∆𝑖𝑙 (α)(g)Ψ𝑖(x ⃗ ⃗) 𝑙 𝑖=1 而左右两边是由线性无关的矢量Ψ𝑖(x ⃗ ⃗)组成的线性组合。如果相等，每个分量都 应该相等，这样的话就有： ∆𝑖𝑙 (α)(g)E𝑖= ∆𝑖𝑙 (α)(g)E′ 对任意i 都成立，但是在𝑖≤𝑙−1时，E𝑖≠E′，这样的话∆𝑖𝑙 (α)(g)必为零。这也就 是说∆𝑖𝑙 (α)(g) = 0对∀g成立，进而表示∆α可约。这个与已知矛盾。 （证毕） 208 这两个定理合在一起， 给了我们一个什么样的图像呢？就是对一个体系， 它 有哈密顿量。 这个哈密顿量的对称性会告诉我们这个系统的系统对称群。 这个系 统对称群有一系列的不等价不可约表示。 对薛定谔方程的一个本征能级， 它上面 的本征态波函数是可以通过相互组合构成这个系统对称群所对应的哈密顿算符 群表示空间的。这个表示空间可以可约也可以不可约，可约时，它可以化为一系 列不可约表示的直和。 但是对某一不可约表示， 承载它的本征态波函数的本征能 量必相同。 其中， 由后者带来的不同本征态之间本征能量相同的这种情况， 我们称为必 然简并，它是由对称性引起的。如某个能级上的表示可约，那这里不同不可约表 示间能量相同的状况我们称为偶然简并。 这些讨论的物理基础是态叠加原理。 在量子力学中， 我们可以用某个本征态 它所承载的不可约表示来标识它。 为了给大家一个你们能记住， 以后也应该能用 上的具体的例子， 我们看一下Si 的能带。 Si 大家都知道晶体结构就是一个fcc 的 晶格上，除了在原点放个Si，沿立方体对角线1/4 处，再放一个。它的布拉菲格 子是面心立方。费米面附近的能带是这个样子： 图4.1 Si 的能带示意图 与我们讨论相关的是我们对这些高对称的k 点的本征态的标记。以Γ 点、 X 点、K 点、L 点以及它们之间的连线为例。由于这个晶体的布里渊区如下： 图4.2 Si 的第一布里渊区示意图 Γ 点、X 点、K 点的点群对称性分别是：Oh、D4h、和C2v。Γ 点与X 点连线上的 点的点群对称性是C4v，X 点与K 点连线就是C1。为什么倒空间中相差整数个倒 格矢的k 点所对应的准粒子激发态相互等价， 我们会在后面第六、 第七节详细介 绍。这里大家先接受一下。 210 按照我们前面讲的Oh = O ⊗{E、I}， O 有两个1 维、 一个2 维、 两个3 维不 可约表示，所以Oh的不可约表示有四个1 维、两个2 维、四个3 维的。D4h = D4 ⊗{E、I}，D4有四个1 维和一个2 维不可约表示，所以D4h的不可约表示是八 个1 维和两个2 维的。C2v与D2同构，它的不可约表示是两个1 维的。C4v与D4同 构，它的不可约表示是四个1 维和一个2 维的。C1只有一个1 维不可约表示。 与之相应，我们看能带上的点，基本也都能反映这些特征。比如Γ 点，它都 是用Oh特征标表中的不可约表示标识的： 图4.3 文献中用到的Oh群特征标表 从Γ 到X 连线上的点记作∆，对称群从Oh变成了C4v。C4v没有三维不可约表 示，特征标表为： 图4.4 文献中用到的C4v群特征标表 与之相应，费米面下本来3 重简并的点分裂为1 个2 重简并（上面）和1 个1 重 简并（下面）的态。 到了X 点，对称群变成了D4h，有八个１维与两个２维不可约表示，特征标 表为： 图4.5 文献中用到的D4h群特征标表 这个时候， 大家可以注意。 费米面以下第二个能带在X 点， 是一个二重简并。但 它承载的不可约表示， 却是两个1 维不可约表示。 这个时候的简并在空间群的点 群概念下是偶然简并20。 从X 点到K 点， 不可约表示全是一维。 在K 点对称性是C2v， 相应的特征标 表为： 图4.6 文献中用到的C2v群特征标表 这里不可约表示用Σ 来标识， 是因为Σ、 U 这些布里渊区的点对称性也都是C2v。 对应的简并也相应的全部消除，所有的能级都不简并，承载一维不可约表示。 20在考虑另外特别滑移面后，此简并也源于对称性。此部分内容超出课程按一个学期的课时规 划，详见Dresselhaus 讲义12.5 节。谢谢刘奇航老师提醒！ 212 到这个地方， 我们量子力学中的本征态与哈密顿算符群对称性的逻辑关系就 讲完了。在这个逻辑关系中，偶然简并并没有被明确禁止。但是当你遇到偶然简 并的情况的时候， 不要理所当然的认为它就是偶然简并。 因为它同时可能意味着 你系统的对称性没有找全。 当你系统对称性找全的时候， 高维的不可约表示在新 的系统对称群中就可以存在了。这时，如果你再回过头看你本征态之间的简并， 它们经常会对应新的系统对称群的不可约表示。 也就是说新的对称性使得可约表 示变成了不可约表示，相应的偶然简并也变成了必然简并。这种情况，在我们量 子力学的发展过程中是经常出现的。在40、50 年代，当时物理学研究的前沿是 求很多量子力学模型的解。 往往你解出了一个好的模型， 你就完成了一个很好的 博士论文。 在当时的这些文章中， 你们会经常看到一些关于accidental degeneracy 和hidden symmetry 的讨论，说的就是这样一个事情。就是你原来认为的偶然简 并， 在你真正理解了这个量子力学系统的对称性之后， 你会发现这个简并其实必 然（对称性升高，其所要求的简并度增加） 。 这个里面最典型的一个例子是氢原子电子的本征态求解。 氢原子大家去想它 的对称性是什么？它与其它含有多个电子的原子去对比，对称性有不同吗？ 最直接地去想，我们可能都会认为没啥不同，都是球对称。相应的系统对称 群是SO(3)群。如果是SO(3)群，我们下一章会讲，它的不可约表示的基是球谐 函数，对一个特定的l，不可约表示的简并度为2l+1。我们原子物理中给出的原 子轨道的分布一般也都是这样的，能量从低到高，一般原子分别是1s、2s、2p、 3s、3p、4s、3d、4p、… 图4.7 SO(3)群能级简并示意图 这个大家应该都很熟悉，也不会有什么疑问。 但是对于氢原子， 如果我们看它的能级分布图的话， 我们会看到这样的现象： 图4.8 氢原子能级简并示意图 这个图和上面那个图就有很大的差别了，因为相同的n 本征能量都相同。 为什么会存在这个差别， 这个在高量课上一般会讲。 氢原子中库仑势是1/r， 这个1/r 很特别，因为早期在天文学的研究中，人们已经知道，在一个具有F ⃗⃗= −𝑘 𝑟2 ⁄ r ̂形式的引力场中， 存在一个Laplace-Range-Lenz 守恒量A ⃗ ⃗⃗= p ⃗⃗× L ⃗⃗−m𝑘r ̂， 其中L ⃗⃗= r ⃗× p ⃗⃗。更具体的这些向量的形式如下图： 214 图4.9 Laplace-Range-Lenz 守恒量示意图 由于这个守恒量的存在，Noether 定理告诉我们它（这个守恒量）必对应一个对 称性。这个导致的结果就是氢原子中，我守恒量比简单球对称体系多，所以我实 际的对称性也比球对称体系高。 我们把这个对称性称为一种动力学对称性， 相应 的氢原子的对称群是SO(4)。 而SO(4)群的简并就符合上面那个图的能级分布了。 对其它原子，因为电子之间相互作用，这个1/r 的对称性不成立，对应的群还是 SO(3)群21。 这一节总结起来就是量子力学中的简并度与对称性息息相关。 我们的研究中， 经常遇到的是因为对称性的升高或降低所引起的能带的合并与劈裂。 偶然简并有 时可以出现， 但它出现的时候我们往往要非常小心， 因为它经常意味着我们对系 统哈密顿量的对称性没有找全。 4.2 微扰引起的能级分裂 有了上一节的理论基础，这一节我们要讲的微扰引起的能级劈裂很好理解。 说的是这样一个事情： 本来系统不是有个哈密顿量H ̂0(x ⃗ ⃗)， 它的系统对称群是GH0， 21关于氢原子的动力学对称性的详细、严格的讨论，请阅读曾谨言老师《量子力学》卷II 的7.1 节。 那一节， 曾老师分氢原子的经典力学描述、 二维氢原子的SO(3)对称性、 三维氢原子的SO(4) 对称性、n 维氢原子的SO(n+1)对称性对这个话题进行了严格的讨论。 它的本征态我们用GH0的不等价不可约表示来标识。 现在我们给系统加了一个微扰H ̂′(x ⃗ ⃗)，它的哈密顿量变成了H ̂(x ⃗ ⃗) = H ̂0(x ⃗ ⃗) + H ̂′(x ⃗ ⃗)，相应的系统对称群变为GH。 显然， 新的本征态就需要用GH的不等价不可约表示来标识了。 我们这部分讨 论要干的事情， 就是在不解薛定谔方程的情况下， 看微扰H ̂′(x ⃗ ⃗)对能级简并情况的 影响。具体情况分两种： 1. GH0的对称性比较高，加入微扰后，对称性降低了。新的系统对称群GH是GH0 的子群。 这种情况下， 原来简并的能级上的态构成GH0的不可约表示， 在加入 微扰之后，它们对应的GH的表示就不一定不可约了。相应的简并不被对称性 保护，从而引起由对称性破缺诱发的能级劈裂。 2. H ̂′(x ⃗ ⃗)的对称性大于等于H ̂0(x ⃗ ⃗)，也就是说H ̂0(x ⃗ ⃗)有的对称性H ̂′(x ⃗ ⃗)都有。这个时 候， GH相对于GH0不发生变化。 原来的必然简并还是必然简并。 但偶然简并的 破缺是可以发生变化的。 作为一个例子，我们来看一下把一个独立原子放到一个简立方的晶格场中， 它的能级会发生什么样的变化？（与之相关的理论在历史上，上个世纪40 年代 之前，在物理学发展过程中曾经起过很重要的作用，有个专门的名字，叫晶格场 理论Crystal Field Theory。这个理论主要研究的是原子轨道在晶格场中的劈裂情 况，手段是量子力学与群论；应用是早期激光器，一般都是把稀土元素或过渡金 属元素作为杂质，掺到透明的晶体中，它的光谱由晶格场中的电子能级决定） 不考虑空间反演对称性。我们假设自旋-轨道耦合很弱，这样在没有晶格场 的时候， 原子轨道可以用n、 l、 m 三个好量子数来标识。 GH0是下一章要讲的SO(3) 群。加入晶格场之后，GH变成了有限点群O。 216 我们要讨论的是H ̂0(x ⃗ ⃗)的本征态， 也就是SO(3)群的不可约表示的基， 在加入 晶格场的微扰之后，发生什么样的变化？其中最重要的，就是要在H ̂0(x ⃗ ⃗)的每个 本征能级上（这里的简并都是必然简并） ，求出P GH表示矩阵。进而求得特征标， 然后参照O 群的特征标表。 把这个表示约化为O 群的不等价不可约表示的直和， 这样就知道了必然简并在这里要发生什么样的劈裂了？ 先看O 群的对称操作，有24 个，分五类，分别是：{E}、8{C3 (1)}、3{C4 (1)2}、 6{C4 (1)}、6{C2 (1)}。我们要看的是它们这些操作在SO(3)群不可约表示表示空间的 表示矩阵的特征标。 我们看s、p、d、f 四个态。其中s 态在SO(3)群中是个一维不可约表示，基 函数的球谐部分为： 𝑌 00(𝜃, 𝜑) = 1 2 √1 𝜋 由于P ̂ g这些转动操作只作用到本征态的角向部分，而径向函数部分至于n、l 有 关，所以我们在求表示矩阵的时候，可以根本就不关心这个径向函数部分。 对s 态，O 群中的任何元素对应的表示矩阵都是1 这个1 阶单位矩阵，所以 特征标都是1。 对p 态，角向部分基函数是： 𝑌 1−1(𝜃, 𝜑) = 1 2 √3 2𝜋e−𝑖𝜑sin 𝜃 𝑌 10(𝜃, 𝜑) = 1 2 √3 2𝜋cos 𝜃 𝑌 11(𝜃, 𝜑) = 1 2 √3 2𝜋e𝑖𝜑sin 𝜃 转动作用到类似的球谐函数基上，得到的表示矩阵的特征标只与转动角度有关。 因此，我们取绕z 轴的转动，对于O 群中的各个类，通过P ̂ g作用到这些基上产生 表示矩阵。这些表示矩阵的特征标分别是： 1. {E}：χ = 1 + 1 + 1 = 3； 2. 8{C3 (1)}：χ = e𝑖2𝜋 3 + 1 + e−𝑖2𝜋 3 = 0； 3. 3{C4 (1)2}：χ = e𝑖𝜋+ 1 + e−𝑖𝜋= −1； 4. 6{C4 (1)}：χ = e𝑖𝜋 2 + 1 + e−𝑖𝜋 2 = 1； 5. 6{C2 (1)}：χ = e𝑖𝜋+ 1 + e−𝑖𝜋= −1。 对d 态、f 态，球谐函数的普遍形式是： 𝑌 𝑙𝑚(𝜃, 𝜑) = (−1)𝑚√(2𝑙+ 1)(𝑙−𝑚)! 4𝜋(𝑙+ 𝑚)! P 𝑙 𝑚(cos 𝜃)e𝑖𝑚𝜑 其中P 𝑙 𝑚(𝑥)为m 阶l 次连带勒让德多项式，微分形式为： P 𝑙 𝑚(𝑥) = (1 −𝑥2) 𝑚 2 1 2𝑙𝑙! d𝑙+𝑚 d𝑥𝑙+𝑚(𝑥2 −1)𝑙 以d 态为例， O 群中的各个类所对应的哈密顿算符作用到d 态波函数的这组基上 产生的表示矩阵的特征标就是： 1. {E}：χ = 1 + 1 + 1 + 1 + 1 = 5； 2. 8{C3 (1)}：χ = e𝑖4𝜋 3 + e𝑖2𝜋 3 + 1 + e−𝑖2𝜋 3 + e−𝑖4𝜋 3 = −1； 3. 3{C4 (1)2}：χ = e𝑖2𝜋+ e𝑖𝜋+ 1 + e−𝑖𝜋+ e−𝑖2𝜋= 1； 4. 6{C4 (1)}：χ = e𝑖𝜋+ e𝑖𝜋 2 + 1 + e−𝑖𝜋 2 + e−𝑖𝜋= −1； 5. 6{C2 (1)}：χ = e𝑖2𝜋+ e𝑖𝜋+ 1 + e−𝑖𝜋+ e−𝑖2𝜋= 1。 f 态可同理求出。这样O 群中的元素在球谐函数下的表示（可约表示）的特 征标表就是： {E} 8{C3 (1)} 3{C4 (1)2} 6{C4 (1)} 6{C2 (1)} s 1 1 1 1 1 218 p 3 0 -1 1 -1 d 5 -1 1 -1 1 f 7 1 -1 -1 -1 表4.1 O 群元素在球谐函数空间的特征标 这个时候，参考O 群的不等价不可约表示特征标表： {E} 8{C3 (1)} 3{C4 (1)2} 6{C4 (1)} 6{C2 (1)} A1 1 1 1 1 1 A2 1 1 1 -1 -1 E 2 -1 2 0 0 T1 3 0 -1 1 -1 T2 3 0 -1 -1 1 表4.2 O 群特征标表 我们就可以得到，s 态还对应A1 这个一维恒等表示；p 态对应T1 这个不可约表 示，也不劈裂；d 态变为：E ⊕T2；f 态变为：A2 ⊕T 1 ⊕T2。 其中d 轨道往E 和T2的劈裂在固体物理中经常用到，特别是关于过渡金属 氧化物电子态的讨论。 在这些固体中， 很多过渡金属处在一个八面体的笼子里面， 它的对称性就是Oh。 再加上空间反演对称性， 这个时候我们可以看到很多文献上 关于Eg、T2g的讨论，说的就是这个事情。 4.3 投影算符与久期行列式的对角化 这一节想给大家讲得具体问题是利用对称化的基函数， 去简化薛定谔方程的 久期行列式。 为了让大家知道如何去对称化一组基， 我们需要去学习一下投影算 符，确切地说是系统对称群的群表示投影算符，算符定义的时候，要用到哈密顿 算符群。在这一节的学习中，投影算符是数学工具，久期行列式的对角化是这里 的物理问题。这个物理问题是量子力学里面的一个常见问题。 除了这个物理问题， 投影算符在群论这门学科本身又有什么用呢？一句话回 答，就是可以简化群的表示。在求群表示过程中，我们是需要一组基的，这组基 我们可以任意选。当然，如果我们任意选的话，我们得到的矩阵群一般是不具备 豆腐块的特征的，我们无法直接通过其矩阵形式判断其是否可约（第二章说过， 可约是线性空间的性质） 。我们可以通过对比特征标来进行这个判断。判断完了 以后， 如果它可约， 我们是需要经过一个相似变换把这个矩阵群变成豆腐块的矩 阵群的。但如何去找到这个相似变换，其实是很难的。 除了相似变换， 另一条路是在我们建立这个群的表示矩阵的时候直接将你给 我的表示空间约化为几个群不变的真子空间的直和。 也就是直接将基函数对称化。 这个群的群表示投影算符干的事情， 就是将你给我的表示空间约化为群不变的真 子空间的直和，也就是将空间中的基函数对称化。 了解完这些目的， 我们就知道在简化群表示以及久期行列式的对角化中， 能 够理解和使用投影算符是至关重要的。 学习过程我们分几步走， 先看投影算符最 基本的意思。 定义4.3 投影算符：线性空间V 上的线性算符𝐏 ̂，若满足𝐏 ̂𝟐= 𝐏 ̂，则称𝐏 ̂是V 上 的一个投影算符。 𝐏 ̂的值域是：𝐑𝒑= 𝐏 ̂𝑽= {𝐳 ⃗∈𝑽|𝐳 ⃗= 𝐏 ̂𝒙 ⃗ ⃗⃗，𝒙 ⃗ ⃗⃗∈𝑽} 𝐏 ̂的核是：𝐍𝒑= {𝐳 ⃗∈𝑽|𝐏 ̂𝐳′ ⃗⃗⃗= 𝟎} 这个定义有些抽象，看一个具体的例子，V 是三维欧式空间，P ̂是对其xy 平 面的投影。 220 图4.10 投影算符，欧式空间 因为：P ̂2𝑥 ⃗= P ̂𝑥 ⃗=z ⃗，所以P ̂是投影算符。z 轴上所有向量都是P ̂的核，xy 平面是它 的值域。 由这个关系，我们很容易知道： 1. 对z ⃗∈R𝑝，有P ̂z ⃗= z ⃗； 2. P ̂是V 上得投影算符，则E ̂ −P ̂也是，且有P ̂(E ̂ −P ̂) = 0； 3. 如存在P ̂，则V=R𝑝⊕N𝑝；同时如果有一个空间V=W 1 ⊕W2，则一定存在一 个相应的投影算符P ̂。 由这个定义，我们还可以知道下面这个定理。 定理4.3 若线性空间V=𝐖𝟏⊕𝐖𝟐⊕⋯⊕𝐖𝒌，则V 上存在投影算符𝐏 ̂𝟏、𝐏 ̂𝟐、 ⋯、𝐏 ̂𝒌，满足： 1. 𝐏 ̂𝒊 𝟐= 𝐏 ̂𝒊； 2. 𝐏 ̂𝒊𝐏 ̂𝒋=0，当𝒊≠𝒋； 3. 𝐏 ̂𝟏+𝐏 ̂𝟐+⋯+ 𝐏 ̂𝒌= 𝐄 ̂； 4. 𝐏 ̂𝒊𝑽= 𝐖𝒊，𝒊= 𝟏、𝟐、⋯、𝐤。 同时，反之，若线性空间V 上存在算符𝐏 ̂𝟏、𝐏 ̂𝟐、⋯、𝐏 ̂𝒌满足上面四个条件，则 V=𝐖𝟏⊕𝐖𝟐⊕⋯⊕𝐖𝒌。 讲完了这两个定义，我们现在来看群表示投影算符，这个其实是一个定理。 定理4.4 设群G 的不可约酉表示为𝐀(𝛂)，𝛂= 𝟏、𝟐、⋯、𝐪，维数为𝐬𝛂。𝐏𝐆为G 对应的算符群（既哈密顿算符群） ，𝐏𝐆= {𝐏 ̂𝐠|𝐠∈𝐆} 。定义算符𝐏 ̂𝒌𝒋 (𝛂) = 𝐬𝛂 𝒏∑ 𝐀𝒌𝒋 (𝛂)∗(𝐠)𝐏 ̂𝐠 𝐠∈𝐆 ，则这些算符满足下列关系： 𝐏 ̂𝒌𝒋 (𝛂)𝐏 ̂𝒊𝒍 (𝛃) = 𝜹𝛂𝛃𝜹𝒊𝒋𝐏 ̂𝒌𝒍 (𝛂) 且𝐏 ̂𝒋𝒋 (𝛂)为投影算符。 证明： P ̂ 𝑘𝑗 (α)P ̂ 𝑖𝑙 (β) = sαsβ 𝑛2 ∑A𝑘𝑗 (α)∗(g)P ̂ g g∈G ∑A𝑖𝑙 (β)∗(g′)P ̂ g′ g′∈G = sαsβ 𝑛2 ∑∑A𝑘𝑗 (α)∗(g)A𝑙𝑖 (β)(g′−1)P ̂ gg′ g′∈G g∈G 令gg′ = g′′，则上式继续等于： = sαsβ 𝑛2 ∑∑A𝑘𝑗 (α)∗(g)A𝑙𝑖 (β)(g′′−1g)P ̂ g′′ g′′∈G g∈G = sαsβ 𝑛2 ∑∑A𝑘𝑗 (α)∗(g) ∑A𝑙𝑚 (β)(g′′−1)A𝑚𝑖 (β)(g)P ̂ g′′ sβ 𝑚=1 g′′∈G g∈G = sα 𝑛∑∑[sβ 𝑛∑A𝑘𝑗 (α)∗(g) g∈G A𝑚𝑖 (β)(g)] g′′∈G A𝑙𝑚 (β)(g′′−1)P ̂ g′′ sβ 𝑚=1 = sα 𝑛∑∑𝛿αβ𝛿𝑘𝑚𝛿 𝑗𝑖 g′′∈G A𝑙𝑚 (β)(g′′−1)P ̂ g′′ sβ 𝑚=1 = 𝛿αβ𝛿 𝑗𝑖[sα 𝑛∑A𝑘𝑙 (β)∗(g′′)P ̂ g′′ g′′∈G ] = 𝛿αβ𝛿 𝑗𝑖P ̂ 𝑘𝑙 (β) 这个是我们要证明的等式，对这样一个等式，取α = β、𝑖= 𝑗= 𝑘= 𝑙时，有： P ̂ 𝑗𝑗 (α)P ̂ 𝑗𝑗 (α) = P ̂ 𝑗𝑗 (α) 因此P ̂ 𝑗𝑗 (α)是投影算符。 （证毕） 222 对这类投影算符，它满足一个性质： 定理4.5 ∑ ∑ 𝐏 ̂𝒊𝒊 (𝛂) 𝐬𝛂 𝒊=𝟏 𝒒 𝛂=𝟏 = 𝐏 ̂𝐞，其中𝐏 ̂𝐞为恒等算符。 证明： 用到不等价不可约酉表示矩阵元正交定理的第二种表述22： ∑∑sβ n A𝑘𝑙 (α)∗(g′)A𝑘𝑙 (α)(g) sα 𝑘,𝑙=1 𝑞 β=1 = 𝛿gg′ 由算符P ̂ 𝑘𝑙 (α)定义，有：P ̂ 𝑘𝑙 (α) = sα n ∑ A𝑘𝑙 (α)∗(g)P ̂ g g∈G 两边乘上A𝑘𝑙 (α)(g′)并对α、k、l 求和，有： ∑∑A𝑘𝑙 (α)(g′)P ̂ 𝑘𝑙 (α) sα 𝑘,𝑙=1 𝑞 α=1 = ∑[∑∑sα n A𝑘𝑙 (α)∗(g)A𝑘𝑙 (α)(g′) sα 𝑘,𝑙=1 𝑞 α=1 ] g∈G P ̂ g = ∑𝛿gg′P ̂ g g∈G = P ̂ g′ 取g′ = e，上式可得： ∑∑A𝑘𝑗 (α)(e)P ̂ 𝑘𝑙 (α) sα 𝑘,𝑙=1 𝑞 α=1 = P ̂ e 22这两种表述的关系可类比特征标两个正交定理，证明也类似。起点是矩阵元第一正交定理的 表述形式： ∑sβ n A𝑘𝑙 (α)∗(g)A𝑘′𝑙′ (β) (g) g∈G = 𝛿αβ𝛿𝑘𝑘′𝛿𝑙𝑙′ 设计矩阵F，矩阵元为F𝑖,𝑗= √ s𝛼 n A𝑘𝑙 (α)(g)，其中𝑖这个指标走遍α、𝑘、𝑙的组合，𝑗这个指标走遍 群元。这样定义的话，和特征标第二正交定理的证明类似，这个矩阵的矩阵元满足： F𝑖,𝑗 ∗= s𝛼 n A𝑘𝑙 (α)∗(g𝑗) = (F+)𝑗,𝑖 这样之前的第一种表述就可以写成∑F𝑖,𝑗(F+)𝑗,𝑖′ 𝑗 = (FF+)𝑖,𝑖′ = 𝛿𝑖𝑖′。进而，F+ = F−。这样还 会有：F+F = E。而最后这个式子的表述形式就是： (F+F)𝑗,𝑗′ = ∑(F+)𝑗,𝑖F𝑖,𝑗′ 𝑖 = ∑∑sβ n A𝑘𝑙 (α)∗(g′)A𝑘𝑙 (α)(g) sα 𝑘,𝑙=1 𝑞 β=1 = 𝛿gg′ 也有教材讲此关系称为完备性定理，这种说法可类比量子力学中本征态波函数完备性定理 的表述形式：∑Ψ𝑖 ∗(r ⃗′)Ψ𝑖(r ⃗) 𝑖 = 𝛿(r ⃗′ −r ⃗)。单论群论不论量子力学的话，不等价不可约表示 的矩阵元和群元的维度都是n。正交得占满这n个维度，自然也完备。 由于A𝑘𝑙 (α)(e) = 𝛿𝑘𝑙，所以上式继续给出： ∑∑𝛿𝑘𝑙P ̂ 𝑘𝑙 (α) sα 𝑘,𝑙=1 𝑞 α=1 = P ̂ e 进而： ∑∑P ̂ 𝑘𝑘 (α) sα 𝑘=1 𝑞 α=1 = P ̂ e （证毕） 关于这个群表示投影算符，有两个比较重要的性质，以定理的形式给出： 定理4.6 有限群不可约酉表示基函数定理I： 设P G = {P ̂ g|g ∈G}是群G 的函数作用算符群 （相当于我们前面介绍的哈密顿算符 群） ， 由它可以定义算符P ̂ 𝑖𝑗 (α)。 这时， 有一个性质， 就是一组基函数φ𝑖 (α)， 𝑖= 1、2、 ⋯、sα构成群G 的第α个不可约酉表示基函数的充要条件是： P ̂ 𝑖𝑗 (α)φ𝑗 (α) = φ𝑖 (α) 这里φ𝑖 (α)称为对称化基函数。 证明： 1. 必要性 （由φ𝑖 (α)， 𝑖= 1、2、⋯、sα构成群G 的第α个不可约酉表示基来证明P ̂ 𝑖𝑗 (α)φ𝑗 (α) = φ𝑖 (α)） 设φ𝑖 (α)，𝑖= 1、2、⋯、sα构成群G 的第α个不可约酉表示基，则： P ̂ gφ𝑘 (α) = ∑A𝑙𝑘 (α)(g)φ𝑙 (α) sα 𝑙=1 两边乘上A𝑖𝑗 (α)∗(g)并对群元求和，有： ∑A𝑖𝑗 (α)∗(g)P ̂ gφ𝑘 (α) g∈G = ∑A𝑖𝑗 (α)∗(g) ∑A𝑙𝑘 (α)(g)φ𝑙 (α) sα 𝑙=1 g∈G 224 = ∑[∑A𝑖𝑗 (α)∗(g)A𝑙𝑘 (α)(g) g∈G ] sα 𝑙=1 φ𝑙 (α) = ∑𝑛 sα 𝛿𝑖𝑙𝛿 𝑗𝑘 sα 𝑙=1 φ𝑙 (α) = 𝑛 sα 𝛿 𝑗𝑘φ𝑖 (α) 由定义P ̂ 𝑖𝑗 (α) = sα 𝑛∑ A𝑖𝑗 (α)∗(g)P ̂ g g∈G ，上式等同于： P ̂ 𝑖𝑗 (α)φ𝑘 (α) = 𝛿 𝑗𝑘φ𝑖 (α) 取𝑗= 𝑘，则有： P ̂ 𝑖𝑗 (α)φ𝑗 (α) = φ𝑖 (α) 2. 充分性（已知P ̂ 𝑖𝑗 (α)φ𝑗 (α) = φ𝑖 (α)，推φ𝑖 (α)，𝑖= 1、2、⋯、sα构成群G 的第α个 不可约酉表示基） 由P ̂ 𝑖𝑗 (α)φ𝑗 (α) = φ𝑖 (α)知： P ̂ gφ𝑖 (α) = P ̂ gP ̂ 𝑖𝑗 (α)φ𝑗 (α) = P ̂ g sα 𝑛∑A𝑖𝑗 (α)∗(g′)P ̂ g′ g′∈G φ𝑗 (α) = ∑sα 𝑛A𝑖𝑗 (α)∗(g′)P ̂ gg′ g′∈G φ𝑗 (α) 取gg′ = g′′，则上式可化为： = ∑sα 𝑛A𝑖𝑗 (α)∗(g−1g′′)P ̂ g′′ g′′∈G φ𝑗 (α) = sα 𝑛∑∑A𝑖𝑘 (α)∗(g−1)A𝑘𝑗 (α)∗(g′′)P ̂ g′′ sα 𝑘=1 g′′∈G φ𝑗 (α) = ∑A𝑖𝑘 (α)∗(g−1) sα 𝑘=1 [sα 𝑛∑A𝑘𝑗 (α)∗(g′′)P ̂ g′′ g′′∈G φ𝑗 (α)] = ∑A𝑘𝑖 (α)(g) sα 𝑘=1 P ̂ 𝑘𝑗 (α)φ𝑗 (α) = ∑A𝑘𝑖 (α)(g) sα 𝑘=1 φ𝑘 (α) 这也就是说φ𝑖 (α)，𝑖= 1、2、⋯、sα构成群G 的第α个不可约酉表示基。 （证毕） 定理4.7 有限群不等价、不可约酉表示的基函数定理II： 有限群不等价、不可约酉表示的基函数φ𝑖 (α)，𝑖= 1、2、⋯、sα，α = 1、2、 ⋯、q满足如下正交关系： (φ𝑖 (α)|φ𝑗 (𝛽)) = 𝛿𝑖𝑗𝛿𝛼𝛽𝑓(𝛼) 其中𝑓(𝛼)与i、j 无关。 证明： 设G 是系统对称群，P G是G 对应的算符群（变换群） 。对G 中元素，有P G中元素 与之对应，且保持乘法规则不变。也就是说，P G是G 的一个表示。 你现在告诉我这个表示是酉表示，那么这时，P ̂ g这个线性变换群P G中的元素就是 酉变换。这些是已知条件。 除了这个已知条件，另外一个已知条件是：φ𝑖 (α)，𝑖= 1、2、⋯、sα与φ𝑗 (β)，𝑗= 1、2、⋯、sβ都是线性变换群P G的表示空间，承载的是群G 的第α个与第β个不 等价不可约酉表示。 由这些已知条件，我们可以得到： (φ𝑖 (α)|φ𝑗 (𝛽)) = (P ̂ gφ𝑖 (α)|P ̂ gφ𝑗 (𝛽)) 因为P ̂ g是酉变换（保内积） 。 而这个式子的右边我们可以写成： (∑ A𝑘𝑖 (α)(g)φ𝑘 (α) sα 𝑘=1 | ∑ A𝑙𝑗 (β)(g)φ𝑙 (𝛽) sβ 𝑙=1 ) 226 = ∑∑A𝑘𝑖 (α)∗(g)A𝑙𝑗 (β)(g) (φ𝑘 (α)|φ𝑙 (𝛽)) sβ 𝑙=1 sα 𝑘=1 也就是说： (φ𝑖 (α)|φ𝑗 (𝛽)) = ∑∑A𝑘𝑖 (α)∗(g)A𝑙𝑗 (β)(g) (φ𝑘 (α)|φ𝑙 (𝛽)) sβ 𝑙=1 sα 𝑘=1 这个式子两边都对g 求和，左边不依赖于g，相当于直接乘上n，而右边有正交 性定理，因此有： n (φ𝑖 (α)|φ𝑗 (𝛽)) = ∑∑n sα 𝛿𝑘𝑙𝛿𝑖𝑗𝛿𝛼𝛽(φ𝑘 (α)|φ𝑙 (𝛽)) sβ 𝑙=1 sα 𝑘=1 = n sα 𝛿𝑖𝑗𝛿𝛼𝛽∑(φ𝑘 (α)|φ𝑘 (𝛽)) sα 𝑘=1 两边n 约掉，有： (φ𝑖 (α)|φ𝑗 (𝛽)) = 1 sα 𝛿𝑖𝑗𝛿𝛼𝛽∑(φ𝑘 (α)|φ𝑘 (𝛽)) sα 𝑘=1 正交性成立。 取： 𝑓(𝛼) = 1 sα ∑(φ𝑘 (α)|φ𝑘 (𝛽)) sα 𝑘=1 这个量显然与i、 j 无关。 这也就是说对一个有限群，不等价不可约酉表示的基函 数相互正交。 （证毕） 在前面两个基函数定理中， 我们讨论的是一个群的不可约表示空间基函数的 性质。如果把哈密顿量的对称性同时考虑进去，我们还有另外一个定理。 定理4.8 若𝛗𝒌 (𝜶)(𝐫 ⃗)是系统对称群的第𝜶个不等价不可约表示的第k 个基，那么 𝐇 ̂(𝐫 ⃗)𝛗𝒌 (𝜶)(𝐫 ⃗)也按照这个群的第𝜶个不等价不可约表示的第k 个基变化。 （换句话说， φ𝑖 (𝛼)(r ⃗)， 𝑖= 1、2、⋯、sα， 形成系统对称群第𝛼个不等价不可约表 示的一组基， 则H ̂(r ⃗)φ𝑖 (𝛼)(r ⃗)， 𝑖= 1、2、⋯、sα， 也形成系统对称群第𝛼个不等价 不可约表示的一组基。这两组基，次序一样） 证明： P ̂ g是哈密顿算符群的一个变换算符，那么，由φ𝑘 (𝛼)(r ⃗)是系统对称群的第𝛼个不等 价不可约表示的第k 个基可知： P ̂ gφ𝑘 (𝛼)(r ⃗) = ∑A𝑙𝑘 (α)(g)φ𝑙 (α)(r ⃗) sα 𝑙=1 由H ̂(r ⃗)P ̂ g = P ̂ gH ̂(r ⃗)又知： P ̂ g[H ̂(r ⃗)φ𝑘 (α)(r ⃗)] = H ̂(r ⃗)P ̂ gφ𝑘 (α)(r ⃗) = H ̂(r ⃗) ∑A𝑙𝑘 (α)(g)φ𝑙 (α)(r ⃗) sα 𝑙=1 = ∑A𝑙𝑘 (α)(g)[H ̂(r ⃗)φ𝑙 (α)(r ⃗)] sα 𝑙=1 这也就是说φ𝑘 (α)(r ⃗)，𝑘= 1、⋯、sα是系统对称群的第α个不等价不可约表示的 基时， H ̂(r ⃗)φ𝑘 (α)(r ⃗)， 𝑘= 1、⋯、sα也构成系统对称群的第α个不等价不可约表示 的基。 （证毕） 上面的这些基函数定理非常有用， 因为它可以让很多问题简化。 一个典型的 例子就是解薛定谔方程的时候对久期行列式的对角化。 这个怎么一个事情呢？就是很多薛定谔方程没有解析解， 我们求解的时候最 重要的一个步骤用一组基来展开波函数，然后对角久期行列式。 我们设这一组基是：φ1(r ⃗)、φ2(r ⃗)、⋯、φ𝑛(r ⃗)、⋯，波函数的展开形式是： 228 Ψ(r ⃗) = ∑𝑐𝑝φ𝑝(r ⃗) ∞ 𝑝=1 这里𝑐𝑝为待定的展开系数。 把这样一个表达式带入薛定谔方程中， 并用各个基函 数去做内积，我们有： |(φ𝑞(r ⃗)|H ̂(r ⃗)φ𝑝(r ⃗)) −E (φ𝑞(r ⃗)|φ𝑝(r ⃗))| = 0 这样一个久期方程了。 为了求解这个久期方程，我们的基函数空间必须做个截断，因为这个p、q 不能一直走下去，这个近似叫截断近似，这个在量子力学里面你们都接触过。当 φ𝑝(r ⃗)走遍从φ1(r ⃗)到φN(r ⃗)的时候，久期行列式就是一个N×N 的行列式。在这个 行列式中，假如基函数本身没有什么对称性，那么它就是一个正常的N×N 的行 列式，解起来计算量随N 的变化规律是N3。但是如果φ𝑝(r ⃗)本身有对称性，那么 根据上面两个定理， 属于不同不等价不可约表示的矩阵元就正交。 这样这个矩阵 就会具备很好的对角化的特征，解起来就很方便了。 举个例子： 已知Φ1(r ⃗)、 ⋯、 Φ6(r ⃗)这个函数组形成波函数的展开空间。 由它们直接形成的 系统对称群的表示是可约的，但如果通过线性组合，我们是可以找到另外六 个线性无关的向量，来承载不可约表示的。如果以原来的基展开波函数，那 么久期行列式就是正常的6×6 行列式。但如果用对称化的基函数，那么久期 行列式就可以简单很多。 如果组合成的六个对称化的波函数是： φ11 1 (r ⃗)、 φ11 2 (r ⃗)、 φ12 2 (r ⃗)、 φ11 3 (r ⃗)、 φ12 3 (r ⃗)、 φ13 3 (r ⃗)， 这里我们用上标代表那个不可约表示， 下标的第一个数代表这个不可 约表示出现的次数，第二个数代表这个不可约表示第几个基。这种情况就是 1、2、3 维不可约表示各出现一次，共六个对称化基。 这时，由我们上面讲到的后两个定理，我们知道矩阵元： K𝑗𝑛,𝑖𝑚 𝑙𝑘 = (φ𝑗𝑛 𝑙(r ⃗)|H ̂(r ⃗)φ𝑖𝑚 𝑘(r ⃗)) −E (φ𝑗𝑛 𝑙(r ⃗)|φ𝑖𝑚 𝑘(r ⃗)) 直接对角，相应的行列式为： | | K11,11 11 0 0 K11,11 22 0 0 0 0 0 0 0 0 0 0 0 0 K12,12 22 0 0 K11,11 33 0 0 0 0 0 0 0 0 0 0 0 0 K12,12 33 0 0 K13,13 33 | | = 0 而如果组合成的六个对称化的波函数是： φ11 1 (r ⃗)、 φ21 1 (r ⃗)、 φ11 2 (r ⃗)、 φ12 2 (r ⃗)、 φ21 2 (r ⃗)、 φ22 2 (r ⃗)，1 维、2 维不可约表示各出现两次，那么这个行列式就不能完全对角 了，但还可以保持比较强的准对角特征。如果我们把它们重新排列顺序： φ11 1 (r ⃗)、φ21 1 (r ⃗)、φ11 2 (r ⃗)、φ21 2 (r ⃗)、φ12 2 (r ⃗)、φ22 2 (r ⃗)，那么行列式为： | | K11,11 11 K11,21 11 K21,11 11 K21,21 11 0 0 0 0 0 0 0 0 0 0 0 0 K11,11 22 K11,21 22 K21,11 22 K21,21 22 0 0 0 0 0 0 0 0 0 0 0 0 K12,12 22 K12,22 22 K22,12 22 K22,22 22 | | = 0 总的来说，利用对称化的基函数还是会让这个久期行列式简单很多。 既然对称化的基函数有这些优点， 那么从一般的基中如何产生对称化的基函 数呢？这个步骤比较规范，我们可以利用的就是前面讲的投影算符P ̂ 𝑖𝑖 (α)。我们把 投影算符作用到这个基函数， 也就是向量上， 得到的就是这个向量在我这个对称 化的基上的投影。有了这个投影，我就可以用： P ̂ 𝑖𝑗 (α)φ𝑗 (α) = φ𝑖 (α) 得到其它的基函数了。 当然， 这样做的前提是我们知道所有不等价不可约表示的矩阵元， 从而可以 230 构造出来P ̂ 𝑖𝑗 (α)。如果我们不知道这个，我们只知道特征标表，那么我们也可以通 过下面的三个简单的步骤来执行： 1. 构造特征标投影算符： P ̂(α) = sα 𝑛∑χ(α)∗(g)P ̂ g g∈G 它和投影算符P ̂ 𝑖𝑖 (α)的关系是： P ̂(α) = ∑P ̂ 𝑖𝑖 (α) sα 𝑖=1 2. 然后我们把这样的一个算符作用到线性空间V 的任意一个向量上。这个向量 记为Ψ，它可以分解为群的不等价不可约表示的基的线性和： Ψ = ∑∑∑𝑎𝑖𝑙 (𝛽)φ𝑖𝑙 (β) 𝑙 𝛽 𝑖 其中i 是某个不等价不可约表示重复出现的index，𝛽是不等价不可约表示的 index，l 是这个不等价不可约表示中基的index。 把P ̂(α)作用到Ψ上，效果为： P ̂(α)Ψ = P ̂(α) ∑∑∑𝑎𝑖𝑙 (β)φ𝑖𝑙 (β) 𝑙 𝛽 𝑖 = ∑∑∑𝑎𝑖𝑙 (β)P ̂(α)φ𝑖𝑙 (β) 𝑙 𝛽 𝑖 = ∑∑∑𝑎𝑖𝑙 (β) ∑P ̂ 𝑗𝑗 (α)φ𝑖𝑙 (β) sα 𝑗=1 𝑙 𝛽 𝑖 = ∑∑∑𝑎𝑖𝑙 (β) ∑𝛿 𝑗𝑙 sα 𝑗=1 𝛿𝛼𝛽φ𝑖𝑙 (β) 𝑙 𝛽 𝑖 = ∑∑∑𝑎𝑖𝑙 (β)𝛿𝛼𝛽φ𝑖𝑙 (β) 𝑙 𝛽 𝑖 = ∑∑𝑎𝑖𝑙 (α)φ𝑖𝑙 (α) 𝑙 𝑖 也就是说你任意给我一个向量，我只把它属于我的某个不等价不可约表示表 示空间的部分取出来。如果我这个不可约表示就出现一次，很好办。如果出 现多次，我需要在进行一个内部的对称化处理。 3. 用P ̂ g对这些向量进行作用，找出其它维度上的独立向量。 我们看几个例子。 例1. D3 群的表示空间为x、y、z 的二次齐次函数空间，基为： φ1 = 𝑥2、φ2 = 𝑦2、φ3 = 𝑧2、φ4 = 𝑥𝑦、φ5 = 𝑦𝑧、φ6 = 𝑥𝑧 试用投影算符的方法将其组合为6 个对称化的基函数，并验证新基下表示的 对称性。 我们这里需要的是D3 群的特征标表： 1{e} 2{d} 3{a} A1 1 1 1 A2 1 1 -1 A3 2 -1 0 解： 第一步是求出D3 群中的元素在三维欧式空间中的表示矩阵以及它们的逆， 因 为我们要操作的线性空间是一个函数空间，我们依赖的基本变换关系是： P ̂ gΨ(r ⃗) = Ψ(g−1r ⃗)。 图4.11 D3 群元在欧式空间的表示 232 A(e) = ( 1 0 0 0 1 0 0 0 1 ) = A−1(e) A(d) = ( −1/2 −√3/2 0 √3/2 −1/2 0 0 0 1 ) = A−1(f) A(f) = ( −1/2 √3/2 0 −√3/2 −1/2 0 0 0 1 ) = A−1(d) A(a) = ( 1/2 √3/2 0 √3/2 −1/2 0 0 0 −1 ) = A−1(a) A(b) = ( −1 0 0 0 1 0 0 0 −1 ) = A−1(b) A(c) = ( 1/2 −√3/2 0 −√3/2 −1/2 0 0 0 −1 ) = A−1(c) 第二步是写出三个不可约表示的特征标投影算符： P ̂(1) = 1 6 (P ̂ e + P ̂ d + P ̂ f + P ̂ a + P ̂ b + P ̂ c) P ̂(2) = 1 6 (P ̂ e + P ̂ d + P ̂ f −P ̂ a −P ̂ b −P ̂ c) P ̂(3) = 2 6 (2P ̂ e −P ̂ d −P ̂ f) 第三步是将这些特征标投影算符作用到基函数上面，其中P ̂(1)作用的结果为： P ̂(1)𝑥2 = 1 6 [𝑥2 + (−1 2 𝑥+ √3 2 𝑦) 2 + (−1 2 𝑥−√3 2 𝑦) 2 + (1 2 𝑥+ √3 2 𝑦) 2 + (−𝑥)2 + (1 2 𝑥−√3 2 𝑦) 2 ] = 1 2 (𝑥2 + 𝑦2) P ̂(1)𝑦2 = 1 2 (𝑥2 + 𝑦2) P ̂(1)𝑧2 = 𝑧2 P ̂(1)𝑥𝑦= 0 P ̂(1)𝑦𝑧= 0 P ̂(1)𝑥𝑧= 0 也就是说这一组基六个函数， 往D3 群的第一个不等价不可约表示的表示空间 做投影，只能生成 1 2 (𝑥2 + 𝑦2)与𝑧2两个向量。 把P ̂(2)作用到这六个函数上，结果是： P ̂(2)𝑥2 = 0 P ̂(2)𝑦2 = 0 P ̂(2)𝑧2 = 0 P ̂(2)𝑥𝑦= 0 P ̂(2)𝑦𝑧= 0 P ̂(2)𝑥𝑧= 0 这六个函数在A2这个不等价不可约表示的表示空间没有投影。 把P ̂(3)作用到这六个函数上，结果是： P ̂(3)𝑥2 = 2 6 [2𝑥2 −(−1 2 𝑥+ √3 2 𝑦) 2 −(−1 2 𝑥−√3 2 𝑦) 2 ] = 1 2 (𝑥2 −𝑦2) P ̂(3)𝑦2 = 2 6 [2𝑦2 −(−√3 2 𝑥−1 2 𝑦) 2 −(√3 2 𝑥−1 2 𝑦) 2 ] = −1 2 (𝑥2 −𝑦2) P ̂(3)𝑧2 = 0 P ̂(3)𝑥𝑦= 𝑥𝑦 P ̂(3)𝑦𝑧= 𝑦𝑧 P ̂(3)𝑥𝑧= 𝑥𝑧 这也就是说这六个函数， 往A3 这个不可约表示上作投影的话， 可以产生4 个 线性无关的向量。 这4 个如何两两配对形成两组承载A3 的基？我们还需要进 行进一步的变换才知道。 我们取其中的 1 2 (𝑥2 −𝑦2)，用P ̂ d作用到它上面，效果是： P ̂ d 1 2 (𝑥2 −𝑦2) = 1 2 (−1 2 𝑥+ √3 2 𝑦) 2 −1 2 (−√3 2 𝑥−1 2 𝑦) 2 = −1 2 [1 2 (𝑥2 −𝑦2)] −√3 2 𝑥𝑦 也就是说 1 2 (𝑥2 −𝑦2)与𝑥𝑦形成一组基。 与之相应， 我们还可以对𝑦𝑧做变换， 知 道它与𝑥𝑧形成一组基。最后对称化的基组就是： φ11 1 = 1 2 (𝑥2 + 𝑦2)、φ21 1 = 𝑧2、φ11 3 = 1 2 (𝑥2 −𝑦2)、φ12 3 = 𝑥𝑦、φ21 3 = 𝑦𝑧、φ22 3 = 𝑥𝑧。 234 对应的表示矩阵为： A(d) = ( 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 −1/2 √3/2 −√3/2 −1/2 0 0 0 0 0 0 0 0 0 0 0 0 −1/2 √3/2 −√3/2 −1/2) 其它类似，用P ̂ g作用到φ11 1 、φ21 1 、φ11 3 、φ12 3 、φ21 3 、φ22 3 的方法推出。 例2. 用投影算符方法求出D3 群的群空间中6 个对称化的基，它们分别承载哪 些不可约表示？ 这道题要用到D3 群的乘法表： e d f a b c e e d f a b c d d f e c a b f f e d b c a a a b c e d f b b c a f e d c c a b d f e 解： 第一步，取D3 群的群空间的六个基： φ1 = e、φ2 = d、φ3 = f、φ4 = a、φ5 = b、φ6 = c 第二步，取特征标投影算符： P ̂(1) = 1 6 (P ̂ e + P ̂ d + P ̂ f + P ̂ a + P ̂ b + P ̂ c) P ̂(2) = 1 6 (P ̂ e + P ̂ d + P ̂ f −P ̂ a −P ̂ b −P ̂ c) P ̂(3) = 2 6 (2P ̂ e −P ̂ d −P ̂ f) 第三步， 将这些特征标投影算符作用到群空间的基上， 对P ̂(1)， 由重排定理， 作用到φ1到φ6的任何一个得到的都是： 1 6 (e + d + f + a + b + c) 它承载D3 的一维恒等表示。在群代数中归一化为23： 𝜑11 1 = 1 √6 (e + d + f + a + b + c) 对P ̂(2)， 它给出的结果是： 1 6 (e + d + f −a −b −c)或 1 6 (a + b + c −e −d −f)。 也就是说它们往D3 群的A2 表示上作投影，投影部分都是 1 6 (e + d + f −a − b −c)这个维度上的向量。在群代数中归一化为： 𝜑11 2 = 1 √6 (e + d + f −a −b −c) 还剩下四个维度，它必然给出两个二维不可约表示。但由： P ̂(α)Ψ = ∑∑𝑎𝑖𝑙 (𝛼)φ𝑖𝑙 (α) 𝑙 𝑖 我们知道像上面这些例子那样我们直接把P ̂(α)作用到Ψ上就得到某个不可约 表示空间中的向量的例子其实是非常幸运的。很多情况下，我们还有一个同 一个不可约表示出现次数的index 的加和。对于剩下的四个维度，如果我们 直接用P ̂(3)作用到φ1到φ6上，我们就会得到类似的情况。 为了处理方便，我们采取的策略是： P ̂(3)(e + a) = 2 6 (2e −d −f + 2a −c −b) 在群代数中归一化为： 𝜑11 3 = 1 2√3 (2e −d −f + 2a −c −b) 然后，用P ̂ d作用到它上面，得： P ̂ d𝜑11 3 = 1 2√3 (2d −f −e + 2c −a −b) 23这里我们在归一的时候，根据的内积定义是：(x|y) = ∑ x∗(g𝑖)y(g𝑖) n 𝑖=1 ，与2.4 节里面讨论到群 函数空间内积的时候的那个定义不一样。 这个定义形式不唯一， 视具体情况而定。 只要在讨论具 体问题开始的时候说请并前后一致就可以。这个说明得谢谢南科大刘奇航老师提醒！ 236 P ̂ d𝜑11 3 与𝜑11 3 线性无关，但并不正交。对它们进行正交化处理，再归一化，可 得： 𝜑12 3 = 1 2 (d −f + c −b) 同理，有： P ̂(3)(e −a) = 2 6 (2e −d −f −2a + c + b) 进而： 𝜑21 3 = 1 2√3 (2e −d −f −2a + c + b) 同样，用P ̂ d作用到它上面，在与𝜑21 3 进行正交归一处理，可得： 𝜑22 3 = 1 2 (d −f + b −c) 4.4 矩阵元定理与选择定则、电偶极跃迁 这一节说的是微扰引起的跃迁， 以及跃迁矩阵元与对称性之间的关系。 也就 是说我系统本身有个H ̂0， 系统处在一系列分立的本征态上， 每个本征态都可以用 系统对称群的一个不等价不可约表示的基来标识。比如Ψ𝛼、Ψ𝛽，它们都是某个 不可约表示的基，对应H ̂0的系统对称群。 现在你给我一个扰动H ̂′，根据微扰理论，这个系统的两个态就可能通过 (Ψα|H ̂′Ψβ)联系起来了。 (Ψα|H ̂′Ψβ)是跃迁矩阵元。 严格意义上， 我可以把它算出 来。怎么算，是量子力学、固体物理这些课程要告诉我们的事情。群论这门课程 要干的事情， 是通过我们已有的群论的知识， 去理解这个矩阵元什么时候必须是 零， 什么时候可以不为零？也就是给我们一个思路去判断， 这个思路是基于对称 性原理的。 怎么去理解这个思路呢？很简单， 就是把我这个跃迁矩阵元中的三个部分分 别当成H ̂0的系统对称群表示基函数： 第一部分是H ̂′， 它可以是一个依赖于坐标的算符。 在这里我们把它看作H ̂0的 哈密顿算符群表示（记为Dv）的基（这个表示不一定不可约） 。 第二部分是Ψ𝛽，它承载的是H ̂0的一个不可约哈密顿算符群表示（记为Dβ） 。 这两部分合起来，H ̂′Ψβ这个函数，对应的，就是Dv与Dβ的直积表示的基函数。 第三个部分是Ψ𝛼， 它承载的是H ̂0的一个不可约哈密顿算符群表示 （记为Dα） 。 这个时候，我们求跃迁矩阵元(Ψ𝛼|H ̂′Ψ𝛽)这个内积，就可以利用不可约表示 基的正交关系来判断。我们要做的事就是先做Dv ⊗Dβ，然后把它分解，看是否 有Dα的成分。如果有，跃迁被对称性允许；没有，被禁止。就这么简单。 为了满足跃迁被对称性允许，我们对H ̂′的对称性有没有什么要求？答案是： 有。 我们可以想一下， 假如这个H ̂′的对称性很高， 以至于对所有H ̂0中的对称操作， 它都不变。那么它承载的是H ̂0的系统对称群的什么表示？ 一维恒等。这样的话Dv ⊗Dβ = Dβ，也就是说H ̂′Ψ𝛽承载的H ̂0的系统对称群 的表示，与Ψ𝛽承载的完全相同！并且对应的表示中的基的index 也完全相同！ 这个时候，如果Ψ𝛼与Ψ𝛽对应的是同一个不可约表示的同一个基的话，这个 跃迁还是被允许的。不然，所以这个跃迁将完全被禁止。 反过来，对微扰H ̂′，如果它的对称性比H ̂0的低，那么H ̂′Ψ𝛽在进行直积后做 直和分解， 包含Ψ𝛼对应的不等价不可约表示的基的几率会大大增加。 因此， 从对 称性的角度， 我们总是希望H ̂′的对称性低， 这样才能尽量多的诱发出系统在H ̂0的 不同本征态之间的跃迁。 有了这个认识，我们来看最常见的一种情况，扰动是个电磁场。电场：E ⃗ ⃗⃗= − 𝜕A ⃗ ⃗⃗ 𝜕t，磁场：B ⃗ ⃗⃗= ∇× A ⃗ ⃗⃗。加入电磁场后，哈密顿量变成了： H ̂(r ⃗) = 1 2m (p ̂ −e c A ⃗ ⃗⃗) 2 + V(r ⃗) 238 = 1 2m p ̂2 + V(r ⃗) −e mc p ̂ ∙A ⃗ ⃗⃗+ e2A2 2mc2 在弱场下， 最后一项是绝对的微扰。 讨论这个跃迁问题， 我们很自然的把H ̂0与H ̂′ 选为： H ̂0 = 1 2m p ̂2 + V(r ⃗) H ̂′ = −e mc p ̂ ∙A ⃗ ⃗⃗ 这个时候，跃迁矩阵元为：(Ψ𝛼|− e mc p ̂ ∙A ⃗ ⃗⃗ Ψ𝛽)。这里p ̂是个动量算符，当我对系 统进行对称操作的时候，它是会变化的。A ⃗ ⃗⃗是外场，你对系统进行对称操作，对 我而言，是你的家务事，与我无关。− e mc是个常数，对称操作跟它更不相关了。 因此， 从对称性的角度来说， 我这个跃迁矩阵元为不为零， 完全由(Ψ𝛼|p ̂|Ψ𝛽) 决定。对(Ψ𝛼|p ̂|Ψ𝛽)这个矢量，我们用 mc 𝑖ℏ[r ̂, H ̂]来代替p ̂，它可以进一步化为： mc 𝑖ℏ(E𝛼−E𝛽)(Ψ𝛼|r ̂|Ψ𝛽) 由于波函数我们一般在坐标表象下写出，这时，算符r ̂等同向量r ⃗。因此，归根结 底，如果我们想用对称性的语言来描述(Ψ𝛼|H ̂′Ψ𝛽)这个外界电磁场所引起的跃迁 矩阵元的话（对应光子吸收过程） ，它的性质与(Ψ𝛼|r ⃗|Ψ𝛽)这个矩阵元完全一样。 这里，由于电偶极算符μ = −er ⃗与它只差一个正电子电荷，我们也把这种跃 迁称为电偶极跃迁。 对于这样的一个电偶极跃迁， 对称性对它的选择定则有什么样的影响呢？我 们可以看一个例子，晶体，对称性是Oh，特征标表是： 图4.12 文献中Oh特征标表截图 对(Ψ𝛼|r ⃗|Ψ𝛽)而言，由上表可知，r ⃗承载的表示是𝑇 1 −，也叫𝑇 1u。这个时候，如果 我的Ψ𝛽为𝑇2 +，也叫𝑇2g。那么这个时候𝑇 1u ⊗𝑇2g对应的特征标就是： E 3C4 2 6C4 1 6C2 1 8C3 I 3iC4 2 6iC4 1 6iC2 1 8iC3 9 1 -1 -1 0 -9 -1 1 1 0 它可以分解为：𝐴2u⨁𝐸u⨁𝑇 1u⨁𝑇2u，也就是上个表中的：𝐴2 −⨁𝐸−⨁𝑇 1 −⨁𝑇2 −。 也就是说在这个电偶极跃迁中，如果初态是一个具有𝑇2g对称性的态，那么 末态的对称性只能是上面这四种。 这里大家有两个地方可以注意： 1. 在这个表中，因为Oh属于立方系，x、y、z 三个轴等价，所以我的吸收对光的 偏振方向没有选择。 如果我把晶体对称性破坏为四方， 比如D4h = D4 ⊗{𝐸, 𝑖}： 240 图4.13 文献中D4h特征标表截图 这个时候大家就会注意到x、y 与z 承载的不可约表示就不一样了。这个时 候，光的偏振方向就会有由对称性诱发的选择性吸收了。具体你要做的，还 是上面的步骤，直积，再分解。对于同一个初态，偏振光沿z 方向的时候， 我用A2作直积；偏振光在xy 平面时，我用E来作。这样允许的末态就会不一 样了。 2. 在𝐴2u⨁𝐸u⨁𝑇 1u⨁𝑇2u这四个态中， 我们有共同的下标u， 代表我的末态是奇宇 称。这个是为什么？ 原因很简单，我的初态是偶宇称，微扰是奇宇称，那末态必须为奇宇称，不 然空间积分为零。反过来，如果我的初态是奇宇称，那么我的末态就必须是 偶宇称。也就是说电偶极跃迁只能发生在两个不同宇称的态之间。 类似讨论有意义的体系是具有中心反演对称性的体系。 这里， 我们可以通过 H ̂′的对称性，展开类似讨论。如果H ̂′对中心反演不变，那么跃迁应该是发生在具 有相同宇称的态之间了。 在上面的讨论中， 有一个很重要的细节， 不知道你们注意到没有？就是我的 哈密顿量里面，变量只有r ⃗，这个电子坐标。这就意味着我们所讨论的光吸收或 辐射对应的必须是电子态之间的跃迁。 由于电偶极算符的奇宇称， 两个电子态之 间的宇称必须相反。 但在实际的情况中， 后来人们发现， 被电偶极跃迁联系起来的两个电子态的 宇称有时是相同的。 这是为什么？这个对应的实际情况一般是跃迁过程中包含了 声子的参与。这样的话我们考虑的跃迁矩阵元就必须是这样一个东西了： (Ψv ′Ψe ′|μ ⃗⃗|ΨvΨe) 我们要做的，就是表示De、Dv、Dμ的直积，然后分解，看是否包含De ′ ⊗Dv ′ 了。 这个时候， 很多在纯电子行为中被禁戒的跃迁就可以发生了。 也就是说不是我们 的对称性分析出了问题，而是实际情况更复杂了。对这种复杂的情况，对称性的 语言依然适用，只不过复杂了一些。 4.5 红外、拉曼谱、和频光谱 声子本身， 除了对电子态之间的跃迁起辅助作用外， 它们也是可以吸收或者 散射电磁波的。 与之相应， 有两种非常常用的实验手段， 红外 （IR） 与拉曼 （Raman） 。 在分析这两种谱的时候，对称性也会帮助我们理解很多东西。 这两个里面，红外比较简单，说的是这样一个事情。我们对一个样品打入一 个连续的、 处于红外波段的光谱。 假设我们入射光的强度随频率的变化是下面图 中的这个实线。那么在每个频率，我们给样品的，是一个以这个频率振荡的电磁 场。 图4.14 红外谱示意图一 在我们的光经过样品的时候， 对于有些频率， 由于它与晶格或者分子的本征 242 振动频率相同， 这个能量会被声子吸收， 从而使得我们在不放样品的时候得到的 谱线是上图中的实线，放了样品后，得到的是虚线。如果把它们的差求出来，就 是： 图4.15 红外谱示意图二 这个谱， 反映的就是我们的样品在红外这个波段， 由于声子振动对电磁场的吸收， 因此叫红外谱。 对于这样的一个吸收，它的选择定则应该怎样理解呢？我们可以这样去想： 我们加入的电磁场 （光子） 的电场强度为E ⃗ ⃗⃗。 对于频率为ω的光子， 它与晶格本身 的处在这个频率上的一个本征振动耦合，从而损失能量，被吸收。假设这个振动 所带来的样品电偶极矩的变化是μ ⃗⃗。 这个μ ⃗⃗是由原子核偏离平衡位置而引起的电荷 重新分布决定的，叫induced dipole moment。那么由于声子与电磁场的耦合而带 来的系统的能量降低就是： H ̂′ = −E ⃗ ⃗⃗∙μ ⃗⃗ 与之相应，这个微扰所带来的跃迁矩阵元就是： (Ψv ′|H ̂′|Ψv) 其中Ψv为振动的基态，就是原子核处于平衡位置的状态，它所对应的H ̂0的系统 对称群的不可约表示是一维恒等表示。 H ̂′ = −E ⃗ ⃗⃗∙μ ⃗⃗，其中E ⃗ ⃗⃗是不依赖于晶体取向的外场（由光子给的） ，对晶体或 分子进行的对称操作对它不起作用。μ ⃗⃗是由振动引起的系统电偶极矩的变化，它 在对称操作下的变换规律与x、y、z 这些函数是一样的。当取一个偏振光只有Ex 的分量的时候，在振动中，只有μx对H ̂′有贡献。与之相应，微扰项所承载的表示 与x 所承载的表示是相同的。 这时，在H ̂′与Ψv作直积的时候，由于Ψv的表示为一维恒等表示，结果就是 只承载一个x 本身可以承载的表示。相应，Ψv ′这个吸收了IR 光子所对应的本征 振动态， 也必须承载这个x 可以承载的不可约表示。 这也就意味着在点群特征标 表中，只有那些x 承载的不可约表示是对应的本征振动可以被激发。 这样的话红外吸收的选择定则就简单了，以具有C2h对称性的晶体为例，它 的特征标表是： C2h(2/𝑚) E C2 σh I 𝑥2、𝑦2、𝑧2、 𝑥𝑦 Rz Ag 1 1 1 1 𝑧 Au 1 1 −1 −1 𝑥𝑧、𝑦𝑧 Rx、Ry Bg 1 −1 −1 1 𝑥、𝑦 Bu 1 −1 1 −1 表4.3 C2h群特征标表 由这个特征标表， 我们可以知道当偏振光沿z 轴时， 它只能激发对称性为Au的本 征振动；当偏振光沿x、y 轴时，它只能激发对称性为Bu的本征振动。 当偏振方向含z 轴分量， 也含x 或y 轴分量的时候， 它可以激发Au或者Bu， 但它怎么都不能激发Ag与Bg。 大家注意一下，这里，对下标有个选择性。u 这个奇宇称可以，g 这个偶宇 称就不行。 这个是为什么呢？原因很简单， 我们选择的这个体系具有中心反演对 称性。因此，这个系统的本征态具有特定的宇称。我们这里激发振动的时候是由 244 基态，也就是承载一维恒等表示的状态向某本征振动态激发。基态是偶宇称，中 间的微扰是奇宇称，所以末态必须是奇宇称。 当我的系统不具备中心反演对称性的时候， 如何去判断一个本征振动是否具 有红外活性呢？答案很简单， 上个例子前半部分的讨论依然成立。 也就是承载x、 y、 z 这些一次齐次函数的不可约表示所对应的本征振动依然具有红外活性。 唯一 不同的地方就是宇称不再是个好量子数了，这个不可约表示不会再有‘u’这样 的下标了。 除了红外， 以后你们不管做不做实验， 在实际科研中都很可能接触到另一类 光谱，拉曼（这个拉曼和玻色一样都是印度人，玻色没得诺奖，拉曼得了，同时 拉曼还是印度科学院的缔造者， 对近代物理学在印度的传播起了非常大的作用） 。 在拉曼谱的理解上应该说对一个真实系统电子能级与振动能级的认识是基础。 大 家可以看这样一个图像： 图4.16 拉曼谱示意图一 对任何多原子系统， 它的能级都可分为电子能级与振动能级。 在能量空间的分布 一般如上图。在电子的基态上，会比较密地分布一些振动能级。它们合在一起占 据了一些能量空间，对应的是电子基态，振动的基态与不同激发态。之后会有一 个很大的禁区， 这个禁区之后才是电子的第一激发态以及它对应的振动基态与不 同激发态。 在红外吸收中， 吸收谱对应的是声子的共振吸收。 系统的跳跃就像左边这个 图描述的。电子始终都处在基态，在吸收过程中，因为原子核动了，电子在其基 态上有个重新分布， 这个重新分布带来一个电偶极矩， 这个电偶极矩与外场耦合 产生共振。 系统的状态， 是从电子基态的声子基态， 调到电子基态的声子激发态。 而拉曼谱，它不是吸收谱，它是散射谱。它对应的物理过程是一束光照到样 品上以后，由于光子本身的电磁场，可以诱发一个样品的极化。我们把这个由光 子的电磁场诱发的样品的极化描述成一个虚的吸收， 也就是说电子从基态跃迁到 了一个虚的激发态。当这个光子被弹性地散射（也就是瑞利散射）的时候，光子 与物质相互作用后只改变动量，不改变能量，系统跳回基态。整个物理过程就像 第二列描述的这样。 （跟瑞利散射相关的最常见的自然现象就是晴天天是蓝的， 官方解释是散射强度与波长的四次方成反比，蓝光散射的多） 。 除了这个弹性散射， 很自然我们还可以想到非弹性散射。 在这个非弹性散射 中，入射光在回到电子基态的时候，会损失或得到一定的能量。损失能量过程对 应的是它回到了电子的基态，声子的激发态。光子的能量损失给声子了。这个过 程叫Stokes 散射。 而得到能量的过程对应的是系统开始处在电子基态， 声子的激 发态。散射完了以后系统回到电子基态，声子基态。与之对应，在这个过程中样 品的声子就把能量给了光子。 当时人们对声子这个东西应该说还没有很深入的理 解，但这个非弹性散射过程的特性，使得拉曼在这个工作做出（1928 年）之后， 很快（1930 年）就得了诺贝尔物理奖。这个背后的原因是什么，其实很值得我们 回味， 可能这个光子能量的改变所蕴藏的对光的粒子性支持， 应该说是很重要的 246 原因，因为那个年代是建立量子力学基本概念的年代，最热的就是这个东西。康 普顿散射也是类似的东西， 光子和电子的非弹性散射， 1923 年的实验， 1927 年就 得了诺奖。现在的科研处在一个什么样的状态，有什么样的特征，这个需要大家 在学习的过程中同样慢慢体会。一句话，物理学发展史上任何重要的东西，重不 重要， 这个要通过时间检验， 看它到底对我们对这个世界的认识增加了多少东西？ 得不得诺奖，有时候看那些年的潮流热什么？ 回到正题，基于这个对拉曼散射过程的描述，在实验里面，理想的拉曼图就 是这样的： 图4.17 拉曼谱示意图二 中间的大峰对应的是瑞利散射。对Anti-Stokes 峰，由于它需要系统开始的时候 就处在振动的激发态，所以低温下不明显，温度高一些比较好测。 而同时我们如果把弹性散射部分扣除， 我们就可以得到纯振动部分。 上面那 个图只给了一个振动模式，实际情况往往是一系列。这一系列的移动，代表的就 是系统的本征振动频率。 图4.18 拉曼谱示意图三。这是另一种画法，右边是长波长，低能部分，与上图 相反 拉曼谱和红外一样，也有个选择定则，可以通过对称性的知识去理解。但和 红外不同的是，在拉曼谱里面，诱导电偶极矩并不是由原子核运动直接产生的。 在红外里面， 吸收信号反映的是声子的本征振动与外场之间的耦合。 也就是说声 子振动产生电偶极矩， 这个电偶极矩直接与光子的电场耦合来产生吸收。 这个过 程是个一阶过程。 而拉曼谱里面， 一般有两束光产生作用。 入射光的作用是产生一个诱导电偶 极矩， 然后由入射光产生的诱导电偶极矩会和散射光的光场耦合。 由于入射光并 不激发系统的本征振动，它要产生偶极矩的话，必须通过一个极化率。这个极化 率记为𝛼 ̿，它是个3 × 3的张量。入射光的电场为E ⃗ ⃗⃗i cos(𝜔t)，𝜔为入射光频率。它 们一起产生的诱导电偶极矩是： 𝜇 ⃗= 𝛼 ̿ ∙E ⃗ ⃗⃗i cos(𝜔t) 这里𝛼 ̿叫拉曼极化率张量（Raman polarizability tensor） 。 它是随着原子核的运动有变化的。 我们把原子核在平衡位置的时候的极化率 记为𝛼 ̿0， 把原子核的运动对它的改变记为∆𝛼 ̿， 那么总的极化率就是𝛼 ̿ = 𝛼 ̿0 + ∆𝛼 ̿， 其中𝛼 ̿0不随时间变化，∆𝛼 ̿会以晶格或分子的振动频率ωv随时间变化，等于 248 ∆𝛼 ̿0 cos(ωvt)，∆𝛼 ̿0是不随时间变化的由晶格振动引起的极化率变化幅度。 这样由它们产生的诱导电偶极矩就分别为： 𝛼 ̿0 ∙E ⃗ ⃗⃗i cos(𝜔t)、 ∆𝛼 ̿ ∙E ⃗ ⃗⃗i cos(𝜔t)。 其中𝜔为入射光频率，极矩整体是： 𝜇 ⃗= 𝛼 ̿ ∙E ⃗ ⃗⃗i cos(𝜔t) = (𝛼 ̿0 + ∆𝛼 ̿) ∙E ⃗ ⃗⃗i cos(𝜔t) = (𝛼 ̿0 + ∆𝛼 ̿0 cos(ωvt)) ∙E ⃗ ⃗⃗i cos(𝜔t) = 𝛼 ̿0 ∙E ⃗ ⃗⃗i cos(𝜔t) + ∆𝛼 ̿0 2 [cos(ω −ωv)t + cos(ω + ωv)t] ∙E ⃗ ⃗⃗i 其中的第一项频率与入射光相同， 对应的是正常的瑞利散射的部分。 第二项与第 三项对应的是拉曼效应中的Stokes 与Anti-Stokes 移动。 这些拉曼所对应的系统哈密顿量的变化是： H ̂′ = −[∆𝛼 ̿0 2 cos(ω ± ωv)t] ∙E ⃗ ⃗⃗i ∙E ⃗ ⃗⃗s E ⃗ ⃗⃗s为散射光的电场强度。这样，拉曼谱中的微扰，所对应的跃迁矩阵元就是： (Ψv ′| −[ ∆𝛼 ̿0 2 cos(ω ± ωv)t] ∙E ⃗ ⃗⃗i ∙E ⃗ ⃗⃗s| Ψv) 我们现在要干的事情， 就是利用对称性， 来分析由这个跃迁矩阵元决定的选择定 则。 和红外一样， 我们的初态|Ψv)是振动基态， 对应一维恒等哈密顿算符群表示。 微扰项−[ ∆𝛼 ̿0 2 cos(ω ± ωv)t] ∙E ⃗ ⃗⃗i ∙E ⃗ ⃗⃗s里面， E ⃗ ⃗⃗i与E ⃗ ⃗⃗s是外场， 对系统的对称操作不变。 我们现在需要知道的，就是∆𝛼 ̿0这个由振动引起的二阶张量承载的是系统对称群 的哪些表示？然后与本征振动|Ψv ′)所对应的系统对称群的表示对应就行了。 而∆𝛼 ̿0这个二阶张量，代表的是由于原子核的运动对系统极化率的影响。它 对系统对称群的对称操作的变换性质与二次函数𝑥2、𝑦2、𝑧2、𝑥𝑦、𝑦𝑧、𝑥𝑧相同。 因此对一个具有特定对称群的分子或晶体， 我们要看它的那些本征振动是有 拉曼活性的？就看它这个振动对应的不可约表示是否承载𝑥2、 𝑦2、 𝑧2、 𝑥𝑦、 𝑦𝑧、 𝑥𝑧这些二次函数基就可以了。还是上面得那个C2h点群，由它的特征标表，我们 就可以看出对应其Ag与Bg不可约表示的本征振动是有拉曼活性的，对应Au与Bu 的没有。 前面我们讲红外的时候说过对应Au与Bu的本征振动是有红外活性的。这里 的下标刚好反过来了。其背后的原因，就是红外的H ̂′是奇宇称的，对应的|Ψv ′)也 要是奇宇称的，所以有红外活性的振动下标都是u。而拉曼的H ̂′是偶宇称的，对 应的|Ψv ′)也要是偶宇称的， 所以有拉曼活性的振动下标都是g。 这两个活性互补。 需要注意的是， 这种互补只对具有中心反演对称性的体系成立。 当系统没有中心 反演对称性时，我们无法通过‘u’ 、 ‘g’这些标示把红外和拉曼活性的振动区分 开。这个时候，某个振动，是允许同时具备拉曼和红外活性的。 与这个直接相关的一个例子就是这些年很流行的一个测表面振动的方法， 叫 和频光谱 （Sum frequency generation） 。 这个技术在近代非线性光学的发展里面很 重要。 其基本特征就是我要测量的是一个三阶过程， 内部包含红外吸收这个一阶 过程与拉曼吸收这个二阶过程， 我们要求这个振动同时具备红外与拉曼活性。 在 液体内部，由于液体本身的均匀性，我们一般认为系统具备中心反演对称性。对 于具备中心反演对称性的系统，由于红外与拉曼的互补，和频信号就会很弱。而 液体表面，由于中心反演对称性的破缺，红外与拉曼不再互补，这样和频信号就 会强很多。因此，和频光谱技术是为数不多的这样一门技术，具备液体表面敏感 这样一个特征[18，19]。 最后一个需要说明的地方是上面在对跃迁矩阵元： (Ψv ′| −[ ∆𝛼 ̿0 2 cos(ω ± ωv)t] ∙E ⃗ ⃗⃗i ∙E ⃗ ⃗⃗s| Ψv) 250 的讨论中，我们假定E ⃗ ⃗⃗i、E ⃗ ⃗⃗s在声子波函数的变化范围内使常数。如果在这个空间 尺度， 这些外场也随r 变化， 那么前面讨论的选择定则失效。 这个不是天方夜谭， 而是通过局域场的调制来破坏一些传统实验手段的局限。针尖增强拉曼散射 （Tip-enhanced Raman Scattering, TERS）技术的发展就是这方面的一个例子。这 方面中国科技大学董振超老师、 罗毅老师团队近期有一些合作的代表性工作， 感 兴趣的同学可以了解一下[20，21]。 4.6 平移不变性与Bloch 定理 在空间群部分，我们讲过晶体的一个重要特性就是原子（离子、或分子）排 列的周期性。这个周期性（平移对称性）可以由点阵来描述。点阵中任意一个格 点可描述为一个正格矢，R ⃗ ⃗⃗𝑙，具体形式为： R ⃗ ⃗⃗𝑙= 𝑙1a ⃗ ⃗1 + 𝑙2a ⃗ ⃗2 + 𝑙3a ⃗ ⃗3 其中𝑙1、 𝑙2、 𝑙3为整数， a ⃗ ⃗1、 a ⃗ ⃗2、 a ⃗ ⃗3为点阵的基矢， 它们构成的平行六面体为原胞。 这个原胞一般不反映晶体点阵的对称性，以fcc 格子为例，它的原胞就是： 图4.19 晶胞与元胞 从这个平行六面体，你看不出任何点阵的对称性。 要想看出这个对称性， 两种方法。 一是取晶胞， 就像上面图中的蓝色立方体， 我的cell 变大了，这个大的cell 可以反映出点阵的对称性；二是对原胞做另一种 取法，Wigner-Seitz Cell。以二维晶体为例，这个cell 的取法是我相对于原点，对 每个格点与原点的连线做平分线，所围出来的最小的面积，比如： 图4.20 Wigner-Seitz Cell 三维情况下，简立方、体心、面心的Wigner-Seitz Cell 分别是： 图4.21 简立方、体心、面心晶体的是空间Wigner-Seitz Cell 252 这些都是实空间的东西。 对于晶体中的元激发， 它们的状态我们一般可以用波矢量来描述， 波矢量对 应的是倒空间。与实空间中的点阵对应，倒空间也有点阵，它们是由b ⃗⃗1、b ⃗⃗2、b ⃗⃗3 的整数线性组合构成的，其中： b ⃗⃗1 = 2π Ω (a ⃗ ⃗2 × a ⃗ ⃗3) b ⃗⃗2 = 2π Ω (a ⃗ ⃗3 × a ⃗ ⃗1) b ⃗⃗3 = 2π Ω (a ⃗ ⃗1 × a ⃗ ⃗2) Ω为实空间中原胞体积。晶体是fcc，对应倒空间点阵是bcc；晶体是bcc，对应 倒空间点阵是fcc。 上面是我们对固体物理中一些基础知识的回顾， 现在看平移对称性。 所谓平 移对称性，指的是将晶体平移R ⃗ ⃗⃗𝑙，系统回到与原来不可分辨状态的属性。当晶体 无穷大时，平移操作{E|R ⃗ ⃗⃗𝑙}无穷多。有限晶体，我们会使用周期性边界条件，取 {E|N1a ⃗ ⃗1} = {E|N2a ⃗ ⃗2} = {E|N3a ⃗ ⃗3} = {E|0}。 这样的话由元素{E|R ⃗ ⃗⃗𝑙}形成的集合是构 成一个群的，因为： １．任意两个平移的乘积仍为一个形势为{E|R ⃗ ⃗⃗𝑙}平移； ２．结合律； ３．{E|R ⃗ ⃗⃗𝑙}逆为{E| −R ⃗ ⃗⃗𝑙}； ４．恒等操作{E|0}。 这个群称为平移群。 现在我们知道了晶体中存在平移群， 这会带来什么后果呢？答案很简单， 就 是布洛赫定理。在讲这个定理之前我们先明确一点，就是平移群是个阿贝尔群。 因此，对于由{E|N1a ⃗ ⃗1} = {E|N2a ⃗ ⃗2} = {E|N3a ⃗ ⃗3} = {E|0}这个周期性边界条件定义 的晶体， 平移群的阶为N1 × N2 × N3， 类的个数也是N1 × N2 × N3， 有N1 × N2 × N3 个一维的不等价不可约表示。 以固体中的电子， 这样一个处在晶格周期场中的量子的粒子为例， 它的元激 发对应某本征态。这个本征态，按照前面系统对称群部分的讨论，它承载固体的 一个哈密顿算符群表示。 这里平移群是固体的系统对称群， 群元为g。 与之相应， 有个哈密顿算符群，群元是P ̂ g。晶体中的电子态是要承载这个哈密顿算符群表示 的。 在求这个表示的过程中， 只需要知道平移群基本生成元{E|a ⃗ ⃗1}、 {E|a ⃗ ⃗2}、 {E|a ⃗ ⃗3} 所对应的P ̂ g的表示矩阵，整个哈密顿算符群表示矩阵就出来了。 以{E|a ⃗ ⃗1}为例， 对应P ̂ gψ(r ⃗) = ψ(r ⃗−a ⃗ ⃗1) = D({E|a ⃗ ⃗1}) ψ(r ⃗)。 由于周期性边界条 件，有DN1({E|a ⃗ ⃗1})ψ(r ⃗) = ψ(r ⃗−N1a ⃗ ⃗1) = ψ(r ⃗)，对∀r ⃗成立。因此有DN1({E|a ⃗ ⃗1}) = 1。 对{E|a ⃗ ⃗2}、 {E|a ⃗ ⃗3}， 同样有DN2({E|a ⃗ ⃗2}) = 1、 DN3({E|a ⃗ ⃗3}) = 1。 这些是对D({E|a ⃗ ⃗1})、 D({E|a ⃗ ⃗2})、D({E|a ⃗ ⃗3})的要求。当D({E|a ⃗ ⃗1})、D({E|a ⃗ ⃗2})、D({E|a ⃗ ⃗3})定下以后，不 可约表示就定下了。 要让DN1({E|a ⃗ ⃗1}) = 1、DN2({E|a ⃗ ⃗2}) = 1、DN3({E|a ⃗ ⃗3}) = 1，我们只需要取 D({E|a ⃗ ⃗1}) = exp [2πi n1 N1]、D({E|a ⃗ ⃗2}) = exp [2πi n2 N2]、D({E|a ⃗ ⃗3}) = exp [2πi n3 N3]，就 满足要求。也就是说平移群的一维不等价、不可约表示，最终可以用 n1 N1、 n2 N2、 n3 N3 这个数组来标识。在这个数组确定后，在这个确定的不等价不可约表示中，平移 操作 R ⃗ ⃗⃗𝑙= 𝑙1a ⃗ ⃗1 + 𝑙2a ⃗ ⃗2 + 𝑙3a ⃗ ⃗3所对应的表示矩阵就是： D({E|R ⃗ ⃗⃗𝑙}) = exp [2πi (n1 N1 𝑙1 + n2 N2 𝑙2 + n3 N3 𝑙3)] 这个时候，引入我们之前关于倒空间的讨论，结合b ⃗⃗1、b ⃗⃗2、b ⃗⃗3与a ⃗ ⃗1、a ⃗ ⃗2、a ⃗ ⃗3 的关系a ⃗ ⃗i ∙b ⃗⃗j = 2πδij，我们就可以用倒空间中的向量： k ⃗⃗= n1 N1 b ⃗⃗1 + n2 N2 b ⃗⃗2 + n3 N3 b ⃗⃗3 254 来标识平移群的一维不等价、不可约表示，而这里的k ⃗⃗，就是倒空间中第一布里 渊区的点。相应的表示就是： Dk ⃗ ⃗⃗({E|R ⃗ ⃗⃗𝑙}) = exp[ik ⃗⃗∙R ⃗ ⃗⃗𝑙] 这样的话我们用k ⃗⃗来标识电子的本征态ψk ⃗ ⃗⃗(r ⃗)， 这个本征态波函数就会具有下 面的特征。当P ̂{E|R ⃗ ⃗⃗l}作用到ψk ⃗ ⃗⃗(r ⃗)上时，一方面有： P ̂{E|R ⃗ ⃗⃗l}ψk ⃗ ⃗⃗(r ⃗) = ψk ⃗ ⃗⃗(r ⃗−R ⃗ ⃗⃗l) 另一方面有： P ̂{E|R ⃗ ⃗⃗l}ψk ⃗ ⃗⃗(r ⃗) = Dk ⃗ ⃗⃗({E|R ⃗ ⃗⃗𝑙})ψk ⃗ ⃗⃗(r ⃗) = exp[ik ⃗⃗∙R ⃗ ⃗⃗𝑙]ψk ⃗ ⃗⃗(r ⃗) 因此： ψk ⃗ ⃗⃗(r ⃗−R ⃗ ⃗⃗l) = exp[ik ⃗⃗∙R ⃗ ⃗⃗l]ψk ⃗ ⃗⃗(r ⃗) 或： ψk ⃗ ⃗⃗(r ⃗+ R ⃗ ⃗⃗l) = exp[−ik ⃗⃗∙R ⃗ ⃗⃗l]ψk ⃗ ⃗⃗(r ⃗) 这个就是Bloch 定理。 由Bloch 定理，我们知道，如果我们把ψk ⃗ ⃗⃗(r ⃗)写成exp[−ik ⃗⃗∙r ⃗]uk ⃗ ⃗⃗(r ⃗)的形式， 那么ψk ⃗ ⃗⃗(r ⃗+ R ⃗ ⃗⃗l)一方面会等于exp[−ik ⃗⃗∙(r ⃗+ R ⃗ ⃗⃗l)]uk ⃗ ⃗⃗(r ⃗+ R ⃗ ⃗⃗l)，另一方面又等于 exp[−ik ⃗⃗∙R ⃗ ⃗⃗l]ψk ⃗ ⃗⃗(r ⃗) = exp[−ik ⃗⃗∙(r ⃗+ R ⃗ ⃗⃗l)]uk ⃗ ⃗⃗(r ⃗) 因此： uk ⃗ ⃗⃗(r ⃗+ R ⃗ ⃗⃗l) = uk ⃗ ⃗⃗(r ⃗) 也就是说处在晶格周期场中元激发的本征态波函数，一定可以写成exp[−ik ⃗⃗∙ r ⃗]uk ⃗ ⃗⃗(r ⃗)（其中uk ⃗ ⃗⃗(r ⃗)为晶格周期函数）的形式。而这里的k ⃗⃗，就是倒空间第一布里 渊区中的点。换句话说，处在晶格周期场中本征元激发，都可以用倒空间第一布 里渊区的点来标识， 其相应的波函数， 具备exp[−ik ⃗⃗∙r ⃗]uk ⃗ ⃗⃗(r ⃗)的特征。 这个是由晶 格的平移对称性决定的。 4.7 布里渊区与晶格对称性 前面的讨论主要关注的是晶体中平移对称性带来的晶体中本征激发的性质。 除了平移，点群空间群部分我们已经说过，晶体中还有转动对称性。这些转动对 称性也会对我们晶体中的本征激发带来很多内在的属性。 其中最重要的， 就是前 面提到标识晶格周期场中本征激发的第一布里渊区的点， 可以通过转动对称性的 折叠，缩小到一个很小的区域，叫不可约（irreducible）布里渊区。下面我们就来 详细解释这个为什么发生？ 出发点是晶体空间群的基本操作，{α|t ⃗}。这里α = E时，t ⃗只能为R ⃗ ⃗⃗𝑙，这个时 候它就是平移群。当α为非恒等转动时，t ⃗可以不为R ⃗ ⃗⃗𝑙。它们所有的组合，形成空 间群。{α|t ⃗}这些元素的逆、乘法满足的规律是： {α|t ⃗}r ⃗= αr ⃗+ t ⃗ {α|t ⃗}{β|s ⃗}r ⃗= {α|t ⃗}(βr ⃗+ s ⃗) = αβr ⃗+ αs ⃗+ t ⃗= {αβ|αs ⃗+ t ⃗}r ⃗ 因此： {α|t ⃗}{β|s ⃗} = {αβ|αs ⃗+ t ⃗} 要想让{α|t ⃗}{β|s ⃗} = {E|0}，需要：{β|s ⃗} = {α−1| −α−1t ⃗}。因此 {α|t ⃗} −1 = {α−1| −α−1t ⃗} 这些是空间群群元的性质。 这些对称元素的存在会对晶体场中的本征激发带来什么影响呢？我们还是 以电子的元激发为例。 {α|t ⃗}为对称操作， P ̂{α|t ⃗}为其对应的函数变换算符。 由本章 第一节的讨论我们知道： P ̂ {α|t ⃗} −1 H ̂P ̂{α|t ⃗} = H ̂ 256 同时， 因为{E|−R ⃗ ⃗⃗𝑙}{α|t ⃗} = {α|−R ⃗ ⃗⃗𝑙+ t ⃗}、 {α|t ⃗}{E| −α−1R ⃗ ⃗⃗𝑙} = {α| −R ⃗ ⃗⃗𝑙+ t ⃗}， 所以： {E| −R ⃗ ⃗⃗𝑙}{α|t ⃗} = {α|t ⃗}{E| −α−1R ⃗ ⃗⃗𝑙} 基于这些性质，我们知道对P ̂{α|t ⃗}、P ̂{E|R ⃗ ⃗⃗𝑙}，有： P ̂{E|−R ⃗ ⃗⃗𝑙}P ̂{α|t ⃗}ψk ⃗ ⃗⃗(r ⃗) = P ̂{α|t ⃗}P ̂{E|−α−1R ⃗ ⃗⃗𝑙}ψk ⃗ ⃗⃗(r ⃗) = P ̂{α|t ⃗}exp[ik ⃗⃗∙(−α−1R ⃗ ⃗⃗𝑙)]ψk ⃗ ⃗⃗(r ⃗) = P ̂{α|t ⃗}exp[−iαk ⃗⃗∙R ⃗ ⃗⃗𝑙]ψk ⃗ ⃗⃗(r ⃗) = exp[−iαk ⃗⃗∙R ⃗ ⃗⃗𝑙]P ̂{α|t ⃗}ψk ⃗ ⃗⃗(r ⃗) 这也就是说P ̂{α|t ⃗}ψk ⃗ ⃗⃗(r ⃗)对于平移群来说，可承载αk ⃗⃗这个第一布里渊区的k ⃗⃗点所对 应的不可约表示。 而另一方面， 由前面的讨论， 我们知道ψαk ⃗ ⃗⃗(r ⃗)本身也承载αk ⃗⃗这个第一布里渊 区的k ⃗⃗点所对应的不可约表示。当band index 一样时，它们必对应相同的线性空 间。因此： ψαk ⃗ ⃗⃗(r ⃗) = λP ̂{α|t ⃗}ψk ⃗ ⃗⃗(r ⃗) 两者都归一的话|λ| = 1。 由这个关系，我们看ψαk ⃗ ⃗⃗(r ⃗)这个本征态的本征能量与ψk ⃗ ⃗⃗(r ⃗)这个本征态的本 征能量存在什么样的关系？答案很简单： Eαk ⃗ ⃗⃗= (ψαk ⃗ ⃗⃗(r ⃗)|H ̂ |ψαk ⃗ ⃗⃗(r ⃗)) = (λP ̂{α|t ⃗}ψk ⃗ ⃗⃗(r ⃗)| H ̂ |λP ̂{α|t ⃗}ψk ⃗ ⃗⃗(r ⃗)) = (P ̂{α|t ⃗}ψk ⃗ ⃗⃗(r ⃗)| H ̂ |P ̂{α|t ⃗}ψk ⃗ ⃗⃗(r ⃗)) = (ψk ⃗ ⃗⃗(r ⃗)|P ̂ {α|t ⃗} −1 H ̂ P ̂{α|t ⃗}|ψk ⃗ ⃗⃗(r ⃗)) = (ψk ⃗ ⃗⃗(r ⃗)|H ̂ |ψk ⃗ ⃗⃗(r ⃗)) = Ek ⃗ ⃗⃗ 也就是说对于空间群，只要存在群元{α|t ⃗}，这里的t ⃗不要求为R ⃗ ⃗⃗𝑙（晶格矢的整数 倍） ， 都可以使αk ⃗⃗与k ⃗⃗所对应的本征态能量相等 （当然， band index ‘n’必须相同） 。 这也是为什么对固体能带，我们最关心的其实不是晶体的点群，而是晶体‘空间 群的点群’ 。 也就是取空间群所有元素{α|t ⃗}， 把α单独拿出来， 形成的转动操作的 集合。布里渊区的转动对称性，是由空间群的点群决定的。 同时，在画能带的时候，我们也不需要把布里渊区所有点都画出来。我们只 需要画不可约的布里渊区就可以了。 以二维系统为例， 如果空间群的点群是D4， 不可约布里渊区就是（阴影部分） ： 图4.22 不可约布里渊区示意图一 如果空间群的点群是C4，不可约布里渊区就是： 图4.23 不可约布里渊区示意图二 空间群的点群对称性越高，不可约布里渊区越小。 258 4.8 时间反演对称性 最后一节我们讲时间反演对称性（之前刚才不管是转动还是平移都是空间 的） 。这一节的内容我们先做一个概念性的整体介绍，这个整体介绍很好理解， 也对你们了解一些与时间反演对称性相关的基本规律有好处。 之后， 我们做一个 更深层次的理论层面的讲解。 先看概念性介绍。 时间反演是改变时间符号的操作， 它对我们系统的主要物 理量带来的变化是：t变为−t、r ⃗不变、p ⃗⃗变为−p ⃗⃗（或者说k ⃗⃗变为−k ⃗⃗） 、L ⃗⃗变为−L ⃗⃗、 σ ⃗ ⃗⃗变为−σ ⃗ ⃗⃗。在无外磁场，且系统没有固有磁序的时候，由于动能正比于p ⃗⃗2、势能 V(r ⃗)不变，所以哈密顿量不发生变化，系统具有时间反演对称性。有外场，或者 系统具有固有磁序的时候， 哈密顿量中多了B ⃗ ⃗⃗∙σ ⃗ ⃗⃗这一项， 在时间反演操作下反号， 而其它项不变，所以总哈密顿量变化，系统不具备时间反演对称性。 对电子本征激发Bloch 态ψn,k ⃗ ⃗⃗,↑(r ⃗)， 它的时间反演态是ψn,−k ⃗ ⃗⃗,↓(r ⃗)。 无外磁场， 且系统没有固有磁序时， 由于哈密顿量具备时间反演对称性， 对于晶体能级能量， 存在： En,k ⃗ ⃗⃗,↑= En,−k ⃗ ⃗⃗,↓ 这个简并是由时间反演对称性要求的。 在时间反演对称性的基础上， 如果系统继续有空间反演对称性， 那么继续有： En,k ⃗ ⃗⃗,↑= En,−k ⃗ ⃗⃗,↑ 这也就是说，当空间与时间反演对称性同时存在，这两个式子一结合，就有： En,k ⃗ ⃗⃗,↑＝En,k ⃗ ⃗⃗,↓ 也就是说同一个波矢的两个不同自旋态相互简并。 同时，由于En,k ⃗ ⃗⃗,↑＝En,k ⃗ ⃗⃗,↓、En,k ⃗ ⃗⃗,↑= En,−k ⃗ ⃗⃗,↓、En,−k ⃗ ⃗⃗,↑= En,k ⃗ ⃗⃗,↓，有： En,k ⃗ ⃗⃗,↑＝En,k ⃗ ⃗⃗,↓= En,−k ⃗ ⃗⃗,↑＝En,−k ⃗ ⃗⃗,↓ 也就是在系统同时具有时间、 空间反演对称性的时候， En,k ⃗ ⃗⃗,σ对k ⃗⃗、 对σ的正负号都 有简并的特征。 因为这个原因，当一个系统既有time-reversal symmetry，又有iversion symmetry 的时候，它的electronic bands 必有spin 简并的特征。在一些关于电子 结构的讨论中，你们应该经常会看到一些类似讨论句子，比如：In order to break the spin degeneracy, one has to break either the timer-reversal or the inversion symmetry，说的就是这个道理。 空间反演对称性如果去除， 时间反演对称性依然要求En,k ⃗ ⃗⃗,↑= En,−k ⃗ ⃗⃗,↓。 前些年 比较热的拓扑绝缘体中， 一个基本特征是具有导电的边缘态， 能带具有如下特征： 图4.24 时间反演要求的能带简并 这里的能带交叉点对应的简并就是被时间反演对称性保护的。 现在看背后更深层次的原理性的东西。 如果你只想理解上面的内容的话， 这 个东西本来可以不讲。 但前两年有细心的同学， 会问我一个关于晶体点群特征标 表（附录A）的问题。对于C3、C4、C6、C3h、C4h、C6h、S4、S6、T这些点群， 有个有意思的情况， 就是我们有时会把两个一维表示放在一起用E 来标识。 根据 我们以前将的习惯，E 一般是用来标识二维不可约表示的。这里为什么要这样处 理呢？这个时候， 如果你再细心点， 你会发现放在一起的两个一维表示是相互共 轭的。相互共轭就意味着它们的表示可以写为D与D∗，背后所对应的物理就是时 260 间反演对称性可以让这两个不被空间的点群对称性要求简并的量子态简并。 而要 理解这个，我们又必须从头说起。 前面说过，时间反演是一种操作。我们可以把它记作T ̂，它做的事情是：t变 为−t、r ⃗不变、p ⃗⃗变为−p ⃗⃗（或者说k ⃗⃗变为−k ⃗⃗） 、L ⃗⃗变为−L ⃗⃗、σ ⃗ ⃗⃗变为−σ ⃗ ⃗⃗。它联系起来的 是两个量子态： 原来的态ψ(r ⃗, t)与它的时间反演共轭态ψ(r ⃗, −t)， 通过： ψ(r ⃗, −t) = T ̂ψ(r ⃗, t)。下面我们来理解T ̂在数学上等效于什么？ 现在先不考虑自旋， 假设系统哈密顿量具有时间反演对称性， 那么这个不考 虑自旋的粒子的含时波函数满足的方程是： iℏ∂ψ(r ⃗, t) ∂t = H ̂(r ⃗, t)ψ(r ⃗, t) t = 0时刻的波函数是定态波函数，可以写成实函数ψ(r ⃗, 0)24。这样的话，沿时间 正轴方向演化的含时波函数就是： ψ(r ⃗, t) = 𝑒−𝑖∫H ̂(r ⃗ ⃗,t′)dt′ t 0 /ℏψ(r ⃗, 0) 含时薛定谔方程的时间轴反向，有ψ(r ⃗, −t)等于： ψ(r ⃗, −t) = 𝑒−𝑖∫ H ̂(r ⃗ ⃗,t′)dt′ −t 0 /ℏψ(r ⃗, 0) 当系统具有时间反演对称性的时候，H ̂(r ⃗, −t′) = H ̂(r ⃗, t′)。因此： ∫H ̂(r ⃗, t′)dt′ −t 0 = −∫H ̂(r ⃗, t′)dt′ t 0 这样的话就有： ψ(r ⃗, −t) = 𝑒𝑖∫H ̂(r ⃗ ⃗,t′)dt′ t 0 /ℏψ(r ⃗, 0) 也就是ψ(r ⃗, −t) = ψ∗(r ⃗, t)。这也就是说在不考虑自旋的时候，时间反演算符T ̂就 24定态波函数满足H ̂(r ⃗)ψ(r ⃗, 0) = Eψ(r ⃗, 0)，E 是实数。对这个方程取转置共轭，有H ̂+(r ⃗)ψ∗(r ⃗, 0) = H ̂(r ⃗)ψ∗(r ⃗, 0) = Eψ∗(r ⃗, 0) 。由这个，可知H ̂(r ⃗)[ψ(r ⃗, 0) + ψ∗(r ⃗, 0)] = E[ψ(r ⃗, 0) + ψ∗(r ⃗, 0)] 。而 ψ(r ⃗, 0) + ψ∗(r ⃗, 0)是实函数。这也就是说定态波函数总可以写成实函数。 等于复数共轭算符K ̂。 考虑自旋，最简单的level，自旋轨道耦合，哈密顿量就变成了： H ̂(r ⃗, t) = 1 2m p ̂2 + V(r ⃗) + 1 4m2c2 σ ⃗ ⃗⃗∙(∇V(r ⃗) × p ̂) 这个时候，为了保证这个哈密顿量在T ̂下不变，就要求T ̂σ ⃗ ⃗⃗= −σ ⃗ ⃗⃗。前两项不变是 在不考虑自旋的时候已经讨论过的，第三项要想不变，p ̂变号了，∇V(r ⃗)没有，所 以σ ⃗ ⃗⃗必须变号。这也就是说时间反演对称性要求T ̂σ ⃗ ⃗⃗= −σ ⃗ ⃗⃗T ̂。 怎么才能让T ̂σ ⃗ ⃗⃗= −σ ⃗ ⃗⃗T ̂呢？我们就需要利用泡利矩阵的性质了。取T ̂ = K ̂σy， 看这样能不能满足T ̂σ ⃗ ⃗⃗= −σ ⃗ ⃗⃗T ̂的要求？σ ⃗ ⃗⃗是：[σx、σy、σz]。其中σx = (0 1 1 0)、 σy = (0 −𝑖 𝑖 0 )、σz = (1 0 0 −1)。T ̂ = K ̂σy作用到它上面的后果是： T ̂σ ⃗ ⃗⃗= K ̂σyσ ⃗ ⃗⃗= K ̂σy[σx、σy、σz] = [K ̂σyσx、K ̂σyσy、K ̂σyσz] 下一步利用到的性质是：σyσx = −σxσy，进而K ̂σyσx = −K ̂σxσy = −σxK ̂σy； σyσy = σyσy，进而K ̂σyσy = −σyK ̂σy；σyσz = −σzσy，进而K ̂σyσz = −K ̂σzσy = −σzK ̂σy。这样综合上面的式子，我们就有： T ̂σ ⃗ ⃗⃗= K ̂σyσ ⃗ ⃗⃗= −[σxK ̂σy、σyK ̂σy、σzK ̂σy] = −σ ⃗ ⃗⃗K ̂σy = −σ ⃗ ⃗⃗T ̂ 也就是T ̂让σ ⃗ ⃗⃗反号了。综合一下，就是说不考虑自旋时T ̂ = K ̂，考虑时T ̂ = K ̂σy。 前面提到的附录A 中点群属于不考虑自旋的情况。这个时候，以C4这个点 群为例，它的特征标表是： C4(4) E C2 C4 C4 3 x2 + y2、z2 Rz、z A 1 1 1 1 x2 −y2、xy B 1 -1 1 -1 (xz、yz) (xz2、yz2) (x、y) (Rx、Ry) E 1 i -1 -i 1 -i -1 i 262 表4.4 C4群特征标表 按理说点群对称性是不要求E 这两个不可约表示简并的，这里时间反演对称性 就起作用了。因为上面那个表示我们记作D，它的基是ψ。由于有时间反演对称 性，这个ψ我们可以把它变作ψ∗。D 对ψ的那些变换也可以对应D∗对ψ∗的变化。 这样， 点群对称性加上时间反演对称性{E ̂、T ̂}， ψ 与ψ∗也就通过对称操作联系起 来了，它们的能量自然简并。 4.9 习题与思考 1. 根据C3v群特征标表，将下面四个函数：1) z、2) xy、3) x2、4) y2形成的线性 空间，约化为C3v群的群不不变子空间的直和； 2. D3群的表示空间为x、 y、 z的六个二次齐次函数组成的六维空间。 试用投影算 符的方法将它们组合为六个对称化的新基， 并写出d、 f 在这组对称化的新基 下的表示矩阵？ 3. 一个杂质原子放到一个晶体中，假设晶体场对称性为O。不考虑自旋轨道耦 合， 结合O 群特征标表讨论原子d 轨道的劈裂情况。 这里要用到下一章的一 点知识， 就是一个角度为α的转动， 在转动群， 也就是原子不考虑自旋轨道耦 合的对称群中，特征标是 sin [(𝑙+1/2)α] sin [α/2] ，对d 轨道𝑙= 2。 4. （接上题）之后，我们对此单晶沿z 方向均匀拉伸，晶体场对称群变成了什 么？这些轨道又会进行什么样的劈裂？ 5. （接上题）之后，我们对此单晶沿y 方向在进行一个不同于z 方向的均匀拉 伸，晶体场对称群又变成了什么？这些轨道又会进行什么样的劈裂？ 6. 根据下图，以a 轴为C2 (1)，b 轴为C2 (2)，c 轴为C2 (3)，定义C3v群。用投影算符 的方法说明从1) z，2) xy 出发，生成的C3v群的表示空间是什么（指出几维， 以及线性无关的基函数）？他们承载的是哪些不可约表示？ 7. 小明生长出来一种晶体，为了对其结构有所了解，他做了一个红外谱实验， 又做了一个拉曼谱实验。他发现在1700cm-1、1750cm-1 的位置（具体数字不 重要） ， 两种谱线都有明显的振动峰。 基于这些观测， 我们在下面四种点群中， 可以排除哪些，为什么？ a) Th b) C3v c) Td d) D3d 8. 晶体结构是fcc，布里渊区如下： 在理解Γ、 X、 K 点的本征电子态时， 分别应基于哪些点群的特征标表来分析？ 从对称性的角度考虑，从Γ点向X 点移动的过程中，简并度一般是升高还是 降低？ 264 第五章 转动群 前面在点群部分我们曾经说过三维实正交群O(3)与三维实特殊正交群SO(3)， 那里的三维实特殊正交群SO(3)就是我们这里要讲的三维转动群，它是三维实正 交群O(3)中行列式为1 的部分。物理中的中心力场问题都与这样的一个群相关， 它是非阿贝尔李群 （群元有连续参数、 参数之间有关系、 关系可有解析函数表达） 这种连续群的一个例子。 学习转动群这一章， 我们的核心任务是三个： 1) SO(3)与SU(2)群是什么， 它 们有怎样的关系？2) SO(3)与SU(2)群的不可约表示是什么？3)在物理系统中有 什么用。 这三个内容我们分三节来讲。 最后加个C-G 系数， 大家以后再用到这门 课程内容的时候会有用。 5.1 SO(3)群与二维特殊酉群SU(2) 我们前面讲过SO(3)群中的元素可以用Ck ⃗ ⃗⃗(𝜓)， 其中k ⃗⃗(𝜃, 𝜑)是转动轴， 𝜓是转 角。当k ⃗⃗为î、ĵ、k ̂中的k ̂时，î、ĵ与之垂直， Ck ⃗ ⃗⃗(𝜓) = ( cos 𝜓 −sin 𝜓 0 sin 𝜓 cos 𝜓 0 0 0 1 ) 而SO(3)群中元素进行的操作，说白了，就是将一个球面转到与其重合的另一个 位置， 且不改变手性。 从球心到球面上的三个向量在转动后夹角不变、 手性不变。 由于这个原因， SO(3)群中的元素可以用欧拉角𝛼、 β、 γ来标记， 记作𝑅(𝛼, 𝛽, 𝛾)。 这些欧拉角怎么定义呢？ 设Oxyz是三维欧式空间中固定的笛卡尔坐标系，𝑅(𝛼, 𝛽, 𝛾)是SO(3)群中的元 素，它可以表示为三个连续转动的乘积。这三个连续转动的定义是： 1. 先绕z 轴转𝛼角，0 ≤𝛼< 2π，此时坐标系由Oxyz变为Ox′y′z′； 2. 再绕y ⃗ ⃗′转𝛽角，0 ≤𝛽≤π，此时Ox′y′z′变为Ox′′y′′z′′； 3. 最后绕z ⃗′′转𝛾角，0 ≤𝛾< 2π，此时Ox′′y′′z′′变为Ox′′′y′′′z′′′。 图5.1 欧拉角 这三个合在一起，就意味着： 𝑅(𝛼, 𝛽, 𝛾) = Cz ⃗⃗′′(𝛾)Cy ⃗ ⃗⃗′(𝛽)Cz ⃗⃗(𝛼) 这里为什么要求0 ≤𝛼< 2π、 0 ≤𝛽≤π、 0 ≤𝛾< 2π呢？大家可以想象一个 球，它有八个象限。𝛽这个转动的作用，是使得上面北半球中的任意一个点，可 以到达南半球。而𝛼、𝛾是使得南北半球内四个象限可以互换。所以𝛽从0 到π就 够了，但为了北极点能到南极点，这个π是闭的。而𝛼、𝛾的2π开的就可以了。 除了这点，还有一点需要说明一下。就是𝛽= 0的时候，𝛼+ 𝛾相同的操作对 应的是同一个SO(3)群中的元素； 当𝛽= π时， 𝛼−𝛾相同的操作对应同一个SO(3) 群中的元素。这也就是说同一个类似的特殊转动，在使用欧拉角描述的时候，存 在多种欧拉角的组合对应同一个转动的情况。不过我们写成SO(3)群的矩阵表示 的时候， 这种多种组合的表示又会归一到同一个矩阵表示， 这个一会儿我们会解 释。 266 这两点说明之后， 我们就来看一下在使用欧拉角表示三维转动的时候， 这个 表示矩阵应该是什么样子？我们取的基是i ⃗、j ⃗、k ⃗⃗，沿x、y、z 方向。之前我们说 了，R(α, β, γ) = Cz ⃗⃗′′(γ)Cy ⃗ ⃗⃗′(β)Cz ⃗⃗(α)，因此要求𝑅(𝛼, 𝛽, 𝛾)，知道Cz ⃗⃗′′(γ)、Cy ⃗ ⃗⃗′(β)、 Cz ⃗⃗(α)在i ⃗、j ⃗、k ⃗⃗下是什么就可以了。 这三个里面最简单的肯定是Cz ⃗⃗(𝛼)， 因为它是绕着k ⃗⃗旋转 𝛼角的操作， 在i ⃗、 j ⃗、 k ⃗⃗下矩阵形式为： Cz ⃗⃗(𝛼) = ( cos 𝛼 −sin 𝛼 0 sin 𝛼 cos 𝛼 0 0 0 1 ) 现在的任务就是要知道Cz ⃗⃗′′(γ)、 Cy ⃗ ⃗⃗′(β)在i ⃗、 j ⃗、 k ⃗⃗下是什么？为了写出这两个矩阵， 我们先进行下面一个简短的讨论。之前做过，就是相似变换。有两组基 (e ⃗ ⃗1, e ⃗ ⃗2, ⋯, e ⃗ ⃗n)与(f ⃗1, f ⃗2, ⋯, f ⃗n)， 我们把前者称为旧基B， 后者称为新基B′， 两者由： (f ⃗1, f ⃗2, ⋯, f ⃗n) = (e ⃗ ⃗1, e ⃗ ⃗2, ⋯, e ⃗ ⃗n)(P) 联系起来。 对于一个线性变换A，它在旧基B下的矩阵为[A]B与它在新基B′下的矩阵形 式[A]B′的关系就是： [A]B = P[A]B′P−1 由这个关系我们知道转动Cy ⃗ ⃗⃗′(β)在坐标系Ox′y′z′下的矩阵为： ( cos 𝛽 0 sin 𝛽 0 1 0 −sin 𝛽 0 cos 𝛽 ) 我们需要求出的， 是它在Oxyz这个旧基下的表示。 我们知道新基Ox′y′z′与旧基Oxyz 的联系，这个联系是： (i ⃗′, j ⃗′, k ⃗⃗′) = (i ⃗, j ⃗, k ⃗⃗) ( cos 𝛼 −sin 𝛼 0 sin 𝛼 cos 𝛼 0 0 0 1 ) 套用上面的关系，Cy ⃗ ⃗⃗′(β)在Oxyz中的表示就是： Cy ⃗ ⃗⃗′(β) = ( cos α −sin α 0 sin α cos α 0 0 0 1 ) ( cos β 0 sin β 0 1 0 −sin β 0 cos β ) ( cos α −sin α 0 sin α cos α 0 0 0 1 ) −1 同理，Cz ⃗⃗′′(γ)在Ox′′y′′z′′下的表示是： ( cos 𝛾 −sin 𝛾 0 sin 𝛾 cos 𝛾 0 0 0 1 ) 而Ox′′y′′z′′的基(i ⃗′′, j ⃗′′, k ⃗⃗′′)与Oxyz的基(i ⃗, j ⃗, k ⃗⃗)的联系是： (i ⃗′′, j ⃗′′, k ⃗⃗′′) = (i ⃗, j ⃗, k ⃗⃗) ( cos α −sin α 0 sin α cos α 0 0 0 1 ) ( cos β 0 sin β 0 1 0 −sin β 0 cos β ) 这样的话Cz ⃗⃗′′(γ)在Oxyz下的表示就是： Cz ⃗⃗′′(γ) = ( cos α −sin α 0 sin α cos α 0 0 0 1 ) ( cos β 0 sin β 0 1 0 −sin β 0 cos β ) ( cos γ −sin γ 0 sin γ cos γ 0 0 0 1 ) ( cos β 0 sin β 0 1 0 −sin β 0 cos β ) −1 ( cos α −sin α 0 sin α cos α 0 0 0 1 ) −1 最后，Cz ⃗⃗′′(γ)Cy ⃗ ⃗⃗′(β)Cz ⃗⃗(α)等于： ( cos α −sin α 0 sin α cos α 0 0 0 1 ) ( cos β 0 sin β 0 1 0 −sin β 0 cos β ) ( cos γ −sin γ 0 sin γ cos γ 0 0 0 1 ) ( cos β 0 sin β 0 1 0 −sin β 0 cos β ) −1 ( cos α −sin α 0 sin α cos α 0 0 0 1 ) −1 ( cos α −sin α 0 sin α cos α 0 0 0 1 ) ( cos β 0 sin β 0 1 0 −sin β 0 cos β ) ( cos α −sin α 0 sin α cos α 0 0 0 1 ) −1 ( cos α −sin α 0 sin α cos α 0 0 0 1 ) = ( cos α −sin α 0 sin α cos α 0 0 0 1 ) ( cos β 0 sin β 0 1 0 −sin β 0 cos β ) ( cos γ −sin γ 0 sin γ cos γ 0 0 0 1 ) = ( cos α cos β cos γ −sin α sin γ −cos α cos β sin γ −sin α cos γ cos α sin β sin α cos β cos γ + cos α sin γ −sin α cos β sin γ + cos α cos γ sin α sin β −sin β cos γ sin β sin γ cos β ) 268 从这个矩阵形式， 大家也可以很容易看出𝛽= 0时， 𝛼+ 𝛾相同对应同一转动； 𝛽= 𝜋时，𝛼−𝛾相同对应同一转动。 现在我们知道了SO(3)群一个元素在三维实空间中用欧拉角是怎么描述的了。 在转动群这一章， 最核心的一个地方， 应该说是利用一个叫二维特殊酉群 （SU(2) 群）与SO(3)群的同态关系，来讨论SO(3)群的不可约表示以及它在具体物理系 统中的应用。因此，在介绍完SO(3)群的群元之后很自然的一个任务就是介绍这 个SU(2)群，以及它与SO(3)群的同态映射关系。 这个SU(2)群叫二维特殊酉群，它是由行列式为1 的二阶幺正矩阵组成的。 由这个条件，如果我们假设其中元素为： u = (𝑎 𝑏 𝑐 𝑑) 其中𝑎、𝑏、𝑐、𝑑∈C（复数） 。那么，由酉群这个限制，我们就会由： (𝑎 𝑏 𝑐 𝑑) (𝑎∗ 𝑐∗ 𝑏∗ 𝑑∗) = E 得到： 𝑎𝑎∗+ 𝑏𝑏∗= 1 𝑐𝑐∗+ 𝑑𝑑∗= 1 𝑎𝑐∗+ 𝑏𝑑∗= 0 以及： 𝑎𝑑−𝑏𝑐= 1 四个独立的限制条件。 再由𝑎𝑐∗+ 𝑏𝑑∗= 0，我们可得： 𝑎∗𝑐+ 𝑏∗𝑑= 0 进而𝑑= −𝑎∗𝑐/𝑏∗，代入𝑎𝑑−𝑏𝑐= 1，有： −𝑎𝑎∗𝑐 𝑏∗−𝑏𝑐= −(𝑎𝑎∗+ 𝑏𝑏∗)𝑐 𝑏∗ = −𝑐 𝑏∗= 1 从而𝑐= −𝑏∗。 这个条件，代入𝑑= −𝑎∗𝑐/𝑏∗，又得：𝑑= 𝑎∗。这样的话u 这个矩阵就简化 为了： ( 𝑎 𝑏 −𝑏∗ 𝑎∗) 其中𝑎𝑎∗+ 𝑏𝑏∗= 1，a、b 为复数。 有这个条件限制的二维么模矩阵是否构成群呢？我们可以取任意的： u1 = ( 𝑎1 𝑏1 −𝑏1 ∗ 𝑎1 ∗) u2 = ( 𝑎2 𝑏2 −𝑏2 ∗ 𝑎2 ∗) 那么： u1u2 = ( 𝑎1𝑎2 −𝑏1𝑏2 ∗ 𝑎1𝑏2 + 𝑏1𝑎2 ∗ −𝑎2𝑏1 ∗−𝑎1 ∗𝑏2 ∗ 𝑎1 ∗𝑎2 ∗−𝑏1 ∗𝑏2) 这个矩阵显然具有u 的形式，同时|u1u2| = |u1||u2| = 1，所以具有封闭性。 除了封闭性，结合律自然成立，有单位矩阵，同时，酉矩阵的逆矩阵还是酉 矩阵（由u+u = E知，(u−1)+u−1 = uu−1 = E） ，因此所有二维幺模酉矩阵构成一 个群。这个群我们称为SU(2)群（二维特殊酉群） 。 现在讲完了SO(3)群群元用欧拉角的表述方式， SU(2)群的特性。 下一个任务 就是要说明这个SU(2)群和SO(3)群的关系。要理解它们之间的联系，其中最关 键的地方就是理解二阶零迹厄米矩阵𝜎与三维实空间中的向量r ⃗的一一对应关系。 这个怎么理解呢？我们需要引入泡利矩阵（前面提到过） 。 泡利矩阵，在量子力学里面大家都学过，有三个，分别是： σx = (0 1 1 0)、σy = (0 −𝑖 𝑖 0 )、σz = (1 0 0 −1) 它们都是二阶、零迹，且厄米的矩阵。同时，它们包含了二阶零迹厄米矩阵的所 有三个维度 （零迹厄米， 说明对角必须为实数， 且合为零； 而非对角要实部相等， 虚部相反） 。 270 这也就意味这如果我们用实的展开系数把上面那三个泡利矩阵进行线性组 合的话，我们可以建立一个二阶零迹厄米矩阵： h = 𝑥σx + 𝑦σy + 𝑧σz = r ⃗∙𝜎 ⃗= ( 𝑧 𝑥−𝑖𝑦 𝑥+ 𝑖𝑦 −𝑧) 其中𝜎 ⃗= σxi ⃗+ σyj ⃗+ σzk ⃗⃗，与三维欧式空间中向量r ⃗的一一对应关系。也就是说一 组x、y、z 对应一个h，也对应一个r ⃗。 之后呢？对于由x、y、z 对应的零迹厄米矩阵h= r ⃗∙𝜎 ⃗，它可以由我们前面 提到的二阶幺正矩阵u 进行u(r ⃗∙𝜎 ⃗)u−1这样一个相似变换。这个相似变换不改变 矩阵的迹，所以u(r ⃗∙𝜎 ⃗)u−1仍然为零迹。同时，它的转置共轭是： [u(r ⃗∙𝜎 ⃗)u−1]+ = (u−1)+(r ⃗∙𝜎 ⃗)+u+ u 是幺正，u+ = u−1；r ⃗∙𝜎 ⃗厄米，(r ⃗∙𝜎 ⃗)+ = r ⃗∙𝜎 ⃗，因此： [u(r ⃗∙𝜎 ⃗)u−1]+ = u(r ⃗∙𝜎 ⃗)u−1 u(r ⃗∙𝜎 ⃗)u−1仍然零迹厄米。 之前我们说过， 零迹厄米矩阵可以与三维实空间中的一个向量联系起来， 因 此我们可以记u(r ⃗∙σ ⃗ ⃗⃗)u−1 = r ⃗′ ∙σ ⃗ ⃗⃗。 这也意味着u(r ⃗∙𝜎 ⃗)u−1中的u， 实际上对应的是 三维实空间的一个变换Ru，它的作用是：r ⃗′ = Rur ⃗。 由于u 是由a、b 决定的，Ru也是由a、b 决定的。这个Ru的确定方式非常 简单，利用：u(r ⃗∙σ ⃗ ⃗⃗)u−1 = r ⃗′ ∙σ ⃗ ⃗⃗，知： ( 𝑎 𝑏 −𝑏∗ 𝑎∗) ( 𝑧 𝑥−𝑖𝑦 𝑥+ 𝑖𝑦 −𝑧) (𝑎∗ −𝑏 𝑏∗ 𝑎) = ( 𝑧′ 𝑥′ −𝑖𝑦′ 𝑥′ + 𝑖𝑦′ −𝑧′ ) 而：( 𝑥′ 𝑦′ 𝑧′ ) = Ru ( 𝑥 𝑦 𝑧 )。因此，由上式决定的( 𝑥′ 𝑦′ 𝑧′ )与( 𝑥 𝑦 𝑧 )之间的关系，可知： Ru = ( 1 2 (𝑎2 + 𝑎∗2 −𝑏2 −𝑏∗2) −𝑖 2 (𝑎2 −𝑎∗2 + 𝑏2 −𝑏∗2) −(𝑎𝑏+ 𝑎∗𝑏∗) 𝑖 2 (𝑎2 −𝑎∗2 −𝑏2 + 𝑏∗2) 1 2 (𝑎2 + 𝑎∗2 + 𝑏2 + 𝑏∗2) 𝑖(𝑎∗𝑏∗−𝑏𝑎) 𝑎∗𝑏+ 𝑏∗𝑎 𝑖(𝑎∗𝑏−𝑏∗𝑎) 𝑎𝑎∗−𝑏𝑏∗ ) 对这个Ru，有两点需要说明： 1. 由于： |r ⃗′ ∙σ ⃗ ⃗⃗| = | z′ x′ −iy′ x′ + iy′ −z′ | = |u(r ⃗∙σ ⃗ ⃗⃗)u−1| = |r ⃗∙σ ⃗ ⃗⃗| = | z x −iy x + iy −z | 所以|r ⃗′| = √(x′2 + y′2 + z′2) = √(x2 + y2 + z2) = |r ⃗|。 也就是说每个SU(2)群中的u，对应的Ru，属于O(3)。 2. 同时，由于Ru是a、b 的连续函数，取a=1、b=0，u 为单位矩阵，行列式为 1。而实正交矩阵的行列式只能为1 或−1，这里已经有了它为1 的情况，同 时它对a、b 连续，不会出现从1 到−1的跳跃。所以在实正交矩阵中，我们 进一步知道|Ru| = 1，也就是说它进一步属于SO(3)。 这也就是说对任意SU(2)群中的元素u，都有一个SO(3)群中的转动与之对 应。这个是我们说明SU(2)与SO(3)群同态对应关系成立的第一步。之后，我们 还需要说明乘法规律不变以及任意SO(3)群中的元素都有SU(2)群中的元素与之 对应才可以： 1. 乘法规律不变。 对∀u、v ∈SU(2)，有SO(3)群中的元素Ru、Rv与之对应，那么uv 所对应的 三维实空间中的转动Ruv，是否等于RuRv？ 我们已知： u(r ⃗∙𝜎 ⃗)u−1 = r ⃗′ ∙𝜎 ⃗= Rur ⃗∙𝜎 ⃗ v(r ⃗∙𝜎 ⃗)v−1 = r ⃗′′ ∙𝜎 ⃗= Rvr ⃗∙𝜎 ⃗ 那么uv(r ⃗∙𝜎 ⃗)(uv)−1一方面等于： uv(r ⃗∙𝜎 ⃗)(uv)−1 = uv(r ⃗∙𝜎 ⃗)v−1u−1 = u(Rvr ⃗∙𝜎 ⃗)u−1 = RuRvr ⃗∙𝜎 ⃗ 另一方面它又直接等于： uv(r ⃗∙𝜎 ⃗)(uv)−1 = Ruvr ⃗∙𝜎 ⃗ 因此：RuRv = Ruv。 2. 任何SO(3)群中的元素都可以找到SU(2)群中的元素与之对应 （也就是满射） 。 272 由我们之前知道的u 与Ru的对应关系，取： u1(𝛼) = (e−𝑖𝛼/2 0 0 e𝑖𝛼/2) 它对应： Ru1(𝛼) = ( cos 𝛼 −sin 𝛼 0 sin 𝛼 cos 𝛼 0 0 0 1 ) 同样，取： v2(𝛽) = ( cos 𝛽 2 −sin 𝛽 2 sin 𝛽 2 cos 𝛽 2 ) 它对应： Rv2(𝛽) = ( cos 𝛽 0 sin 𝛽 0 1 0 −sin 𝛽 0 cos 𝛽 ) 取： u1(𝛾) = (e−𝑖𝛾/2 0 0 e𝑖𝛾/2) Ru1(𝛾) = ( cos 𝛾 −sin 𝛾 0 sin 𝛾 cos 𝛾 0 0 0 1 ) 这样的话u1(𝛼)v2(𝛽)u1(𝛾)这个SU(2)群中的元素，就会对应： ( cos 𝛼 −sin 𝛼 0 sin 𝛼 cos 𝛼 0 0 0 1 ) ( cos 𝛽 0 sin 𝛽 0 1 0 −sin 𝛽 0 cos 𝛽 ) ( cos 𝛾 −sin 𝛾 0 sin 𝛾 cos 𝛾 0 0 0 1 ) 这个SO(3)群中的转动。而这里u1(𝛼)v2(𝛽)u1(𝛾)等于： (e−𝑖𝛼/2 0 0 e𝑖𝛼/2) ( cos 𝛽 2 −sin 𝛽 2 sin 𝛽 2 cos 𝛽 2 ) (e−𝑖𝛾/2 0 0 e𝑖𝛾/2) = ( cos 𝛽 2 e−𝑖(𝛼+𝛾)/2 −sin 𝛽 2 e−𝑖(𝛼−𝛾)/2 sin 𝛽 2 e𝑖(𝛼−𝛾)/2 cos 𝛽 2 e𝑖(𝛼+𝛾)/2 ) 由于SO(3)群中的任意一个转动都可以用一组欧拉角描述。同时需要说明的 是：在𝛽= 0时，𝛼+ 𝛾相同的组合对应同一转动；𝛽= π时，𝛼−𝛾相同的组合 对应同一转动这个性质在： ( cos 𝛽 2 e−𝑖(𝛼+𝛾)/2 −sin 𝛽 2 e−𝑖(𝛼−𝛾)/2 sin 𝛽 2 e𝑖(𝛼−𝛾)/2 cos 𝛽 2 e𝑖(𝛼+𝛾)/2 ) 这个公式中也有体现。 结合这上面的三点 （任意SU(2)群元有SO(3)群元与之对应， 乘法关系不变， 满射） ，SU(2)与SO(3)同态。其中的同态核是SO(3)中的： ( 1 0 0 0 1 0 0 0 1 ) 对应的SU(2)群中的元素。根据上面那个： ( cos 𝛼 −sin 𝛼 0 sin 𝛼 cos 𝛼 0 0 0 1 ) ( cos 𝛽 0 sin 𝛽 0 1 0 −sin 𝛽 0 cos 𝛽 ) ( cos 𝛾 −sin 𝛾 0 sin 𝛾 cos 𝛾 0 0 0 1 ) 与 ( cos 𝛽 2 e−𝑖(𝛼+𝛾)/2 −sin 𝛽 2 e−𝑖(𝛼−𝛾)/2 sin 𝛽 2 e𝑖(𝛼−𝛾)/2 cos 𝛽 2 e𝑖(𝛼+𝛾)/2 ) 的对应关系， 我们知道𝛼+ 𝛾= 0， 𝛽= 0对应的SO(3)群中元素是单位矩阵， SU(2) 群中元素也是单位矩阵。同时SU(2)群中取𝛼+ 𝛾= 2π，𝛽= 0，元素为： (−1 0 0 −1) 根据上面的对应关系，给出的SO(3)群中的元素仍然为： ( 1 0 0 0 1 0 0 0 1 ) 这也说明同态核是： {(1 0 0 1) ，(−1 0 0 −1)} 再由同态核定理， 同态核{E、 -E}为SU(2)群的不变子群， 且其中任意一个陪 274 集{u、 -u}对应SO(3)群中的同一个转动Ru。 （这个其实我们从最初Ru作为a、 b 函 数的那个表达式也可以看出，a、b 同取负号，Ru不变） 5.2SO(3)群与SU(2)群的不可约表示 现在如果我们把第一节的内容做个总结的话，基本上是下面三句话： 1. SU(2)群与SO(3)群存在2 对1 的同态映射关系， SU(2)群的两个元素u与−u对 应SO(3)群中的一个转动Ru； 2. 如已知SU(2)群中的元素u，可由Ru的表达式求出Ru； 3. 如已知SO(3)群中的元素R(α、β、γ)，也可由： ( cos 𝛼 −sin 𝛼 0 sin 𝛼 cos 𝛼 0 0 0 1 ) ( cos 𝛽 0 sin 𝛽 0 1 0 −sin 𝛽 0 cos 𝛽 ) ( cos 𝛾 −sin 𝛾 0 sin 𝛾 cos 𝛾 0 0 0 1 ) 对应 ( cos 𝛽 2 e−𝑖(𝛼+𝛾)/2 −sin 𝛽 2 e−𝑖(𝛼−𝛾)/2 sin 𝛽 2 e𝑖(𝛼−𝛾)/2 cos 𝛽 2 e𝑖(𝛼+𝛾)/2 ) 求出u与−u。 在已知这个SO(3)与SU(2)的关系以后，下一个任务，很自然，就是求它们 的不等价、不可约表示。怎么求？这个其实是困扰了我好几年的一个问题，因为 不同的教材给出的做法不是很统一。我这里采取的是我认为最严格的一个做法， follow 的是马中骐老师的那本英文版的教材。它的基本思路是这样的，我们要取 一个线性空间作为表示空间，这个线性空间我们记为ℒj，它是一个函数空间。函 数空间的基为ψm j (x ⃗ ⃗)， 其中x ⃗ ⃗为SU(2)群这个线性变换群所对应的线性空间中的向 量，由(x1 x2)表示。SU(2)群中的线性变换u 作用到这个向量(x1 x2)上，得到新的向 量(x′1 x′2)。两者之间的联系是： (x′1 x′2) = u (x1 x2) = ( 𝑎 𝑏 −𝑏∗ 𝑎∗) (x1 x2) 同时，既然我们说线性空间ℒj由j 来标识，那么对于一个特定的j（非负整数或 半整数） ，就会存在不同ψm j (x ⃗ ⃗)，这些ψm j (x ⃗ ⃗)通过线性组合形成线性空间。这里， 我们取ψm j (x ⃗ ⃗)的形式为： ψm j (x ⃗ ⃗) = ψm j (x1，x2) = (−1)𝑗−𝑚 √(𝑗+ 𝑚)! (𝑗−𝑚)! x1 𝑗−𝑚x2 𝑗+𝑚 m取值范围是从j逐次减一到−j。 这样根据前面讲的函数空间变换规则，u所对应的函数变换算符P ̂ u作用到 ψm j (x1，x2)上，结果应该是： P ̂ uψm j (x ⃗ ⃗) = ψm j (u−1x ⃗ ⃗) 我们唯一需要做的， 就是将ψm j (u−1x ⃗ ⃗)按∑ ψm′ j (x ⃗ ⃗) m′ Am′m j (u)展开来确定矩阵的列， 从而产生表示矩阵。其中m′、m 的取值是从j 到−j，排列也是按这个来排列。 明确了这些，我们就按这个步骤来看SU(2)群的元素u 在线性空间ℒj中的表 示矩阵是什么了？前面说过由于SO(3)群中的转动与SU(2)群中的元素{u、-u}的 对应关系，u 本身可以用欧拉角描述为： u = ( cos 𝛽 2 e−𝑖(𝛼+𝛾) 2 −sin 𝛽 2 e−𝑖(𝛼−𝛾) 2 sin 𝛽 2 e 𝑖(𝛼−𝛾) 2 cos 𝛽 2 e 𝑖(𝛼+𝛾) 2 ) = (e−𝑖𝛼/2 0 0 e𝑖𝛼/2) ( cos 𝛽 2 −sin 𝛽 2 sin 𝛽 2 cos 𝛽 2 ) (e−𝑖𝛾/2 0 0 e𝑖𝛾/2) 当u对应绕z轴转𝛼角的转动R(α、0、0)时， u = (e−𝑖𝛼/2 0 0 e𝑖𝛼/2)。 把u记为(e ⃗ ⃗3，𝛼)， 这时， (e ⃗ ⃗3，𝛼) −1 = (e 𝑖𝛼 2 0 0 e−𝑖𝛼 2 ) 276 对应的变换 P ̂ (e ⃗⃗3，𝛼)ψm j (x ⃗ ⃗) = ψm j ((e ⃗ ⃗3，𝛼) −1 x ⃗ ⃗) = ψm j (e 𝑖𝛼 2 x1，e−𝑖𝛼 2 x2) = (−1)𝑗−𝑚 √(𝑗+ 𝑚)! (𝑗−𝑚)! x1 𝑗−𝑚x2 𝑗+𝑚e−𝑖𝛼𝑚= ψm j (x1，x2)e−𝑖𝑚𝛼= ψm j (x ⃗ ⃗)e−𝑖𝑚𝛼 而P ̂ (e ⃗⃗3，𝛼)ψm j (x ⃗ ⃗) = ∑ ψm′ j (x ⃗ ⃗) m′ Am′m j ((e ⃗ ⃗3，𝛼)) ，因此相应的Am′m j ((e ⃗ ⃗3，𝛼))为： Am′m j ((e ⃗ ⃗3，𝛼)) = δm′me−𝑖𝑚𝛼 同样， 对与(e−𝑖𝛾 2 0 0 e 𝑖𝛾 2 )这个绕z 轴转𝛾的操作R(0、0、𝛾)， 把u记为(e ⃗ ⃗3，𝛾)， 也有： Am′m j ((e ⃗ ⃗3，𝛾)) = δm′me−𝑖𝑚𝛾 最后剩下的是对( cos 𝛽 2 −sin 𝛽 2 sin 𝛽 2 cos 𝛽 2 )这个绕y 轴转𝛽角的操作R(0、𝛽、0)，把u 记为(e ⃗ ⃗2，𝛽)，情况会稍微复杂些，因为： (e ⃗ ⃗2，𝛽) −1 = ( cos 𝛽 2 sin 𝛽 2 −sin 𝛽 2 cos 𝛽 2 ) 这样的话 P ̂ (e ⃗⃗2，𝛽)ψm j (x ⃗ ⃗) = ψm j ((e ⃗ ⃗2，𝛽) −1 x ⃗ ⃗) = ψm j (cos 𝛽 2 x1 + sin 𝛽 2 x2，−sin 𝛽 2 x1 + cos 𝛽 2 x2) = (−1)𝑗−𝑚 √(𝑗+ 𝑚)! (𝑗−𝑚)! (cos 𝛽 2 x1 + sin 𝛽 2 x2) 𝑗−𝑚 (−sin 𝛽 2 x1 + cos 𝛽 2 x2) 𝑗+𝑚 这时，利用二项式定理： (x + y)𝑛= ∑ 𝑛! 𝑟! (𝑛−𝑟)! x𝑟y𝑛−𝑟 𝑛 𝑟=0 P ̂ (e ⃗⃗2，𝛽)ψm j (x ⃗ ⃗)继续等于： (−1)𝑗−𝑚 √(𝑗+ 𝑚)! (𝑗−𝑚)! {∑ (𝑗−𝑚)! 𝑟! (𝑗−𝑚−𝑟)! (cos 𝛽 2 x1) 𝑗−𝑚−𝑟 (sin 𝛽 2 x2) 𝑟 𝑗−𝑚 𝑟=0 } {∑ (𝑗+ 𝑚)! 𝑟′! (𝑗+ 𝑚−𝑟′)! (−sin 𝛽 2 x1) 𝑗+𝑚−𝑟′ (cos 𝛽 2 x2) 𝑟′ 𝑗+𝑚 𝑟′=0 } = (−1)𝑗−𝑚∑∑ √(𝑗+ 𝑚)! (𝑗−𝑚)! 𝑟! (𝑗−𝑚−𝑟)! 𝑟′! (𝑗+ 𝑚−𝑟′)! 𝑗+𝑚 𝑟′=0 (cos 𝛽 2) 𝑗−𝑚−𝑟+𝑟′ (sin 𝛽 2) 𝑗+𝑚−𝑟′+𝑟 𝑗−𝑚 𝑟=0 (−1)𝑗+𝑚−𝑟′(x1)2𝑗−𝑟−𝑟′(x2)𝑟+𝑟′ 令𝑚′ = 𝑟+ 𝑟′ −𝑗， 则x1 2𝑗−𝑟−𝑟′x2 𝑟+𝑟′ = x1 𝑗−𝑚′x2 𝑗+𝑚′， 𝑟′ = 𝑗+ 𝑚′ −𝑟， 𝑗+ 𝑚−𝑟′ = 𝑗+ 𝑚− (𝑗+ 𝑚′ −𝑟) = 𝑟+ 𝑚−𝑚′，进而上式继续等于： ∑∑ √(𝑗+ 𝑚)! (𝑗−𝑚)! (𝑗+ 𝑚′)! (𝑗−𝑚′)! 𝑟! (𝑗−𝑚−𝑟)! (𝑗+ 𝑚′ −𝑟)! (𝑟+ 𝑚−𝑚′)! (−1)𝑟 𝑗−𝑚 𝑟=0 (cos 𝛽 2) 2𝑗−𝑚−2𝑟+𝑚′ (sin 𝛽 2) 𝑚+2𝑟−𝑚′ −𝑗 𝑚′=𝑗 (−1)𝑗−𝑚′ √(𝑗+ 𝑚′)! (𝑗−𝑚′)! x1 𝑗−𝑚′x2 𝑗+𝑚′ = ∑Am′m j ((e ⃗ ⃗2，𝛽)) ψm′ j (x ⃗ ⃗) −𝑗 𝑚′=𝑗 这个里面−1的指数从(𝑗−𝑚) + (𝑗+ 𝑚−𝑟′) = 2𝑗−𝑟′变为 𝑟+ 𝑗−𝑚′。这个变 化的原因是：𝑟′ = 𝑗+ 𝑚′ −𝑟，因此2𝑗−𝑟′ = 2𝑗−(𝑗+ 𝑚′ −𝑟) = 𝑟+ 𝑗−𝑚′。指 数不发生变化。 这个表达式中，Am′m j ((e ⃗ ⃗2，𝛽))等于： ∑ √(𝑗+ 𝑚)! (𝑗−𝑚)! (𝑗+ 𝑚′)! (𝑗−𝑚′)! 𝑟! (𝑗−𝑚−𝑟)! (𝑗+ 𝑚′ −𝑟)! (𝑟+ 𝑚−𝑚′)! (−1)𝑟 𝑗−𝑚 𝑟=0 (cos 𝛽 2) 2𝑗−𝑚−2𝑟+𝑚′ (sin 𝛽 2) 𝑚+2𝑟−𝑚′ 这里对加和指标𝑟的要求是： 1. 𝑟≥0；2. 𝑟≤𝑗−𝑚；3. 𝑟≤𝑗+ 𝑚′；4. 𝑟≥𝑚′ −𝑚。 而完整的一个SU(2)群中的元素 278 u = (e−𝑖𝛼/2 0 0 e𝑖𝛼/2) ( cos 𝛽 2 −sin 𝛽 2 sin 𝛽 2 cos 𝛽 2 ) (e−𝑖𝛾/2 0 0 e𝑖𝛾/2) 所对应的ℒj中的变换矩阵就是： Am′m j (𝛼，𝛽，𝛾) = {Aj ((e ⃗ ⃗3，𝛼)) Aj ((e ⃗ ⃗2，𝛽)) Aj ((e ⃗ ⃗3，𝛾))} m′m = e−𝑖m′𝛼Am′m j ((e ⃗ ⃗2，𝛽)) e−𝑖m𝛾 这里，m′、m 的取值是从j 到−j，排列也是按这个顺序来排列。 这里，当u变为−u时，表示矩阵要么不变，要么反号。怎么理解？我们分特 殊情况（三种）与一般情况展开讨论。 特殊情况一，(e−𝑖𝛼 2 0 0 e 𝑖𝛼 2 ) 与(e−𝑖𝛾 2 0 0 e 𝑖𝛾 2 ) 不变，( cos 𝛽 2 −sin 𝛽 2 sin 𝛽 2 cos 𝛽 2 ) 变成了 −( cos 𝛽 2 −sin 𝛽 2 sin 𝛽 2 cos 𝛽 2 ) 。这时，对e−𝑖m′𝛼Am′m j ((e ⃗ ⃗2，𝛽)) e−𝑖m𝛾的变化，关键看 Am′m j ((e ⃗ ⃗2，𝛽))。由于矩阵元表达式中cos 𝛽 2的次数是2𝑗−𝑚−2𝑟+ 𝑚′，sin 𝛽 2的 次数是 𝑚+ 2𝑟−𝑚′，两者的和是2𝑗。因此，当𝑗为整数时，这个表示为偶表示； 当𝑗为半整数时，这个表示为奇表示。 特殊情况二和三，(e−𝑖𝛼 2 0 0 e 𝑖𝛼 2 )与(e−𝑖𝛾 2 0 0 e 𝑖𝛾 2 )中的一个反号，另一个不变， ( cos 𝛽 2 −sin 𝛽 2 sin 𝛽 2 cos 𝛽 2 )不变。 这时e−𝑖m′𝛼Am′m j ((e ⃗ ⃗2，𝛽)) e−𝑖m𝛾的变化还是𝑗为整数时不 变，当𝑗为半整数时反号。 一般情况，u 变为−u不能分解为上面三项中一项反号。但这种情况可以通 过一个相似变换的方式和上面的特殊情况联系起来，ug = αusα−1。ug代表一般 情况，us代表上面的特殊情况。ug变成−ug实际上是α(−us)α−1。这时： A(−ug) = A(α(−us)α−1) = A(α)A(−us)A(α)−1 这样，根据上面特殊情况同理。当𝑗为整数时，A(−us) = A(us)，因此A(−ug) = A(ug)；当𝑗为半整数时，A(−us) = −A(us)，A(−ug) = −A(ug)。不管怎样，都是 当𝑗为整数时，表示为偶表示；当𝑗为半整数时，表示为奇表示。 下面看一下简单的几个具体情况中表示矩阵元是什么： 1. 当𝑗= 0时， 这时𝑚与𝑚′只能为零， 𝑟也只能为零， 对应的基组就一个基函数， 是1 这个常数，表示矩阵也是1 阶的，为： A00 0 (𝛼，𝛽，𝛾) = 1 一维恒等表示。 2. 当𝑗= 1/2时， 第一个矩阵元： A1/2,1/2 1 2 (𝛼，𝛽，𝛾) = cos 𝛽 2 e−𝑖(𝛼+𝛾) 2 这里𝑗= 1/2、m = 1/2、m′ = 1/2，加和指标𝑟的要求是： 1. 𝑟≥0；2. 𝑟≤𝑗−𝑚= 0；3. 𝑟≤𝑗+ 𝑚′ = 1；4. 𝑟≥𝑚′ −𝑚= 0，综合就 是𝑟= 0。因此， A1/2,1/2 1 2 (𝛼，𝛽，𝛾) = e−𝑖𝛼 2 cos 𝛽 2 e−𝑖𝛾 2 = cos 𝛽 2 e−𝑖(𝛼+𝛾) 2 第二个矩阵元： A1/2,− 1/2 1/2 (𝛼，𝛽，𝛾) = −sin 𝛽 2 e−𝑖(𝛼−𝛾) 2 这里𝑗= 1/2、m = −1/2、m′ = 1/2，加和指标𝑟的要求是： 1. 𝑟≥0；2. 𝑟≤𝑗−𝑚= 1；3. 𝑟≤𝑗+ 𝑚′ = 1；4. 𝑟≥𝑚′ −𝑚= 1，综合就 是𝑟= 1。因此， A1/2,− 1/2 1 2 (𝛼，𝛽，𝛾) = e−𝑖𝛼 2 (−sin 𝛽 2) e 𝑖𝛾 2 = −sin 𝛽 2 e−𝑖(𝛼−𝛾) 2 280 第三个矩阵元： A−1/2,1/2 1/2 (𝛼，𝛽，𝛾) = sin 𝛽 2 e 𝑖(𝛼−𝛾) 2 这里𝑗= 1/2、m = 1/2、m′ = −1/2，加和指标𝑟的要求是： 1. 𝑟≥0；2. 𝑟≤𝑗−𝑚= 0；3. 𝑟≤𝑗+ 𝑚′ = 0；4. 𝑟≥𝑚′ −𝑚= −1，综合 就是𝑟= 0。因此， A−1/2,1/2 1 2 (𝛼，𝛽，𝛾) = e 𝑖𝛼 2 sin 𝛽 2 e−𝑖𝛾 2 = sin 𝛽 2 e 𝑖(𝛼−𝛾) 2 第四个矩阵元： A−1/2,−1/2 1/2 (𝛼，𝛽，𝛾) = cos 𝛽 2 e 𝑖(𝛼+𝛾) 2 这里𝑗= 1/2、m = −1/2、m′ = −1/2，加和指标𝑟的要求是： 1. 𝑟≥0；2. 𝑟≤𝑗−𝑚= 1；3. 𝑟≤𝑗+ 𝑚′ = 0；4. 𝑟≥𝑚′ −𝑚= 0，综合就 是𝑟= 0。因此， A−1/2,−1/2 1 2 (𝛼，𝛽，𝛾) = e−𝑖𝛼 2 cos 𝛽 2 e−𝑖𝛾 2 = cos 𝛽 2 e 𝑖(𝛼+𝛾) 2 这四个矩阵元放在一起形成的矩阵是( cos 𝛽 2 e−𝑖(𝛼+𝛾) 2 −sin 𝛽 2 e−𝑖(𝛼−𝛾) 2 sin 𝛽 2 e 𝑖(𝛼−𝛾) 2 cos 𝛽 2 e 𝑖(𝛼+𝛾) 2 )， 刚好就 是SU(2)这个群中的矩阵本身。 3. 𝑗更大时，还是按照这个规则来，只是矩阵元的产生过程更复杂一些。 当𝑗走遍所有的正的整数与半整数时， Aj给出SU(2)群的所有不等价、 不可约 酉表示。 前面我们说过， 前两节的核心任务是理解： 1) SO(3)与SU(2)群是什么， 它们 有怎样的关系？2) SO(3)与SU(2)群的不可约表示是什么？现在我们第一点是完 全知道了，第二点知道了SU(2)群的不可约表示，还剩SO(3)群的没有说。要说 这部分，用到的知识就是SO(3)与SU(2)群的关系。 我们分两个方面来理解这种关系。 一方面， 在SU(2)群与SO(3)群的对应上， 我们知道SU(2)群中的u与−u都对应SO(3)群中的转动Ru， 对应关系基于欧拉角。 同时，在讲SU(2)群的不可约表示A𝑗的时候，我们也说了当𝑗为整数的时候，有 A𝑗(u) = A𝑗(−u)，这也就意味着存在这样一个关系： 图5.2 SO(3)与SU(2)群及其不可约表示的关系示意图一 也就是说当𝑗为整数时，A𝑗也是SO(3)群的表示。同时当𝑗走遍所有整数的时候， A𝑗给出所有SO(3)群的不等价不可约酉表示。这样的话它们的不等价不可约表示 的情况就清楚了。 5.3 双群与自旋半奇数粒子的旋量波函数 上面那些讨论给我们的信息里面，直接和真实的物理系统建立联系的是 SO(3)群，它对应的物理系统是中心力场。如果我们只是为了这个目的，那么我 们可以去想一下，上面关于SU(2)群的讨论有很多是不必要的，虽然我们利用它 给出了SO(3)群的不可约表示A𝑗（𝑗为整数） 。 为什么要对它进行这么多的讨论？应该说本质上的原因是SU(2)可以描述一 个自旋1/2 的费米子系统在转动操作下波函数的自旋部分变换的性质。 其中最基 本的一个就是对这样一个系统， 你在三维实空间转2π角的时候， 它的波函数不回 到其本身，而是多了一个负号。其中，三维空间波函数回到了它本身，但电子自 282 旋内禀空间的并没有。与之相应，我们在描述这类系统时，也就不能用SO(3)群 了， 而是要用它的双群SOD(3)。 同时， 当系统的对称性由中心力场降低为分子或 晶体中的点群的时候，我们描述它的对称性的工具，也不能是前面讲的点群了， 而是要用点群的双群。 在我们的日常的研究中， 一个最常见的问题就是在考虑自旋轨道耦合的时候， 我们的能级或能带如何劈裂？下面， 我们就会以这几个概念 （SOD(3)、 点群的双 群、旋轨耦合引起的能级劈裂）为重点，来讲一下前两节的内容在这类物理系统 中的应用。 先看SO(3)群的双群SOD(3)。这里的基础是一个与上节最后相似的关系，不 过对应的是𝑗为半整数的情况。 这个时候， 由于A𝑗(−u) = −A𝑗(u)， 这样A𝑗矩阵群、 SU(2)群、SO(3)群的关系就变成了这样： 图5.3 SO(3)与SU(2)群及其不可约表示的关系示意图二 这个时候一个SO(3)群中的转动R(α, β, γ)会对应相差一个负号的两个SU(2) 群中的元素u与−u，同样也对应相差一个负号的两个矩阵A𝑗(u)与A𝑗(−u) = −A𝑗(u)。 如果我们把“表示”这个概念中“一个群元对应一个矩阵”弱化为“一个群 元对应相差一个负号的两个矩阵” ；同时把保持乘法规则不变这个规定弱化为保 持乘法规则在相差一个负号的情况下不变， 也就是： A𝑗(u1)A𝑗(u2) = ±A𝑗(u1u2)， 这时我们可以把A𝑗(u)，在𝑗为半整数时对SO(3)群的表示称为一个双值表示。 这是处理SO(3)群中的元素u 与A𝑗(u)这个矩阵在𝑗为半整数时对应关系的一 种手段。和它差不多，我们还可以采取另外一个手段，就是利用SO(3)群中元素 R(α, β, γ) = R(α + 2π, β, γ)这样一个特征， 把绕某轴转2π角的操作定义为一个新的 非恒等操作E ̅。这个E ̅ ≠E，但E ̅E ̅ =E。这个时候，我们再把每个SO(3)群中的元 素乘上E ̅，得到一个新的元素集合，每个SO(3)群中的元素都唯一的对应这个集 合中的元素。 现在把这个集合与SO(3)放在一起，形成一个新的集合。这个新的集合，在 与SO(3)群相同的乘法规则下，是形成一个群的。这个群，就与SU(2)同构，相 应的，在𝑗为半整数时的矩阵群{A𝑗(u)}，也就很自然的形成了它的一个表示。 那么我们这里的处理是保持“表示”本身的定义不变，但是把SO(3)群扩大 了一倍。这样形成的一个群，称为SO(3)群的双群SOD(3)。也就是说为了描述前 面两个图的差别，我们可以做两个处理。一个是用双值表示这个概念，一个是用 双群。双值表示就是一个概念，很多教材会提到，但实用价值不大；双群的实用 价值很大。 需要注意的是在我这个SOD(3)中，SO(3)群元素的结合形成一个子集，但它 不再是子群了。因为我规定转2π不等于不转，也就是我两个SO(3)群中的元素， 各转3π/2，乘完的元素转3π，它就不属于SO(3)这个集合了。也正是因为这个原 因，SOD(3)并不是SO(3)与{E，E ̅}的直积，相应的SOD(3)的表示就不再是SO(3) 284 的表示乘上{E，E ̅}的一维恒等于一维非恒等表示那么简单了25。 这样的一个双群，在物理系统中，具体对应电子的自旋态。我们学过量子力 学的都知道电子并不是一个简单的具有三个自由度的粒子， 它还有一个自由度必 须有自旋来描述。自旋，按照曾谨言老师《量子力学》这本书第一册、第八章、 第一小节、第三部分的第一段话，是这样描述的“自旋这个力学量虽然有角动量 的性质，但与轨道角动量不同，它并无经典对应（当ℏ趋近于零时，自旋效应自 然消失） 。自旋的系统理论属于相对论量子力学的范围，它是电子场在空间转动 下的特性的反映。 在非相对论量子力学中， 可以唯象地根据实验上反映出来的自 旋的特点，选择适当的数学工具来描述它。 ”很惭愧我自己到现在都没有学过相 对论量子力学，就课程讲授而言，我们权且把自旋理解为这样一个东西：它是电 子的内禀属性；它有相应的角动量与磁矩；在任何一个方向，它都有两个分立的 值。 由于自旋轨道耦合的原因， 对一个单电子问题中的单电子， 它的总的角动量 J ̂，就是其轨道角动量L ̂与自旋角动量S ̂的矢量和： J ̂ = L ̂ + S ̂ 当我们选定一个特定轴 （比如z） 的时候， 电子本征态波函数是一个旋量波函数， 形式是： Ψ 𝑗𝑚(r ⃗, 𝑡) = ( Ψ 𝑗𝑚(r ⃗, ℏ/2, 𝑡) Ψ 𝑗𝑚(r ⃗, −ℏ/2, 𝑡)) 其中Ψ 𝑗𝑚(r ⃗, ℏ/2, 𝑡)为该本征态自旋在z 轴投影为ℏ/2的的空间依赖部分， Ψ 𝑗𝑚(r ⃗, −ℏ/2, 𝑡)为该本征态自旋在z 轴投影为−ℏ/2的空间依赖部分。 25这个也是SU(2)群与SOD(3)群同构，但它们都不与O(3)群同构的原因，虽然它们和SO(3)都有 2 对1 的关系。对SU(2)与SO(3)，它们可以有欧拉角通过连续的变换联系起来。SOD(3)也具备 这样的特征。但对于O(3)群，反演操作I 不可能由转动的连续操作得到。 在该本征态下，系统自旋向上的几率为：∫|Ψ𝑗𝑚(r ⃗, ℏ/2, 𝑡)| 2 𝑑r ⃗，自旋向下的 几率为：∫|Ψ𝑗𝑚(r ⃗, −ℏ/2, 𝑡)| 2 𝑑r ⃗。总的波函数归一条件是：∫[|Ψ𝑗𝑚(r ⃗, ℏ/2, 𝑡)| 2 + |Ψ𝑗𝑚(r ⃗, −ℏ/2, 𝑡)| 2] 𝑑r ⃗= 1 这里的两个好量子数是𝑗、𝑚，对应的力学量期待值是： J ̂2Ψ𝑗𝑚= 𝑗(𝑗+ 1)ℏ2Ψ𝑗𝑚 J ̂𝑧Ψ𝑗𝑚= 𝑚ℏΨ𝑗𝑚 J ̂为总角动量， J ̂𝑧为它在z 方向的投影。 就好量子数取值而言𝑗= 1/2，3/2，5/2， ⋯；对一个特定的𝑗，𝑚= −𝑗，−𝑗+ 1，⋯，𝑗−1，𝑗。 对于这样一个旋量波函数，如果我们把一个绕z 轴转动𝛼角的操作P ̂ z,𝛼= e−𝑖 ℏ𝛼J ̂𝑧作用到它上面，效果就是： P ̂ z,𝛼Ψ𝑗𝑚(r ⃗, 𝑡) = e−𝑖 ℏ𝛼J ̂𝑧Ψ𝑗𝑚(r ⃗, 𝑡) = e−𝑖 ℏ𝛼𝑚ℏΨ𝑗𝑚(r ⃗, 𝑡) = e−𝑖𝛼𝑚Ψ𝑗𝑚(r ⃗, 𝑡) 由于𝑗为半奇数，𝑚也是半奇数。这也就意味着我系统转2𝜋角的时候，本征态波 函数反号。 只有转4𝜋的时候才回到本身。 这个就和SO(3)群的双群SOD(3)就对应 起来了。 在不考虑自旋轨道耦合的时候， 由于总的角动量为整数， 所以转2𝜋角之后系 统到了本身。系统的对称性就是SO(3)或点群，但考虑了自旋轨道耦合之后，由 于总自旋变为了半奇数， 系统的对称性就变成了SOD(3)或双点群。 相应， 在本征 态标注的环节，在不考虑自旋轨道耦合的时候，系统的群要么是SO(3)群，要么 是某个点群，它的本征态对应的是SO(3)或者这个点群的不可约表示。在考虑了 自旋轨道耦合之后，由于SOD(3)并不是SO(3)与{E，E ̅}的直积，或者说点群双群 不是点群与{E， E ̅}的直积， 那么原来对应不可约表示的本征态现在对应的就不再 是不可约表示了，相应的能带或能级就会发生劈裂。 286 在原子物理中， 这种自旋轨道耦合效应带来的一个直接后果是在研究原子光 谱的时候，我们需要用总的角动量去理解这些原子谱。历史上，也正是由于要解 释这些原子光谱的需求， 才导致了人们发现电子自旋这个内禀属性 （Uhlenbeck， Goudsmit，Kronig 这些人在薛定谔方程、狄拉克方程提出前的工作） 。 就我们利用这样一个对称性来理解物性而言， 我这里举两个例子， 就是晶体 里面考虑了自旋轨道耦合后， 能带会发生怎样的变化？ （这个你们以后都会遇到） 例1. 系统本身是D2群，它有四个类，E、c2𝑥、c2𝑦、c2𝑧。对应4 个一维不可约 表示。考虑旋轨耦合，对它的双群D2 D，由于E ̅的引入，多了四个元素，现 在八个元素是：E、c2𝑥、c2𝑦、c2𝑧、E ̅、E ̅c2𝑥、E ̅c2𝑦、E ̅c2𝑧，这里E ̅为绕z 轴 转2𝜋的操作。它们的阶是1、4、4、4、2、4、4、4。 对c2𝑧，它代表绕z 轴逆时针转𝜋角的操作；E ̅c2𝑧，代表绕z 轴转3𝜋角的操 作，它们不相等。但是由于c2𝑥的存在，(c2𝑥)−1E ̅c2𝑧(c2𝑥)代表绕z 轴负方 向转3𝜋，也就是z 轴转𝜋的操作c2𝑧。即(c2𝑥)−1E ̅c2𝑧(c2𝑥) = c2𝑧，c2𝑧与E ̅c2𝑧 同类。 对c2𝑥，它代表绕x 轴逆时针转𝜋角的操作；E ̅c2𝑥，代表绕x 轴转3𝜋角的操 作，它们不相等。注意，这个E ̅我的严格的定义是绕某轴。这个“某” ，我 是可以选取的。但是由于c2𝑧的存在，(c2𝑧)−1E ̅c2𝑥(c2𝑧)代表绕x 轴负方向 转3𝜋，也就是x 轴转𝜋的操作c2𝑥。即(c2𝑧)−1E ̅c2𝑥(c2𝑧) = c2𝑥，c2𝑥与E ̅c2𝑥同 类。 对c2𝑦、E ̅c2𝑦，同理。这样我的D2 D就有五个类。{E}、{E ̅}、{c2𝑥、E ̅c2𝑥}、 {c2𝑦、E ̅c2𝑦}、{c2𝑧、E ̅c2𝑧}。相应于D2群的四个一维表示，D2 D的Burnside 方程就是： 12 + 12 + 12 + 12 + 22 = 8 它就有四个一维不可约表示，一个二维不可约表示。 D2群的特征标表是： 1{E} 1{c2𝑥} 1{c2𝑦} 1{c2𝑧} A1 1 1 1 1 A2 1 1 -1 -1 A3 1 -1 1 -1 A4 1 -1 -1 1 表5.1 D2群的特征标表 D2 D群的特征标表是： 1{E} 1{E ̅} 2{c2𝑥} 2{c2𝑦} 2{c2𝑧} A1 1 1 1 1 1 A2 1 1 1 -1 -1 A3 1 1 -1 1 -1 A4 1 1 -1 -1 1 A5 2 -2 0 0 0 表5.2 D2 D群的特征标表 现在考虑一个具有D2群对称性的晶体本征态，在引入自旋轨道耦合后的 能带变化情况。 在引入旋轨耦合前， 总体波函数的空间依赖部分是ψ(r)， 对称群为D2， 表 示是D2的A1到A4中间的一个， 由于这些不可约表示是一维的， 轨道部分是 288 单重态。 而自旋部分， 可容纳↑、 ↓两个态 （自旋双重态） ， 对称群是SU(2)， 以↑、↓两个态为基，表示就是SU(2)这个矩阵群，具体形式是： ( cos 𝛽 2 e−𝑖(𝛼+𝛾)/2 −sin 𝛽 2 e−𝑖(𝛼−𝛾)/2 sin 𝛽 2 e𝑖(𝛼−𝛾)/2 cos 𝛽 2 e𝑖(𝛼+𝛾)/2 ) 要看考虑自旋轨耦合后的本征能级变化， 说白了， 就是做ψ(r)承载的D2的 不可约表示与电子自旋承载的SU(2)群的二维不可约表示的直积，然后往 D2 D的不可约表示上做投影即可。如ψ(r)承载的D2群的不可约表示是A1， {E}、 {E ̅}、 {c2𝑥、E ̅c2𝑥}、 {c2𝑦、E ̅c2𝑦}、 {c2𝑧、E ̅c2𝑧}对应的D2群的特征标分 别是1、1、1、1、1。而它们承载的SU(2)群的不可约表示的特征标分别是 2、-2、0、0、0。这样的话直积表示的特征标就是2、-2、0、0、0。这个 时候对应的情况是在考虑旋轨耦合之前，轨道部分承载D2的不可约表示 A1，自旋部分简并得坐着自旋向上、向下两个电子，考虑旋轨耦合之后， 这两个电子的本征态变成了D2 D的不可约表示A5所对应的本征态。 如ψ(r)承载的D2群的不可约表示是A2， A3， 或A4， {E}、 {E ̅}、 {c2𝑥、E ̅c2𝑥}、 {c2𝑦、E ̅c2𝑦}、{c2𝑧、E ̅c2𝑧}对应的D2群的特征标分别是1、1、1、-1、-1， 1、1、-1、1、-1，或1、1、-1、-1、1。而它们承载的SU(2)群的不可约表 示的特征标还是2、-2、0、0、0。这样的话直积表示的特征标也还是2、 -2、0、0、0。对应的情况和上一段的讨论很类似，考虑旋轨耦合之前，轨 道部分承载D2的不可约表示A2、 A3、 或A4， 自旋部分简并得坐着自旋向上、 向下两个电子， 考虑旋轨耦合之后， 这两个电子的本征态变成了D2 D的不可 约表示A5所对应的本征态。 还是二重简并， 但这个时候的那个态自旋与空 间部分就不再分立了。 我们需要说明的是在上面的讨论中，由于c2𝑥的存在，使得c2𝑦与E ̅c2𝑦、c2𝑧与 E ̅c2𝑧成为同一类元素。由于c2𝑦的存在，使得c2𝑥与E ̅c2𝑥也成为同一类元素。这些 元素同类的条件都是有一个二次轴与这个二次轴相互垂直。 这个时候由于我们上 面的分析，每个二次转动与它乘上E ̅之后形成的元素属于同一类。 例2. 系统本身是D4群， 它有五个类， {E}、 {C4 1、C4 3}、 {C4 2}、 {C2 (1)、C2 (3)}、{C2 (2)、 C2 (4)}。对应4 个一维不可约表示、1 个二维不可约表示。对它的双群D4 D， 由于E ̅的引入，多了八个元素，现在16 个元素。C4 1代表z 轴正向逆时针转 π/2；E ̅C4 3代表z 轴正向逆时针转7π/2，相当于z 轴反向逆时针转π/2，因 此C4 1与E ̅C4 3同类。同理，C4 2与E ̅C4 2一类；C4 3与E ̅C4 1一类；C2 (1)、E ̅C2 (1)、C2 (3)、 E ̅C2 (3)一类； C2 (2)、 E ̅C2 (2)、 C2 (4)、 E ̅C2 (4)一类。 再加上{E}是一类， {E ̅}是一类， 一共七类。 例3. O 群有24 个元素5 个类：{E}、{3C4 2}、{6C4}、{6C2}、{8C3}。在加入了E ̅ 后，{E}、{6C4}、{8C3}都是在乘上E ̅后，与原来{E}、{6C4}、{8C3}这个集 合进行重组， 给出： {E}、{E ̅}、 {3C4、3E ̅C4 3}、 {3C4 3、3E ̅C4}、{4C3、4E ̅C3 2}、 {4E ̅C3、4C3 2}六个类。对{3C4 2}，由于这三个二次轴相互垂直，{3E ̅C4 2}与它 们同类，在OD中，这个类是：{3C4 2、3E ̅C4 2}。同样，{6C2}也是，在OD中， 这个类是：{6C2、6E ̅C2}。 综合起来， OD群就有48 个元素， {E}、 {3C4、3E ̅C4 3}、 {4C3、4E ̅C3 2}、{E ̅}、 {3C4 3、3E ̅C4}、{4E ̅C3、4C3 2}、{3C4 2、3E ̅C4 2}、{6C2、6E ̅C2}八个类。 290 对应的Burnside 定理就是： 12 + 12 + 22 + 32 + 32 + 22 + 22 + 42 = 48 OD群的不可约表示特征标表是： 1{E} 1{E ̅} 6{C4 2} 6{C4} 6{E ̅C4} 12{C2} 8{C3} 8{E ̅C3} Γ 1 1 1 1 1 1 1 1 1 Γ 2 1 1 1 -1 -1 -1 1 1 Γ 12 2 2 2 0 0 0 -1 -1 Γ 15 3 3 -1 1 1 -1 0 0 Γ 25 3 3 -1 -1 -1 1 0 0 Γ 6 2 -2 0 √2 −√2 0 1 -1 Γ 7 2 -2 0 −√2 √2 0 1 -1 Γ 8 4 -4 0 0 0 0 -1 1 表5.3 OD群特征标表 上面的讨论是对应纯转动点群， 在考虑反演操作的时候， 如考虑的非纯转 动点群包含I 操作， 那么对应的非纯转动点群的双群Oh D的特征标表， 按照 我们前面讲的，就是利用上个表格，与{E、I}的一维不可约表示做直积。 相应的不可约表示的标识， 也会由Γ1变成Γ1 +、 Γ1 −这样的偶、 奇宇称态。 Γ2 变成Γ2 +、Γ2 −；Γ12变成Γ12 + 、Γ12 −，以此类推。 如果不包含I，就是利用同构关系依照某个纯转动点群双群的特征标表去 理解物性。 这些是铺垫性讨论， 现在回到能带与能级劈裂这个话题本身， 还是刚才那 句话， 在不考虑双群的时候， Oh不可约表示的每个维度可以坐两个电子； 而考虑了旋轨耦合带来的双群，Oh D不可约表示的每个维度就坐一个电子。 以我们之前讲过的原子轨道在晶格场中的劈裂作为例子。之前我们说过， 一个3d 过渡金属原子，在一个具备Oh对称性的晶体中，d 轨道会劈裂为 Eg与T2g， 这个E 与T 是对点群不可约表示的一种标识， 对应二维与三维， g 代表这个态是偶宇称的。 有些文献上， 在Wigner 的标识习惯中， Eg与T2g会被写为Γ12 + 与Γ25 + 。 Γ12 + 是 个二维表示，上面坐四个电子，Γ25 + 是个三维表示，上面坐六个电子。这 个只是一种习惯，没有什么复杂的。 现在考虑自旋轨道耦合，Γ12 + 与Γ25 + 这两个态就会发生劈裂了。对于电子这 样的自旋1/2 的费米子，它劈裂的规则是按上面讨论的，由这个自旋1/2 费米子所对应SU(2)群二维表示与Γ12 + 、Γ25 + 作直积，然后再往Oh D的不可约 表示做直和分解得到了。 结果，往往是这样的： 292 图5.4 旋轨耦合引起的能级劈裂示意图 这是一个例子。 对于具有Oh点群对称性的晶体， 它的能带在没有考虑旋轨 耦合的时候我们可以这样标识： 图5.5 未考虑旋轨耦合的能带示意图 考虑了旋轨耦合以后，就会是： 图5.6 考虑旋轨耦合的能带示意图 为什么这样？为简单起见， 我们先忽略空间反演操作I， 解释为什么Γ25会 变成Γ7和Γ8的直和。之后g、u（或者正、负）这些代表宇称的指标直接加 上就可以了。 对于O 群的Γ25的本征态，OD群的下面这些类的特征标分为： 1{E} 1{E ̅} 6{C4 2} 6{C4} 6{E ̅C4} 12{C2} 8{C3} 8{E ̅C3} 3 3 -1 -1 -1 1 0 0 这些类在自旋空间，SU(2)群的不可约表示下的特征标根据： ( cos 𝛽 2 e−𝑖(𝛼+𝛾)/2 −sin 𝛽 2 e−𝑖(𝛼−𝛾)/2 sin 𝛽 2 e𝑖(𝛼−𝛾)/2 cos 𝛽 2 e𝑖(𝛼+𝛾)/2 ) 又分别等于： 1{E} 1{E ̅} 6{C4 2} 6{C4} 6{E ̅C4} 12{C2} 8{C3} 8{E ̅C3} 2 -2 0 √2 −√2 0 1 −1 294 这两个表示做直乘，结果是： 1{E} 1{E ̅} 6{C4 2} 6{C4} 6{E ̅C4} 12{C2} 8{C3} 8{E ̅C3} 6 -6 0 −√2 √2 0 0 0 这个结果，刚好分解为下面两个OD群的不可约表示的直和： Γ 7 2 -2 0 −√2 √2 0 1 -1 Γ 8 4 -4 0 0 0 0 -1 1 因为这个原因， 我们在文献中看到的就是前面介绍的东西。 理解这些东西， 套路是用点群与自旋的SU(2)做直积，然后往点群双群做分解。需要的就 是点群的特征标表与双群的特征标表。32 种晶体点群的特征标表我们在 附录A 中给出，对应双群的请参考文献 [5, 22-24]。 这些东西为什么重要？因为旋轨耦合是一种相对论效应， 也是我们在研究电 子结构的时候非常关注的一个课题。 大家去注意近期凝聚态物理的一些新的进展， 比如拓扑绝缘体、Weyl Semimetal，旋轨耦合在里面都起了最为关键的作用。这 些问题应该说是近些年赋予了能带论很多新的内容， 你们从事研究的时候很可能 触及，因为这个原因，我相信这些会对你们中的很多人挺有用的。以后具体做研 究的时候，更多细节需要你们在文献[5, 22-24]中挖掘。 5.4 Clebsch-Gordan 系数 C-G 系数这个概念最早是由两个德国数学家Alfred Clebsh（1833-1872）与 Paul Gordan （1837-1912） 提出。 他们想解决的问题是两个球谐函数相乘后， 如何 再用球谐函数进行加和展开?26 后来，人们发现它可以描述量子力学中的角动量 耦合。而这个物理问题的根，又是群论中的不可约表示直积与直和分解问题。因 为这个原因，这部分内容是《群论》课程需要介绍的。我们不展开，之说简单规 则，出于以后使用的时候方便的考虑。 分解情况很简单，由𝑗1标识的非耦合表象空间维度是2𝑗1 + 1，由𝑗2标识的非 耦合表象空间维度是2𝑗2 + 1。 两者直积空间的维度是(2𝑗1 + 1) × (2𝑗2 + 1)。 耦合 后，𝑗的取值从𝑗1 −𝑗2到𝑗1 + 𝑗2，每个𝑗的维度是2𝑗+ 1，总维度数还是(2𝑗1 + 1) × (2𝑗2 + 1)。 这个问题是一个从(2𝑗1 + 1) × (2𝑗2 + 1)维空间向(2𝑗1 + 1) × (2𝑗2 + 1) 维空间的约化，约化完了以后每个维度由𝑗、𝑚进行标识。我们需要知道的，是 |𝑗, 𝑚⟩这个本征态是怎么由|𝑗1, 𝑚1⟩与|𝑗2, 𝑚2⟩的乘积通过线性组合组成的？这里 要用到的式子很简单，就是： 𝜓𝑗𝑚= ∑( 𝑗1 𝑗2 𝑚1 𝑚2| 𝑗 𝑚) 𝜓𝑚1 𝑗1 𝜓𝑚2 𝑗2 𝑚1,𝑚2 其中( 𝑗1 𝑗2 𝑚1 𝑚2| 𝑗 𝑚)对𝑗2 = 1/2的形式最简单也最常用到，这里直接给出： ( 𝑗1 1/2 𝑚1 𝑚2 | 𝑗 𝑚) 𝑗 𝑚2 = 1/2 𝑚2 = −1/2 𝑗1 + 1/2 ( 𝑗1 + 𝑚+ 1 2 2𝑗1 + 1 ) 1 2 ( 𝑗1 −𝑚+ 1 2 2𝑗1 + 1 ) 1 2 26略作展开讨论：这是一个纯数学问题，他们两个人也是数学家。但德国科学在19 世纪-20 世纪 的发展中，数学家、物理学家、化学家的交流使得科学在这个阶段产生了质变。Alfred Clebsh 除 了学术成就，1868 年他与数学物理学家Carl Neumann（此诺伊曼的研究与磁通量单位的那个 Weber 和温度与熵的那个Clausius 有交集， 非我们更熟悉的那个量子力学的诺伊曼） 一起建立的 Mathematische Annalen 杂志对后来的数学与物理发展很关键。Paul Gordan 被称为不变量理论之 王，他是雅可比的学生。C-G 系数这个早期数学上的处理，在量子力学发展起来之后，真正发挥 出了价值。 296 𝑗1 −1/2 −( 𝑗1 −𝑚+ 1 2 2𝑗1 + 1 ) 1 2 ( 𝑗1 + 𝑚+ 1 2 2𝑗1 + 1 ) 1 2 表5.4 𝑗2 = 1/2系统C-G 系数表 对𝑗2是整数或其它半整数的情况， 请参考曾谨言老师 《量子力学》 第一卷第九章。 第六章 置换群 置换群之所以在物理中重要， 一个重要的原因是真实的物理系统拥有这样的 对称性，比如全同粒子系统。另外，在置换理论发展过程中发展起来的杨算符方 法在近代物理的发展过程中也起到了非常重要的作用。 与此同时， 所有的有限群， 均同构于置换群的子群（第一章的凯莱定理） 。结合这些，置换群是我们《群论 一》的重要组成部分。 单单把这段的第一点展开，就可以牵扯出很多内容。最直接的一个，就是我 们前面讲到的点群与空间群、 转动群描述的都是一个单粒子本征态的对称性； 而 这里的置换群， 其实描述的是一个由全同粒子构成的多体系统的对称性。 以多电 子体系为例 （这个电子体系可以是一个原子中多电子， 也可以是分子或固体中多 电子体系） ，其多粒子本征态就需要用置换群的不可约表示来标识。类似研究在 1926 年薛定谔方程针对氢原子这个单电子体系提出之后一度是一个热点（量子 体系由单电子向多电子体系过渡再自然不过） 。其中最直接的过渡就是除氢以外 的其它原子体系。 类似体系既有完全转动群， 又有我们这里要讲的置换群对称性。 到了上世纪40 年代，人们在方面的研究又有了一些扩展，对象是晶体或者分子 配位场中的过渡金属原子。 对称性由完全转动群下降为某点群， 但置换对称性保 留。这方面的理论叫配位场理论Ligand Field Theory（或称晶体场理论，Crystal Field Theory）。再后来，凝聚态物理中的非线性光学问题与弹性理论发展过 程中，置换群理论也有一定程度的应用。 本课程中，我们如下四部分内容：1)n 阶置换群；2)杨盘及其引理；3)置换群 的不可约表示；4)多电子原子体系波函数。其中杨盘及其引理部分牵扯到许多证 明，由于时间限制，课上主要讲解这个证明的基本逻辑，具体证明过程也放在附 298 录D 中。配位场理论感兴趣的同学可参考文献，凝聚态体系非线性光学问题 与弹性理论中的置换可参考Dresselhaus 教材第18 章。 最后要说明的一点是由于笔者的背景并不是理论物理， 因此对置换群在粒子 物理中多粒子体系的应用没有什么体会， 建议选修 《群论二》 的同学参考文献[6、 9-11、17]了解更多内容。 6.1 n 阶置换群 （首先是置换的定义） 定义6.1 将n 个数字{𝟏、𝟐、⋯、𝐧}的排列𝒂𝟏，𝒂𝟐，⋯，𝒂𝐧 （注意， 是 “排列” ， “组合”只有一个）映为排列𝒃𝟏，𝒃𝟐，⋯，𝒃𝐧的操作，称为一个n 阶置换，记 为s，s 的形式为： 𝐬= ( 𝒂𝟏，𝒂𝟐，⋯，𝒂𝐧 𝒃𝟏，𝒃𝟐，⋯，𝒃𝐧 ) 这个置换干的事情， 就是把𝒂𝟏变为𝒃𝟏， 𝒂𝟐变为𝒃𝟐， ⋯， 𝒂𝐧变为𝒃𝐧。 它取决于诸对 数码的对换，与诸对数码的排列顺序无关，比如： ( 𝒂𝟏，𝒂𝟐，𝒂𝟑，⋯，𝒂𝐧 𝒃𝟏，𝒃𝟐，𝒃𝟑，⋯，𝒃𝐧 ) = ( 𝒂𝟏，𝒂𝟑，𝒂𝟐，⋯，𝒂𝐧 𝒃𝟏，𝒃𝟑，𝒃𝟐，⋯，𝒃𝐧 ) 只要配对相同即可。 定义6.2 置换群：定义两个置换r、s 的乘积rs 为先执行置换s，再执行置换r， 则在此乘法规则下所有的n 阶置换的集合构成一个群，这个群就称为n 阶置换 群或n 阶对称群，记为𝐒𝐧。 在这个群中，单位元是恒等置换。对s = ( 𝑎1，𝑎2，⋯，𝑎n 𝑏1，𝑏2，⋯，𝑏n )，逆元为s−1 = ( 𝑏1，𝑏2，⋯，𝑏n 𝑎1，𝑎2，⋯，𝑎n )。置换的乘法满足封闭性与结合律，Sn群的阶为n！ 。 定义6.3 轮换：一种特殊的置换( 𝒆𝟏，𝒆𝟐，⋯，𝒆𝐦 𝒆𝟐，𝒆𝟑，⋯，𝒆𝟏 )称为轮换，记为(𝒆𝟏，𝒆𝟐， ⋯，𝒆𝐦)，轮换数码的个数m 称为轮换的阶。 关于轮换，它的性质包括： 1. 轮换内的数码作轮换，仍代表同一个轮换，即： (𝑒1，𝑒2，⋯，𝑒m) = (𝑒2，𝑒3，⋯，𝑒m，𝑒1) = (𝑒m，𝑒1，𝑒2，⋯，𝑒m−1) 2. 两个轮换(𝑒1，𝑒2，⋯，𝑒m)与(𝑓 1，𝑓 2，⋯，𝑓 n)若没有公共数码，则称它们 相互独立，相互独立的轮换之间的乘法满足交换律，即： (𝑒1，𝑒2，⋯，𝑒m)(𝑓 1，𝑓 2，⋯，𝑓 n) = ( 𝑒1，𝑒2，⋯，𝑒m，𝑓 1，𝑓 2，⋯，𝑓 n 𝑒2，𝑒3，⋯，𝑒1，𝑓 2，𝑓 3，⋯，𝑓 1 ) = ( 𝑓 1，𝑓 2，⋯，𝑓 n，𝑒1，𝑒2，⋯，𝑒m 𝑓 2，𝑓 3，⋯，𝑓 1，𝑒2，𝑒3，⋯，𝑒1 ) = (𝑓 1，𝑓 2，⋯，𝑓 n)(𝑒1，𝑒2，⋯，𝑒m) 3. 任意的n 阶置换总可以分解为相互独立的轮换的乘积，比如： ( 1，2，3，4，5，6 4，2，6，5，1，3 ) = (1，4，5)(2)(3，6) 你要做的很简单，就是先盯上第一个数，看它变到几，再盯上那个数，依次 类推，这样总能找到一个轮换； 在找到这个轮换之后，取不属于这个轮换的第一个数，重复上面操作。依次 类推。这样任何一个置换都可以分解为轮换的乘积。 4. 轮换的逆，就是反过来，比如： (𝑒1，𝑒2，⋯，𝑒m) −1 = (𝑒m，𝑒m−1，⋯，𝑒1) 5. 二阶轮换(𝑒1，𝑒2)称为对换，任意一个m 阶轮换都可以写成m −1个对换的 300 乘积。因为： (𝑒1，𝑒2，⋯，𝑒m) = ( 𝑒1，𝑒2，𝑒3，⋯，𝑒m 𝑒2，𝑒3，𝑒4，⋯，𝑒1 ) = ( 𝑒1，𝑒2，𝑒3，⋯，𝑒m 𝑒2，𝑒1，𝑒3，⋯，𝑒m 𝑒2，𝑒1，𝑒3，⋯，𝑒m 𝑒2，𝑒3，𝑒4，⋯，𝑒1 ) = (𝑒1，𝑒3，⋯，𝑒m)(𝑒1，𝑒2) = (𝑒1，𝑒4，⋯，𝑒m)(𝑒1，𝑒3)(𝑒1，𝑒2) = ⋯= (𝑒1，𝑒m)(𝑒1，𝑒m−1) ⋯(𝑒1，𝑒3)(𝑒1，𝑒2) 6. 上面的逻辑是一个任意置换可以写成轮换的乘积，而任意一个轮换可以写成 对换的乘积。但是在这个轮换分解为对换的过程中，对换的对象不一定是相 邻的数。针对这个，第6 个性质说的是对∀(𝑒1，𝑒k)，有： (𝑒1，𝑒k) = (𝑒2，𝑒k)(𝑒1，𝑒2)(𝑒2，𝑒k) 因为： (𝑒2，𝑒k)(𝑒1，𝑒2)(𝑒2，𝑒k) = ( 𝑒1，𝑒2，𝑒k 𝑒1，𝑒k，𝑒2 𝑒1，𝑒k，𝑒2 𝑒2，𝑒k，𝑒1 𝑒2，𝑒k，𝑒1 𝑒k，𝑒2，𝑒1) = ( 𝑒1，𝑒2，𝑒k 𝑒k，𝑒2，𝑒1 ) = (𝑒1，𝑒k) 这样一个性质，结合3、5 两点，就一个把任意一个置换分解为相临对换的乘 积了。比如： ( 1，2，3，4 3，2，4，1 ) = (1，3，4) = (1，4)(1，3) = (2，4)(1，2)(2，4)(2，3)(1，2)(2，3) = (3，4)(2，3)(3，4)(1，2)(3，4)(2，3)(3，4)(2，3)(1，2)(2，3) 基于上面的介绍，我们可以给出这一部分的第一个定理。 定理6.1 具有相同轮换结构的置换构成置换群𝐒𝐧的一个类。 （相同的轮换结构这样的规定有两个意思， 既指它们有相同个数的轮换因子， 又 指各轮换因子中数码个数也完全相同） 证明： （分两个方面， 一是共轭的置换有相同的轮换结构； 二是具有相同轮换结构的置 换共轭） 1. 先证共轭置换有相同的轮换结构。取： ∀s = ( 1，2，⋯，n 𝑐1，𝑐2，⋯，𝑐n ) ∈Sn 由 ∀t ∈Sn，可产生一个s 的共轭置换， t = ( 1，2，⋯，n 𝑑1，𝑑2，⋯，𝑑n ) 为了求出tst−1是多少，我们把(1，2，⋯，n)重排为(𝑐1，𝑐2，⋯，𝑐n)，这样的 话t 也可以写成： t = ( 𝑐1，𝑐2，⋯，𝑐n 𝑓 1，𝑓 2，⋯，𝑓 n ) 同样，t−1也可以写为： ( 𝑓 1，𝑓 2，⋯，𝑓 n 𝑐1，𝑐2，⋯，𝑐n ) 或 302 ( 𝑑1，𝑑2，⋯，𝑑n 1，2，⋯，n ) 而tst−1就是： ( 𝑑1，𝑑2，⋯，𝑑n 1，2，⋯，n 1，2，⋯，n 𝑐1，𝑐2，⋯，𝑐n 𝑐1，𝑐2，⋯，𝑐n 𝑓 1，𝑓 2，⋯，𝑓 n ) = ( 𝑑1，𝑑2，⋯，𝑑n 𝑓 1，𝑓 2，⋯，𝑓 n ) 这也就是说s 的共轭元素tst−1是由t 对s 的上下两行( 1，2，⋯，n 𝑐1，𝑐2，⋯，𝑐n )同时作置 换得到的。这里t 既可以写成( 1，2，⋯，n 𝑑1，𝑑2，⋯，𝑑n )对s 上面那行操作，也可写成 ( 𝑐1，𝑐2，⋯，𝑐n 𝑓 1，𝑓 2，⋯，𝑓 n )对s 下面那行操作。 最终的结果是：( 𝑑1，𝑑2，⋯，𝑑n 𝑓 1，𝑓 2，⋯，𝑓 n )。 现在假设s 有k 个独立的轮换因子，s = s1s2 ⋯sk，那么其共轭tst−1可写为： ts1t−1ts2t−1 ⋯tskt−1。对s 的第i 个轮换因子，我们看tsit−1的效果： si = (𝑠1，𝑠2，⋯，𝑠m) = ( 𝑠1，𝑠2，⋯，𝑠m，𝑠m+1，𝑠m+2，⋯，𝑠n 𝑠2，𝑠3，⋯，𝑠1，𝑠m+1，𝑠m+2，⋯，𝑠n ) 任意取一个t，它是： ( 𝑠1，𝑠2，⋯，𝑠m，𝑠m+1，𝑠m+2，⋯，𝑠n 𝑡1，𝑡2，⋯，𝑡𝑚，𝑡m+1，𝑡m+2，⋯，𝑡n ) 由置换与队的排列顺序无关这个特点，我们知道t 也可写为： ( 𝑠2，⋯，𝑠m，𝑠1，𝑠m+1，𝑠m+2，⋯，𝑠n 𝑡2，⋯，𝑡𝑚，𝑡1，𝑡m+1，𝑡m+2，⋯，𝑡n ) 这样的话， 由前面讨论内容， tsit−1就是利用上面两式对si的上下行分别进行置换， 也就是： ( 𝑡1，𝑡2，⋯，𝑡𝑚，𝑡m+1，𝑡m+2，⋯，𝑡n 𝑡2，⋯，𝑡𝑚，𝑡1，𝑡m+1，𝑡m+2，⋯，𝑡n ) = (𝑡1，𝑡2，⋯，𝑡m) 也就是说tsit−1与si是同阶轮换。对其它轮换因子，做同样操作，依次类推，这样 ts1t−1ts2t−1 ⋯tskt−1就与s1s2 ⋯sk有相同的轮换结构。也就是tst−1与s 有相同的 轮换结构（共轭置换的轮换结构相同） 。 2. 现在看第二部分，具有相同轮换因子的置换共轭，取两个这样的置换： s = (𝑎1，𝑎2，⋯，𝑎n1)(𝑏1，𝑏2，⋯，𝑏n2) ⋯(𝑐1，𝑐2，⋯，𝑐n𝑙) r = (𝑑1，𝑑2，⋯，𝑑n1)(𝑒1，𝑒2，⋯，𝑒n2) ⋯(𝑓 1，𝑓 2，⋯，𝑓 n𝑙) 这个时候，一定∃t ∈Sn， t = ( 𝑎1，𝑎2，⋯，𝑎n1 𝑑1，𝑑2，⋯，𝑑n1 ) ( 𝑏1，𝑏2，⋯，𝑏n2 𝑒1，𝑒2，⋯，𝑒n2 ) ⋯( 𝑐1，𝑐2，⋯，𝑐n𝑙 𝑓 1，𝑓 2，⋯，𝑓 n𝑙 ) 使得： tst−1 = ( 𝑑1，𝑑2，⋯，𝑑n1，𝑒1，𝑒2，⋯，𝑒n2，𝑓 1，𝑓 2，⋯，𝑓 n𝑙 𝑎1，𝑎2，⋯，𝑎n1，𝑏1，𝑏2，⋯，𝑏n2，𝑐1，𝑐2，⋯，𝑐n𝑙 𝑎1，𝑎2，⋯，𝑎n1，𝑏1，𝑏2，⋯，𝑏n2，𝑐1，𝑐2，⋯，𝑐n𝑙 𝑎2，𝑎3，⋯，𝑎1，𝑏2，𝑏3，⋯，𝑏1，𝑐2，𝑐3，⋯，𝑐1 𝑎2，𝑎3，⋯，𝑎1，𝑏2，𝑏3，⋯，𝑏1，𝑐2，𝑐3，⋯，𝑐1 𝑑2，𝑑3，⋯，𝑑1，𝑒2，𝑒3，⋯，𝑒1，𝑓 2，𝑓 3，⋯，𝑓 1 ) = (𝑑1，𝑑2，⋯，𝑑n1，𝑒1，𝑒2，⋯，𝑒n2，𝑓 1，𝑓 2，⋯，𝑓 n𝑙 𝑑2，𝑑3，⋯，𝑑1，𝑒2，𝑒3，⋯，𝑒1，𝑓 2，𝑓 3，⋯，𝑓 1 ) = (𝑑1，𝑑2，⋯，𝑑n1)(𝑒1，𝑒2，⋯，𝑒n2) ⋯(𝑓 1，𝑓 2，⋯，𝑓 n𝑙) = r 具有相同轮结构的置换必共轭。 结合这两点，我们现在就知道了置换群的类与轮换结构存在一一对应的关系。 （证毕） 304 这个性质是置换群的一个关键的性质。 正是因为它的存在， 我们才可以用杨 图 （Young Diagram） 来分析置换群。 与这个性质相关的置换群的其它性质还包括： 1. 可以由轮换分解来划分置换群的类别。这个很显然，因为每个置换都可以写 成轮换的乘积，而轮换结构与类之间存在一一对应的关系。 这个轮换分解我们标记为： (𝛾)， (𝛾) = (1𝛾1，2𝛾2，3𝛾3，⋯，n𝛾n)， 即该类中 有𝛾1个一阶轮换，𝛾2个二阶轮换，⋯，𝛾n个n 阶轮换。 且由于变换对象只有n 个，所以： 𝛾1 + 2𝛾2 + ⋯+ n𝛾n = n 这里𝛾1、𝛾2、⋯、𝛾n为非负整数。 2. Sn中类(𝛾)里的置换群元个数为： n! (1𝛾1𝛾1!)(2𝛾2𝛾2!) ⋯(n𝛾n𝛾n!) 这是因为我们一共有n 个空要去填，填法是n!种。但是对其中的一个m 阶轮 换， (𝑒1，𝑒2，⋯，𝑒m) = (𝑒2，𝑒3，⋯，𝑒1) = ⋯= (𝑒m，𝑒1，⋯，𝑒m−1) 这m 种填法代表同样的置换。这样如果有𝛾m个m 阶置换的话，类似的重复 会出现m𝛾m次。 同时对这𝛾m个m 阶置换，由于它们没有公共因子，排列的前后顺序不影响结 果，类似重复又会出现𝛾m!次，所以上面的分母是那个样子。 3. 根据这样一个轮换分解的标记(𝛾) = (1𝛾1，2𝛾2，3𝛾3，⋯，n𝛾n)，我们来定义 杨图，它的标记方式是：[𝜆] = [𝜆1，𝜆2，⋯，𝜆n]，其中： 𝜆1 = 𝛾1 + 𝛾2 + ⋯+ 𝛾n 𝜆2 = 𝛾2 + ⋯+ 𝛾n ⋮ 𝜆n = 𝛾n 这样的话𝜆n ≤𝜆n−1 ≤⋯≤𝜆1。且由： 𝛾1 + 2𝛾2 + ⋯+ n𝛾n = n 我们知： 𝜆1 + 𝜆2 + ⋯+ 𝜆n = n 4. 这样的话Sn的分类就可以用杨图来表示了。它就是n 个小方格，排列方式为 第一行、第二行、⋯、第n 行分别是𝜆1、𝜆2、⋯、𝜆n个小方格，它们的第一列 靠左对其。 例：S3可由[𝜆]写成、[2，1]、[1，1，1]。杨图为： 图6.1 杨图示意图一 对杨图而言， 如果一个杨图可由另一个杨图通过转置得到， 则称这两个杨图 共轭。如果一个杨图转置后不变，则称其自轭。杨图共轭与元素共轭是两回事。 6.2 杨盘及其引理 上一节我们讲的杨图， 是针对置换群分类的一个工具。 这节课我们要讲的杨 盘（Young Tableau） ，针对的是置换群的不等价不可约表示。 其中最核心的地方， 是杨盘定理。 这个定理证明起来比较麻烦， 有７个引理， 分别都需要证明。细走这些步骤，对于不学《群论二》的同学，更多的是时间上 的浪费。对于学《群论二》的同学，你们可参考附录中的内容进行理解。在课堂 306 上，我们着重于讲解几个重要的概念：1) 什么是杨盘？2) 什么是杨算符？3) 什 么是本质幂等元？4) 杨盘定理说的是什么？5)它是怎么证明的 （正文是思路， 细 节见附录）？下面，我们会根据这个逻辑展开讲解。 定义6.4 杨盘：将数字1、2、⋯、n 分别填到𝐒𝐧的杨图的n 个小方格中，得到的 就是杨盘（填完了数的杨图） 。 例：S6的杨图[3、2、1]的两个杨盘Ta、Tb。 图6.2 杨图示意图二 由这个定义我们知道杨盘可以具有下面的性质： 1. 由一个杨图可以得到n！个杨盘； 2. 一个杨盘中的数字可以由其行与列来确定； 3. 同一个杨图的不同杨盘Ta、Tb，可通过一个置换相互转换。将Ta、Tb 中的数 按照从左到右，从上到下的顺序排成有序对𝑎1，𝑎2，⋯，𝑎n；𝑏1，𝑏2，⋯， 𝑏n。则杨盘Ta 到杨盘Tb 的置换为： s = ( 𝑎1，𝑎2，⋯，𝑎n 𝑏1，𝑏2，⋯，𝑏n ) 以上图为例： s = ( 1，2，5，3，4，6 1，4，6，5，2，3 ) = (1)(2，4)(5，6，3) Tb = (1)(2，4)(5，6，3)Ta 4. 由一个杨盘T 可以定义行置换R(T)与列置换C(T)。 R(T)是保持杨盘T 的各行中的数字还在其相对行上的所有置换p ̂的集合{p ̂}； C(T)是保持杨盘T 的各列中的数字还在其相对列上的所有置换q ̂的集合{q ̂}。 显然R(T)、C(T)都是Sn这个置换群的子群，他们有唯一的公共元素s0；若杨 盘T 对应的杨图为[𝜆] = [𝜆1，𝜆2，⋯，𝜆n]，则R(T)的阶为：𝜆1! 𝜆2! ⋯𝜆n!； C(T)的阶为𝜆 ̃1! 𝜆 ̃2! ⋯𝜆 ̃n!，其中[𝜆 ̃] = [𝜆 ̃1，𝜆 ̃2，⋯，𝜆 ̃n]为杨图[𝜆]的共轭。 5. 由行、列置换p ̂、q ̂可以定义两个算符P ̂(T)与Q ̂(T)： P ̂(T) = ∑p ̂ p ̂∈R(T) Q ̂(T) = ∑𝛿qq ̂ q ̂∈C(T) 其中𝛿q = 1如果q ̂为偶置换， 𝛿q = −1如果q ̂为奇置换。 这里偶置换与奇置换指 的是在置换本身化为对换的乘积后，对换的个数是偶数个还是奇数个？ 举个例子，前面的杨盘Ta。对它来说： R(Ta) = { (1)，(1，2)，(1，5)，(2，5)，(1，2，5)，(1，5，2)； (3，4)，(3，4)(1，2)，(3，4)(1，5)，(3，4)(2，5)， (3，4)(1，2，5)，(3，4)(1，5，2) } C(Ta) = { (1)，(1，3)，(1，6)，(3，6)，(1，3，6)，(1，6，3)； (2，4)，(2，4)(1，3)，(2，4)(1，6)，(2，4)(3，6)， (2，4)(1，3，6)，(2，4)(1，6，3) } 而P ̂(Ta)就是 R(Ta)中所有操作的和。Q ̂(Ta)就是 C(Ta)中所有操作乘上它的奇 偶性的和。 对(2，4)(1，3，6)这个操作而言， (1，3，6)是两个对换， (2，4) 是一个对换，所以总的对换数是3，这个操作的奇偶性为奇。 由这个P ̂(T)与Q ̂(T)的定义，我们知道它们是置换群中群元的线性组合，这个 很容易让我们联想到前面讲过的一个概念：群代数。后面，我们会通过杨算 308 符把P ̂(T)、Q ̂(T)与群代数联系起来。 6. 杨盘所具备的最后一个性质是同一个杨图的不同杨盘，其对应的行置换群相 互同构，列置换群也相互同构。 这个性质很直接，以： 图6.3 杨图示意图三 为例，R(Ta)、R(Tb)同构，C(Ta)、C(Tb)同构，看一下就知道了，因为就是变 换对象变了一下。 有了杨图、杨盘这些概念，下一个是杨算符。 定义6.5 杨算符：它是杨盘T 的算符𝐏 ̂(𝐓)与𝐐 ̂(𝐓)的乘积，形式为： 𝐄 ̂(𝐓) = 𝐏 ̂(𝐓)𝐐 ̂(𝐓) = ∑ ∑𝜹𝐪 𝐪 ̂∈𝐂(𝐓) 𝐩 ̂𝐪 ̂ 𝐩 ̂∈𝐑(𝐓) 显然杨算符是群空间𝐑𝐒𝐧中的一个矢量。一个杨盘有一个杨算符。 与杨算符相关的性质有很多，先说两个： 1. 若p ̂、p ̂′ ∈R(T)，q ̂、q ̂′ ∈C(T)，且p ̂q ̂ = p ̂′q ̂′，则必有：p ̂＝p ̂′、q ̂ = q ̂′。这个 性质的证明用到的很重要的一点就是对R(T)、C(T)这两个这两个Sn的子群， 它们的交集只有单位元素s0， 这样的话由p ̂q ̂ = p ̂′q ̂′可以得出： p ̂′−1p ̂ = q ̂′q ̂−1， 这个等式左边∈R(T)，右边∈C(T)，所以只能有p ̂′−1p ̂ = q ̂′q ̂−1 = s0，这样的 话就只能有：p ̂＝p ̂′、q ̂ = q ̂′。 2. 同时由上面一点我们也很容易知道对E ̂(T) = ∑ ∑ 𝛿q q ̂∈C(T) p ̂q ̂ p ̂∈R(T) 的加和，一 定不会出现不同p ̂、q ̂产生相同的乘积，进而由于𝛿q的变号所带来的p ̂q ̂相互抵 消的情况。因此，E ̂(T)一定是非零的群空间中的向量。 现在是讲完了杨图、杨盘、和杨算符的概念了。再说一个幂等元，我们就可 以把杨盘定理的内容讲出来了。这个幂等元，和之前讲的投影算符相关。 定义6.6 幂等元： 在群代数𝐑𝐆中， 满足𝐞 ⃗⃗𝟐= 𝐞 ⃗⃗的元素𝐞 ⃗⃗， 称为幂等元。 而满足𝐞 ⃗⃗𝟐= 𝝀𝐞 ⃗⃗的元素𝐞 ⃗⃗，称为本质幂等元。 对本质幂等元，你可以这样理解：只要满足e ⃗ ⃗2 = 𝜆e ⃗ ⃗，那么这个e ⃗ ⃗只要乘上一 个常数， 它就是幂等元， 或者说它本质上就是一个幂等元。 而这个常数， 是𝜆−1， 因为：(𝜆−1e ⃗ ⃗)2 = 𝜆−2e ⃗ ⃗2 = 𝜆−2𝜆e ⃗ ⃗= 𝜆−1e ⃗ ⃗。 现在看这个幂等元的定义是不是和前面讲的投影算符P ̂ i 2 = P ̂ i有相似的地方？ 它们之间确实存在着重要的联系， 这个联系是： 群代数RG中左正则变换L(G)的群 不变的子空间及其投影算符与群代数RG中的幂等元一一对应27。总结为下面一个 定理。 定理6.2 群代数𝐑𝐆中有多少个幂等元，群空间就有多少个对左正则变换不变的 子空间。相应的，就有多少个投影算符。 要证明这个定理，需要另一个定理6.3。这个定理比较好证，我们先证这个，然 后回到6.2。 定理6.3 对群G 在表示空间V 上的表示𝐀(𝐠)，若V 可分解为𝐖𝟏⨁𝐖𝟐⨁⋯⨁𝐖𝐤 这k 个子空间的直和，则其中𝐖𝐢为群不变的子空间的充要条件为𝐖𝐢对应的投影 27在课程中研究群本身有多少不等价、不可约表示以及它们性质的时候，我们会借助群代数以及 正则表示作为工具来分析。 这个在第二章的正交性、 完备性定理中已有体现。 在这些性质清楚后， 具体的不可约表示表示空间可以任意。但只要牵扯到针对不等价、不可约表示自身性质的分析， 正则表示这些还是基本工具，比如这里。 310 算符𝐏 ̂𝐢与任意一个g 对应的𝐀(𝐠)互易，即𝐀(𝐠)𝐏 ̂𝐢= 𝐏 ̂𝐢 𝐀(𝐠)，对∀𝐠∈𝐆成立。 证明： 必要性， 由Wi为群不变的子空间， 推Wi对应的投影算符P ̂ i与任意一个g 对应的A(g) 互易。 由于P ̂ i为Wi对应的投影算符， 满足： P ̂ i 2 = P ̂ i， P ̂ iP ̂ j = 0 （当i ≠j） ， P ̂ iV = Wi， P ̂ ix ⃗ ⃗i = x ⃗ ⃗i，所以： A(g)P ̂ ix ⃗ ⃗i = A(g)x ⃗ ⃗i = P ̂ iA(g)x ⃗ ⃗i 对∀x ⃗ ⃗i成立。由于x ⃗ ⃗i的任意性，A(g)P ̂ i = P ̂ i A(g)。 充分性，若V 上存在投影算符P ̂ i并且有：A(g)P ̂ i = P ̂ i A(g)，对∀g ∈G成立，则对 ∀x ⃗ ⃗i ∈Wi，由P ̂ ix ⃗ ⃗i = x ⃗ ⃗i，有： A(g)x ⃗ ⃗i＝A(g)P ̂ ix ⃗ ⃗i＝P ̂ i(A(g)x ⃗ ⃗i) ∈Wi 所以Wi为群不变的子空间。 （证毕） 现在回到定理6.2。 证明： 要证明在群代数RG中， 幂等元与左正则变换不变子空间的一一对应关系， 需要说 明两点： 1. 若Wi = P ̂ iRG为正则变换G 不变的子空间，则存在一个幂等元与之对应； 2. 若有一个幂等元，则存在一个与之对应的G 不变的RG的子空间与投影算符。 先看第一点： 有Wi = P ̂ iRG为正则变换G 不变的子空间，P ̂ i为投影算符。这时，群代数RG中有 这样一个向量e ⃗ ⃗i = P ̂ ig ⃗ ⃗0（g ⃗ ⃗0为RG中单位元素） ，这个向量是幂等元。 为什么呢？这个就要用到我们刚才介绍的那个性质了。因为Wi为对左正则变换 L(G)而言是G 不变的子空间，所以有L(g)P ̂ i = P ̂ iL(g)对∀g ∈G成立，也就是gP ̂ i = P ̂ ig对∀g ∈G成立。 这样的话对 ∀x ⃗ ⃗= ∑𝑥kg ⃗ ⃗k k ∈RG 有 P ̂ ix ⃗ ⃗= ∑𝑥kP ̂ ig ⃗ ⃗k k g ⃗ ⃗0 = ∑𝑥kg ⃗ ⃗kP ̂ i k g ⃗ ⃗0 = ∑𝑥kg ⃗ ⃗k k e ⃗ ⃗i = x ⃗ ⃗e ⃗ ⃗i 由P ̂ ix ⃗ ⃗= x ⃗ ⃗e ⃗ ⃗i，进而 P ̂ i 2x ⃗ ⃗= P ̂ iP ̂ ix ⃗ ⃗= P ̂ ix ⃗ ⃗e ⃗ ⃗i = x ⃗ ⃗e ⃗ ⃗ie ⃗ ⃗i 而P ̂ i为投影算符，P ̂ i 2 = P ̂ i，P ̂ ix ⃗ ⃗= x ⃗ ⃗e ⃗ ⃗i，所以P ̂ i 2x ⃗ ⃗= P ̂ ix ⃗ ⃗。这样就有： x ⃗ ⃗e ⃗ ⃗i = x ⃗ ⃗e ⃗ ⃗ie ⃗ ⃗i 对∀x ⃗ ⃗∈RG成立。这样的话： e ⃗ ⃗i = e ⃗ ⃗i 2 对Wi = P ̂ iRG这个对正则变换G 不变的子空间，我们就找到了与之对应的幂等元 e ⃗ ⃗i，它等于P ̂ ig ⃗ ⃗0。 再看第二点： 设e ⃗ ⃗i ∈RG为幂等元，要找与之对应的G 不变的子空间以及相应的投影算符。 定义算符P ̂ i，为P ̂ ix ⃗ ⃗= x ⃗ ⃗e ⃗ ⃗i，e ⃗ ⃗i就是我们已知的幂等元，x ⃗ ⃗为RG中任意向量。 由这个定义，知： P ̂ i 2x ⃗ ⃗= P ̂ i(P ̂ ix ⃗ ⃗) = P ̂ ix ⃗ ⃗e ⃗ ⃗i = x ⃗ ⃗e ⃗ ⃗ie ⃗ ⃗i = x ⃗ ⃗e ⃗ ⃗i = P ̂ ix ⃗ ⃗ 对∀x ⃗ ⃗∈RG成立。因此P ̂ i 2 = P ̂ i，P ̂ i为投影算符。 这个投影算符作用到群代数上形成的子空间是G 不变的子空间。这个好证，因 为对任意置换群群元g ⃗ ⃗k，有： P ̂ ig ⃗ ⃗kx ⃗ ⃗= g ⃗ ⃗kx ⃗ ⃗e ⃗ ⃗i 312 而g ⃗ ⃗kP ̂ ix ⃗ ⃗= g ⃗ ⃗kx ⃗ ⃗e ⃗ ⃗i。因此P ̂ ig ⃗ ⃗kx ⃗ ⃗= g ⃗ ⃗kP ̂ ix ⃗ ⃗对任意x ⃗ ⃗成立，P ̂ ig ⃗ ⃗k = g ⃗ ⃗kP ̂ i。再由定理6.3，知 P ̂ i作用到群代数上形成的子空间是G 不变的子空间。这个与P ̂ i是投影算符合在一 起，就是我们前面说的第二个方面。 两个方面都证完了，自然就有群代数RG对左正则表示G 不变的子空间及其投影 算符与群代数RG中的幂等元一一对应。 （证毕） 这里这个幂等元是：P ̂ ig ⃗ ⃗0。我们要求P ̂ iRG为G 不变的RG的子空间，并不要求 它承载不可约表示。也就是说它承载表示，但这个表示不一定不可约，幂等元与 表示对应，不一定与不可约表示对应。 与不可约表示对应的幂等元称为本原幂等元。 它的特点是它对应的群代数中 群不变的子空间，为不可约的群不变的子空间。换句话说，它是最基本的， “本 原”的。幂等元与RG的G 不变的子空间对应，本原幂等元与RG的G 不变的不可 约的子空间对应。本原是其中最小的部分。 现在这些概念的积累说完了， 我们看杨盘定理说的是什么？实际上， 它是用 7 个引理来说明三句话。 定理6.4（杨盘定理） 1. 杨盘T 的杨算符𝐄 ̂(𝐓)可给出置换群群空间𝐑𝐒𝐧中的一个本原幂等元𝐄 ̂(𝐓)/𝛉， 其中𝛉为一个常数。也就是说一个杨盘给出置换群在其群空间𝐑𝐒𝐧中的一个不 可约表示； 2. 同一个杨图的不同杨盘给出的不可约表示相互等价； 3. 不同杨图的杨盘给出的不可约表示不等价。 这七个引理分别是： 引理6.1 设𝐓、𝐓′是由置换 𝐫 联系起来的杨盘，𝐓′ = 𝐫 𝐓，如果置换 𝐬 作用在𝐓 上，使得𝐓(𝐢，𝐣)数字变到𝐬𝐓中的(𝐢′， 𝐣′)处，则𝐬′ = 𝐫𝐬𝐫−𝟏也会使得𝐓′(𝐢，𝐣)中的 数字变到𝐬′𝐓′的(𝐢′， 𝐣′)处。 用图来理解，这个引理说的就是： 图6.4 杨盘定理引理示意图一 其中， r = ( 1，2，3，4，5，6 1，5，3，2，4，6 ) s = ( 1，2，3，4，5，6 1，3，2，4，5，6 ) rsr−1 = ( 1，5，3，2，4，6 1，3，5，2，4，6 ) 而rsr−1T′，等于： 314 图6.5 杨盘定理引理示意图二 s干的事情，是将T 的2、3 互换；s′干的事情，是将T′的3、5 互换。2、3 在T 中 的位置，和3、5 在T′中的位置，是相同的。 由这个引理，我们还可以知道： T′ = rT时，有R(T′) = rR(T)r−1，C(T′) = rC(T)r−1，P ̂(T′) = rP ̂(T)r−1，Q ̂(T′) = rQ ̂(T)r−1，E ̂(T′) = rE ̂(T)r−1。 这些引理（到6.7）与性质的详细证明，均见附录。 引理6.2 设𝐩 ̂、𝐪 ̂是杨盘T 的行、列置换，则T 中位于同一行的任意两个数字不 可能出现在𝐓′ = 𝐩 ̂𝐪 ̂𝐓的同一列中；反之，若𝐓′ = 𝐫𝐓时，T 中位于同一行的任意 两个数字都不出现在𝐓′的同一列中， 则杨盘T 存在行、 列置换𝐩 ̂、 𝐪 ̂， 使𝐫= 𝐩 ̂𝐪 ̂。 这两个引理， 说的都是同一杨图的不同杨盘的性质， 结合杨盘定理本身内容， 我们知道它们是在后面说明同一个杨图的不同杨盘给出的不可约表示等价的时 候会有用。 引理6.3 设杨盘𝐓和𝐓′分别属于杨图[𝛌]、[𝛌′]，且[𝛌]>[𝛌′] 这里， [λ]>[λ′]是指[𝜆] = [𝜆1，𝜆2，⋯，𝜆n]， [𝜆′] = [𝜆′1，𝜆′2，⋯，𝜆′n]， 第一个 不等于零的𝜆𝑖−𝜆′𝑖，一定满足𝜆𝑖> 𝜆′𝑖，比如： 图6.6 杨盘定理引理示意图三 则存在两个数码位于T的同一行与T′的同一列。 引理6.4 若有两个数字，位于杨盘𝐓的同一行与杨盘𝐓′的同一列，则它们的杨算 符𝐄 ̂(𝐓′)𝐄 ̂(𝐓) = 𝟎。 （这两个引理对应两个不同杨图的杨盘的性质， 它们的杨算符E ̂(T′)E ̂(T) = 0。 这 个我们把它放到杨盘定理全部内容的背景下， 说的就是不同杨图的杨盘给出的杨 算符对应的群空间中的子空间相互正交。 相应， 它们的不可约表示相互不等价。 ） 引理6.5 设置换群𝐒𝐧的群代数𝐑𝐒𝐧中的矢量𝐱 ⃗ ⃗= ∑ 𝐱𝐧𝐬 𝐬∈𝐒𝐧 ， 𝐓为𝐒𝐧的杨盘。 若∀𝐩 ̂ ∈ 𝐑(𝐓)，∀𝐪 ̂ ∈𝐂(𝐓)，𝐩 ̂𝐱 ⃗ ⃗𝐪 ̂ = 𝛅𝐪𝐱 ⃗ ⃗，则𝐱 ⃗ ⃗与𝐓盘的杨算符𝐄 ̂(𝐓)相差一个常数因子，即 𝐱 ⃗ ⃗= 𝛉𝐄 ̂(𝐓)，常数𝛉与𝐱 ⃗ ⃗有关。 引理6.6 杨盘𝐓的杨算符𝐄 ̂(𝐓)是置换群𝐒𝐧的群代数𝐑𝐒𝐧中的一个本质的本原幂等 元，不变子空间𝐑𝐒𝐧𝐄 ̂(𝐓)是置换群𝐒𝐧的一个不可约表示的表示空间，其维数是 𝐧！的因子。 这两个引理很明显6.6 是重点，6.5 是为了证6.6，而引理6.6 说的就是杨盘 定理的前半部分，杨盘T的杨算符E ̂(T)是其置换群群代数的本质的本原幂等元， RSnE ̂(T)给出置换群Sn的一个不可约表示。 而引理6.1 到6.4， 合在一起，是为了说明杨盘定理的后半部分。我们把这个 316 后半部分归纳为引理6.7。 引理6.7 置换群𝐒𝐧的同一个杨图的不同杨盘， 给出的不可约表示是等价的， 不同 杨盘给出的该置换群的不等价、不可约的表示。 引理6.6 加上引理6.7， 就给出了杨盘定理。 再结合杨图个数等于置换群类的 个数进而等于不等价不可约表示数， 我们就最终这一部分要传达的信息归结为下 面三句： 1. 一个置换群的不等价、不可约表示数等于其杨图的个数； 2. 从一个杨图，我们可以基于其杨盘来求置换群的不可约表示； 3. 因为一个杨图的杨盘有很多，这个不可约表示有很多的等价形式。 具体怎么求这个不等价、 不可约表示？操作过程中需要再理解一个定义， 一 个定理。 定义6.7 （标准盘，Standard Young Tableau）杨盘中，每行、每列，数字从左到 右，从上到下都是逐渐增加的盘，就叫标准盘。 以三阶循环群为例，杨图有： 图6.7 三阶循环群杨图 它们的标准盘有： 图6.8 三阶循环群标准盘 定理6.5 杨图[𝛌]对应的不可约表示的维度，等于其标准盘的个数28。 以上面的三阶置换群为例， 三个杨图所对应的不可约表示的维度， 分别是1、 2、1。它们的平方和刚好满足Burnside 定理。 还以这个三阶置换群为例， 杨图[2，1]的不可约表示怎么求呢？我们从标准 盘： 图6.9 杨图[2,1]的标准盘 出发，它所对应的R(T)、C(T)分别是： R(T) = {(1)、(1，2)} C(T) = {(1)、(1，3)} 杨算符： E ̂(T) = {(1) + (1，2)} {(1) −(1，3)} = (1) + (1，2) −(1，3) −(1，2)(1，3) = (1) + (1，2) −(1，3) −(1，3，2) 不可约表示空间为RS3E ̂(T)，维度是2。因此，我们要确定它的两个基，再作表示 28一个杨图可以容纳的杨盘的个数是n!，其中标准盘的个数等于其对应的不可约表示的维度。 318 矩阵。如何确定这两个基呢？我们可以做： (1)E ̂(T) = (1) + (1，2) −(1，3) −(1，3，2) = E ̂(T) (1，2)E ̂(T) = (1，2) {(1) + (1，2) −(1，3) −(1，3，2)} = (1，2) + (1) −(1，2)(1，3) −(1，2)(1，3，2) = (1，2) + (1) −(1，3，2) −(1，3) = E ̂(T) (1，3)E ̂(T) = (1，3) {(1) + (1，2) −(1，3) −(1，3，2)} = (1，3) + (1，2，3) −(1) −(2，3) (2，3)E ̂(T) = (2，3) {(1) + (1，2) −(1，3) −(1，3，2)} = (2，3) + (2，3)(2，1) −(2，3)(1，3) −(2，3)(1，2)(1，3) = (2，3) + (2，1，3) −(3，2)(3，1) −(2，3)(2，1)(1，3) = (2，3) + (1，3，2) −(3，1，2) −(2，1，3)(1，3) = (2，3) + (1，3，2) −(1，2，3) −(1，3，2)(1，3) = (2，3) + (1，3，2) −(1，2，3) −(1，2)(1，3)(1，3) = (2，3) + (1，3，2) −(1，2，3) −(1，2) = −{(1) + (1，2) −(1，3) −(1，3，2)} −{(1，3) + (1，2，3) −(1) −(2，3)} = −E ̂(T) −(1，3)E ̂(T) (1，2，3)E ̂(T) = (1，3)(1，2)E ̂(T) = (1，3)E ̂(T) (1，3，2)E ̂(T) = (2，1，3)E ̂(T) = (2，3)(2，1)E ̂(T) = (2，3)(1，2)E ̂(T) = (2，3)E ̂(T) = −E ̂(T) −(1，3)E ̂(T) 因此RS3E ̂(T)的两个基是：E ̂(T)、(1，3)E ̂(T)。以它们为基，我们可求出这个三 阶置换群的表示矩阵。以群元(1，3，2)为例： (1，3，2)E ̂(T) = −E ̂(T) −(1，3)E ̂(T) (1，3，2)(1，3)E ̂(T) = (1，2)(1，3)(1，3)E ̂(T) = (1，2)E ̂(T) = E ̂(T) 因此表示矩阵为： (−1 1 −1 0) 其它群元的表示矩阵求法类似。 6.3 多电子原子本征态波函数 现在开始基于前面的置换群基础理论讲应用。 具体例子是置换对称性允许的 全同粒子体系（比如𝑛个电子）本征态波函数。 我们用到的例子是多电子原子的本征态。 在这个例子中， 除了电子置换对称 性，系统对称性还包含：1. 原子体系本身的SO(3)对称性、2. 电子自旋的SU(2) 对称性。讨论中，我们会先从置换群对称性出发，推出的承载置换群不可约表示 的多体波函数。这个多体波函数不一定会同时承载SO(3)与SU(2)的不可约表示。 但由于此类系统同时具备这三种对称性， 由置换群对称性推出的承载置换群不可 320 约表示的波函数在进行线性组合后， 也可构成同时承载SO(3)与SU(2)的不可约表 示形式（这个相当于线性空间内内部结构的调整） 。这种组合对两电子体系很简 单，但对更多电子体系会比较麻烦。因此，在两电子体系的讨论中，我们会详细 说明在得到置换群的不可约表示本征态后如何线性组合同时得到SO(3)与SU(2) 群的不可约表示。 在三电子及以上电子数的例子中， 由于这节的主要内容是置换 群， 我们会将讨论重点放在置换对称性， 不针对SO(3)与SU(2)群的不可约表示进 行特殊讨论。 这里要用到的群论知识主要是一个n 阶置换群 （群元个数是n!） 根据Burnside 定理得到的各不等价、 不可约表示的维度以及各个置换群的特征标表， 用我们本 章前一节的内容可以求得，但过程会很复杂。这一节，我们用到的时候会把它们 当作已知条件给出。 群 类数 n! = ∑𝑙𝑖 2 𝑖 S1 1 1! = 1 = 12 S2 2 2! = 2 = 12 + 12 S3 3 3! = 6 = 12 + 12 + 22 S4 5 4! = 24 = 12 + 12 + 22 + 32 + 32 S5 7 5! = 120 = 12 + 12 + 42 + 42 + 52 + 52 + 62 S6 11 6! = 720 = 12 + 12 + 52 + 52 + 52 + 52 + 92 + 92 + 102 + 102 + 162 S7 15 7! = 5040 = 12 + 12 + 62 + 62 + 142 + 142 + 142 + 142 + 152 + 152 + 212 + 212 + 352 + 352 + 202 S8 22 8! = 40320 = 12 + 12 + 72 + 72 + 142 + 142 + 202 + 202 + 212 + 212 + 282 + 282 + 352 + 352 + 562 + 562 + 642 + 642 + 702 + 702 + 422 + 902 ⋯ ⋯ ⋯ 表6.1 置换群不等价、不可约表示维度 同时， 对两电子体系， 在考虑SO(3)与SU(2)对称性的时候我们还会用到上一章最 后一节的一些内容 （具体而言就是C-G 系数， 其对应的物理问题是角动量耦合） 。 但前面说过， 这一节重点讨论的是置换对称性， SO(3)与SU(2)对称性也只是在方 便讨论的时候详细讨论。 上面是对要用到的群论知识的简单介绍。 除此之外， 在本节的讨论中还有两 点需要说明。 第一点是在本节描述多电子波函数的时候， 我们把它描述为无相互 作用的多体系统。也就是说我们用下面的单体哈密顿量： H ̂(𝑥 ⃗𝑖) = 𝑝̂𝑖 2 2m + V(𝑥 ⃗𝑖) 确定单电子态𝜓0(𝑥 ⃗𝑖)、𝜓1(𝑥 ⃗𝑖)、⋯。其中𝑥 ⃗𝑖是第i 个电子的坐标，包含空间部分 （𝑟 ⃗ 𝑖， 三个连续分量） 与自旋部分 （两个状态α或β， 也就是自旋向上↑或自旋向下 ↓两个分立值） 。电子之间由于其全同性，允许交换。同时，电子是费米子，其波 函数必须交换反对称。 我们会根据这个限制， 说明在一个无相互作用的n 电子系 统中，其n 阶置换群的对称性会允许或禁止什么样的多体波函数存在？ 除了上面提到的第一点说明， 第二点是电子的空间坐标与自旋坐标严格意义 上有耦合。为简单起见，我们忽略旋轨耦合，将多电子波函数的自旋部分与轨道 部分分开处理。 也就是说电子间的轨道角动量𝑙 ⃗𝑖耦合为𝐿 ⃗ ⃗= ∑ 𝑙 ⃗ 𝑖 n 𝑖=1 ， 自旋角动量𝑠 ⃗𝑖 耦合为𝑆 ⃗= ∑ 𝑠 ⃗𝑖 n 𝑖=1 ，但𝐿 ⃗ ⃗与𝑆 ⃗之间的耦合我们不考虑，在此基础上讨论置换。在粒 子物理的很多例子中，人们也会采用类似处理，把自由度分开。先讨论各个自由 度本身的置换对称性， 然后合在一起让其满足玻色子或费米子的性质要求。 最典 型的一个粒子就是标准模型中人们对∆++重子（Baryon）的描述。它是一个自旋 3/2 粒子，由三个夸克（Quark）组成。三个夸克排列一样，自旋部分交换对称。 同时，它的空间部分与味（Flavor）部分也交换对称。如果只有这三个自由度的 话，就和它本身的费米子属性矛盾了。这个问题在上世纪60 年代曾经困扰了人 322 们一段时间。 后来的处理方式是引入色 （Color） 这个量子数， 系统在这个自由度 下处在交换反对称的单重态，由此拯救此系统中的费米统计。 我们讨论两电子原子与三电子原子。 更复杂情况按讨论规则展开。 先看两电 子系统，它的置换对称群是S2，特征标表如下： 1{E} 1{A} Γ 1 s 1 1 Γ 1 a 1 −1 Γ permut(𝜓1𝜓1) 1 1 ⇒Γ 1 s Γ permut(𝜓1𝜓2) 2 0 ⇒Γ 1 s ⊕Γ 1 a 表6. 2 二阶置换群特征标表 这个表的前三行是前面经常用到的特征标表的正常内容。 后面两行的意思是如果 两个电子占据的态是𝜓1𝜓1（或𝜓1𝜓2）的时候，在由𝜓1𝜓1（或𝜓1𝜓2）形成的线性 空间中， 二阶置换群S2的表示特征标Γ permut(𝜓1𝜓1) （或Γ permut(𝜓1𝜓2)） 是什么？ 以及它可以分解为哪些不可约表示的直和？更具体来说，就是permut 代表置换 permutation。Γ permut(𝜓1𝜓1)代表当我们可以置换的两个电子分别处在𝜓1态与𝜓1 态时置换群的表示。Γ permut(𝜓1𝜓2)代表当我们可以置换的两个电子一个处在𝜓1 态另一个处在𝜓2态时候置换群的表示。由于Γ permut(𝜓1𝜓1)与Γ permut(𝜓1𝜓2)可能 是可约表示，后两行的最后一列代表它们可以约化为哪些不可约表示的直积？ 在原子环境下，单电子态分别是1s、2s、2p、3s、3p、3d、4s…。这些单电 子态在能量轴的不连续分布如下： 图6.10 原子中的单电子轨道示意图 由于忽略了电子之间的相互作用， 两个单电子态组成的双电子态直接就是这个双 电子系统本征态。我们对这个态的唯一要求是费米统计。在下面讨论中，我们会 先确定要用到哪两个单电子态？然后讨论它们形成的双电子系统的双电子波函 数的情况。 当然， 对双电子波函数的描述， 还是将自旋部分与轨道部分分开讨论。 先看自旋部分， 两个电子态分别是自旋向上↑或自旋向下↓， 记为α或β。 当两 个电子的自旋状态相同（即都是α或都是β）的时候，根据表6.2 倒数第二行，置 换群表示空间是一维的。 这个一维空间承载二阶置换群S2的一维恒等表示Γ 1 s。它 对应的自旋构型分两种，α1α2、β1β2。 当两个电子的自旋处在的状态不同， 也就是一个自旋向上、 一个自旋向下的 时候，根据表6.2 最后一行，由αβ这种状态可形成由α1β2、α2β1组成的二维表示 空间（下标1、2 代表是哪个电子） 。这个表示空间承载二维表示，它可约化为二 阶置换群的一个一维恒等与一维非恒等的直和Γ 1 s ⊕Γ 1 a。其中承载一维恒等表示 Γ 1 s的正交归一基是(α1β2 + α2β1)/√2，承载一维非恒等表示Γ 1 a的正交归一基是 (α1β2 −α2β1)/√2。由于电子自旋只有两种状态，这四种情况（α1α2、β1β2、 (α1β2 + α2β1)/√2、(α1β2 −α2β1)/√2）就对应了两电子体系自旋部分的所有可 324 能。其中α1α2、β1β2、(α1β2 + α2β1)/√2承载二阶置换群的一维恒等表示Γ 1 s，置 换对称；(α1β2 −α2β1)/√2承载置换的一维非恒等表示Γ 1 a，置换反对称。与此同 时，根据第五章第四节内容（C-G 系数展开） ，前三个态刚好对应一个自旋S = 1 系统的自旋三重态，最后一个态对应自旋S = 0的自旋单重态。它们承载SU(2)的 三维不可约表示与一维不可约表示。这样，自旋的部分的两个对称性：SU(2)对 称性、S2对称性，在这个多体波函数自旋部分的描述中就同时梳理清楚了。这样 的自旋部分两体波函数可同时反映SU(2)对称性与S2对称性。 再看轨道部分，如果两个电子都占1s 轨道，那么轨道角动量耦合𝐿= 0。同 时两个电子的轨道部分波函数，根据表6.2，承载置换群的交换对称表示Γ 1 s。这 个时候， 费米统计要求自旋部分只能选承载Γ 1 a的自旋单重态。 在原子物理的语言 中， 不考虑旋轨耦合与电子间相互作用的时候， 多电子波函数经常用 𝐿𝑀 𝑆𝑧 2𝑆+1 来标 识。根据前面的对称性讨论，当两个电子都坐在1s 轨道时，允许的双电子态就 只能有S0 0 1 ， 也可简单标记为S 1 。 这里的这个S代表𝐿= 0对应的双电子轨道态 （不 同电子轨道之间的耦合已经考虑） 。 如果一个电子处在1s 轨道、另一个处在2s 轨道，轨道角动量耦合的𝐿依然 为零。但由于𝜓1s态与𝜓2s态不同，和前面自旋部分讨论一样，两体波函数的轨道 部分有两种可能：(𝜓1s(𝑟 ⃗ 1)𝜓2s(𝑟 ⃗ 2) + 𝜓1s(𝑟 ⃗ 2)𝜓2s(𝑟 ⃗ 1))/√2 、(𝜓1s(𝑟 ⃗ 1)𝜓2s(𝑟 ⃗ 2) − 𝜓1s(𝑟 ⃗ 2)𝜓2s(𝑟 ⃗ 1))/√2，分别承载二阶置换群的交换对称表示Γ 1 s与交换对称反表示 Γ 1 a。 当轨道部分是(𝜓1s(𝑟 ⃗ 1)𝜓2s(𝑟 ⃗ 2) + 𝜓1s(𝑟 ⃗ 2)𝜓2s(𝑟 ⃗ 1))/√2时， 由于交换对称， 自旋 部分必须交换反对称，对应单重态。总体两体波函数用 𝐿𝑀 𝑆𝑧 2𝑆+1 来标识就是S0 0 1 ， 简记为S 1 。当轨道部分是(𝜓1s(𝑟 ⃗ 1)𝜓2s(𝑟 ⃗ 2) −𝜓1s(𝑟 ⃗ 2)𝜓2s(𝑟 ⃗ 1))/√2时，由于交换反 对称，自旋部分必须交换对称，对应三重态（S = 1） 。综合轨道与自旋部分，两 体波函数用 𝐿𝑀 𝑆𝑧 2𝑆+1 来标识就是S0 −1 3 、S0 0 3 、S0 1 3 ，共同标记为S 3 。两者综合，如 果一个电子处在1s 轨道、另一个处在2s 轨道，允许的双电子态就是S 1 、S 3 。 如果一个是s 态、一个是p 态，轨道角动量耦合的𝐿是1，用P来标识。由于 𝜓𝑛s态与𝜓𝑛′p态不同，两体波函数的轨道部分有两种可能：(𝜓𝑛s(𝑟 ⃗ 1)𝜓𝑛′p(𝑟 ⃗ 2) + 𝜓𝑛s(𝑟 ⃗ 2)𝜓𝑛′p(𝑟 ⃗ 1)) /√2、 (𝜓𝑛s(𝑟 ⃗ 1)𝜓𝑛′p(𝑟 ⃗ 2) −𝜓𝑛s(𝑟 ⃗ 2)𝜓𝑛′p(𝑟 ⃗ 1)) /√2， 分别承载二阶置 换群的交换对称表示Γ 1 s与交换反对称表示Γ 1 a。当轨道部分是(𝜓𝑛s(𝑟 ⃗ 1)𝜓𝑛′p(𝑟 ⃗ 2) + 𝜓𝑛s(𝑟 ⃗ 2)𝜓𝑛′p(𝑟 ⃗ 1)) /√2时， 由于交换对称， 自旋部分必须交换反对称， 对应单重态。 总体两体波函数用 𝐿𝑀 𝑆𝑧 2𝑆+1 来标识就是P −1 0 1 、P 0 0 1 、P 1 0 1 ，共同标记为P 1 。当轨道 部分是(𝜓𝑛s(𝑟 ⃗ 1)𝜓𝑛′p(𝑟 ⃗ 2) −𝜓𝑛s(𝑟 ⃗ 2)𝜓𝑛′p(𝑟 ⃗ 1)) /√2时， 由于交换反对称， 自旋部分必 须交换对称， 对应三重态 （S = 1） 。 总体两体波函数用 𝐿𝑀 𝑆𝑧 2𝑆+1 来标识就是P −1 −1 3 、 P 0 −1 3 、P 1 −1 3 、P −1 0 3 、P 0 0 3 、P 1 0 3 、P −1 1 3 、P 0 1 3 、P 1 1 3 ，共同标记为P 3 。 两个电子都处在p 态，轨道角动量耦合的𝐿可以是0、1、2，用S、P、D来标 识。 𝜓𝑛p态有三种选择 （𝜓𝑛p1、 𝜓𝑛p0、 𝜓𝑛p−1） ， 𝜓𝑛′p态同样三种选择 （𝜓𝑛′p1、 𝜓𝑛′p0、 𝜓𝑛′p−1） 。但讨论要分𝑛= 𝑛′与𝑛≠𝑛′两个情况展开。 当𝑛= 𝑛′时， 两体波函数基的选择有： 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2)、 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)、 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2)、𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2)、𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)、𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2)、 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2)、𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)、𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2)九种（3 乘3） 。 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2)、 𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)、 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2)三种为单粒子轨道相同 的情况，对应表6.2 中倒数第二行，承载二阶置换群一维对称恒等表示。而 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)、𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2)、𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2)、𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2)、 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2)、𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)六种为轨道不同情况，根据电子置换规则 与表6.2， 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)与𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2)形成一个二阶置换群二维表示空间。 326 其中，(𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2) + 𝜓𝑛p1(𝑟 ⃗ 2)𝜓𝑛p0(𝑟 ⃗ 1)) /√2承载一维不可约交换对称表示， (𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2) −𝜓𝑛p1(𝑟 ⃗ 2)𝜓𝑛p0(𝑟 ⃗ 1)) /√2承载一维不可约交换反对称表示。 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2)与𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2)、𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2)与𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2) 分别形成的二阶置换群二维表示空间的情况与前例类似。 也就是说由不考虑置换 对称性的基形成的9 维空间，在进行线性变换后，可整理出六个维度（3 个相同 单粒子轨道的情况，3 个不同单粒子轨道的情况）承载交换对称的一维恒等不可 约表示，三个维度（都是不同单粒子轨道的情况）承载交换反对称的一维非恒等 不可约表示。置换对称性的要求不会造成维度浪费。 通过上述处理， 我们可以找出承载二阶置换群不可约表示的多体轨道波函数 的形式。 但与自旋部分不同的是这些承载二阶置换群不可约表示的波函数并不反 映SO(3)群的对称性。 要想让这些反映置换群对称性的波函数同时也反映SO(3)群 的对称性，我们还需要再进行一些线性操作。具体而言，就是： 1. Ψ(𝐿= 2, 𝑀= 2) = 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2) 2. Ψ(𝐿= 2, 𝑀= 1) = (𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2) + 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)) /√2 3. Ψ(𝐿= 2, 𝑀= 0) = [2𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2) + (𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2) + 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2))] /√6 4. Ψ(𝐿= 2, 𝑀= −1) = (𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2) + 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)) /√2 5. Ψ(𝐿= 2, 𝑀= −2) = 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2) 承载一维交换对称恒等表示Γ 1 s的同时，它们还承载SO(3)群的五维不可约表示， 对应𝐿= 2，也就是D轨道。除了这五个维度，剩下的四个维度中，有三个： 1. Ψ(𝐿= 1, 𝑀= 1) = (𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2) −𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)) /√2 2. Ψ(𝐿= 1, 𝑀= 0) = (𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2) −𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2)) /√2 3. Ψ(𝐿= 1, 𝑀= −1) = (𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2) −𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2)) /√2 就二阶置换群对称性而言，承载一维交换反对称非恒等表示Γ 1 a。就SO(3)群对称 性而言，承载三维不可约表示，对应𝐿= 1，也就是P轨道。最后的一个维度： 1. Ψ(𝐿= 0, 𝑀= 0) = −[𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛p0(𝑟 ⃗ 2) −(𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛p−1(𝑟 ⃗ 2) + 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛p1(𝑟 ⃗ 2))] /√3 就二阶置换群对称性而言，承载一维交换对称恒等表示Γ 1 s。就SO(3)群对称性而 言， 承载一维不可约表示， 对应𝐿= 0， 也就是S轨道。 这样的化轨道波函数部分， 我们也同时按二阶置换群S2、SO(3)进行了对称化的处理。 现在我们把轨道部分与自旋部分合起来， 当轨道部分是Ψ(𝐿= 0, 𝑀= 0)， 也 就是S态时，自旋部分只能是交换反对称的单重态。对应两体波函数用 𝐿𝑀 𝑆𝑧 2𝑆+1 来 标识就是S0 0 1 。 当轨道部分是Ψ(𝐿= 1, 𝑀= 0、± 1)， 也就是P态时， 自旋部分只 能是交换对称的三重态。 对应两体波函数用 𝐿𝑀 𝑆𝑧 2𝑆+1 来标识就是P −1 −1 3 、P −1 0 3 、P −1 1 3 、 P 0 −1 3 、P 0 0 3 、P1 3 、P 1 −1 3 、P 1 0 3 、P 1 1 3 ， 共同标记为P 3 。 当轨道部分是Ψ(𝐿= 2, 𝑀= 0、± 1、± 2)， 也就是D态时， 自旋部分只能是交换反对称的单重态。 对应两体 波函数用 𝐿𝑀 𝑆𝑧 2𝑆+1 来标识就是D−2 0 1 、D−1 0 1 、D0 0 1 、D1 0 1 、D2 0 1 ，共同标记为D 1 。 当占据态是p2的𝑛≠𝑛′时，分析与上面p2(𝑛= 𝑛′)类似。不同的是两体9 组 基由于𝑛≠𝑛′，可通过置换再产生9 个维度（共18 个） 。具体而言，就是相对于 前面9 个维度的情况相比，将交换对称与交换反对称的情况补全。以𝐿= 2的情 况为例，就会从原来的5 个交换对称的维度，变成10 个既包含交换对称又包含 交换反对称的维度。其中： 1. Ψs(𝐿= 2, 𝑀= 2) = (𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛′p1(𝑟 ⃗ 2) + 𝜓𝑛p1(𝑟 ⃗ 2)𝜓𝑛′p1(𝑟 ⃗ 1)) /√2 2. Ψs(𝐿= 2, 𝑀= 1) 328 = (𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛′p1(𝑟 ⃗ 2) + 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛′p0(𝑟 ⃗ 2)) /2 + (𝜓𝑛p0(𝑟 ⃗ 2)𝜓𝑛′p1(𝑟 ⃗ 1) + 𝜓𝑛p1(𝑟 ⃗ 2)𝜓𝑛′p0(𝑟 ⃗ 1)) /2 3. Ψs(𝐿= 2, 𝑀= 0) = {[2𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛′p0(𝑟 ⃗ 2) + (𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛′p−1(𝑟 ⃗ 2) + 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛′p1(𝑟 ⃗ 2))] + [2𝜓𝑛p0(𝑟 ⃗ 2)𝜓𝑛′p0(𝑟 ⃗ 1) + (𝜓𝑛p1(𝑟 ⃗ 2)𝜓𝑛′p−1(𝑟 ⃗ 1) + 𝜓𝑛p−1(𝑟 ⃗ 2)𝜓𝑛′p1(𝑟 ⃗ 1))]} /2√3 4. Ψs(𝐿= 2, 𝑀= −1) = (𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛′p−1(𝑟 ⃗ 2) + 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛′p0(𝑟 ⃗ 2)) /2 + (𝜓𝑛p0(𝑟 ⃗ 2)𝜓𝑛′p−1(𝑟 ⃗ 1) + 𝜓𝑛p−1(𝑟 ⃗ 2)𝜓𝑛′p0(𝑟 ⃗ 1)) /2 5. Ψs(𝐿= 2, 𝑀= −2) = (𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛′p−1(𝑟 ⃗ 2) + 𝜓𝑛p−1(𝑟 ⃗ 2)𝜓𝑛′p−1(𝑟 ⃗ 1)) /√2 本身是承载SO(3)群的𝐿= 2不可约表示的态，它们同时又交换对称。而： 1. Ψa(𝐿= 2, 𝑀= 2) = (𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛′p1(𝑟 ⃗ 2) −𝜓𝑛p1(𝑟 ⃗ 2)𝜓𝑛′p1(𝑟 ⃗ 1)) /√2 2. Ψa(𝐿= 2, 𝑀= 1) = (𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛′p1(𝑟 ⃗ 2) + 𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛′p0(𝑟 ⃗ 2)) /2 −(𝜓𝑛p0(𝑟 ⃗ 2)𝜓𝑛′p1(𝑟 ⃗ 1) + 𝜓𝑛p1(𝑟 ⃗ 2)𝜓𝑛′p0(𝑟 ⃗ 1)) /2 3. Ψa(𝐿= 2, 𝑀= 0) = {[2𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛′p0(𝑟 ⃗ 2) + (𝜓𝑛p1(𝑟 ⃗ 1)𝜓𝑛′p−1(𝑟 ⃗ 2) + 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛′p1(𝑟 ⃗ 2))] −[2𝜓𝑛p0(𝑟 ⃗ 2)𝜓𝑛′p0(𝑟 ⃗ 1) + (𝜓𝑛p1(𝑟 ⃗ 2)𝜓𝑛′p−1(𝑟 ⃗ 1) + 𝜓𝑛p−1(𝑟 ⃗ 2)𝜓𝑛′p1(𝑟 ⃗ 1))]} /2√3 4. Ψa(𝐿= 2, 𝑀= −1) = (𝜓𝑛p0(𝑟 ⃗ 1)𝜓𝑛′p−1(𝑟 ⃗ 2) + 𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛′p0(𝑟 ⃗ 2)) /2 −(𝜓𝑛p0(𝑟 ⃗ 2)𝜓𝑛′p−1(𝑟 ⃗ 1) + 𝜓𝑛p−1(𝑟 ⃗ 2)𝜓𝑛′p0(𝑟 ⃗ 1)) /2 5. Ψa(𝐿= 2, 𝑀= −2) = (𝜓𝑛p−1(𝑟 ⃗ 1)𝜓𝑛′p−1(𝑟 ⃗ 2) −𝜓𝑛p−1(𝑟 ⃗ 2)𝜓𝑛′p−1(𝑟 ⃗ 1)) /√2 在承载SO(3)群的𝐿= 2不可约表示的态的同时， 又承载置换群的一维非恒等交换 反对称表示。这样的话，每个𝐿对应的轨道就可以即和自旋单重态结合，又和自 旋三重态结合了。𝐿= 1、𝐿= 0的情况类似。 总结一下， 在保证总体波函数的交换反对称的前提下， 我们就可以知道允许 下述两体态的存在： 构型（表示空间维度） 态 不可约表示 最终允许的态 αα（一维） 𝑆z = 1 Γ 1 s ββ（一维） 𝑆z = −1 Γ 1 s αβ（二维） 𝑆z = 0 Γ 1 s ⊕Γ 1 a s2 𝐿= 0 Γ 1 s S 1 1s2s 𝐿= 0 Γ 1 s⨁Γ 1 a S 1 、S 3 sp 𝐿= 1 Γ 1 s⨁Γ 1 a P 1 、P 3 p2(𝑛= 𝑛′) 𝐿= 0 Γ 1 s S 1 p2(𝑛= 𝑛′) 𝐿= 1 Γ 1 a P 3 p2(𝑛= 𝑛′) 𝐿= 2 Γ 1 s D 1 p2(𝑛≠𝑛′) 𝐿= 0 Γ 1 s⨁Γ 1 a S 1 、S 3 p2(𝑛≠𝑛′) 𝐿= 1 Γ 1 s⨁Γ 1 a P 1 、P 3 p2(𝑛≠𝑛′) 𝐿= 2 Γ 1 s⨁Γ 1 a D 1 、D 3 表6.3 双电子体系交换反对称态 有些教材会在讲解两电子系统的时候说情况很简单， 跳过很多步骤。 根据这个分 析，我们应该知道即使是对这样一个简单的两体系统，将对称性分析清楚，其实 并不简单。 330 三电子系统的情况类似， 但会更复杂。 它的置换对称群是S3， 特征标表如下： 1{E} 3{A、B、C} 2{D、E} S3 (1)(2)(3) (1，2)(3)、 (2，3)(1) 、 (3，1)(2)、 (1，2，3)、 (1，3，2) Γ 1 s 1 1 1 Γ 1 a 1 −1 1 Γ 2 2 0 −1 Γ permut(𝜓1𝜓1𝜓1) 1 1 1 ⇒Γ 1 s Γ permut(𝜓1𝜓1𝜓3) 3 1 0 ⇒Γ 1 s ⊕Γ 2 Γ permut(𝜓1𝜓2𝜓3) 6 0 0 ⇒Γ 1 s ⊕Γ 1 a ⊕2Γ 2 表6.4 三阶置换群特征标表 这里前两行将D3群与S3群做了一个同构分析。3-5 行是正常的特征标表内容。最 后三行的内容与表6.2 类似， 就是如果三个电子占据的态是𝜓1𝜓1𝜓1 （或𝜓1𝜓1𝜓3、 𝜓1𝜓2𝜓3）的时候，在由𝜓1𝜓1𝜓1（或𝜓1𝜓1𝜓3、𝜓1𝜓2𝜓3）形成的线性空间中，三 阶置换群S3 的表示特征标Γ permut(𝜓1𝜓1𝜓1) （或Γ permut(𝜓1𝜓1𝜓3) 、 Γ permut(𝜓1𝜓2𝜓3)） 。 最后三行表示的特征标如何确定呢？对由𝜓1𝜓1𝜓1确定的表示，很显然线性 空间是一维的。S3群中任意一个元素作用到这个基上，都是这个向量本身，所以 特征标都是1。这个表示也是一维恒等对称表示Γ 1 s。 由𝜓1𝜓1𝜓3 可以形成一个三维线性空间，基为𝜓1(𝑥 ⃗1)𝜓1(𝑥 ⃗2)𝜓3(𝑥 ⃗3) 、 𝜓1(𝑥 ⃗2)𝜓1(𝑥 ⃗3)𝜓3(𝑥 ⃗1)、 𝜓1(𝑥 ⃗1)𝜓1(𝑥 ⃗3)𝜓3(𝑥 ⃗2)。 当S3群中元素作用到这三个基上时， (1)(2)(3) 的表示矩阵是三维单位矩阵，特征标是3 。(1，2)(3) 使 𝜓1(𝑥 ⃗1)𝜓1(𝑥 ⃗2)𝜓3(𝑥 ⃗3)变成其本身， 𝜓1(𝑥 ⃗2)𝜓1(𝑥 ⃗3)𝜓3(𝑥 ⃗1)变成𝜓1(𝑥 ⃗1)𝜓1(𝑥 ⃗3)𝜓3(𝑥 ⃗2)， 𝜓1(𝑥 ⃗1)𝜓1(𝑥 ⃗3)𝜓3(𝑥 ⃗2)变成𝜓1(𝑥 ⃗2)𝜓1(𝑥 ⃗3)𝜓3(𝑥 ⃗1)，矩阵为： ( 1 0 0 0 0 1 0 1 0 ) 特征标为1 ；(1，2，3) 把𝜓1(𝑥 ⃗1)𝜓1(𝑥 ⃗2)𝜓3(𝑥 ⃗3) 变成𝜓1(𝑥 ⃗2)𝜓1(𝑥 ⃗3)𝜓3(𝑥 ⃗1) ， 𝜓1(𝑥 ⃗2)𝜓1(𝑥 ⃗3)𝜓3(𝑥 ⃗1) 变成𝜓1(𝑥 ⃗3)𝜓1(𝑥 ⃗1)𝜓3(𝑥 ⃗2) ，𝜓1(𝑥 ⃗1)𝜓1(𝑥 ⃗3)𝜓3(𝑥 ⃗2) 变成 𝜓1(𝑥 ⃗1)𝜓1(𝑥 ⃗2)𝜓3(𝑥 ⃗3)，表示矩阵为： ( 0 0 1 1 0 0 0 1 0 ) 特征标为0。显然这是一个可约表示，它可以约化为：Γ 1 s ⊕Γ 2。 基于𝜓1𝜓2𝜓3 的置换群表示空间是六维的，基于𝜓1(𝑥 ⃗1)𝜓2(𝑥 ⃗2)𝜓3(𝑥 ⃗3)、 𝜓1(𝑥 ⃗1)𝜓2(𝑥 ⃗3)𝜓3(𝑥 ⃗2) 、 𝜓1(𝑥 ⃗2)𝜓2(𝑥 ⃗1)𝜓3(𝑥 ⃗3) 、 𝜓1(𝑥 ⃗2)𝜓2(𝑥 ⃗3)𝜓3(𝑥 ⃗1) 、 𝜓1(𝑥 ⃗3)𝜓2(𝑥 ⃗1)𝜓3(𝑥 ⃗2)、𝜓1(𝑥 ⃗3)𝜓2(𝑥 ⃗2)𝜓3(𝑥 ⃗1)做表示。很容易得到其特征标为6、 0、0。这个表示也可约，它可以约化为：Γ 1 s ⊕Γ 1 a ⊕2Γ 2。 这些是特征标表给我们的信息， 现在来看波函数。 先看自旋部分， 三个电子， 每个电子自旋两个状态，一共是8 个状态。其中ααα、βββ各占一个，ααβ、αββ 各占三个。根据特征标表，ααα、βββ给出的两个状态都承载交换群的一维恒等 表示。ααβ给出的三个承载Γ 1 s ⊕Γ 2。其中，承载Γ 1 s表示的基是： (α1α2β3 + α1β2α3 + β1α2α3)/√3 它承载一维恒等表示。另外两个基： (α1α2β3 + e𝑖2𝜋/3α1β2α3 + e𝑖4𝜋/3β1α2α3)/√3 (α1α2β3 + e𝑖4𝜋/3α1β2α3 + e𝑖2𝜋/3β1α2α3)/√3 332 给出的三类的特征标是2、0、−1，对应不可约表示Γ 2。 αββ的情况类似，也是三个维度，其中： (α1β2β3 + β1α2β3 + β1β2α3)/√3 承载一维恒等表示。 (α1β2β3 + e𝑖2𝜋/3β1α2β3 + e𝑖4𝜋/3β1β2α3)/√3 (α1β2β3 + e𝑖4𝜋/3β1α2β3 + e𝑖2𝜋/3β1β2α3)/√3 承载Γ 2。 总结一下，就是自旋部分8 个多体态，其中α1α2α3、(α1α2β3 + α1β2α3 + β1α2α3)/√3、 (α1β2β3 + β1α2β3 + β1β2α3)/√3、 β1β2β3四个维度对应一维置换恒 等表示，它们四个刚好也形成𝑆= 3/2 对应的自旋四重态。(α1α2β3 + e𝑖2𝜋/3α1β2α3 + e𝑖4𝜋/3β1α2α3)/√3 、(α1α2β3 + e𝑖4𝜋/3α1β2α3 + e𝑖2𝜋/3β1α2α3)/√3 都是𝑆z = 1/2态，承载Γ 2表示。(α1β2β3 + e𝑖2𝜋/3β1α2β3 + e𝑖4𝜋/3β1β2α3)/√3、 (α1β2β3 + e𝑖4𝜋/3β1α2β3 + e𝑖2𝜋/3β1β2α3)/√3都是𝑆z = −1/2态， 承载Γ 2。 总的来说， 自旋部分的8 个维度可分解为：4Γ 1 s⨁2Γ 2。 再看轨道部分。如果三个电子都处在同一个s 轨道，比如1s。这样，它们轨 道部分的多体波函数根据表6.4 倒数第三行，承载置换群的Γ 1 s表示。而Γ 1 s ⊗ (4Γ 1 s⨁2Γ 2)怎么都不可能有Γ 1 a的成分，所以这种情况不可能发生。这个分析可以 说是Pauli 不相容原理的一种群论表达。 当两个电子处在同一个s 轨道 （比如1s） ， 另一个电子处在另一个s 轨道 （比 如2s）时，根据表6.4 倒数第二行，它们形成的表示空间承载Γ 1 s⨁Γ 2。自旋部分 是4Γ 1 s⨁2Γ 2。Γ 2与Γ 2直积，可分解为Γ 1 s⨁Γ 1 a⨁Γ 2，有Γ 1 a的情况。因此这种构型可以 被置换对称性允许。 我们需要注意的是虽然允许， 置换对称性在这里已经帮助我 们排除了很多构型。自旋部分维度为8，轨道部分维度为3，严格意义上三体波 函数有24 个维度。 这里， 因为2Γ 2 ⊗Γ 2包含2Γ 1 a。 这24 个构型空间的维度只有两 个是可以形成合格的三体波函数的。它们都对应𝑆= 1/2的自旋双重态。这个态 用 𝐿𝑀 𝑆𝑧 2𝑆+1 来标识的话，形式就是S 2 。 三电子体系的其它构型， 分析类似。 前面几个轨道置换对称允许的构型如下： 构型 态 不可约表示 最终允许的态 |1 2 , ± 1 2⟩ 自旋双重态S = 1 2 Γ 2 |3 2 , ± 3 2⟩ 自旋四重态S = 3 2 Γ 1 s s3 𝐿= 0 Γ 1 s 无 1s22s 𝐿= 0 Γ 1 s⨁Γ 2 S 2 s2p 𝐿= 1 Γ 1 s⨁Γ 2 P 2 sp2 𝐿= 0 Γ 1 s⨁Γ 2 S 2 sp2 𝐿= 1 Γ 1 a⨁Γ 2 P 2 、P 4 sp2 𝐿= 2 Γ 1 s⨁Γ 2 D 2 表6.5 三电子体系交换反对称态 更多电子体系，也是用同样的方法分析，需要用到的置换群特征标表。更多情况 请参考Dresselhaus 那本教材的第17 章。我们这里的说明相对于她们的讲义更详 细，但覆盖面小很多。 334 参考文献 E. Noether "Invariante Variationsprobleme". Nachrichten von der Gesellschaft der Wissenschaften zu Göttingen, Mathematisch-Physikalische Klasse 1918, 235–257 (1918) ( E. P. Wigner, Gruppentheorie und ihre Anwendung auf die Quanten mechanik der Atomspektren (in German). Braunschweig, Germany: Friedrich Vieweg und Sohn. page 251–254 (1931) E. P. Wigner, E. P. Group Theory and its Application to the Quantum Mechanics of Atomic Spectra. translation from German by J. J. Griffin. New York: Academic Press. page 233–236, ISBN 978-0-1275-0550-3 (1959) H. Weyl, The Theory of Groups and Quantum Mechanics, Dover Publications, New York (1950) (Note: In 1929 Hermann Weyl's The theory of groups and quantum mechanics was published in German. A second German edition of the work was published by Hirzel in Leipzig in 1931. An English translation of the book was published by Dover Publications, New York, in 1950) M. S. Dresselhaus, G. Dresselhaus, A. 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Hou, Chemical mapping of a single molecule by plasmon-enhanced Raman scattering, Nature 498, 82 (2013) Y. Zhang, Y. Luo, Y. Zhang, Y. J. Yu, Y. M. Kuang, L. Zhang, Q. S. Meng, Y. Luo, J. L. Yang, Z. C. Dong, and J. G. Hou, Visualizing coherent intermolecular dipole-dipole coupling in real space, Nature 531, 623 (2016) G. F. Koster, J. O. Dimmock, R. G. Wheeler, and H. Statz, Properties of the thirty-two point groups, MIT press, Cambridge, 1964 S. C. Miller and W. H. Love, Tables of irreducible representations of space groups and co-representations of magnetic space groups, Pruett press, Denver, 1967 336 Bilbao Crystallographic Server, University of the Basque Country, Bilbao, Basque Country, Spain Oleksandr Ney, Magnetism and dynamics of oxide interfaces (electronic theory), Dissertation (i.e. Ph.D Thesis), Martin-Luther-University Halle-Wittenberg, 2003, 附录A 晶体点群的特征标表 这部分我们先按晶系的分类给出32 种晶体点群的特征标表。需要提前说明 的是在不等价不可约表示的标识中， 我们采用原子分子物理领域常用的标识规则， 用A、B 表示一维不可约表示，E 表示二维，T 表示三维。对于有些点群，比如 C3、C4、C5、C6、C3h、C4h、C5h、C6h、S4、S6、T等，它们会存在两个一维表 示互为共轭不等价的情况。也就是说，仅仅依据这些点群的对称性以及Burnside 定理，这两个不可约表示所对应的本征态不简并（对称性不要求它们简并） 。但 是由于这两个一维表示相互共轭又不等价，当系统存在时间反演对称性的时候， 时间反演对称性要求它们相互简并。 也就是说这种简并并不是点群对称性要求的， 是额外的时间反演对称性要求的。 由于时间反演对称性在很多实际非磁的量子体 系中存在， 在这里的特征标表中 （包括文献上可以找到的绝大部分特征标表中） ， 人们都习惯上把它们放在一起，用二维表示E 来表示。稍微详细一些的讨论见 4.8 节结尾部分。这里展示的特征标表主要参考Dresselhaus 的教材以及下面这个 网站： 三斜系（S2、C1）： S2(1 ̅) E I x2、y2、z2、xy、xz、yz Rx、Ry、Rz Ag 1 1 x、y、z Au 1 −1 C1(1) E A 1 338 单斜系（C2h、C2、C1h）： C2h(2/m) E C2 σh I x2、y2、z2、xy Rz Ag 1 1 1 1 Z Au 1 1 −1 −1 yz、xz Rx、Ry Bg 1 −1 −1 1 x、y Bu 1 −1 1 −1 C2(2) E C2 x2、y2、z2、xy Rz、z A 1 1 xz、yz x、y、Rx、Ry B 1 −1 C1h(m) E σh x2、y2、z2、xy Rz、x、y A′ 1 1 xz、yz Rx、Ry、z A′′ 1 −1 正交系（D2h、D2、C2v）： D2h(2/m2/m 2/m) = D2⨂S2 E C2z C2y C2x I IC2z IC2y IC2x x2、y2、z2 Ag 1 1 1 1 1 1 1 1 xy Rx B1g 1 1 −1 −1 1 1 −1 −1 xz Ry B2g 1 −1 1 −1 1 −1 1 −1 yz Rz B3g 1 −1 −1 1 1 −1 −1 1 xyz Au 1 1 1 1 −1 −1 −1 −1 z3、z(x2 −y2) x B1u 1 1 −1 −1 −1 −1 1 1 yz2、y(3x2 −y2) y B2u 1 −1 1 −1 −1 1 −1 1 xz2、x(x2 −3y2) z B3u 1 −1 −1 1 −1 1 1 −1 D2(222) E C2z C2y C2x x2、y2、z2 A1 1 1 1 1 xy Rz、z B1 1 1 −1 −1 xz Rx、x B2 1 −1 1 −1 yz Ry、y B3 1 −1 −1 1 C2v(2mm) E C2 σv σv′ x2、y2、z2 Z A1 1 1 1 1 xy Rz A2 1 1 −1 −1 xz Rx、x B1 1 −1 1 −1 yz Ry、y B2 1 −1 −1 1 四方系（ D4h、C4、S4、D4、C4v、C4h、D2d）： D4h(4/mmm) = D4⨂S2 E 2C4 1 C4 2 2C2 (1) 2C2 (2) I 2IC4 1 IC4 2 2IC2 (1) 2IC2 (2) x2 + y2、z2 A1g 1 1 1 1 1 1 1 1 1 1 Rz A2g 1 1 1 −1 −1 1 1 1 −1 −1 x2 −y2 B1g 1 −1 1 1 −1 1 −1 1 1 −1 xy B2g 1 −1 1 −1 1 1 −1 1 −1 1 (xz、yz) (Rx、 Ry ) Eg 2 0 −2 0 0 2 0 −2 0 0 340 A1u 1 1 1 1 1 −1 −1 −1 −1 −1 z3 Z A2u 1 1 1 −1 −1 −1 −1 −1 1 1 Xyz B1u 1 −1 1 1 −1 −1 1 −1 −1 1 z(x2 −y2) B2u 1 −1 1 −1 1 −1 1 −1 1 −1 (xz2、yz2) (x(x2 −3y2)、 y(3x2 −y2) ) (x、y) Eu 2 0 −2 0 0 −2 0 2 0 0 C4(4) E C4 C4 2 C4 3 x2 + y2、z2 Rz、z A 1 1 1 1 x2 −y2、xy B 1 −1 1 −1 (xz、yz) (xz2、yz2) (x、y) (Rx、Ry) E 1 i −1 −i 1 −i −1 i （ 基组， 以 (x、y)为例， 代表当系统具备时间反演对称性时， 后两个一维不可 约表示简并，对应的二维基组。下面的特征标表类似处理。） S4(4 ̅) E C4 2 IC4 1 IC4 3 x2 + y2、z2 Rz A 1 1 1 1 z B 1 1 −1 −1 (xz、yz) (xz2、yz2) (x、y) (Rx、Ry) E 1 −1 i −i 1 −1 −i i D4(422) E C4 2 2C4 1 2C2 (1) 2C2 (2) x2 + y2、z2 A1 1 1 1 1 1 Rz、z A2 1 1 1 −1 −1 x2 −y2 B1 1 1 −1 1 −1 Xy B2 1 1 −1 −1 1 (xz、yz) (x、y) (Rx、Ry) E 2 −2 0 0 0 C4v(4mm) E C4 2 2C4 1 2σv 2σd x2 + y2、z2 z A1 1 1 1 1 1 Rz A2 1 1 1 −1 −1 x2 −y2 B1 1 1 −1 1 −1 xy B2 1 1 −1 −1 1 (xz、yz) (x、y) (Rx、Ry) E 2 −2 0 0 0 342 C4h(4/m) = C4⨂S2 E C4 1 C4 2 C4 3 I IC4 1 σh IC4 3 x2 + y2、z2 Rz Ag 1 1 1 1 1 1 1 1 x2 −y2、xy Bg 1 −1 1 −1 1 −1 1 −1 (xz、yz) (Ry、Rz) Eg 1 I −1 −i 1 i −1 −i 1 −i −1 i 1 −i −1 i z3 z Au 1 1 1 1 −1 −1 −1 −1 xyz、z(x2 −y2) Bu 1 −1 1 −1 −1 1 −1 1 (xz2、yz2) (x、y) Eu 1 I −1 −i −1 −i 1 i 1 −i −1 i −1 i 1 −i D2d(4 ̅2m) E C2 2S4 2C2 (1) 2σd x2 + y2、z2 A1 1 1 1 1 1 Rz A2 1 1 1 −1 −1 x2 −y2 B1 1 1 −1 1 −1 xy z B2 1 1 −1 −1 1 (xz、yz) (x、y) (Rx、Ry) E 2 −2 0 0 0 三方系（ D3d、S6、C3、C3v、D3）： D3d(3 ̅m) E 2C3 3C2 I 2IC3 3IC2 x2 + y2、z2 A1g 1 1 1 1 1 1 Rz A2g 1 1 −1 1 1 −1 (xz、yz) (x2 −y2、xy) (Rx、Ry) Eg 2 −1 0 2 −1 0 A1u 1 1 1 −1 −1 −1 z A2u 1 1 −1 −1 −1 1 (x、y) Eu 2 −1 0 −2 1 0 S6(3 ̅) E C3 C3 2 I IC3 IC3 2 x2 + y2、z2 Rz Ag 1 1 1 1 1 1 (x2 −y2、xy) (xz、yz) (Rx、 Ry) Eg 1 ε ε∗ 1 ε ε∗ 1 ε∗ ε 1 ε∗ ε z3、x(x2 −3y2) Z Au 1 1 1 −1 −1 −1 z3 (xz、yz) (x、y) Eu 1 ε ε∗ −1 -ε -ε∗ 1 ε∗ ε −1 -ε∗ -ε （上表中ε = exp(2πi/3)） C3(3) E C3 C3 2 x2 + y2、z2 Rz、z A 1 1 1 (xz、yz) (x2 −y2、xy) (x、y) (Rx、Ry) E 1 ε ε∗ 1 ε∗ ε 344 C3v(3m) E 2C3 3σv x2 + y2、z2 z A1 1 1 1 Rz A2 1 1 −1 (x2 −y2、xy) (xz、yz) (x、y) (Rx、Ry) E 2 −1 0 D3(32) E 2C3 3C2 x2 + y2、z2 A1 1 1 1 Rz、z A2 1 1 −1 (x2 −y2、xy) (xz、yz) (x、y) (Rx、Ry) E 2 −1 0 六角系（ D6h、C6、C3h、C6h、C6v、D6、D3h）： D6h(6 m 2 m 2 m) = D6⨂S2 E C2 2C3 2C6 3C2 (1) 3C2 (2) I IC2 2IC3 2IC6 3IC2 (1) 3IC2 (2) x2 + y2、z2 A1g 1 1 1 1 1 1 1 1 1 1 1 1 Rz A2g 1 1 1 1 −1 −1 1 1 1 1 −1 −1 B1g 1 −1 1 −1 1 −1 1 −1 1 −1 1 −1 B2g 1 −1 1 −1 −1 1 1 −1 1 −1 −1 1 (xz、yz) (Rx、 Ry ) E1g 2 −2 −1 1 0 0 2 −2 −1 1 0 0 (x2 −y2、xy) E2g 2 2 −1 −1 0 0 2 2 −1 −1 0 0 A1u 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 z A2u 1 1 1 1 −1 −1 −1 −1 −1 −1 1 1 B1u 1 −1 1 −1 1 −1 −1 1 −1 1 −1 1 B2u 1 −1 1 −1 −1 1 −1 1 −1 1 1 −1 (x、y) E1u 2 −2 −1 1 0 0 −2 2 1 −1 0 0 E2u 2 2 −1 −1 0 0 −2 −2 1 1 0 0 C6(6) E C6 C3 C2 C3 2 C6 5 x2 + y2、z2 Rz、z A 1 1 1 1 1 1 B 1 −1 1 −1 1 −1 (xz、yz) (x、y) (Rx、Ry) E′ 1 ε ε2 ε3 ε4 ε5 1 ε5 ε4 ε3 ε2 ε (x2 −y2、xy) E′′ 1 ε2 ε4 1 ε2 ε4 1 ε4 ε2 1 ε4 ε2 （上表中：ε = ei2π/6） 346 C3h(S3) E C3 C3 2 σh S3 σhC3 2 x2 + y2、z2 Rz、z A 1 1 1 1 1 1 B 1 1 1 −1 −1 −1 (x2 −y2、xy) (x、y) E′ 1 ε ε2 1 ε ε2 1 ε2 ε 1 ε2 ε (xz、yz) (Rx、Ry) E′′ 1 ε ε2 −1 −ε −ε2 1 ε2 ε 1 −ε2 −ε （上表中：ε = ei2π/3） C6h(6 ̅) = C6⨂S2 E C6 C3 C2 C3 2 C6 5 I IC6 IC3 IC2 IC3 2 IC6 5 x2 + y2、z2 Ag 1 1 1 1 1 1 1 1 1 1 1 1 Bg 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 (xz、yz) (Rx，Ry) E1g 1 ε ε2 ε3 ε4 ε5 1 ε ε2 ε3 ε4 ε5 1 ε5 ε4 ε3 ε2 ε 1 ε5 ε4 ε3 ε2 ε (x2 −y2、xy) E2g 1 ε2 ε4 1 ε2 ε4 1 ε2 ε4 1 ε2 ε4 1 ε4 ε2 1 ε4 ε2 1 ε4 ε2 1 ε4 ε2 Z Au 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 Bu 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 (x、y) E1u 1 ε ε2 ε3 ε4 ε5 -1 -ε -ε2 -ε3 -ε4 -ε5 1 ε5 ε4 ε3 ε2 ε -1 -ε5 -ε4 -ε3 -ε2 -ε (x、y) E2u 1 ε2 ε4 1 ε2 ε4 -1 -ε2 -ε4 -1 -ε2 -ε4 1 ε4 ε2 1 ε4 ε2 -1 -ε4 -ε2 -1 -ε4 -ε2 （上表中：ε = ei2π/6） C6v(6mm) E C2 2C3 2C6 3σd 3σv x2 + y2、z2 z A1 1 1 1 1 1 1 Rz A2 1 1 1 1 -1 -1 B1 1 -1 1 -1 -1 1 B2 1 -1 1 -1 1 -1 (xz、yz) (x、y) (Rx、Ry) E1 2 -2 -1 1 0 0 (x2 −y2、xy) E2 2 2 -1 -1 0 0 D6(622) E C2 2C3 2C6 3C2 (1) 3C2 (2) x2 + y2、z2 A1 1 1 1 1 1 1 Rz、z A2 1 1 1 1 -1 -1 B1 1 -1 1 -1 1 -1 B2 1 -1 1 -1 -1 1 (xz、yz) (x、y) (Rx、Ry) E1 2 -2 -1 1 0 0 (x2 −y2、xy) E2 2 2 -1 -1 0 0 348 D3h(6 ̅m2) = D3 ⊗σh E σh 2C3 2S3 3C2 ′ 3σv x2 + y2、z2 A1 1 1 1 1 1 1 Rz A2 1 1 1 1 -1 -1 A1 ′ 1 -1 1 -1 1 -1 z A2 ′ 1 -1 1 -1 -1 1 (x2 −y2、xy) (x、y) E1 2 2 -1 -1 0 0 (xz、yz) (Rx、Ry) E2 ′ 2 -2 -1 1 0 0 立方系（ Oh、T、O、Th、Td）： Oh(4/m3 ̅2/m) = O⨂S2 E 3C4 2 6C4 6C2 ′ 8C3 I 3IC4 2 6IC4 6IC2 ′ 8IC3 x2 + y2 + z2 A1g 1 1 1 1 1 1 1 1 1 1 A2g 1 1 -1 -1 1 1 1 -1 -1 1 (2z2 −x2 −y2、x2 −y2) Eg 2 2 0 0 -1 2 2 0 0 -1 (Rx、Ry、 Rz) T 1g 3 -1 1 -1 0 3 -1 1 -1 0 (x、y、z) T2g 3 -1 -1 1 0 3 -1 -1 1 0 A1u 1 1 1 1 1 -1 -1 -1 -1 -1 z A2u 1 1 -1 -1 1 -1 -1 1 1 -1 Eu 2 2 0 0 -1 -2 -2 0 0 1 (x、y、z) T 1u 3 -1 1 -1 0 -3 1 -1 1 0 T2u 3 -1 -1 1 0 -3 1 1 -1 0 T(23) E 3C2 4C3 4C3 ′ x2 + y2 + z2 A 1 1 1 1 (x2 −y2、2z2 −x2 −y2) E 1 1 ε ε2 1 1 ε2 Ε (yz、zx、xy) (Rx、Ry、Rz) (x、y、z) T 3 -1 0 0 （上表中：ε = ei2π/3） 350 O(432) E 8C3 3C4 2 6C2 ′ 6C4 x2 + y2 + z2 A1 1 1 1 1 1 A2 1 1 1 -1 -1 (x2 −y2、2z2 −x2 −y2) E 2 -1 2 0 0 (Rx、Ry、Rz) (x、y、z) T 1 3 0 -1 -1 1 (xy、yz、zx) T2 3 0 -1 1 -1 Th(2/m3 ̅) = T⨂S2 E 3C2 4C3 4C3 ′ I 3IC2 4IC3 4IC3 ′ x2 + y2 + z2 Ag 1 1 1 1 1 1 1 1 (x2 −y2、2z2 −x2 −y2) Eg 1 1 ε ε2 1 1 ε ε2 1 1 ε2 ε 1 1 ε2 ε (yz、zx、xy) (Rx、Ry、 Rz) Tg 3 -1 0 0 3 -1 0 0 Au 1 1 1 1 -1 -1 -1 -1 Eu 1 1 ε ε2 -1 -1 -ε -ε2 1 1 ε2 ε -1 -1 -ε2 -ε (x、y、z) Tu 3 -1 0 0 -3 1 0 0 （上表中：ε = ei2π/3） Td(4 ̅3m) E 8C3 3C2 6σd 6S4 x2 + y2、z2 A1 1 1 1 1 1 A2 1 1 1 -1 -1 (x2 −y2、2z2 −x2 −y2) E 2 -1 2 0 0 (yz、zx、xy) (Rx、Ry、Rz) T 1 3 0 -1 -1 1 (x、y、z) T2 3 0 -1 0 -1 352 附录B 空间群情况说明 本附录是对第三章基于下表讨论简单空间群的情况的详细介绍， 以及230 种 空间群所属晶格系统情况的简单说明29。 晶系 点群 布拉菲格子 晶格系统 简单空间群 三斜 2 (C1、S2) 1 三斜 2 单斜 3 (C2、C1h、C2h) 2 、 单斜 6 正交 3 (D2、C2v、D2h) 4 、 、 、 正交 12 四方 7(C4、S4、C4h、 D4、C4v、D4h D2d) 2 、 四方 14 三方 Trigonal 5 (C3、D3、D3d、 S6、C3v) 1 菱方 Rhombohedra l 5 1 六角 Hexagonal 5 7 六角 Hexagonal 7(C6、C3h、C6h、 D6、D3h、C6v、 D6h) 29注意，这里用的名词是晶格系统，不是晶系，因为讨论的是空间群 立方 5 (T、Td、O、 Th、Oh) 3 、 、 立方 15 共7 种 共32 种 共14 种 共7 种 共66 种 什么是针对简单群的详细介绍呢？就是上面的算法给出66 种简单群，实际 上是73 种，我们把多余的七种抠出来。 73 种简单空间群， 三斜晶格系统含2 种、 单斜晶格系统含6 种、 正交晶格系 统含13 种、四方晶格系统含16 种、菱方晶格系统含5 种、六角晶格系统含16 种、立方晶格系统含15 种。对照上图，我们知道相对于简单组合多出的几个分 别是： 正交晶格系统多出1 种、 四方晶格系统多出2 种、 六角晶格系统多出4 种。 其中，正交晶格系统多出的一种是C2v点群与面心晶格组合的时候，组合不 知一种， 而是三种， 但C2v的对称性使得其中两个等价， 最终可以出现两种情况。 这样正交晶格系统中简单空间群的总数就是3 乘4 加1，共13 种。 四方晶格系统多出的两种是D2d与简单、体心晶格组合的时候，垂直方向的 反射面（连带平分其的二次轴）选取也各有两种情况。这样总数正交晶格系统中 简单空间群的个数就是7 乘2 加2，共16 种。 六角晶格系统的情况复杂。 简单组合的时候， 三方晶系贡献5 种简单空间群， 六角晶系贡献7 种简单空间群，共12 种。多出的四种分别是：三方晶系中的D3 依据水平方向2 次轴的选取多贡献1 种，C3v依据垂直方向发射面的选取多贡献 1 种， 三方晶系中的D3d依据垂直方向的反射面 （连带平分其的二次轴） 选取多贡 献1 种，以及六角晶系中的D3h依据其母群D6群的二次轴的选取多贡献一种。 理解到这里， 晶系是点群概念、 晶格系统是空间群概念这句话笔者想表达的 354 意思就基本清楚了。230 种空间群要想推出来，笔者想都没有想过。熊夫利他们 确实太厉害了！ 从实用的角度， 大家知道所有详细内容都在Bilbao 的那个服务器 上，并会用就可以了。 附录C 晶体点群的双群的特征标表 本附录很大程度上文献的全部以及文献（也就是Dresselhaus 教材）的 附录D30。和附录A 类似，我们还是按晶系展开讨论。为了与文献一致，在 表示点群双群的不可约表示的时候， 我们不再像附录A 那样采用A、 B、 E、 T 这 些符号，而是采用Γ加下标的方式。 其它文献中的特征标表有的使用转动反射来标识点群群元， 有的使用转动反 演来标识点群群元。这里，为了和讲义主体中点群划分的讨论一致，我们多使用 转动反演。对于占用空间比较大，为了省空间，我们也会使用转动反射。它们之 间的关系是： S3 = IC6 −1、 S3 −1 = IC6、 S4 = IC4 −1、 S4 −1 = IC4、 S6 = IC3 −1、 S6 −1 = IC3。 1. 三斜系（S2 D、C1 D）： S2 D E I E ̅ E ̅I Γ 1 + 1 1 1 1 Γ 1 − 1 −1 1 −1 Γ 2 + 1 1 −1 −1 Γ 2 − 1 −1 −1 1 C1 D E E ̅ Γ 1 1 1 Γ 2 1 −1 30 其中文献的附录D 主要参考的是文献与文献。 356 2. 单斜系（C2h D 、C2 D、C1h D ）： C2h D E C2 σh I E ̅ E ̅C2 E ̅σh E ̅I Γ 1 + 1 1 1 1 1 1 1 1 Γ 1 − 1 1 −1 −1 1 1 −1 −1 Γ 2 + 1 −1 −1 1 1 −1 −1 1 Γ 2 − 1 −1 1 −1 1 −1 1 −1 Γ 3 + 1 𝑖 𝑖 1 −1 −𝑖 −𝑖 −1 Γ 3 − 1 𝑖 −𝑖 −1 −1 −𝑖 𝑖 1 Γ 4 + 1 −𝑖 −𝑖 1 −1 𝑖 𝑖 −1 Γ 4 − 1 −𝑖 𝑖 −1 −1 𝑖 −𝑖 1 C2 D E C2 E ̅ E ̅C2 C1h D E σh E ̅ E ̅σh Γ 1 1 1 1 1 Γ 2 1 −1 1 −1 Γ 3 1 𝑖 −1 −𝑖 Γ 4 1 −𝑖 −1 𝑖 3. 正交系（D2h D 、D2 D、C2v D ）： D2h D E E ̅ {C2z E ̅C2z} {C2y E ̅C2y} {C2x E ̅C2x} I E ̅I {IC2z IE ̅C2z} {IC2y IE ̅C2y} {IC2x IE ̅C2x} Γ 1 + 1 1 1 1 1 1 1 1 1 1 Γ 2 + 1 1 −1 1 −1 1 1 −1 1 −1 Γ 3 + 1 1 1 −1 −1 1 1 1 −1 −1 Γ 4 + 1 1 −1 −1 1 1 1 −1 −1 1 Γ 1 − 1 1 1 1 1 −1 −1 −1 −1 −1 Γ 2 − 1 1 −1 1 −1 −1 −1 1 −1 1 Γ 3 − 1 1 1 −1 −1 −1 −1 −1 1 1 Γ 4 − 1 1 −1 −1 1 −1 −1 1 1 −1 Γ 5 + 2 −2 0 0 0 2 −2 0 0 0 Γ 5 − 2 −2 0 0 0 −2 2 0 0 0 D2 D E E ̅ {C2z、E ̅C2z} {C2y、E ̅C2y} {C2x、E ̅C2x} C2v D E E ̅ {C2、E ̅C2} {σv、E ̅σv} {σv′、E ̅σv′} Γ 1 1 1 1 1 1 Γ 2 1 1 −1 1 −1 Γ 3 1 1 1 −1 −1 Γ 4 1 1 −1 −1 1 Γ 5 2 −2 0 0 0 4. 四方系（ D4h D 、C4 D、S4 D、D4 D、C4v D 、C4h D 、D2d D ）： D4h D E E ̅ 2C4 1 2E ̅C4 1 {C4 2 E ̅C4 2} {2C2 (1) 2E ̅C2 (1)} {2C2 (2) 2E ̅C2 (2)} I E ̅I 2IC4 3 2E ̅IC4 3 {IC4 2 IE ̅C4 2} {2IC2 (1) 2E ̅IC2 (1)} {2IC2 (2) 2E ̅IC2 (2)} Γ 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Γ 2 + 1 1 1 1 1 −1 −1 1 1 1 1 1 −1 −1 Γ 3 + 1 1 −1 −1 1 1 −1 1 1 −1 −1 1 1 −1 Γ 4 + 1 1 −1 −1 1 −1 1 1 1 −1 −1 1 −1 1 Γ 5 + 2 2 0 0 −2 0 0 2 2 0 0 −2 0 0 Γ 1 − 1 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 Γ 2 − 1 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 1 1 Γ 3 − 1 1 −1 −1 1 1 −1 −1 −1 1 1 −1 −1 1 Γ 4 − 1 1 −1 −1 1 −1 1 −1 −1 1 1 −1 1 −1 Γ 5 − 2 2 0 0 −2 0 0 −2 −2 0 0 2 0 0 Γ 6 + 2 −2 √2 −√2 0 0 0 2 −2 √2 −√2 0 0 0 Γ 7 + 2 −2 −√2 √2 0 0 0 2 −2 −√2 √2 0 0 0 Γ 6 − 2 −2 √2 −√2 0 0 0 −2 2 −√2 √2 0 0 0 Γ 7 − 2 −2 −√2 √2 0 0 0 −2 2 √2 −√2 0 0 0 358 C4 D E E ̅ C4 E ̅C4 C4 2 E ̅C4 2 C4 3 E ̅C4 3 S4 D E E ̅ IC4 E ̅IC4 C4 2 E ̅C4 2 IC4 3 E ̅IC4 3 Γ 1 1 1 1 1 1 1 1 1 Γ 2 1 1 −1 −1 1 1 −1 −1 Γ 3 1 1 I I −1 −1 −i −i Γ 4 1 1 −i −i −1 −1 i i Γ 5 1 −1 ω −ω i −i −ω3 ω Γ 6 1 −1 −ω3 ω3 −i i ω −ω3 Γ 7 1 −1 −ω ω i −i ω3 −ω Γ 8 1 −1 ω3 −ω3 −i i −ω ω3 上表中ω = exp [πi/4]。 D4 D E E ̅ 2C4 2E ̅C4 {C4 2 E ̅C4 2} {2C2 (1) 2E ̅C2 (1)} {2C2 (2) 2E ̅C2 (2)} C4v D E E ̅ 2C4 2E ̅C4 {C4 2 E ̅C4 2} {2σv 2E ̅σv} {2σd 2E ̅σd} D2d D E E ̅ 2IC4 3 2E ̅IC4 3 {C4 2 E ̅C4 2} {2C2 (1) 2E ̅C2 (1)} {2σd 2E ̅σd} Γ 1 1 1 1 1 1 1 1 Γ 2 1 1 1 1 1 −1 −1 Γ 3 1 1 −1 −1 1 1 −1 Γ 4 1 1 −1 −1 1 −1 1 Γ 5 2 2 0 0 −2 0 0 Γ 6 2 −2 √2 −√2 0 0 0 Γ 7 2 −2 −√2 √2 0 0 0 C4h D E E ̅ C4 E ̅C4 C4 2 E ̅C4 2 C4 3 E ̅C4 3 I E ̅I IC4 E ̅IC4 σh E ̅σh IC4 3 E ̅IC4 3 Γ 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Γ 2 + 1 1 −1 −1 1 1 −1 −1 1 1 −1 −1 1 1 −1 −1 Γ 3 + 1 1 i i −1 −1 −i −i 1 1 i i −1 −1 −i −i Γ 4 + 1 1 −i −i −1 −1 I i 1 1 −i −i −1 −1 I i Γ 1 − 1 1 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 −1 Γ 2 − 1 1 −1 −1 1 1 −1 −1 −1 −1 1 1 −1 −1 1 1 Γ 3 − 1 1 i i −1 −1 −i −i −1 −1 −i −i 1 1 i i Γ 4 − 1 1 −i −i −1 −1 i i −1 −1 I i 1 1 −i −i Γ 5 + 1 −1 ω −ω i −i −ω3 ω3 1 −1 Ω −ω I −i −ω3 ω3 Γ 6 + 1 −1 −ω3 ω3 −i i Ω −ω 1 −1 −ω3 ω3 −i i ω −ω Γ 7 + 1 −1 −ω ω i −i ω3 −ω3 1 −1 −ω ω I −i ω3 −ω3 Γ 8 + 1 −1 ω3 −ω3 −i i −ω ω 1 −1 ω3 −ω3 −i i −ω ω Γ 5 − 1 −1 ω −ω i −i −ω3 ω3 −1 1 −ω ω −i i ω3 −ω3 Γ 6 − 1 −1 −ω3 ω3 −i i ω −ω −1 1 ω3 −ω3 i −i −ω ω Γ 7 − 1 −1 −ω ω i −i ω3 −ω3 −1 1 ω −ω −i i −ω3 ω3 Γ 8 − 1 −1 ω3 −ω3 −i i −ω ω −1 1 −ω3 ω3 i −i ω −ω 上表中ω = exp [πi/4]。 5. 三方系（ D3d D 、S6 D、C3 D、C3v D 、D3 D）： D3d D E E ̅ 2C3 2E ̅C3 3C2 3E ̅C2 I E ̅I 2IC3 2 2E ̅IC3 2 3IC2 3E ̅IC2 Γ 1 + 1 1 1 1 1 1 1 1 1 1 1 1 Γ 2 + 1 1 1 1 −1 −1 1 1 1 1 −1 −1 Γ 3 + 2 2 −1 −1 0 0 2 2 −1 −1 0 0 Γ 1 − 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 Γ 2 − 1 1 1 1 −1 −1 −1 −1 −1 −1 1 1 Γ 3 − 2 2 −1 −1 0 0 −2 −2 1 1 0 0 Γ 4 + 2 −2 1 −1 0 0 2 −2 1 −1 0 0 Γ 5 + 1 −1 −1 1 i −i 1 −1 −1 1 i −i Γ 6 + 1 −1 −1 1 −i i 1 −1 −1 1 −i i Γ 4 − 2 −2 1 −1 0 0 −2 2 −1 1 0 0 Γ 5 − 1 −1 −1 1 i −i −1 1 1 −1 −i i Γ 6 − 1 −1 −1 1 −i i −1 1 1 −1 i −i 360 S6 D E E ̅ C3 E ̅C3 C3 2 E ̅C3 2 I E ̅I IC3 E ̅IC3 IC3 2 E ̅IC3 2 Γ 1 + 1 1 1 1 1 1 1 1 1 1 1 1 Γ 2 + 1 1 ω2 ω2 −ω −ω 1 1 ω2 ω2 −ω −ω Γ 3 + 1 1 −ω −ω ω2 ω2 1 1 −ω −ω ω2 ω2 Γ 1 − 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 Γ 2 − 1 1 ω2 ω2 −ω −ω −1 −1 −ω2 −ω2 ω ω Γ 3 − 1 1 −ω −ω ω2 ω2 −1 −1 ω ω −ω2 −ω2 Γ 4 + 1 −1 ω −ω −ω2 ω2 1 −1 ω −ω −ω2 ω2 Γ 5 + 1 −1 −ω2 ω2 ω −ω 1 −1 −ω2 ω2 ω −ω Γ 6 + 1 −1 −1 1 −1 1 1 −1 −1 1 −1 1 Γ 4 − 1 −1 ω −ω −ω2 ω2 −1 1 −ω ω ω2 −ω2 Γ 5 − 1 −1 −ω2 ω2 ω −ω −1 1 ω2 −ω2 −ω ω Γ 6 − 1 −1 −1 1 −1 1 −1 1 1 −1 1 −1 上表中ω = exp [πi/3]。 C3 D E E ̅ C3 E ̅C3 C3 2 E ̅C3 2 Γ 1 1 1 1 1 1 1 Γ 2 1 1 ω2 ω2 −ω −ω Γ 3 1 1 −ω −ω ω2 ω2 Γ 4 1 −1 ω −ω −ω2 ω2 Γ 5 1 −1 −ω2 ω2 ω −ω Γ 6 1 −1 −1 1 −1 1 上表中ω = exp [πi/3]。 D3 D E E ̅ 2C3 2E ̅C3 3C2 (1) 3E ̅C2 (1) C3v D E E ̅ 2C3 2E ̅C3 3𝜎v 3E ̅𝜎v Γ 1 1 1 1 1 1 1 Γ 2 1 1 1 1 −1 −1 Γ 3 2 2 −1 −1 0 0 Γ 4 2 −2 1 −1 0 0 Γ 5 1 −1 −1 1 i −i Γ 6 1 −1 −1 1 −i i 6. 六角系（ D6h D 、C6 D、C3h D 、C6h D 、C6v D 、D6 D、D3h D ）： C6h D E E ̅ C6 C6 ̅̅̅ C3 C3 ̅̅̅ C2 C2 ̅̅̅ C3 2 C3 2 ̅̅̅ C6 5 C6 5 ̅̅̅ I I̅ S3 −1 S3 −1 ̅ ̅ ̅ ̅ ̅ S6 −1 S6 −1 ̅ ̅ ̅ ̅ ̅ 𝜎h 𝜎h ̅̅̅ S6 S6 ̅ ̅ ̅ S3 S3 ̅ ̅ ̅ Γ 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Γ 2 + 1 1 −ω2 −ω2 ω4 ω4 1 1 −ω2 −ω2 ω4 ω4 1 1 −ω2 −ω2 ω4 ω4 1 1 −ω2 −ω2 ω4 ω4 Γ 3 + 1 1 ω4 ω4 −ω2 −ω2 1 1 ω4 ω4 −ω2 −ω2 1 1 ω4 ω4 −ω2 −ω2 1 1 ω4 ω4 −ω2 −ω2 Γ 4 + 1 1 −1 −1 1 1 −1 −1 1 1 −1 −1 1 1 −1 −1 1 1 −1 −1 1 1 −1 −1 Γ 5 + 1 1 ω2 ω2 ω4 ω4 −1 −1 −ω2 −ω2 −ω4 −ω4 1 1 ω2 ω2 ω4 ω4 −1 −1 −ω2 −ω2 −ω4 −ω4 Γ 6 + 1 1 −ω4 −ω4 −ω2 −ω2 −1 −1 ω4 ω4 ω2 ω2 1 1 −ω4 −ω4 −ω2 −ω2 −1 −1 ω4 ω4 ω2 ω2 Γ 1 − 1 1 1 1 1 1 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 Γ 2 − 1 1 −ω2 −ω2 ω4 ω4 1 1 −ω2 −ω2 ω4 ω4 −1 −1 ω2 ω2 −ω4 −ω4 −1 −1 ω2 ω2 −ω4 −ω4 Γ 3 − 1 1 ω4 ω4 −ω2 −ω2 1 1 ω4 ω4 −ω2 −ω2 −1 −1 −ω4 −ω4 ω2 ω2 −1 −1 −ω4 −ω4 ω2 ω2 Γ 4 − 1 1 −1 −1 1 1 −1 −1 1 1 −1 −1 −1 −1 1 1 −1 −1 1 1 −1 −1 1 1 Γ 5 − 1 1 ω2 ω2 ω4 ω4 −1 −1 −ω2 −ω2 −ω4 −ω4 −1 −1 −ω2 −ω2 −ω4 −ω4 1 1 ω2 ω2 ω4 ω4 Γ 6 − 1 1 −ω4 −ω4 −ω2 −ω2 −1 −1 ω4 ω4 ω2 ω2 −1 −1 ω4 ω4 ω2 ω2 1 1 −ω4 −ω4 −ω2 −ω2 Γ 7 + 1 −1 ω −ω ω2 −ω2 i −i −ω4 ω4 −ω5 ω5 1 −1 ω −ω ω2 −ω2 i −i −ω4 ω4 −ω5 ω5 Γ 8 + 1 −1 −ω5 ω5 −ω4 ω4 −i I ω2 −ω2 ω −ω 1 −1 −ω5 ω5 −ω4 ω4 −i i ω2 −ω2 ω −ω Γ 9 + 1 −1 −ω ω ω2 −ω2 −i i −ω4 ω4 ω5 −ω5 1 −1 −ω ω ω2 −ω2 −i i −ω4 ω4 ω5 −ω5 Γ 10 + 1 −1 ω5 −ω5 −ω4 ω4 i −i ω2 −ω2 −ω ω 1 −1 ω5 −ω5 −ω4 ω4 I −i ω2 −ω2 −ω ω Γ 11 + 1 −1 −i I −1 1 i −i −1 1 i −i 1 −1 −i i −1 1 i −i −1 1 i −i Γ 12 + 1 −1 i −i −1 1 −i i −1 1 −i i 1 −1 i −i −1 1 −i i −1 1 −i i Γ 7 − 1 −1 ω −ω ω2 −ω2 i −i −ω4 ω4 −ω5 ω5 −1 1 −ω ω −ω2 ω2 −i i ω4 −ω4 ω5 −ω5 362 Γ 8 − 1 −1 −ω5 ω5 −ω4 ω4 −i I ω2 −ω2 ω −ω −1 1 ω5 −ω5 ω4 −ω4 I −i −ω2 ω2 −ω Ω Γ 9 − 1 −1 −ω ω ω2 −ω2 −i I −ω4 ω4 ω5 −ω5 −1 1 ω −ω −ω2 ω2 i −i ω4 −ω4 −ω5 ω5 Γ 10 − 1 −1 ω5 −ω5 −ω4 ω4 i −i ω2 −ω2 −ω ω −1 1 −ω5 ω5 ω4 −ω4 −i i −ω2 ω2 ω −ω Γ 11 − 1 −1 −i I −1 1 i −i −1 1 i −i −1 1 i −i 1 −1 −i i 1 −1 −i i Γ 12 − 1 −1 i −i −1 1 −i I −1 1 −i i −1 1 −i i 1 −1 i −i 1 −1 i −i C6h D E E ̅ C6 C6 ̅̅̅ C3 C3 ̅̅̅ C2 C2 ̅̅̅ C3 2 C3 2 ̅̅̅ C6 5 C6 5 ̅̅̅ I I̅ S3 −1 S3 −1 ̅ ̅ ̅ ̅ ̅ S6 −1 S6 −1 ̅ ̅ ̅ ̅ ̅ 𝜎h 𝜎h ̅̅̅ S6 S6 ̅ ̅ ̅ S3 S3 ̅ ̅ ̅ 上表中ω = exp [πi/6]。由于空间限制，对于非单位群元A，我们使用A ̅来表示E ̅A。 D6h D E E ̅ {C2 E ̅C2} 2C3 2E ̅C3 2C6 2E ̅C6 {3C2 (1) 3E ̅C2 (1)} {3C2 (2) 3E ̅C2 (2)} I E ̅ I {IC2 IE ̅C2} 2IC3 2IE ̅C3 2IC6 2IE ̅C6 {3IC2 (1) 3IE ̅C2 (1)} {3IC2 (2) 3IE ̅C2 (2)} Γ 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Γ 2 + 1 1 1 1 1 1 1 −1 −1 1 1 1 1 1 1 1 −1 −1 Γ 3 + 1 1 −1 1 1 −1 −1 1 −1 1 1 −1 1 1 −1 −1 1 −1 Γ 4 + 1 1 −1 1 1 −1 −1 −1 1 1 1 −1 1 1 −1 −1 −1 1 Γ 5 + 2 2 −2 −1 −1 1 1 0 0 2 2 −2 −1 −1 1 1 0 0 Γ 6 + 2 2 2 −1 −1 −1 −1 0 0 2 2 2 −1 −1 −1 −1 0 0 Γ 1 − 1 1 1 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 −1 −1 Γ 2 − 1 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 −1 −1 1 1 Γ 3 − 1 1 −1 1 1 −1 −1 1 −1 −1 −1 1 −1 −1 1 1 −1 1 Γ 4 − 1 1 −1 1 1 −1 −1 −1 1 −1 −1 1 −1 −1 1 1 1 −1 Γ 5 − 2 2 −2 −1 −1 1 1 0 0 −2 −2 2 1 1 −1 −1 0 0 Γ 6 − 2 2 2 −1 −1 −1 −1 0 0 −2 −2 −2 1 1 1 1 0 0 Γ 7 + 2 −2 0 1 −1 √3 −√3 0 0 2 −2 0 1 −1 √3 −√3 0 0 Γ 8 + 2 −2 0 1 −1 −√3 √3 0 0 2 −2 0 1 −1 −√3 √3 0 0 Γ 9 + 2 −2 0 −2 2 0 0 0 0 2 −2 0 −2 2 0 0 0 0 Γ 7 − 2 −2 0 1 −1 √3 −√3 0 0 −2 2 0 −1 1 −√3 √3 0 0 Γ 8 − 2 −2 0 1 −1 −√3 √3 0 0 −2 2 0 −1 1 √3 −√3 0 0 Γ 9 − 2 −2 0 −2 2 0 0 0 0 −2 2 0 2 −2 0 0 0 0 D6h D E E ̅ {C2 E ̅C2} 2C3 2E ̅C3 2C6 2E ̅C6 {3C2 (1) 3E ̅C2 (1)} {3C2 (2) 3E ̅C2 (2)} I E ̅ I {IC2 IE ̅C2} 2IC3 2IE ̅C3 2IC6 2IE ̅C6 {3IC2 (1) 3IE ̅C2 (1)} {3IC2 (2) 3IE ̅C2 (2)} 364 C6 D E E ̅ C6 E ̅C6 C3 E ̅C3 C2 E ̅C2 C3 2 E ̅C3 2 C6 5 E ̅C6 5 C3h D E E ̅ IC6 E ̅IC6 C3 E ̅C3 σh E ̅σh C3 2 E ̅C3 2 IC6 5 E ̅IC6 5 Γ 1 1 1 1 1 1 1 1 1 1 1 1 1 Γ 2 1 1 −ω2 −ω2 ω4 ω4 1 1 −ω2 −ω2 ω4 ω4 Γ 3 1 1 ω4 ω4 −ω2 −ω2 1 1 ω4 ω4 −ω2 −ω2 Γ 4 1 1 −1 −1 1 1 −1 −1 1 1 −1 −1 Γ 5 1 1 ω2 ω2 ω4 ω4 −1 −1 −ω2 −ω2 −ω4 −ω4 Γ 6 1 1 −ω4 −ω4 −ω2 −ω2 −1 −1 ω4 ω4 ω2 ω2 Γ 7 1 −1 ω −ω ω2 −ω2 i −i −ω4 ω4 −ω5 ω5 Γ 8 1 −1 −ω5 ω5 −ω4 ω4 −i i ω2 −ω2 ω −ω Γ 9 1 −1 −ω ω ω2 −ω2 −i i −ω4 −ω4 ω5 −ω5 Γ 10 1 −1 ω5 −ω5 −ω4 ω4 i −i ω2 −ω2 −ω ω Γ 11 1 −1 −i i −1 1 i −i −1 1 i −i Γ 12 1 −1 i −i −1 1 −i i −1 1 −i i 上表中ω = exp [πi/6]。 D6 D E E ̅ {C2 E ̅C2} 2C3 2E ̅C3 2C6 2E ̅C6 {3C2 (1) 3E ̅C2 (1)} {3C2 (2) 3E ̅C2 (2)} C6v D E E ̅ {C2 E ̅C2} 2C3 2E ̅C3 2C6 2E ̅C6 {3IC2 (1) 3IE ̅C2 (1)} {3IC2 (2) 3IE ̅C2 (2)} D3h D E E ̅ {IC2 E ̅IC2} 2C3 2E ̅C3 2C6 2E ̅C6 {3C2 (1) 3E ̅C2 (1)} {3IC2 (2) 3IE ̅C2 (2)} Γ 1 1 1 1 1 1 1 1 1 1 Γ 2 1 1 1 1 1 1 1 −1 −1 Γ 3 1 1 −1 1 1 −1 −1 1 −1 Γ 4 1 1 −1 1 1 −1 −1 −1 1 Γ 5 2 2 −2 −1 −1 1 1 0 0 Γ 6 2 2 2 −1 −1 −1 −1 0 0 Γ 7 2 −2 0 1 −1 √3 −√3 0 0 Γ 8 2 −2 0 1 −1 −√3 √3 0 0 Γ 9 2 −2 0 −2 2 0 0 0 0 7. 立方系（ Oh、T、O、Th、Td）： Oh D E E ̅ 8C3 8E ̅C3 {3C4 2 3E ̅C4 2 6C4 6E ̅C4 {6C2 ′ 6E ̅C2 ′ } I I̅ 8IC3 8IE ̅C3 {3IC4 2 3IE ̅C4 2 6IC4 6IE ̅C4 {6IC2 ′ 6IE ̅C2 ′ } Γ 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Γ 2 + 1 1 1 1 1 −1 −1 −1 1 1 1 1 1 −1 −1 −1 Γ 3 + 2 2 −1 −1 2 0 0 0 2 2 −1 −1 2 0 0 0 Γ 4 + 3 3 0 0 −1 1 1 −1 3 3 0 0 −1 1 1 −1 Γ 5 + 3 3 0 0 −1 −1 −1 1 3 3 0 0 −1 −1 −1 1 Γ 1 − 1 1 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 −1 Γ 2 − 1 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 −1 1 1 1 Γ 3 − 2 2 −1 −1 2 0 0 0 −2 −2 1 1 −2 0 0 0 Γ 4 − 3 3 0 0 −1 1 1 −1 −3 −3 0 0 1 −1 −1 1 Γ 5 − 3 3 0 0 −1 −1 −1 1 −3 −3 0 0 1 1 1 −1 Γ 6 + 2 −2 1 −1 0 √2 −√2 0 2 −2 1 −1 0 √2 −√2 0 Γ 7 + 2 −2 1 −1 0 −√2 √2 0 2 −2 1 −1 0 −√2 √2 0 Γ 8 + 4 −4 −1 1 0 0 0 0 4 −4 −1 1 0 0 0 0 Γ 6 − 2 −2 1 −1 0 √2 −√2 0 −2 2 −1 1 0 −√2 √2 0 Γ 7 − 2 −2 1 −1 0 −√2 √2 0 −2 2 −1 1 0 √2 −√2 0 366 Γ 8 − 4 −4 −1 1 0 0 0 0 −4 4 1 −1 0 0 0 0 Oh D E E ̅ {3C2 3E ̅C2} 4C3 4E ̅C3 4C3 −1 4E ̅C3 −1 I I̅ {3IC2 3IE ̅C2} 4IC3 4IE ̅C3 4IC3 −1 4E ̅IC3 −1 Γ 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Γ 2 + 1 1 1 ω ω ω2 ω2 1 1 1 ω ω ω2 ω2 Γ 3 + 1 1 1 ω2 ω2 ω ω 1 1 1 ω2 ω2 ω ω Γ 4 + 3 3 −1 0 0 0 0 3 3 −1 0 0 0 0 Γ 1 − 1 1 1 1 1 1 1 −1 −1 −1 −1 −1 −1 −1 Γ 2 − 1 1 1 ω ω ω2 ω2 −1 −1 −1 −ω −ω −ω2 −ω2 Γ 3 − 1 1 1 ω2 ω2 ω ω −1 −1 −1 −ω2 −ω2 −ω −ω Γ 4 − 3 3 −1 0 0 0 0 −3 −3 1 0 0 0 0 Γ 5 + 2 −2 0 1 −1 1 −1 2 −2 0 1 −1 1 −1 Γ 6 + 2 −2 0 ω −ω ω2 −ω2 2 −2 0 ω −ω ω2 −ω2 Γ 7 + 2 −2 0 ω2 −ω2 ω −ω 2 −2 0 ω2 −ω2 ω −ω Γ 5 − 2 −2 0 1 −1 1 −1 −2 2 0 −1 1 −1 1 Γ 6 − 2 −2 0 ω −ω ω2 −ω2 −2 2 0 −ω ω −ω2 ω2 Γ 7 − 2 −2 0 ω2 −ω2 Ω −ω −2 2 0 −ω2 ω2 −ω ω TD E E ̅ {3C2 3E ̅C2} 4C3 4E ̅C3 4C3 ′ 4E ̅C3 ′ Γ 1 1 1 1 1 1 1 1 Γ 2 1 1 1 ω ω ω2 ω2 Γ 3 1 1 1 ω2 ω2 ω ω Γ 4 3 3 −1 0 0 0 0 Γ 5 2 −2 0 1 −1 1 −1 Γ 6 2 −2 0 ω −ω ω2 −ω2 Γ 7 2 −2 0 ω2 −ω2 ω −ω 上面两个表中ω = exp [2πi/3]。 OD E E ̅ 8C3 8E ̅C3 {3C4 2 3E ̅C4 2} 6C4 6E ̅C4 {6C2 ′ 6E ̅C2 ′ } Td D E E ̅ 8C3 8E ̅C3 {3C2 3E ̅C2} 6S4 6E ̅S4 {6σd 6E ̅σd} Γ 1 1 1 1 1 1 1 1 1 Γ 2 1 1 1 1 1 −1 −1 −1 Γ 3 2 2 −1 −1 2 0 0 0 Γ 4 3 3 0 0 −1 1 1 −1 Γ 5 3 3 0 0 −1 −1 −1 1 Γ 6 2 −2 1 −1 0 √2 −√2 0 Γ 7 2 −2 1 −1 0 −√2 √2 0 Γ 8 4 4 −1 1 0 0 0 0 368 附录D 置换群部分相关定理与引理证明 补充定理1 （本原幂等元判别定理） ： 幂等元e ⃗ ⃗i为本原幂等元的充要条件为： e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗i = λe ⃗ ⃗i对∀x ⃗ ⃗∈RG成立，其中λ为常数。 证明： 必要性，设e ⃗ ⃗i为本原幂等元，对应投影算符P ̂ i，子空间Wi = P ̂ iRG为群不变的不可 约的子空间。由上个定理，知L(g)P ̂ i = P ̂ iL(g)对∀g ∈G成立。 这时， 对∀x ⃗ ⃗∈RG， 定义一个与x ⃗ ⃗相关的算符A ̂， 这个算符作用到群空间中向量y ⃗ ⃗上 的效果是A ̂y ⃗ ⃗= y ⃗ ⃗e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗i。 这样的话，当A ̂作用到L(g)y ⃗ ⃗上的时候，就有： A ̂(L(g)y ⃗ ⃗) = L(g)y ⃗ ⃗e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗i 而A ̂y ⃗ ⃗= y ⃗ ⃗e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗i，所以进一步有： A ̂(L(g)y ⃗ ⃗) = L(g)(A ̂y ⃗ ⃗) 由于y ⃗ ⃗为RG中任意向量，所以A ̂L(g) = L(g)A ̂对∀g ∈G成立。 对这个y ⃗ ⃗，可分为属于本原幂等元e ⃗ ⃗i所对应的子空间Wi的部分y ⃗ ⃗1与不属于Wi的部 分y ⃗ ⃗2。 由于A ̂与L(g)都是线性算符， A ̂L(g) = L(g)A ̂对y ⃗ ⃗1、 y ⃗ ⃗2都是成立的。 同时A ̂y ⃗ ⃗= y ⃗ ⃗e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗i也可分解为A ̂y ⃗ ⃗1 = y ⃗ ⃗1e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗i与A ̂y ⃗ ⃗2 = y ⃗ ⃗2e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗i两个部分。 先看y ⃗ ⃗1 部分，由于A ̂L(g) = L(g)A ̂ ，由舒尔引理二可知A ̂ 在Wi 上对应的矩阵只能是常数矩阵。这样的话： A ̂y ⃗ ⃗1 = λy ⃗ ⃗1 = λP ̂ iy ⃗ ⃗1 再看y ⃗ ⃗2，它不属于Wi，由 A ̂y ⃗ ⃗2 = y ⃗ ⃗2e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗ i 结合正文定理6.2 的证明中我们说过的幂等元e ⃗ ⃗i与它对应的投影算符P ̂ i的关系 P ̂ iy ⃗ ⃗= y ⃗ ⃗e ⃗ ⃗i，可知上式右边等于： y ⃗ ⃗2e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗ i = (P ̂ iy ⃗ ⃗2)x ⃗ ⃗e ⃗ ⃗ i = P ̂ i ((P ̂ iy ⃗ ⃗2)x ⃗ ⃗) 由于P ̂ iy ⃗ ⃗2 = 0，所以A ̂y ⃗ ⃗2 = P ̂ i(0x ⃗ ⃗) = 0 = λP ̂ iy ⃗ ⃗2，其中λ为任意复数。 这个A ̂y ⃗ ⃗2 = λP ̂ iy ⃗ ⃗2与前面A ̂y ⃗ ⃗1 = λP ̂ iy ⃗ ⃗1结合，有A ̂y ⃗ ⃗＝λP ̂ iy ⃗ ⃗。再由y ⃗ ⃗的任意性，可知A ̂ = λP ̂ i。 这个时候，再结合A ̂的定义，即对∀y ⃗ ⃗∈RG，有： A ̂y ⃗ ⃗＝y ⃗ ⃗e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗ i 以及： A ̂y ⃗ ⃗＝λP ̂ iy ⃗ ⃗＝λy ⃗ ⃗e ⃗ ⃗ i＝y ⃗ ⃗λe ⃗ ⃗ i 可得： y ⃗ ⃗e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗ i＝y ⃗ ⃗λe ⃗ ⃗ i 这个等式，同样是对∀y ⃗ ⃗∈RG成立的，因此e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗ i＝λe ⃗ ⃗ i。必要性得证。由Wi为不可 约表示空间可得e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗ i＝λe ⃗ ⃗ i。这里不可约在舒尔引理二的应用中起了关键作用。 充分性，由e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗ i＝λe ⃗ ⃗ i推e ⃗ ⃗ i对应的Wi为不可约表示空间。反证，设e ⃗ ⃗ i不是本原幂 等元，它可以继续分为e ⃗ ⃗ i1+e ⃗ ⃗ i2，于是： e ⃗ ⃗ ie ⃗ ⃗ i1e ⃗ ⃗ i = (e ⃗ ⃗ i1 + e ⃗ ⃗ i2)e ⃗ ⃗ i1(e ⃗ ⃗ i1 + e ⃗ ⃗ i2) = e ⃗ ⃗i1 3 + e ⃗ ⃗i1 2 e ⃗ ⃗ i2 + e ⃗ ⃗ i2e ⃗ ⃗i1 2 + e ⃗ ⃗ i2e ⃗ ⃗ i1e ⃗ ⃗ i2 = e ⃗ ⃗ i1 这个等式成立的原因是e ⃗ ⃗ i1、e ⃗ ⃗ i2为e ⃗ ⃗ i继续分解的两个部分，它们都是幂等元，且 相互正交。因此，等式的第一项三次方等于一次方（幂等元性质） ，后三项等于 零（正交性质） 。理解正交性质一点，大家可参考P ̂ iy ⃗ ⃗= y ⃗ ⃗e ⃗ ⃗ i这个幂等元与投影算 符的关系式。 换句话说， e ⃗ ⃗ i是群代数中这样一个向量， 它与任何的一个向量相乘， 结果就是P ̂ i所对应的群不变子空间中的向量。 P ̂ i1与P ̂ i2的子空间相互正交， e ⃗ ⃗ i1、 e ⃗ ⃗ i2 也相互正交。 而e ⃗ ⃗ix ⃗ ⃗e ⃗ ⃗ i＝λe ⃗ ⃗ i，所以上式还等于λe ⃗ ⃗ i，进而e ⃗ ⃗ i1 = λe ⃗ ⃗ i。 370 这样的话，可以继续有：e ⃗ ⃗i1 2 = λ2e ⃗ ⃗i 2。同时，e ⃗ ⃗i1 2 = e ⃗ ⃗ i1 = λe ⃗ ⃗ i，所以λ要么为零，要 么为一。 λ为零时， e ⃗ ⃗ i1 = λe ⃗ ⃗ i = 0， e ⃗ ⃗ i = e ⃗ ⃗ i2； λ为一时， e ⃗ ⃗ i1 = λe ⃗ ⃗ i = e ⃗ ⃗ i，e ⃗ ⃗ i2 = 0。 不管怎样，都是说e ⃗ ⃗ i不能继续分，是个本原幂等元。充分性同样得证。 引理6.1 设𝐓、𝐓′是由置换 𝐫 联系起来的杨盘，𝐓′ = 𝐫 𝐓，如果置换 𝐬 作用在𝐓 上，使得𝐓(𝐢，𝐣)数字变到𝐬𝐓中的(𝐢′， 𝐣′)处，则𝐬′ = 𝐫𝐬𝐫−𝟏也会使得𝐓′(𝐢，𝐣)中的 数字变到𝐬′𝐓′的(𝐢′， 𝐣′)处。 （正文部分讲过，这个引理要说明的关系就是下面这个图。 杨盘sT中各个数相对杨盘T中的变化与杨盘s′T′中各个数相对于杨盘T′中各个数 的变化完全一样。 ） 证明： 把杨盘T、T′、sT、s′T′中的数均按从左到右、从上到下的顺序排列，记为：{t1， t2，⋯，tn}、 {t1 ′ ，t2 ′ ，⋯，tn ′ }、 {st1，st2，⋯，stn}、 {s′t1 ′ ，s′t2 ′ ，⋯，s′tn ′ }。 由上图关系，知 r = ( t1，t2，⋯，tn t1 ′ ，t2 ′ ，⋯，tn ′ ) s = ( t1，t2，⋯，tn st1，st2，⋯，stn ) s′ = ( t1 ′ ，t2 ′ ，⋯，tn ′ s′t1 ′ ，s′t2 ′ ，⋯，s′tn ′ ) 同时， 由于s′ = rsr−1， 这意味着它干的事是把s的上下两行分别用r置换。 上面那 行是把t1，t2，⋯，tn变成了t1 ′ ，t2 ′ ，⋯，tn ′ ，下面那行是把st1，st2，⋯，stn 变成了s′t1 ′ ，s′t2 ′ ，⋯，s′tn ′ ，所以r其实有两种写法，分别是： ( t1，t2，⋯，tn t1 ′ ，t2 ′ ，⋯，tn ′ ) 、( st1，st2，⋯，stn s′t1 ′ ，s′t2 ′ ，⋯，s′tn ′ ) 比较这两个等价的写法，我们就知道，当左边第i列的数码ti在右边的位置为第j 列，也就是ti = stj时，一定有：ti ′ = s′tj ′。也就是上面图中说的规律。 这个引理还有个推论： 设T′ = rT， 则有R(T′) = rR(T)r−1， C(T′) = rC(T)r−1， P ̂(T′) = rP ̂(T)r−1， Q ̂(T′) = rQ ̂(T)r−1，E ̂(T′) = rE ̂(T)r−1。 证明： 还是基于上面那张图。对∀r ∈Sn，取∀p ̂ ∈C(T)，把这个q ̂理解为上面图中的s， 只不过它干的事情只是将T中同列的数码相互置换。这样的话，由于上图中s对T 的置换在相对位置上完全等同于rsr−1对rT的置换，所以如果q ̂是对杨图T的列置 换的话，rq ̂r−1就是对杨图T′ = rT的等同的列置换。这种等同是一对一的关系， 所以在q ̂走遍 C(T)中所有元素的时候rq ̂r−1也走遍 C(T′)中所有元素。最终的效果 就是C(T′) = rC(T)r−1。 对R(T′) = rR(T)r−1 ，逻辑是完全类似的。而对P ̂(T′) = rP ̂(T)r−1 、Q ̂(T′) = rQ ̂(T)r−1、 E ̂(T′) = rE ̂(T)r−1也一样， 只不过这里置换的集合变成了置换的线性叠 加罢了。 372 引理6.2 设𝐩 ̂、𝐪 ̂是杨盘T 的行、列置换，则T 中位于同一行的任意两个数字不 可能出现在𝐓′′ = 𝐩 ̂𝐪 ̂𝐓的同一列中；反之，若𝐓′′ = 𝐫𝐓时，T 中位于同一行的任意 两个数字都不出现在𝐓′的同一列中， 则杨盘T 存在行、 列置换𝐩 ̂、 𝐪 ̂， 使𝐫= 𝐩 ̂𝐪 ̂。 证明： 还是基于引理6.1 那张图，p ̂ ∈R(T)，q ̂ ∈C(T)，T′′ = p ̂q ̂T，证T 中位于同一行 的任意两个数字不可能出现在T′′的同一列中。 令T′ = p ̂T，q ̂′ = p ̂q ̂p ̂−1。现在讨论的内容与上面那张图的对应关系是这里的T对 应图中的T，这里的p ̂对应图中的r，这里的T′对应图中的T′，这里的q ̂对应图中的 s， 这里的q ̂′ = p ̂q ̂p ̂−1对应图中的s ̂′ = rsr−1， 这里的T′′ = p ̂q ̂T = p ̂q ̂p ̂−1p ̂T = q ̂′T′ 对应图中的s ̂′T′。 这个定理说白了，说的是左上角那个图中T中同一行的任意两个数字，不可能通 过取r = p ̂、s = q ̂的方式，由上面图中显示的变换，变换到右下角T′′ = q ̂′T′的同 一列中。 怎么理解？很容易，由上个引理推论中的讨论，q ̂是T的列置换，q ̂′也是T′的列置 换。因此q ̂′不可能将T′中位于同行的两个数码换到T′′的同一列中。而另一方面 T′ = p ̂T，所以T′的行数码与T的行数码是相同的。所以T中的行数码在经历了p ̂这 个行置换变成T′，再经历q ̂′ = p ̂q ̂p ̂−1这个列置换变成T′′ = p ̂q ̂T后，不可能处在T′′ 的同一列中。这个是这个引理第一部分说的事情。 反过来， T′′ = rT， 若T中同一行的任意两个数码都不出现在T′′的同一列中， 要证 r = p ̂q ̂。由已知条件T中同一行的任意两个数码都不出现在T′′的同一列中，知我 们总可以用行置换p ̂ ∈R(T)对杨图T操作， 使得结果T′ = p ̂T与T′′的各列数码相同， 只是每列中各个数码的上下位置可以不同。 这样的话， 我们可以在这个基础上对 T′进行一个列置换q ̂′，调整每一列中各数码的行，使得q ̂′ T′与T′完全相同。如果 取q ̂′ = p ̂q ̂p ̂−1，这样的话q ̂′ T′就是p ̂q ̂p ̂−1p ̂T = p ̂q ̂T，它与T′′ = rT完全相同。这样 就一定有r = p ̂q ̂。问题得证。 前面也说过，引理6.1 与6.2 说的是同一个杨图的杨盘的性质。 引理6.3 设杨盘𝐓和𝐓′分别属于杨图[𝛌]、[𝛌′]，且[𝛌]>[𝛌′] （这里， [λ]>[λ′]是指[𝜆] = [𝜆1，𝜆2，⋯，𝜆n]， [𝜆′] = [𝜆′1，𝜆′2，⋯，𝜆′n]， 第一 个不等于零的𝜆𝑖−𝜆′𝑖，一定满足𝜆𝑖> 𝜆′𝑖，比如： ） 则存在两个数码位于T的同一行与T′的同一列。 证明： 反证法，设杨盘T中任意两个同行的数码均处在T′的不同列中。这样的话要想让 T中第一行的𝜆1个数字出现在T′的不同列中， 需要𝜆′1 ≥𝜆1。 而已知条件是[λ]>[λ′]， 所以𝜆′1只能等于𝜆1。这样处理完以后，T′的第一行被T第一行的数字占满。再看 第二行， 同样道理， 也有𝜆′2 = 𝜆2。 第二行也被占满， 类推， 最终会有[𝜆′] = [λ]。 这与已知[λ]>[λ′]矛盾， 因此假设不成立， 当[λ]>[λ′]时， T和T′必存在两个数码位 于T的同一行与T′的同一列。 引理6.4 若有两个数字，位于杨盘𝐓的同一行与杨盘𝐓′的同一列，则它们的杨算 符𝐄 ̂(𝐓′)𝐄 ̂(𝐓) = 𝟎。 证明： 374 设数字a1、a2是位于杨盘T的同一行与杨盘T′的同一列的两个数码，则有对换t = (a1，a2)，这个对换既属于杨盘T的行置换R(T)，又属于杨盘T′的列置换C(T′)。 而R(T)、 C(T′)又同时为Sn的子群， 且t2 = s0， t为奇置换， δt = −1。 由重排定理， 我们知道： tP ̂(T) = t ∑p ̂ p ̂∈R(T) = P ̂(T) Q ̂(T′)t = ∑δqq ̂ q ̂∈C(T′) δtδtt = δt ∑δqδtq ̂ q ̂∈C(T′) t = δt ∑ δqtqt ̂ qt ̂∈C(T′) = δtQ ̂(T′) 这样的话就会有： Q ̂(T′)P ̂(T)＝Q ̂(T′)s0P ̂(T)＝Q ̂(T′)ttP ̂(T)＝δtQ ̂(T′)P ̂(T)＝−Q ̂(T′)P ̂(T) 因此Q ̂(T′)P ̂(T) = 0。 对于这个式子，再左乘P ̂(T′)、右乘Q ̂(T)，就会有： P ̂(T′)Q ̂(T′)P ̂(T)Q ̂(T) = 0 进而： E ̂(T′)E ̂(T) = 0。 得证。 这样的话，结合引理6.3 与6.4，我们就知道当T、T′属于不同杨图[λ]、[λ′]，且 [λ]>[λ′]时，有E ̂(T′)E ̂(T) = 0。 引理6.5 设置换群𝐒𝐧的群代数𝐑𝐒𝐧中的矢量𝐱 ⃗ ⃗= ∑ 𝐱𝐬𝐬 𝐬∈𝐒𝐧 ， 𝐓为𝐒𝐧的杨盘。若∀𝐩 ̂ ∈ 𝐑(𝐓)，∀𝐪 ̂ ∈𝐂(𝐓)，𝐩 ̂𝐱 ⃗ ⃗𝐪 ̂ = 𝛅𝐪𝐱 ⃗ ⃗，则𝐱 ⃗ ⃗与𝐓盘的杨算符𝐄 ̂(𝐓)相差一个常数因子，即 𝐱 ⃗ ⃗= 𝛉𝐄 ̂(𝐓)，常数𝛉与𝐱 ⃗ ⃗有关，为𝐱 ⃗ ⃗中𝐬𝟎的系数。 （这个引理说的是群代数RSn中对一个特定的T满足p ̂x ⃗ ⃗q ̂ = δqx ⃗ ⃗的向量x ⃗ ⃗的性质， 它 在证明下一个引理时会用到。课上如果讲这个附录，时间又紧，从逻辑关系上这 个证明的过程可以略过） 证明： 分两步，第一步证Sn中不能写成p ̂q ̂形式的群元s一定可以表示为p ̂sq ̂的形式，即 s=p ̂sq ̂，其中p ̂ ∈R(T)，q ̂ ∈C(T)。 令T′ = sT，由于s不具备p ̂q ̂的形式，由引理6.2 的逆否命题，知至少存在两个数 码a1、a2，位于T的同一行，T′的同一列。 取t = (a1，a2)，有t ∈R(T) ∩C(T′)，t2 = s0。由于T = s−1T′，由引理6.1 可知 t ∈C(T′)时s−1ts ∈C(T)。这样的话，如果我们取p ̂ = t、q ̂ = s−1ts，则 p ̂sq ̂ = tss−1ts = t2s = s 第二步， 由p ̂x ⃗ ⃗q ̂ = δqx ⃗ ⃗来求x ⃗ ⃗。 由于x ⃗ ⃗= ∑ xss s∈Sn ， 只需定出展开系数xs即可。 这样 的话，一方面 p ̂x ⃗ ⃗q ̂ = p ̂ ∑xss s∈Sn q ̂ = ∑xsp ̂s s∈Sn q ̂ 另一方面 δqx ⃗ ⃗= ∑δqxss s∈Sn 要想p ̂x ⃗ ⃗q ̂ = δqx ⃗ ⃗成立，需∑ xsp ̂s s∈Sn q ̂ = ∑ δqxss s∈Sn 。 当s不具备p ̂q ̂的形式时， 由于p ̂sq ̂ = s， 上式左边这个s上的分量是δqxs， 右边这个 s上的分量是xs， 由于q ̂ = s−1ts为奇置换， 所以δq = −1。 因此xs = −xs， xs = 0。 当s具备p ̂q ̂ 的形式时，取该p ̂、q ̂ 代入p ̂x ⃗ ⃗q ̂ = δqx ⃗ ⃗，并看该p ̂q ̂ 分量的系数。由 ∑ xsp ̂s s∈Sn q ̂ = ∑ δqxss s∈Sn 这个等式，知左边的p ̂q ̂系数是xs0，而右边的p ̂q ̂项系数 是δqxpq，于是xs0 = δqxpq。取θ = xs0，有xpq = δqθ。对不同的具备p ̂q ̂的形式的 s，x ⃗ ⃗在它上面的分量xs = xpq = δqθ，其中θ与x ⃗ ⃗有关。 两者综合起来，就是对于对∀p ̂ ∈R(T)、∀q ̂ ∈C(T)，满足p ̂x ⃗ ⃗q ̂ = δqx ⃗ ⃗的RSn中的矢 量x ⃗ ⃗，其分量满足： 376 xs = { 0，当s 不具备p ̂q ̂的形式 δqθ，当s 具备p ̂q ̂的形式 这样的话： x ⃗ ⃗= ∑xss s∈Sn = ∑xpqp ̂q ̂ p∈R(T) q∈C(T) = ∑δqθp ̂q ̂ p∈R(T) q∈C(T) = θE ̂(T) 其中θ与x ⃗ ⃗有关。引理得证。 引理6.6 杨盘𝐓的杨算符𝐄 ̂(𝐓)是置换群𝐒𝐧的群代数𝐑𝐒𝐧中的一个本质的本原幂等 元，不变子空间𝐑𝐒𝐧𝐄 ̂(𝐓)是置换群𝐒𝐧的一个不可约表示的表示空间，其维数是 𝐧！的因子。 证明： 第一步，证E ̂(T)就是群代数RSn中幂等元（利用引理6.5） 。 对∀p ̂ ∈R(T)、∀q ̂ ∈C(T)，由重排定理，有： p ̂E ̂(T)2q ̂ = p ̂P ̂(T)Q ̂(T)P ̂(T)Q ̂(T)q ̂ = P ̂(T)Q ̂(T)P ̂(T)δqQ ̂(T) = δqE ̂(T)2 结合引理6.5，我们知道E ̂(T)2一定具备θE ̂(T)的形式。这样的话E ̂(T)2 = θE ̂(T)， E ̂(T)为本质幂等元。其中θ，由前面的讨论，知为E ̂(T)2中s0的系数，待定（因为 除了s0，阶为2 的群元也会有贡献） 。E ̂(T)/θ为幂等元。 第二步，确定θ，证E ̂(T)/θ对应的群代数RSn中群不变的子空间RSnE ̂(T)/θ维数是 n! 的因子。 由正文部分的讨论，一个幂等元E ̂(T)/θ对应一个投影算符P ̂，关系是：对∀x ⃗ ⃗∈ RSn，有P ̂x ⃗ ⃗= x ⃗ ⃗E ̂(T)/θ。 对这个算符P ̂，取RSn的基为s0、s1、⋯、sn!−1，则表示矩阵对角元满足： P jj = (P ̂sj)sj = (sjE ̂(T)/θ)sj 其中E ̂(T)贡献的， 必为s0。 而E ̂(T)的s0分量的系数是1， 所以P jj = 1/θ。 这样算符 P ̂的迹就是n!/θ。 线性变换后， 取RSn的基v1、 v2、⋯、vf、vf+1、⋯、vn!， 其中v1、 v2、⋯、vf为 RSnE ̂(T)/θ所对应的群不变的子空间W的基。 这时， 对1 ≤j ≤f， 有vj ∈RSnE ̂(T)/θ， 因此P ̂vj = vj，它们所对应的(P ̂vj)j = 1。而对j > f，P ̂vj = vjE ̂(T)/θ ∈W，它在vj 上的分量为零。 所以(P ̂vj)j = 0。 这样的话算符P ̂的迹就是f。 而这段的讨论与上面 一段的讨论差的就是一个线性变换，不改变矩阵的迹，所以f = n!/θ。由于f必为 整数，而θ为E ̂(T)2中s0的系数，也必为整数，f必为n!的因子。 第三步， 证幂等元E ̂(T)/θ为本原幂等元 （利用本原幂等元判别定理以及引理6.5） 。 对∀x ⃗ ⃗∈RSn，p ̂ ∈R(T)，q ̂ ∈C(T)，有 p ̂ ((E ̂(T) θ ) x ⃗ ⃗(E ̂(T) θ )) q ̂ = (p ̂E ̂(T) θ ) x ⃗ ⃗(E ̂(T)q ̂ θ ) = (E ̂(T) θ ) x ⃗ ⃗δq (E ̂(T) θ ) = δq ((E ̂(T) θ ) x ⃗ ⃗(E ̂(T) θ )) 这样结合引理6.5， 就有( E ̂(T) θ ) x ⃗ ⃗( E ̂(T) θ ) = μE ̂(T)， 其中μ为( E ̂(T) θ ) x ⃗ ⃗( E ̂(T) θ )中s0系数。 由前面讲到的本原幂等元判据定理，知E ̂(T)/θ为本原幂等元，RSnE ̂(T)/θ为Sn的 不可约表示空间。引理得证。 由这个引理， 我们也知一个杨盘T， 可求出一个本原幂等元E ̂(T)/θ， 从而得到n阶 置换群Sn的一个不可约表示。 引理6.7 置换群𝐒𝐧同一个杨图的不同杨盘给出的不可约表示是等价的，不同杨图 的杨盘给出的不可约表示是不等价的。 证明： 对Sn，群代数RSn是群元素算符的不变空间，对应的表示是正则表示，现在我们 考虑左正则表示L(g)，g ∈Sn。 378 设有两个杨盘T与T′，杨算符分别是E ̂(T)、E ̂(T′)，它们是本质幂等元，对应的不 可约表示为A(g)、A′(g)，表示空间为W = RSnE ̂(T)、W′ = RSnE ̂(T′)，群空间向它 们的表示空间得投影算符是P ̂、P′ ̂。设E ̂2(T) = θE ̂(T)、E ̂2(T′) = θ′E ̂(T′)，取其相 应幂等元为e ⃗ ⃗= θ−1E ̂(T)、e ⃗ ⃗′ = θ′−1E ̂(T′)。 第一步， 我们需要先说明： 杨算符E ̂(T)、 E ̂(T′)所对应的不可约表示等价的充要条 件是至少存在一个群代数RSn中的元素c ⃗，使得E ̂(T)c ⃗E ̂(T′) ≠0。 先看充分性，由至少存在一个群代数RSn中的元素c ⃗，使得E ̂(T)c ⃗E ̂(T′) ≠0，来推 E ̂(T)、E ̂(T′)所对应的不可约表示等价。 若E ̂(T)c ⃗E ̂(T′) ≠0，可定义映射P ̂′′：W →W′，操作规则是对∀w ⃗ ⃗ ⃗⃗∈W， P ̂′′w ⃗ ⃗ ⃗⃗= w ⃗ ⃗ ⃗⃗e ⃗ ⃗c ⃗e ⃗ ⃗′。这里，由于w ⃗ ⃗ ⃗⃗e ⃗ ⃗c ⃗e ⃗ ⃗′ = P′ ̂(w ⃗ ⃗ ⃗⃗e ⃗ ⃗c ⃗) ∈W′，所以定义的P ̂′′作用到W中向量时，得 到的新的向量属于W′。我们需要证明P ̂′′是从W到W′的一一满映射。 这个证明很简单。 先定义P ̂′′作用到W上得到的向量的合集是W′′， W′′是W′的子集， 只要证明它非空，且群不变，就可以利用W′是不可约表示的表示空间得到W′′等 于W′。 这个是第一步满映射， 之后再证明是单射， 两者结合， 就是一一满映射了。 具体过程先看满映射， W′′ = P ̂′′W = {w ⃗ ⃗ ⃗⃗e ⃗ ⃗c ⃗e ⃗ ⃗′|w ⃗ ⃗ ⃗⃗∈W}， 对∀w ⃗ ⃗ ⃗⃗′′ = w ⃗ ⃗ ⃗⃗e ⃗ ⃗c ⃗e ⃗ ⃗′ ∈W′′， 有： L(g)w ⃗ ⃗ ⃗⃗′′ = gw ⃗ ⃗ ⃗⃗′′ = gw ⃗ ⃗ ⃗⃗e ⃗ ⃗c ⃗e ⃗ ⃗′ = (L(g)w ⃗ ⃗ ⃗⃗)e ⃗ ⃗c ⃗e ⃗ ⃗′ ∈W′′ 所以W′′是群不变的子空间。同时对e ⃗ ⃗∈W，有e ⃗ ⃗e ⃗ ⃗c ⃗e ⃗ ⃗′ = e ⃗ ⃗c ⃗e ⃗ ⃗′ ≠0，所以W′′非空。 结合W′′是不可约表示的表示空间，W′′ = W′。满映射成立。 单射：若不同w ⃗ ⃗ ⃗⃗1、w ⃗ ⃗ ⃗⃗2 ∈W对应P ̂′′w ⃗ ⃗ ⃗⃗1 = P ̂′′w ⃗ ⃗ ⃗⃗2，则P ̂′′(w ⃗ ⃗ ⃗⃗1 −w ⃗ ⃗ ⃗⃗2) = 0，进而(w ⃗ ⃗ ⃗⃗1 − w ⃗ ⃗ ⃗⃗2)e ⃗ ⃗c ⃗e ⃗ ⃗′ = 0 。由于w ⃗ ⃗ ⃗⃗1 −w ⃗ ⃗ ⃗⃗2 不为零，所以e ⃗ ⃗c ⃗e ⃗ ⃗′ = 0 ，与已知矛盾（已知是 E ̂(T)c ⃗E ̂(T′) ≠0，因此c ⃗≠0、e ⃗ ⃗c ⃗e ⃗ ⃗′ ≠0） 。因此P ̂′′不光是满映射，还是单射。存在 逆P ̂′′−1。 这个时候，因为对∀w ⃗ ⃗ ⃗⃗∈W，有: P ̂′′L(g)w ⃗ ⃗ ⃗⃗= P ̂′′(gw ⃗ ⃗ ⃗⃗) = (gw ⃗ ⃗ ⃗⃗)e ⃗ ⃗c ⃗e ⃗ ⃗′ = g(w ⃗ ⃗ ⃗⃗e ⃗ ⃗c ⃗e ⃗ ⃗′) = L(g)P ̂′′w ⃗ ⃗ ⃗⃗ 左边，L(g)是作用在w ⃗ ⃗ ⃗⃗的线性空间W的；右边，L(g)是作用在P ̂′′w ⃗ ⃗ ⃗⃗的线性空间W′ 的。同时这个等式对∀w ⃗ ⃗ ⃗⃗∈W成立。因此，写成表示的形式，就有： P ̂′′A(g) = A′(g)P ̂′′ 而P ̂′′−1存在，所以A′(g) = P ̂′′A(g)P ̂′′−1，A与A′等价。充分性得证。 必要性， 由A与A′等价来证至少存在一个群代数RSn中的元素c ⃗， 使得E ̂(T)c ⃗E ̂(T′) ≠ 0。既然等价，一定存在P ̂′′，使得对∀g ∈Sn，∀w ⃗ ⃗ ⃗⃗∈W，有： P ̂′′A(g)w ⃗ ⃗ ⃗⃗= A′(g)P ̂′′w ⃗ ⃗ ⃗⃗ 即 P ̂′′gw ⃗ ⃗ ⃗⃗= gP ̂′′w ⃗ ⃗ ⃗⃗ 由于P ̂′′为非奇异线性算符， 可通过线性组合使得对∀x ⃗ ⃗∈RSn， 有P ̂′′x ⃗ ⃗w ⃗ ⃗ ⃗⃗= x ⃗ ⃗P ̂′′w ⃗ ⃗ ⃗⃗。 定义c ⃗= P ̂′′e ⃗ ⃗，由于P ̂′′为非奇异线性算符，所以c ⃗≠0。对这样定义的c ⃗，有： c ⃗= P ̂′′e ⃗ ⃗e ⃗ ⃗= e ⃗ ⃗P ̂′′e ⃗ ⃗= e ⃗ ⃗c ⃗ 同时由于c ⃗∈W′，还存在c ⃗e ⃗ ⃗′ = P ̂′c ⃗= c ⃗，进而： c ⃗= e ⃗ ⃗c ⃗e ⃗ ⃗′ 由于c ⃗≠0，所以e ⃗ ⃗c ⃗e ⃗ ⃗′ ≠0，进而E ̂(T)c ⃗E ̂(T′) ≠0。 现在是证明了杨算符E ̂(T)、E ̂(T′)所对应的不可约表示等价的充要条件是至少存 在一个群代数RSn 中的元素c ⃗，使得E ̂(T)c ⃗E ̂(T′) ≠0 。下面看杨盘T 与T′ 属于与不属于同一个杨图时，会发生什么事情？ 当杨盘T与T′属于同一个杨图时，由前面的讨论，必存在r ∈Sn，且r ⃗≠0，使得 T′ = rT ，进而E ̂(T′) = rE ̂(T)r−1。 此时，有r−1 ∈RSn ，使得E ̂(T)r−1E ̂(T′) = r−1E ̂2(T′) = θr−1E ̂(T′) 。这个 E ̂(T)r−1E ̂(T′) ≠0，因为不然的话就会有E ̂(T′) = 0。正文中，定义6.5 讨论过， 杨算符一定不为零。 380 这样的话由上半部分的讨论，结合E ̂(T)r−1E ̂(T′) ≠0，就知道E ̂(T)、E ̂(T′)所对应 的不可约表示等价。 另一种情况， 就是T与T′属于不同杨图[λ]、[λ′]。 不同的两个杨图可以通过之前的 讨论定义大小，不失一般性，取[λ] > [λ′]。这样的话对∀s ∈Sn，我们知道杨盘sT 的杨算符是sE ̂(T)s−1。 这个时候，由引理6.4，不同杨图的杨盘对应的杨算符满足E ̂(T′)sE ̂(T)s−1 = 0。 两边乘上s，有E ̂(T′)sE ̂(T) = 0。这个时候，由s的一般性，可知对∀x ⃗ ⃗∈RSn，都有 E ̂(T′)x ⃗ ⃗E ̂(T) = 0。 这样， 同样结合上面的讨论， 知道T与T′对应的不可约表示相互 之间不等价。 这七个引理结合在一起，就给出了杨盘定理的全部内容。
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--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | | | | --- --- | | | | | | | | | | | | | | | | | | --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Complex Components Abs[z] | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | Introduction to the complex components The study of complex numbers and their characteristics has a long history. It all started with questions about how to understand and interpret the solution of the simple quadratic equation . It was clear that . But it was not clear how to get –1 from something squared. This problem was intensively discussed in the 16th, 17th, and 18th centuries. As a result, mathematicians proposed a special symbol—the imaginary unit , which is represented by : | | | | L. Euler (1755) introduced the word "complex" (1777) and first used the letter for denoting . Later, C. F. Gauss (1831) introduced the name "imaginary unit" for . Accordingly, and and the above quadratic equation has two solutions as is expected for a quadratic polynomial: | | | | The imaginary unit was interpreted in a geometrical sense as the point with coordinates in the Cartesian (Euclidean) ,‐plane with the vertical -axis upward and the origin . This geometric interpretation established the following representations of the complex number through two real numbers and as: | | | | | | | | where is the distance between points and , and is the angle between the line connecting the points and and the positive -axis direction (the so-called polar representation). The last formula lead to the basic relations: | | | | | | | | | | | | | | | | which describe the main characteristics of the complex number —the so-called modulus (absolute value) , the real part , the imaginary part , and the argument . A new era in the theory of complex numbers and functions of complex arguments (analytic functions) arose from the investigations of L. Euler (1727, 1728). In a letter to Goldbach (1731) L. Euler introduced the notation ⅇ for the base of the natural logarithm ⅇ⩵2.71828182…, and he proved that ⅇ is irrational. Later on L. Euler (1740–1748) found a series expansion for , which lead to the famous very basic formula, connecting exponential and trigonometric functions: | | | | This is known as the Euler formula (although it was already derived by R. Cotes in 1714). The Euler formula allows presentation of the complex number , using polar coordinates in the more compact form: | | | | It also expressed the logarithm of complex numbers through the formula: | | | | Taking into account that the cosine and sine have period , it follows that has period : | | | | Generically, the logarithm function is the multivalued function: | | | | For specifying just one value for the logarithm and one value of the argument φ for a given complex number , the restriction π < φ ≤ π for the argument φ is generally used. During the 18th and 19th centuries many mathematicians worked on building the theory of the functions of complex variables, which was called the theory of analytic functions. Today this is a widely used theory, not only for the above‐mentioned four complex components (absolute value, argument, real and imaginary parts), but for complimentary characteristics of a complex number such as the conjugate complex number and the signum (sign) . J. R. Argand (1806, 1814) introduced the word "module" for the absolute value, and A. L. Cauchy (1821) was the first to use the word "conjugate" for complex numbers in the modern sense. Later K. Weierstrass (1841) introduced the notation ❘z❘ for the absolute value. It was shown that the set of complex numbers and the set of real numbers have basic properties in common—they both are fields because they satisfy so-called field axioms. Complex and real numbers exhibit commutativity under addition and multiplication described by the formulas: | | | | | | | | Complex and real numbers also have associativity under addition and multiplication described by the formulas: | | | | | | | | and distributivity described by the formulas: | | | | | | | | (The set of rational numbers also satisfies all of the previous field axioms and is also a field. This set is countable, which means that each rational number can be numerated and placed in a definite position with a corresponding integer number . But the set of rational numbers does not include so-called irrational numbers like or . The set of irrational numbers is much larger and cannot be numerated. The sets of all real and complex numbers form uncountable sets.) The great success and achievements of the complex number theory stimulated attempts to introduce not only the imaginary unit in the Cartesian (Euclidean) plane , but a similar special third unit in Cartesian (Euclidean) three-dimensional space , which can be used for building a similar theory of (hyper)complex numbers : | | | | Unfortunately, such an attempt fails to fulfill the field axioms. Further generalizations to build the so‐called quaternions and octonions are needed to obtain mathematically interesting and rich objects. Definitions of complex components The complex components include six basic characteristics describing complex numbersabsolute value (modulus) , argument (phase) , real part , imaginary part , complex conjugate , and sign function (signum) . It is impossible to define real and imaginary parts of the complex number through other functions or complex characteristics. They are too basic, so their symbols can be described by simple sentences, for example, " gives the real part of the number ," and " gives the imaginary part of the number ." All other complex components are defined by the following formulas: | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Geometrically, the absolute value (or modulus) of a complex number is the Euclidean distance from to the origin, which can also be described by the formula: | | | | Geometrically, the argument of a complex number is the phase angle (in radians) that the line from 0 to makes with the positive real axis. So, the complex number can be presented by the formulas: | | | | | | | | | | | | Geometrically, the real part of a complex number is the projection of the complex point on the real axis. So, the real part of the complex number can be presented by the formulas: | | | | | | | | Geometrically, the imaginary part of a complex number is the projection of complex point on the imaginary axis. So, the imaginary part of the complex number can be presented by the formulas: | | | | | | | | Geometrically, the complex conjugate of a complex number is the complex point , which is symmetrical to with respect to the real axis. So, the conjugate value of the complex number can be presented by the formulas: | | | | | | | | Geometrically, the sign function (signum) is the complex point that lays on the intersection of the unit circle and the line from 0 to (if ). So, the conjugate value of the complex number can be presented by the formulas: | | | | | | | | A quick look at the complex components Here is a quick look at the graphics for the complex components of the complex components over the complex ‐plane. The empty graphic indicates that the function value is not real. | | | | | | | | | | | | | | | | | | | | | | | | Connections within the group of complex components and with other function groups Representations through more general functions All six complex component functions , , , , , and cannot be easily represented by more generalized functions because most of them are analytic functions of their arguments. But sometimes such representations can be found through Meijer G functions, for example: | | | | Representations through other functions All six complex components , , , , , and satisfy numerous internal relations of the type , where and are different complex components and is a basic arithmetic operation or (composition of) elementary functions. The most important of these relations are represented in the following table: | | | | Other internal relations between complex components of the type , where and are different complex components that also exist. Some of them are shown here: | | | | Here are some more formulas of the last type: | | | | | | | | | | | | | | | | | | | | | | | | | | | | (here is the Heaviside theta function, also called the unit step function). The first table can be rewritten using the notations : | | | | The best-known properties and formulas for complex components Real values for real arguments For real values of argument , the values of all six complex components , , , , , and are real. Simple values at zero The six complex components , , , , , and have the following values for the argument : | | | | | | | | | | | | | | | | | | | | | | | | is not a uniquely defined number. Depending on the argument of , the limit can take any value in the interval . Specific values for specialized variable The six complex components , , , , , and have the following values for some concrete numeric arguments: | | | | Restricted arguments have the following formulas for the six complex components , , , , , and : | | | | The values of complex components , , , , , and at any infinity can be described through the following: | | | | Analyticity All six complex components , , , , , and are not analytical functions. None of them fulfills the Cauchy–Riemann conditions and as such the value of the derivative depends on the direction. The functions , , , and are real‐analytic functions of the variable (except, maybe, ). The real and the imaginary parts of and are real‐analytic functions of the variable . Sets of discontinuity The four complex components , , , and are continuous functions in . The function has discontinuity at point . The function is a single‐valued, continuous function on the ‐plane cut along the interval , where it is continuous from above. Its behavior can be described by the following formulas: | | | | | | | | Periodicity All six complex components , , , , , and do not have any periodicity. Parity and symmetry All six complex components , , , , , and have mirror symmetry: | | | | | | | | | | | | The absolute value is an even function. The four complex components , , , and are odd functions. The argument is an odd function for almost all : | | | | | | | | | | | | | | | | | | | | Homogeneity The six complex components , , , , , and have the following homogeneity properties: | | | | | | | | | | | | | | | | | | | | | | | | Scale symmetry Some complex components have scale symmetry: | | | | | | | | Series representations The functions and with real have the following series expansions near point : | | | | | | | | | | | | | | | | | | | | | | | | Integral representations The function with real has the following contour integral representation: | | | | Limit representations The functions and with real have the following limit representations: | | | | | | | | | | | | The last two representations are sometimes called generalized Padé approximations. Transformations The values of all complex components , , , , , and at the points , , –ⅈ z, , , , , and are given by the following identities: | | | | | | | | The values of all complex components , , , , , and at the points , , and are described by the following table: | | | | Some complex components can be easily evaluated in more general cases of the points including symbolic sums and products of , , for example: | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | The previous tables and formulas can be modified or simplified for particular cases when some variables become real or satisfy special restrictions, for example: | | | | | | | | | | | | Taking into account that complex components have numerous representations through other complex components and elementary functions such as the logarithm, exponential function, or the inverse tangent function, all of the previous formulas can be transformed into different equivalent forms. Here are some of the resulting formulas for the power function : | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Similar identities can be derived for the exponent functions, such as: | | | | | | | | | | | | Some arithmetical operations involving complex components or elementary functions of complex components are: | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Complex characteristics The next two tables describe all the complex components applied to all complex components , , , , , and at the points and : | | | | | | | | Differentiation The derivatives of five complex components , , , , and at the real point can be interpreted in a real‐analytic or distributional sense and are given by the following formulas: | | | | | | | | | | | | | | | | | | | | where is the Dirac delta function. It is impossible to make a classical, direction-independent interpretation of these derivatives for complex values of variable because the complex components do not fulfill the Cauchy-Riemann conditions. Indefinite integration The indefinite integrals of some complex components at the real point can be represented by the following formulas: | | | | | | | | Definite integration The definite integrals of some complex components in the complex plane can also be represented through complex components, for example: | | | | | | | | | | | | | | | | | | | | Some definite integrals including absolute values can be easily evaluated, for example (in the Hadamard sense of integration, the next identity is correct for all complex values of ): | | | | Integral transforms Fourier integral transforms of the absolute value and signum functions and can be evaluated through generalized functions: | | | | | | | | | | | | | | | | | | | | | | | | Laplace integral transforms of these functions can be evaluated in a classical sense and have the following values: | | | | | | | | Differential equations The absolute value function for real satisfies the following simple first-order differential equation understandable in a distributional sense: | | | | In a similar manner: | | | | | | | | | | | | All six complex components , , , , and satisfy numerous inequalities. The best known are so-called triangle inequalities for absolute values: | | | | | | | | Some other inequalities can be described by the following formulas: | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | The six complex components , , , , , and have the set of zeros described by the following formulas: | | | | | | | | | | | | | | | | | | | | | | | | Applications of complex components All six complex components are used throughout mathematics, the exact sciences, and engineering. | | | | | | | | | | |
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https://www.nature.com/articles/s41598-025-11308-z
Peculiar proteome of dark-cultivated Euglena gracilis | Scientific Reports Your privacy, your choice We use essential cookies to make sure the site can function. We also use optional cookies for advertising, personalisation of content, usage analysis, and social media. By accepting optional cookies, you consent to the processing of your personal data - including transfers to third parties. Some third parties are outside of the European Economic Area, with varying standards of data protection. See our privacy policy for more information on the use of your personal data. Manage preferences for further information and to change your choices. Accept all cookies Skip to main content Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). 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Advertisement View all journals Search Search Search articles by subject, keyword or author Show results from Search Advanced search Quick links Explore articles by subject Find a job Guide to authors Editorial policies Log in Explore content Explore content Research articles News & Comment Collections Subjects Follow us on Facebook Follow us on Twitter Sign up for alerts RSS feed About the journal About the journal About Scientific Reports Contact Journal policies Guide to referees Calls for Papers Editor's Choice Journal highlights Open Access Fees and Funding Publish with us Publish with us For authors Language editing services Open access funding Submit manuscript Sign up for alerts RSS feed nature scientific reports articles article Peculiar proteome of dark-cultivated Euglena gracilis Download PDF Download PDF Article Open access Published: 16 July 2025 Peculiar proteome of dark-cultivated Euglena gracilis Adriana Paprčková1na1, Katarína Klubicová2na1, Eva Ürgeová1, Maksym Danchenko2, Peter Baráth3, Olha Lakhneko2, Juraj Krajčovič1& … Ľubica Uváčková1 Show authors Scientific Reportsvolume 15, Article number:25721 (2025) Cite this article 1224 Accesses Metrics details Abstract Euglena gracilis is a flagellate photosynthetic microalga that, thanks to its metabolic adaptability, can grow under both autotrophic and heterotrophic conditions. This adaptability makes euglena an interesting species for applied biotechnology. We focused on the proteome of E. gracilis cultivated in Cramer-Myers medium (supplemented with ethanol) in dark and light conditions. Cultures grown in the light showed a characteristic green coloration, while cultures incubated in the dark were bright yellow. When cultured in the dark, microalga showed reduced concentration of chlorophylls (a,b, and total) and carotenoids compared to cells cultured in the light. Conversely, there was an increase in proline content in the dark compared to light cultivation. Using proteomic approach, we revealed 162 differentially accumulated proteins in light- and dark-grown cells classified into 12 functional groups. Notably, alterations in the metabolism of fatty acids and amino acids, secondary metabolism, and accumulation of stress- and detoxification-related proteins in microalgal cells cultivated in darkness with ethanol as a carbon source may help euglena adapt to these conditions. Based on our results and literature, we hypothesize that vitamin B12 potentially plays an important role in light/dark metabolic switch, similarly as in bacteria. Similar content being viewed by others New insights into raceway cultivation of Euglena gracilis under long-term semi-continuous nitrogen starvation Article Open access 02 May 2023 The effect of light intensity on microalgae biofilm structures and physiology under continuous illumination Article Open access 11 January 2024 Light cues drive community-wide transcriptional shifts in the hypersaline South Bay Salt Works Article Open access 17 March 2025 Introduction The phylum Euglenozoa represents a widely branched group of unicellular eukaryotic protists1."). The order Euglenida includes a species-rich group of individually living flagellates characterized by diverse and even unique properties. Among the characteristic features of these protists is the presence of a pellicle composed of parallel protein bands and microtubules, which are located under the cytoplasmic membrane. They ensure cellular plasticity during movement. Type species _Euglena gracilis_ (thereupon euglena) is one of the most highly studied eukaryotes because of the ease it can be cultured and its high potential for biotechnology2."). Inside the cytosol, euglena has a photosensitive receptor—stigma (eyespot) containing derivatives of β-carotene and other carotenoid pigments, which filter light and concentrate it on the paraflagellar body while participating in phototaxis. It reacts to the direction and intensity of light using a photoreceptor located inside the dorsal flagellum38 "). The adaptability of the metabolism allows it to grow in different cultivation conditions, such as autotrophy (using sunlight), heterotrophy (using an external carbon source), and mixotrophy (combining both modes)4."). Euglena is metabolically versatile thanks to its genome, encoding numerous enzymes and enabling it to utilize various organic compounds as a source of carbon and energy. Most of metabolically active enzymes are regulated post-transcriptionally, allowing cells to adapt quickly to environmental changes2."),5."). Euglena cells change their shape in response to different nutrition and growth conditions. In light-adapted cells, mitochondria are more randomly distributed in the cytosol, with a tendency to localize near the plastids, which may be related to local oxygen production in the plastids62 ."). Chloroplasts are easily lost and become permanently bleached under various cultivation conditions. In dark-adapted euglena, the network of mitochondria is located near the surface of the cell and is adjacent to other organelles6 (Springer, 2017). ."). These cells contain also undeveloped proplastids, and, after 3 days of exposure to light, they again become developed green chloroplasts7."). In different cultivation conditions, euglena shows various levels of light harvesting and photosynthesis, which is related to its metabolic pathway. Euglena grown in the light shows an elevated level of pyruvate and a low level of UDP-glucose, glycerate 1-3-bisphosphate, 2-phosphoglycerate, and ATP compared to microalga cultivated in the dark. Euglena grows much faster in the light (mixotrophy) than in the dark (heterotrophy)4."),8.") and even faster than under autotrophic conditions due to the limited amount of CO 29."). Furthermore, microalgae show strong adaptability to various abiotic stresses, constantly fine-tuning their cellular mechanisms to cope with adverse conditions and achieve homeostasis. They accumulate stress metabolites that are closely related to the changes in their metabolic pathways10."). Light is the most critical environmental factor for the growth of microalgae. High light intensity can significantly increase lipid accumulation in microalgae11."). One of the first metabolic responses to abiotic stress is inhibition of growth and protein synthesis. As stress becomes more severe, energy metabolism (sugars, lipids, photosynthesis) is affected. Thus, in response to stress, gradual complex changes occur in the metabolism12."). Polar lipids play an essential role in the structure and fluidity of membranes and are modified by changes in their environment. If the environmental stress acting on the microalgae is mild, only the length of the fatty acid chains and the degree of their deposition will change. On the other hand, if microalgae are exposed to many stresses, the ratio of various polar lipids is modified with higher energy requirements136 ."). Chlorophyll a is the most widespread pigment in photosynthetic organisms. Chlorophyll b is the second most abundant pigment in eukaryotic microalgae and plants. In addition to Chl a and Chl b, other types of Chls are also known in microalgae c, d, and f. An enhanced Chl a/Chl b ratio is associated with higher photosynthetic activity and faster cell growth. In unfavorable conditions, the Chl content decreases14."). Both low light intensity and too high light intensity can have a limiting effect on the growth of microalgae15."). Light intensity directly affects the rate of photosynthesis. With excessive illumination, microalgae can experience photooxidative damage to the photosynthetic apparatus and photoinhibition—a reduction in the efficiency and speed of photosynthesis16."). During photoinhibition, increased production of reactive oxygen species (ROS) can occur in cells. In tiny amounts, they fulfill a signaling function and trigger various anti-stress mechanisms, but their high concentrations lead to damage to cellular structures and the transformation of physiological and biochemical processes. As a result of the stress response, there are changes in the composition and ratio of Chls, carotenoids, lipids, and fatty acids17."). Erickson et al.16."). reported that high cell density prevents light penetration, which, in turn, reduces the intensity of photosynthesis. In photosynthetic organisms, carotenoids participate in light harvesting and prevent photooxidative damage18."). Carotenoids are divided into carotenes, oxygen-free pigments, and xanthophylls, oxygen-containing pigments19."). The most common carotenoids in microalgae are β-carotene and xanthophylls—lutein, violaxanthin, and zeaxanthin17."). Proline is a multifunctional amino acid and an effective osmoprotectant, which can have a significant role not only in plant development but also in responses to biotic and abiotic stress20."). The amount of proline reflects a certain physiological state through which we can assess the resilience of plants to stress. It helps microalgae to cope with stress conditions by adjusting the cellular osmotic pressure. Proline contributes to ROS detoxification, membrane integrity, and protein stabilization21."). Free proline, as well as its end groups in polypeptides, can directly react with hydrogen peroxide and singlet oxygen, neutralizing them. The antioxidant effect of proline is also related to its ability to protect protein-lipid complexes of membranes and indirectly reduce the intensity of lipid peroxidation by inactivating hydroxyl radicals and other ROS. During stress, proline begins to accumulate, its biosynthesis increases, and degradation decreases22."). The transcriptome datasets, coupled with a draft proteome of E. gracilis grown in light and dark conditions, have been generated recently at the descriptive level23."). The main goal of our work was to monitor the effect of light and dark cultivation of euglena with ethanol as a carbon source on the proteome. This approach can provide a global perspective on the comparison of mixotrophic versus heterotrophic culturing modes. We complemented proteomic data by determining the photosynthetic pigments and proline concentration (stress marker in plants). Materials and methods Experimental material and cultivation conditions Euglena gracilis strain Z (Pringsheim strain Z, SAG 1224-5/25 Collection of Algae, Göttingen, Germany) was used in this study. Approximately 2 × 10 6 cells were diluted in Erlenmeyer flasks containing 50 mL of liquid Cramer and Myers (CM) medium24.") supplemented with ethanol (final concentration 0.8%25.")), and pH adjusted to 4.9. The cultures were statically cultivated for 7 days in a cultivation chamber at 23°C under constant light at an illumination intensity of 30 µmol photons∙m−2∙s−1 and the second part of the cultures was grown under the same conditions in the dark. After 7 days of cultivation, the cells were pelleted by centrifugation at 3350 × _g_, 10 min at 24°C, and stored at −80°C until analysis. Light microscopy All culture samples were observed under a binocular biological microscope CX23 (Olympus, Hamburg, Germany) at 400-fold magnification. We used 100 µL of formaldehyde to immobilize the cells. Likewise, to observe and compare changes in cellular motility, we observed cells without the addition of formaldehyde in both types of cultures. Protein extraction and digestion Proteins were extracted in five independent experiments from both samples (dark and light), according to26."). Briefly, frozen cells were ground in liquid nitrogen with mortar and pestle. Then 5 mL of extraction buffer (0.1 M Tris–HCl pH 8.8; 10 mM EDTA; 0.9 M sucrose), 20 µL of 2-mercaptoethanol, and 5 mL of basic phenol, pH 8.8 were added. The mixture was placed on a swinging shaker, stirred for 30 min at 4°C, and then centrifuged at 6000 × _g_ at 4°C for 15 min. Proteins from the phenol phase were precipitated using five volumes of ice-cold 0.1 M ammonium acetate in 100% methanol and incubated overnight at − 20°C. The precipitate was collected by centrifugation for 20 min, 4000 × _g_ at 4°C. Finally, the pellet was washed 2 times with 0.1 M ammonium acetate in methanol, 2 times with ice-cold 80% acetone, and once with 70% ethanol. Protein digestion with trypsin on beads was performed according to27."). Protein samples were solubilized in SDT buffer (4% sodium dodecyl sulfate, SDS; 0.1 M dithiothreitol; 0.1 M Tris-HCl, pH 7.6) at 1000 rpm mixing, room temperature. Protein concentration was measured using Pierce reducing agent-compatible BCA assay (Thermo Scientific, Waltham, MA USA). 60 µg of protein samples were alkylated with 20 mM iodoacetamide in the dark for 30 min at 24°C and 600 rpm on thermoshaker TS-100 C (Biosan, Riga, Latvia) and quenched with 5 mM dithiothreitol. Next, the mixture (1:1) of beads was prepared: Sera-Mag Carboxylate-Modified Beads (Cytiva, Marlborough, MA, USA), hydrophilic solids 50 µg/µL, and hydrophobic solids 50 µg/µL. On the magnetic rack MagneSphere (Promega, Madison, WI, USA), the beads were pelleted for 1 min, and the supernatant was removed by pipette. Off the rack, the beads were reconstituted in bidistilled water and pipette mixed. The prepared stock of beads was 20 µg/µL. Following, 30 µL (60 µg) of the bead stock was added to the whole volume of the alkylated/quenched protein sample and pipette-mixed to homogenize the beads and lysate. For binding, 70 µL of 100% ethanol was added to the suspension to achieve 50% final ethanol concentration. The tubes were incubated in the thermoshaker for 10 min at 24°C and 1000 rpm. On the magnetic rack, the beads were pelleted for 2 min; then, the supernatant was removed and discarded by pipette. Bond proteins were washed 5 times with 140 µL of fresh 80% ethanol. For digestion, 10 µL of 0.1 µg/µL trypsin (Promega) and 50 µL 50 mM ammonium bicarbonate were added, making the final digestion volume 60 µL. The samples were incubated in the thermoshaker for 18 h at 37°C and 1000 rpm. On the magnetic rack, the beads were pelleted for 2 min, and the supernatant was recovered into a fresh 2 mL tube, followed by a second elution with 60 µL of 50 mM ammonium bicarbonate. The peptide concentration was measured with a microvolume spectrophotometer DS-11 FX+ (Denovix, Wilmington, DE, USA). Identification and quantification of proteins by mass spectrometry Quantitative proteome analysis was done as described earlier28.") 500 ng of purified peptides per sample was loaded onto a trap column (PepMap100 C18, 300 μm × 5 mm, 2-µm particle size, Dionex, Germering, Germany) and separated with an EASY-Spray PepMap RSLC C18 analytical column having integrated nanospray emitter (75 μm × 500 mm, 5-µm particle size) (Thermo Scientific) on Ultimate 3000 RSLCnano system (Dionex) in a 120-min gradient (3–43% of 80% acetonitrile with 0.1% formic acid), and flow rate 250 nL/min. Eluted peptides were sprayed into Orbitrap Elite mass spectrometer (Thermo Scientific), equipped with EASY-Spray ion source, and spectral datasets were collected in the data-dependent mode using Top15 strategy to select precursor ions. Precursors were measured in the mass range 300–1700 m/z with a resolution 120,000, and fragments were obtained by the HCD mechanism with a normalized collision energy 25 and resolution 15,000. Obtained datasets were processed by MaxQuant v1.6.17.0 with a built-in Andromeda search engine and the following parameters: (i) carbamidomethylation cysteine as permanent and oxidation methionine as variable modifications; (ii) 20 ppm peptide tolerance in the first search, 4.5 ppm in the main search upon recalibration, and 20 ppm fragment tolerance; (iii) 1% peptide and protein false discovery rates based on reverse decoy database search; (iv) match between the runs and label-free quantification (LFQ intensities). The search was performed against euglena translated transcriptome referenced in23."), available in the supporting proteomic dataset at ebi.ac.uk/pride/archive/projects/PXD009998 (36,526 sequences). The statistical analysis was performed using Perseus v1.6.15.0. Output protein Group table from MaxQuant was filtered for reverse proteins, contaminants, and low-confidence proteins. After the log 2 transformation of the LFQ intensities, only proteins with 4 or more valid values out of 5 biological replicates in any experimental group were retained. Consequently, the missing values were imputed from the normal distribution. Principal component analysis (PCA) was used to evaluate sources of variability among samples and replicates. Next, the Student’s test was performed with permutation correction for multiple testing with a q-value threshold at 0.01. Protein sequences were functionally annotated using BLASTP with an e-value cut-off at 2 × 10−4 querying NCBI non-redundant protein database. Not annotated proteins were additionally searched for predicted protein domains and families using InterPro (ebi.ac.uk/interpro/)29."). We used eggNOG-mapper for functional enrichment30.") and KEGG-mapper to obtain metabolic pathway map31."). Determination of the concentration of photosynthetic pigments and proline The concentration of photosynthetic pigments was determined in 5 independent replicates from each culture (light and dark). From 1 mL of sample, cells were collected by centrifugation at 5800 × g for 5 min and intensively mixed for 2 min in 1mL of 80% (v/v) acetone. After centrifugation at 10,000 × g for 5 min, the concentration of photosynthetic pigments, chlorophylls and carotenoids, was determined spectrophotometrically (Spectroquant UV/VIS Spectrophotometer Pharo 300, Merck, Darmstadt, Germany) at three wavelengths 470 nm, 645 nm, and 662 nm. The concentrations were calculated according to32."). The proline concentration was determined according to33.") in 3 replicates. We added 1.5 mL of 95% ethanol to each frozen cell sample and thoroughly homogenized with mortar and pestle. After centrifugation for 10 min at 2650 × _g_ room temperature, we added a reaction mixture containing 1% ninhydrin, 20% ethanol, and 60% acetic acid. After intensive mixing for 1 min and incubation at 95°C for 20 min, we determined the concentration of proline spectrophotometrically based on the calibration curve by measuring the absorbance at 520 nm. Statistical analysis was done in Excel (Microsoft) using Student’s test with a p-value threshold at 0.05. Results and discussion Different phenotypes in light versus darkness As a result of the changed cultivation conditions (light versus darkness), we observed the growth of euglena cells in both types of cell cultures but with different coloring. Microalgal cells cultivated in the light showed a typical green coloration, while euglena cells grown in the dark were bright, without the typical green coloration (Fig.1A-B). Green microalgae absorb light energy through the main photosynthetic pigment chlorophyll a (b) in the range of 450–475 nm and 630–675 nm34."). Fig. 1 Phenotype of Euglena gracilis. Cultures in light (A) and dark (B) conditions after 7 days of incubation in Cramer-Meyers medium. Cells cultured in the light showed a characteristic green color, while cells cultured in the dark were yellowish. Microscopy images with 400 × magnification in light (C) and dark (D) cultivation conditions. Cells cultured in the dark changed shape from oblong to spherical. Full size image Based on microscopic observations, we can conclude, cells cultured in the dark moved significantly slower than those cultured in the light (data not shown). There were visible also significant changes in morphology. Cells cultured in the dark changed shape from oblong to spherical. In the light-grown sample, we noted a higher cellular density (Fig.1C-D). According to the biological clock, euglena changes its shape twice per day, enacting a circadian rhythm in cell shape. The mean cell length of the population increases to a maximum in the middle of the light period when photosynthetic capacity is highest, and then decreases for the remainder of the 24-hour period. The population becomes spherical by the end of the 24-hour period when the cycle reinitiates35."). Opposing content of photosynthetic pigments and proline In cells cultivated in the dark, the concentration of total Chl, Chls a/b, and carotenoids significantly decreased compared to microalgal cells cultivated in the light. Chlorophyll a was reduced at least 40-fold, Chl b 3.3-fold, total Chl 24-fold, and carotenoids 3.8-fold in euglena cells cultivated in the dark (Fig.2A-B). Photosynthetic pigments are essential compounds for energy absorption and conversion in phototrophic organs. An optimal amount of these pigments in the thylakoid membrane is necessary for a balanced energy state and vitality of phototrophs. Cultivation in the dark affected the efficiency of photosynthesis and, thus, cell survival. Azizullah et al.36."). reported that Chl _b_ is more resistant to environmental stresses in euglena than Chl _a_ and carotenoids. The ratio Chl _a_/Chl _b_ is relatively stable under standard conditions and usually around 3 in the plants37."). However, when plants are exposed to stress, there can be alterations in this ratio. In our experiments, the ratio Chl _a_/Chl _b_ (16.6) was influenced by (light) cultivation conditions. Fig. 2 Concentrations of chlorophylls n = 7 (A), carotenoids n = 7 (B), and proline n = 3 (C) in Euglena gracilis cultivated in light (light grey) and dark (black) conditions after 7 days of incubation in Cramer-Meyers medium. Full size image Middepogu et al.38.") observed a significant decrease in Chl _a_/Chl _b_ content in _Chlorella pyrenoidosa_ at different concentrations of titanium dioxide nanoparticles. Chlorophyll content decreased with increasing nanoparticle concentration. The shielding effect of nanoparticles can affect photosynthetic productivity due to the reduction of light penetration into algal cells39."). During cultivation in the dark, there is a significant decrease in the content of Chl in euglena cells. Subsequently, the cells acquire a yellow color, primarily attributed to the presence of carotenoids40."). We observed an increase in proline content in cultures maintained in the dark compared to cultures grown in the light (Fig.2C). In the case of cultivation in the light, the proline concentration was 5.6 µg/mL, and in the case of cultivation in the dark, 11.4 µg/mL. As a result of the changed culture conditions, there was a 2-fold increase in proline in cells cultured in the dark compared to the light. Also interesting are the results in Scenedesmus sp. exposed to high salt concentration, where a significant increase in proline content was recorded; however, decreased with prolonged exposure to stress, suggesting a possible dynamic regulation of its metabolism41."). Mitochondria in euglena are known to contain a higher level of proline42 247–314 (Academic, 1989)."). Mitochondrial proline catabolism is linked to oxidative respiration and provides energy for resuming growth after stress43."). Light-cultivated microalga showed more heterogenic proteome profile We identified 2449 proteins by mass spectrometry (Table S1). At the 99% confidence level (q-value is 0.01), we revealed 162 differentially abundant proteins (Table S2), classified into 12 functional groups: primary metabolism (23), energy (62), protein destination and storage (13), transcription (3), cell growth and division (4), protein synthesis (11), disease/defence (19), transporters (7), signal transduction (6), secondary metabolism (3), unclear characterization (3). Due to the genetic and metabolic specificity of euglena, some potentially important proteins still remain annotated/uncharacterized (8). Overall, 102 proteins accumulated in the light and 60 in the dark (Table S3). Based on the PCA analysis (Fig.3), 1st component explained 62% of the total variance in the data, clearly separating experimental groups. Samples cultivated in the dark showed high homogeneity of the overall protein profile. The group of samples cultivated in the light showed higher heterogeneity aligned with component 2, explaining only 12% of data variance. Fig. 3 Principal component analysis of quantified proteins showing separation and distribution of the microalgal samples cultivated in light (green, L1-L5) and dark (blue, D1-D5) conditions. The figure shows 5 independent samples for each variant. The first component, explaining a major share of the total data variance, aligned with different experimental conditions—cultivation in light versus darkness. The second component explained much lower proteome-level variance between biological replicates, particularly in light cultivation. Full size image Due to the high specificity of euglena genome, we used several tools for the functional annotation of proteins, first we used BLASTP search against the NCBI database (Table S3), InterPro for functional domains, and eggNOG-mapper (Table S4). Finally, we created metabolic maps based on KEGG annotations30."),31.") (Fig.4). Fig. 4 KEGG-reconstruct assisted visualization of proteomic map of fatty acid metabolism in dark and in light cultivated Euglena gracilis. Green arrows represent proteins identified in our study. Full size image The highest difference we observed in the functional groups related to the energy, metabolism of fatty acids and amino acids, defence and antioxidant mechanism (Fig.5). Fig. 5 Functional distribution of differentially abundant proteins accumulated in light (light grey) and dark (black) cultivated Euglena gracilis. Full size image When cultivated in the dark with oxygen availability, euglena switches to a heterotrophic way of life, where it relies on various organic substances from the environment as a source of carbon and energy instead of sunlight. In anaerobic or hypoxic conditions, euglena degraded and converted paramylon, the storage polysaccharide, into wax esters, process called wax ester fermentation443 . PMID: 28429326."). High variability was observed even in the level of ability to tolerate hypoxia; some strains can just survive, a few can grow heterotrophically even under anoxia in the dark45."). Dark cultivation caused the accumulation of numerous defense-related proteins in microalga We also observed the accumulation of proteins active in stress and detoxification in euglena cultured in the dark. The most represented proteins are 3 entries of glutathione-S-transferases and 2 glutathione transferase (super)family (Table S3), Hsp70 (2 entries more abundant in the dark). The regulation of ROS is essential for improving plant stress tolerance and is mediated by an antioxidant defense system consisting of several antioxidant enzymes and non-enzymatic antioxidants. Enzymatic antioxidants in microalgae include ascorbate peroxidase (APX), and glutathione-S-transferase. Non-enzymatic antioxidants consist of metabolites such as vitamin C (AsA), glutathione, carotenoids, phenolics, and proline46."). APX has the ability to synthesize AsA in plants and algae47."). Water-soluble thiol, glutathione, increases the tolerance of plants to various abiotic stresses, such as cultivation in the dark, and helps keeping ROS under control48."). Hsp70 (2 entries more abundant in the dark) is a vital chaperone protein involved in protecting cells from stress conditions and helping to maintain protein functions. It binds to damaged and normal proteins and protects them from aggregation and degradation. Hsp70 helps proteins to fold correctly during synthesis or after exposure to stress49 gene family in pumpkin (Cucurbita moschata) rootstock under drought stress suggested the potential role of these chaperones in stress tolerance. Int. J. Mol. Sci. 23, 1918. (2022)."). In the dark, we identified 7 stress response proteins, while in the light, only 2. Almost the same ratio of identified proteins between dark and light is in the subgroup detoxification, 8:2. Peroxiredoxins (1 accumulated in the dark and 1 in the light) are vital components of the cellular defense against oxidative stress. They catalyze the reduction of peroxides and reduce the level of ROS, thereby protecting and maintaining cellular homeostasis. They may be involved in various cellular processes, including the protection of cell membranes, the regulation of growth, and stress response50."). Fine details of ROS metabolism still remain unclear in euglena and should be a focus of deep investigation. Essential changes in amino acid and fatty acid metabolism caused by the absence of light Among the proteins accumulated in dark-cultured microalga, 8 proteins were involved in the metabolism of amino acids, fatty acids, and sugars (Table S3). We focus on discussing plausible functional implications of the modified amino acids and fatty acids metabolism and secondary metabolism. Amino acid metabolism is crucial for various cellular processes, including DNA and RNA synthesis, energy flows, and signaling pathways. Amino acids can be degraded to tricarboxylic acid cycle intermediates through oxidation. The metabolism of alanine, aspartic acid, and glutamate not only contributes to protein synthesis but also generates intermediates for various metabolic processes. Changes in the metabolism of these amino acids upon exposure to stress can have significant consequences for energy metabolism, redox balance, and production of key metabolites51."). Aspartate semialdehyde dehydrogenase, accumulated in dark-grown cells, catalyzes the conversion of aspartate semialdehyde to aspartate, a key step in the biosynthesis of the amino acids, lysine, and tyrosine. This enzyme is essential for maintaining the balance of amino acids in euglena cells. Another protein, more abundant in dark conditions, L-aspartate oxidase, catalyzes the conversion of L-aspartate to oxaloacetate in the presence of oxygen, releasing hydrogen52."). In green algal and plant cells, aromatic amino acids (phenylalanine, tyrosine, and tryptophan) are primarily produced in plastids, but the shikimate pathway functions in both chloroplasts and cytosol. The preferred pathway depends on the conditions of cultivation. The cytosolic pathway dominates when cultured in the dark, and the plastidic pathway is preferential when cultured in the light. Tryptophan is synthesized from chorismate by a series of reactions via anthranilate9."). In the dark, we revealed the accumulation of anthranilate phosphoribosyltransferase, which catalyzes the second step in tryptophan synthesis. It ensures the transfer of the phosphoribosyl group to anthranilate to form phosphoribosyl anthranilate. Tryptophan synthesis is essential for proper growth and physiological processes53."). Another essential process affected by dark cultivation is fatty acid metabolism (Fig.4). In dark cultivated euglena, accumulated proteins active in this process showed chloroplastic localization. In euglena grown in dark accumulated proteins active in fatty acid metabolism localized in mitochondria (L-3-hydroxyacyl-CoA dehydrogenase subunit precursor, acetyl-CoA acetyltransferase (2 entries), known also as thiolases). Mitochondria in euglena are very interesting organelles such as they are facultatively anaerobic and allow to generate ATP both with and without oxygen. One of five systems for fatty acid synthesis occurs in mitochondria – wax ester fermentation42. However, the enzymes that accumulated in dark are also active in the β-oxidation of fatty acids in euglena. Together with cytosolic acyl coenzyme A thioester hydrolase isoform X4, which also accumulated in dark, we found 3 of 4 enzymes active in this pathway. Therefore, we interpret that β-oxidation is vital for euglena grown in heterotrophic conditions. Fatty acid β-oxidation is an important catabolic process providing energy for cells. In the darkness, photosynthesis is not feasible, so cells must rely on alternative energy sources. Activation of acyl-CoA by the enzyme acyl-CoA synthetase initiates β-oxidation in the peroxisomal membrane. The basic reaction of β-oxidation of fatty acids requires a cyclic enzymatic reaction in four steps, namely oxidation, hydration, dehydrogenation, and cleavage of acetyl-CoA. The products of this pathway are acetyl-CoA, NADH, and FAD62 ."). Secondary metabolism potentially regulates light/dark metabolic switch We found enzymes active in secondary metabolism accumulated in euglena grown in the dark. Notably, we found a protein (EG_transcript_16043) that was not functionally annotated in the previous study23."). However, due to increased data availability, InterPro analysis now showed similarity of this sequence to corrinoid adenosyltransferase PduO-type. This enzyme, accumulated in cells grown in dark, catalyzes the conversion of cobalamin (vitamin B 12) into its coenzyme form, adenosylcobalamin (AdoB 12). This may lead to an increased level of available vitamin B 12 coenzyme form, which has various regulatory functions in bacteria. As a cofactor, it influences the activity of several enzymes such as isomerases, methyltransferases, and reductases54."). Adenosylcobalamin, as a part of a specific transcriptional repressor, is involved in down-regulation of a light-inducible promoter in _Myxococcus xynthus_55."). Finally, photoreceptors use adenosylcobalamin to sense light and mediate light-dependent gene regulation in _Thermus thermophilus_56."),57."). Therefore, we hypothesize that accumulation of this enzyme in dark-grown cells may represent one of mechanisms involved in light/dark metabolic switch in euglena. However, the exact mechanism should be examined. Finally, spermidine/spermine synthase, active in polyamine (spermidine and spermine) synthesis, also accumulated in dark-grown cells. Polyamines are crucial for cell survival and function. They play essential roles in various cellular processes such as cell growth, cell division, and adaptation to environmental conditions, especially through the protection of cells from oxidative stress585 (1981)."),59."). Several energy-related proteins accumulated in dark conditions During cultivation in the dark, we revealed 4 accumulated proteins that belong to the energy category. The main cellular function of β-phosphoglucomutase, exclusive to bacteria and protists, is to catalyze the conversion of β-D-glucose 1-phosphate to β-D-glucose 6-phosphate, which is a universal source of cellular energy and leads to the generation of ATP and NADPH through glycolysis and the pentose phosphate pathway. It also links glycolysis and gluconeogenesis60.") 2,3-biphosphoglycerate-independent phosphoglycerate mutase (gpmA) catalyzes the reversible conversion of 3-phosphoglycerate to 2-phosphoglycerate in glycolysis61."). The enzyme glucose-6-phosphate 1-dehydrogenase belongs to the pentose phosphate pathway and catalyzes the dehydrogenation of glucose 6-phosphate, contributing to intracellular redox homeostasis62. "). Aldo-keto reductases comprise a large group of NADPH-dependent enzymes that catalyze redox reactions involved in detoxification, biosynthesis, and intermediate metabolism. Aldo-keto reductases reduce aldehydes, ketones, ketosterols, monosaccharides, and prostaglandins and oxidize trans-dihydrodiols of aromatic hydrocarbons and hydroxysteroids. Plants are associated with the aldo/keto reductase family four, related to the detoxification of aldehydes/ketones and xenobiotics to ascorbic acid biosynthesis63."). Conclusion Biochemical parameters and proteome profiles of dark-cultivated cells of Euglena gracilis compared with light-grown ones indicated that cultures experienced environmental stress. Notably, enhanced proline content and accumulation of defense-related proteins in dark conditions corroborated such idea. Particularly, we detected glutathione-S-transferases, peroxiredoxins, and heat-shock proteins among defense-related proteins. As a part of amino acid metabolism, we highlighted 4 proteins, more abundant in the dark, involved in the metabolism of aspartic acid, tryptophan, and purines. Amino acid metabolism is involved in energy metabolism in plants and is crucial for various cellular processes, including DNA and RNA synthesis, energy flow, and signaling pathways. During cultivation in the dark, changes in amino acid metabolism might occur because the lack of light affects energy balance and cellular processes. Particular attention deserves fatty acid metabolism. In the dark, we revealed an accumulation of protein involved in the β-oxidation of fatty acids. Under dark culture conditions, photosynthesis does not occur, requiring cells to rely on alternative energy sources. Euglena can oxidize fatty acids to produce ATP, which is essential for survival. We hypothesize that this process is a vital adaptation to heterotrophic cultivation in the dark. Our proteomic results indicate a putative role of adenosylcobalamin in the regulation of light/dark metabolic switch in euglena, similarly as in bacteria. 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The PRIDE database at 20 years: 2025 update. Nucleic Acids Res.6, 53. (2025). ArticleGoogle Scholar Download references Acknowledgements This research was supported by VEGA 1/0230/24 and KEGA 012UCM-4/2025. Author information Author notes 1. Adriana Paprčková and Katarína Klubicová contributed equally. Authors and Affiliations Department of Biology, Institute of Biology and Biotechnology, Faculty of Natural Sciences, University of Ss. Cyril and Methodius in Trnava, Nám. J. Herdu 2, Trnava, SK-917 01, Slovakia Adriana Paprčková,Eva Ürgeová,Juraj Krajčovič&Ľubica Uváčková Institute of Plant Genetics and Biotechnology, Plant Science and Biodiversity Centre, Slovak Academy of Sciences, Akademická 2, Nitra, SK-950 07, Slovakia Katarína Klubicová,Maksym Danchenko&Olha Lakhneko Institute of Chemistry, Slovak Academy of Sciences, Dúbravská cesta 9, Bratislava, SK-845 38, Slovakia Peter Baráth Authors 1. Adriana PaprčkováView author publications Search author on:PubMedGoogle Scholar 2. Katarína KlubicováView author publications Search author on:PubMedGoogle Scholar 3. Eva ÜrgeováView author publications Search author on:PubMedGoogle Scholar 4. Maksym DanchenkoView author publications Search author on:PubMedGoogle Scholar 5. Peter BaráthView author publications Search author on:PubMedGoogle Scholar 6. Olha LakhnekoView author publications Search author on:PubMedGoogle Scholar 7. Juraj KrajčovičView author publications Search author on:PubMedGoogle Scholar 8. Ľubica UváčkováView author publications Search author on:PubMedGoogle Scholar Contributions Author ContributionsConceptualization, LU; methodology, LU; validation, LU, AP; formal analysis, LU, AP, EU; investigation AP, LU, OL and MD; resources JK; data curation LU, PB, EU; writing—original draft preparation, AP and LU; writing - review and editing, LU, EU, JK, KK and MD; visualization, AP, LU, and MD; supervision, LU; project administration JK; funding acquisition JK. 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Copy to clipboard Provided by the Springer Nature SharedIt content-sharing initiative Keywords Adenosylcobalamin Β-oxidation of fatty acids Mixotrophy Photosynthetic pigments Proline Stress Subjects Cell biology Plant sciences Plant stress responses Download PDF Sections Figures References Abstract Introduction Materials and methods Results and discussion Conclusion Data availability References Acknowledgements Author information Ethics declarations Additional information Electronic supplementary material Rights and permissions About this article Advertisement Fig. 1 View in articleFull size image Fig. 2 View in articleFull size image Fig. 3 View in articleFull size image Fig. 4 View in articleFull size image Fig. 5 View in articleFull size image Vesteg, M. et al. Comparative molecular cell biology of phototrophic Euglenids and parasitic trypanosomatids sheds light on the ancestor of euglenozoa. Biol. Rev.94, 1701. (2019). ArticlePubMedGoogle Scholar O’Neill, E. C. et al. 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https://openstax.org/books/university-physics-volume-3/pages/5-3-time-dilation
Skip to Content Go to accessibility page Keyboard shortcuts menu Log in University Physics Volume 3 5.3 Time Dilation University Physics Volume 3 5.3 Time Dilation Search for key terms or text. Learning Objectives By the end of this section, you will be able to: Explain how time intervals can be measured differently in different reference frames. Describe how to distinguish a proper time interval from a dilated time interval. Describe the significance of the muon experiment. Explain why the twin paradox is not a contradiction. Calculate time dilation given the speed of an object in a given frame. The analysis of simultaneity shows that Einstein’s postulates imply an important effect: Time intervals have different values when measured in different inertial frames. Suppose, for example, an astronaut measures the time it takes for a pulse of light to travel a distance perpendicular to the direction of his ship’s motion (relative to an earthbound observer), bounce off a mirror, and return (Figure 5.4). How does the elapsed time that the astronaut measures in the spacecraft compare with the elapsed time that an earthbound observer measures by observing what is happening in the spacecraft? Examining this question leads to a profound result. The elapsed time for a process depends on which observer is measuring it. In this case, the time measured by the astronaut (within the spaceship where the astronaut is at rest) is smaller than the time measured by the earthbound observer (to whom the astronaut is moving). The time elapsed for the same process is different for the observers, because the distance the light pulse travels in the astronaut’s frame is smaller than in the earthbound frame, as seen in Figure 5.4. Light travels at the same speed in each frame, so it takes more time to travel the greater distance in the earthbound frame. Figure 5.4 (a) An astronaut measures the time Δτ for light to travel distance 2D in the astronaut’s frame. (b) A NASA scientist on Earth sees the light follow the longer path 2s and take a longer time Δt. (c) These triangles are used to find the relationship between the two distances D and s. Time Dilation Time dilation is the lengthening of the time interval between two events for an observer in an inertial frame that is moving with respect to the rest frame of the events (in which the events occur at the same location). To quantitatively compare the time measurements in the two inertial frames, we can relate the distances in Figure 5.4 to each other, then express each distance in terms of the time of travel (respectively either Δt or Δτ) of the pulse in the corresponding reference frame. The resulting equation can then be solved for Δt in terms of Δτ. The lengths D and L in Figure 5.4 are the sides of a right triangle with hypotenuse s. From the Pythagorean theorem, s2=D2+L2. The lengths 2s and 2L are, respectively, the distances that the pulse of light and the spacecraft travel in time Δt in the earthbound observer’s frame. The length D is the distance that the light pulse travels in time Δτ in the astronaut’s frame. This gives us three equations: 2s=cΔt;2L=vΔt;2D=cΔτ. Note that we used Einstein’s second postulate by taking the speed of light to be c in both inertial frames. We substitute these results into the previous expression from the Pythagorean theorem: s2(cΔt2)2==D2+L2(cΔτ2)2+(vΔt2)2. Then we rearrange to obtain (cΔt)2−(vΔt)2=(cΔτ)2. Finally, solving for Δt in terms of Δτ gives us Δt=Δτ1−(v/c)2−−−−−−−−√. 5.1 This is equivalent to Δt=γΔτ, where γ is the relativistic factor (often called the Lorentz factor) given by γ=11−v2c2−−−−−√ 5.2 and v and c are the speeds of the moving observer and light, respectively. Note the asymmetry between the two measurements. Only one of them is a measurement of the time interval between two events—the emission and arrival of the light pulse—at the same position. It is a measurement of the time interval in the rest frame of a single clock. The measurement in the earthbound frame involves comparing the time interval between two events that occur at different locations. The time interval between events that occur at a single location has a separate name to distinguish it from the time measured by the earthbound observer, and we use the separate symbol Δτ to refer to it throughout this chapter. Proper Time The proper time interval Δτ between two events is the time interval measured by an observer for whom both events occur at the same location. The equation relating Δt and Δτ is truly remarkable. First, as stated earlier, elapsed time is not the same for different observers moving relative to one another, even though both are in inertial frames. A proper time interval Δτ for an observer who, like the astronaut, is moving with the apparatus, is smaller than the time interval for other observers. It is the smallest possible measured time between two events. The earthbound observer sees time intervals within the moving system as dilated (i.e., lengthened) relative to how the observer moving relative to Earth sees them within the moving system. Alternatively, according to the earthbound observer, less time passes between events within the moving frame. Note that the shortest elapsed time between events is in the inertial frame in which the observer sees the events (e.g., the emission and arrival of the light signal) occur at the same point. This time effect is real and is not caused by inaccurate clocks or improper measurements. Time-interval measurements of the same event differ for observers in relative motion. The dilation of time is an intrinsic property of time itself. All clocks moving relative to an observer, including biological clocks, such as a person’s heartbeat, or aging, are observed to run more slowly compared with a clock that is stationary relative to the observer. Note that if the relative velocity is much less than the speed of light (v<<c), then v2/c2 is extremely small, and the elapsed times Δt and Δτ are nearly equal. At low velocities, physics based on modern relativity approaches classical physics—everyday experiences involve very small relativistic effects. However, for speeds near the speed of light, v2/c2 is close to one, so 1−v2/c2−−−−−−−−√ is very small and Δt becomes significantly larger than Δτ. Half-Life of a Muon There is considerable experimental evidence that the equation Δt=γΔτ is correct. One example is found in cosmic ray particles that continuously rain down on Earth from deep space. Some collisions of these particles with nuclei in the upper atmosphere result in short-lived particles called muons. The half-life (amount of time for half of a material to decay) of a muon is 1.52 μs when it is at rest relative to the observer who measures the half-life. This is the proper time interval Δτ. This short time allows very few muons to reach Earth’s surface and be detected if Newtonian assumptions about time and space were correct. However, muons produced by cosmic ray particles have a range of velocities, with some moving near the speed of light. It has been found that the muon’s half-life as measured by an earthbound observer (Δt) varies with velocity exactly as predicted by the equation Δt=γΔτ. The faster the muon moves, the longer it lives. We on Earth see the muon last much longer than its half-life predicts within its own rest frame. As viewed from our frame, the muon decays more slowly than it does when at rest relative to us. A far larger fraction of muons reach the ground as a result. Before we present the first example of solving a problem in relativity, we state a strategy you can use as a guideline for these calculations. Problem-Solving Strategy Relativity Make a list of what is given or can be inferred from the problem as stated (identify the knowns). Look in particular for information on relative velocity v. Identify exactly what needs to be determined in the problem (identify the unknowns). Make certain you understand the conceptual aspects of the problem before making any calculations (express the answer as an equation). Decide, for example, which observer sees time dilated or length contracted before working with the equations or using them to carry out the calculation. If you have thought about who sees what, who is moving with the event being observed, who sees proper time, and so on, you will find it much easier to determine if your calculation is reasonable. Determine the primary type of calculation to be done to find the unknowns identified above (do the calculation). You will find the section summary helpful in determining whether a length contraction, relativistic kinetic energy, or some other concept is involved. Note that you should not round off during the calculation. As noted in the text, you must often perform your calculations to many digits to see the desired effect. You may round off at the very end of the problem solution, but do not use a rounded number in a subsequent calculation. Also, check the answer to see if it is reasonable: Does it make sense? This may be more difficult for relativity, which has few everyday examples to provide experience with what is reasonable. But you can look for velocities greater than c or relativistic effects that are in the wrong direction (such as a time contraction where a dilation was expected). Example 5.1 Time Dilation in a High-Speed Vehicle The Hypersonic Technology Vehicle 2 (HTV-2) is an experimental rocket vehicle capable of traveling at 21,000 km/h (5830 m/s). If an electronic clock in the HTV-2 measures a time interval of exactly 1-s duration, what would observers on Earth measure the time interval to be? Strategy Apply the time dilation formula to relate the proper time interval of the signal in HTV-2 to the time interval measured on the ground. Solution Identify the knowns: Δτ=1s;v=5830m/s. Identify the unknown: Δt. Express the answer as an equation: Δt=γΔτ=Δτ1−v2c2−−−−−√. 4. Do the calculation. Use the expression for γ to determine Δt from Δτ: Δt=1s1−(5830m/s3.00×108m/s)2√=1.000000000189s=1s+1.89×10−10s. Significance The very high speed of the HTV-2 is still only 10-5 times the speed of light. Relativistic effects for the HTV-2 are negligible for almost all purposes, but are not zero. Example 5.2 What Speeds are Relativistic? How fast must a vehicle travel for 1 second of time measured on a passenger’s watch in the vehicle to differ by 1% for an observer measuring it from the ground outside? Strategy Use the time dilation formula to find v/c for the given ratio of times. Solution Identify the known: ΔτΔt=11.01. 2. Identify the unknown: v/c. 3. Express the answer as an equation: ΔtΔτΔt(ΔτΔt)2vc====γΔτ=11−v2/c2√Δτ1−v2/c2−−−−−−−−√1−v2c21−(Δτ/Δt)2−−−−−−−−−−−√. 4. Do the calculation: vc=1−(1/1.01)2−−−−−−−−−−−√=0.14. Significance The result shows that an object must travel at very roughly 10% of the speed of light for its motion to produce significant relativistic time dilation effects. Example 5.3 Calculating Δt for a Relativistic Event Suppose a cosmic ray colliding with a nucleus in Earth’s upper atmosphere produces a muon that has a velocity v=0.950c. The muon then travels at constant velocity and lives 2.20 μs as measured in the muon’s frame of reference. (You can imagine this as the muon’s internal clock.) How long does the muon live as measured by an earthbound observer (Figure 5.5)? Figure 5.5 A muon in Earth’s atmosphere lives longer as measured by an earthbound observer than as measured by the muon’s internal clock. As we will discuss later, in the muon’s reference frame, it travels a shorter distance than measured in Earth’s reference frame. Strategy A clock moving with the muon measures the proper time of its decay process, so the time we are given is Δτ=2.20μs. The earthbound observer measures Δt as given by the equation Δt=γΔτ. Because the velocity is given, we can calculate the time in Earth’s frame of reference. Solution Identify the knowns: v=0.950c,Δτ=2.20μs. Identify the unknown: Δt. Express the answer as an equation. Use: Δt=γΔτ with γ=11−v2c2−−−−−√. 4. Do the calculation. Use the expression for γ to determine Δt from Δτ Δt=γΔτ=11−v2c2√Δτ=2.20μs1−(0.950)2√=7.05μs. Remember to keep extra significant figures until the final answer. Significance One implication of this example is that because γ=3.20 at 95.0% of the speed of light (v=0.950c), the relativistic effects are significant. The two time intervals differ by a factor of 3.20, when classically they would be the same. Something moving at 0.950c is said to be highly relativistic. Example 5.4 Relativistic Television A non-flat screen, older-style television display (Figure 5.6) works by accelerating electrons over a short distance to relativistic speed, and then using electromagnetic fields to control where the electron beam strikes a fluorescent layer at the front of the tube. Suppose the electrons travel at 6.00×107m/s through a distance of 0.200m from the start of the beam to the screen. (a) What is the time of travel of an electron in the rest frame of the television set? (b) What is the electron’s time of travel in its own rest frame? Figure 5.6 The electron beam in a cathode ray tube television display. Strategy for (a) (a) Calculate the time from vt=d. Even though the speed is relativistic, the calculation is entirely in one frame of reference, and relativity is therefore not involved. Solution Identify the knowns: v=6.00×107m/s;d=0.200m. 2. Identify the unknown: the time of travel Δt. 3. Express the answer as an equation: Δt=dv. 4. Do the calculation: t=0.200m6.00×107m/s=3.33×10−9s. Significance The time of travel is extremely short, as expected. Because the calculation is entirely within a single frame of reference, relativity is not involved, even though the electron speed is close to c. Strategy for (b) (b) In the frame of reference of the electron, the vacuum tube is moving and the electron is stationary. The electron-emitting cathode leaves the electron and the front of the vacuum tube strikes the electron with the electron at the same location. Therefore we use the time dilation formula to relate the proper time in the electron rest frame to the time in the television frame. Solution Identify the knowns (from part a): Δt=3.33×10−9s;v=6.00×107m/s;d=0.200m. 2. Identify the unknown: τ. 3. Express the answer as an equation: ΔtΔτ==γΔτ=Δτ1−v2/c2√Δt1−v2/c2−−−−−−−−√. 4. Do the calculation: Δτ=(3.33×10−9s)1−(6.00×107m/s3.00×108m/s)2−−−−−−−−−−−−−−−√=3.26×10−9s. Significance The time of travel is shorter in the electron frame of reference. Because the problem requires finding the time interval measured in different reference frames for the same process, relativity is involved. If we had tried to calculate the time in the electron rest frame by simply dividing the 0.200 m by the speed, the result would be slightly incorrect because of the relativistic speed of the electron. Check Your Understanding 5.2 What is γ if v=0.650c? The Twin Paradox An intriguing consequence of time dilation is that a space traveler moving at a high velocity relative to Earth would age less than the astronaut’s earthbound twin. This is often known as the twin paradox. Imagine the astronaut moving at such a velocity that γ=30.0, as in Figure 5.7. A trip that takes 2.00 years in her frame would take 60.0 years in the earthbound twin’s frame. Suppose the astronaut travels 1.00 year to another star system, briefly explores the area, and then travels 1.00 year back. An astronaut who was 40 years old at the start of the trip would be 42 when the spaceship returns. Everything on Earth, however, would have aged 60.0 years. The earthbound twin, if still alive, would be 100 years old. The situation would seem different to the astronaut in Figure 5.7. Because motion is relative, the spaceship would seem to be stationary and Earth would appear to move. (This is the sensation you have when flying in a jet.) Looking out the window of the spaceship, the astronaut would see time slow down on Earth by a factor of γ=30.0. Seen from the spaceship, the earthbound sibling will have aged only 2/30, or 0.07, of a year, whereas the astronaut would have aged 2.00 years. Figure 5.7 The twin paradox consists of the conflicting conclusions about which twin ages more as a result of a long space journey at relativistic speed. The paradox here is that the two twins cannot both be correct. As with all paradoxes, conflicting conclusions come from a false premise. In fact, the astronaut’s motion is significantly different from that of the earthbound twin. The astronaut accelerates to a high velocity and then accelerates opposite to the motion to view the star system. To return to Earth, she again accelerates and decelerates. The spacecraft is not in a single inertial frame to which the time dilation formula can be directly applied. That is, the astronaut twin changes inertial references. The earthbound twin does not experience these accelerations and remains in the same inertial frame. Thus, the situation is not symmetric, and it is incorrect to claim that the astronaut observes the same effects as her twin. The lack of symmetry between the twins will be still more evident when we analyze the journey later in this chapter in terms of the path the astronaut follows through four-dimensional space-time. In 1971, American physicists Joseph Hafele and Richard Keating verified time dilation at low relative velocities by flying extremely accurate atomic clocks around the world on commercial aircraft. They measured elapsed time to an accuracy of a few nanoseconds and compared it with the time measured by clocks left behind. Hafele and Keating’s results were within experimental uncertainties of the predictions of relativity. Both special and general relativity had to be taken into account, because gravity and accelerations were involved as well as relative motion. Check Your Understanding 5.3 a. A particle travels at 1.90×108m/s and lives 2.10×10−8s when at rest relative to an observer. How long does the particle live as viewed in the laboratory? b. Spacecraft A and B pass in opposite directions at a relative speed of 4.00×107m/s. An internal clock in spacecraft A causes it to emit a radio signal for 1.00 s. The computer in spacecraft B corrects for the beginning and end of the signal having traveled different distances, to calculate the time interval during which ship A was emitting the signal. What is the time interval that the computer in spacecraft B calculates? Previous Next Order a print copy Citation/Attribution This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute OpenStax. Attribution information If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution: Access for free at If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution: Access for free at Citation information Use the information below to generate a citation. We recommend using a citation tool such as this one. Authors: Samuel J. Ling, Jeff Sanny, William Moebs Publisher/website: OpenStax Book title: University Physics Volume 3 Publication date: Sep 29, 2016 Location: Houston, Texas Book URL: Section URL: © Jul 8, 2025 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.
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https://stackoverflow.com/questions/77508686/number-of-overtakes-in-a-race
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Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Number of overtakes in a race Ask Question Asked 1 year, 10 months ago Modified1 year, 10 months ago Viewed 227 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Im given an input array racers of length N where each racer has it own speed, (1 <= speed[i] <= 10^8) and starting position. Racers move on a infinitly long track in the same direction.(from left to right) Racers array is sorted in ascending order based on positions meaning: 1 <= positions < positions < ... < positions[n] < 10^9. I want to calculate how many overtakes would occur until first racer reaches position position, racers.Position < position < racers[n].Position. Overtake occurs when racer A's position is equal to racer B's position( it can happen when racer A's position had a smaller value than racer B's position but racer A's speed is greater than racer B's speed). My idea looks like this: ```csharp static uint OvertakingCountAtPosition(Racer[] racers, double position, uint racerCount) { var overtakingCount = 0u; for (int i = 0; i < racerCount; i++) { var time1 = (position - racers[i].Position) / racers[i].Speed; for (int j = i + 1; j < racerCount; j++) { var time2 = (position - racers[j].Position) / racers[j].Speed; if (0.0001d >= time1 - time2) { overtakingCount++; } } } return overtakingCount; } ``` Basically, I choose one racer and calculate the time he needed to go to from his position to the given position. Then, I go trough the rest of the array and find time needed for other racers in front of him to reach target position. When I compare two racers, I check if second racer needed more or equal time to go to target position then first racer. If that is true, I can say that there was an overtake. Once an overtake occured, those two racers wont be able to overtake each other again. I think problem can be solved in a better way but for now im looking for a working solution. c# arrays algorithm Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Nov 18, 2023 at 21:45 ProgrammerProgrammer asked Nov 18, 2023 at 20:55 ProgrammerProgrammer 11 1 1 bronze badge 11 What language is the shown code written in?Some programmer dude –Some programmer dude 2023-11-18 20:58:06 +00:00 Commented Nov 18, 2023 at 20:58 Code is written in c#Programmer –Programmer 2023-11-18 20:59:11 +00:00 Commented Nov 18, 2023 at 20:59 3 More importantly maybe: what's your question? Where are you stuck? And if you are, what is the outcome, the expected outcome and steps you've performed trying to correct it?Maarten Bodewes –Maarten Bodewes 2023-11-18 20:59:59 +00:00 Commented Nov 18, 2023 at 20:59 I think my question is pretty clear. I need to find number of overtakes in a race. Im stuck since my code doesn't work and I am not quite sure whats wrong. Expected outcome is that number. I tried covering some edge cases but I am not sure what those are and whether I might miss something.Programmer –Programmer 2023-11-18 21:02:49 +00:00 Commented Nov 18, 2023 at 21:02 " I need to find number of overtakes in a race". No that's your assignment. Explain your results and how they differ and what you've done to debug your code.Maarten Bodewes –Maarten Bodewes 2023-11-18 21:05:49 +00:00 Commented Nov 18, 2023 at 21:05 |Show 6 more comments 2 Answers 2 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. I will explain the suggested algorithm in terms of an example. I have written calculations in Ruby but readers unfamiliar with that language can view it as pseudo-code. It should be a straightforward exercise to convert it to C#. Data Suppose csharp position = 6 and the racers are as follows: csharp racers = [ { :pos=>2, :speed=>2.1 }, { :pos=>3, :speed=>2.0 }, { :pos=>4, :speed=>1.0 }, { :pos=>5, :speed=>0.2 }, { :pos=>7, :speed=>3.0 }, { :pos=>8, :speed=>2.0 }, { :pos=>9, :speed=>1.8 } ] Step 1: Partition racers between those who have not yet reached position = 6 and those who have csharp before, after = racers.partition { |racer| racer[:pos] < position } before #=> [{:pos=>2, :speed=>2.1}, # {:pos=>3, :speed=>2.0}, # {:pos=>4, :speed=>1.0}, # {:pos=>5, :speed=>0.2}] after #=> [{:pos=>7, :speed=>3.0}, # {:pos=>8, :speed=>2.0}, # {:pos=>9, :speed=>1.8}] Order must be preserved in the sense that before + after must equal racers. Step 2: Determine time it takes each "before" racer to reach position csharp times_to_reach_position = before.map do |h| (position - h[:pos])/h[:speed] end #=> [1.9047619047619047, 1.5, 2.0, 5.0] Step 3: Determine time until first "before" car reaches position csharp time = times_to_reach_position.min #=>1.5 Step 4: Determine position of each car at time time and save positional order ```csharp def m4(arr) arr.each_with_index do |h,i| h[:new_pos] = h[:pos] + time h[:speed] h[:order] = i end end m4(before) #=> [{:pos=>2, :speed=>2.1, :new_pos=>5.15, :order=>0}, # {:pos=>3, :speed=>2.0, :new_pos=>6.0, :order=>1}, # {:pos=>4, :speed=>1.0, :new_pos=>5.5, :order=>2}, # {:pos=>5, :speed=>0.2, :new_pos=>5.3, :order=>3}] m4(after) #=> [{:pos=>7, :speed=>3.0, :new_pos=>11.5, :order=>0}, # {:pos=>8, :speed=>2.0, :new_pos=>11.0, :order=>1}, # {:pos=>9, :speed=>1.8, :new_pos=>11.7, :order=>2}] ``` Step 5: Sort before and after by :new_pos ```csharp def m5(arr) # sort in place arr.sort_by! { |h| h[:new_pos] } end m5(before) #=> [{:pos=>2, :speed=>2.1, :new_pos=>5.15, :order=>0}, # {:pos=>5, :speed=>0.2, :new_pos=>5.3, :order=>3}, # {:pos=>4, :speed=>1.0, :new_pos=>5.5, :order=>2}, # {:pos=>3, :speed=>2.0, :new_pos=>6.0, :order=>1}] m5(after) #=> [{:pos=>8, :speed=>2.0, :new_pos=>11.0, :order=>1}, # {:pos=>7, :speed=>3.0, :new_pos=>11.5, :order=>0}, # {:pos=>9, :speed=>1.8, :new_pos=>11.7, :order=>2}] ``` Step 6: Compute total number of overtakes ```csharp def m6(arr) count = 0 arr.each_with_index { |h,i| count += [i-h[:order], 0].max } count end m6(before) + m6(after) #=> 3 ``` As seen the car originally at position 3 passed cars originally in positions 4 and 5 and the car originally at position 7 passed the car originally at position 8. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Nov 19, 2023 at 6:45 answered Nov 19, 2023 at 2:58 Cary SwovelandCary Swoveland 111k 6 6 gold badges 69 69 silver badges 105 105 bronze badges Comments Add a comment This answer is useful 0 Save this answer. Show activity on this post. If I understand correctly, since you just need an overtakeCount you don't necessarily need to calculate times, since its infinite track you just need to check the position of the racer and if the racer in front is slower. solving for total overtakes in a race (not until first racer reaches position position) could look something along the lines: ```csharp //loop the following while racers are not in final position for (int r1 = 0; r1 < racers.Length; r1++) { for (int r2 = 0; r2 < racers.Length; r2++) { if (racers[r2].position == racers[r1].position - 1) { if (racers[r1].speed > racers[r2].speed) { racers[r1].position--; racers[r2].position++; overtakes++; } } } } ``` Edit: I decided to improve on my previous solution since you don't appear to need a step by step iteration, for the sake of example chose the types I found most convenient for the sample but they are easily changed. The change is simple, I just check the relative position instead of checking if is right in front: (also tidy up the code) ```csharp for (int r1 = 0; r1 < racers.Length; r1++) { for (int r2 = 0; r2 < racers.Length; r2++) { if (racers[r1].speed <= racers[r2].speed) continue; if(racers[r1].startingPosition <= racers[r2].startingPosition) continue; overtakes++; } } ``` sample See the results in a graph x is position y is time Note: I suggest you edit the question to more clearly elaborate the issue, not just drop an assignment with a proposed solution but explain step by step what is wrong with your solution, what you expect a solution to look like and overall simplify the question, we don't really need to know the max speed is specifically 10^8 for example speed ranges from 1 to maxSpeed its more readable and would work the same in any algorithm. Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Nov 19, 2023 at 16:11 answered Nov 19, 2023 at 1:10 BarretoBarreto 420 4 4 silver badges 18 18 bronze badges Comments Add a comment Your Answer Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. To learn more, see our tips on writing great answers. Draft saved Draft discarded Sign up or log in Sign up using Google Sign up using Email and Password Submit Post as a guest Name Email Required, but never shown Post Your Answer Discard By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions c# arrays algorithm See similar questions with these tags. 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https://engoo.com/blog/language-tips/how-to-use-mind/
How to Use "Mind" Like a Native Speaker | Engoo Blog Engoo Blog Login Register Home CategoryExpand child menu Study Language Tips Travel & Fun Parents & Kids Business & Work Engoo Search for: Engoo►Blog►Language Tips How to Use "Mind" Like a Native Speaker David May 23, 2025 Table of Contents Toggle As a Noun As a Verb “Mind” in Idioms Have something in mind Give a piece of one’s mind Never mind Keep something in mind Don’t mind ~ / Don’t pay any mind ~ Come to mind Mind over matter On one’s mind Out of one’s mind Put one’s mind to something Have an open mind Wrap-up Although similar, "mind" is not the same as "brain." In fact, it has a deeper meaning and can be used in many more types of situations. To help you learn how to use "mind" like a native speaker, let's take a closer look at each of its forms as well as some example sentences and idioms. As a Noun In its noun form, "mind" refers to the part of us that allows us to think, feel and experience the world. On the other hand, "brain" refers to the physical part of our bodies that processes information. Clear your mind and try to relax. Psychologists study how the human mind works. When she heard his name, many memories came to her mind. In some cases, "mind" can mean "expert" or "intelligent person." Many of the world's greatest minds gathered at the science conference. As a Verb When used as a verb, "mind" has a few meanings. Perhaps the most common is "to care or be bothered by," which is mostly used in questions and in the negative form. Do you mind if I sit here? My apartment is near a major road, but I don't mind the noise from the traffic. It can also mean "pay attention to" or "take care of." This usage sounds slightly formal in American English. Mind the time; you don't want to miss your train. Can you mind the children while I go to the post office? That was a rude thing to say. Mind your manners! Also, announcements on London's train system tell riders to "mind the gap" between the train and the platform. In other words, "Pay attention to the empty space so you don't fall and hurt yourself." “Mind” in Idioms "Mind" appears in quite a few English idioms that are used often in daily conversations. Have something in mind "Having something in mind" means thinking about something as a suggestion or as a preference. Let's go out to eat tonight. B OK. What did you have in mind? Italian. I'd like to watch a movie, but I don't have anything particular in mind. Give a piece of one’s mind When you "give someone a piece of your mind," you confront them directly about something they've done that you don't like. Jack is always late for meetings. When he finally gets here, I'm going to give hima piece of my mind! Jane gaveher ex-boyfriend a piece of her mind before she dumped him. Related article:English Slang for Online Dating Never mind This common phrase means "It's not important" or "Forget about it." Can I borrow your pen? Oh, never mind – I have one here. A What were you going to say? B Never mind – it wasn't important. Remember that on social media and in text messages, "never mind" is often shortened to "NVM" or "nvm." Keep something in mind This phrase means "Please remember," and it's often used before introducing important information. People sometimes say "bear in mind" instead. Keep in mind that temperatures can drop very low at night, so please pack warm clothing. English is common in tourist areas, but not in rural towns, so please bear that in mind. Don’t mind ~ / Don’t pay any mind~ We use these expressions as a way to say, "Don't worry/think about ~." Don't mind my puppy. He's very friendly around new people. The painter continues to make art that he feels is authentic, withoutpaying any mind to his critics. Come to mind "Come to mind" means "to remember" or "to think about." He asked me for restaurant recommendations, but nothing came to mind. What comes to mind when I say "perfect vacation spot"? Mind over matter We use this idiom to say that what we think about something can have a greater effect on us than the reality of the situation. She finished the marathon despite being physically exhausted, proving that success came down to mind over matter. On one’s mind If something is "on your mind," it means you've been thinking about it. Cassie has been on my mind recently. I should call her and see how she's been. You look like you have a lot on your mind. Wanna talk about it? Out of one’s mind This idiom refers to being crazy or not thinking clearly. You're wearing a t-shirt and shorts in this cold weather? Are you out of your mind?? Put one’s mind to something "Putting one's mind to something" means focusing one's attention and energy in order to complete a task. I told my daughter that she can accomplish anything if she puts her mind to it. Have an open mind People with "open minds" are willing to accept new and different ideas and opinions. The movie is very experimental, so please try to have anopen mind while you watch it. All of my friends are very kind and open-minded. Wrap-up Because of the deep meaning of "mind," you might hear it often in daily conversations, each time with a different nuance. Use this article as a guide to help you get comfortable with its usages —then practice them yourself in a one-on-one lesson with an Engoo tutor! If you'd like to learn more English related to how we use our minds, be sure to read Using Your Head: 15 Different Ways to Say "Think" how to use idioms what's the difference Share on: Related Post How to Talk About the Economy in English May 7, 2025 Business & Work 7 Alternatives for "Love" to Use on Valentine's Day [Updated] February 7, 2025 Language Tips English Idioms and Expressions for the Summer [Updated] July 18, 2025 Travel & Fun Testimonials Materials Our Tutors FAQ Register © 2025 DMM.com LLC. All Rights Reserved. Want to speak English? Ask us how!
8460
https://www.aapm.org/pubs/reports/detail.asp?docid=84
AAPM Reports - Supplement to the 2004 update of the AAPM Task Group No. 43 Report Encrypted | Login Home | Directory | Career Services | Continuing Education | BBS | Contact Publications Login AAPM Public & Media International Medical Physicist Membership Students Meetings Education Quality & Safety Government Affairs Publications Medical Physics Journal Journal of Applied Clinical Medical Physics Newsletter WPSC Newsletter e-News Research Spotlight Physics Today CT Protocols MRI Homogeneity Testing Resources Medical Physics Practice Guidelines Radiation and Medical Imaging Communications Guide ACR-AAPM Practice Parameters and Technical Standards Annals of the ICRP Online ICRU Publications Online NCRP Publications QUANTEC HyTEC Reports Ad Hoc Reports Medical Physics Electronic Content (MPEC) Author Archives Books Surveys (Professional & Research) Copyright and Permissions Career Services Corporate Affiliates Links of Interest Advertising Opportunities Report No.084S - Supplement to the 2004 update of the AAPM Task Group No. 43 Report (2007) Category: Reports Since publication of the 2004 update to the American Association of Physicists in Medicine (AAPM) Task Group No. 43 Report (TG-43U1), several new low-energy photon-emitting brachy- therapy sources have become available. Many of these sources have satisfied the AAPM prerequi- sites for routine clinical use as of January 10, 2005, and are posted on the Joint AAPM/RPC Brachytherapy Seed Registry. Consequently, the AAPM has prepared this supplement to the 2004 AAPM TG-43 update. This paper presents the AAPM-approved consensus datasets for these sources, and includes the following 125I sources: Amersham model 6733, Draximage model LS-1, Implant Sciences model 3500, IBt model 1251L, IsoAid model IAI-125A, Mentor model SL-125/ SH-125, and SourceTech Medical model STM1251. The Best Medical model 2335 103Pd source is also included. While the methodology used to determine these data sets is identical to that published in the AAPM TG-43U1 report, additional information and discussion are presented here on some questions that arose since the publication of the TG-43U1 report. Specifically, details of interpola- tion and extrapolation methods are described further, new methodologies are recommended, and example calculations are provided. Despite these changes, additions, and clarifications, the overall methodology, the procedures for developing consensus data sets, and the dose calculation formal- ism largely remain the same as in the TG-43U1 report. Thus, the AAPM recommends that the consensus data sets and resultant source-specific dose-rate distributions included in this supplement be adopted by all end users for clinical treatment planning of low-energy photon-emitting brachy- therapy sources. Adoption of these recommendations may result in changes to patient dose calcu- lations, and these changes should be carefully evaluated and reviewed with the radiation oncologist prior to implementation of the current protocol. Medical Physics, 34, 2187-2205 (2007) Altmetrics for this report Brachytherapy Subcommittee Workgroup on Low Energy Brachytherapy Source Dosimetry This is a supplement to Report No. 084 - Update of AAPM Task Group No. 43 Report: A revised AAPM protocol for brachytherapy dose (2004) Jeffrey F. Williamson, Mark J. Rivard, Wayne M. Butler, Larry A. DeWerd, M. Saiful Huq, Geoffrey S. Ibbott, Ali S. Meigooni, Christopher S. Melhus, Michael G. Mitch, Ravinder Nath Committee Responsible: Brachytherapy Subcommittee Last Review Date: DISCLAIMER I Disagree I Agree Show list of all AAPM Reports » AAPM is the American Association of Physicists in Medicine. 1631 Prince Street Alexandria, VA 22314 Phone 571-298-1300 Fax 571-298-1301 Send general questions to 2025.aapm@aapm.org AAPM's Privacy Policy Use of the site constitutes your acceptance to its terms and conditions. AAPM is a scientific, educational, and professional nonprofit organization devoted to the discipline of physics in medicine. The information provided in this website is offered for the benefit of its members and the general public, however, AAPM does not independently verify or substantiate the information provided on other websites that may be linked to this site.
8461
https://www.youtube.com/watch?v=DHRTdr-gE30
2.4.2 Vertical Throw with Air Resistance (Graph) xmphysics 43800 subscribers 9 likes Description 427 views Posted: 21 Oct 2023 This video complements the lecture notes published at xmphysics.com A-level Physics Learning Resources Created by Mr Chua Kah Hean ( Transcript: all right students um we'll be sketching the graphs for a vertical throw of a ping pong ball starting from the instant the ball leaves the hand to the instant it returns to the hand so we've done it before isn't it isn't it just a straight line graph with a constant gradients of 9.81 m/s square no the VT graph is different this time because this time is a ping pong ball for a ping pong ball air resistance is probably not negligible so on the way up the ball encounters a downward air resistance in the same direction as the weight so the two will add up to give a net force that's larger than mg so the acceleration is going to be more than 9.81 m/s squ downward what about when a ball is falling when a ball is falling it also encounters air resistance but this time the air resistance is going to be upward in opposite direction to mg so it's a subtraction now that means the net force is going to be smaller than mg and the acceleration is also uh smaller than 9.81 m/s Square okay so let's give it another shot so on the way up the acceleration is stronger than 9.1 m/ second Square so the graph is steeper then on the way down um since acceleration is weaker than G so the graph is also uh not as steep so is this a correct graph wrong huh wrong because it shouldn't be straight because on the way up as V decreases continuously the air resistance also decreases continuously so the net force decreases continuously and so does the acceleration on the way down uh it's the same thing huh so on the way down as as the ball uh speeds up the velocity increases continuously the air resistance will increase continuously and the acceleration will decrease continuously so the gradient of the VT graph should also change continuously so it should be a curve it shouldn't be straight lines and what about when the boy is at the peak you know there's always something special at a peak so at a peak the ball is stationary so if you're stationary there's no air resistance and the net force will be just mg and the acceleration should be exactly equals to 9.1 m/s Square when you're at the top so what does that mean that means your VT graph should look like this it starts off steeper than G flattens out as it rises and the downward air resistance force weakens to become as steep as G when it's at a peak where the air resistance is zero and then it continues to flatten out as it falls and as the air resistance strengthens so how do we know where to end the graph remember the area under the VT graph represents displacement so if we are ending the graph where the ball returns to the same level same height level then the positive area which represents the rise height must be matched by the negative area which represents the for height just agation all right just agation all right let's take a moment to Marvel at our exquisitely drawn graph just by admiring the graph we can tell that it takes the ball longer to fall down than for it to go up this kind of makes sense H because the ball decelerates sharply on the way up but accelerates leisurely on the way down because the air resistance uh help you to decelerate on the way up but it but it prevents you from accelerating on the way down also the ball returns to the hand at a speed that's lower than when it left the hand again uh this uh this makes sense because there's K lost to air resistance all right we're done with the VT graph let's move on to the at graph which should be easy so if air resistance is negligible then the acceleration would have been a constant downward 9.81 m/s square with a resistance the acceleration is still going to be downward but it starts off stronger than 9.81 becomes 9.81 at a peak and then weaker than 9.81 after that as for the St graph if the air resistance is negligible then it would have been this symmetrical quadratic curve with air resistance it loses a symmetry as mentioned just now it takes longer to calm down than to go up and pay attention to the gradients huh so the gradient at the start of the graph should be drawn steeper than the gradient uh at the end because these gradients correspond to the speed at which the ping pong ball leaves the hand and at which it returns to the hand all right that's all TAA
8462
https://math-gpt.org/problems/question/solve-the-simultaneous-linear-inequations-x2-8x-20-and-4x-2-x2
To solve the simultaneous linear inequations x^2 - 8x - 20 | MathGPT To solve the simultaneous linear inequations x^2 - 8x - 20 and 4x - 2x^2 , we need to analyze each expression separat... Question: To solve the simultaneous linear inequations x^2 - 8x - 20 and 4x - 2x^2 , we need to analyze each expression separately.1. Solve for x in x^2 - 8x - 20 = 0 We can use the quadratic formula: x = -b b^2 - 4ac2a where a = 1 , b = -8 , and c = -20 . First, calculate the discriminant: b^2 - 4ac = (-8)^2 - 4 1 (-20) = 64 + 80 = 144. Now applying the quadratic formula: x = 8 1442 1 = 8 122. This gives us two solutions: x_1 = 202 = 10, x_2 = -42 = -2. Thus, the roots of the equation x^2 - 8x - 20 = 0 are x = 10 and x = -2 .2. Analyze the second expression 4x - 2x^2 To solve the inequality 4x - 2x^2 , we first determine where the expression is zero: 4x - 2 = 0 x = 24 = 12. We also need to consider where the expression is undefined, which is when the denominator is zero: x^2 = 0 x = 0. 3. Determine the intervals for 4x - 2x^2 > 0 We can test the sign of the expression 4x - 2x^2 over the intervals determined by x = -2 , x = 0 , x = 12 , and x = 10 , which are (-, -2) , (-2, 0) , (0, 12) , (12, 10) , and (10, ) . Testing a point in each interval: - For x = -3 in (-, -2) : 4(-3) - 2(-3)^2 = -12 - 29 < 0. - For x = -1 in (-2, 0) : 4(-1) - 2(-1)^2 = -4 - 21 < 0. - For x = 14 in (0, 12) : 4(14) - 2(14)^2 = 1 - 2116 < 0. - For x = 1 in (12, 10) : 4(1) - 21^2 = 4 - 21 > 0. - For x = 11 in (10, ) : 4(11) - 211^2 > 0. 4. Combine the results: The expression 4x - 2x^2 > 0 is satisfied in the intervals ( 12, 10 ) and (10, ) .Thus, the simultaneous inequalities give us the solution intervals combined:x ( 12, 10 ) (10, ). Answer: To solve the simultaneous inequalities we solve each inequality separately and then find the intersection of their solutions. For the first inequality: Factoring the quadratic expression, we get: The cri... Unlock with MathGPT tical points are and . Testing intervals gives us: For the second inequality: We factor and analyze the sign changes: The critical points are (from the denominator) and (from the numerator). Testing the intervals: For , the expression is positive. For , the expression is negative. For , the expression is positive. Intersection of solutions: The first inequality’s solutions are and . The second inequality’s solution is or . The intersection (where both inequalities are satisfied) is: Therefore, considering both inequalities, the solution is: Thus, the combined solution for both inequalities, considering the restrictions from the first inequality, is: This simplifies to: MathGPTNext Problem ;
8463
https://medium.com/@pbercker/the-problem-with-liars-a-probability-puzzle-29b0a3d70cbc
The Problem with Liars — a probability puzzle | by Pascal Bercker | Medium Sitemap Open in app Sign up Sign in Write Search Sign up Sign in Mastodon The Problem with Liars — a probability puzzle Pascal Bercker Follow 4 min read · Mar 2, 2023 1 Listen Share Another puzzle I found in Ruma Falk’s book of probability problems: In effect, Dick is (indirectly) saying that P is true. We want to know the probability that P is true, given that Tom said it and Dick confirmed it. Not even knowing what P is, we can set a uniform prior for its being true or false. That will be our initial hypothesis. The evidence we have is the fact that Tom says it’s true and Dick also confirms (indirectly) it’s true. This is the simple network for this hypothesis: What we want to say is that if P is true, there’s only a 1/3 chance that Tom will say it’s true, and a 2/3 chance he will say it’s false (hence lying about P). We will do likewise for Dick. We can fill-in in these two equations: There’s a 1/3 chance that what Tom says is true We compile the equations: When we ask what is the probability that Tom was speaking the truth is in effect to ask for the probability that P is true given that Tom says “P” and that Dick agrees (or confirms) that what Tom says is true. The prior probability was of course .50, but now we have two pieces of evidence to enter, and we do now (in green), one by one. First for Tom: Notice that the probability that P is true is now only 1/3 Note that by entering this finding, we are NOT saying that P is true, but rather that it’s true that Tom said P was true. That’s the evidence we have along with the fact that Tom lies 2/3 of the time. Notice that “Proposition P” is true and has been updated, down to 33% from our prior of 50%. This is exactly what we would expect. But we are not done. We also need to enter the other finding that by saying that what Tom said is true, Dick is asserting that P is true as well. So we enter that finding as well: The probability that Tom spoke the truth is 20% Notice that we have learned something, namely that P is more likely to be false than true now that Tom and Dick have spoken. Get Pascal Bercker’s stories in your inbox Join Medium for free to get updates from this writer. Subscribe Subscribe I always find it instructive to find different ways of doing the same thing. There’s a simpler more direct way of doing this. We can suppose that Tom’s speaking somehow causes Dick to speak as well. So our model looks like this: Before entering the equations We know that the prior probability that Tom speaks truly is 1/3. But now we need to think about the conditional probability that Dick speaks truly given what Tom says. The prior for Tom, and the conditional probability for Dick The second equation gives us the conditional probability table (CPT): This table essentially says that if Tom speaks the truth, there’s only a 1/3 chance that Dick will confirm it as true, and if Tom spoke falsely, there is once again a 1/3 chance that Dick will confirm it as such (meaning there’s a 2/3 chance that Dick will lie and say that it’s true). Once compiled we have the following updated network: Notice the posterior probability for Dick Since what we want is the posterior probability that Tom spoke truly, we enter our evidence that Dick confirmed that what Tom said was true (in green): As expected, the probability Tom spoke truly is 20% We can do all this in terms of Bayes’s theorem: This confirms our results H is the hypothesis that Tom spoke truly, and P(H), the prior, is 1/3. We are interested in the posterior, meaning P(H/e), the probability that Tom spoke truly given the evidence (e) that Dick confirmed that Tom spoke truly. For next time: The problem of three liars: Tom, Dick, and Harry are three liars. Each of them tells the truth only a third of the time (the probability that each is lying is 2/3, every time). Tom makes a statement, and Harry tells us that Dick said that Tom was speaking the truth. What is the probability that Tom was telling the truth? Bayes Theorem Logic Puzzle Liars 1 1 Follow Written by Pascal Bercker ------------------------- 125 followers ·150 following Philosophy | philosophy of science | logic | Bayesian networks and logic puzzles | keen on the use of Bayesian networks to solve Smullyan-type logic puzzles Follow No responses yet Write a response What are your thoughts? Cancel Respond More from Pascal Bercker Pascal Bercker I Had 6 Eggs. I Broke 2. I Fried 2. I Ate 2. How Many Eggs Are Left? What Says ChatGPT4? ---------------------------------------------------------------------------------------- ### This is a classic little riddle, but I’ve never seen it before. Mar 13, 2024 Pascal Bercker Debunking Wielicky Again — The Would-be Climate Skeptic ------------------------------------------------------- ### Dr. Wielicki is at it again. I follow him to get a better perspective on what Climate Science skeptics are on about. Sometimes he makes an… Feb 21 Pascal Bercker WHERE IS MY BAG? A SIMPLE BAYESIAN NETWORK APPROACH --------------------------------------------------- ### This simple little problem comes from Judea Pearl’s book The Book of Why: The New Science of Cause and Effect. His discussion is used to… Sep 4 Pascal Bercker Two Coins — One Fair — One Biased — Solved with a Bayesian Network ------------------------------------------------------------------ ### This puzzle comes from Cut the Knot (see link below) which has a wealth of probability puzzles that frequently offers two or more… Jul 21, 2024 See all from Pascal Bercker Recommended from Medium In Math Games by BL An Advanced Indian Math Exam Question ------------------------------------- ### An infinite product 6d ago 1 In Puzzle Sphere by Avinash Kumar Singh, Ph.D. A Family of Truths and Lies — Can You Crack This Logic Puzzle? -------------------------------------------------------------- ### A mysterious statement, a curious village, and one person you can be sure about. May 15 5 In Data Science Collective by Calvin Hui The Maths Behind Principal Component Analysis (PCA) --------------------------------------------------- ### Illustrate the connection between maximum variance and minimum reconstruction error + Demo how to compress an image using PCA Sep 4 1 In Graphicmaths by Martin McBride Root 2 is irrational — proof by the carpets theorem --------------------------------------------------- ### In the article Root 2 is irrational — proof by prime factors, we used proof by contradiction to prove that the square root of 2 is… Aug 20 1 AmeerSaleem Why the Multivariate Gaussian distribution isn’t as scary as you might think ---------------------------------------------------------------------------- ### Explaining how the Multivariate Gaussian’s parameters and probability density function are a natural extension one-dimensional version. May 12 In ThinkArt by Nnamdi Samuel An Interview Puzzle That Broke the Internet in 2020 --------------------------------------------------- ### Don’t Trust the Math — Find the Pattern Instead 6d ago 8 See more recommendations Help Status About Careers Press Blog Privacy Rules Terms Text to speech
8464
https://content.one.lumenlearning.com/collegealgebra/chapter/equations-of-lines-learn-it-1/
College Algebra Book Contents Navigation Introduction Glossary of Terms Students: Additional Lumen Resources Teaching Resources Evidence-Based Teaching Practices Algebra Essentials: Cheat Sheet Algebra Essentials: Background You'll Need 1 Algebra Essentials: Background You'll Need 2 Algebra Essentials: Background You'll Need 3 Introduction to Real Numbers: Learn It 1 Introduction to Real Numbers: Learn It 2 Introduction to Real Numbers: Learn It 3 Introduction to Real Numbers: Learn It 4 Introduction to Real Numbers: Learn It 5 Introduction to Real Numbers: Learn It 6 Introduction to Real Numbers: Apply It 1 Introduction to Real Numbers: Fresh Take Exponents and Scientific Notation: Learn It 1 Exponents and Scientific Notation: Learn It 2 Exponents and Scientific Notation: Learn It 3 Exponents and Scientific Notation: Learn It 4 Exponents and Scientific Notation: Learn It 5 Exponents and Scientific Notation: Learn It 6 Exponents and Scientific Notation: Learn It 7 Exponents and Scientific Notation: Apply It 1 Exponents and Scientific Notation: Fresh Take Radicals and Rational Exponents: Learn It 1 Radicals and Rational Exponents: Learn It 2 Radicals and Rational Exponents: Learn It 3 Radicals and Rational Exponents: Learn It 4 Radicals and Rational Exponents: Learn It 5 Radicals and Rational Exponents: Apply It 1 Radicals and Rational Exponents: Fresh Take Algebra Essentials: Get Stronger Polynomial and Rational Expressions: Cheat Sheet Polynomial and Rational Expressions: Background You'll Need 1 Polynomial and Rational Expressions: Background You'll Need 2 Polynomial Basics: Learn It 1 Polynomial Basics: Learn It 2 Polynomial Basics: Learn It 3 Polynomial Basics: Learn It 4 Polynomial Basics: Learn It 5 Polynomial Basics: Apply It 1 Polynomial Basics: Fresh Take Factoring Polynomials: Learn It 1 Factoring Polynomials: Learn It 2 Factoring Polynomials: Learn It 3 Factoring Polynomials: Learn It 4 Factoring Polynomials: Apply It 1 Factoring Polynomials: Fresh Take Rational Expressions: Learn It 1 Rational Expressions: Learn It 2 Rational Expressions: Learn It 3 Rational Expressions: Learn It 4 Rational Expressions: Apply It 1 Rational Expressions: Fresh Take Polynomial and Rational Expressions: Get Stronger Linear Equations and Inequalities: Cheat Sheet Linear Equations and Inequalities: Background You'll Need 1 Linear Equations and Inequalities: Background You'll Need 2 Linear Equations and Inequalities: Background You'll Need 3 Graphing and Analyzing Linear Equations: Learn It 1 Graphing and Analyzing Linear Equations: Learn It 2 Graphing and Analyzing Linear Equations: Learn It 3 Graphing and Analyzing Linear Equations: Learn It 4 Graphing and Analyzing Linear Equations: Learn It 5 Graphing and Analyzing Linear Equations: Apply It 1 Graphing and Analyzing Linear Equations: Fresh Take Equations of Lines: Learn It 1 Equations of Lines: Learn It 2 Equations of Lines: Learn It 3 Equations of Lines: Learn It 4 Equations of Lines: Learn It 5 Equations of Lines: Learn It 6 Equations of Lines: Apply It 1 Equations of Lines: Fresh Take Modeling with Linear Equations: Learn It 1 Modeling with Linear Equations: Learn It 2 Modeling with Linear Equations: Learn It 3 Modeling with Linear Equations: Apply It 1 Modeling with Linear Equations: Fresh Take Linear Inequalities: Learn It 1 Linear Inequalities: Learn It 2 Linear Inequalities: Learn It 3 Linear Inequalities: Apply It 1 Linear Inequalities: Fresh Take Linear Equations and Inequalities: Get Stronger Non-Linear Equations: Cheat Sheet Non-Linear Equations: Background You'll Need 1 Non-Linear Equations: Background You'll Need 2 Non-Linear Equations: Background You'll Need 3 Quadratic Equations: Learn It 1 Quadratic Equations: Learn It 2 Quadratic Equations: Learn It 3 Quadratic Equations: Learn It 4 Quadratic Equations: Apply It 1 Quadratic Equations: Fresh Take Other Types of Equations: Learn It 1 Other Types of Equations: Learn It 2 Other Types of Equations: Learn It 3 Other Types of Equations: Learn It 4 Other Types of Equations: Learn It 5 Other Types of Equations: Apply It 1 Other Types of Equations: Apply It 2 Other Types of Equations: Fresh Take Applications of Non-Linear Equations: Learn It 1 Applications of Non-Linear Equations: Learn It 2 Applications of Non-Linear Equations: Learn It 3 Applications of Non-Linear Equations: Apply It 1 Applications of Non-Linear Equations: Fresh Take Non-Linear Equations Get Stronger Function Basics: Cheat Sheet Function Basic: Background You'll Need 1 Function Basic: Background You'll Need 2 Function Basic: Background You'll Need 3 Introduction to Functions: Learn It 1 Introduction to Functions: Learn It 2 Introduction to Functions: Learn It 3 Introduction to Functions: Learn It 4 Introduction to Functions: Learn It 5 Introduction to Functions: Learn It 6 Introduction to Functions: Apply It 1 Introduction to Functions: Fresh Take Domain and Range: Learn It 1 Domain and Range: Learn It 2 Domain and Range: Learn It 3 Domain and Range: Learn It 4 Domain and Range: Learn It 5 Domain and Range: Apply It 1 Domain and Range: Fresh Take Rates of Change and Behavior of Graphs: Learn It 1 Rates of Change and Behavior of Graphs: Learn It 2 Rates of Change and Behavior of Graphs: Learn It 3 Rates of Change and Behavior of Graphs: Learn It 4 Rates of Change and Behavior of Graphs: Apply It 1 Rates of Change and Behavior of Graphs: Fresh Take Function Basic: Get Stronger Algebraic Operations on Functions: Cheat Sheet Algebraic Operations on Functions: Background You'll Need 1 Algebraic Operations on Functions: Background You'll Need 2 Algebraic Operations on Functions: Background You'll Need 3 Combinations and Compositions of Functions: Learn It 1 Combinations and Compositions of Functions: Learn It 2 Combinations and Compositions of Functions: Learn It 3 Combinations and Compositions of Functions: Learn It 4 Combinations and Compositions of Functions: Apply It 1 Combinations and Compositions of Functions: Fresh Take Transformations of Functions: Learn It 1 Transformations of Functions: Learn It 2 Transformations of Functions: Learn It 3 Transformations of Functions: Learn It 4 Transformations of Functions: Learn It 5 Transformations of Functions: Learn It 6 Transformations of Functions: Apply It 1 Transformations of Functions: Fresh Take Inverse Functions: Learn It 1 Inverse Functions: Learn It 2 Inverse Functions: Learn It 3 Inverse Functions: Learn It 4 Inverse Functions: Apply It 1 Inverse Functions: Fresh Take Algebraic Operations on Functions: Get Stronger Linear Functions: Cheat Sheet Linear Functions: Background You'll Need 1 Linear Functions: Background You'll Need 2 Linear Functions: Background You'll Need 3 Introduction to Linear Functions: Learn It 1 Introduction to Linear Functions: Learn it 2 Introduction to Linear Functions: Learn It 3 Introduction to Linear Functions: Learn It 4 Introduction to Linear Functions: Learn It 5 Introduction to Linear Functions: Apply It 1 Introduction to Linear Functions: Fresh Take Graphs of Linear Functions: Learn It 1 Graphs of Linear Functions: Learn It 2 Graphs of Linear Functions: Learn It 3 Graphs of Linear Functions: Learn It 4 Graphs of Linear Functions: Learn It 5 Graphs of Linear Functions: Learn It 6 Graphs of Linear Functions: Apply It 1 Graphs of Linear Functions: Fresh Take Fitting Linear Models to Data: Learn It 1 Fitting Linear Models to Data: Learn It 2 Fitting Linear Models to Data: Learn It 3 Fitting Linear Models to Data: Learn It 4 Fitting Linear Models to Data: Learn It 5 Fitting Linear Models to Data: Learn It 6 Fitting Linear Models to Data: Apply It 1 Fitting Linear Models to Data: Fresh Take Linear Functions: Get Stronger Quadratic Functions: Cheat Sheet Quadratic Functions: Background You'll Need 1 Quadratic Functions: Background You'll Need 2 Introduction to Quadratic Functions and Parabolas: Learn It 1 Introduction to Quadratic Functions and Parabolas: Learn It 2 Introduction to Quadratic Functions and Parabolas: Learn It 3 Introduction to Quadratic Functions and Parabolas: Learn It 4 Introduction to Quadratic Functions and Parabolas: Learn It 5 Introduction to Quadratic Functions and Parabolas: Apply It 1 Introduction to Quadratic Functions and Parabolas: Fresh Take Complex Numbers and Operations: Learn It 1 Complex Numbers and Operations: Learn It 2 Complex Numbers and Operations: Learn It 3 Complex Numbers and Operations: Learn It 4 Complex Numbers and Operations: Learn It 5 Complex Numbers and Operations: Apply It 1 Complex Numbers and Operations: Fresh Take Analysis of Quadratic Functions: Learn It 1 Analysis of Quadratic Functions: Learn It 2 Analysis of Quadratic Functions: Learn It 3 Analysis of Quadratic Functions: Apply It 1 Analysis of Quadratic Functions: Fresh Take Quadratic Functions: Get Stronger Power and Polynomial Functions: Cheat Sheet Power and Polynomial Functions: Background You’ll Need 1 Power and Polynomial Functions: Background You’ll Need 2 Power and Polynomial Functions: Background You’ll Need 3 Introduction to Power and Polynomial Functions: Learn It 1 Introduction to Power and Polynomial Functions: Learn It 2 Introduction to Power and Polynomial Functions: Learn It 3 Introduction to Power and Polynomial Functions: Learn It 4 Introduction to Power and Polynomial Functions: Learn It 5 Introduction to Power and Polynomial Functions: Learn It 6 Introduction to Power and Polynomial Functions: Apply It 1 Introduction to Power and Polynomial Functions: Fresh Take Graphs of Polynomial Functions: Learn It 1 Graphs of Polynomial Functions: Learn It 2 Graphs of Polynomial Functions: Learn It 3 Graphs of Polynomial Functions: Learn It 4 Graphs of Polynomial Functions: Learn It 5 Graphs of Polynomial Functions: Learn It 6 Graphs of Polynomial Functions: Apply It 1 Graphs of Polynomial Functions: Fresh Take Dividing Polynomials: Learn It 1 Dividing Polynomials: Learn It 2 Dividing Polynomials: Learn It 3 Dividing Polynomials: Apply 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The only power of the variable is 1. Linear equations in one variable may take the form [latex]ax + b = 0[/latex] and are solved using basic algebraic operations. An equation will always contain an equal sign with an expression on each side. Think of an equal sign as meaning “the same as.” Some examples of equations are [latex]y = mx + b[/latex], [latex]\Large\frac{3}{4}\normalsize x = v + {3} - r[/latex], and [latex]2(6-d) + f(3+k) = \Large\frac{1}{4}\normalsize d[/latex]. We can classify linear equations in one variable as one of three types: The following figure shows how coefficients, variables, terms, and expressions all come together to make equations. In the equation [latex]2x-3^2=10x[/latex], the variable is [latex]x[/latex], a coefficient is [latex]10[/latex], a term is [latex]10x[/latex], and an expression is [latex]2x-3^2[/latex]. Solving linear equations involves the fundamental properties of equality and basic algebraic operations. Some equations can be solved quickly in your head. For example, what is the value of [latex]y[/latex] in the equation [latex]2y=6[/latex]? You can easily determine that [latex]y=3[/latex] by dividing both sides by [latex]2[/latex]. Other equations are more complicated. Solving [latex]\displaystyle 4\left(\frac{1}{3}\normalsize t+\frac{1}{2}\normalsize\right)=6[/latex] without writing anything down is difficult! That is because this equation contains not just a variable but also fractions and terms inside parentheses. This is a multi-step equation,one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules. Remember that you can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The addition property of equality and the multiplication property of equality explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you will keep both sides of the equation equal. [latex]\begin{array}{r}\,\,3x+5x+4-x+7=\,\,\,88\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x+11=\,\,\,88\end{array}[/latex] The equation is now in the form [latex]ax+b=c[/latex], so we can solve as before. Subtract [latex]11[/latex] from both sides. [latex]\begin{array}{r}7x+11\,\,\,=\,\,\,88\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-11\,\,\,\,\,\,\,-11}\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7x\,\,\,=\,\,\,77\end{array}[/latex] Divide both sides by [latex]7[/latex]. [latex]\begin{array}{r}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{7x}\,\,\,=\,\,\,\underline{77}\7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7\,\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,=\,\,\,11\end{array}[/latex] Some equations may have the variable on both sides of the equal sign, as in this equation: [latex]4x-6=2x+10[/latex]. To solve this equation, we need to “move” one of the variable terms. This can make it difficult to decide which side to work with. It does not matter which term gets moved, [latex]4x[/latex] or [latex]2x[/latex]; however, to avoid negative coefficients, you can move the smaller term. [latex]\begin{array}{r}4x-6=2x+10\,\\underline{-2x\,\,\,\,\,\,\,\,\,\,\,\,-2x\,\,\,\,\,\,\,\,\,\,\,\,\,}\2x-6=10\end{array}[/latex] Now add [latex]6[/latex] to both sides to isolate the term with the variable. [latex]\begin{array}{r}2x-6=10\\underline{\,\,\,\,+6\,\,\,+6}\2x=16\end{array}[/latex] Now divide each side by [latex]2[/latex] to isolate the variable [latex]x[/latex]. [latex]\begin{array}{c}\Large\frac{2x}{2}\normalsize=\Large\frac{16}{2}\\\normalsize{x=8}\end{array}[/latex] Previous/next navigation
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https://www.math.dartmouth.edu/~carlp/rangeofs6.pdf
The range of the sum-of-proper-divisors function Florian Luca Mathematical Institute, UNAM Juriquilla 76230 Santiago de Quer´ etaro, M´ exico and School of Mathematics, University of the Witwatersrand P. O. Box Wits 2050, South Africa fluca@matmor.unam.mx Carl Pomerance Department of Mathematics Dartmouth College Hanover, NH 03755-3551, USA carl.pomerance@dartmouth.edu Abstract Answering a question of Erd˝ os, we show that a positive proportion of even numbers are in the form s(n), where s(n) = σ(n) −n, the sum of proper divisors of n. 2000 Mathematics Subject Classification: Primary 11A25, Secondary 11N37 1 Introduction For a positive integer n, let s(n) = σ(n) −n, the sum of the proper divisors of n. The function s has been studied since antiquity; it may be the first function ever defined by mathematicians. Beginning with Pythagoras, we have looked for cycles in the dynamical system formed when iterating s. There are still a number of unsolved problems connected with this dynamical system: Are there infinitely many cycles? Examples of cycles are 6 →6 and 220 →284 →220; about 12 million are known. Does the set of numbers involved in some cycle have asymptotic density 0? We know the upper density is bounded above by about 0.002. Is there an unbounded orbit? The least starting value in question is n = 276. (For references on these questions, see .) 1 Perhaps a more basic question with the function s is to identify its image: What numbers are of the form s(n)? Note that if p, q are different primes then s(pq) = p+q+1. Not many even numbers are of this form, but a slightly stronger version of Goldbach’s conjecture (every even number starting with 8 is the sum of two different primes) implies that every odd number starting with 9 is in the range of s. Since s(2) = 1, s(4) = 3, and s(8) = 7, while s(n) = 5 has no solutions, it then follows from this slightly stronger Goldbach conjecture that every odd number except 5 is in the range of s. Moreover, this slightly stronger form of Goldbach’s conjecture is known to be usually true. There are many papers in this line, a recent survey is . So, almost all odd numbers (in the sense of asymptotic density) are of the form s(n). In a short beautiful paper, Erd˝ os looked at the even values of s, showing that a positive proportion of even numbers are missed. He raises the issue of whether the asymptotic density of even values exists, saying that is not even known if the lower density is positive. Similar questions are raised for the function sϕ(n) := n −ϕ(n), where ϕ is Euler’s function. Again, almost all odd numbers are attained by sϕ, but even less is known about even values, compared with s(n). In fact, the Erd˝ os argument that shows that s misses a positive proportion of even values fails for sϕ. These thoughts were put in a more general context in . There the following conjecture is formulated. Conjecture 1. If A is a set of natural numbers of asymptotic density 0, then s−1(A) also has asymptotic density 0. If this is true, one consequence would be that the set of even values of s does not have density 0. Indeed, if A is the set of even numbers in the range of s, then s−1(A) = {n even : n, n/2 not squares} ∪{n2 : n odd}, so s−1(A) has asymptotic density 1 2. Thus, if Conjecture 1 is true, then A does not have asymptotic density 0. In this paper we prove the following theorem. Theorem 1. The set of even numbers of the form s(n) for some integer n has positive lower density. Essentially the same proof will show that even numbers of the form n −ϕ(n) comprise a set of positive lower density. It is hoped that the methods in this paper can be of help in proving Conjecture 1. 2 It seems likely that the asymptotic density of even numbers in the range of s exists. In some numerical work in it appears that the even numbers in the range have density about 1 3 and the density of even numbers missing is about 1 6. In it is shown that the lower density of the set of even numbers missing from the range is at least 0.06. The proof of Theorem 1 that we present is effective, but we have made no effort towards finding some explicit lower bound for the lower density of even values of s. 2 Notation and lemmas We have the letters p, q, r, with or without dashes or subscripts representing prime numbers. We let τ(n) denote the number of positive divisors of n. We say a positive integer n is deficient if s(n) < n. We let P(n) denote the largest prime factor of n when n > 1, and we let P(1) = 1. We say a positive integer n is z-smooth if P(n) ≤z. For each prime p and natural number n, we let vp(n) denote the exponent of p in the prime factorization of n. For each large number n, let y = y(n) = log log n/ log log log n. Lemma 1. On a set of asymptotic density 1 we have (1) p2a | σ(n) for every prime power pa ≤y, (2) P(gcd(n, σ(n)) ≤y, (3) σ(n)/ gcd(n, σ(n)) is divisible by every prime p ≤y, (4) and every prime factor of s(n)/ gcd(n, σ(n)) exceeds y. Proof. (1) Let x be large, let y = y(x), and let d be an integer with 1 < d ≤y. The integers n ≤x with d2 | σ(n) include all n ≤x which are precisely divisible (i.e., divisible to just the first power) by two different primes p1, p2 in the residue class −1 (mod d). The number of n ≤x which do not have 3 this property is, by the sieve, ≪x    1 + X p≤x p ≡−1 (mod d) 1 p     Y p≤x p ≡−1 (mod d)  1 −1 p + 1 p2  ≪x log log x ϕ(d) exp  −log log x ϕ(d)  ≤x log log x ϕ(d) exp  −log log x d  ≤ ( x ϕ(d), if 1 2y < d ≤y, x ϕ(d) log log x, if d ≤1 2y. Letting d run over primes and powers of primes, we see that the number of integers n ≤x which do not have the property in (1) is ≪x/ log y = o(x) as x →∞. (2) In [6, Theorem 8], it is shown that on a set of asymptotic den-sity 1, gcd(n, ϕ(n)) is the largest divisor of n supported on the primes at most log log n. Virtually the same proof establishes the analogous result for gcd(n, σ(n)), so that for almost all n, gcd(n, ϕ(n)) = gcd(n, σ(n)). (Also see [3, 5, 7, 11].) That the assertion (2) usually holds, it suffices to note that the number of n ≤x divisible by a prime in (y, log log x] is o(x) as x →∞. (3) Let y = y(x), where x is large. This assertion will follow from (1) for n ≤x if for each prime power pa with pa ≤y < pa+1, we have p2a ∤n. But, the number of n ≤x which fail to have this condition is at most x X p≤y 1 y = x y π(y) = o(x), x →∞. (4) For this part, we have seen that we may assume that for each prime p ≤y, we have vp(σ(n)) > vp(n). Thus, vp(s(n)) = vp(n) = vp(gcd(n, σ(n)) for such primes p. Lemma 2. The set of deficient numbers n for which s(n) is non-deficient has asymptotic density 0. This result follows from [5, Theorem 5.1] and the continuity of the dis-tribution function for σ(n)/n. Lemma 3. On a set of integers n of asymptotic density 1 we have τ(s(n)) = (log n)log 2+o(1) as n →∞. 4 This result follows from the estimates in . We remark that our proof does not depend on this lemma, we could have used the weaker inequality τ(s(n)) ≤no(1) which holds for all n as n →∞, but we thought it good to highlight some other recent research concerning the statistical study of s(n). Lemma 4. On a set of integers n of asymptotic density 1 we have X r|σ(n) r>(log log n)2 1 r ≤1. This follows by the method of proof of [2, Lemma 5]. 3 Proof of the theorem In this section we prove Theorem 1. Proof. We identify a set of integers A such that every member of s(A) is even and s(A) has positive lower density. We shall pile on a number of conditions for A to satisfy. For our initial choice for A, we take the set of even deficient numbers. This set has a positive density, see . Let x be large; we study A(x) := A ∩[1, x]. We assume that each member n of A(x) is of the form n = pm, p ∈  x 2m, x m i , m = qℓ= qrk, k ≤x1/60, r ∈(x1/15, x1/12], q ∈(x7/20, x11/30]. So n = pm = pqℓ= pqrk. Note that n, m, ℓ, k are all even deficient numbers, each running through a positive proportion of numbers to their respective bounds: n ≤x, m ≤x7/15, ℓ≤x1/10, and k ≤x1/60. We assume that each of these 4 variables satisfy the properties in the lemmas. We also assume that k has no prime factors in (y(k), y(x)]. Let y = y(x). For each y-smooth integer d, let Ad(x) denote the subset of A(x) consisting of those members n with largest y-smooth divisor equal to d. There is some number c and a set D ⊆[1, yc] of y-smooth numbers d such that X d∈D 1 d ≫log y, #Ad(x) ≫ x d log y, where the latter inequality holds uniformly for d ∈D. 5 For d ∈D and a positive integer s, let rd(s) denote the number of representations of s in the form s(n) for n ∈Ad(x). Clearly, X s rd(s) = #Ad(x) ≫ x d log y uniformly for all d ∈D. Note too that if d ̸= d′, then we cannot have both rd(s), rd′(s) > 0. Indeed, by Lemma 1, if rd(s) > 0, then d is the largest y-smooth divisor of s. We will show that X s rd(s)2 ≪ x d log y (1) uniformly for each d ∈D, so that from Cauchy’s inequality, it will follow that #s(A(x)) = X d∈D #s(Ad(x)) ≥ X d∈D (P s rd(s))2 P s rd(s)2 ≫ X d∈D x d log y ≫x. The sum P s rd(s)2 counts solutions to s(n) = s(n′) for n, n′ ∈Ad(x), with n = pm, n′ = p′m′. Suppose that m = m′. From the equation ps(m) + σ(m) = p′s(m′) + σ(m′) (2) and m > 1 (which implies that s(m) > 0), we deduce that p = p′. This situation contributes P s rd(s) to P s rd(s)2, which is easily seen to be ≪ x/(d log y). Thus, we may assume that m ̸= m′. By Lemma 1, we have gcd(m, σ(m)) = gcd(m′, σ(m′)) = d, so that d | (s(m), s(m′)). Write gcd(s(m), s(m′)) = dh. By Lemma 1, every prime factor of h exceeds y. We have from (2), ps(m) dh −p′ s(m′) dh = σ(m′) −σ(m) dh . (3) For fixed m, m′, we count the number of pairs of primes p, p′ that satisfy this equation. Note that σ(m) ̸= σ(m′), since if they would be equal, we would then get from (2) that ps(m) = p′s(m′), and since min{p, p′} > max{m, m′} > max{s(m), s(m′)}, we would get that s(m) = s(m′), so m = m′, which is false. Let u, u′ be the integral solution of the linear equation (3) in p, p′ with u > 0 and minimal. Then p = u + s(m′) dh t and p′ = u′ + s(m) dh t 6 are both primes and 0 ≤t ≤(x/m)/(s(m′)/dh) = xdh/(ms(m′)). Let A = s(m) dh × s(m′) dh × |σ(m) −σ(m′)| dh =: A1A2A3, say. By the sieve, the number of such p ≤x/m is ≪ xdh ms(m′)(log(xdh/ms(m′)))2 A ϕ(A) ≪ xdh mm′(log x)2 A1 ϕ(A1) A2 ϕ(A2) A3 ϕ(A3), (4) where the above inequality follows because ms(m′) ≤mm′ ≤x14/15 and s(m′) ≫m′. Since s(m)/(dh) and s(m′)/(dh) are deficient, it follows that A1 ϕ(A1) ≪1, A2 ϕ(A2) ≪1 However, A3/ϕ(A3) is not small. In fact, by Lemma 1, we may assume that A3 is divisible by all primes ≤y = y(x), so log y ≪A3/ϕ(A3) ≪ log log x. Write A3 = A3,1A3,2A3,3, where A3,1 is the largest divisor with P(A3,1) ≤(log log x)2 and A3,2 is the largest divisor of what remains with P(A3,2) ≤log x. Since A3 has O(log x/ log log x) distinct prime factors, it follows that A3,3/ϕ(A3,3) ∼1 as x →∞and so A1A2A3 ϕ(A1)ϕ(A2)ϕ(A3) ≪ A3 ϕ(A3) ≪ A3,2 ϕ(A3,2) log y. (5) Let A′ 3,2 be the largest divisor of A3,2 which is coprime to σ(m). By Lemma 4, we may assume that A3,2/ϕ(A3,2) ≪A′ 3,2/ϕ(A′ 3,2). From (4), we now have the problem of showing that for d ∈D, x log y (log x)2 X m,m′ dhA′ 3,2 mm′ϕ(A′ 3,2) ≪ x d log y, (6) where dh = gcd(s(m), s(m′)). We first sum over m, m′ with h > x1/3, showing that the contribution to (6) is small. With m = qℓand h | s(m), we have s(m) = qs(ℓ) + σ(ℓ) ≡0 (mod h). (7) In addition, h and σ(ℓ) are coprime. Indeed, if some prime π | gcd(h, σ(ℓ)), then π = q or π | s(ℓ). In the latter case, π | ℓ, so π | n. But π | σ(ℓ) implies that π | σ(n), so we have a contradiction to our assumption that the properties in Lemma 1 hold. If π = q, since π | σ(ℓ), we again get 7 π | gcd(n, σ(n)), a contradiction. So, given h, ℓwe have from (7) that q is in a fixed coprime residue class modulo h; say q ≡ah,ℓ (mod h). Similarly, we have m′ = q′ℓ′ and q′ ≡ah,ℓ′ (mod h). Since h | gcd(s(m), s(m′)), (2) implies that h | σ(m) −σ(m′)), so that m ≡m′ (mod h). With (7) we get that ℓσ(ℓ) s(ℓ) ≡−qℓ= −m ≡−m′ = −q′ℓ′ ≡ℓ′σ(ℓ′) s(ℓ′) (mod h), which implies s(ℓ′)ℓσ(ℓ) −s(ℓ)ℓ′σ(ℓ′) ≡0 (mod h). (8) The absolute value of the left-hand side is < 2 max{ℓ3, ℓ′3} < 2x3/10. Thus, for h > x1/3, then it must be the case that the integer in the left–hand side of the above congruence must be the zero integer. We thus get that ℓσ(ℓ) s(ℓ) = ℓ′σ(ℓ′) s(ℓ′) , or equivalently, ℓ2 s(ℓ) + ℓ= ℓ′2 s(ℓ′) + ℓ′. (9) For us, gcd(ℓ, s(ℓ)) = gcd(ℓ′, s(ℓ′)) = d. Further, by property (3) in Lemma 1, d rad(d) | gcd(σ(ℓ), σ(ℓ′)), where rad(d) is the largest squarefree divisor of d. Hence, gcd(ℓ2, s(ℓ)) = d and the same is true for gcd(ℓ′2, s(ℓ′)). Putting ℓ= dλ, ℓ′ = dλ′, we get that dλ2 s(ℓ)/d − dλ′2 s(ℓ′)/d = ℓ−ℓ′, and the two fractions appearing in the left-hand side above are reduced. So, their denominators must be equal, that is, s(ℓ)/d = s(ℓ′)/d, therefore s(ℓ) = s(ℓ′). Now equation (9) gives ℓ2 + ℓs(ℓ) = ℓ′2 + ℓ′s(ℓ), and since the function t2 + ts(ℓ) is increasing in t, this gives ℓ= ℓ′. Thus, in the case h > x1/3, we must have ℓ= ℓ′ and the congruence classes ah,ℓ, ah,ℓ′ of q and q′ modulo h are the same. Summing the expression in (6) over m, m′ where h | gcd(s(m), s(m′)), h > x1/3, and using the maximal order of A′ 3,2/ϕ(A′ 3,2) , we have dx log log x (log x)2 X m,m′,h h mm′ = dx log log x (log x)2 X q,q′,ℓ,h h qq′ℓ2 . 8 Since ℓ= ℓ′ and m ̸= m′, we have q ̸= q′; assume that q > q′. Since q ≡q′ ≡ah,ℓ(mod h), the sum of 1/q above is O((log x)/h) and the sum of 1/q′ is O(1), even forgetting that q, q′ are prime. Thus, the above sum reduces to dx log log x (log x)2 X ℓ,h log x ℓ2 ≤dx log log x log x X ℓ τ(s(m′)) ℓ2 . The sum of 1/ℓ2 is O(x−1/10), so by Lemma 3, we have the estimate x9/10(log x)O(1) = O  x d log y  , which is consistent with (6). We now turn to values of h with h ≤x1/3. Since s(m′) is deficient, s(m′)/ϕ(s(m′)) ≪1, so that A′ 3,2/ϕ(A′ 3,2) ≪A′′ 3,2/ϕ(A′′ 3,2), where A′′ 3,2 is the largest divisor of A′ 3,2 coprime to s(m′). Fix m′, h with h | s(m′) and consider numbers m that can arise. As noted before, m ≡σ(m) ≡σ(m′) ≡m′ (mod h). Since h | s(m) and gcd(m, σ(m)) = d, we have gcd(m, h) = gcd(σ(m), h) = 1. Thus, the above congruences, rewritten as qrk ≡(q + 1)(r + 1)σ(k) ≡m′ (mod h), determine qr (mod h) and q + r (mod h). Hence, there are at most τ(h) pairs a, b such that q ≡a (mod h) and r ≡b (mod h). Fix one of these pairs a, b. Define f(m) = X π|σ(m)−σ(m′) (log log x)2<π≤log x π ∤hσ(m) 1 π, where π runs over primes. Note that if f(m) ≤1, then A′′ 3,2/ϕ(A′′ 3,2) ≪1. Say m = qℓand π have π | σ(m) −σ(m′) and π ∤σ(m). Since qσ(ℓ) = −σ(ℓ) + σ(m) ≡−σ(ℓ) + σ(m′) (mod π), if ℓ, π are fixed, then q is in a residue class modulo π, say cπ,ℓ(mod π). To summarize, with m′, h, ℓ, a, b fixed, if m = qℓ= qrk has π | A′′ 3,2, we have q ≡cπ,ℓ(mod π), q ≡a (mod h), r ≡b (mod h). Since π ∤h, the two 9 congruences for q may be combined to put q in a single residue class modulo πh. Thus, using h ≤x1/3 and π ≤log x, X m f(m) m ≪ X π 1 π X k 1 k X r 1 r X q 1 q ≪ X π 1 π2h X k 1 k X r 1 r. To estimate P r 1 r we consider two ranges for h. Since r ≡b (mod h), we have X r 1 r ≪ ( log log x h + log x x1/15 , if h > x1/20, 1 h, if h ≤x1/20. Thus, in the case that h > x1/20, we have X m f(m) m ≪ X π 1 π2h X k 1 k log log x h + log x x1/15  ≪ X π log x log log x π2h2d log y + X π (log x)2 π2hdx1/15 ≪ log x h2d log y log log x + (log x)2 hdx1/15 , (10) while in the case h ≤x1/20, a similar calculation shows that X m f(m) m ≪ X π 1 π2h X k 1 kh ≪ X π log x π2h2d log y ≪ log x h2d log y(log log x)2 . (11) The expression in (6) for h ≤x1/3 can be dealt with as follows. Fix m′, h, a, b. Since A′ 3,2/ϕ(A′ 3,2) ≪1 or log log x/ log y depending on whether f(m) ≤1 or f(m) > 1, x log y (log x)2 X m dhA′ 3,2 mm′ϕ(A′ 3,2) ≪x log y (log x)2 dh m′  X f(m)≤1 1 m + X f(m)>1 log log x m log y   ≤x log y (log x)2 dh m′ X m 1 m + x log log x (log x)2 dh m′ X m f(m) m = S1 + S2, say. First assume that x1/20 < h ≤x1/3. Writing m = qrk and with m′, h, a, b 10 fixed, we have X m 1 m ≪ X k 1 k X r 1 r X q 1 q ≪1 h X k 1 k X r 1 r ≪1 h X k 1 k log log x h + log x x1/15  ≪log x log log x h2d log y + (log x)2 hdx1/15 log y. Thus, summing over choices for m′, h, a, b, X m′,h,a,b S1 ≪x log y (log x)2 X m′,h,a,b dh m′ log x log log x h2d log y + (log x)2 hdx1/15 log y  ≤ x log x X m′,h τ(h) log log x hm′ + τ(h) log x m′x1/15  ≪x log log x log x X m′,h τ(h) x1/20m′ . Now X h|s(m′) τ(h) ≤τ(s(m′))2 ≤(log x)1.4, using Lemma 3. Thus, X m′,h,a,b S1 ≪x19/20(log x)O(1) X m′ 1 m′ ≤x19/20(log x)O(1), which is consistent with our goal in (6). For S2 we use our estimate (10) for the sum of f(m)/m and summing over choices for m′, h, a, b, we get that X m′,h,a,b S2 ≪x log log x (log x)2 X m′,h dhτ(h) m′  log x h2d log y log log x + (log x)2 hdx1/15  = x log x log y X m′,h τ(h) hm′ + x14/15 log log x X m′,h τ(h) m′ ≪ x log x log y (log x)1.4 x1/20 X m′ 1 m′ + x14/15 log log x(log x)1.4 X m′ 1 m′ ≤x19/20(log x)O(1), which is also consistent with (6). 11 It remains to consider the case h ≤x1/20. For a given choice of m′, h, a, b, we have X m 1 m ≪ X k 1 k X q 1 q X r 1 r ≪1 h2 X k 1 k ≪ log x dh2 log y. Since P h τ(h)/h ≤(P h 1/h)2 < 4, we have X m′,h,a,b S1 ≪x log y (log x)2 X m′,h,a,b dh m′ log x dh2 log y ≤ x log x X m′,h τ(h) hm′ ≪ x log x X m′ 1 m′ ≪ x d log y. This is again consistent with the goal in (6). Finally, we use (11) to see that X m′,h,a,b S2 ≪x log log x (log x)2 X m′,h dhτ(h) m′ log x h2d log y(log log x)2 ≪ x log x log y log log x X m′,h τ(h) hm′ ≪ x log x log y log log x X m′ 1 m′ ≪ x d(log y)2 log log x, which also is in line with (6) This calculation allows us to conclude that (6) holds, which we have seen then implies our theorem. This completes the proof. 4 Acknowledgements Research of F. L. on this project was carried on while he visited the Mathe-matics Department of Dartmouth College in Spring 2014. F. L. thanks the Mathematics Department of Dartmouth College for their hospitality. We thank Paul Pollack for his interest in this paper. 12 References Y.-G. Chen and Q.-Q. Zhao, Nonaliquot numbers, Publ. Math. De-brecen 78 (2011), no. 2, 439–442. J.-M. de Koninck and F. Luca, On the composition of the Euler func-tion and the sum of divisors function, Colloq. Math. 108 (2007), no. 1, 31–51. P. Erd˝ os, On perfect and multiply perfect numbers, Ann. Mat. Pura Appl. 42 (1956), 253–258. P. Erd˝ os, ¨ Uber die Zahlen der Form σ(n) −n und n −ϕ(n), Elem. Math. 11 (1973), 83–86. P. Erd˝ os, A. Granville, C. Pomerance, and C. Spiro, On the normal behavior of the iterates of some arithmetic functions Analytic Num-ber Theory, Proc. Conf. in honor of Paul T. Bateman, B. C. Berndt, et al. eds., Birkhauser, Boston, 1990, pp. 165–204. P. Erd˝ os, F. Luca, and C. Pomerance, On the proportion of numbers coprime to a given integer, Anatomy of integers, 47–64, CRM Proc. Lecture Notes, 46, Amer. Math. Soc., Providence, RI, 2008. I. K´ atai and M. V. Subbarao, Some further remarks on the iterates of the ϕ and the σ-functions, Annales Univ. Sci. Budapest, Sect. Comp. 26 (2006), 51–63. M. Kobayashi, On the density of abundant numbers, PhD thesis, Dartmouth College, 2010. M. Kobayashi, P. Pollack, and C. Pomerance, On the distribution of sociable numbers, J. Number Theory 129 (2009), 1990–2009. J. Pintz, Recent results on the Goldbach conjecture, Elementare und analytische Zahlentheorie, 220–254, Schr. Wiss. Ges. Johann Wolf-gang Goethe Univ. Frankfurt am Main, 20, Franz Steiner Verlag Stuttgart, Stuttgart, 2006. P. Pollack, On the greatest common divisor of a number and its sum of divisors, Michigan Math. J. 60 (2011), 199–214. C. Pomerance and H.-S. Yang, Variant of a theorem of Erd˝ os on the sum-of-proper-divisors function, Math. Comp. 83 (2014), 1903–1913. 13 L. Troupe, On the number of prime factors of values of the sum-of-proper-divisors function. arXiv 1405.3587. 14
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https://www.youtube.com/watch?v=ZfCmNPkzcyQ
Graph x-y-1=0 . graph of lines in the Cartesian plane , graph of linear functions Mate316 533000 subscribers Description 217 views Posted: 1 Aug 2024 how to graph a line graph of linear functions line graph Graph of f(x) linear function graph lines by tabulating m316 Transcript: here the any number or here any number for example 0 x is 0 0 - Y is - y - 1 = 0 equation Y is - one okay here here any number or here any number for example y zero x - 0 is x - 1 = 0 equation x is 1 X is zero therefore in y y is min -1 negative positive minus one is here point in y because X is zero Y is zero therefore in x x is 1 positive negative positive one is here point in X because Y is zero okay
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https://oeis.org/A003586
A003586 - OEIS login The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!) A003586 3-smooth numbers: numbers of the form 2^i3^j with i, j >= 0. 356 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 243, 256, 288, 324, 384, 432, 486, 512, 576, 648, 729, 768, 864, 972, 1024, 1152, 1296, 1458, 1536, 1728, 1944, 2048, 2187, 2304, 2592, 2916, 3072, 3456, 3888 (list; graph; refs; listen; history; text; internal format) OFFSET 1,2 COMMENTS This sequence is easily confused with A033845, which gives numbers of the form 2^i3^j with i, j >= 1. Don't simply say "numbers of the form 2^i3^j", but specify which sequence you mean. - N. J. A. Sloane, May 26 2024 These numbers were once called "harmonic numbers", see Lenstra links. - N. J. A. Sloane, Jul 03 2015 Successive numbers k such that phi(6k) = 2k. - Artur Jasinski, Nov 05 2008 Where record values greater than 1 occur in A088468: A160519(n) = A088468(a(n)). - Reinhard Zumkeller, May 16 2009 Also numbers that are divisible by neither 6k - 1 nor 6k + 1, for all k > 0. - Robert G. Wilson v, Oct 26 2010 Also numbers m such that the rooted tree with Matula-Goebel number m has m antichains. The Matula-Goebel number of a rooted tree can be defined in the following recursive manner: to the one-vertex tree there corresponds the number 1; to a tree T with root degree 1 there corresponds the t-th prime number, where t is the Matula-Goebel number of the tree obtained from T by deleting the edge emanating from the root; to a tree T with root degree m>=2 there corresponds the product of the Matula-Goebel numbers of the m branches of T. The vertices of a rooted tree can be regarded as a partially ordered set, where u<=v holds for two vertices u and v if and only if u lies on the unique path between v and the root. An antichain is a nonempty set of mutually incomparable vertices. Example: m=4 is in the sequence because the corresponding rooted tree is \/=ARB (R is the root) having 4 antichains (A, R, B, AB). - Emeric Deutsch, Jan 30 2012 A204455(3a(n)) = 3, and only for these numbers. - Wolfdieter Lang, Feb 04 2012 The number of terms less than or equal to n is Sum_{i=0..floor(log_2(n))} floor(log_3(n/2^i) + 1), or Sum_{i=0..floor(log_3(n))} floor(log_2(n/3^i) + 1), which requires fewer terms to compute. - Robert G. Wilson v, Aug 17 2012 Named 3-friables in French. - Michel Marcus, Jul 17 2013 In the 14th century Levi Ben Gerson proved that the only pairs of terms which differ by 1 are (1,2), (2,3), (3,4), and (8,9); see A235365, A235366, A236210. - Jonathan Sondow, Jan 20 2014 Range of values of A000005(n) (and also A181819(n)) for cubefree numbers n. - Matthew Vandermast, May 14 2014 A036561 is a permutation of this sequence. - L. Edson Jeffery, Sep 22 2014 Also the sorted union of A000244 and A007694. - Lei Zhou, Apr 19 2017 The sum of the reciprocals of the 3-smooth numbers is equal to 3. Brief proof: 1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + ... = (Sum_{k>=0} 1/2^k) (Sum_{m>=0} 1/3^m) = (1/(1-1/2)) (1/(1-1/3)) = (2/(2-1)) (3/(3-1)) = 3. - Bernard Schott, Feb 19 2019 Also those integers k for which, for every prime p > 3, p^(2k) - 1 == 0 (mod 24k). - Federico Provvedi, May 23 2022 For n>1, the exponents’ parity {parity(i), parity(j)} of one out of four consecutive terms is {odd, odd}. Therefore, for n>1, at least one out of every four consecutive terms is a Zumkeller number (A083207). If for the term whose parity is {even, odd}, even also means nonzero, then this term is also a Zumkeller number (as is the case with the last of the four consecutive terms 1296, 1458, 1536, 1728). - Ivan N. Ianakiev, Jul 10 2022 Except the initial terms 2, 3, 4, 8, 9 and 16, these are numbers k such that k^6 divides 6^k. Except the initial terms 2, 3, 4, 6, 8, 9, 16, 18 and 27, these are numbers k such that k^12 divides 12^k. - Mohammed Yaseen, Jul 21 2022 In music theory, a comma is a ratio, close to 1 (typically less than 1.04), between two natural numbers divisible by only small primes (typically single digit). In this sequence, a(131) / a(130) = 531441 / 524288 ~ 1.013643 is the Pythagorean comma (A221363), the difference between 12 perfect fifths and 7 octaves. - Hal M. Switkay, Mar 23 2025 REFERENCES J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 654 pp. 85, 287-8, Ellipses Paris 2004. S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962, p. xxiv. R. Tijdeman, Some applications of Diophantine approximation, pp. 261-284 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003. LINKS Lei Zhou, Table of n, a(n) for n = 1..10000 (first 501 terms from Franklin T. Adams-Watters) R. Blecksmith, M. McCallum and J. L. Selfridge, 3-smooth representations of integers, Amer. Math. Monthly, 105 (1998), 529-543. Thierry Bousch, La Tour de Stockmeyer, Séminaire Lotharingien de Combinatoire 77 (2017), Article B77d. Benoit Cloitre, a(n)/((1/sqrt(6))exp(sqrt(2log(2)log(3)n))) for 0<n<10^5 Natalia da Silva, Serban Raianu, and Hector Salgado, Differences of Harmonic Numbers and the abc-Conjecture, arXiv:1708.00620 [math.NT], 2017. Emeric Deutsch, Rooted tree statistics from Matula numbers, arXiv:1111.4288 [math.CO], 2011. David Eppstein, Making Change in 2048, arXiv:1804.07396 [cs.DM], 2018. F. Goebel, On a 1-1-correspondence between rooted trees and natural numbers, J. Combin. Theory, B 29 (1980), 141-143. I. Gutman and A. Ivic, On Matula numbers, Discrete Math., 150, 1996, 131-142. I. Gutman and Yeong-Nan Yeh, Deducing properties of trees from their Matula numbers, Publ. Inst. Math., 53 (67), 1993, 17-22. A. M. Hinz, S. Klavžar, U. Milutinović, and C. Petr, The Tower of Hanoi - Myths and Maths, Birkhäuser 2013. See page 252. Book's website H. W. Lenstra Jr., Harmonic Numbers H. W. Lenstra, Jr., Harmonic Numbers and the ABC-conjecture, Abstract of talk, May 30, 2001 [Annotated scanned copy] D. Matula, A natural rooted tree enumeration by prime factorization, SIAM Rev. 10 (1968) 273. D. J. Mintz, 2,3 sequence as a binary mixture, Fib. Quarterly, Vol. 19, No 4, Oct 1981, pp. 351-360. I. Peterson, Medieval Harmony Raphael Schumacher, The Formulas for the Distribution of the 3-Smooth, 5-Smooth, 7-Smooth and all other Smooth Numbers, arXiv preprint arXiv:1608.06928 [math.NT], 2016. Eric Weisstein's World of Mathematics, Smooth Number FORMULA An asymptotic formula for a(n) is roughly a(n) ~ 1/sqrt(6)exp(sqrt(2log(2)log(3)n)). - Benoit Cloitre, Nov 20 2001 A061987(n) = a(n + 1) - a(n), a(A084791(n)) = A084789(n), a(A084791(n) + 1) = A084790(n). - Reinhard Zumkeller, Jun 03 2003 Union of powers of 2 and 3 with n such that psi(n) = 2n, where psi(n) = nProduct_(1 + 1/p) over all prime factors p of n = A001615(n). - Lekraj Beedassy, Sep 07 2004; corrected by Franklin T. Adams-Watters, Mar 19 2009 a(n) = 2^A022328(n)3^A022329(n). - N. J. A. Sloane, Mar 19 2009 The characteristic function of this sequence is given by Sum_{n >= 1} x^a(n) = Sum_{n >= 1} moebius(6n)x^n/(1 - x^n). - Paul D. Hanna, Sep 18 2011 a(n) = A007694(n+1)/2. - Lei Zhou, Apr 19 2017 MAPLE A003586 := proc(n) option remember; if n = 1 then 1; else for a from procname(n-1)+1 do numtheoryfactorset minus {2, 3} ; if % = {} then return a; end if; end do: end if; end proc: # R. J. Mathar, Feb 28 2011 with(numtheory): for i from 1 to 23328 do if(i/phi(i)=3)then print(i/6) fi od; # Gary Detlefs, Jun 28 2011 MATHEMATICA a = 1; j = 1; k = 1; n = 100; For[k = 2, k <= n, k++, If[2a[k - j] < 3^j, a[k] = 2a[k - j], {a[k] = 3^j, j++}]]; Table[a[i], {i, 1, n}] ( Hai He (hai(AT)mathteach.net) and Gilbert Traub, Dec 28 2004 ) aa = {}; Do[If[EulerPhi[6 n] == 2 n, AppendTo[aa, n]], {n, 1, 1000}]; aa ( Artur Jasinski, Nov 05 2008 ) fQ[n_] := Union[ MemberQ[{1, 5}, # ] & /@ Union@ Mod[ Rest@ Divisors@ n, 6]] == {False}; fQ = True; Select[ Range@ 4000, fQ] ( Robert G. Wilson v, Oct 26 2010 ) powerOfTwo = 12; Select[Nest[Union@Join[#, 2#, 3#] &, {1}, powerOfTwo-1], # < 2^powerOfTwo &] ( Robert G. Wilson v and T. D. Noe, Mar 03 2011 ) fQ[n_] := n == 3 EulerPhi@ n; Select[6 Range@ 4000, fQ]/6 ( Robert G. Wilson v, Jul 08 2011 ) mx = 4000; Sort@ Flatten@ Table[2^i3^j, {i, 0, Log[2, mx]}, {j, 0, Log[3, mx/2^i]}] ( Robert G. Wilson v, Aug 17 2012 ) f[n_] := Block[{p2, p3 = 3^Range[0, Floor@ Log[3, n] + 1]}, p2 = 2^Floor[Log[2, n/p3] + 1]; Min[ Select[ p2p3, IntegerQ]]]; NestList[f, 1, 54] ( Robert G. Wilson v, Aug 22 2012 ) Select[Range@4000, Last@Map[First, FactorInteger@#] <= 3 &] ( Vincenzo Librandi, Aug 25 2016 ) Select[Range, Max[FactorInteger[#]]<4&] ( Harvey P. Dale, Jan 11 2017 ) PROG (PARI) test(n)=for(p=2, 3, while(n%p==0, n/=p)); n==1; for(n=1, 4000, if(test(n), print1(n", "))) (PARI) list(lim)=my(v=List(), N); for(n=0, log(lim\1+.5)\log(3), N=3^n; while(N<=lim, listput(v, N); N<<=1)); vecsort(Vec(v)) \ Charles R Greathouse IV, Jun 28 2011 (PARI) is_A003586(n)=n<5||vecmax(factor(n, 5)[, 1])<5 \ M. F. Hasler, Jan 16 2015 (PARI) list(lim)=my(v=List(), N); for(n=0, logint(lim\=1, 3), N=3^n; while(N<=lim, listput(v, N); N<<=1)); Set(v) \ Charles R Greathouse IV, Jan 10 2018 (Haskell) import Data.Set (Set, singleton, insert, deleteFindMin) smooth :: Set Integer -> [Integer] smooth s = x : smooth (insert (3x) $ insert (2x) s') where (x, s') = deleteFindMin s a003586_list = smooth (singleton 1) a003586 n = a003586_list !! (n-1) -- Reinhard Zumkeller, Dec 16 2010 (SageMath) def isA003586(n) : return not any(d != 2 and d != 3 for d in prime_divisors(n)) @CachedFunction def A003586(n) : if n == 1 : return 1 k = A003586(n-1) + 1 while not isA003586(k) : k += 1 return k [A003586(n) for n in (1..55)] # Peter Luschny, Jul 20 2012 (Python) from itertools import count, takewhile def aupto(lim): pows2 = list(takewhile(lambda x: x<lim, (2i for i in count(0)))) pows3 = list(takewhile(lambda x: x<lim, (3i for i in count(0)))) return sorted(cd for c in pows2 for d in pows3 if cd <= lim) print(aupto(104)) # Michael S. Branicky, Jul 08 2022 (Python) from sympy import integer_log def A003586(n): def bisection(f, kmin=0, kmax=1): while f(kmax) > kmax: kmax <<= 1 while kmax-kmin > 1: kmid = kmax+kmin>>1 if f(kmid) <= kmid: kmax = kmid else: kmin = kmid return kmax def f(x): return n+x-sum((x//3i).bit_length() for i in range(integer_log(x, 3)+1)) return bisection(f, n, n) # Chai Wah Wu, Sep 15 2024 (Python) # faster for initial segment of sequence import heapq from itertools import islice def A003586gen(): # generator of terms v, oldv, h, psmooth_primes, = 1, 0, , [2, 3] while True: v = heapq.heappop(h) if v != oldv: yield v oldv = v for p in psmooth_primes: heapq.heappush(h, vp) print(list(islice(A003586gen(), 65))) # Michael S. Branicky, Sep 17 2024 (C++) // Returns A003586<= threshold without approximations nor sorting include std::forward_listA003586(const int threshold) { std::forward_list sequence; auto start_it = sequence.before_begin(); for (int i = 1; i <= threshold; i = 2) { for (int inc = 1; std::next(start_it) != sequence.end() && inc <= i; inc = 3) ++start_it; auto it = start_it; for (int j = 1; i j <= threshold; j = 3) { sequence.emplace_after(it, i j); for (int inc = 1; std::next(it) != sequence.end() && inc <= i; inc = 2) ++it; } } return sequence; } // Eben Gino Lester, Apr 17 2025 (Magma) [n: n in [1..4000] | PrimeDivisors(n) subset [2, 3]]; // Bruno Berselli, Sep 24 2012 CROSSREFS Cf. A051037, A002473, A051038, A080197, A080681, A080682, A117221, A105420, A062051, A117222, A117220, A090184, A131096, A131097, A186711, A186712, A186771, A088468, A061987, A080683 (p-smooth numbers with other values of p), A025613 (a subsequence). Cf. also A235365, A235366, A236210, A036561. Cf. also A000244, A007694. - Lei Zhou, Apr 19 2017 Cf. A191475 (successive values of i), A191476 (successive values of j), A022330 (indices of the pure terms 2^i), A022331 (indices of the pure terms 3^j). - N. J. A. Sloane, May 26 2024 Cf. A221363. Sequence in context: A301704A083854A275199 A114334A262609A018690 Adjacent sequences: A003583A003584A003585 A003587A003588A003589 KEYWORD nonn,easy,nice,changed AUTHOR Paul Zimmermann, Dec 11 1996 EXTENSIONS Deleted claim that this sequence is union of 2^n (A000079) and 3^n (A000244) sequences -- this does not include the terms which are not pure powers. - Walter Roscello (wroscello(AT)comcast.net), Nov 16 2008 STATUS approved LookupWelcomeWikiRegisterMusicPlot 2DemosIndexWebCamContributeFormatStyle SheetTransformsSuperseekerRecents The OEIS Community Maintained by The OEIS Foundation Inc. Last modified September 29 04:57 EDT 2025. Contains 388827 sequences. License Agreements, Terms of Use, Privacy Policy
8468
https://www.tiger-algebra.com/en/solution/combination-without-repetition/c%285%2C3%29/
Copyright Ⓒ 2013-2025 tiger-algebra.com This site is best viewed with Javascript. If you are unable to turn on Javascript, please click here. Solution - Combinations without repetition Other Ways to Solve Step-by-step explanation 1. Find the number of terms in the set n represents the total number of items in the set: c(n,k) c(5,3) n=5 2. Find the number of items selected from the set k represents the number of items selected from the set: c(n,k) c(5,3) k=3 3. Calculate the combinations using the formula Plug n (n=5) and k (k=3) into the combination formula: C(n,k)=n!k!(n-k)! C(5,3)=5!3!(5-3)! C(5,3)=5!3!·2! C(5,3)=5·4·3!3!·2! C(5,3)=5·42! C(5,3)=5·42·1 C(5,3)=10 There are 10 ways that 3 items chosen from a set of 5 can be combined. How did we do? Why learn this Combinations and permutations If you have 2 types of crust, 4 types of toppings, and 3 types of cheese, how many different pizza combinations can you make? If there are 8 swimmers in a race, how many different sets of 1st, 2nd, and 3rd place winners could there be? What are your chances of winning the lottery? All of these questions can be answered using two of the most fundamental concepts in probability: combinations and permutations. Though these concepts are very similar, probability theory holds that they have some important differences. Both combinations and permutations are used to calculate the number of possible combinations of things. The most important difference between the two, however, is that combinations deal with arrangements in which the order of the items being arranged does not matter—such as combinations of pizza toppings—while permutations deal with arrangements in which the order the items being arranged does matter—such as setting the combination to a combination lock, which should really be called a permutation lock because the order of the input matters. What these two concepts have in common, is that they both help us understand the relationships between sets and the items or subsets that make up those sets. As the examples above illustrate, this can be used to better understand many different types of situations. Combinations and permutations Terms and topics Related links Latest Related Drills Solved Copyright Ⓒ 2013-2025 tiger-algebra.com
8469
https://pmc.ncbi.nlm.nih.gov/articles/PMC8951218/
ESHRE guideline: endometriosis - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer | PMC Copyright Notice Hum Reprod Open . 2022 Feb 26;2022(2):hoac009. doi: 10.1093/hropen/hoac009 Search in PMC Search in PubMed View in NLM Catalog Add to search ESHRE guideline: endometriosis† Christian M Becker Christian M Becker 1 Nuffield Department of Women’s and Reproductive Health, Endometriosis CaRe Centre, University of Oxford, Oxford, UK Find articles by Christian M Becker 1,✉, Attila Bokor Attila Bokor 2 Department of Obstetrics and Gynecology, Semmelweis University, Budapest, Hungary Find articles by Attila Bokor 2, Oskari Heikinheimo Oskari Heikinheimo 3 Department of Obstetrics & Gynecology, University of Helsinki and Helsinki University Hospital, Helsinki, Finland Find articles by Oskari Heikinheimo 3, Andrew Horne Andrew Horne 4 EXPPECT Centre for Endometriosis and Pelvic Pain, MRC Centre for Reproductive Health, University of Edinburgh, Edinburgh, UK Find articles by Andrew Horne 4, Femke Jansen Femke Jansen 5 EndoHome—Endometriosis Association Belgium, Belgium Find articles by Femke Jansen 5, Ludwig Kiesel Ludwig Kiesel 6 Department of Gynecology and Obstetrics, University Hospital Muenster, Muenster, Germany Find articles by Ludwig Kiesel 6, Kathleen King Kathleen King 7 Individual Endometriosis Advocate, Ireland Find articles by Kathleen King 7, Marina Kvaskoff Marina Kvaskoff 8 Paris-Saclay University, UVSQ, Univ. Paris-Sud, Inserm, Gustave Roussy, “Exposome and Heredity” Team, CESP, Villejuif, France Find articles by Marina Kvaskoff 8, Annemiek Nap Annemiek Nap 9 Department of Gynaecology and Obstetrics, Radboudumc, Nijmegen, The Netherlands Find articles by Annemiek Nap 9, Katrine Petersen Katrine Petersen 10 Pain Management Centre, UCLH, London, UK Find articles by Katrine Petersen 10, Ertan Saridogan Ertan Saridogan 11 Department of Obstetrics and Gynaecology, University College London Hospital, London, UK 12 Elizabeth Garrett Anderson Institute for Women’s Health, University College London, London, UK Find articles by Ertan Saridogan 11,12, Carla Tomassetti Carla Tomassetti 13 Department of Obstetrics and Gynaecology, Leuven University Fertility Center, University Hospitals Leuven, Leuven, Belgium 14 Faculty of Medicine, Department of Development and Regeneration, LEERM (Lab of Endometrium, Endometriosis and Reproductive Medicine), KU Leuven, Leuven, Belgium Find articles by Carla Tomassetti 13,14, Nehalennia van Hanegem Nehalennia van Hanegem 15 Department of Reproductive Medicine and Gynecology, University Medical Center Utrecht, Utrecht, The Netherlands Find articles by Nehalennia van Hanegem 15, Nicolas Vulliemoz Nicolas Vulliemoz 16 Department of Woman Mother Child, Fertility Medicine and Gynaecological Endocrinology, Lausanne University Hospital, Lausanne, Switzerland Find articles by Nicolas Vulliemoz 16, Nathalie Vermeulen Nathalie Vermeulen 17 European Society of Human Reproduction and Embryology, Strombeek-Bever, Belgium Find articles by Nathalie Vermeulen 17; ESHRE Endometriosis Guideline Group 2 Author information Article notes Copyright and License information 1 Nuffield Department of Women’s and Reproductive Health, Endometriosis CaRe Centre, University of Oxford, Oxford, UK 2 Department of Obstetrics and Gynecology, Semmelweis University, Budapest, Hungary 3 Department of Obstetrics & Gynecology, University of Helsinki and Helsinki University Hospital, Helsinki, Finland 4 EXPPECT Centre for Endometriosis and Pelvic Pain, MRC Centre for Reproductive Health, University of Edinburgh, Edinburgh, UK 5 EndoHome—Endometriosis Association Belgium, Belgium 6 Department of Gynecology and Obstetrics, University Hospital Muenster, Muenster, Germany 7 Individual Endometriosis Advocate, Ireland 8 Paris-Saclay University, UVSQ, Univ. Paris-Sud, Inserm, Gustave Roussy, “Exposome and Heredity” Team, CESP, Villejuif, France 9 Department of Gynaecology and Obstetrics, Radboudumc, Nijmegen, The Netherlands 10 Pain Management Centre, UCLH, London, UK 11 Department of Obstetrics and Gynaecology, University College London Hospital, London, UK 12 Elizabeth Garrett Anderson Institute for Women’s Health, University College London, London, UK 13 Department of Obstetrics and Gynaecology, Leuven University Fertility Center, University Hospitals Leuven, Leuven, Belgium 14 Faculty of Medicine, Department of Development and Regeneration, LEERM (Lab of Endometrium, Endometriosis and Reproductive Medicine), KU Leuven, Leuven, Belgium 15 Department of Reproductive Medicine and Gynecology, University Medical Center Utrecht, Utrecht, The Netherlands 16 Department of Woman Mother Child, Fertility Medicine and Gynaecological Endocrinology, Lausanne University Hospital, Lausanne, Switzerland 17 European Society of Human Reproduction and Embryology, Strombeek-Bever, Belgium ESHRE Pages content is not externally peer reviewed. The manuscript has been approved by the Executive Committee of ESHRE. 2 The members of the ESHRE Endometriosis Guideline Group are given in the Appendix. ✉ Correspondence address. Nuffield Department of Women’s and Reproductive Health, Endometriosis CaRe Centre, John Radcliffe Hospital, University of Oxford, Level 3, Women’s Centre, Oxford OX3 9DU, UK. E-mail: christian.becker@wrh.ox.ac.uk; guidelines@eshre.eu Received 2022 Feb 10; Collection date 2022. © The Author(s) 2022. Published by Oxford University Press on behalf of European Society of Human Reproduction and Embryology. This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits non-commercial re-use, distribution, and reproduction in any medium, provided the original work is properly cited. For commercial re-use, please contact journals.permissions@oup.com PMC Copyright notice PMCID: PMC8951218 PMID: 35350465 Abstract STUDY QUESTION How should endometriosis be diagnosed and managed based on the best available evidence from published literature? SUMMARY ANSWER The current guideline provides 109 recommendations on diagnosis, treatments for pain and infertility, management of disease recurrence, asymptomatic or extrapelvic disease, endometriosis in adolescents and postmenopausal women, prevention and the association with cancer. WHAT IS KNOWN ALREADY Endometriosis is a chronic condition with a plethora of presentations in terms of not only the occurrence of lesions, but also the presence of signs and symptoms. The most important symptoms include pain and infertility. STUDY DESIGN, SIZE, DURATION The guideline was developed according to the structured methodology for development of ESHRE guidelines. After formulation of key questions by a group of experts, literature searches and assessments were performed. Papers published up to 1 December 2020 and written in English were included in the literature review. PARTICIPANTS/MATERIALS, SETTING, METHODS Based on the collected evidence, recommendations were formulated and discussed within specialist subgroups and then presented to the core guideline development group (GDG) until consensus was reached. A stakeholder review was organized after finalization of the draft. The final version was approved by the GDG and the ESHRE Executive Committee. MAIN RESULTS AND THE ROLE OF CHANCE This guideline aims to help clinicians to apply best care for women with endometriosis. Although studies mostly focus on women of reproductive age, the guideline also addresses endometriosis in adolescents and postmenopausal women. The guideline outlines the diagnostic process for endometriosis, which challenges laparoscopy and histology as gold standard diagnostic tests. The options for treatment of endometriosis-associated pain symptoms include analgesics, medical treatments and surgery. Non-pharmacological treatments are also discussed. For management of endometriosis-associated infertility, surgical treatment and/or medically assisted reproduction are feasible. While most of the more recent studies confirm previous ESHRE recommendations, there are five topics in which significant changes to recommendations were required and changes in clinical practice are to be expected. LIMITATIONS, REASONS FOR CAUTION The guideline describes different management options but, based on existing evidence, no firm recommendations could be formulated on the most appropriate treatments. Also, for specific clinical issues, such as asymptomatic endometriosis or extrapelvic endometriosis, the evidence is too scarce to make evidence-based recommendations. WIDER IMPLICATIONS OF THE FINDINGS The guideline provides clinicians with clear advice on best practice in endometriosis care, based on the best evidence currently available. In addition, a list of research recommendations is provided to stimulate further studies in endometriosis. STUDY FUNDING/COMPETING INTEREST(S) The guideline was developed and funded by ESHRE, covering expenses associated with the guideline meetings, with the literature searches and with the dissemination of the guideline. The guideline group members did not receive payments. C.M.B. reports grants from Bayer Healthcare and the European Commission; Participation on a Data Safety Monitoring Board or Advisory Board with ObsEva (Data Safety Monitoring Group) and Myovant (Scientific Advisory Group). A.B. reports grants from FEMaLE executive board member and European Commission Horizon 2020 grant; consulting fees from Ethicon Endo Surgery, Medtronic; honoraria for lectures from Ethicon; and support for meeting attendance from Gedeon Richter; A.H. reports grants from MRC, NIHR, CSO, Roche Diagnostics, Astra Zeneca, Ferring; Consulting fees from Roche Diagnostics, Nordic Pharma, Chugai and Benevolent Al Bio Limited all paid to the institution; a pending patent on Serum endometriosis biomarker; he is also Chair of TSC for STOP-OHSS and CERM trials. O.H. reports consulting fees and speaker’s fees from Gedeon Richter and Bayer AG; support for attending meetings from Gedeon-Richter, and leadership roles at the Finnish Society for Obstetrics and Gynecology and the Nordic federation of the societies of obstetrics and gynecology. L.K. reports consulting fees from Gedeon Richter, AstraZeneca, Novartis, Dr KADE/Besins, Palleos Healthcare, Roche, Mithra; honoraria for lectures from Gedeon Richter, AstraZeneca, Novartis, Dr KADE/Besins, Palleos Healthcare, Roche, Mithra; support for attending meetings from Gedeon Richter, AstraZeneca, Novartis, Dr KADE/Besins, Palleos Healthcare, Roche, Mithra; he also has a leadership role in the German Society of Gynecological Endocrinology (DGGEF). M.K. reports grants from French Foundation for Medical Research (FRM), Australian Ministry of Health, Medical Research Future Fund and French National Cancer Institute; support for meeting attendance from European Society for Gynaecological Endoscopy (ESGE), European Congress on Endometriosis (EEC) and ESHRE; She is an advisory Board Member, FEMaLe Project (Finding Endometriosis Using Machine Learning), Scientific Committee Chair for the French Foundation for Research on Endometriosis and Scientific Committee Chair for the ComPaRe-Endometriosis cohort. A.N. reports grants from Merck SA and Ferring; speaker fees from Merck SA and Ferring; support for meeting attendance from Merck SA; Participation on a Data Safety Monitoring Board or Advisory Board with Nordic Pharma and Merck SA; she also is a board member of medical advisory board, Endometriosis Society, the Netherlands (patients advocacy group) and an executive board member of the World Endometriosis Society. E.S. reports grants from National Institute for Health Research UK, Rosetrees Trust, Barts and the London Charity; Royalties from De Gruyter (book editor); consulting fees from Hologic; speakers fees from Hologic, Johnson & Johnson, Medtronic, Intuitive, Olympus and Karl Storz; Participation in the Medicines for Women’s Health Expert Advisory Group with Medicines and Healthcare Products Regulatory Agency (MHRA); he is also Ambassador for the World Endometriosis Society. C.T. reports grants from Merck SA; Consulting fees from Gedeon Richter, Nordic Pharma and Merck SA; speaker fees from Merck SA, all paid to the institution; and support for meeting attendance from Ferring, Gedeon Richter and Merck SA. The other authors have no conflicts of interest to declare. DISCLAIMER This guideline represents the views of ESHRE, which were achieved after careful consideration of the scientific evidence available at the time of preparation. In the absence of scientific evidence on certain aspects, a consensus between the relevant ESHRE stakeholders has been obtained. Adherence to these clinical practice guidelines does not guarantee a successful or specific outcome, nor does it establish a standard of care. Clinical practice guidelines do not replace the need for application of clinical judgement to each individual presentation, nor variations based on locality and facility type. ESHRE makes no warranty, express or implied, regarding the clinical practice guidelines and specifically excludes any warranties of merchantability and fitness for a particular use or purpose (Full disclaimer available atwww.eshre.eu/guidelines.). Keywords: endometriosis, guideline, fertility, pelvic pain, adolescent, surgery, ESHRE guideline WHAT DOES THIS MEAN FOR PATIENTS? Endometriosis is a chronic condition with a large impact not only on the patient’s quality of life, but also on social contacts and work. Endometriosis is characterized mainly by symptoms of pain (often linked to the menstrual cycle) and infertility. Often, endometriosis is subdivided according to the type and location of the lesions into peritoneal endometriosis, deep endometriosis and endometrioma. The current paper summarizes the ESHRE guideline on endometriosis, providing clinicians with evidence-based recommendations on the diagnosis and management of endometriosis-associated symptoms, including medical treatment, surgery and assisted reproduction. Information and recommendations are also provided on other topics related to endometriosis, such as prevention, pregnancy and cancer. Introduction Endometriosis is a disease characterized by the presence of endometrium-like epithelium and/or stroma outside the endometrium and myometrium, usually with an associated inflammatory process (International Working Group of AAGL, ESGE, ESHRE and WES et al., 2021). The exact prevalence of endometriosis is unknown, but estimates range from 2% to 10% within the general female population and up to 50% in infertile women (Eskenazi and Warner, 1997; Meuleman et al., 2009; Zondervan et al., 2020). The ESHRE Guideline for the Diagnosis and Treatment of Endometriosis (2005) and the ESHRE Guideline: Management of women with endometriosis (2013) have been a reference point for best clinical care in endometriosis for years (Kennedy et al., 2005; Dunselman et al., 2014). Based on continuous new research and developments, it was considered that the last recommendations formulated in 2013/2014 required a revision. Materials and methods The guideline was developed according to a well-documented methodology that is universal to ESHRE guidelines (Vermeulen et al., 2019). The core guideline development group (GDG) was composed of past members of the guideline group from 2013 and additional experts selected from applicants to a call for experts. All other European experts applying to the call were included as subgroup members, assisting a core group member preparing the guideline on a certain topic. The GDG included two patient representatives, and five patient organizations were represented in the subgroups. Forty-two key questions were formulated by the GDG, of which seven were answered as narrative questions, and 35 as PICO (Patient, Intervention, Comparison, Outcome) questions. For each PICO question, databases (PubMed/MEDLINE and the Cochrane library) were searched from inception to 1 December 2020, limited to studies written in English. From the literature searches, studies were selected based on the PICO questions, assessed for quality and summarized in evidence tables. GDG subgroup meetings were organized, face-to-face and online, for presentation and discussion of the evidence and draft recommendations by the assigned core group member. Proposed recommendations by the subgroups were then discussed in core group meetings until a consensus was reached. Each recommendation was labelled as strong or weak and a grade was assigned based on the strength of the supporting evidence (High ⊕⊕⊕⊕, Moderate ⊕⊕⊕◯, Low ⊕⊕◯◯ and Very low ⊕◯◯◯). Good practice points (GPPs) based on clinical expertise were added where relevant to clarify the recommendations or to provide further practical advice. ‘Research only’ recommendations were also made, and those interventions should be applied only within the context of research, with appropriate precautions and ethical approval. Strong recommendations should be used as a recommendation to be applied for most patients, while weak recommendations require discussion and shared decision-making (Fig. 1). Figure 1. Open in a new tab Suggested interpretation of strong and weak recommendations by patients, clinicians and health care policy makers. For the narrative questions, a similar literature search was conducted. Collected data were summarized in a narrative summary and conclusions were formulated. In case of insufficient data to provide recommendations in reply to a PICO question, a conclusion was also added. For clarity, these conclusions are labelled ‘conclusion, not recommendation’ in the current paper. The guideline draft and an invitation to participate in the stakeholder review were published on the ESHRE website between 24 June and 15 August 2021. All comments were processed by the core group, either by adapting the content of the guideline and/or by replying to the reviewer. The review process was summarized in the review report, which is published on the ESHRE website (www.eshre.eu/Guidelines). Overall, 56.5% of the 253 comments resulted in an adaptation or correction in the guideline text. This guideline will be considered for update 4 years after publication, with an intermediate assessment of the need for updating 2 years after publication. Results Key questions and recommendations The scope of the ESHRE guideline on endometriosis is to provide guidance on the management of endometriosis; either diagnosed or strongly suspected. In line with endometriosis research, terminology and discussion, the guideline is focused on females and menstruation. The GDG recognizes that there are individuals living with endometriosis who are transgender, who do not menstruate, who do not have a uterus or who do not identify with the terms used in the literature. Throughout, the term ‘women with endometriosis’ is used, but this is not intended to isolate, exclude or diminish any individual’s experience nor to discriminate against any group. Diagnosis of endometriosis The recommended diagnostic process for endometriosis is summarized in Fig. 2. Figure 2. Open in a new tab The recommended diagnostic process for endometriosis. DE, deep endometriosis; US, ultrasound. Can clinical symptoms predict the presence of endometriosis? The GDG recommends that clinicians should consider the diagnosis of endometriosis in individuals presenting with the following cyclical and non-cyclical signs and symptoms: dysmenorrhoea, deep dyspareunia, dysuria, dyschezia, painful rectal bleeding or haematuria, shoulder tip pain, catamenial pneumothorax, cyclical cough/haemoptysis/chest pain, cyclical scar swelling and pain, fatigue and infertility (Forman et al., 1993; Eskenazi et al., 2001; Ballard et al., 2008; Nnoaham et al., 2012).GPP Open in a new tab Does the use of symptom diaries or questionnaires compared to traditional history taking lead to improved or earlier diagnosis of endometriosis? As no recommendation could be made, the following conclusion was formulated. Although currently no evidence exists that a symptom diary/questionnaire/app reduces the time to diagnosis or leads to earlier diagnosis, the GDG considers their potential benefit in complementing the traditional history taking process as it aids in objectifying pain and empowering women to demonstrate their symptoms (conclusion, not recommendation). Does clinical examination of symptomatic women reliably predict the presence of endometriosis? Clinical examination, including vaginal examination where appropriate, should be considered to identify deep nodules or endometriomas in patients with suspected endometriosis, although the diagnostic accuracy is low (Ripps and Martin, 1992; Nezhat et al., 1994; Koninckx et al., 1996; Eskenazi et al., 2001; Chapron et al., 2002; Condous et al., 2007; Bazot et al., 2009; Khawaja et al., 2009; Hudelist et al., 2011; Paulson and Paulson, 2011). Strong recommendation ⊕○○○ In women with suspected endometriosis, further diagnostic steps, including imaging, should be considered even if the clinical examination is normal. Strong recommendation ⊕⊕○○ Open in a new tab Are medical technologies reliable in diagnosing endometriosis and establishing the extent of the disease? Clinicians should not use measurement of biomarkers in endometrial tissue, blood, menstrual or uterine fluids to diagnose endometriosis (Mol et al., 1998; May et al., 2010; May et al., 2011; Liu et al., 2015; Cosar et al., 2016; Gupta et al., 2016; Hirsch et al., 2016; Nisenblat et al., 2016a; Vanhie et al., 2019; Moustafa et al., 2020). Strong recommendation ⊕⊕⊕○ Clinicians are recommended to use imaging (ultrasound (US) or MRI) in the diagnostic work-up for endometriosis, but they need to be aware that a negative finding does not exclude endometriosis, particularly superficial peritoneal disease (Bazot et al., 2009; Manganaro et al., 2012; Guerriero et al., 2014; Thomeer et al., 2014; Nisenblat et al., 2016b; Moura et al., 2019). Strong recommendation ⊕⊕○○ In patients with negative imaging results or where empirical treatment was unsuccessful or inappropriate, the GDG recommends that clinicians consider offering laparoscopy for the diagnosis and treatment of suspected endometriosis.GPP The GDG recommends that laparoscopic identification of endometriotic lesions is confirmed by histology although negative histology does not entirely rule out the disease.GPP Open in a new tab Does diagnostic laparoscopy compared to empirical medical treatment result in better symptom management in women suspected of endometriosis? As there is no evidence of superiority of either approach (Chapron et al., 1998; Byrne et al., 2018; Bafort et al., 2020), the GDG concluded that both diagnostic laparoscopy and imaging combined with empirical treatment (hormonal contraceptives or progestogens) can be considered in women suspected of endometriosis. Pros and cons should be discussed with the patient (conclusion, not recommendation). Is long-term monitoring of women with endometriosis beneficial in preventing adverse outcomes (recurrence, complications, malignancy)? Follow-up and psychological support should be considered in women with confirmed endometriosis, particularly deep and ovarian endometriosis, although there is currently no evidence of benefit of regular long-term monitoring for early detection of recurrence, complications, or malignancy (Pittaway, 1990; Matalliotakis et al., 1994; Chen et al., 1998). Weak recommendation ⊕○○○ The appropriate frequency and type of follow-up or monitoring is unknown and should be individualized based on previous and current treatments, and severity of the disease and symptoms.GPP Open in a new tab Does early diagnosis of endometriosis versus late diagnosis lead to better quality of life? Although no adequate studies exist to support the benefits of early versus late diagnosis, the GDG recommends that in symptomatic women, attempts should be made to relieve symptoms, either by empirical treatment or after a diagnosis of endometriosis (conclusion, not recommendation). Treatment of endometriosis-associated pain The recommendations for treatment of pain symptoms linked to endometriosis are summarized in Fig. 3. Figure 3. Open in a new tab Summary of the recommendations for treatment of pain symptoms linked to endometriosis. NSAID, non-steroidal anti-inflammatory. Are analgesics effective for symptomatic relief of painful symptoms associated with endometriosis? Women may be offered non-steroidal anti-inflammatory drugs (NSAIDs) or other analgesics (either alone or in combination with other treatments) to reduce endometriosis-associated pain (Brown et al., 2017). Weak recommendation ⊕○○○ Open in a new tab Are hormone therapies effective for painful symptoms associated with endometriosis? It is recommended to offer women hormone treatment (combined hormonal contraceptives, progestogens, GnRH agonists or GnRH antagonists) as one of the options to reduce endometriosis-associated pain. Strong recommendation ⊕⊕⊕○ The GDG recommends that clinicians take a shared decision-making approach and take individual preferences, side effects, individual efficacy, costs and availability into consideration when choosing hormone treatments for endometriosis-associated pain.GPP It is recommended to prescribe women a combined hormonal contraceptive (oral, vaginal ring or transdermal) to reduce endometriosis-associated dyspareunia, dysmenorrhoea and non-menstrual pain (Brown et al., 2018; Jensen et al., 2018; Grandi et al., 2019). Strong recommendation ⊕⊕○○ Women suffering from endometriosis-associated dysmenorrhoea can be offered the continuous use of a combined hormonal contraceptive pill (Hee et al., 2013; Zorbas et al., 2015; Muzii et al., 2016b). Weak recommendation ⊕⊕○○ It is recommended to prescribe women progestogens to reduce endometriosis-associated pain (Momoeda et al., 2009; Brown et al., 2012; Petraglia et al., 2012; Andres et al., 2015; Dragoman and Gaffield, 2016). Strong recommendation ⊕⊕○○ The GDG recommends that clinicians take the different side effect profiles of progestogens into account when prescribing them.GPP It is recommended to prescribe women a levonorgestrel-releasing intrauterine (LNG-IUS) system or an etonogestrel-releasing subdermal implant to reduce endometriosis-associated pain (Lan et al., 2013; Margatho et al., 2020). Strong recommendation ⊕⊕⊕○ It is recommended to prescribe women GnRH agonists to reduce endometriosis-associated pain, although evidence is limited regarding dosage or duration of treatment (Brown et al., 2010; Tang et al., 2017). Strong recommendation ⊕⊕○○ The GDG recommends that GnRH agonists are prescribed as second-line (e.g. if hormonal contraceptives or progestogens have been ineffective) due to their side effect profile.GPP Clinicians should consider prescribing combined hormonal add-back therapy alongside GnRH agonist therapy to prevent bone loss and hypo-oestrogenic symptoms (Wu et al., 2014; Sauerbrun-Cutler and Alvero, 2019). Strong recommendation ⊕⊕⊕○ It can be considered to prescribe women GnRH antagonists to reduce endometriosis-associated pain, although evidence is limited regarding dosage or duration of treatment (Taylor et al., 2017; Donnez et al., 2020; Osuga et al., 2021). Weak recommendation ⊕⊕⊕○ The GDG recommends that GnRH antagonists are prescribed as second-line (e.g. if hormonal contraceptives or progestogens have been ineffective) owing to their side effect profile.GPP In women with endometriosis-associated pain refractory to other medical or surgical treatment, it is recommended to prescribe aromatase inhibitors, as they reduce endometriosis-associated pain. Aromatase inhibitors may be prescribed in combination with oral contraceptives, progestogens, GnRH agonists or GnRH antagonists (Ferrero et al., 2011; Almassinokiani et al., 2014; Agarwal and Foster, 2015). Strong recommendation ⊕⊕○○ Open in a new tab Is surgery effective for treatment of pain associated with endometriosis? It is recommended to offer surgery as one of the options to reduce endometriosis-associated pain (Sutton et al., 1994; Franck et al., 2018; Arcoverde et al., 2019; Bafort et al., 2020). Strong recommendation ⊕⊕○○ When surgery is performed, clinicians may consider excision instead of ablation of endometriosis to reduce endometriosis-associated pain (Wright et al., 2005; Healey et al., 2014; Pundir et al., 2017).Weak recommendation ⊕⊕○○ Open in a new tab It can be concluded that laparoscopic uterosacral nerve ablation (LUNA) is not beneficial as an additional procedure to conventional laparoscopic surgery for endometriosis, as it offers no additional benefit over surgery alone (Proctor et al., 2005). Presacral neurectomy (PSN) is beneficial for treatment of endometriosis-associated midline pain as an adjunct to conventional laparoscopic surgery, but it should be stressed that PSN requires a high degree of skill and is associated with an increased risk of adverse effects such as intraoperative bleeding, and postoperative constipation, urinary urgency and painless first stage of labour (Miller et al., 2020) (conclusion, not recommendation). When performing surgery in women with ovarian endometrioma, clinicians should perform cystectomy instead of drainage and coagulation, as cystectomy reduces recurrence of endometrioma and endometriosis-associated pain (Hart et al., 2008; Candiani et al., 2020).Strong recommendation ⊕⊕○○ When performing surgery in women with ovarian endometrioma, clinicians can consider both cystectomy and CO 2 laser vaporization, as both techniques appear to have similar recurrence rates beyond the first year after surgery. Early post-surgical recurrence rates may be lower after cystectomy (Muzii et al., 2005, 2016a; Mossa et al., 2010; Porpora et al., 2010; Carmona et al., 2011; Shaltout et al., 2019). Weak recommendation ⊕○○○ When performing surgery for ovarian endometrioma, specific caution should be used to minimize ovarian damage (Busacca et al., 2006; Muzii et al., 2015; Muzii et al., 2016a; Shaltout et al., 2019; Younis et al., 2019). Strong recommendation ⊕○○○ Clinicians can consider performing surgical removal of deep endometriosis, as it may reduce endometriosis-associated pain and improves quality of life (Stepniewska et al., 2010; De Cicco et al., 2011; Meuleman et al., 2011; Byrne et al., 2018; Arcoverde et al., 2019; Bendifallah et al., 2020). Weak recommendation ⊕⊕○○ The GDG recommends that women with deep endometriosis are referred to a centre of expertise.GPP The GDG recommends that patients undergoing surgery, particularly for deep endometriosis, are informed of potential risks, benefits and long-term effect on quality of life.GPP Open in a new tab Owing to the heterogeneity of patient populations, surgical approaches, preferences and techniques, the GDG decided not to make any conclusions or recommendations on the techniques to be applied for treatment of pain associated with deep endometriosis (conclusion, not recommendation). Clinicians can consider hysterectomy (with or without removal of the ovaries) with removal of all visible endometriosis lesions, in those women who no longer wish to conceive and failed to respond to more conservative treatments. Women should be informed that hysterectomy will not necessarily cure the symptoms or the disease. Weak recommendation ⊕⊕○○ When a decision is made whether to remove the ovaries, the long-term consequences of early menopause and possible need for hormone replacement therapy should be considered.GPP The GDG recommends that when hysterectomy is performed, a total hysterectomy is preferred (Namnoum et al., 1995; Sandström et al., 2020; Shakiba et al., 2008).GPP Open in a new tab Is there a subgroup of women with confirmed endometriosis who respond better to surgery than others? There are currently no prognostic markers that can be used to select patients that would benefit from surgery. Such markers would need to be assessed prior to surgery and predict a clinically meaningful improvement of pain symptoms. In the absence of prognostic markers, no recommendation could be formulated (conclusion, not recommendation). Are medical therapies effective as an adjunct to surgical therapy? It is not recommended to prescribe preoperative hormone treatment to improve the immediate outcome of surgery for pain in women with endometriosis (Chen et al., 2020). Strong recommendation ⊕⊕○○ Women may be offered postoperative hormone treatment to improve the immediate outcome of surgery for pain in women with endometriosis if not desiring immediate pregnancy (Tanmahasamut et al., 2012; Chen et al., 2020). Weak recommendation ⊕⊕○○ Open in a new tab Are surgical therapies more effective than medical therapies for women with endometriosis with pain symptoms? The GDG recommends that clinicians take a shared decision-making approach and take individual preferences, side effects, individual efficacy, costs and availability into consideration when choosing between hormone treatments and surgical treatments for endometriosis-associated pain.GPP Open in a new tab What non-medical management strategies are effective for symptoms associated with endometriosis (pain and quality of life)? The GDG recommends that clinicians discuss non-medical strategies to address quality of life and psychological well-being of women managing symptoms of endometriosis. However, no recommendations can be made for any specific non-medical intervention (Chinese medicine, nutrition, electrotherapy, acupuncture, physiotherapy, exercise and psychological interventions) to reduce pain or improve quality of life measures in women with endometriosis, as the potential benefits and harms are unclear.GPP Open in a new tab Treatment of endometriosis-associated infertility The recommendations for treatment of endometriosis-associated infertility are summarized in Fig. 4. Figure 4. Open in a new tab Summary of the recommendations on treatment of endometriosis-associated infertility. EFI, endometriosis fertility index; MAR, medically assisted reproduction. Are hormone/medical therapies effective for treatment of endometriosis-associated infertility? In infertile women with endometriosis, clinicians should not prescribe ovarian suppression treatment to improve fertility (Hughes et al., 2007). Strong recommendation ⊕⊕○○ Women seeking pregnancy should not be prescribed postoperative hormone suppression with the sole purpose to enhance future pregnancy rates (Chen et al., 2020). Strong recommendation ⊕⊕○○ Those women who cannot attempt to or decide not to conceive immediately after surgery may be offered hormone therapy as it does not negatively impact their fertility and improves the immediate outcome of surgery for pain (Chen et al., 2020). Weak recommendation ⊕⊕○○ In infertile women with endometriosis, clinicians should not prescribe pentoxifylline, other anti-inflammatory drugs or letrozole outside ovulation-induction to improve natural pregnancy rates (Alborzi et al., 2011; Lu et al., 2012). Strong recommendation ⊕○○○ Open in a new tab In women with endometriosis, is surgery effective to increase the chance of natural pregnancy? Operative laparoscopy could be offered as a treatment option for endometriosis-associated infertility in revised American Society for Reproductive Medicine (rASRM) stage I/II endometriosis as it improves the rate of ongoing pregnancy (Jin and Ruiz Beguerie, 2014; Bafort et al., 2020; Hodgson et al., 2020). Weak recommendation ⊕⊕○○ Clinicians may consider operative laparoscopy for the treatment of endometrioma-associated infertility as it may increase their chance of natural pregnancy, although no data from comparative studies exist (Dan and Limin, 2013; Alborzi et al., 2019; Candiani et al., 2020). Weak recommendation ⊕○○○ Although no compelling evidence exists that operative laparoscopy for deep endometriosis improves fertility, operative laparoscopy may represent a treatment option in symptomatic patients wishing to conceive (Meuleman et al., 2011; Vercellini et al., 2012; Iversen et al., 2017). Weak recommendation ⊕○○○ The GDG recommends that the decision to perform surgery should be guided by the presence or absence of pain symptoms, patient age and preferences, history of previous surgery, presence of other infertility factors, ovarian reserve and the estimated endometriosis fertility index (EFI).GPP Open in a new tab Which patients need treatment with assisted reproduction technology after surgery? While no recommendation could be formulated, the GDG concluded that women should be counselled of their chances of becoming pregnant after surgery. To identify patients that may benefit from ART after surgery, the EFI should be used as it is validated, reproducible and cost-effective. The results of other fertility investigations, such as their partner’s sperm analysis, should be taken into account (conclusion, not recommendation). Is medically assisted reproduction effective for infertility associated with endometriosis? In infertile women with rASRM stage I/II endometriosis, clinicians may perform IUI with ovarian stimulation, instead of expectant management or IUI alone, as it increases pregnancy rates (Nulsen et al., 1993; Tummon et al., 1997; Omland et al., 1998). Weak recommendation ⊕○○○ Although the value of IUI in infertile women with rASRM stage III/IV endometriosis with tubal patency is uncertain, the use of IUI with ovarian stimulation could be considered (van der Houwen et al., 2014). Weak recommendation ⊕○○○ ART can be performed for infertility associated with endometriosis, especially if tubal function is compromised, if there is male factor infertility, in case of low EFI and/or if other treatments have failed (Harb et al., 2013; Hamdan et al., 2015b; Senapati et al., 2016; Murta et al., 2018; Muteshi et al., 2018; Alshehre et al., 2021). Weak recommendation ⊕⊕○○ A specific protocol for ART in women with endometriosis cannot be recommended. Both GnRH antagonist and agonist protocols can be offered based on patients’ and physicians’ preferences as no difference in pregnancy or live birth rate has been demonstrated (Pabuccu et al., 2007; Rodriguez-Purata et al., 2013; Bastu et al., 2014; Kolanska et al., 2017; Drakopoulos et al., 2018). Weak recommendation ⊕○○○ Women with endometriosis can be reassured regarding the safety of ART since the recurrence rates are not increased compared to those women not undergoing ART (Benaglia et al., 2008; Somigliana et al., 2019). Weak recommendation ⊕⊕⊕○ In women with endometrioma, clinicians may use antibiotic prophylaxis at the time of oocyte retrieval, although the risk of ovarian abscess formation following follicle aspiration is low.GPP Open in a new tab Are medical therapies effective as an adjunct to medically assisted reproduction for endometriosis-associated infertility? The extended administration of GnRH agonist prior to ART treatment to improve live birth rate in infertile women with endometriosis is not recommended, as the benefit is uncertain (Georgiou et al., 2019; Cao et al., 2020; Kaponis et al., 2020). Strong recommendation ⊕○○○ There is insufficient evidence to recommend prolonged administration of the combined oral contraceptives (COC)/progestogens as a pre-treatment to ART to increase live birth rates (de Ziegler et al., 2010). Weak recommendation ⊕○○○ Open in a new tab Are surgical therapies effective as an adjunct prior to medically assisted reproduction for endometriosis-associated infertility? Clinicians are not recommended to routinely perform surgery prior to ART to improve live birth rates in women with rASRM stage I/II endometriosis, as the potential benefits are unclear (Opoien et al., 2011; Hamdan et al., 2015b). Strong recommendation ⊕⊕○○ Clinicians are not recommended to routinely perform surgery for ovarian endometrioma prior to ART to improve live birth rates, as the current evidence shows no benefit and surgery is likely to have a negative impact on ovarian reserve (Coccia et al., 2014; Hamdan et al., 2015a; Nickkho-Amiry et al., 2018; Şükür et al., 2021). Strong recommendation ⊕⊕○○ Surgery for endometrioma prior to ART can be considered to improve endometriosis-associated pain or accessibility of follicles.GPP The decision to offer surgical excision of deep endometriosis lesions prior to ART should be guided mainly by pain symptoms and patient preference as its effectiveness on reproductive outcome is uncertain owing to lack of randomized studies (Bianchi et al., 2009; Soriano et al., 2016; Bendifallah et al., 2017; Breteau et al., 2020). Strong recommendation ⊕○○○ Open in a new tab What non-medical management strategies are effective for infertility associated with endometriosis? Regarding non-medical strategies on infertility, there is no clear evidence that any non-medical interventions for women with endometriosis will be of benefit to increase the chance of pregnancy. No recommendation can be made to support any non-medical interventions (nutrition, Chinese medicine, electrotherapy, acupuncture, physiotherapy, exercise and psychological interventions) to increase fertility in women with endometriosis. The potential benefits and harms are unclear (conclusion, not recommendation). Is endometriosis an indication for fertility preservation (ovarian tissue/oocytes)? In case of extensive ovarian endometriosis, clinicians should discuss the pros and cons of fertility preservation with women with endometriosis. The true benefit of fertility preservation in women with endometriosis remains unknown (Cobo et al., 2020; Kim et al., 2020). Strong recommendation ⊕○○○ Open in a new tab What is the impact of endometriosis on pregnancy and obstetric outcomes? Patients should not be advised to become pregnant with the sole purpose of treating endometriosis, as pregnancy does not always lead to improvement of symptoms or reduction of disease progression (Leeners et al., 2018). Strong recommendation ⊕○○○ Endometriomas may change in appearance during pregnancy. In case of finding an atypical endometrioma during US in pregnancy, it is recommended to refer the patient to a centre with appropriate expertise (Leone Roberti Maggiore et al., 2016). Strong recommendation ⊕○○○ Open in a new tab Complications related directly to pre-existing endometriosis lesions are rare, but probably under-reported. Such complications may be related to their decidualization, adhesion formation/stretching and endometriosis-related chronic inflammation. Although rare, they may represent life-threatening situations that may require surgical management (Leone Roberti Maggiore et al., 2016; Leone Roberti Maggiore et al., 2017; Lier et al., 2017; Glavind et al., 2018). Clinicians should be aware that there may be an increased risk of first trimester miscarriage and ectopic pregnancy in women with endometriosis (Leone Roberti Maggiore et al., 2016; Santulli et al., 2016; Kohl Schwartz et al., 2017; Saraswat et al., 2017; Yong et al., 2020). Strong recommendation ⊕⊕○○ Clinicians should be aware of endometriosis-associated complications in pregnancy, although these are rare. As these findings are based on low/moderate quality studies, these results should be interpreted with caution and currently do not warrant increased antenatal monitoring or dissuade women from becoming pregnant (Leone Roberti Maggiore et al., 2016; Lalani et al., 2018; Perez-Lopez et al., 2018; Horton et al., 2019). Strong recommendation ⊕⊕○○ Open in a new tab The recommendations and information on endometriosis and pregnancy are summarized in Fig. 5. Figure 5. Open in a new tab Summary of the recommendations and information on endometriosis and pregnancy. Endometriosis recurrence Is there a role for secondary prevention of recurrence of disease and painful symptoms in patients treated for endometriosis? When surgery is indicated in women with an endometrioma, clinicians should perform ovarian cystectomy, instead of drainage and electrocoagulation, for the secondary prevention of endometriosis-associated dysmenorrhoea, dyspareunia and non-menstrual pelvic pain. However, the risk of reduced ovarian reserve should be taken into account. Strong recommendation ⊕⊕○○ Clinicians should consider prescribing the postoperative use of a LNG-IUS system (52 mg) or a combined hormonal contraceptive for at least 18–24 months for the secondary prevention of endometriosis-associated dysmenorrhoea (Seracchioli et al., 2009; Lee et al., 2018; Song et al., 2018; Chen et al., 2020; Zakhari et al., 2021). Strong recommendation ⊕⊕○○ After surgical management of ovarian endometrioma in women not immediately seeking conception, clinicians are recommended to offer long-term hormone treatment (e.g. combined hormonal contraceptives) for the secondary prevention of endometrioma and endometriosis-associated related symptom recurrence. Strong recommendation ⊕○○○ For the prevention of recurrence of deep endometriosis and associated symptoms, long-term administration of postoperative hormone treatment can be considered. Weak recommendation ⊕○○○ Clinicians can perform ART in women with deep endometriosis, as it does not seem to increase endometriosis recurrence per se (Somigliana et al., 2019). Weak recommendation ⊕⊕⊕○ Open in a new tab How should patients with reoccurring endometriosis or recurring symptoms be managed? Is repetitive surgery effective for symptoms associated with endometriosis? Any hormone treatment or surgery can be offered to treat recurring pain symptoms in women with endometriosis (Candiani et al., 1991; Hornstein et al., 1997; Vercellini et al., 2002; Razzi et al., 2007; Muzii et al., 2015; Abdou et al., 2018; Koshiba et al., 2018; Lee et al., 2018). Weak recommendation ⊕○○○ Open in a new tab Endometriosis and adolescence Which diagnostic procedures should be applied in adolescents with possible endometriosis? In adolescents, clinicians should take a careful history to identify possible risk factors for endometriosis, such as a positive family history, obstructive genital malformations, early menarche or short menstrual cycle (Geysenbergh et al., 2017). Strong recommendation ⊕○○○ Clinicians may consider endometriosis in young women presenting with (cyclical) absenteeism from school, or with use of oral contraceptives for treatment of dysmenorrhoea (Chapron et al., 2011). Weak recommendation ⊕○○○ In adolescents, clinicians should take a careful history and consider the following symptoms as suggestive of the presence of endometriosis: chronic or acyclical pelvic pain, particularly combined with nausea, dysmenorrhoea, dyschezia, dysuria, dyspareunia; cyclical pelvic pain (Greene et al., 2009; Treloar et al., 2010; Vicino et al., 2010; Yang et al., 2012; DiVasta et al., 2018). Strong recommendation ⊕○○○ Open in a new tab In the absence of evidence for adolescents specifically, the recommendations for clinical examination in adults can be applied. The GDG recommends that before performing vaginal examination and/or rectal examination in adolescents, the acceptability should be discussed with the adolescent and her caregiver, taking into consideration the patient’s age and cultural background.GPP Transvaginal US is recommended to be used in adolescents in whom it is appropriate, as it is effective in diagnosing ovarian endometriosis. If a transvaginal scan is not appropriate, MRI, transabdominal, transperineal or transrectal scan may be considered (Yang et al., 2012; Brosens et al., 2013; Martire et al., 2020). Strong recommendation ⊕⊕○○ Serum biomarkers (e.g. CA-125) are not recommended for diagnosing or ruling out endometriosis in adolescents (Seckin et al., 2018; Sasamoto et al., 2020). Strong recommendation ⊕⊕○○ In adolescents with suspected endometriosis where imaging is negative and medical treatments (with NSAIDs and/or hormonal contraceptives) have not been successful, diagnostic laparoscopy may be considered (Vicino et al., 2010; Shah and Missmer, 2011; Yang et al., 2012; Brosens et al., 2013). Weak recommendation ⊕⊕○○ Open in a new tab Should diagnosis of endometriosis in adolescents be confirmed by histology? If a laparoscopy is performed, clinicians should consider taking biopsies to confirm the diagnosis histologically, although negative histology does not entirely rule out the disease (Janssen et al., 2013). Strong recommendation ⊕⊕○○ Open in a new tab What is the best treatment for adolescents with (suspected) endometriosis? In adolescents with severe dysmenorrhoea and/or endometriosis-associated pain, clinicians should prescribe hormonal contraceptives or progestogens (systemically or via LNG-IUS) as first-line hormonal hormone therapy because they may be effective and safe. However, it is important to note that some progestogens may decrease bone mineral density (Davis et al., 2005; Yoost et al., 2013; Ebert et al., 2017). Strong recommendation ⊕○○○ The GDG recommends clinicians consider NSAIDs as treatment for endometriosis-associated pain in adolescents with (suspected) endometriosis, especially if first-line hormone treatment is not an option.GPP In adolescents with laparoscopically confirmed endometriosis and associated pain in whom hormonal contraceptives or progestogen therapy failed, clinicians may consider prescribing GnRH agonists for up to 1 year, as they are effective and safe when combined with add-back therapy (DiVasta et al., 2015; Gallagher et al., 2017; Gallagher et al., 2018). Weak recommendation ⊕⊕○○ The GDG recommends that in young women and adolescents, if GnRH agonist treatment is considered, it should be used only after careful consideration and discussion of potential side effects and potential long-term health risks with a practitioner in a secondary or tertiary care setting.GPP In adolescents with endometriosis, clinicians may consider surgical removal of endometriosis lesions to manage endometriosis-related symptoms. However, symptom recurrence rates may be considerable, especially when surgery is not followed by hormone treatment (Roman, 2010; Tandoi et al., 2011; Yeung et al., 2011; Lee et al., 2017). Weak recommendation ⊕○○○ The GDG recommends that if surgical treatment is indicated in adolescents with endometriosis, it should be performed laparoscopically by an experienced surgeon, and, if possible, complete laparoscopic removal of all present endometriosis should be performed.GPP In adolescents with endometriosis, clinicians should consider postoperative hormone therapy, as this may suppress recurrence of symptoms (Doyle et al., 2009; Seo et al., 2017). Strong recommendation ⊕○○○ Open in a new tab Is endometriosis in adolescents an indication for fertility preservation (ovarian tissue/oocytes)? The GDG recommends that adolescents with endometriosis are informed of the potential detrimental effect of ovarian endometriosis and surgery on ovarian reserve and future fertility.GPP Fertility preservation options exist and the GDG recommends that adolescents are informed about them, although the true benefit, safety and indications in adolescents with endometriosis remain unknown.GPP Open in a new tab Endometriosis and menopause Is endometriosis still active during menopause and, if so, how should the symptoms be treated? The GDG concluded that clinicians should be aware that endometriosis can still be active/symptomatic after menopause (conclusion, not recommendation). Is surgical/medical treatment effective and safe in women with a history of endometriosis? Clinicians may consider surgical treatment for postmenopausal women presenting with signs of endometriosis and/or pain to enable histological confirmation of the diagnosis of endometriosis (Redwine, 1994; Clayton et al., 1999; Morotti et al., 2012; Sun et al., 2013; Behera et al., 2006). Weak recommendation ⊕○○○ The GDG recommends that clinicians acknowledge the uncertainty towards the risk of malignancy in postmenopausal women. If a pelvic mass is detected, the work-up and treatment should be performed according to national oncology guidelines.GPP For postmenopausal women with endometriosis-associated pain, clinicians may consider aromatase inhibitors as a treatment option especially if surgery is not feasible (Polyzos et al., 2011; Pavone and Bulun, 2012). Weak recommendation ⊕○○○ Open in a new tab Is hormone treatment effective and safe for relief of menopausal symptoms in women with a history of endometriosis? Clinicians may consider combined menopausal hormone therapy for the treatment of postmenopausal symptoms in women (both after natural and surgical menopause) with a history of endometriosis (Matorras et al., 2002; Gemmell et al., 2017). Weak recommendation ⊕⊕○○ Clinicians should avoid prescribing oestrogen-only regimens for the treatment of vasomotor symptoms in postmenopausal women with a history of endometriosis, as these regimens may be associated with a higher risk of malignant transformation (Gemmell et al., 2017). Strong recommendation ⊕⊕○○ The GDG recommends that clinicians continue to treat women with a history of endometriosis after surgical menopause with combined oestrogen–progestogen at least up to the age of natural menopause.GPP Open in a new tab Are women with endometriosis at higher risk of experiencing menopause-related major health concerns? Clinicians should be aware that women with endometriosis who have undergone an early bilateral salpingo-oophorectomy as part of their treatment have an increased risk of diminished bone density, dementia and cardiovascular disease. It is also important to note that women with endometriosis have an increased risk of cardiovascular disease, irrespective of whether they have had an early surgical menopause (conclusion, not recommendation). Extrapelvic endometriosis How reliable is imaging for diagnosing extrapelvic endometriosis? Clinicians should be aware of symptoms of extrapelvic endometriosis, such as cyclical shoulder pain, cyclical spontaneous pneumothorax, cyclical cough or nodules which enlarge during menses.GPP It is advisable to discuss diagnosis and management of extrapelvic endometriosis in a multidisciplinary team in a centre with sufficient expertise.GPP Open in a new tab Does treatment for extrapelvic endometriosis relieve symptoms? For abdominal extrapelvic endometriosis, surgical removal is the preferred treatment, when possible, to relieve symptoms. Hormone treatment may also be an option when surgery is not possible or acceptable (Horton et al., 2008; Zhu et al., 2017; Andres et al., 2020; Hirata et al., 2020). Weak recommendation ⊕○○○ For thoracic endometriosis, hormone treatment can be offered. If surgery is indicated, it should be performed in a multidisciplinary manner involving a thoracic surgeon and/or other relevant specialists (Joseph and Sahn, 1996; Ceccaroni et al., 2013; Nezhat et al., 2014; Gil and Tulandi, 2020; Andres et al., 2020; Vigueras Smith et al., 2021; Ciriaco et al., 2022). Weak recommendation ⊕○○○ Open in a new tab Asymptomatic endometriosis Is treatment beneficial for incidental finding of asymptomatic endometriosis? The GDG recommends that clinicians should inform and counsel women about any incidental finding of endometriosis.GPP Clinicians should not routinely perform surgical excision/ablation for an incidental finding of asymptomatic endometriosis at the time of surgery (Moen and Stokstad, 2002). Strong recommendation ⊕○○○ Clinicians should not prescribe medical treatment in women with incidental finding of endometriosis. Strong recommendation ⊕○○○ Open in a new tab Is long term monitoring of women with asymptomatic endometriosis beneficial in preventing adverse outcomes? Routine US monitoring of asymptomatic endometriosis can be considered (Maouris, 1991; Alcázar et al., 2005; Pearce et al., 2012; Serati et al., 2013). Weak recommendation ⊕○○○ Open in a new tab Primary prevention of endometriosis Is there a role for primary prevention of endometriosis? Although there is no direct evidence of benefit in preventing endometriosis in the future, women can be advised of aiming for a healthy lifestyle and diet, with reduced alcohol intake and regular physical activity (Hansen and Knudsen, 2013; Parazzini et al., 2013a,b; Bravi et al., 2014; Ricci et al., 2016; Harris et al., 2018; Nodler et al., 2020; Qiu et al., 2020). Weak recommendation ⊕⊕○○ The usefulness of hormonal contraceptives for the primary prevention of endometriosis is uncertain (Vercellini et al., 2011). Weak recommendation ⊕⊕○○ Genetic testing in women with suspected or confirmed endometriosis should only be performed within a research setting.RESEARCH-ONLY Open in a new tab Endometriosis and cancer Are patients with endometriosis at increased risk of cancer? Clinicians should inform women with endometriosis requesting information on their risk of developing cancer that endometriosis is not associated with a significantly higher risk of cancer overall (Fig. 6). Although endometriosis is associated with a higher risk of ovarian, breast and thyroid cancers in particular, the increase in absolute risk compared with women in the general population is low (Kvaskoff et al., 2021). Strong recommendation ⊕⊕○○ Open in a new tab Figure 6. Open in a new tab Infographic on the absolute risk of developing cancer in a woman’s lifetime. What information could clinicians provide to women with endometriosis regarding their risk of developing cancer? The GDG recommends that clinicians reassure women with endometriosis with regards to their cancer risk and address their concern to reduce their risk by recommending general cancer prevention measures (avoiding smoking, maintaining a healthy weight, exercising regularly, having a balanced diet with high intakes of fruits and vegetables and low intakes of alcohol, and using sun protection).GPP Open in a new tab Are somatic mutations in deep endometriosis of patients without cancer predictive for ovarian cancer development and/or progression? Based on the limited literature and controversial findings, there is little evidence that somatic mutations in patients with deep endometriosis may be predictive of development and/or progression of ovarian cancer (conclusion, not recommendation). Does the use of hormone treatments increase the risk of cancer? Clinicians should reassure women with endometriosis about the risk of malignancy associated with the use of hormonal contraceptives (Smith et al., 2003; Zucchetto et al., 2009; Gierisch et al., 2013; Havrilesky et al., 2013; Braganza et al., 2014; Berlanda et al., 2016; Wentzensen et al., 2016; Butt et al., 2018; Michels et al., 2018). Strong recommendation ⊕○○○ Open in a new tab Should women with endometriosis be monitored for detection of malignancy? In women with endometriosis, clinicians should not systematically perform cancer screening beyond the existing population-based cancer screening guidelines (Kvaskoff et al., 2021). Strong recommendation ⊕⊕○○ Clinicians can consider cancer screening according to local guidelines in individual patients that have additional risk factors, e.g. strong family history, specific germline mutations.GPP Open in a new tab Does surgery for endometriosis change the future risk of cancer? Clinicians should be aware that there is epidemiological data, mostly on ovarian endometriosis, showing that complete excision of visible endometriosis may reduce the risk of ovarian cancer. The potential benefits should be weighed against the risks of surgery (morbidity, pain and ovarian reserve) (Rossing et al., 2008; Melin et al., 2013; Haraguchi et al., 2016). Strong recommendation ⊕⊕○○ Open in a new tab Discussion This paper provides an overview of recommendations for diagnosis of endometriosis and treatment of associated symptoms during different stages of life. In addition, guidance is provided on the possible connection with development of cancer, and with regards to prevention. Overall, 109 recommendations have been formulated, 79 supported by research data and 30 GPPs based primarily on clinical expertise. The guidelines are based on the best available evidence or, where data of sufficient quality were absent, on recommendations by the GDG (GPPs). The current guideline and recommendations are an update of the ESHRE endometriosis guidelines published in 2013 and 2005 (Kennedy et al., 2005; Dunselman et al., 2014). The key questions and topics covered in the guideline of 2013 were updated based on data published between 2013 and 2021, where available, and in accordance with changes in clinical practice. The latter applied, for example, to the oral use of danazol and anti-progestogens as a medical treatment and to LUNA, PSN and anti-adhesion agents as surgical interventions. These interventions are still discussed in the guideline, but no longer discussed in recommendations for clinical practice. While most of the more recent studies confirm previous ESHRE recommendations, there are five topics in which significant changes in clinical practice are to be expected. The first change, primarily based on clinical practice rather than published data, is the evolution in the diagnostic process. While previously a laparoscopy was regarded as the diagnostic gold standard, it is now only recommended in patients with negative imaging results and/or where empirical treatment was unsuccessful or inappropriate. Secondly, studies on GnRH antagonist treatments support their use as an additional (second-line) treatment option. Thirdly, recent data indicate that postoperative medical treatment may be beneficial towards pain management and support a recommendation to offer it to women not desiring immediate pregnancy. Fourthly, the extended administration of GnRH agonist prior to ART treatment to improve live birth rate in infertile women with endometriosis (ultralong protocol) is no longer recommended because of unclear benefits. Finally, the EFI was added as a step in the treatment as it can support decision-making for the most appropriate option to achieve pregnancy after surgery. In addition to the topics discussed in the previous guideline, the current guideline addresses highly important previous gaps in clinical management, with an additional chapter on adolescent endometriosis, information on pregnancy and fertility preservation, and extended information on endometriosis in menopause, as well as data on the link between endometriosis and cancer. Despite our best efforts to provide clear guidance on the management of endometriosis using all available evidence, there is still an urgent need for more research both to achieve more clarity on the most appropriate diagnostic and treatment options, and to answer very basic questions as to the natural course of the disease. This guideline provides 30 recommendations for research written to inspire researchers and hopefully also facilitate funding for endometriosis studies (Supplementary Data). In summary, the 2022 ESHRE Guideline: Endometriosis is a comprehensive update of the existing evidence and should assist healthcare professionals in their decision making and patients in their understanding of the management suggestions. Active involvement and input by patient representatives at all stages was central to the success of this endeavour. As such, the guideline was created by medical professionals, patient representatives and specialists in epidemiology and guideline methodology. The detailed guideline document and a patient-friendly version can be accessed via the ESHRE website ( Supplementary data Supplementary data are available at Human Reproduction Open online. Data availability The full guideline and supporting data (literature report, evidence tables) are available on www.eshre.eu/guidelines. Supplementary Material hoac009_Supplementary_Data Click here for additional data file. (67.5KB, pdf) Acknowledgements The authors would like to thank the experts that participated in the stakeholder review for their constructive and helpful comments. Authors’ roles The guideline core group was chaired by C.M.B. N.V. provided methodological support. All other authors contributed equally to writing the guideline. All authors have revised and approved the final version. The collaborators supported the guideline core group for the individual chapters. Funding The guideline was developed and funded by ESHRE, covering expenses associated with the guideline meetings, with the literature searches and with the dissemination of the guideline. The guideline group members did not receive payment. Conflict of interest C.M.B. reports grants from Bayer Healthcare and the European Commission; Participation on a Data Safety Monitoring Board or Advisory Board with ObsEva (Data Safety Monitoring Group) and Myovant (Scientific Advisory Group). A.B. reports grants from FEMaLE executive board member and European Commission Horizon 2020 grant; consulting fees from Ethicon Endo Surgery, Medtronic; honoraria for lectures from Ethicon; and support for meeting attendance from Gedeon Richter; A.H. reports grants from MRC, NIHR, CSO, Roche Diagnostics, Astra Zeneca, Ferring; Consulting fees from Roche Diagnostics, Nordic Pharma, Chugai and Benevolent Al Bio Limited all paid to the institution; a pending patent on Serum endometriosis biomarker; he is also Chair of TSC for STOP-OHSS and CERM trials. O.H. reports consulting fees and speaker’s fees from Gedeon Richter and Bayer AG; support for attending meetings from Gedeon-Richter, and leadership roles at the Finnish Society for Obstetrics and Gynecology and the Nordic federation of the societies of obstetrics and gynecology. L.K. reports consulting fees from Gedeon Richter, AstraZeneca, Novartis, Dr KADE/Besins, Palleos Healthcare, Roche, Mithra; honoraria for lectures from Gedeon Richter, AstraZeneca, Novartis, Dr KADE/Besins, Palleos Healthcare, Roche, Mithra; support for attending meetings from Gedeon Richter, AstraZeneca, Novartis, Dr KADE/Besins, Palleos Healthcare, Roche, Mithra; he also has a leadership role in the German Society of Gynecological Endocrinology (DGGEF). M.K. reports grants from French Foundation for Medical Research (FRM), Australian Ministry of Health, Medical Research Future Fund and French National Cancer Institute; support for meeting attendance from European Society for Gynaecological Endoscopy (ESGE), European Congress on Endometriosis (EEC) and ESHRE; She is an advisory Board Member, FEMaLe Project (Finding Endometriosis Using Machine Learning), Scientific Committee Chair for the French Foundation for Research on Endometriosis and Scientific Committee Chair for the ComPaRe-Endometriosis cohort. A.N. reports grants from Merck SA and Ferring; speaker fees from Merck SA and Ferring; support for meeting attendance from Merck SA; Participation on a Data Safety Monitoring Board or Advisory Board with Nordic Pharma and Merck SA; she also is a board member of medical advisory board, Endometriosis Society, the Netherlands (patients advocacy group) and an executive board member World Endometriosis Society. E.S. reports grants from National Institute for Health Research UK, Rosetrees Trust, Barts and the London Charity; Royalties from De Gruyter (book editor); consulting fees from Hologic; speakers fees from Hologic, Johnson & Johnson, Medtronic, Intuitive, Olympus and Karl Storz; Participation in the Medicines for Women’s Health Expert Advisory Group with Medicines and Healthcare Products Regulatory Agency (MHRA); he is also Ambassador for the World Endometriosis Society. C.T. reports grants from Merck SA; Consulting fees from Gedeon Richter, Nordic Pharma and Merck SA; speaker fees from Merck SA, all paid to the institution; and support for meeting attendance from Ferring, Gedeon Richter, Merck SA. The other authors have no conflicts of interest to declare. Appendix Members of the ESHRE Endometriosis Guideline Group Signe AltmäeDepartment of Biochemistry and Molecular Biology, Faculty of Sciences, University of Granda, Spain; Div. Obstetrics and Gynaecology, CLINTEC, Karolinska Institutet, Sweden. Baris AtaKoc University School of Medicine, Turkey Elizabeth BallThe Royal London Hospital, Bartshealth NHS Trust and Queen Mary University of London, London, UK; City University London, London, UK Fabio BarraAcademic Unit of Obstetrics and Gynecology, IRCCS Ospedale Policlinico San Martino, Genoa, Italy; Department of Neurosciences, Rehabilitation, Ophthalmology, Genetics, Maternal and Child Health (DiNOGMI), University of Genoa, Genoa, Italy. Ercan BastuDepartment of Obstetrics and Gynecology, University of Massachusetts Chan Medical School, USA Alexandra Bianco-AnilEndoFrance, French patients’ Association, France Ulla Breth KnudsenAarhus University, Aarhus, Denmark Réka BrubelSemmelweis University, Faculty of Medicine, Budapest, Hungary Julia CambitziPain Management Centre, University College London Hospitals (UCLH), London, UK Astrid CantineauUniversity of Groningen, University Medical Center Groningen, Groningen, The Netherlands Ying CheongUniversity of Southampton, Complete Fertility Southampton, Southampton, UK Angelos Daniilidis2nd University Department of Obstetrics & Gynecology, Hippokratio General Hospital, Aristotle University of Thessaloniki, Greece Bianca De BieEndometriose Stichting, The Netherlands Caterina ExacoustosDepartment of Surgical Sciences, Obstetric and Gynecological Unit, University of Rome ‘Tor Vergata’, Rome, Italy Simone FerreroAcademic Unit of Obstetrics and Gynecology, IRCCS Ospedale Policlinico San Martino, Genoa, Italy; Department of Neurosciences, Rehabilitation, Ophthalmology, Genetics, Maternal and Child Health (DiNOGMI), University of Genoa, Genoa, Italy Tarek GelbayaUniversity Hospitals of Leicester, Leicester, UK Josepha Goetz-CollinetEndoFrance, French patients’ Association, France Gernot HudelistHospital St. John of God Vienna, Vienna, Austria Munawar HussainSouthend University Hospital, UK Tereza Indrielle-KellyBurton and Derby hospitals NHS Trust, Burton on Trent, UK Shaheen KhazaliCentre for Endometriosis and Minimally Invasive Gynaecology (CEMIG) at The HCA Lister Hospital, Chelsea, London, UK Sujata Lalit KumarDepartment of Women’s and Children’s Health, Karolinska Institutet, Stockholm, Sweden; Stockholm IVF, Stockholm, Sweden Umberto Leone Roberti MaggioreDepartment of Gynecologic Oncology, IRCCS National Cancer Institute, Milan, Italy Jacques WM MaasMaastricht University Medical Centre, Department of Obstetrics and Gynecology and GROW—School for Oncology and Developmental Biology, Maastricht, the Netherlands. Helen McLaughlinEndometriosis advocate, London; Endometriosis UK José MetelloCIRMA, Hospital Garcia de Orta, Almada; GINEMED, MaloClinics, Lisboa, Portugal Velja MijatovicAcademic Endometriosis Center Amsterdam UMC, Amsterdam, The Netherlands Yasaman MiremadiThe Austrian Society of Sterility, Fertility and Endocrinology, Austria; TFP kinderwunschklinik Wien, Austria Charles MuteshiUniversity of Oxford, Oxford, UK Michelle NisolleUniversity of Liege/CHR Citadelle, Liege, Belgium Engin OralDepartment of Obstetrics and Gynecology, Bezmialem Vakif University Medical Faculty, Istanbul, Turkey George PadosAristotle University of Thessaloniki, 1st Dept. OB-GYN, ‘Papageorgiou’ General Hospital, Thessaloniki and Centre for Endoscopic Surgery ‘DIAVALKANIKO’ hospital, Thessaloniki, Greece Dana ParadesEndometriosis Association, Finland Nicola PluchinoDivision of Gynecology and Obstetrics, University Hospital of Geneva, Geneva, Switzerland Prasanna Raj SupramaniamEndometriosis CaRe Centre Oxford, Nuffield Department of Women’s and Reproductive Health, University of Oxford, UK; Oxford University Hospitals NHS Foundation Trust, Oxford, UK Maren SchickInstitute of Medical Psychology, Center for Psychosocial Medicine, University Hospital Heidelberg, Heidelberg, Germany Beata SeeberDepartment of Gynecologic Endocrinology and Reproductive Medicine, Medical University of Innsbruck, Innsbruck, Austria Renato SeracchioliDivision of Gynecology and Human Reproduction Physiopathology, IRCCS Azienda Ospedaliero-Universitaria di Bologna, S. Orsola Hospital, University of Bologna, Bologna, Italy Antonio Simone LaganàDepartment of Obstetrics and Gynecology, ‘Filippo Del Ponte’ Hospital, University of Insubria, Varese, Italy Andreas StavroulisEndometriosis and Fertility Center, Cyprus; American Medical Center, Nicosia, Cyprus Linda TebacheUniversity of Liege/CHR Citadelle, Liege, Belgium Gürkan UncuUludag University, Bursa, Turkey Uschi Van den BroeckLeuven University Fertility Center (LUFC), University Hospitals Leuven, Leuven, Belgium Arno van PeperstratenUniversity Medical Center Utrecht, The Netherlands Attila VereczkeyVersys Clinics Human Reproduction Institute, Budapest, Hungary Albert WolthuisUniversity Hospitals Leuven, Leuven, Belgium Pınar Yalçın BahatIstanbul Saglik Bilimleri University Kanuni Sultan Suleyman Training and Research Hospital, Istanbul, Turkey Chadi YazbeckCherest Fertility Center—Reprogynes Institute, Paris, France; Foch University Hospital, Dept Obstetrics Gynecology and Reproductive Medicine, Suresnes, France Open in a new tab Contributor Information ESHRE Endometriosis Guideline Group: Signe Altmäe, Baris Ata, Elizabeth Ball, Fabio Barra, Ercan Bastu, Alexandra Bianco-Anil, Ulla Breth Knudsen, Réka Brubel, Julia Cambitzi, Astrid Cantineau, Ying Cheong, Angelos Daniilidis, Bianca De Bie, Caterina Exacoustos, Simone Ferrero, Tarek Gelbaya, Josepha Goetz-Collinet, Gernot Hudelist, Munawar Hussain, Tereza Indrielle-Kelly, Shaheen Khazali, Sujata Lalit Kumar, Umberto Leone Roberti Maggiore, Jacques W M Maas, Helen McLaughlin, José Metello, Velja Mijatovic, Yasaman Miremadi, Charles Muteshi, Michelle Nisolle, Engin Oral, George Pados, Dana Parades, Nicola Pluchino, Prasanna Raj Supramaniam, Maren Schick, Beata Seeber, Renato Seracchioli, Antonio Simone Laganà, Andreas Stavroulis, Linda Tebache, Gürkan Uncu, Uschi Van den Broeck, Arno van Peperstraten, Attila Vereczkey, Albert Wolthuis, Pınar Yalçın Bahat, and Chadi Yazbeck References Abdou AM, Ammar IMM, Alnemr AAA, Abdelrhman AA.. 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(67.5KB, pdf) Data Availability Statement The full guideline and supporting data (literature report, evidence tables) are available on www.eshre.eu/guidelines. 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Textbooks :: Mathspace Book a Demo Topics 9.P o l y g o n s 9.0 1 A n g l e s o f p o l y g o n s 9.0 2 P a r a l l e l o g r a m s LessonPractice 9.0 3 S p e c i a l p a r a l l e l o g r a m s 9.0 4 T r a p e z o i d s Log inSign upBook a Demo United States of AmericaVirginia Geometry 9.02 Parallelograms LessonPractice Ideas Explore properties of parallelograms Apply properties of parallelograms Explore properties of parallelograms Quadrilateral A polygon with exactly four sides and four vertices Parallelogram A quadrilateral with both pairs of opposite sides parallel. Consecutive angles are angles of a polygon that share a side. Exploration Drag the points to change the quadrilateral and use the checkboxes to explore the applet. Loading interactive... Use the applet to complete the following sentences: A quadrilateral is a parallelogram if and only if its opposite sides are ⬚⬚⬚. A quadrilateral is a parallelogram if and only if its opposite angles are ⬚⬚⬚. In a parallelogram, consecutive angles will be ⬚⬚⬚. A quadrilateral is a parallelogram if and only if its diagonals ⬚⬚⬚ each other. We have many tools in our mathematical tool box to help with proofs now, for example: The diagonal of a parallelogram is a transversal between a pair of parallel lines The other diagonal of a parallelogram is also a transversal between a pair of parallel lines If we extend the sides of a parallelogram, one pair of sides can be seen as transversals of the other pair of parallel sides. We can utilize congruent triangle theorems since the diagonals of a parallelogram break it into triangles: Side-side-side, or SSS: The two triangles have three pairs of congruent sides Side-angle-side, or SAS: The two triangles have two pairs of congruent sides, and the angles between these sides are also congruent Angle-side-angle, or ASA: If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent Angle-Angle-Side, or AAS: If two angles and the non-included side of one triangle are congruent to the corresponding parts of another triangle, the triangles are congruent Examples Example 1 Consider the quadrilateral shown: a If A B C D ABCD A BC D is a parallelogram, prove the opposite sides are congruent. Worked Solution Create a strategy Start by stating what is given and the definition of a parallelogram to build the proof. Apply the idea To prove: Opposite sides of a parallelogram are congruent| | Statements | Reasons | --- | 1. | A B C D ABCD A BC D is a parallelogram | Given | | 2. | A B‾∥D C‾and A D‾∥B C‾\overline{AB}\parallel \overline{DC} \text{ and } \overline{AD}\parallel \overline{BC}A B∥D C and A D∥BC | Definition of a parallelogram | | 3. | ∠B A C≅∠D C A and∠A C B≅∠C A D\angle{BAC}\cong\angle{DCA} \text{ and } \angle{ACB}\cong\angle{CAD}∠B A C≅∠D C A and∠A CB≅∠C A D | Alternate interior angles theorem | | 4. | A C‾≅A C‾\overline{AC}\cong \overline{AC}A C≅A C | Reflexive property of congruence | | 5. | △A B C≅△C D A\triangle{ABC}\cong \triangle{CDA}△A BC≅△C D A | ASA congruence theorem | | 6. | A B‾≅C D‾and B C‾≅D A‾\overline{AB}\cong\overline{CD} \text{ and } \overline{BC}\cong \overline{DA}A B≅C D and BC≅D A | CPCTC | Watch question walkthrough b Prove that if opposite sides of a quadrilateral are congruent, then it is a parallelogram. Worked Solution Create a strategy Construct a diagonal from B B B to D D D. Include congruence markings based on what is given. Apply the idea To prove: If opposite sides of a quadrilateral are congruent, then it is a parallelogram| | Statements | Reasons | --- | 1. | A B‾≅C D‾and B C‾≅D A‾\overline{AB}\cong\overline{CD} \text{ and } \overline{BC}\cong \overline{DA}A B≅C D and BC≅D A | Given | | 2. | B D‾≅B D‾\overline{BD}\cong \overline{BD}B D≅B D | Reflexive property of congruence | | 3. | △A B D≅△C D B\triangle{ABD}\cong \triangle{CDB}△A B D≅△C D B | SSS congruence theorem | | 4. | ∠A B D≅∠C D B\angle ABD \cong \angle CDB∠A B D≅∠C D B and ∠A D B≅∠C B D\angle ADB \cong \angle CBD∠A D B≅∠CB D | CPCTC | | 5. | A B‾∥C D‾\overline{AB} \parallel \overline {CD}A B∥C D | Since ∠A B D≅∠C D B\angle ABD \cong \angle CDB∠A B D≅∠C D B by converse of alternate interior angles theorem | | 6. | A D‾∥B C‾\overline{AD} \parallel \overline {BC}A D∥BC | Since ∠A D B≅∠C B D\angle ADB \cong \angle CBD∠A D B≅∠CB D by converse of alternate interior angles theorem | | 7. | A B C D ABCD A BC D is a parallelogram | Definition of a parallelogram | Watch question walkthrough c Verify the opposite sides of a parallelogram are congruent using constructions. Worked Solution Create a strategy Construct a copy of the opposite sides to two consecutive sides to show that the parallelogram's opposite sides are congruent. Apply the idea If the other end of the compass lines up with C C C, then A B‾\overline{AB}A B is congruent to C D‾\overline{CD}C D. Now, let's check the other pair of sides. If the other end of the compass lines up with C C C, then A D‾\overline{AD}A D is congruent to B C‾\overline{BC}BC. Example 2 Consider the quadrilateral shown: a If A B C D ABCD A BC D is a parallelogram, prove that the diagonals bisect each other. Worked Solution Create a strategy Use the alternate interior angles theorem, congruence theorems, and the definition of a bisector to help prove that the diagonals bisect each other. Apply the idea Watch question walkthrough b Prove that if the diagonals of a quadrilateral bisect each other, then it is a parallelogram. Worked Solution Create a strategy Draw congruence markings using the given diagram to support starting a proof. Apply the idea To prove: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram| | Statements | Reasons | --- | 1. | A P‾≅C P‾\overline{AP} \cong \overline {CP}A P≅CP and B P‾≅D P‾\overline{BP} \cong \overline {DP}BP≅D P | Given | | 2. | ∠A P B≅∠C P D\angle APB \cong \angle CPD∠A PB≅∠CP D and ∠A P D≅∠B P C\angle APD \cong \angle BPC∠A P D≅∠BPC | Vertical angles theorem | | 3. | △A B P≅△C D P\triangle {ABP} \cong \triangle{CDP}△A BP≅△C D P and △A P D≅△C P B\triangle {APD} \cong \triangle {CPB}△A P D≅△CPB | SAS congruence theorem | | 4. | A B‾≅C D‾\overline{AB} \cong \overline{CD}A B≅C D and A D‾≅B C‾\overline{AD} \cong \overline{BC}A D≅BC | CPCTC | | 5. | ∠A B P≅∠C D P\angle ABP \cong \angle CDP∠A BP≅∠C D P and ∠A D P≅∠C B P\angle ADP \cong \angle CBP∠A D P≅∠CBP | CPCTC | | 6. | A B‾∥C D‾\overline{AB} \parallel \overline{CD}A B∥C D and A D‾∥B C‾\overline{AD} \parallel \overline{BC}A D∥BC | Since ∠A B P≅∠C D P\angle ABP \cong \angle CDP∠A BP≅∠C D P and ∠A D P≅∠C B P\angle ADP \cong \angle CBP∠A D P≅∠CBP by converse of alternate interior angles theorem | | 7. | A B C D ABCD A BC D is a parallelogram | Definition of a parallelogram | Watch question walkthrough c Verify that both halves of each diagonal are congruent using constructions. Worked Solution Create a strategy Construct a copy of one half of the diagonals for each diagonal to show that both halves of each diagonal in a parallelogram are congruent. Apply the idea If the other end of the compass lines up with P P P, then A P‾\overline{AP}A P is congruent to C P‾\overline{CP}CP. Now, let's check the other diagonal. If the other end of the compass lines up with P P P, then B P‾\overline{BP}BP is congruent to D P‾\overline{DP}D P. Example 3 Use geometric constructions to show that A B C D ABCD A BC D is a parallelogram if its opposite angles are congruent. Worked Solution Create a strategy Construct a copy of the opposite angles to two consecutive angles to show that if the quadrilateral's opposite angles are congruent, it must be a parallelogram. Apply the idea If the constructed copy of ∠A\angle{A}∠A lines up with ∠C\angle{C}∠C, then ∠A\angle{A}∠A is congruent to ∠C\angle{C}∠C. If the constructed copy of ∠B\angle{B}∠B lines up with ∠D\angle{D}∠D, then ∠B\angle{B}∠B is congruent to ∠D\angle{D}∠D. Idea summary We can use the definition of a parallelogram, theorems about congruency, and transversals to prove properties of parallelograms. Apply properties of parallelograms Parallelograms have special properties regarding side lengths, angles, and diagonals. We can use these properties to find unknown angles or sides of parallelograms, or to prove that a quadrilateral is a parallelogram. Parallelogram opposite sides theorem A quadrilateral is a parallelogram if and only if its opposite sides are congruent. Parallelogram opposite angles theorem A quadrilateral is a parallelogram if and only if its opposite angles are congruent. Parallelogram consecutive angles theorem If a quadrilateral is a parallelogram, then its consecutive angles are supplementary Example: ∠A D C\angle ADC∠A D C and ∠D A B\angle DAB∠D A B are supplementary. Parallelogram diagonals theorem A quadrilateral is a parallelogram if and only if its diagonals bisect each other. We may use these properties to solve problems when we are told that a diagram is a parallelogram. Examples Example 4 Find the missing parts of the parallelograms. a Given parallelogram P Q R S PQRS PQRS, find R S RS RS. Worked Solution Create a strategy Since we know P Q R S PQRS PQRS is a parallelogram, we want to use the theorems about parallelograms to determine R S RS RS. Apply the idea Opposite sides of a parallelogram are congruent so R S‾≅P Q‾\overline{RS} \cong \overline{PQ}RS≅PQ​. R S=P Q=2.41 RS = PQ = 2.41 RS=PQ=2.41 Watch question walkthrough b Given parallelogram D E F G DEFG D EFG, find m∠D G F m \angle DGF m∠D GF. Worked Solution Create a strategy The two labeled angles, ∠D E F\angle DEF∠D EF and ∠E D G\angle EDG∠E D G, are consecutive angles. Since D E F G DEFG D EFG is a parallelogram, the consecutive angles are supplementary. We want to write an equation relating the two labeled angles and then solve for x x x. Once we solve for x x x, we then want to use the theorem that states that opposite angles of a parallelogram are congruent. Using this theorem, we know that ∠D E F≅∠D G F\angle DEF \cong \angle DGF∠D EF≅∠D GF. We want to substitute the value we solved for x x x and into 5 x 5x 5 x and evaluate m∠D E F m \angle DEF m∠D EF as this will be the same as m∠D G F m \angle DGF m∠D GF. Apply the idea (5 x)+(2 x+5)\displaystyle \left(5x\right)+\left(2x+5\right)(5 x)+(2 x+5)=\displaystyle ==180\displaystyle 180 180 Consecutive angles are supplementary 7 x+5\displaystyle 7x+5 7 x+5=\displaystyle ==180\displaystyle 180 180 Combine like terms 7 x\displaystyle 7x 7 x=\displaystyle ==175\displaystyle 175 175 Subtract 5 5 5 from both sides of equation x\displaystyle x x=\displaystyle ==25\displaystyle 25 25 Divide both sides of equation by 7 7 7 Since we know that ∠D E F≅∠D G F\angle DEF \cong \angle DGF∠D EF≅∠D GF, we know that ∠D G F=5 x\angle DGF = 5x∠D GF=5 x Substituting 25 25 25 for x x x and evaluating, we get 5(25)=125 5\left(25\right)=125 5(25)=125. m∠D G F=125°m\angle DGF = 125 \degree m∠D GF=125° Watch question walkthrough Example 5 Determine whether or not each of the given quadrilaterals is a parallelogram. a Worked Solution Create a strategy We know that if a quadrilateral is a parallelogram, its opposite angles are congruent and its consecutive angles are supplementary. Use the polygon angle sum theorem to find the missing angle and determine if the quadrilateral satisfies the conditions of a parallelogram. Apply the idea By the polygon angle sum theorem, we know that the sum of the angles of an quadrilateral must be (4−2)⋅180°=2⋅180°=360°\left(4-2\right) \cdot 180 \degree = 2 \cdot 180 \degree = 360 \degree(4−2)⋅180°=2⋅180°=360°. Let the missing angle be x x x. For the given quadrilateral, we have 79+101+101+x\displaystyle 79 + 101 + 101\ + \ x 79+101+101+x=\displaystyle ==360\displaystyle 360 360 Polygon angle sum theorem 281+x\displaystyle 281 \ + \ x 281+x=\displaystyle ==360\displaystyle 360 360 Combine like terms x\displaystyle x x=\displaystyle ==79\displaystyle 79 79 Subtract 281 281 281 from both sides Since the unknown angle in the quadrilateral is 79°79 \degree 79°, we know that the opposite angles are congruent and therefore the quadrilateral is a parallelogram. Reflect and check We can also use that consecutive angles are supplementary in a parallelogram, so 101°+x=180°{101 \degree + \ x = 180 \degree}101°+x=180° and the unknown angle must be 79°79 \degree 79°, so we have that opposite angles are congruent and therefore the quadrilateral is a parallelogram. Watch question walkthrough b Worked Solution Apply the idea Since 42 m≠30 m 42 \text{ m} \neq 30 \text{ m}42 m=30 m, the diagonals of the quadrilateral do not bisect each other and the quadrilateral is not a parallelogram. Reflect and check If the segments were instead the same length along each diagonal, we could use the fact that if a quadrilateral is a parallelogram, then its diagonals bisect each other. Watch question walkthrough c Worked Solution Create a strategy We aren't given information about the angles or diagonals of the quadrilateral, so we rely on determining if the quadrilateral meets the criteria: If a quadrilateral is a parallelogram, then its opposite sides are congruent. Apply the idea Since the opposite sides of the quadrilateral are congruent, the quadrilateral is a parallelogram. Watch question walkthrough Example 6 Solve for the unknown variables in the diagram that make the quadrilateral a parallelogram. Worked Solution Create a strategy We can use the fact that the diagonals of the parallelogram form transversals, so we can use the alternate interior angles theorem to state that 32=4 z 32 = 4z 32=4 z. We know that diagonals of a parallelogram bisect each other, so 4 y+6=5 y−5 4y + 6 = 5y - 5 4 y+6=5 y−5. Since opposite sides of a parallelogram are congruent, 4 x=16 4x = 16 4 x=16. Apply the idea We have 32\displaystyle 32 32=\displaystyle ==4 z\displaystyle 4z 4 z Alternate interior angles theorem 8\displaystyle 8 8=\displaystyle ==z\displaystyle z z Divide both sides by 4 4 4 4 y+6\displaystyle 4y+6 4 y+6=\displaystyle ==5 y−5\displaystyle 5y-5 5 y−5 Diagonals of a parallelogram bisect each other 4 y+11\displaystyle 4y+11 4 y+11=\displaystyle ==5 y\displaystyle 5y 5 y Add 5 5 5 to both sides 11\displaystyle 11 11=\displaystyle ==y\displaystyle y y Subtract 4 y 4y 4 y from both sides 4 x\displaystyle 4x 4 x=\displaystyle ==16\displaystyle 16 16 Opposite sides of a parallelogram are congruent x\displaystyle x x=\displaystyle ==4\displaystyle 4 4 Divide both sides by 4 4 4 Watch question walkthrough Idea summary Use the following about quadrilaterals to solve problems involving parallelograms: A quadrilateral is a parallelogram if and only if its opposite sides are congruent A quadrilateral is a parallelogram if and only if its opposite angles are congruent In a parallelogram, consecutive angles will be supplementary A quadrilateral is a parallelogram if and only if its diagonals bisect each other Outcomes G.PC.1 The student will prove and justify theorems and properties of quadrilaterals, and verify and use properties of quadrilaterals to solve problems, including the relationships between the sides, angles, and diagonals. G.PC.1a Solve problems, using the properties specific to parallelograms, rectangles, rhombi, squares, isosceles trapezoids, and trapezoids. G.PC.1b Prove and justify that quadrilaterals have specific properties, using coordinate and algebraic methods, such as the slope formula, the distance formula, and the midpoint formula. G.PC.1c Prove and justify theorems and properties of quadrilaterals using deductive reasoning. G.PC.1d Use congruent segment, congruent angle, angle bisector, perpendicular line, and/or parallel line constructions to verify properties of quadrilaterals. What is Mathspace About Mathspace
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https://jordanbell.info/LaTeX/mathematics/kolmogorov-extension/kolmogorov-extension.pdf
The Kolmogorov extension theorem Jordan Bell June 21, 2014 1 σ-algebras and semirings If X is a nonempty set, an algebra of sets on X is a collection A of subsets of X such that if {Ai} ⊂A is finite then S i Ai ∈A , and if A ∈A then X \ A ∈A . An algebra A is called a σ-algebra if {Ai} ⊂A being countable implies that S i Ai ∈A . If X is a set and G is a collection of subsets of X, we denote by σ(G ) the smallest σ-algebra containing G , and we say that σ(G ) is the σ-algebra generated by G . Later we will also use the following notion. If X is a nonempty set, a semiring of sets on X is a collection S of subsets of X such that (i) ∅∈S , (ii) if A, B ∈S then A ∩B ∈S , and (iii) if A, B ∈S then there are pairwise disjoint S1, . . . , Sn ∈S such that A \ B = Sn i=1 Si; we do not demand that this union itself belong to S . We remark that a semiring on X need not include X. If S is a semiring of sets and µ0 : S →[0, ∞], we say that µ0 is finitely ad-ditive if {Si} ⊂S being finite, pairwise disjoint, and satisfying S i Si ∈S im-plies that µ0 (S i Si) = P i µ0(Si), and countably additive if {Si} ⊂S being countable, pairwise disjoint, and satisfying S i Si ∈S implies that µ0 (S i Si) = P i µ0(Si). If G is a collection of subsets of X, the algebra generated by G is the smallest algebra containing G . We shall use the following lemma in the proof of Lemma 10. Lemma 1. If S is a semiring on a set X and X ∈S , then the algebra generated by S is equal to the collection of finite unions of members of S . For a bounded countably additive function, the Carath´ eodory extension theorem states the following.1 Theorem 2 (Carath´ eodory extension theorem). Suppose that X is a nonempty set, that S is a semiring on X, and that µ0 : S →[0, 1] is countably additive. Then there is one and only one measure on σ(S ) whose restriction to S is equal to µ0. 1Ren´ e L. Schilling, Measures, Integrals and Martingales, p. 37, Theorem 6.1. If we had not specified that µ0 : S →[0, 1] but rather had talked about µ0 : S →[0, ∞], then the Carath´ eodory extension theorem shows that there is some extension of µ0 to σ(S ), but this extension need not be unique. 1 2 Product σ-algebras Suppose that X is a set, that {(Yi, Mi) : i ∈I} is a family of measurable spaces, and that fi : X →Yi are functions. The smallest σ-algebra on X such that each fi is measurable is called the σ-algebra generated by {fi : i ∈I}. This is analogous to the initial topology induced by a family of functions on a set. Calling this σ-algebra M and supposing that σ(Gi) = Mi for each i ∈I, we check then that M = σ {f −1 i (A) : i ∈I, A ∈Gi}  . (1) Suppose that {(Xi, Mi) : i ∈I} is a family of measurable spaces. Let X = Y i∈I Xi, the cartesian product of the sets Xi, and let πi : X →Xi be the projection maps. The product σ-algebra on X is the σ-algebra M generated by {πi : i ∈I}, and is denoted M = O i∈I Mi. This is analogous to the product topology on a cartesian product of topological spaces, which has the initial topology induced by the family of projection maps. For H ⊂I, we define XH = Y i∈H Xi. Thus, XI = X and X∅= {∅}, and for G ⊂H, XH = XG × XH\G. For H ⊂I, let MH = O i∈H Mi, the product σ-algebra on XH. Thus, MI = M and M∅= {∅, {∅}}, and for G ⊂H we have MH = MG ⊗MH\G. For G ⊂H, we define PH,G : XH →XG to be the projection map: an element of XH is a function x on H such that x(i) ∈Xi for all i ∈H, and PH,G(x) is the restriction of x to G. Lemma 3. For G ⊂H, PH,G : (XH, MH) →(XG, MG) is measurable. If F is a finite subset of I and A ∈MF , we call A × XI\F ∈M an F-cylinder set. Cylinder sets for the product σ-algebra are analogous to the usual basic open sets for the product topology. Lemma 4. The collection of all cylinder sets is an algebra of sets on Q i∈I Xi, and this collection generates the product σ-algebra N I∈I Mi. 2 The product σ-algebra can in fact be generated by a smaller collection of sets. (The following collection of sets is not a minimal collection of sets that generates the product σ-algebra, but it is smaller than the collection of all cylinder sets and it has the structure of a semiring, which will turn out to be useful.) An intersection of finitely many sets of the form A × MI{t}, A ∈Mt, is called a product cylinder. Lemma 5. The collection of all product cylinders is a semiring of sets on Q i∈I Xi, and this collection generates the product σ-algebra N i∈I Mi. 3 Borel σ-algebras If (X, τ) is a topogical space, the Borel σ-algebra on X is σ(τ), and is denoted BX. A member of BX is called a Borel set. Lemma 6. If X is a topological space and G is a countable subbasis for the topology of X, then BX = σ(G ). A separable metrizable space is second-countable, so we can apply the fol-lowing theorem to such spaces. Theorem 7. Suppose that Xi, i ∈N, are second-countable topological spaces and let X = Q i∈N Xi, with the product topology. Then BX = O i∈N BXi. Proof. For each i ∈N, let Gi be a countable subbasis for the topology of Xi. Because Gi is a subbasis for the topology of Xi for each i, we check that G is a subbasis for the product topology of X, where G = {π−1 i (A) : i ∈N, A ∈Gi}. Because each Gi is countable and N is countable, G is countable. Hence by Lemma 6, BX = σ(G ). On the other hand, for each i ∈N we have by Lemma 6 that BXi = σ(Gi), and so by (1), O i∈N BXi = σ(G ). 3 4 Product measures If {(Xi, Mi, µi) : 1 ≤i ≤n} are σ-finite measure spaces, let X = Qn i=1 Xi and M = Nn i=1 Mi. It is a fact that there is a unique measure µ on M such that for Ai ∈Mi, µ n Y i=1 Ai ! = n Y i=1 µi(Ai), and µ is a σ-finite measure. We write µ = Qn i=1 µi and call µ the product measure.2 5 Compact classes If X is a set and C is a collection of subsets of X, we say that C is a compact class if every countable subset of C with the finite intersection property has nonempty intersection. We remind ourselves that a collection E of sets is said to have the finite intersection property if for any finite subset F of E we have T A∈F A ̸= ∅. Usually one speaks about a collection of sets having the finite intersection property in the following setting: A topological space Y is compact if and only if every collection of closed sets that has the finite intersection property has nonempty intersection. We will employ the following lemma in the proof of the Kolmogorov extension theorem.3 Lemma 8. Suppose that C 0 is a compact class of subsets of a set X and let C be the collection of countable intersections of finite unions of members of C 0. C is the smallest collection of subsets of X containing C 0 that is closed under finite unions and countable intersections, and C is itself a compact class. We state the following result that gives conditions under which a finitely additive functions on an algebra of sets is in fact countably additive,4 and then use it to prove an analogous result for semirings. Lemma 9. Suppose that A is an algebra of sets on a set X, and that µ0 : A → [0, ∞) is finitely additive and µ0(X) < ∞. If there is a compact class C ⊂A such that µ0(A) = sup{µ0(C) : C ∈C and C ⊂A}, A ∈A , then µ0 is countably additive. The following lemma gives conditions under which a finitely additive function on a semiring of sets is in fact countably additive.5 2Gerald B. Folland, Real Analysis: Modern Techniques and Their Applications, second ed., pp. 64–65, and p. 31, Theorem 1.14. 3V. I. Bogachev, Measure Theory, volume I, p. 50, Proposition 1.12.4. 4Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitch-hiker’s Guide, third ed., p. 378, Theorem 10.13. 5Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitch-hiker’s Guide, third ed., p. 521, Lemma 15.25. 4 Lemma 10. Suppose that S is a semiring of sets on X with X ∈S and that µ0 : S →[0, ∞) is finitely additive and µ0(X) < ∞. If there is a compact class C ⊂S such that µ0(A) = sup{µ0(C) : C ∈C and C ⊂A}, A ∈S , then µ0 is a countably additive. Proof. Let Cu be the collection of finite unions of members of C . Cu is a subset of the compact class produced in Lemma 8, hence is itself a compact class. Let A be the collection of finite unions of members of S , which by Lemma 1 is the algebra generated by S . Because C ⊂S ⊂A and A is closed under finite unions, it follows that Cu ⊂A . Because S is a semiring, it is a fact that if A1, . . . , An, A ∈S , then there are pairwise disjoint S1, . . . , Sm ∈S such that A \ Sn i=1 Ai = Sm i=1 Si.6 Thus, if A1, . . . , An ∈S , defining Ei = Ai \ Si−1 j=1 Aj, with E1 = A1 \ ∅= A1, the sets E1, . . . , En are pairwise disjoint, and for each i there are pairwise disjoint Si,1, . . . , Si,ai ∈S such that Ei = Sai j=1 Si,j. Then the sets Si,j, 1 ≤i ≤n, 1 ≤j ≤ai are pairwise disjoint and their union is equal to Sn i=1 Ai. This shows that any element of A can be written as a union of pairwise disjoint elements of S . Furthermore, because S is a semiring, if A1, . . . , AN ∈S , there are pairwise disjoint S1, . . . , Sk ∈S such that for each 1 ≤i ≤k there is some 1 ≤n ≤N such that Si ∈An, and for each 1 ≤n ≤N, there is a subset F ⊂{1, . . . , k} such that An = S i∈F Si.7 Let E ∈A and suppose that E = SN n=1 An, where A1, . . . , AN ∈S are pairwise disjoint, and that E = SM m=1 Bm, where B1, . . . , BM ∈S are pairwise disjoint. There are pairwise disjoint S1, . . . , Sk ∈S such that for each 1 ≤i ≤k there is some 1 ≤n ≤N or 1 ≤m ≤M such that, respectively, Si ∈An or Si ∈Bm, and for each 1 ≤n ≤N there is some subset F ⊂{1, . . . , k} such that An = S i∈F Si, and for each 1 ≤m ≤M there is some subset F ⊂{1, . . . , k} such that Bm = S i∈F Si. It follows that E = Sk i=1 Si, and because µ0 is finitely additive, N X n=1 µ0(An) = k X i=1 µ0(Si) = M X m=1 µ0(Bm). Therefore, for E ∈A it makes sense to define µ(E) = n X n=1 µ0(Ai), where A1, . . . , An are pairwise disjoint elements of S whose union is equal to E. Also, µ(X) = µ0(X) < ∞. 6Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitch-hiker’s Guide, third ed., p. 134, Lemma 4.7. 7Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitch-hiker’s Guide, third ed., p. 134, Lemma 4.8. 5 We shall now show that the function µ : A →[0, ∞) is finitely additive. If E1, . . . , EN ∈A are pairwise disjoint, for each n there are pairwise disjoint An,1, . . . , An,an ∈S such that En = San j=1 An,j, and there are pairwise disjoint S1, . . . , Sk ∈S such that for each 1 ≤i ≤k, there is some 1 ≤n ≤N and some 1 ≤j ≤an such that Si ∈An,j, and for each 1 ≤n ≤N and each 1 ≤j ≤an there is some subset F ⊂{1, . . . , k} such that An,j = S i∈F Si. It follows that SN n=1 En = Sk i=1 Si, and µ N [ n=1 En ! = µ k [ i=1 Si ! = k X i=1 µ0(Si) = N X n=1 an X j=1 µ0(An,j) = N X n=1 µ(En), showing that µ is finitely additive. For E = Sn i=1 Ai ∈A with pairwise disjoint A1, . . . , An ∈S , let ϵ > 0, and for each 1 ≤i ≤n let Ci ∈C with µ0(Ci) > µ0(Ai) + ϵ n and Ci ⊂Ai. Then C = Sn i=1 Ci ∈Cu. As A1, . . . , An are pairwise disjoint and Ci ⊂Ai, C1, . . . , Cn are pairwise disjoint, so because µ is finitely additive on A , µ(C) = n X i=1 µ(Ci) = n X i=1 µ0(Ci) > n X i=1  µ0(Ai) + ϵ n  = µ(E) + ϵ. Lemma 9 tells us now that µ : A →[0, ∞) is countably additive, and therefore µ0, its restriction to the semiring S , is countably additive. 6 Kolmogorov consistent families Suppose that {(Xi, Mi) : i ∈I} is a family of measurable spaces. The collection D of all finite subsets of I, ordered by set inclusion, is a directed set. Suppose that for each F ∈D, µF is a probability measure on MF ; we defined the notation MF in §2 and we use that here. We say that the family of measures {µF : F ∈D} is Kolmogorov consistent if whenever F, G ∈D with F ⊂G, it happens that PG,F ∗µG = µF , where f∗µ denotes the pushforward of µ by f, i.e. f∗µ = µ ◦f −1. It makes sense to talk about PG,F ∗µG because PG,F : (XG, MG) →(XF , MF ) is measurable, as stated in Lemma 3. We are now prepared to prove the Kolmogorov extension theorem.8 Theorem 11 (Kolmogorov extension theorem). Suppose that {(Xi, Mi) : i ∈I} is a family of measurable spaces and suppose that for each F ∈D, µF is a probability measure on MF . If the family of probability measures {µF : F ∈D} is Kolmogorov consistent and if for each i ∈I there is a compact class Ci ⊂Mi satisfying µi(A) = sup{µi(C) : C ∈Ci and C ⊂A}, A ∈Mi, (2) then there is a unique probability measure on MI such that for each F ∈D, the pushforward of µ by the projection map PI,F : XI →XF is equal to µF . 8Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitch-hiker’s Guide, third ed., p. 522, Theorem 15.26. 6 Proof. Define C 0 = {C × XI{i} : i ∈I, C ∈Ci}. We shall show that C 0 is a compact class. Suppose {Cn × XI{in} : n ∈N, Cn ∈Cin} ⊂C 0 has empty intersection. For each i ∈I, let Qi = \ in=i Cn, and if there are no such in, then Qi = Xi. Then \ n∈N Cn × XI{in} = Y i∈I Qi. Because this intersection is equal to ∅, one of the factors in the product is equal to ∅. (For some purposes one wants to keep track of where the axiom of choice is used, so we mention that concluding that some factor of any empty cartesian product is itself empty is equivalent to the axiom of choice). No Xi is empty, so this empty Qi must be of the form T in=i Cn for which at least one in is equal to i. But if in = i then Cn ∈Ci, and because Ci is a compact class, T in=i Cn = ∅ implies that there are finitely many a1, . . . , aN such that TN n=1 Can = ∅, and this yields TN n=1 Can × XI{ian} = ∅. We have thus proved that if an intersection of countably many members of C 0 is empty then some intersection of finitely many of these is empty, showing that C 0 is a compact class. Let C 1 be the smallest collection of subsets of XI containing C that is closed under finite unions and countable intersections, and by Lemma 8 we know that C 1 is a compact class; we use the notation C 1 because presently we will use a subset of C 1. Let A be the collection of all cylinder sets of the product σ-algebra MI. Explicitly, A = {A × XI\F : F ∈D, A ∈MF }. Suppose F, G ∈D, F ⊂G, A ∈MF , B ∈MG, and that A × XI\F = B × XI\G. It follows that B = A × XG\F , and then using PG,F ∗µG = µF we get µG(B) = µG(A × XG\F ) = µG(P −1 G,F (A)) = µF (A). Therefore it makes sense to define µ0 : A →[0, 1] as follows: for A×XI\F ∈A , µ0(A × XI\F ) = µF (A). Let F1, . . . , Fn ∈D and A1 ∈MF1, . . . , An ∈MFn, and suppose that A1 × XI\F1, . . . , An × XI\Fn ∈G are pairwise disjoint. With F = Sn j=1 Fj ∈D, n [ j=1 Aj × XI\Fj =   n [ j=1 Aj × XF \Fj  × XI\F , 7 which is an F-cylinder set. Then, µ0   n [ j=1 Aj × XI\Fj   = µF   n [ j=1 Aj × XF \Fj   = n X j=1 µF (Aj × XF \Fj) = n X j=1 µ0(Aj × XI\Fj), showing that µ0 : A →[0, 1] is finitely additive. Let G be the collection of all product cylinder sets of the product σ-algebra MI. Explicitly, G = ( \ i∈F Ai × XI{i} : F ∈D, and Ai ∈Mi for i ∈F ) . It is apparent that C 0 ⊂G . Let C be the intersection of C 1 and G . A subset of a compact class is a compact class, so C is a compact class, and C 0 ⊂C . Suppose that E ∈G : there is some F ∈D and Ai ∈Mi for each i ∈F such that E = \ i∈F Ai × XI{i} = Y i∈F Ai ! × XI\F . Take n = |F|, and let ϵ > 0. Then, for each i ∈F, by (2) there is some Ci ∈Ci such that Ci ⊂Ai and µi(Ai) < µi(Ci) + ϵ n, and we set C = \ i∈F Ci × XI{i} = Y i∈F Ci ! × XI\F , which is a finite intersection of members of C 0 and hence belongs to C 1, and which visibly belongs to G , and hence belongs to C . We have E \ C =  [ i∈F (Ai \ Ci) × Y j∈F {i} Aj  × XI\F ⊂ [ i∈F (Ai \ Ci) × XI{i}. Both E \ C and the above union are cylinder sets so it makes sense to apply µ0 8 to them, and because µ0 is finitely additive, µ0(E \ C) ≤ X i∈F µ0((Ai \ Ci) × XI{i}) = X i∈F µi(Ai \ Ci) = X i∈F µi(Ai) −µi(Ci) < X i∈F ϵ n = ϵ. Hence µ0(E) −µ0(C) = µ0(E \ C) < ϵ, i.e. µ0(E) < µ0(C) + ϵ. Thus, we have proved that for each ϵ > 0, there is some C ∈C such that C ⊂E and µ0(E) < µ0(C) + ϵ, which means that µ0(E) = sup{µ0(C) : C ∈C and C ⊂E}. Using the restriction of µ0 : A →[0, 1] to the semiring G and the compact class C , the conditions of Lemma 10 are satisfied, and therefore the restriction of µ0 to G is countably additive. By Lemma 5, MI = σ(G ). Because the restriction of µ0 to the semiring G is countably additive, we can apply the Carath´ eodory extension theorem, which tells us that there is a unique measure µ on σ(G ) = MI whose restriction to G is equal to the restriction of µ0 to G . Check that the restriction of µ to A is equal to µ0. For F ∈D and A ∈MF , PI,F ∗µ(A) = µ(P −1 I,F (A)) = µ(A × XI\F ) = µ0(A × XI\F ) = µF (A), showing that PI,F ∗µ = µF . Certainly µ(XI) = 1, namely, µ is a probability measure. If ν is a probability measure on XI whose pushforward by PI,F is equal to µF for each F ∈D, then check that the restriction of ν to G is equal to the restriction of µ0 to G , and then by the assertion of uniqueness in Carath´ eodory’s theorem, ν = µ, completing the proof. If X is a Hausdorff space, we say that a Borel measure µ on X is tight if for every A ∈BX, µ(A) = sup{µ(K) : K is compact and K ⊂A}. A Polish space is a topological space that is homeomorphic to a complete separable metric space, and it is a fact that a finite Borel measure on a Polish space is tight.9 In particular, any Borel probability measure on a Polish space is tight. We use this in the proof of the following version of the Kolmogorov extension theorem, which applies for instance to the case where Xi = R for each i ∈I, with I any index set. 9Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A Hitch-hiker’s Guide, third ed., p. 438, Theorem 12.7. 9 Corollary 12. Suppose that {Xi : i ∈I} is a family of Polish spaces and suppose that for each F ∈D, µF is a Borel probability measure on XF . If the family of measures {µF : F ∈D} is Kolmogorov consistent, then there is a unique probability measure on MI = N i∈I BXi such that for each F ∈D, the pushforward of µ by the projection map PI,F : XI →XF is equal to µF . Proof. For each i ∈I, let Ci be the collection of all compact subsets of Xi. In any topological space, check that a collection of compact sets is a compact class. The fact that µi is a Borel probability measure on a Polish space then implies that it is tight, which we can write as µi(A) = sup{µi(C) : K ∈Ci and K ⊂A}, A ∈Mi. Therefore the conditions of Theorem 11 are satisfied, so the claim follows. If the index set I in the above corollary is countable, then by Theorem 7 the product σ-algebra N i∈I BXi is equal to the Borel σ-algebra of the product Q t∈T Xi, so that the probability measure µ on the product σ-algebra is in this case a Borel measure. 10
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http://www.ae.metu.edu.tr/~ae451/critical_performance_parameters.pdf
AE 451 Aeronautical Engineering Design I Estimation of Critical Performance Parameters Prof. Dr. Serkan Özgen Dept. Aerospace Engineering Fall 2015 Airfoil selection • The airfoil effects the cruise speed, takeoff and landing distances, stall speed, handling qualities and overall aerodynamic efficiency during all phases of flight. • The airfoil may be separated into: – Thickness distribution, influences profile drag, – Zero-thickness camber line, influences lift and drag due to lift. • Upper surface of an airfoil or wing produces roughly 2/3 of total lift. • Zero-lift angle of attack is roughly equal to the percent camber of the airfoil (in deg). 2 Design lift coefficient • This is the lift coefficient at which the airfoil has the best L/D. • The airplane should be designed such that it flies the design mission at or near the design lift coefficient to maximize the aerodynamic efficiency. 3 Airfoil geometry 4 Stall • Some airfoils show a gradual reduction in lift during stall, while others show a violent loss of lift with a rapid change in pitching moment. 5 Stall • Thick airfoils (round leading edge, t/c>14%) stall starting from the trailing edge. At around α=10o, the boundary-layer begins to separate starting at the tariling edge and moving forward as the angle of attack is further increased. The loss of lift is gradual, pitching moment does not change significantly. • Moderately thick airfoils (6%<t/c<14%) stall from the leading edge. Flow separates over the nose at a very low angle of attack, but immediately reattaches, so the effect is initially small. At some higher α, the flow does not reattach and the airfoil stalls almost immediately. Lift and pitching moment vary violently. 6 Stall • Thin airfoils (t/c<6%) stall from the leading edge and the flow reattaches immediately. As α is increased, the bubble contnues to stretch towards the trailing edge as the angle of attack is increased. At α where the bubble stretches all the way to the trailing edge, cl,max is reached. Beyond that α, the flow is separated over the entire airfoil, so stall occurs. The loss of lift is smooth, but large changes in pitching moment are observed. • Twisting the wing such that the tip airfoils have a reduced angle of attck compared to the root (washout) can cause the wing to stall at the root first. 7 Stall • Different airfoil sections may be used at the root and the tip, with a tip airfoil that stalls at a higher angle of attack. This produces good flow over the ailerons for roll control (aileron authority) at an angle of attack where the root has stalled. • Stall characteristics for thinner airfoils may be improved with leading edge devices like slots, slats, leading edge flaps. • Wing stall is directly related to airfoil stall only for high aspect ratio, unswept wings. For low aspect ratio, swept wings, 3-D effects dominate stall characteristics and airfoil stall characteristics can be ignored. • Horizontal tail or canard size is directly related to the magnitude of the wing pitching moment to be balanced. 8 Thickness ratio • Airfoil thickness ratio has a direct effect on drag, maximum lift, stall characteristics and structural weight. • A wing with a fairly high AR, moderate sweep, large nose radius provides a higher stall angle and a higher CL,max. • For a wing with low AR, swept wings, a sharper leading edge provides greater CL,maxdue to the formation of vortices behind the leading edge. • Wing structural weight ~1/ 𝑡/𝑐 •halving the thickness ratio increases the empty weight of the airplane by 6%. 9 Thickness ratio 10 Thickness ratio 11 Thickness ratio 12 Thickness ratio • For initial selection of the thickness ratio, historical trends can be used. Supercritical airfoils can be chosen 10% thicker. 13 Thickness ratio • In subsonic airplanes, the root airfoil can be 20-60% thicker than the tip airfoil without effecting the drag due to fuselage effects. This thicker root should not extend beyond 30% of span. • This results in a structural weight reduction as well as more volume for fuel and landing gear. • Each airfoil is designed for a certain Reynolds number. Use of an airfoil at greatly different Reynolds numbers produce section characteristics much different than expected. This is especially true for laminar flow airfoils. 14 Wing geometry • The reference or trapezoidal wing is the basic geometry to begin the layout. • The leading edge sweep is important for supersonic flight. In order to reduce drag, it is important to sweep the wing leading edge behind the Mach cone. 15 Wing geometry • The quarter chord sweep is related to the subsonic flight since the lift produced by a wing is proportional to the component of the freestream velocity vector perpendicular to the quarter chord line. 16 Wing geometry • For a complete trapezoidal wing, the aerodynamic center is at the quarter chord point of the mean aerodynamic chord. • In supersonic flow, the aerodynamic center moves approximately back to 40% of the mean aerodynamic chord. 17 Aspect ratio • For finite aspect ratio wing, tip vortices lower the pressure difference between the upper and lower surfaces. This reduces the lift near the wingtip. • The tip vortices reduce the effective angle of attack of the wing, more so at the wingtips. • A high aspect ratio wing has wingtips further apart compared to an equal area wing with low AR. Therefore, the amount of wing effected by the wingtip is less for a high aspect ratio wing and the strength of the wingtip vortex is reduced. loss of lift and induced drag is less for high aspect ratio wing. 𝐿/𝐷)𝑚𝑎𝑥~ 𝐴𝑅 𝑊 𝑤𝑖𝑛𝑔~ 𝐴𝑅 18 Lift to drag ratio • L/D is a measure of overall aerodynamic efficiency. – Subsonic speeds: L/D=L/D(wing span,wetted area) – Supersonic speeds: L/D=L/D(wing span, wetted area, Mach) • Drag components at subsonic speeds: – Induced drag or drag due to lift is a function of the wing span – Parasite drag or zero lift drag is a function of total surface area exposed to air L/D is a function of “wetted aspect ratio”=b2/Swet 19 Wetted aspect ratio 20 Aspect ratio • Due to reduced effective angle of attack of the wingtips, a low AR wing will stall at a higher angle of attack compared to a high aspect ratio wing. • This is why tails have low AR compared to wings. • This ensures adequate control even when the wing stalls. 21 Aspect ratio 22 Wing sweep • Wing sweep is used primarily to reduce the adverse effects of transonic and supersonic flow. • The leading edge sweep must be such that it is behind the Mach cone. • Theoretically, the shock wave formation on a swept wing is determined by the air velocity in a direction perpendicular to the leading edge of the wing. • In the transonic flow regime, wing sweep is determined by the requirement for a high critical Mach number, Mcrit. This requires subsonic airflow over the airfoil measured perpendicular to the leading edge, thus a swept wing. 23 Wing sweep 24 Wing sweep • The exact wing sweep selection depends on the selected airfoil, thickness ratio, taper ratio, etc. 25 Wing sweep • Wing sweep improves lateral stability (roll). A swept wing has a natural dihedral effect 10o sweep ≈ 1o dihedral. • It may be necessary to use zero or negative dihedral on a swept wing in order to avoid a stiff airplane. • The wing sweep and aspect ratio together have a strong effect on the pitch-up characteristics • Pitch-up is a highly undesirable tendency of some aircraft near the stall angle to suddenly and uncontrollably increase the angle of attack. 26 Wing sweep 27 Taper ratio • An elliptical wing will produce the lowest induced drag but is difficult and more costly to produce. • A tapered wing is almost equally efficient in terms of induced drag. 28 Taper ratio •There are two competing considerations: ― Smaller the taper ratio, lighter the wing structure. If λ is less, more lift will be produced at the wingroot center of pressure moves towards the wing root and the moment arm from the wingroot to the center of pressure decreases and the bending moment at the root decreasing the need for heavier structure. ― Wings with low λ show undesirable stall characteristics. Separation at the root has two advantages: Turbulent flow trailing downstream from the root causes buffeting as it flows over the tail, giving a strong stall warning to the pilot. The wingtips have attached flow so the ailerons will be more efficient. 29 Taper ratio • Low sweep wing: typically have taper ratios around 0.4-0.5. • High sweep wings: have taper ratios around 0.2-0.3. • A swept wing will direct the air outward towards the wingtips. • This loads up the wingtips creating more lift there compared to an equivalent unswept wing. • In order to restore the elliptic lift distribution, it is necessary to reduce the taper ratio. 30 Taper ratio 31 Twist • Wing twist is used to prevent tip stall and to revise the lift distribution to approximate an elliptical one. • Typically, wings are twisted between 0o-5o. • Geometric twist is the actual change in airfoil angle of incidence measured with respect to the root airfoil. • A wing with a tip airfoil at a negative angle compared to the root airfoil has «washout». For such a wing, the root will stall before the tip, which improves aileron control at high α and tends to reduce wing rock. • If a wing has linear twist, the twist angle changes in proportion to the distance from the wingroot. 32 Twist • Aerodynamic twist = 𝛼𝐿=0,𝑟𝑜𝑜𝑡−𝛼𝐿=0,𝑡𝑖𝑝 • Optimizing the lift distribution by twisting the wing will be valid only for one geometric angle of attack. • The more twist required to produce an elliptic lift distribution at the design lift coefficient, the worse the wing will perform at other lift coefficients. • For this reason, high amounts of twist (>5o) should be avoided. • Typically, 3o twist provides adequate stall characteristics. 33 Thrust-to-weight ratio and wing loading • Thrust-to-weight ratio (T/W) and wing loading (W/S) are the two most important parameters effecting aircraft performance. Optimization of these parameters forms a major part of conceptual design. • Wing loading and thrust-to-weight ratio are not independent of each other. Takeoff distance, maximum velocity, rate of climb and maximum load factor are dependent on both T/W and W/S. • A good approach would be to guess one parameter and calculate the other to meet various performance characteristics. • Most of the time T/W appears as the guessed parameter because statistical norms are more meaningful and scatter is less among airplanes of a given class. 34 Thrust-to-weight ratio • For propeller-driven airplanes, P/W or W/P (power loading) is a more convenient definition. 𝑇 𝑊= 𝜂𝑝 𝑉 ∞ 𝑃 𝑊= 550𝜂𝑝 𝑉 ∞ ℎ𝑝 𝑊 • For jet-powered airplanes, 𝑇𝑊 𝑜= 𝑎𝑀𝑚𝑎𝑥 𝐶 , • For propeller-powered airplanes, 𝑃𝑊 𝑜= 𝑎𝑉 𝑚𝑎𝑥 𝐶 35 Thrust- and power-to weight ratio 36 Thrust- and power-to weight ratio 37 Thrust-to-weight ratio • T/W directly effects the performance of an airplane. An airplane with a high T/W will: ― Accelerate more quickly, ― Climb more rapidly, ― Reach a higher maximum speed, ― Sustain a higher turn rate, ― Consume more fuel, which will increase the takeoff gross weight. • T/W varies throughout the flight as fuel is consumed. • Engine thrust varies with altitude and velocity. • T/W usually refers to sea-level static thrust (𝑉 ∞= 0), at design takeoff gross weight 𝑊 𝑜and maximum thrust setting. 38 Thrust matching 𝑇 𝑊𝑐𝑟𝑢𝑖𝑠𝑒 = 1 𝐿𝐷𝑚𝑎𝑥 • Weight of the airplane at the beginning of the cruise is the takeoff weight minus the fuel burned during takeoff and climb. • Thrust during cruise is also different from the takeoff value. • For jet aircraft, optimum cruise altitude: 30 000-40 000 ft (best specific fuel consumption), • For jet aircraft, optimum thrust setting: 70-100% of the continuous non-ab thrust. 39 Thrust matching • High by-pass ratio turbofans, optimum thrust=20-25% takeoff thrust. • Low by-pass ratio turbofans, optimum thrust=40-70% takeoff thrust. 40 Thrust matching • Piston-powered airplanes, optimum power setting=75% takeoff power. • Turboprop powered airplanes, optimum power setting=60-80% takeoff power. 41 Wing loading • Wing loading effects: ― Stall speed, ― Climb rate, ― Takeoff and landing distances, ― Maneuvrability, etc. • Wing loading and thrust-to-weight ratio must be optimized together . 42 Wing loading – stall speed 𝑊= 𝐿= 1 2 𝜌∞𝑉 𝑠𝑡𝑎𝑙𝑙 2 𝐶𝐿,𝑚𝑎𝑥𝑆⇒ 𝑊𝑆= 1 2 𝜌∞𝑉 𝑠𝑡𝑎𝑙𝑙 2 𝐶𝐿,𝑚𝑎𝑥 • Maximum lift coefficient depends on: ― Wing geometry, ― Airfoil shape, ― Flap geometry and span, ― Leading edge slat or flap geometry, ― Reynolds number, texture and interference with other components of the airplane. 43 Wing loading – stall speed • During landing, flaps will be deployed to maximum, • During takeoff, they will be partially deployed. 𝐶𝐿,𝑚𝑎𝑥,𝑡𝑜≈0.8𝐶𝐿,𝑚𝑎𝑥,𝑙𝑎𝑛𝑑𝑖𝑛𝑔 • For AR>5, 𝐶𝐿,𝑚𝑎𝑥≈0.9𝑐𝑙,𝑚𝑎𝑥 44 Wing loading – takeoff distance 45 Wing loading – takeoff distance • Ground roll: actual distance travelled before the wheels leave the ground, 𝑉 𝐿𝑂= 1.1𝑉 𝑠𝑡𝑎𝑙𝑙. • Obstacle clearing distance: distance required from brake release until the airplane has reached some specified altitude, hOB=50 ft (military and small civilian airplanes), hOB=35 ft for civilian transport airplanes. • Decision speed: the speed at which the distance to stop after an engine failure exactly equals the distance to continue the takeoff on the remaining engines. • Balanced field length: is the distance required to takeoff and clear the specified obstacle when one engine fails exactly at the decision speed. 46 Wing loading – takeoff distance • Factors effecting takeoff distance: ― T/W and W/S, ― Aerodynamic drag, ― Rolling resistance. • Takeoff parameter: 𝑇𝑂𝑃= 𝑊/𝑆 𝜎𝐶𝐿,𝑡𝑜 𝑇𝑊, jet engines, 𝑇𝑂𝑃= 𝑊/𝑆 𝜎𝐶𝐿,𝑡𝑜 𝑏ℎ𝑝𝑊, propeller engines. • Density ratio, 𝜎= 𝜌 𝜌𝑆𝐿, 𝐶𝐿,𝑡𝑜= 𝐶𝐿,𝑚𝑎𝑥 1.21 𝑉 𝐿𝑂= 1.1𝑉 𝑠𝑡𝑎𝑙𝑙 • Lift coefficient during takeoff may be limited by the maximum taildown angle. 47 Wing loading – takeoff distance • Jet-powered airplanes: 𝑊 𝑆= 𝑇𝑂𝑃𝜎𝐶𝐿,𝑡𝑜𝑇/𝑊 • Propeller-powered airplanes: 𝑊 𝑆= 𝑇𝑂𝑃𝜎𝐶𝐿,𝑡𝑜ℎ𝑝/𝑊 48 Wing loading – landing distance 49 Wing loading – landing distance • Landing ground roll: actual distance the aircraft travels from the time the wheels touch the runway, to the time the aircraft comes to a complete stop. • Landing field length: includes clearing a 50 ft obstacle while the aircraft is still at approach speed. • For military aircraft, 𝑉 𝑎𝑝𝑝= 1.2𝑉 𝑠𝑡𝑎𝑙𝑙, • For civilian aircraft, 𝑉 𝑎𝑝𝑝= 1.3𝑉 𝑠𝑡𝑎𝑙𝑙. 𝑠𝑔= 80 𝑊 𝑆 1 𝜎𝐶𝐿,𝑚𝑎𝑥 , 𝑠= 𝑠𝑔+ 𝑠𝑎. sa =1000 ft (airliners, 3o glideslope) =600 ft (general aviation, power-off approach) 50 Wing loading – landing distance • If the aircraft is equipped with a thrust reverser or reversible pitch propellers, multiply the ground portion of the distance by 0.66. • For commercial aircraft, multiply total landing distance by 1.67 to provide the required safety margin. • For propeller-powered airplanes, Wlanding=0.85Wo, • For jet aircraft, Wlanding=0.85 Wo. • Military requirements, Wlanding=We+Wc+Wp+0.5Wf . 51 Wing loading – cruise • For propeller aircraft, range is maximized when L/D=L/D)max or when induced drag = parasite drag. 𝑞∞𝑆𝐶𝐷𝑜= 𝑞∞𝑆𝐾𝐶𝐿 2 = 𝑞∞𝑆 𝐶𝐿 2 𝜋𝐴𝑅𝑒 e: Oswald span efficiency factor is a function of taper ratio and aspect ratio. 𝑊= 𝐿= 1 2 𝜌∞𝑉 ∞ 2𝐶𝐿𝑆⇒𝐶𝐿= 𝑊/𝑆 𝑞∞ Substituting above yields: 𝑊𝑆= 𝑞∞𝜋𝐴𝑅𝑒𝐶𝐷𝑜 52 Wing loading – cruise • For jet aircraft, range is maximized when L/D=0.866L/D)max or when parasite drag = 3induced drag. 𝑞∞𝑆𝐶𝐷𝑜= 3𝑞∞𝑆𝐾𝐶𝐿 2 = 3𝑞∞𝑆 𝐶𝐿 2 𝜋𝐴𝑅𝑒 e: Oswald span efficiency factor is a function of taper ratio and aspect ratio. 𝑊= 𝐿= 1 2 𝜌∞𝑉 ∞ 2𝐶𝐿𝑆⇒𝐶𝐿= 𝑊/𝑆 𝑞∞ Substituting above yields: 𝑊𝑆= 𝑞∞ 𝜋𝐴𝑅𝑒𝐶𝐷𝑜 3 53 Wing loading – loiter • For propeller aircraft, endurance is maximized when L/D=0.866L/D)max or when induced drag = 3parasite drag. 3𝑞∞𝑆𝐶𝐷𝑜= 𝑞∞𝑆𝐾𝐶𝐿 2 = 𝑞∞𝑆 𝐶𝐿 2 𝜋𝐴𝑅𝑒 e: Oswald span efficiency factor is a function of taper ratio and aspect ratio. 𝑊= 𝐿= 1 2 𝜌∞𝑉 ∞ 2𝐶𝐿𝑆⇒𝐶𝐿= 𝑊/𝑆 𝑞∞ Substituting above yields: 𝑊𝑆= 𝑞∞3𝜋𝐴𝑅𝑒𝐶𝐷𝑜 54 Wing loading – loiter • For jet aircraft, endurance is maximized when L/D=L/D)max or when induced drag = parasite drag. 𝑞∞𝑆𝐶𝐷𝑜= 𝑞∞𝑆𝐾𝐶𝐿 2 = 𝑞∞𝑆 𝐶𝐿 2 𝜋𝐴𝑅𝑒 e: Oswald span efficiency factor is a function of taper ratio and aspect ratio. 𝑊= 𝐿= 1 2 𝜌∞𝑉 ∞ 2𝐶𝐿𝑆⇒𝐶𝐿= 𝑊/𝑆 𝑞∞ Substituting above yields: 𝑊𝑆= 𝑞∞𝜋𝐴𝑅𝑒𝐶𝐷𝑜 55 Oswald span efficiency factor 56 Wing loading - loiter •For initial estimates: • Piston-props: Vloiter=80-120 knots • Jet airplanes: Vloiter=150-200 knots • Turboprops: Vloiter=150-200 knots 57 Estimation of CDo 𝐶𝐷𝑜= 𝑆𝑤𝑒𝑡 𝑆 𝐶 𝑓𝑒 • Cfe: equivalent skin friction coefficient is a function of the Reynolds number, Re. 58 Wing loading – instantaneous turn • An aircraft designed for air-to-air combat (dogfight) must be capable of high turn rate. • An aircraft with a higher turn rate will be able to maneuver behind the other. A turn rate superiority of 2o/s is significant. 59 Wing loading – instantaneous turn • There are two important turn rates: ― Sustained turn: turn rate at which the thrust of the aircraft is just sufficient to maintain velocity and altitude in the turn (T=D); thrust available is the limit. For a level turn: 𝜓= 𝑔𝑛2−1 𝑉 ∞ , 𝑛= 𝐿 𝑊= 1 2 𝜌∞𝑉 ∞ 2 𝑆 𝑊𝐶𝐿 ― Instantaneous turn is limited by the maximum lift, stall or CL,max is the limit. • The speed at which the maximum lift is equal to the allowable structural load factor is the «corner speed» and provides the maximum turn rate for a given altitude. • Modern fighters have a corner speed around 300-350 knots. 60 Wing loading – instantaneous turn 𝑛= 𝜓𝑉 ∞ 𝑔 2 + 1 Solving for W/S: 𝑊𝑆= 1 2 𝜌∞𝑉 ∞ 2 𝐶𝐿,𝑚𝑎𝑥 𝑛 CL,max≈0.6-0.8 for a fighter with a simple trailing edge flap, CL,max≈1.0-1.5 for a fighter with leading and trailing edge flaps. 61 Wing loading – sustained turn • An aircraft will probably not be able to maintain speed and altitude while turning at the maxium instantaneous turn rate. • Sustained turn rate is usually specified in terms of the maximum load factor at a given flight condition that the aircraft can sustain, e.g. 4-5g at M=0.9 at 30000 ft. 𝑇= 𝐷, 𝐿= 𝑛𝑊⇒𝑛= 𝑇 𝑊 𝐿 𝐷 • Load factor in a sustained turn increases when T/W and L/D increases. 62 Wing loading – sustained turn • Equating thrust available and drag yields: 𝑊𝑆= 𝑇/𝑊∓ 𝑇/𝑊2 −4𝑛2𝐶𝐷𝑜𝐾 2𝑛2𝐾/𝑞∞ 𝑇 𝑊≥2𝑛𝐾𝐶𝐷𝑜 63 Wing loading – climb and glide 𝑅 𝐶= 𝑒𝑥𝑐𝑒𝑠𝑠𝑝𝑜𝑤𝑒𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑇−𝐷 𝑊 ⇒𝑅/𝐶 𝑉 ∞ = 𝐺= 𝑇−𝐷 𝑊 ⇒𝐷 𝑊= 𝑇 𝑊−𝐺 𝐷 𝑊= 𝑞∞𝑆𝐶𝐷𝑜+ 𝑞∞𝑆𝐶𝐿 2 𝑊 • Equating the two yields: 𝑊 𝑆= 𝑇𝑊−𝐺∓ 𝑇𝑊−𝐺2 −4𝐾𝐶𝐷𝑂 2𝐾/𝑞∞ 𝑇 𝑊−𝐺≥4𝐾𝐶𝐷𝑜⇒ 𝑇 𝑊≥𝐺+ 4𝐾𝐶𝐷𝑜; T/W must be greater than the climb gradient. 64 Wing loading – climb and glide • Takeoff flap setting, ∆𝐶𝐷𝑜≈0.02, ∆𝑒≈−5% • Landing flap setting, ∆𝐶𝐷𝑜≈0.07, ∆𝑒≈−10% • Landing gear down, ∆𝐶𝐷𝑜≈0.02 • The above equation can also be used to obtain the wing loading corresponding to a glide angle, T/W=0 with a negative G. 65 Wing loading – maximum ceiling • The same equation can be used to find the absolute ceiling (G=0), service ceiling (R/C=100 ft/min) or combat ceiling (R/C=500 ft/min). 66 Wing loading • Remember: ― For the wing loadings estimated above, choose the lowest one to ensure that the wing is large enough for all flight conditions. Convert all the wing loadings calculated to takeoff conditions. ― A low wing loading (large wing) will always increase aircraft weight and cost. ― When W/S is selected, T/W should be rechecked to ensure that all requirements are still met. 67
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https://www.teacherspayteachers.com/browse/elementary/4th-grade/free?search=column%20addition
Column Addition | TPT Log InSign Up Cart is empty Total: $0.00 View Wish ListView Cart Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Adult education Resource type Student practice Independent work packet Worksheets Assessment Graphic organizers Task cards Flash cards Teacher tools Classroom management Teacher manuals Outlines Rubrics Syllabi Unit plans Lessons Activities Games Centers Projects Laboratory Songs Clip art Classroom decor Bulletin board ideas Posters Word walls Printables Seasonal Holiday Black History Month Christmas-Chanukah-Kwanzaa Earth Day Easter Halloween Hispanic Heritage Month Martin Luther King Day Presidents' Day St. Patrick's Day Thanksgiving New Year Valentine's Day Women's History Month Seasonal Autumn Winter Spring Summer Back to school End of year ELA ELA by grade PreK ELA Kindergarten ELA 1st grade ELA 2nd grade ELA 3rd grade ELA 4th grade ELA 5th grade ELA 6th grade ELA 7th grade ELA 8th grade ELA High school ELA Elementary ELA Reading Writing Phonics Vocabulary Grammar Spelling Poetry ELA test prep Middle school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep High school ELA Literature Informational text Writing Creative writing Writing-essays ELA test prep Math Math by grade PreK math Kindergarten math 1st grade math 2nd grade math 3rd grade math 4th grade math 5th grade math 6th grade math 7th grade math 8th grade math High school math Elementary math Basic operations Numbers Geometry Measurement Mental math Place value Arithmetic Fractions Decimals Math test prep Middle school math Algebra Basic operations Decimals Fractions Geometry Math test prep High school math Algebra Algebra 2 Geometry Math test prep Statistics Precalculus Calculus Science Science by grade PreK science Kindergarten science 1st grade science 2nd grade science 3rd grade science 4th grade science 5th grade science 6th grade science 7th grade science 8th grade science High school science By topic Astronomy Biology Chemistry Earth sciences Physics Physical science Social studies Social studies by grade PreK social studies Kindergarten social studies 1st grade social studies 2nd grade social studies 3rd grade social studies 4th grade social studies 5th grade social studies 6th grade social studies 7th grade social studies 8th grade social studies High school social studies Social studies by topic Ancient history Economics European history Government Geography Native Americans Middle ages Psychology U.S. History World history Languages Languages American sign language Arabic Chinese French German Italian Japanese Latin Portuguese Spanish Arts Arts Art history Graphic arts Visual arts Other (arts) Performing arts Dance Drama Instrumental music Music Music composition Vocal music Special education Speech therapy Social emotional Social emotional Character education Classroom community School counseling School psychology Social emotional learning Specialty Specialty Career and technical education Child care Coaching Cooking Health Life skills Occupational therapy Physical education Physical therapy Professional development Service learning Vocational education Other (specialty) Column Addition 370+results Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Sort by: Relevance Relevance Rating Rating Count Price (Ascending) Price (Descending) Most Recent Search 4th grade Subject Supports Free Format All filters (2) Filters 4th grade Free Clear all Grade Elementary Preschool Kindergarten 1st grade 2nd grade 3rd grade 4th grade 5th grade Middle school 6th grade 7th grade 8th grade High school 9th grade 10th grade 11th grade 12th grade Higher education Adult education More Not grade specific Subject Art Art history Graphic arts Visual arts English language arts Balanced literacy Close reading Creative writing Grammar Handwriting Informational text Library skills Literature Phonics & phonological awareness Poetry Reading Reading strategies Spelling Vocabulary Writing Writing-essays Writing-expository Other (ELA) Math Algebra Applied math Arithmetic Basic operations Calculus Decimals Fractions Geometry Graphing Math test prep Measurement Mental math Numbers Order of operations Place value Other (math) Performing arts Instrumental music Music Vocal music Physical education More Science Anatomy Computer science - 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The two-digit addition sums are set out on squared paper to help support children with the layout and lining up the columns correctly. Related products:Addition: Two-digit column addition using base ten (no regrouping) COLORAddition: Two-digit column addition using base ten (no regrouping) B&WPart-Part-Whole Addition and Subtraction Jigsaw 1 st - 5 th Basic Operations, Math, Numbers FREE Rated 4.5 out of 5, based on 2 reviews 4.5(2) Log in to Download Wish List Column or Vertical Addition with Regrouping lesson plans, worksheets and more Created by Save Teachers Sundays These plans worksheets and other teaching resources cover column or vertical addition with regrouping This is a zipped folder containing a PDF version and an editable Word / PowerPoint / Excel of each of the following files: - a more detailed 2 page lesson plan (suitable for trainees) - a less detailed 1 page lesson plan - a 15 slide PowerPoint with calculations to use to model how to do column addition (with squared paper), a list of key things to remember to do and visual explanations of why 2 nd - 4 th Basic Operations, Math FREE Rated 5 out of 5, based on 5 reviews 5.0(5) Log in to Download Wish List Get more with resources under $5 See all Blank MAB Column Addition and Subtraction Worksheets DIY Tutor $1.50 Original Price $1.50 Rated 4.85 out of 5, based on 38 reviews 4.9(38) Column Addition Tic Tac Toe Game: Adding Three 2-digit Numbers The Learning LO-Down $3.00 Original Price $3.00 Rated 5 out of 5, based on 29 reviews 5.0(29) 2- and 3-Digit Addition with Regrouping (ones column only) keepingitcreative $2.50 Original Price $2.50 Rated 5 out of 5, based on 10 reviews 5.0(10) Column Addition With And Without ReGrouping Tarsia Puzzles Saving The Teachers $3.00 Original Price $3.00 Rated 5 out of 5, based on 3 reviews 5.0(3) Column Addition 3-Digit Numbers. (With Regrouping) Created by Sarin Art & Work Addition skills is a huge and important subject. These worksheets are a great way to help your students master 3 digit addition with regrouping and without regrouping. This worksheet includes 150 calculations to solve to build your student‘s confidence. Parents, you can use this resource to support your child's learning in the comfort of your own home. Simply download and print, and use the answer sheet afterwards to quickly check your little one's work. How to use these worksheets:Math Centers 3 rd - 6 th Arithmetic, Math Test Prep, Numbers Also included in:Balancing Equations : Make The Equation Balanced : Single-Double Digit Addition. FREE Log in to Download Wish List Math Addition Worksheets, Adding in columns (no carrying),grade 1 Created by NEWORIKA Welcome to our 2 Digit Addition Worksheets page.Take a look at our double digit addition worksheets to help your child learn and practice their addition skills with regrouping. Using these sheets will help your child to: use column addition to add up two 2-digit numbers together where regrouping of Ones into a Ten is needed;use column addition to add up two 2-digit numbers together where regrouping of Tens into a Hundred is needed;set out a 2 digit column addition;The sheets have been split into K - 6 th Calculus, Math, Other (Math) FREE Log in to Download Wish List COLUMN ADDITION FUN (3 DIGIT) Created by A&A WHIZ KIDS STUDIO COLUMN ADDITION FUN (3 DIGIT) ,Let's download for practice skill. K - 5 th Basic Operations, Numbers, Other (Math) FREE Log in to Download Wish List Math Addition Worksheets, Adding in columns Vertical (no carrying) Created by NEWORIKA Adding Three 2-Digit Numbers in Columns - Vertical Addition Worksheets - No PrepYou'll get 20 worksheets that you can use for short assessments, morning work or warm up exercises. Students will be able to easily practise addition while having fun! It'll be a useful resource during your class that will boost your teaching efficiency! ------------------------------------------------------------------------------------------------------ Thank you for stopping by!You can contact me with questions ab K - 6 th Calculus, Math, Numbers FREE Log in to Download Wish List Math Operations Practice: Column Addition Created by Ready for Tomorrow This worksheet scaffolds the method for column addition, whilst building children up towards independent use of the method. It contains 3 sections. The first has no regrouping. The second has regrouping, with numbers that total less than 100. And the final part has regrouping with numbers that total less than 1000. 2 nd - 4 th Math FREE Rated 5 out of 5, based on 1 reviews 5.0(1) Log in to Download Wish List Turkey Addition | Free | Column Addition Created by Life with Small Fries Use this simple printable near Thanksgiving for practice or assessment of column addition. Answer key included. You may also like my Free Turkey Multiplication printable. 2 nd - 5 th Basic Operations, Math FREE Log in to Download Wish List 5th Grade Everyday Math Algorithm Poster - Column Addition Created by Doug Hinkle Post this step-by-step explanation of Column Addition to help your students remember the steps for this EDM Algorithm. My fifth graders find it to be very beneficial as a reference. 4 th - 6 th Arithmetic, Basic Operations FREE Rated 5 out of 5, based on 2 reviews 5.0(2) Log in to Download Wish List Column addition 3 digits exchanging twice Created by RebeccaTheMathLady Single worksheet: Column addition 3 digits exchanging twice. No answers. This worksheet is one of a series of worksheets for column addition which are all available on this website and which were created to support this video about column addition by RebeccaTheMathsLady. This worksheet was developed by Jeff Kutcher from Jeff's Notebooks and he is kindly allowing it to be freeshared here. Thank you Jeff! There is a link from the worksheet to another resource on Jeff's TPT site that will 2 nd - 4 th Basic Operations, Math CCSS 2.NBT.B.6 , 2.NBT.B.7 , 3.NBT.A.3 +1 FREE Log in to Download Wish List Column addition 3 digits no exchanging Created by RebeccaTheMathLady Single worksheet: Column addition 3 digits no exchanging. No answers. This worksheet is one of a series of worksheets for column addition which are all available on this website and which were created to support this video about column addition by RebeccaTheMathsLady. This worksheet was developed by Jeff Kutcher from Jeff's Notebooks and he is kindly allowing it to be freeshared here. Thank you Jeff! There is a link from the worksheet to another resource on Jeff's TPT site that will gen 2 nd - 4 th Basic Operations, Math CCSS 2.NBT.B.6 , 2.NBT.B.7 , 3.NBT.A.2 +1 FREE Log in to Download Wish List Column addition 3 digits exchanging once Created by RebeccaTheMathLady Single worksheet: Column addition 3 digits exchanging once. No answers. This worksheet is one of a series of worksheets for column addition which are all available on this website and which were created to support this video about column addition by RebeccaTheMathsLady. This worksheet was developed by Jeff Kutcher from Jeff's Notebooks and he is kindly allowing it to be freeshared here. Thank you Jeff! There is a link from the worksheet to another resource on Jeff's TPT site that will g 2 nd - 4 th Basic Operations, Math CCSS 2.NBT.B.6 , 2.NBT.B.7 , 3.NBT.A.2 +1 FREE Log in to Download Wish List Column addition 2 digits exchanging once Created by RebeccaTheMathLady Single worksheet: Column addition 2 digits exchanging once. No answers. This worksheet is one of a series of worksheets for column addition which are all available on this website and which were created to support this video about column addition by RebeccaTheMathsLady. This worksheet was developed by Jeff Kutcher from Jeff's Notebooks and he is kindly allowing it to be freeshared here. Thank you Jeff! There is a link from the worksheet to another resource on Jeff's TPT site that will g 2 nd - 4 th Basic Operations, Math CCSS 2.NBT.B.6 , 2.NBT.B.7 , 3.NBT.A.2 +1 FREE Log in to Download Wish List Column Addition - Find Someone Who cooperative learning activity Created by What A Lark Teaching Students walk around the room trying to find a person who can complete each column addition calculation. Students enjoy the mobility and sociability of the strategy and allows the students to peer check the answers. 1 st - 4 th Arithmetic, Basic Operations, Math CCSS 1.NBT.C.4 , 2.NBT.B.5 , 2.NBT.B.6 +4 FREE Log in to Download Wish List Column addition 3 digits exchanging three times Created by RebeccaTheMathLady This worksheet is one of a series of worksheets for column addition which are all available on this website and which were created to support this video about column addition by RebeccaTheMathsLady. This worksheet was developed by Jeff Kutcher from Jeff's Notebooks and he is kindly allowing it to be freeshared here. Thank you Jeff! There is a link from the worksheet to another resource on Jeff's TPT site that will generate new worksheets on this topic (with and without answer sheets) but 2 nd - 4 th Basic Operations, Math CCSS 2.NBT.B.6 , 2.NBT.B.7 , 3.NBT.A.2 +1 FREE Log in to Download Wish List Addition of four digit numbers in columns freebie worksheet Created by Tigerlearn A simple format, 15 question, worksheet for students to practise their addition in columns by adding 4 digit numbers together. Please leave a review if you download this free resource - it really helps my store! 2 nd - 7 th Math FREE Log in to Download Wish List Column addition 2 digits no exchanging Created by RebeccaTheMathLady This worksheet is one of a series of worksheets for column addition which are all available on this website and which were created to support this video about column addition by RebeccaTheMathsLady. This worksheet was developed by Jeff Kutcher from Jeff's Notebooks and he is kindly allowing it to be freeshared here. Thank you Jeff! There is a link from the worksheet to another resource on Jeff's TPT site that will generate new worksheets on this topic (with and without answer sheets) but 2 nd - 4 th Basic Operations, Math CCSS 2.NBT.B.6 , 2.NBT.B.7 , 3.NBT.A.2 +1 FREE Log in to Download Wish List Column Addition Worksheet Created by Made by Matt A quick column addition worksheet to help your learners practice their skills. 3 rd - 6 th Basic Operations, Math, Numbers FREE Log in to Download Wish List Column addition 2 digits exchanging twice Created by RebeccaTheMathLady Single worksheet: Column addition 2 digits exchanging twice. No answers. This worksheet is one of a series of worksheets for column addition which are all available on this website and which were created to support this video about column addition by RebeccaTheMathsLady. This worksheet was developed by Jeff Kutcher from Jeff's Notebooks and he is kindly allowing it to be freeshared here. Thank you Jeff! There is a link from the worksheet to another resource on Jeff's TPT site that will 2 nd - 4 th Basic Operations, Math CCSS 2.NBT.B.6 , 2.NBT.B.7 , 3.NBT.A.2 +1 FREE Log in to Download Wish List Column addition with carrying/regrouping 2 Digits Created by Amy Donnelly Grade 2-3 Year 3-4 P5 Column addition (2 digit) Editable Ideal for using manipulatives/concrete materials 2 nd - 4 th Math FREE Log in to Download Wish List Column Addition Practice Created by Jacqueline Clinton Enjoy this worksheet to have your students practice column addition! 2 nd - 4 th Basic Operations, Math CCSS MP1 FREE Log in to Download Wish List Column Addition and Subtraction - Reasoning and Skill Drill Created by Lauren Grant Differentiated worksheets (secure and working above secure). A skill drill to begin with followed by reasoning style questions including missing number questions. 2 nd - 5 th Arithmetic, Basic Operations, Numbers FREE Log in to Download Wish List Column addition Created by Roberts Part One of my column addition worksheets Please see Dr Doom's addition for all my work. 3 rd - 5 th Basic Operations, Math FREE Log in to Download Wish List Two digit addition/subtraction with three addends template Created by Muinteoir A template for two digit addition or subtraction with three addends. Purchase the template and then you can insert the numbers into the grids. This page includes a tens and ones column as well as a carrying box. This template will help students when they are learning about adding multiple numbers with regrouping. 2 nd - 4 th Arithmetic, Basic Operations, Math CCSS 2.NBT.B.6 FREE Rated 4.77 out of 5, based on 47 reviews 4.8(47) Log in to Download Wish List 1 2 3 4 5 Showing 1-24 of 370+results TPT is the largest marketplace for PreK-12 resources, powered by a community of educators. Facebook Instagram Pinterest Twitter About Who we are We're hiring Press Blog Gift Cards Support Help & FAQ Security Privacy policy Student privacy Terms of service Tell us what you think Updates Get our weekly newsletter with free resources, updates, and special offers. 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8474
https://pressbooks.bccampus.ca/chbe220/chapter/pressure-definition-absolute-gauge-pressure/
Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 22 Pressure Definition; Absolute & Gauge Pressure Learning Objectives By the end of this section, you should be able to: Calculate pressure using its definition Understand the difference between absolute & gauge pressure Physical Properties Physical properties are properties that can be measured without changing the molecular structure of the substance. The three main physical properties we will introduce in this lecture are pressure, temperature, and volume: Pressure – the amount of force exerted per area in a system Temperature – a measure of the average kinetic energy of a system Volume – the space occupied by a system To see more about how these are related for gases, there is an interesting gas property simulator available online: Pressure [latex]^{}[/latex] When a gas molecule elastically collides with the wall of a container, it exerts a force on the wall. These forces are the source of pressure in a gas. In a sample of gas in a container, the randomness of the molecular motion causes the number of collisions to fluctuate in a given time. However, because a huge number of molecules collide with the wall in a short time, the number of collisions on the scales of time and space we measure fluctuates by only a tiny, usually unobservable fraction from the average. As the number of molecules increases, the number of collisions, and thus the pressure, increases. If the average velocity of the molecules is higher, each collision exerts a larger force on the wall, therefore the gas pressure is higher. Image obtained from OpenStax University Physics Volume 2 / CC BY 4.0 Calculation and Units: [latex]Pressure (P)=\frac{force}{area}[/latex] [latex]\text{units for } P= \frac{ML}{t^2}L^{-2}=\frac{M}{LT^2}[/latex] General notation: M – any units of mass (g, kg, lbs…) L – any units of length (m, cm, in…) t – any units of time (s, min, hr…) Common units for pressure: [S.I.] [latex]\frac{kg}{ms^2}=\frac{N}{m^2}=Pa[/latex] (i.e. Pascal) [cgs] [latex]\frac{g}{cms^2} = \frac{dyne}{cm^2}[/latex] [American] [latex]\frac{lbm}{fts^2}[/latex] Other common pressure units: [latex]1k!Pa (kilo-Pascal) = 10^3 Pa[/latex] [latex]1M!Pa (mega-Pascal) = 10^6 Pa[/latex] [latex]1bar= 10^5 Pa = 10^5 N/m^2[/latex] [latex]1 atm = 1.013×10^5 Pa = 760 mmHg[/latex] (millimetres of mercury column at 0 °C) = [latex]14.7 psi[/latex] (i.e. [latex]lbf/in^2[/latex] pound force per square inch) Exercise: Pressure Calculation What is the force exerted from a column of air on us in mass per square metre? Take atmosphere pressure = [latex]1.013×10^5 Pa[/latex] Solution We can use Newton’s second law [latex]F=ma[/latex] to find mass, with [latex]a[/latex] being the acceleration of gravity. Since the mass we are looking for is “per square metre”, we can also divide the force by area(in [latex]m^2[/latex]), which gives us pressure on the left side of the equation. [latex]mass\;(per\;m^2)=\frac{P}{a}=\frac{1.013×10^5Pa}{9.81\frac{m}{s^2}}=\frac{1.013×10^5\frac{kg}{ms^2}}{9.81\frac{m}{s^2}}=10326\frac{kg}{m^2}[/latex] Measuring Pressure (Manometers) [latex]^{}[/latex] A manometer can be used to determine gas pressures. The manometer is essentially a U-shaped tube containing some kind of fluid with known density, and one side is connected to the region of interest while the reference pressure is applied to the other. The difference in liquid level represents the applied pressure. Manometer fluid – mercury was used originally (hence mmHg), but we now have a wide variety of fluids of various densities. Manometer types: open-ended manometer: [latex]P_{ref} = P_{atm}[/latex] sealed-end manometer: [latex]P_{ref} = \text{whatever it is set to (generally ~0, vacuum)}[/latex] Right side image obtained from Apply Science Concepts to Trades Applications / CC BY 4.0 The difference in heights of the liquid columns correlates to the difference in the pressure of the gases. The pressure of the gas in the vessel is calculated by: [latex]P=P_{ref}+\rho g(h_{ref}-h)[/latex] Exercise: Manometer Calculation Compressed air has been used to transmit and store power since the 1870s. Cities such as Paris, Dresden and Buenos Aires had compressed air lines to transmit power to homes and businesses. The technology is now being applied to energy storage in former salt mines underground. One such plant in McIntosh, Alabama stores gas at 1100 psi. If a sealed end manometer with a fluid with a density of 3450 [latex]\frac{𝑘𝑔}{𝑚^3}[/latex] was used to measure this pressure, what would the height of the fluid column be? (take [latex]g=9.8m/s^2[/latex]) Solution Manipulate h from the equation: [latex]P=P_{ref}+\rho g(h_{ref}-h)[/latex] \begin{align}h& = \frac{P-P_{0}}{\rho g}\h& = \frac{1100psi×\frac{101325Pa}{14.696psi}}{3450\frac{kg}{m^3}×9.8\frac{m}{s^2}}\&=\frac{7.58×10^6Pa}{3.381×10^4\frac{kg}{m^2s^2}} \& = 2.242×10^2 \frac{Pa}{\frac{kg}{m^2s^2}}×\frac{1\frac{kg}{ms^2}}{1Pa}\& = 2.24×10^2m\end{align} height = [latex]2.2410^2m[/latex] or 224 m, very high pressure, likely difficult to measure with a manometer. Gauge vs. Absolute Pressure The absolute pressure is the actual pressure at the point of interest. The absolute pressure is 0 in a vacuum and cannot be negative. Gauge pressure is defined to be the difference between absolute pressure and atmospheric pressure: [latex]gauge\; pressure\; =\; absolute\; pressure\; – \;atmospheric\; pressure[/latex] when gauge pressure is 0, absolute pressure = atmospheric pressure Most measuring devices measure gauge pressure. Many measuring devices use a flexible membrane between a chamber of known pressure (the outside chamber, which is connected to the atmosphere) and a vessel with the pressure we want to measure. If there is a difference in pressure on both sides of the membrane, the membrane will expand or contract. Through this, we can relate the membrane’s position to the gauge pressure of the vessel. The units often used on the devices include: [latex]psi[/latex] = pounds per square inch [latex]inH!g[/latex] = inches of mercury Exercise: Reading Absolute and Gauge Pressures If the gauge in the image reads 5 psi at the orange mark and the atmospheric pressure is 14.7 psi, which of the following are the gauge and absolute pressures? A – gauge = -5 psi, absolute = 9.7 psi B – gauge = 5 psi, absolute = 9.7 psi C – gauge = -5 psi, absolute = 19.7 psi D – gauge = 5 psi, absolute = 19.7 psi If the gauge in the image reads 10 inHg at the purple mark and the atmospheric pressure is 29.9 inHg, which of the following are the gauge and absolute pressures? A – gauge = -10 inHg, absolute = 39.9 inHg B – gauge = 10 inHg, absolute = 19.9 inHg C – gauge = -10 inHg, absolute = 19.9 inHg D – gauge = 10 inHg, absolute = 39.9 inHg Solution 1. Answer: D The arrow is on the “pressure” side, which indicates that the gauge pressure is positive. The absolute pressure is the sum of absolute pressure and gauge pressure.\begin{align}absolute\;pressure& =gauge\;pressure+atmospheric\;pressure\& = 5psi+14.7psi\&=19.7 psi\end{align} 2. Answer: C The arrow is on the “vacuum” side, which indicates that the gauge pressure is negative. The absolute pressure is the sum of absolute pressure and gauge pressure.\begin{align}absolute\;pressure& =gauge\;pressure+atmospheric\;pressure\& =-10inH!g+29.9inH!g\&=19.9inH!g\end{align} References OpenStax University Physics Volume 2. 2016. 2.2 Pressure, Temperature, and RMS Speed. [online] [Accessed 11 May 2020]. Line D – Organizational Skills Competency D-2: Apply Science Concepts to Trades Applications. 2015. Manometer. [online] [Accessed 11 May 2020]. License Foundations of Chemical and Biological Engineering I Copyright © 2020 by Jonathan Verrett is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License, except where otherwise noted. 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8475
https://fiveable.me/key-terms/college-bio/polycistronic-mrna
Polycistronic mRNA - (General Biology I) - Vocab, Definition, Explanations | Fiveable | Fiveable new!Printable guides for educators Printable guides for educators. Bring Fiveable to your classroom ap study content toolsprintablespricing my subjectsupgrade All Key Terms General Biology I Polycistronic mRNA 🔬general biology i review key term - Polycistronic mRNA Citation: MLA Definition Polycistronic mRNA is a type of messenger RNA that carries the genetic information for multiple genes and can be translated into several different proteins. This feature is particularly common in prokaryotes, where the arrangement of genes within operons allows for coordinated expression of related functions. The ability to produce multiple proteins from a single mRNA molecule is efficient for prokaryotic cells, as it facilitates simultaneous regulation and expression of genes involved in similar biological pathways. 5 Must Know Facts For Your Next Test In prokaryotes, polycistronic mRNA is often produced from operons, which allows related genes to be expressed together in response to environmental changes. Each coding region within a polycistronic mRNA has its own ribosome binding site, which facilitates the translation of each protein independently. Polycistronic mRNA plays a critical role in metabolic pathways by enabling the efficient production of multiple enzymes that are needed simultaneously. While polycistronic mRNA is characteristic of prokaryotes, some examples exist in certain eukaryotic systems, but they are relatively rare. The presence of polycistronic mRNA allows prokaryotic cells to respond rapidly to changes in their environment by quickly adjusting the expression of multiple genes. Review Questions How does polycistronic mRNA contribute to the efficiency of gene regulation in prokaryotic cells? Polycistronic mRNA enhances gene regulation efficiency in prokaryotic cells by allowing multiple genes to be transcribed and translated simultaneously from a single mRNA molecule. This arrangement facilitates coordinated expression of functionally related proteins, which is crucial when cells need to respond quickly to environmental changes. By using operons to cluster related genes together, prokaryotic cells can effectively manage their metabolic needs with fewer regulatory steps. Compare and contrast polycistronic and monocistronic mRNA regarding their roles in gene expression across different organisms. Polycistronic mRNA typically found in prokaryotes allows for the simultaneous expression of multiple genes, which is efficient for coordinating responses to environmental conditions. In contrast, monocistronic mRNA, predominant in eukaryotes, encodes only a single protein per transcript. This difference means that eukaryotic gene regulation often involves more complex mechanisms, including processing and splicing events that do not occur with polycistronic mRNA in prokaryotes. Evaluate the impact of polycistronic mRNA on metabolic regulation in prokaryotic organisms and its implications for biotechnology. Polycistronic mRNA significantly impacts metabolic regulation in prokaryotic organisms by enabling the simultaneous production of enzymes necessary for specific metabolic pathways. This ability allows bacteria to adapt quickly to changing environments or nutrient availability. In biotechnology, harnessing the principles of polycistronic gene expression can enhance recombinant protein production and metabolic engineering strategies, making it possible to design microbial systems that efficiently produce valuable bioproducts or biofuels. Related terms Operon: A cluster of genes under the control of a single promoter, allowing for the coordinated regulation of gene expression in prokaryotes. Monocistronic mRNA:A type of messenger RNA that encodes only a single protein, typical in eukaryotic organisms. Transcription:The process by which genetic information from DNA is copied into RNA, serving as the first step in gene expression. 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8476
https://www.retinagroupflorida.com/retina-conditions/metamorphopsia
Metamorphopsia | Retina Group of Florida™ Skip To Main NavigationSkip To ContentSkip To Footer New Vitreoretinal Surgery Fellowship Program is now available at Retina Group of Florida. Learn more. For Patients For Doctors Careers Language: en Select your language English Español Arabic Chinese Creole French Hebrew Italian Portuguese Russian Bill Pay Locations Contact Us East Coast(954) 776-6880 Gulf Coast(727) 862-3090 Naples(239) 325-3970 Treasure Coast(561) 784-3788 West Coast(941) 924-0303 Contact For Patients Contact For Doctors Appointments About Us Practice Overview Our Physicians Our Team Patient Reviews Careers Insurance Locations The RGF Blog Our Physicians Meet the Physicians Find a Doctor East Coast Physicians Gulf Coast Physicians Naples Physician Treasure Coast Physicians West Coast Physicians Conditions Our Conditions Macular Degeneration / AMD Diabetic Retinopathy Flashes & Floaters Retinal Tears and Detachments Macular Hole & Macular Pucker Artery & Vein Occlusions Uveitis Other Conditions Cataracts Central Serous Retinopathy Cystoid Macular Edema Endophthalmitis HIV Infection and the Eye Intraocular Tumors Low Vision Metamorphopsia Ocular Migraine Ocular Toxoplasmosis Retinal Vascular Disease Vitreous Hemorrhage Treatments Diagnostic Tools Retina Surgery Scleral Buckle Vitrectomy Eye Injections Intravitreal Injections Avastin (Bevacizumab) Eylea (Aflibercept) Lucentis (Ranibizumab) Retinal Implant Surgery Iluvien (Fluocinolone Acetonide Implant) Laser Treatments Laser Treatment (Photocoagulation) Photodynamic Laser Treatment (PDT) Research Research at RGF Current Clinical Trials Presentations & Publications Resources Your First Visit Request an Appointment Patient Portal Insurance LocationsCareersBill PaySearch We speak Español and English. Language: en Select your language English Español Arabic Chinese Creole French Hebrew Italian Portuguese Russian (954) 776-6880East Coast (727) 862-3090Gulf Coast (239) 325-3970Naples (561) 784-3788Treasure Coast (941) 924-0303West Coast For PatientsContact For DoctorsContact Our Conditions Home Our Conditions Metamorphopsia Metamorphopsia Metamorphopsia is a symptom that occurs in many macular and retinal conditions, including age-related macular degeneration (AMD), macular puckers, macular edema, and retinal detachment. With metamorphopsia, straight lines and linear objects appear wavy, distorted, or rounded. In some cases, faces can even appear distorted. Metamorphopsia can also include micropsia and macropsia, which is when objects appear smaller or larger, respectively. Metamorphopsia is commonly diagnosed using an Amsler grid. Amsler grids can also be used at home to help patients monitor their vision health, detect any visual disruptions, and report their symptoms to their ophthalmologist. Our Conditions Our Conditions Macular Degeneration / AMD Diabetic Retinopathy Flashes & Floaters Retinal Tears and Detachments Macular Hole & Macular Pucker Artery & Vein Occlusions Uveitis Other Conditions Cataracts Central Serous Retinopathy Cystoid Macular Edema Cytomegalovirus Retinitis Endophthalmitis HIV Infection and the Eye Intraocular Tumors Low Vision Metamorphopsia Ocular Migraine Ocular Toxoplasmosis Retinal Vascular Disease Vitreous Hemorrhage We speak Español and English. 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Venice Ave. Suite #201 Venice, FL 34285 Information • Directions Phone:941-412-0303 Fax:941-412-0309 Naples NaplesMore Info Personalized Retina Care of Naples 3467 Pine Ridge Rd, 103 Naples, FL 34109 Information • Directions Phone:239-325-3970 Fax:239-325-3971 Physicians Scott R. Anagnoste, M.D. Mario del Cid, M.D. Mandeep Singh Dhalla, M.D. Matthew Donovan, M.D. Darin R. Goldman, M.D. Larry S. Halperin, M.D. Kevin T. Kelly, M.D. Robert A. Lalane III, M.D. Dan P. Montzka, M.D. John Niffenegger, M.D. Renata Picciani, M.D. Elizabeth Richter, M.D., Ph.D. Linnet Rodriguez, M.D. Ryan Shields, M.D. Gary Shienbaum, M.D. George Skopis, M.D. Katia E. Taba, M.D. Barry S. Taney, M.D. W. Scott Thompson, M.D. Eduardo Uchiyama, M.D. Thomas G. Ward, D.O. Keye Wong, M.D. Jie Sun Conditions Macular Degeneration / AMD Diabetic Retinopathy Flashes & Floaters Retinal Tears and Detachments Macular Hole & Macular Pucker Artery & Vein Occlusions Uveitis Other Conditions Cataracts Central Serous Retinopathy Cystoid Macular Edema Cytomegalovirus Retinitis Endophthalmitis HIV Infection and the Eye Intraocular Tumors Low Vision Metamorphopsia Ocular Migraine Ocular Toxoplasmosis Retinal Vascular Disease Vitreous Hemorrhage About Us Our Physicians Conditions Treatments Research Resources About Us Practice Overview Our Physicians Our Team Patient Reviews Careers Insurance Locations The RGF Blog Our Physicians Meet the Physicians Find a Doctor East Coast Physicians Gulf Coast Physicians Naples Physician Treasure Coast Physicians West Coast Physicians Conditions Our Conditions Macular Degeneration / AMD Diabetic Retinopathy Flashes & Floaters Retinal Tears and Detachments Macular Hole & Macular Pucker Artery & Vein Occlusions Uveitis Other Conditions Cataracts Central Serous Retinopathy Cystoid Macular Edema Endophthalmitis HIV Infection and the Eye Intraocular Tumors Low Vision Metamorphopsia Ocular Migraine Ocular Toxoplasmosis Retinal Vascular Disease Vitreous Hemorrhage Treatments Diagnostic Tools Retina Surgery Scleral Buckle Vitrectomy Eye Injections Intravitreal Injections Avastin (Bevacizumab) Eylea (Aflibercept) Lucentis (Ranibizumab) Retinal Implant Surgery Iluvien (Fluocinolone Acetonide Implant) Laser Treatments Laser Treatment (Photocoagulation) Photodynamic Laser Treatment (PDT) Research Research at RGF Current Clinical Trials Presentations & Publications Resources Your First Visit Request an Appointment Patient Portal Insurance Retina Group of Florida Retina Group of Florida is a leading retina-only ophthalmology practice that has been providing state-of-the-art retinal, macular, and vitreoretinal care for over 40 years. With 23 convenient locations throughout Broward County, Palm Beach County, Pinellas County, and the Gulf Coast area, including Boca Raton, Boynton Beach, Brooksville, Clearwater, Delray Beach, Fort Pierce, Fort Lauderdale, Hollywood, Hollywood - West, Inverness, Jupiter, Lakewood Ranch, Margate, Miramar, Naples, Plantation, Port Charlotte, Port St. Lucie, Sarasota, Stuart, Trinity, Venice, and Wellington. We're easily accessible to patients across all of Southeast and Southwest Florida and beyond. Our retina specialists and vitreoretinal surgeons have been recognized as some of the top ophthalmologists in the country by prominent publications such as Castle Connolly and the New York Times. Using the most sophisticated equipment and techniques, we diagnose and treat the full spectrum of vitreoretinal conditions, including dry and wet age-related macular degeneration (AMD), diabetic retinopathy, retinal tears, retinal detachment, macular holes, macular puckers (epiretinal membranes), and more. We are also actively involved in retina clinical trials and operate a fully staffed retina research center so that we can provide our patients with the most advanced retinal care available. Follow RGF © 2025 Retina Group of Florida. All rights reserved. Privacy PolicyAccessibilitySitemapDo Not Sell My Personal InformationDesign by IV Interactive Welcome Dr. Jie Sun to RGF We’re happy to welcome Dr. Jie Sun to the RGF team! Dr. Jie is an accomplished retina specialist with over two decades of experience in clinical care, academic research, and healthcare administration. She will be seeing patients in our Sarasota, LWR, Venice, and Port Charlotte locations. Request an appointment today! Get to know Dr. Sun Welcome Dr. Jie Sun to RGF We’re happy to welcome Dr. Jie Sun to the RGF team! Dr. Jie is an accomplished retina specialist with over two decades of experience in clinical care, academic research, and healthcare administration. She will be seeing patients in our Sarasota, LWR, Venice, and Port Charlotte locations. Request an appointment today! Get to know Dr. Sun This website uses cookies to enhance user experience and to analyze performance and traffic on our website. We also share information about your use of our site with our social media, advertising and analytics partners. Accept Cookies Do Not Sell My Personal Information Do Not Sell My Personal Information When you visit our website, we store cookies on your browser to collect information. The information collected might relate to you, your preferences or your device, and is mostly used to make the site work as you expect it to and to provide a more personalized web experience. 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8477
https://www.reddit.com/r/learnmath/comments/189bd7n/does_the_range_of_a_function_take_into_account/
Does the range of a function take into account domain restrictions? : r/learnmath Skip to main contentDoes the range of a function take into account domain restrictions? : r/learnmath Open menu Open navigationGo to Reddit Home r/learnmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to learnmath r/learnmath r/learnmath Post all of your math-learning resources here. Questions, no matter how basic, will be answered (to the best ability of the online subscribers). 398K Members Online •2 yr. ago Thefunkycow21 Does the range of a function take into account domain restrictions? When determining the range of a function, should we take into account the restrictions placed on the domain? Or do we simply state the range for the function assuming any input is possible. For example, the function f(x) = −2log_3(𝑥^2 − 5) + 1 has the domain x < -sqrt(5), x > sqrt(5). Should the range then be y < f(sqrt(5)), or would it just be y as any real number. Thanks in advance for the help. Read more Share Related Answers Section Related Answers Best resources for learning calculus online Tips for mastering algebra concepts quickly How to approach geometry proofs effectively Understanding the fundamentals of statistics Effective study techniques for math exams New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community Top Posts Reddit reReddit: Top posts of December 2, 2023 Reddit reReddit: Top posts of December 2023 Reddit reReddit: Top posts of 2023 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
8478
https://link.springer.com/chapter/10.1007/978-1-4471-1915-9_5
Advertisement Heat Conduction in a Cylinder 184 Accesses Abstract The equation of heat conduction in cylindrical coordinates (r, θ, z) is expressed as follows: The case where temperature depends only on r, z and t is considered. There is thus a symmetry of revolution about the Oz axis. The first equation becomes where r is the radial abscissa. This is a preview of subscription content, log in via an institution to check access. Access this chapter Institutional subscriptions Preview Unable to display preview. Download preview PDF. Unable to display preview. Download preview PDF. Explore related subjects References Wilson AH. Phil Mag 1984; 39: 48 Google Scholar Carslaw HS, Jaeger JC. Conduction of heat in solids, 2nd edn, Clarendon Press, Oxford, 1978, p 207 Google Scholar Adda Y, Philibert J. La diffusion dans les solides, Tome I, Presses Univ. de France, Paris, 1966, p 166 Google Scholar Download references Author information Authors and Affiliations Laboratoire de Chimie des Matériaux et Chimie Industrielle, Faculté des Sciences et Techniques, Université de Saint-Etienne, 23 rue du Docteur Paul Michelon, 42023, Saint-Etienne Cédex 2, France J.-W. Vergnaud PhD & J. Bouzon PhD Search author on:PubMed Google Scholar Search author on:PubMed Google Scholar Rights and permissions Reprints and permissions Copyright information © 1992 Springer-Verlag London Limited About this chapter Cite this chapter Vergnaud, JW., Bouzon, J. (1992). Heat Conduction in a Cylinder. In: Cure of Thermosetting Resins. Springer, London. Download citation DOI: Publisher Name: Springer, London Print ISBN: 978-1-4471-1917-3 Online ISBN: 978-1-4471-1915-9 eBook Packages: Springer Book Archive Share this chapter Anyone you share the following link with will be able to read this content: Sorry, a shareable link is not currently available for this article. Provided by the Springer Nature SharedIt content-sharing initiative Keywords These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves. Publish with us Policies and ethics Access this chapter Institutional subscriptions Search Navigation Discover content Publish with us Products and services Our brands 13.223.188.199 Not affiliated © 2025 Springer Nature
8479
https://wordsinasentence.com/anomie-in-a-sentence/
Anomie in a Sentence 🔉 Definition of Anomie social instability caused by erosion of standards and values Examples of Anomie in a sentence Carl claims that the children of gangsters are prone to anomie because they were never raised with a sense of right and wrong. 🔉 As society’s standards weaken and people are affected by anomie, natural societal bonds that we take for granted begin to corrode. 🔉 When Josh moved to the high-rise section of his city, he began to feel that the locals were more effected by anomie and alienation than the suburbanites. 🔉 Growing up in a society devoid of moral guidance, Corey's feelings of anomie led him to join a dangerous gang. 🔉 Trish’s feelings of anomie began to develop within her as she watched the society that she grew up in move away from their established norms and traditions. 🔉 Other words in the Seclusion category WATCH our daily vocabulary videos and LEARN new words in a fun and exciting way! SUBSCRIBE to our YouTube channel to keep video production going! Visit VocabularyVideos.com to watch our FULL library of videos.
8480
https://wumbo.net/examples/derive-trigonometric-identities/
Derive the Trigonometric Identities This example shows how to derive the trigonometric identities using algebra and the right triangle definitions of the trigonometric functions. The identities can also be derived using the unit circle or the complex plane. The identities that this example derives are summarized below: Derive Pythagorean Identity Derive Sum of Two Angles Identities Derive Difference of Two Angles Identities Derive Double Angle Identities (Algebra) Derive Half Angle Identities (Algebra) Derive Pythagorean Identity To derive the Pythagorean identity the lengths of the adjacent and opposite sides of the right triangle are defined in terms of the cosine and sine of the angle of the right triangle. Then, the lengths are substituted into Pythagorean’s Theorem. Let’s start with the triangle formed by the unit circle, which visualizes all right triangles of hypotenuse one. Note, that the adjacent side corresponds to the x-component of the right triangle and the opposite side corresponds to the y-component of the right triangle. The two components form the (x,y) point along the circumference of the circle. Then, using the definitions of the trigonometric functions cosine and sine, we can substitute the variables which represent the adjacent side (x), the opposite side (y), and hypotenuse (1) of the right triangle into the equations. cos(θ)=hypotenuseadjacent​=1x​=x sin(θ)=hypotenuseopposite​=1y​=y After simplifying the equations, the adjacent side corresponds directly with the cosine function and the opposite side corresponds with the sine function for a given angle. 2. Next, recall the equation for Pythagorean’s Theorem which relates the squares of the sides together as shown below: c2=a2+b2 After substituting the corresponding variables to convert the theorem into the Cartesian Coordinate System we are left with a familiar equation, the equation of a circle. 12=x2+y2 Then, by substituting the corresponding sine and cosine function above, which we found to correspond to the x and y components of the triangle, we get Pythagorean’s identity. 1=cos2(θ)+sin2(θ) Derive Sum of Two Angles Identities (Right Triangle) This example derives the sum of two angles identities using the right triangle definitions of the functions sine and cosine. The right triangle definitions of sine and cosine are shown below. sin(θ)= hypotenuse opposite ​ cos(θ)= hypotenuse adjacent ​ Steps Start by drawing a right triangle with an angle α+β and hypotenuse of 1 as shown below. The geometry of this triangle will be used to derive the identities. Solve for the lengths of the adjacent and opposite sides by substituting AB, BC and AC=1 into the definitions of sine and cosine. BC=sin(α+β) AB=cos(α+β) Label these lengths in the figure. Draw a line parallel to AB and use the corresponding angle theorem to label the corresponding angles α and β. Draw a right triangle with the angle β starting at the point C that shares the hypotenuse AC. Then draw two more right triangles that complete the rectangular shape. The geometry of this shape can be used to represent the lengths of the adjacent and opposite sides of the original right triangle. BC=DE+EF AB=DC−FA Substitute the sine and cosine of the angle α+β from above. This gives us the general form of the identities, next we will find the unknown lengths. sin(α+β)=DE+EF cos(α+β)=DC−FA 2. Find the adjacent and opposite lengths of the right triangle △CEA. Substitute the adjacent side CE, opposite side EA and hypotenuse AC=1 into the definitions of sine and cosine and solve for the adjacent and opposite sides. CE=cos(β) EA=sin(β) Label these lengths in the figure. 3. Find the adjacent and opposite lengths of the right triangle △CDE. Substitute the adjacent side CD, opposite side DE and hypotenuse CE=cos(β) into the definitions of sine and cosine and solve for the adjacent and opposite sides. CD=cos(α)cos(β) DE=sin(α)cos(β) Label these lengths in the figure. 4. Observe that the angle ∠AEF is equal to α, because it is complementary to the angle ∠CED which is complementary to α. Find the adjacent and opposite lengths of the right triangle △EFA. Substitute the adjacent side EF, opposite side FA and hypotenuse EA=sin(β) into the definitions of sine and cosine and solve for the adjacent and opposite sides. EF=cos(α)sin(β) FA=sin(α)sin(β) Label these lengths in the figure. Substitute the unknown lengths into the equation from the end of step 1. BC=DE+EF AB=DC−FA This gives us the sum of two angles identities. sin(α+β)=sin(α)cos(β)+cos(α)sin(β) cos(α+β)=cos(α)cos(β)−sin(α)sin(β) Derive Difference of Two Angles Identities This example shows how to derive the difference of two angles identities using the right triangle definitions of the functions sine and cosine. The definitions are shown below. sin(θ)= hypotenuse opposite ​ cos(θ)= hypotenuse adjacent ​ Steps Start by drawing a right triangle with an angle of α−β and a hypotenuse of 1. The angles α (alpha) and β (beta) are also drawn. Solve for the lengths of the adjacent and opposite side. Substitute the hypotenuse AC=1, adjacent side AB and opposite side BC into the definitions of sine and cosine. BC=sin(α−β) AB=cos(α−β) Label these lengths in the figure. The goal is to represent these side-lengths in terms of the sine and cosine of the angles α and β. To achieve this goal, draw another right triangle of hypotenuse 1 with the angle of β on top of the first right triangle. This allows us to represent the length of AB as the sum of the lengths FE and ED. And to represent the length BC as the difference of the lengths FE and ED. BC=AF−CD AB=FE+ED Substituting the expressions from above gives us the starting point for the identities. In the next steps, we will solve for the unknown lengths using the definitions of sine and cosine. sin(α−β)=AF−CD cos(α−β)=FE+ED 2. Solve for the adjacent and opposite sides of the right triangle △AEC illustrated below. Substitute the hypotenuse AC=1, adjacent side AE and opposite side EC into the definitions of sine and cosine, then solve for the adjacent and opposite side. EC=sin(β) AE=cos(β) Label the side lengths in the figure. 3. Solve for the adjacent and opposite sides of the right triangle △EFA illustrated below. From the corresponding angle theorem we know that ∠AEF is the same as α. Substitute the hypotenuse AE=cos(β), adjacent side EF and opposite side FA into the definitions of sine and cosine, then solve for the adjacent and opposite side. EF=sin(α)cos(β) FA=cos(α)cos(β) Label these lengths in the figure. 4. Solve for the adjacent and opposite sides of the right-triangle △CDE illustrated below. We know that ∠ECD is equal α, because it is complementary to ∠CED which is complementary angle to ∠AEF. Substitute the hypotenuse CE=sin(β), adjacent side CD and opposite side DE into the definitions of sine and cosine and solve for the adjacent and opposite side. DE=sin(α)sin(β) CD=cos(α)sin(β) Label these lengths into the illustration below. This gives us all the unknown lengths in the figure. 5. Substitute the lengths into the equation from the end of step 1. sin(α−β)=AF−CD cos(α−β)=FE+ED This give us the difference of two angles identities. sin(α−β)=sin(α)cos(β)−cos(α)sin(β) cos(α−β)=cos(α)cos(β)+sin(α)sin(β) Derive Double Angle Identities (Algebra) This example derives the double angle identities using algebra and the sum of two angles identities. Steps Start with the sum of two angles identities. sin(α+β)=sin(α)cos(β)+cos(α)sin(β) cos(α+β)=cos(α)cos(β)−sin(α)sin(β) 2. Substitute α=θ and β=θ into the identities. This is same as saying the angle α (alpha) is equal to β (beta). sin(θ+θ)=sin(θ)cos(θ)+cos(θ)sin(θ) cos(θ+θ)=cos(θ)cos(θ)−sin(θ)sin(θ) 3. Combine the arguments on the left and simplify the expressions on the right. This gives us the double angle identities. sin(2θ)=2sin(θ)cos(θ) cos(2θ)=cos2(θ)−sin2(θ) 4. Optionally, the Pythagorean identity, shown below, can be used to calculate the two double-angle identity variations. cos2(θ)+sin2(θ)=1 Subtract sin2(θ) from both sides. cos2(θ)=1−sin2(θ) Substitute this expression into the identity from step 3 and combine like terms. This gives us the first variant. cos(2θ)=1−2sin2(θ) The second variant is found by subtracting cos2(θ) from both sides of the pythagorean identity. sin2(θ)=1−cos2(θ) Substitute this expression into the identity from step 3 and combine like terms. This gives us the second variant. cos(2θ)=2cos2(θ)−1 Derive Half Angle Identities (Algebra) This example derives the half-angle identities using algebra and the double angles identities. Steps Start with the double angle identities. cos(2θ)=1−2sin2(θ) cos(2θ)=2cos2(θ)−1 2. Transform the equations by substituting the half-angle of α (alpha) in for θ (theta). θ=21​α cos(2(21​α))=1−2sin2(21​α) cos(2(21​α))=2cos2(21​α)−1 3. Simplify the left side of the equations. cos(α)=1−2sin2(21​α) cos(α)=2cos2(21​α)−1 4. Add 2sin2(21​α) to both sides of the first equation in step 3. Then subtract cos(α) from both sides. 2sin2(21​α)=1−cos(α) Divide both sides by two. sin2(21​α)=21−cos(α)​ Take the square root of both sides. sin(21​α)=21−cos(α)​​ This gives us the first half-angle identity. 5. Rearrange the second equation from step 3 so that the half angle is on the left side. 2cos2(21​α)−1=cos(α) Add 1 to both sides. 2cos2(21​α)=cos(α)+1 Divide both sides by two. cos2(21​α)=2cos(α)+1​ Take the square root of both sides. cos(21​α)=2cos(α)+1​​ This gives us the second half-angle identity. 6. In conclusion, the two half angle idenities are given below. sin(21​α)=21−cos(α)​​ cos(21​α)=2cos(α)+1​​ Derive the Trigonometric Identities (Unit Circle) Example Derive the Trigonometric Identities (Complex Plane) Example
8481
https://www.khanacademy.org/science/chemistry/chemical-equilibrium
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8482
https://www.chegg.com/homework-help/questions-and-answers/let-f-x-1-f-x-f-4-calculate-x-f-4-h-f-4-h-x-4-express-numbers-exact-form-use-symbolic-nota-q56631384
Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Let f(x) = 1 f(x) - f(4) Calculate x' and f(4 + h) - f(4) h X-4 (Express numbers in exact form. Use symbolic notation and fractions where needed.) f(x) - f(4) X-4 $(4 + h) - f(4) h Take the appropriate limit to find f'(4). (Give an exact answer. Use symbolic notation and fractions where needed.) f'(4) = This AI-generated tip is based on Chegg's full solution. Sign up to see more! To start solving the problem, write and express using the given function. Fractional calculus is a branch of mathematical analysis that studies the several distinct opportuni... Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
8483
https://www.doubtnut.com/qna/1380341
Find the domain and range of the following functions: f(x)=√log2(x2+1). More from this Exercise Step by step video & image solution for Find the domain and range of the following functions: f(x) = sqrt(log_2(x^2 +1)). by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams. Similar Questions Find the domain and range of the function f(x)=√2−x+√1+x Domain and range of the function f(x)=cos(sqrt(1-x^(2))) are Domain of the function ,f(x)=√sin−1(log2x) is Find the domain and range of the function f(x)=1√x−5 Find the domain and range of the function f(x)=1√x−5 Find the domain and range of the function: f(x)=√9−x2 Find the domain and range of the following functions. f(x)=√x−1 Find the domain of the function f(x)=√−logx+42(log2(2x−13+x) Recommended Questions Find the domain and range of the following functions: f(x) = sqrt(log2... Find the domain and range of the following real functions:(i) f(x)=-|x... Find the domain and range of the following functions: f(x)=sqrt(log(2)... Find the domain and range of the function f(x)=(1)/sqrt(x-5) Find the domain and range of each of the following real functions: f(... वास्तविक फलन f(x)=sqrt(x-1) का प्रान्त (domain ) व परिसर ( Range)... Find domain of the function f(x) = sqrt(-log((x+4)/2)(log2(2x-1)/(3... Find the domain and range of the function f(x) = sqrt(x-1) Find the domain and range of the function f(x) = sqrt(16 - x^(2)) Exams Free Textbook Solutions Free Ncert Solutions English Medium Free Ncert Solutions Hindi Medium Boards Resources Doubtnut is No.1 Study App and Learning App with Instant Video Solutions for NCERT Class 6, Class 7, Class 8, Class 9, Class 10, Class 11 and Class 12, IIT JEE prep, NEET preparation and CBSE, UP Board, Bihar Board, Rajasthan Board, MP Board, Telangana Board etc NCERT solutions for CBSE and other state boards is a key requirement for students. Doubtnut helps with homework, doubts and solutions to all the questions. It has helped students get under AIR 100 in NEET & IIT JEE. Get PDF and video solutions of IIT-JEE Mains & Advanced previous year papers, NEET previous year papers, NCERT books for classes 6 to 12, CBSE, Pathfinder Publications, RD Sharma, RS Aggarwal, Manohar Ray, Cengage books for boards and competitive exams. Doubtnut is the perfect NEET and IIT JEE preparation App. Get solutions for NEET and IIT JEE previous years papers, along with chapter wise NEET MCQ solutions. Get all the study material in Hindi medium and English medium for IIT JEE and NEET preparation Contact Us
8484
https://www.fei.vsb.cz/export/sites/fei/470/.content/galerie-souboru/zaverecnePrace/mgr/pos220.pdf
V ˇ SB - Technical University of Ostrava Faculty of Electrical Engineering and Computer Science Department of Applied Mathematics Minimizing quadratic functions with separable quadratic constraints master thesis 2010 Luk´ aˇ s Posp´ ıˇ sil I declare I elaborated this thesis by myself. All literary sources and publications I have used had been cited. Ostrava, May 5, 2010 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . R´ ad bych na tomto m´ ıstˇ e podˇ ekoval pˇ redevˇ s´ ım Prof. RNDr. Zdeˇ nku Dost´ alovi, DrSc. za pomoc a vynikaj´ ıc´ ı motivuj´ ıc´ ı veden´ ı m´ e pr´ ace, zejm´ ena za myˇ slenku nov´ eho algoritmu. Podˇ ekov´ an´ ı si zaslouˇ z´ ı i Doc. RNDr. Radek Kuˇ cera, Ph.D. za poskytnut´ ı zdrojov´ ych k´ od˚ u ”konkurenˇ cn´ ıho” algoritmu. D´ ale dˇ ekuji m´ emu bratru Mgr. Michalovi Posp´ ıˇ silovi. Jeho pomoc s anglick´ ym jazykem dala vˇ etˇ sinˇ e vˇ et v t´ eto pr´ aci smysl. Za mor´ aln´ ı a finanˇ cn´ ı pomoc a podporu dˇ ekuji pˇ r´ atel˚ um a rodinˇ e, zejm´ ena mamince - d´ ıky n´ ı jsem, v´ ım a chci vˇ edˇ et. Speci´ aln´ ı podˇ ekov´ an´ ı patˇ r´ ı m´ e milovan´ e Veronice za nov´ y smysl. Abstract This thesis deals with the application of Dual problem in quadratic programming and in-troduces algorithms for solving minimizing problem of quadratic function subject to set pre-scribed by quadratic constraint functions. We proceed from simple observations to a new algorithm which was never presented before. Quadratic constraints are characteristic for contact problems with Coulomb friction. Keywords: Dual problem, Inverse dual problem, quadratic function, PDP Abstrakt Tato pr´ ace popisuje vyuˇ zit´ ı Du´ aln´ ı ´ ulohy v kvadratick´ em programov´ an´ ı a pˇ redstavuje algo-ritmy pro minimalizaci kvadratick´ e funkce vzhledem k mnoˇ zinˇ e popsan´ e vazebn´ ımi kvadrat-ick´ ymi funkcemi. Od pozorov´ an´ ı jednoduch´ ych algoritm˚ u pˇ rech´ az´ ı k algoritmu nov´ emu, kter´ y zat´ ım nebyl nikde publikov´ an. Kvadratick´ e vazby jsou charakteristick´ e pro kontaktn´ ı ´ ulohy s Coulombovsk´ ym tˇ ren´ ım. Kl´ ıˇ cov´ a slova: Du´ aln´ ı ´ uloha, Inverzn´ ı du´ aln´ ı ´ uloha, kvadratick´ a funkce, PDP CONTENTS 1. Contents 1 Introduction 2 2 Motivation 3 3 Problem definition 4 3.1 Quadratic function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 3.2 Separated quadratic constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 3.3 Minimizing function subject to constraint set . . . . . . . . . . . . . . . . . . . 6 4 Behind the new algorithm 8 4.1 Lagrange function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 4.2 Analytical solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 4.3 Conjugate gradient method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 4.4 MPRGP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 5 KKT system and dual problem 15 5.1 KKT system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 5.2 Dual problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 5.3 Inverse dual problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 6 Lagrange multipliers 19 6.1 Lagrange multiplier as linear penalty . . . . . . . . . . . . . . . . . . . . . . . . 19 6.2 Lagrange multipliers sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 6.3 Constant Update of Lagrange multipliers . . . . . . . . . . . . . . . . . . . . . 22 7 Simple update Lagrange methods 25 7.1 Linear constraint update . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 7.2 Adaptive linear constraint update . . . . . . . . . . . . . . . . . . . . . . . . . 26 7.3 Bisection method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 7.4 Numerical tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 8 Projected Dual problem (PDP) algorithm 32 8.1 Projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 8.2 Idea of PDP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 8.3 Inequality constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 8.4 Radius scaling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 8.5 Numerical tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 9 Conclusion 46 10 References 47 Appendix 48 1 INTRODUCTION 2. 1 Introduction In my master thesis I try to show application of Dual problem in minimizing quadratic func-tions with separable quadratic constraints solutions. This problem arises in problems with Coulomb friction. The motivation example is presented in Chapter 2. Formulation of minimizing problem can be found in Chapter 3. In this chapter are given also graphs of quadratic function and quadratic constraint set of one constraint problem. I introduce Lagrange function in Chapter 4 and also its utilization in analytical solution of a simle two dimensional problem. From presented example one can see the point of using KKT conditions. In this chapter, I introduce two numerical algorithms used later - Conju-gate gradient method and Modified proportioning with reduced gradient projections. These algorithms are introduced without detail analysis. Further implementation can be found in Appendix. In Chapter 5, I examine KKT conditions for more dimensional problems. Using simple modifications we can infer Dual problem and Inverse dual problem - two key components of new algorithm. In Chapter 6, I try to find meaning of Lagrange multiplier. Due to my observations, it can be regarded as linear penalty, increasing of which we can attract aproximations to boundary of constraint set. Simple algorithms, which use first KKT condition and linear update, are introduced in Chapter 7. Their convergence depends on the choice of input data. These algorithms are helpful in costruction of a main algorithm. Finally, I used all previous observations to introduce the new pretentious algorithm in Chapter 8 - Projected Dual Problem method (PDP). This algorithm uses both of KKT con-ditions and update Lagrange multipliers in the best way - it uses projection to boundary of quadratic constraint set. Numerical tests are also presented. 2 MOTIVATION 3. 2 Motivation We start with motivation example. This problem consists of solving minimizing problem of quadratic function with linear inequalities and quadratic inequality constraints. But in this thesis, I try to solve simpler problem only with quadratic constraints. Example 2.0.1 Let us consider the steel brick lying on a rigid foundation as it is shown in figure.1 The brick occupies in the reference configuration the domain ω ⊂R3, whose bound-ary ∂ω is split into three nonempty disjoint parts γu, γp, and γc with different boundary conditions: zero displacements γu, surface tractions γp and contact conditions γc (i.e., the nonpenetration and the effect of friction). The elastic behavior of the brick is described by Lam´ e equations that, after finite element discretization, lead to a symmetric positive definite stiffness matrix K ∈R3nc×3nc and to a load vector f ∈R3nc. Moreover, we introduce full rank matrices N, T1, T2 ∈Rmc×3nc projecting displacements at contact nodes to normal and tangential directions, respectively, and we denote B = ( N T , T T 1 , T T 2 )T ∈R3nc×3nc. Here, we shall use the dual formulation in terms of contact stresses. We start with the contact problem with Tresca friction that reads as    minimize 1 2λT Qλ −λT h, subject to λν,i ≥0, λ2 t1,i + λ2 t2,i ≤r2 i , i = 1, . . . , mc λ = ( λT ν , λT t1, λT t2 )T , λν, λt1, λt2 ∈Rmc, where Q = BK−1BT , h = BK−1f, and ri ≥0 are given slip bound values at contact nodes. Let us point out that λν and λt1, λt2 represent normal and tangential contact stresses, respectively. 1This example was introduced and numericaly solved in . More details about model problem can be found in . 3 PROBLEM DEFINITION 4. 3 Problem definition In this chapter I formulate minimizing problem and show how quadratic function and set prescribed by quadratic function looks. 3.1 Quadratic function Definition 3.1.1 (Quadratic function definition) The quadratic function has prescription f(x) def = 1 2xT Ax −bT x (1) where • n ∈N is problem dimension • f : R2n →R • A ∈R2n × R2n is symetric positive definite matrix • b ∈R2n is vector of right sides Theorem 3.1.1 (Quadratic function gradient) Gradient of function defined by equation (1) is ▽f = Ax −b Remark: Minimum of 3.1.1 without constraints is equal to solution of system ▽f = o, respectively Ax = b. That is the reason, why we called b the vector of right sides. ≈ Proof: Let us consider improvement x + αv of point x, where x, v ∈Rn, α ∈R Then f(x + αv) −f(x) = (1 2(x + αv)T A(x + αv) −bT (x + αv) ) − (1 2xT Ax −bT x ) = = αxT Av −αbT v + 1 2α2vT Av = 1 2α2vT Av + α(Ax −b)T v 3 PROBLEM DEFINITION 5. • ¯ x = min f(x) ⇒A¯ x = b Necessary condition of min f(x) is ▽f = o ▽f(x) = Ax −b ⇒Ax −b = o ⇒Ax = b • ¯ x = min f(x) ⇐A¯ x = b A¯ x = b ⇒A¯ x −b = o f(¯ x + αv) −f(¯ x) = 1 2α2vT Av ≥0, ∀α ∈R∀v ∈Rn (A is positive definite) ⇒f(¯ x + αv) ≥f(¯ x), ∀α ∈R∀v ∈Rn □ Theorem 3.1.2 (Existence of minimum) Function f(x) given by equation (1) has one minimum. System ▽f = o has only one solution. 3.2 Separated quadratic constraints Definition 3.2.1 (Constraint function) Let us define n quadratic constraint functions gi(x) def = x2 2i−1 + x2 2i −r2 i , i = 1, 2, . . . n (2) where • n ∈N is number of constraint functions, • gi : R2n →R, • r ∈Rn is vector of radii. If we choose firm g(x) = 0 then the geometric representation of set described by quadratic function is circle with radius r. 3 PROBLEM DEFINITION 6. Definition 3.2.2 (Constraint set) Quadratic constraint functions define equality constraint set ΩE def = {x ∈R2n : gi(x) = 0, i = 1, 2, . . . n}. (3) We can also define inequality constraint set ΩI def = {x ∈R2n : gi(x) ≤0, i = 1, 2, . . . n}. (4) 3.3 Minimizing function subject to constraint set Definition 3.3.1 (Minimizing problem) Unconstrained problem: Find ¯ x def = min x∈R2n f(x) (5) Equality problem: Find ¯ xE def = min x∈ΩE f(x) (6) Inequality problem: Find ¯ xI def = min x∈ΩI f(x) (7) Example 3.3.1 Find solution of Equality problem defined by equation 6 with A = [ 2 −1 −1 2 ] , b = [ 3 4 ] , r = 1 Geometrically, quadratic function f(x) is a modified elliptic paraboloid. 3 PROBLEM DEFINITION 7. −10 −5 0 5 10 −10 −5 0 5 10 −200 0 200 400 Isolines of this function (curves with same function value) are depicted in the following figure. −10 −5 0 5 10 −10 −8 −6 −4 −2 0 2 4 6 8 10 The geometric representation of equality constraint set ΩE of (6) is a circle with centre at [0, 0] and radius r = 1. If we combine isolines and constraint we can estimate the probable location of minimum, as plotted in the next figure: −6 −4 −2 0 2 4 6 −5 −4 −3 −2 −1 0 1 2 3 4 5 x1 x2 4 BEHIND THE NEW ALGORITHM 8. 4 Behind the new algorithm 4.1 Lagrange function Definition 4.1.1 (Lagrange function) Lagrange function has prescription L(x, λ) def = f(x) + λg(x) where • f(x) : Rn →R is cost function • g(x) : Rn →R is constraint function • λ ∈R is Lagrange multiplier Theorem 4.1.1 (bounded local extremes subject to equality constraint set) Let • f, g : Rn →R be C1 in open set Ω⊂Rn, n > 1 • grad g(x) ̸= (0, . . . , 0) for each x ∈Ω • ΩE def = {x ∈Ω: g(x) = 0}. Then 1. (Necessary condition of existence of local bounded extreme) If f has in c ∈Ωlocal extreme subject to set ΩE, there exists λ ∈R such that c is a stationary point of L(x) = f(x) + λg(x), x ∈Ω. 2. (Sufficient condition of existence of local bounded extreme) Let c ∈Ωbe a stationary point of function L(x) = f(x) + λg(x) for some λ ∈R, let f and g have in c continuous second partial derivatives and let d2Lc (for given λ) be positive definite quadratic form. Then f has in c local minimum subject to ΩE. 4 BEHIND THE NEW ALGORITHM 9. Theorem 4.1.2 (Sufficient condition subject to inequality constraint set) Let f, g, L be same functions as in Definition 4.1.1. Let ΩI def = {x ∈Ω: g(x) ≤0}. If • c ∈Ωis a stationary point of function L(x) = f(x) + λg(x) for some λ ≥0, • f and g have in c continuous second partial derivatives, • d2Lc (for given λ) is positive definite quadratic form then f has in c local minimum subject to ΩI. These theorems (i.e., sufficient conditions) gives us manual how to find bounded local extremes subject to equality and inequality constraint set. Definition 4.1.2 (Lagrange function for more constraints) Lagrange function for problems with m constraints has prescription L(x, λ) def = f(x) + m ∑ i=1 λigi(x) where • f(x) : Rn →R is cost function • m ≥1 is number of constraints • gi(x) : Rn →R is one of m constraint functions • λ ∈Rm is vector of Lagrange multipliers 4 BEHIND THE NEW ALGORITHM 10. 4.2 Analytical solution In analytical solution we will proceed with standart ”bounded local extremes” search algo-rithm. At first we consider Equality problem (see Definition 3.3.1). We assume a simple problem with one quadratic constraint. Assume Lagrange function for one condition L(x, λ) = f(x) + λ.g(x). For saddle point of this Lagrange function applies min x∈Ωf(x) = min x∈R2 L(x, λ) = ¯ x. In saddle point also holds constraint condition g(¯ x) = 0 Derivative of Lagrange function in saddle point has zero value (it is stacionary point of Lagrange function), so our task is to compute derivative of L(x, λ) and set it equal to zero. Since Lagrange function is a function of two variables, we have to compute partial deriva-tives and solve system of two equations. I.) ▽xL(x, λ) = o II.) ▽λL(x, λ) = o These conditions are also called ”Karush-Kuhr-Tucker conditions” (alias ”KKT system”, see Chapter 5). So I.) ▽xL(x, λ) = ▽xf(x) + λ▽xg(x) = Ax −b + 2λx II.) ▽λL(x, λ) = ▽λf(x) + ▽λλg(x) = g(x) and derived KKT system is I.) Ax −b + 2λx = o (8) II.) g(x) = 0 (9) For Inequality problem, we simply modify second condition II.) g(x) ≤0 Example 4.2.1 Consider Inequality problem with input data A = [ 2 −1 −1 2 ] , b = [ 3 4 ] , r = 1. 4 BEHIND THE NEW ALGORITHM 11. So our problem is to find ¯ x = def = min x∈ΩI f(x) where ΩI def = {x ∈R2 : g(x) ≤0} and one constraint is defined g(x) def = x2 1 + x2 2 −r2 Left-hand side of first KKT equation (8) (we consider λ as parameter λ ∈R, λ ≥0) has the form [ 2 −1 −1 2 ] . [ x1 x2 ] − [ 3 4 ] + [ 2λx1 2λx2 ] = [ 2x1 −x2 −3 + 2λx1 −x1 + 2x2 −4 + 2λx2 ] Thus (8) is transformed to the next system (2 + 2λ)x1 −x2 = 3 −x1 + (2 + 2λ)x2 = 4. Using Kramer formulas we obtain D = 2 + 2λ −1 −1 2 + 2λ = (2 + 2λ)2 −(−1)2 = 3 + 8λ + 4λ2 D1 = 3 −1 4 2 + 2λ = 3(2 + 2λ) −(−1).4 = 10 + 6λ D2 = 2 + 2λ 3 −1 4 = 4.(2 + 2λ) −3.(−1) = 11 + 8λ and parametric solution is (for common case refer to Dual task in Chapter 5) x1 = D1 D = 10 + 6λ 3 + 8λ + 4λ2 x2 = D2 D = 11 + 8λ 3 + 8λ + 4λ2 Now consider the constraint function g(x) = x2 1 + x2 2 −1 = ( 10 + 6λ 3 + 8λ + 4λ2 )2 + ( 11 + 8λ 3 + 8λ + 4λ2 )2 −1 and put it equal to zero ( 10 + 6λ 3 + 8λ + 4λ2 )2 + ( 11 + 8λ 3 + 8λ + 4λ2 )2 = 1 (10 + 6λ)2 + (11 + 8λ)2 (3 + 8λ + 4λ2)2 = 1 4 BEHIND THE NEW ALGORITHM 12. (10 + 6λ)2 + (11 + 8λ)2 = (3 + 8λ + 4λ2)2 (100 + 120λ + 36λ2) + (121 + 176λ + 64λ2) = (3 + 8λ + 4λ2)(3 + 8λ + 4λ2) 221 + 296λ + 100λ2 = 9 + 48λ + 88λ2 + 64λ3 + 16λ4 0 = 16λ4 + 64λ3 −12λ2 −248λ −212 This polynom has 4 roots - two of them are real, one is positive. Value of it is approxi-mately λ = 1.9877, so the solution after substitution is ¯ x = [ 0.6318 0.7751 ] . Example 4.2.2 Consider Inequality problem with input data A = [ 4 −1 −1 2 ] , b = [ 1 1 ] , r = 1. So our problem is to find ¯ x = def = min x∈ΩI f(x) where ΩI def = {x ∈R2 : g(x) ≤0} and one constraint is defined g(x) def = x2 1 + x2 2 −r2 First, we refer (8): [ 4 −1 −1 2 ] . [ x1 x2 ] − [ 1 1 ] + [ 2λx1 2λx2 ] = [ 4x1 −x2 −1 + 2λx1 −x1 + 2x2 −1 + 2λx2 ] = ( 0 0 ) Now solve system (4 + 2λ)x1 −x2 = 1 −x1 + (2 + 2λ)x2 = 1 using Kramer formulas D = 4 + 2λ −1 −1 2 + 2λ = (4 + 2λ)(2 + 2λ) −(−1)2 = 7 + 12λ + 4λ2 D1 = 1 −1 1 2 + 2λ = (2 + 2λ) −(−1) = 3 + 2λ D2 = 4 + 2λ 1 −1 1 = (4 + 2λ) −(−1) = 5 + 2.λ 4 BEHIND THE NEW ALGORITHM 13. Parametric solution is x1 = D1 D = 3 + 2λ 7 + 12λ + 4λ2 x2 = D2 D = 5 + 2λ 7 + 12λ + 4λ2 . Constraint function g(x) = x2 1 + x2 2 −1 = ( 3 + 2λ 7 + 12λ + 4λ2 )2 + ( 5 + 2λ 7 + 12λ + 4λ2 )2 −1 is set to zero and solved (3 + 2λ)2 + (5 + 2λ)2 = (7 + 12λ + 4λ2)2 (9 + 12λ + 4λ2) + (25 + 20λ + 4λ2) = 16λ4 + 96λ3 + 200λ2 + 168λ + 49 0 = 16λ4 + 96λ3 + 192λ2 + 136λ + 15 This polynom has two real roots, but all of them are negative, because minimum of orig-inal problem naturaly satisfies quadratic inequality constraint. Thus we search for λ ≥0 and because founded λ < 0, we simply choose λ = 0 and get L(x, 0) = f(x) + 0.g(x) = f(x) ⇒min L(x, λ) = min f(x) We can find minimum of this Inequality problem using simple minimalization algorithm without constraints. ¯ x = min x∈ΩI f(x) = min x∈R2 f(x) ⇔▽xf(¯ x) = 0 We solve equation A¯ x −b = o A¯ x = b using Gauss-Jordan elimination method we have [ 4 −1 1 −1 2 1 ] ∼ [ 4 −1 1 −4 8 4 ] ∼ [ 4 −1 1 0 7 5 ] ∼ [ 28 −7 7 0 7 5 ] ∼ ∼ [ 28 0 12 0 7 5 ] ∼ [ 1 0 3 7 0 1 5 7 ] We obtain the solution ¯ x = 1 7 [ 3 5 ] which really satisfies constraint g(¯ x) = ¯ x2 1 + ¯ x2 2 −r2 = (3 7 )2 + (5 7 )2 −1 = 9 + 25 −49 49 = −15 49 < 0 ⇒¯ x ∈ΩI 4 BEHIND THE NEW ALGORITHM 14. 4.3 Conjugate gradient method The Conjugate gradient method is iterative method for solving system Ax = b where A is symmetric positive definite matrix and b is the vector of right sides. Remark: The CG method is also used to find minimum of quadratic function with SPD matrix. (see Remark after Theorem 3.1.1) ≈ More information about Conjugate Gradient method can be found in . 4.4 MPRGP Modified proportioning with reduced gradient projections (MPRGP) is iterative method for minimizing quadratic cost function f(x) = 1 2xT Ax −bT x subject to linear inequalities l ∈Rm ∀i = 1, . . . m : xi ≥li More information about MPRGP can be found in . 5 KKT SYSTEM AND DUAL PROBLEM 15. 5 KKT system and dual problem During analytical solution of minimizing problem in Section 4.2 we deduce that in minimum of Lagrange function are accomplished two equations implied from partial derivatives of this function. In this section we try to generalize these equations and then we make some observations which we use into Dual problem definition. In whole section we consider minimizing problem with quadratic function f : R2n →R and n quadratic constraints which bind together succesively pairs of components of vector of variables x. 5.1 KKT system 5.1.1 Minimum of Lagrange function We consider Lagrange function (see Definition 4.1.1) L(x, λ) = f(x) + n ∑ i=1 λigi(x) λ ∈Rn, L : R2n+n →R and express KKT conditions in saddle point of L(x, λ) using Theorem ?? ▽xL(x, λ) = ▽f(x) + n ∑ i=1 λi ▽gi(x) = o2n (10) ▽λL(x, λ) = g(x) = on (11) Remark: on denote zero vector of n components. ≈ 5.1.2 Duplication of Lagrange multipliers Now consider first KKT condition (10) ▽f(x) + n ∑ i=1 λi ▽gi(x) = o2n. At first we express gradient of quadratic function ▽f(x) = Ax −b (12) and gradient of separable quadratic constraints ▽gi(x) =            ∂gi ∂x1 . . . ∂gi ∂x2i−1 ∂gi ∂x2i . . . ∂gi ∂x2n            =           0 . . . 2x2i−1 2x2i . . . 0           ∈R2n (13) 5 KKT SYSTEM AND DUAL PROBLEM 16. Then we substitute (12) and (13) into (10). We obtain ▽f(x) + n ∑ i=1 λi ▽gi(x) = Ax −b + n ∑ i=1 ( λi(0, . . . , 2x2i−1, 2x2i, . . . , 0)T ) = = Ax −b + 2 ( 2n ∑ i=1 λ⌈i 2 ⌉xi ) = Ax −b + 2diag(˜ λ)x where ˜ λ def = (λ1, λ1, λ2, λ2, . . . λn, λn)T ∈R2n. (14) Hence Ax −b + 2 diag(˜ λ)x = o2n. (15) 5.2 Dual problem Let us assume modified first KKT condition (15) and express variable x: Ax −b + 2 diag(˜ λ)x = o2n Ax + 2 diag(˜ λ)x = b (A + 2 diag(˜ λ))x = b x = (A + 2 diag(˜ λ))−1b (16) We call equation (16) Dual problem. It represents relation between variable x and corre-sponding Lagrange multipliers λ (supposing the first KKT condition to be accomplished). If we have λ, we can simply solve (16) with Conjugate gradient method (see Section 4.3) to get solution x. Definition 5.2.1 (Dual problem solution) We say that a pair (x, λ) solve Dual problem, if equation x = (A + 2 diag(. . . , λi, λi, . . .))−1b (17) is fulfilled. 5.3 Inverse dual problem 5.3.1 Inverse problem Now we consider situation, when we have approximation x and our task is to find corre-sponding Lagrange multipliers λ that (x, λ) solve Dual problem (17) as good as possible. 5 KKT SYSTEM AND DUAL PROBLEM 17. Remark: In Dual problem dimension of vector of Langrange multipliers λ is half of dimen-sion of variable x. That is reason why not for all x ∈R2n exists corresponding ˜ λ. ≈ From λ we require that ˜ λ given by (14) satisfies (16) as good as possible: • ∀i = 1, . . . n : ˜ λ2i−1 = ˜ λ2i = λi • equation (17) from Dual problem is accomplished as good as possible 5.3.2 Error function At first we express ˜ λ from Dual problem (16) Ax + 2. diag(˜ λ)x = b 2. diag(˜ λ)x = (b −Ax) 2.diag(x)˜ λ = (b −Ax) (18) From equation (18) we can derive error function, which describes distance of approxi-mate solution ˜ λ to exact solution of dual problem equation (16). err def = 2.diag(x)˜ λ −(b −Ax) (19) Our aim is to have err as small as possible ∥err∥2 = errT err Substitute and compose errT err = ( 2.diag(x)˜ λ −(b −Ax) )T ( 2.diag(x)˜ λ −(b −Ax) ) = = 4.˜ λT diag(x)2˜ λ −4.˜ λT diag(x)(b −Ax) + (b −Ax)T (b −Ax) = = 4. 2n ∑ i=1 ( ˜ λ2 i x2 i ) −4. 2n ∑ i=1 ( ˜ λixi[b −Ax]i ) + (b −Ax)T (b −Ax) (20) But we know ˜ λ2i−1 = ˜ λ2i = λi, so we can write (20) as follows 4. n ∑ i=1 ( λ2 i (x2 2i−1 + x2 2i) ) −4. n ∑ i=1 (λi(x2i−1[b −Ax]2i−1 + x2i[b −Ax]2i)) + (b −Ax)T (b −Ax) = = 4.λT diag(. . . , x2 2i−1+x2 2i, . . .)λ−4.(. . . , x2i−1[b−Ax]2i−1+x2i[b−Ax]2i, . . .)λ+(b−Ax)T (b−Ax) 5 KKT SYSTEM AND DUAL PROBLEM 18. 5.3.3 Final simplification If we denote Q def = 8. diag(. . . , x2 2i−1 + x2 2i, . . .) (21) q def = 4.(. . . , x2i−1[b −Ax]2i−1 + x2i[b −Ax]2i, . . .)T (22) our next task is to minimize ∥err∥2 = 1 2λT Qλ −qT λ + (b −Ax)T (b −Ax). (23) Further work depends on original problem formulation (due to Definition 3.3.1) • Equality problem λ = min λ∈R ∥err∥2 • Inequality problem λ = min λ≤R ∥err∥2 5.3.4 Minimum of error function without constraints Let us consider Equality problem from Definition 3.3.1. We look for λ ∈Rn minimizing error function (23). So we have to find roots of first deriva-tive. At first we compute first derivative ∂ [ ∥err∥2] ∂λi = Qλ −q Remark: We used remark after Theorem (3.1.1) for minimizing quadratic function. Q is symmetric positive definite matrix and ∂(b −Ax)T (b −Ax) ∂λ = 0. ≈ Now we put first derivative of error function equal to zero λ = Q−1q and substitute (21) and (22) λ = 1 8.     ... 0 1 x2 2i−1+x2 2i 0 ...    .4.     . . . x2i−1[b −Ax]2i−1 + x2i[b −Ax]2i . . .    . We express prescription for i-th element of λ λi = 1 2(x2 2i−1 + x2 2i) (x2i−1[b −Ax]2i−1 + x2i[b −Ax]2i) . (24) Thence Inverse Dual problem for Equality problem (see Definition 3.3.1) can be solved using equation (24). But for Inequality problem we use MPRGP algorithm (see Section 4.4). 6 LAGRANGE MULTIPLIERS 19. 6 Lagrange multipliers In this chapter we describe Lagrange multipliers and their relation to a proper parameter. We consider Equality problem (see Definition 3.3.1). In our exploration figures will be very useful. 6.1 Lagrange multiplier as linear penalty 6.1.1 Linear penalty introduction Let us consider function ˜ fv(x) given by ˜ fv(x) def = f(x) + vT g(x) (25) where v ∈Rn is appropriately chosen constant vector with positive components. We refer to v as the linear penalty parameter. We can derive these properties: • if x ∈∂ΩI, then g(x) = o, thus ˜ fv(x) = f(x) + vT g(x) = f(x) + vT o = f(x), • if x ∈ΩI \ ∂ΩI, then −c ≤g(x) < o (meaning −ci ≤gi(x) < 0, ∀i = 1 . . . n), hence ˜ fv(x) = f(x) + vT g(x) < f(x) • if x ∈R2n \ ΩI then g(x) > o (meaning gi(x) > 0, ∀i = 1 . . . n), therefore ˜ fv(x) = f(x) + vT g(x) > f(x). Due to the observations we can say that linear penalty modifies value subject to constraints -it increases values in R2n \ ΩI and decreases in ΩI \ ∂ΩI. Example 6.1.1 Let us have specific values of Equality problem with one constraint A = [ 2 −1 −1 2 ] , b = [ 1 1 ] , r = 1 and draw isolines of original function f(x) and function with linear penalty ˜ fv(x) subject to one quadratic constraint. We try some different values of linear penalty parameter v. 6 LAGRANGE MULTIPLIERS 20. −5 0 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x1 x2 v = 0 −5 0 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x1 x2 v = 0.2 −5 0 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x1 x2 v = 0.5 −5 0 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x1 x2 v = 100 Example 6.1.2 Another values can be A = [ 2 −1 −1 2 ] , b = [ 4 20 ] , r = 1 But now, for more illustrative example, we put a cross into figure on coordinates where, for concrete value of v, the real minimum of ˜ fv(x) is. We try to set v0 = 0, vi+1 = vi + 0.25, i = 0, 1..20. 6 LAGRANGE MULTIPLIERS 21. −5 0 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x1 x2 v0 We can also demostrate difference between value in minimum of original function f and function ˜ f using specific vi. 0 2 4 6 8 10 12 14 16 18 20 −180 −160 −140 −120 −100 −80 −60 −40 −20 0 vi value with penalty original 6 LAGRANGE MULTIPLIERS 22. 6.1.2 Lagrangie multipliers as linear penalty parameter For next consideration we will need prescription of common Lagrange function L(x, λ) def = f(x) + λT g(x) Linear penalty has the same rules as Lagrange function, so logically, we can consider the vector of Lagrange multipliers as linear penalty parameter. From Example 6.1.2 we can note that if we increase Lagrange multiplier, the minimum of L(x, λ) will be more closer to the center of the circle defined by constraint function g(x). Written in limit form lim λ→∞(arg min L(x, λ)) →o where convergence λ →∞means ∀i = 1 . . . n : λi →∞. 6.2 Lagrange multipliers sequence Let us consider a simple two-dimensional problem. That means, we have only one constraint and also only one Langrange multiplier. We already tried to find minimum of quadratic function, but this minimum is not from ΩE (see Definition 3.2.2). There exists ¯ λ which is efficient to construct function L(x, y) which minimum is in this set. At this point x, the first KKT condition is accomplished (it is the minimum of Lagrange function at all) also the second (this x is from ΩE). We refer to this point as the ¯ x. Let us get back to Example 6.1.2. In fact, we construct a sequence of Lagrange multipliers λ1 < λ2 < . . . < ¯ λ < . . . < ∞ and we stepwise by substitute members of this sequence to Lagrange function. Mini-mum of this function started to move towards ΩE, but it didn’t stop in ΩE, but it continues to zero point o (to the centre of the circle described by quadratic constraint). Now our task is to find ¯ λ corresponding to minimum ¯ x of Lagrange function in ΩE. 6.3 Constant Update of Lagrange multipliers We simply try to put some values of Langrange multipliers into Dual problem. Since we want to show how Dual problem works, we choose simple equidistant arithmetic progres-sion with convenient ϵ ∈R difference. λk+1 = λk + ϵ (26) 6 LAGRANGE MULTIPLIERS 23. Listing 1: Constant update Lagrangian method 1 lambdas = 0: epsilon : lambda max ; 2 3 for i =1:( length ( lambdas ) ) 4 xi ( i , : ) = cg (A + 2 ∗diag ( [ lambdas ( i ) , lambdas ( i ) ] ) , b , x0 , e ) ; 5 end Example 6.3.1 Consider input data A = ( 2 −1 −1 2 ) , b = ( −5 6 ) , r = 1, ϵ = 0.1, λ max = 5 If we try to plot aproximations xk, we get something like this: −5 0 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x1 x2 and in case that we evaluate quadratic function and quadratic constraint: 0 1 2 3 4 5 −11 −10 −9 −8 −7 −6 −5 −4 λ f 6 LAGRANGE MULTIPLIERS 24. 0 1 2 3 4 5 −1 0 1 2 3 4 5 6 7 λ g 6.3.1 Sequence of Update Lagrange algorithm aproximation using different input data Example 6.3.2 Let us consider testing data A = ( 2 1 1 2 ) , r = 1, ϵ = 0.1, λ max = 2 and let us try to plot sequence of minima for different right side vectors. We choose b ∈{−5, . . . , 5} × {−5, . . . , 5}. Output: −5 0 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x y 7 SIMPLE UPDATE LAGRANGE METHODS 25. 7 Simple update Lagrange methods We consider Equality problem (see Definition 3.3.1) with one quadratic constraint. From previous observations in Chapter 6 we know how to move aproximations towards equality constraint set ΩE. But we do not know how to stop this progression. In this chapter we try some simple algorithms which solve this problem. 7.1 Linear constraint update Let us consider prescription λk+1 = λk + ρ.g(xk) where ρ is sufficiently small real constant. This prescription tries to update Lagrange multiplier using sofisticated method - size of up-date is adequate to distance of actual aproximation from ΩE. Using this prescription we construct algorithm: • input – A ∈R2 × R2 - SPD matrix – b ∈R2 - right side vector – r ∈R - radius of boundary – e ∈R - precision of algorithm – x0 ∈R2 - initial approximation – λ0 = 0 - initial approximation of Lagrange multiplier – k = 0 - iterator • while xT k .x −r > e do – xk+1 = cg(A + 2 ∗diag([λk, λk]), b, xk, e) – λk+1 = λk + ρ.(xT k+1.xk+1 −r) – k = k + 1 where cg(A, b, x0, e) is implemented algorithm of Conjugated gradient method, see Sec-tion 4.3. 7 SIMPLE UPDATE LAGRANGE METHODS 26. Example 7.1.1 Let us choose the following input data A = [ 2 −1 −1 2 ] , b = [ 4 20 ] , x0 = [ 1 1 ] , e = 10−4, r = 1. Using different constant coeficients ρ, algorithm find solution subject to precision using dif-ferent number of iterations: ρ # of iterations 0.020 2779 0.021 2340 0.022 5000+ 0.023 1791 0.023 1791 0.024 1872 0.025 1602 0.026 1542 0.027 1503 0.028 1843 0.029 1600 0.03 1232 0.031 1528 0.0311 922 0.0312 623 0.03121 2295 0.03122 1105 0.03123 735 0.03124 653 0.03125 541 0.03126 407 0.03127 200 0.03128 -0.0313 -0.032 -0.04+ -7.2 Adaptive linear constraint update We modify previous algorithm - we find adequate coeficient ρ by testing and making shorter in every iteration. • input – A ∈R2 × R2 - SPD matrix – b ∈R2 - right side vector – r ∈R - radius of boundary 7 SIMPLE UPDATE LAGRANGE METHODS 27. – e ∈R - precision of algorithm – x0 ∈R2 - initial approximation – λ0 = 0 - initial approximation of Lagrange multiplier – ρ = 1 - initial update coefficient – k = 0 - iterator • while xT k .xk −r > e do – try to update: λtest = λk + ρ.(xT k+1.xk+1 −r) – compute testing aproximation: xk+1 = cg(A + 2 ∗diag([λtest, λtest]), b, xk, e) – while xT test.xtest −r < e ∗ρ = ρ 2 ∗try to update: λtest = λk + ρ.(xT k+1.xk+1 −r) ∗compute testing aproximation: xtest = cg(A + 2 ∗diag([λtest, λtest]), b, xk, e) – λk+1 = λtest – xk+1 = xtest Example 7.2.1 Consider input data A = [ 2 −1 −1 2 ] , b = [ 1 1 ] , r = 1, x0 = [ 5 3 ] , eps = 10−4 Output of this algorithm: −5 0 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x y 7 SIMPLE UPDATE LAGRANGE METHODS 28. 0 20 40 60 80 100 120 140 160 180 −1 −0.5 0 0.5 1 1.5 2 it f 0 20 40 60 80 100 120 140 160 180 0 2 4 6 8 10 12 it g 7.3 Bisection method In this algorithm we try to find λmax using bisection method. There exists sufficient by large λmax such that g(x(λmax)) < 0 7 SIMPLE UPDATE LAGRANGE METHODS 29. Then our solution ¯ x = x(¯ λ) with ¯ λ ∈(0, λmax). We search this ¯ λ using Bisection method with stop condition |g(x(¯ λ))| < ϵ where ϵ > 0 is required precision. Listing 2: bisect 1 % f i n d any lambda max 2 lambda max = 0; 3 x max = cg (A + 2 ∗diag ( [ lambda max , lambda max ] ) , b , x 00 , e ) ; 4 while ( x max ’∗x max −c ) > e 5 lambda max = lambda max + 1; % try to i n c r e a s e 6 x max = cg (A + 2 ∗diag ( [ lambda max , lambda max ] ) , b , x 00 , e ) ; 7 end 8 9 % i n i t i a l i z a t i o n 10 a b i s e c t = 0; % lower e s t i m a t i o n 11 b b i s e c t = lambda max ; % upper e s t i m a t i o n 12 s b i s e c t = ( a b i s e c t +b b i s e c t )/2; % p i v o t 13 x = cg (A + 2 ∗diag ( [ s bisect , s b i s e c t ] ) , b , x 00 , e ) ; 14 15 % main i t e r a t i o n s 16 while abs ( x ’∗x −r ) > e 17 % compute new i n t e r v a l 18 i f x ’∗x −r > 0 19 a b i s e c t = s b i s e c t ; 20 else 21 b b i s e c t = s b i s e c t ; 22 end 23 24 % compute new p i v o t 25 s b i s e c t = ( a b i s e c t + b b i s e c t )/2; 26 x = cg (A + 2 ∗diag ( [ s bisect , s b i s e c t ] ) , b , x 00 , e ) ; 27 end 7 SIMPLE UPDATE LAGRANGE METHODS 30. Output of this algorithm: −5 0 5 −5 −4 −3 −2 −1 0 1 2 3 4 5 x y 0 2 4 6 8 10 12 14 −7.8 −7.6 −7.4 −7.2 −7 −6.8 −6.6 −6.4 −6.2 −6 it f 7 SIMPLE UPDATE LAGRANGE METHODS 31. 0 2 4 6 8 10 12 14 −0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 it g 7.4 Numerical tests A1 = [ 2 1 1 2 ] , A2 = [ 2 −1 −1 2 ] , A3 = [ 3 1 1 1 ] , A4 = [ 2 2 2 3 ] b1 = [ −5 6 ] , b2 = [ 4 5 ] , b3 = [ −3 −3 ] , b4 = [ 0 4 ] input adaptive ULM bisection method A b x1 x2 λ it x1 x2 λ it A1 b1 -0.654301 0.756363 3.399414 523 -0.654208 0.756261 3.399414 12 A1 b2 0.585708 0.810689 1.723145 52 0.585623 0.810529 1.723145 12 A1 b3 -1.000000 -1.000000 0.621338 8 -0.707101 -0.707101 0.621338 12 A1 b4 -0.224782 0.974464 1.167847 70 -0.224741 0.974410 1.167847 14 A2 b1 -0.616196 0.787549 2.418091 73 -0.616200 0.787545 2.418091 13 A2 b2 0.645064 0.764311 2.693481 170 0.644944 0.764177 2.693481 13 A2 b3 -0.039247 -0.039247 1.621338 6 -0.707101 -0.707101 1.621338 13 A2 b4 0.224782 0.974464 1.167847 70 0.224741 0.974410 1.167847 14 A3 b1 -0.578076 0.816043 3.530762 926 -0.578042 0.815980 3.530762 13 A3 b2 0.438290 0.898793 2.037598 192 0.438317 0.898819 2.037598 11 A3 b3 -0.432857 -0.901445 0.923828 121 -0.432894 -0.901480 0.923828 9 A3 b4 -0.158926 0.987283 1.606201 252 -0.158925 0.987305 1.606201 13 A4 b1 -0.697204 0.721933 3.635254 1000 -0.694557 0.719450 3.635254 13 A4 b2 0.641243 0.767328 0.922363 31 0.641220 0.767341 0.922363 11 A4 b3 -0.948984 -0.315123 0.248535 34 -0.949028 -0.315105 0.248535 11 A4 b4 -0.419434 0.907905 1.164917 183 -0.419347 0.907851 1.164917 14 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 32. 8 Projected Dual problem (PDP) algorithm In the most important chapter of this thesis we introduce new algorithm for solving Equality and Inequality problems. At first we introduce projection to boundary of set and then use observations in previous chapters to construct PDP algorithm. 8.1 Projection Our following problem is to find the nearest Px ∈R2 to x, which satisfy II. KKT condition (11) in the best way. Definition 8.1.1 (Projection) ∀x ∈R2 \ {o} : Px def = r ∥x∥2 x Remark: We simply normalize vector of actual iteration x and then extend it to r, thus g(Px) = 0 ⇔Px ∈ΩE, ∂ΩI. ≈ 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 33. Theorem 8.1.1 For every iteration xk ∈R2 is Pxk from (8.1.1) the nearest point accomplishing II. KKT condition (11). ∀xk ∈R2∀y ∈R2 : (g(y) = 0 ∧y ̸= Pxk) ⇒(∥xk −Pxk∥2 < ∥xk −y∥2) Proof: For projection holds ∥xk∥2 = ∥Pxk∥2 + ∥xk −Pxk∥2 so ∥xk −Pxk∥2 = ∥xk∥2 −∥Pxk∥2 = ∥xk∥2 −r and because g(y) = 0 ⇒∥y∥2 = r, we have ∥xk∥2 −r = ∥xk∥2 −∥y∥2 = ∥(xk −y) + y∥2 −∥y∥2 For every norm ∥x + y∥≤∥x∥+ ∥y∥, so we can write ∥(xk −y) + y∥2 −∥y∥2 ≤∥xk −y∥2 + ∥y∥2 −∥y∥2 = ∥xk −y∥ Equality is possible, only if xk −y = −y ⇒xk = 0. For this point, projection (8.1.1) is not defined. So we can say ∥x −Pxk∥2 < ∥xk −y∥2 □ Definition 8.1.2 (Projection in more dimensions) For every iteration x ∈R2n \ {x ∈R2n : ∥(x2i−1, x2i)∥2 ̸= 0, i = 1, 2, . . . n} we define projection Px = (P(x1, x2), . . . P(x2i−1, x2i), . . . P(x2n−1, x2n))T 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 34. 8.2 Idea of PDP Previous algorithms in Chapter 7 (except Bisection method) update Lagrange multipliers from previous iteration by multiple of value of quadratic contraint in this iteration. Now we try to compute this update using more sofisticated process - we use Lagrange multiplier corresponding to projection of previous iteration to boundary of constraint set. We will use update prescription λk+1 = λk + ˙ λk where • λk is Lagrange multiplier from previous iteration • λk+1 is Lagrange multiplier corresponding to next iteration • ˙ λk is update 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 35. The algorithm consists of these steps: • Initialization Find minimum of quadratic function without constraints x0 = min x∈R2n f(x) using CG method. Set λ0 = 0. • KKT conditions accomplishment The algorithm is over, if both of KKT conditions are accomplished due to precision. Because first KKT condition is accomplished in every iteration (every next iteration is computed using dual problem solver), we simply test accomplishment of second KKT condition. • Projection computation Compute projection of actual iteration using Definition 8.1.2. • Update computation Minimize inverse dual problem function - find Lagrange multiplier corresponding to projection using: – CG algorithm for finding ˙ λk ∈R without confidement - if original problem is with equality constraints (or use prescription from Section 5.3.4), – MPGRP algorithm for finding ˙ λk > 0 - if original problem is with inequality con-straints. • Lagrange multipliers update Compute next Lagrange multiplier by updating λk+1 = λk + ˙ λk • Next aproximation computation Find next aproximation xk+1 corresponding to λk+1, using Dual problem definition 5.2.1 - use CG algorithm. 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 36. 8.3 Inequality constraints 8.3.1 Characterization Denote (xk, λk) k-th iteration (solution aproximation and corresponding vector of Langrange multipliers in k-th iteration, this pair solve Dual problem, see Definition 5.2.1 ). Next iteration (xk+1, λk+1) can be expressed from previous one. We compute first iteration (x0, λ0): • λ0 = 0 • x0 = min x∈R2n f(x) is minimum without constraints (can be computed using CG method, see Section 4.3) In each iteration we compute pair (xk+1, λk+1) using this method: (We denote for simplicity (xk, λk) = (x, λ)) • projection of previous iteration Pxk = [. . . ri ∥(x2i−1, x2i)∥2 (x2i−1, x2i) . . .]T (projection in more dimensions see Definition 8.1.2) • Langrange multiplier from projection Q = 8. diag(. . . , (Px)2 2i−1 + (Px)2 2i, . . .) q = 4.(. . . , (Px)2i−1[b −A(Px)]2i−1 + (Px)2i[b −A(Px)]2i, . . .)T ˙ λk = min λ≥0 1 2λT Qλ −qT λ (minimum of Invert Dual problem error function with constraint λ ≥0, see equa-tion (23) in Section 5.3.3) for solving this problem, we use minimalization algorithm MPRGP, see Chapter 4.4. • update Lagrange multipliers λk+1 = λk + ˙ λk • compute next iteration using new multipliers xk+1 = (A + 2diag(. . . , [λk+1]i, [λk+1]i, . . .))−1b (for solving this system can be used CG method see Chapter 4.3) 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 37. 8.3.2 Algorithm in Matlab Main algorithm: Listing 3: pdp ineq 1 % i n i t i a l i z a t i o n 2 k = 0 ; 3 x k = cg (A, b , x 00 , eps ) ; 4 lambda k = zeros ( length ( x k ) , 1 ) ; 5 6 % main i t e r a t i o n s 7 while ˜ is in omega ( x k , r , eps ) 8 % p r o j e c t i o n 9 Px k = projection ( x k , r , eps ) ; 10 11 % f i n d update 12 lambda dot k = get lambda (A + 2 ∗diag ( lambda k ) , b , r , Px k , eps \ ) ; 13 14 % update lagrange m u l t i p l i e r s 15 lambda k = lambda k + lambda dot k ; 16 17 % f i n d next aproximation using Dual problem 18 x k = cg (A + 2∗diag ( lambda k ) , b , x k , eps ) ; 19 20 % i n c r e a s e i t e r a t i o n counter 21 k = k + 1; 22 end Stop condition: Listing 4: is in omega 1 function [ return value ] = is in omega ( x , r , eps ) 2 return value=true ; 3 for i =1:( length ( x )/2) 4 i f ( ˜ s a t i s f y q u a d r a t i c c o n s t r a i n ( x ((2∗i −1):(2∗i ) ) , r ( i ) , eps ) ) 5 x (2∗i −1)ˆ2 + x (2∗i ) ˆ 2 −r ( i ) ˆ 2 6 return value = f a l s e ; 7 end 8 end 9 end 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 38. Verify condition: Listing 5: satisfy condition 1 function [ return value ] = s a t i s f y q u a d r a t i c c o n s t r a i n ( x , r , eps ) 2 i f x ( 1 ) ˆ 2 + x ( 2 ) ˆ 2 −r ˆ2 <= eps 3 return value = true ; 4 else 5 return value = f a l s e ; 6 end 7 end Projection: Listing 6: projection 1 function [ x ] = projection ( x , r , eps ) 2 for i = 1 : ( length ( x )/2) % f o r a l l c o n s t r a i n t s 3 x couple = x ((2∗i −1):(2∗i ) ) ; 4 % compute p r o j e c t i o n to a c t u a l boudary 5 x ((2∗i −1):(2∗i ) ) = ( r ( i ) ) / ( sqrt ( x couple (1)ˆ2+ x couple ( 2 ) ˆ 2 ) ) . . . 6 ∗x couple ; 7 end 8 end Update computation: Listing 7: compute update 1 function [ lambda out ] = get lambda (A, b , r , x , eps ) 2 reziduum = b− A∗x ; 3 Q = zeros ( length ( x )/2 , length ( x ) / 2 ) ; 4 q = zeros ( length ( x ) / 2 , 1 ) ; 5 for i = 1: length ( x)/2 6 Q( i , i ) = 8∗( x (2∗i −1)ˆ2 + x (2∗i ) ˆ 2 ) ; 7 q ( i ) = 4∗( x (2∗i −1)∗reziduum (2∗i −1) + x (2∗i )∗reziduum (2∗i ) ) ; 8 end 9 % compute s o l u t i o n using MPRGP 10 lambda = mprgp(Q, q , zeros ( length ( x )/2 ,1) , eps ) ; 11 12 lambda out = zeros ( length ( lambda ) ∗2 , 1 ) ; 13 for i = 1: length ( lambda ) 14 lambda out (2∗i −1) = lambda ( i ) ; 15 lambda out (2∗i ) = lambda ( i ) ; 16 end 17 end 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 39. 8.4 Radius scaling We shall remind minimizing problem with separable inequality quadratic constraints (??): Find ¯ x ∈R2n such that ¯ x def = min x∈Ωf(x) f(x) def = 1 2xT Ax −bT x Ωdef = {x ∈R2n : gi(x) ≤0, i = 1, 2, . . . n} gi(x) def = x2 2i−1 + x2 2i −r2 i (27) where • n ∈N is problem dimension, resp. number of constraint functions • f : R2n →R is quadratic function • A ∈R2n × R2n is symetric positive definite matrix • b ∈R2n is vector of right-hand sides • r ∈Rn is vector of radii Definition 8.4.1 (Identity of radius) We say that problem 27 has identical radius r = ρ ∈R if ∀i = 1, . . . n : ri = ρ Let us consider constraint function gi(x) ≤0. We try to identity its radius gi(x) ≤0 x2 2i−1 + x2 2i −r2 i ≤0 x2 2i−1 + x2 2i ≤r2 i ρ2x2 2i−1 + ρ2x2 2i ≤ρ2r2 i ρ2 r2 i x2 2i−1 + ρ2 r2 i x2 2i ≤ρ2 ˜ x2 2i−1 + ˜ x2 2i ≤ρ2 where we used subsitution ˜ x2i−1 = ρ ri x2i−1 ˜ x2i = ρ ri x2i (28) 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 40. We can use this substitution to whole vector x: ˜ x = Rx, R def = diag( ρ r1 , ρ r1 , . . . ρ rn , ρ rn ) (29) Then x ∈Ωis equivalent to Rx ∈{x ∈R2n : ˜ gi(x) ≤0, i = 1, 2, . . . n}, where ˜ gi(x) def = x2 2i−1 + x2 2i −ρ2 (30) Now we can express f(˜ x) using the previous substitution f(˜ x) = f(Rx) = 1 2(Rx)T A(Rx) −bT (Rx) = 1 2xT RARx −(Rb)T x = 1 2xT ˜ Ax −˜ bT x where ˜ A = RAR ˜ b = Rb (31) Theorem 8.4.1 (Problems equivalency) Solution of problem 27 denoted as ¯ x is equivalent (after substitution ¯ x = R−1¯ ˜ x) to solution of problem with identical radius: Find ¯ ˜ x ∈R2n such that ¯ ˜ x def = min x∈Ω ˜ f(x) ˜ f(x) def = 1 2xT ˜ Ax −˜ bT x Ωdef = {x ∈R2n : ˜ gi(x) ≤0, i = 1, 2, . . . n} ˜ gi(x) def = x2 2i−1 + x2 2i −ρ2 8.5 Numerical tests Example 8.5.1 Let us consider input data: A = ( 2 −1 −1 2 ) , b = ( 2 3 ) , r = ( 1 ) , eps = 0.0001 Algorithm is over in one iteration. 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 41. Example 8.5.2 Let us consider input data: A =     4 −1 −1 0 −1 4 −1 −1 −1 −1 4 −1 0 −1 −1 4    , b = (1, 1, −20, 50)T , r = (1, 1)T , eps = 0.0001 Algorithm is over in two iterations. Iteration progress and progress of constraint functions values: −1 −0.5 0 0.5 1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 progress of iterations due to g1(x) = x1 2 + x2 2 − 12 = 0 x1 x2 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −0.701 −0.7 −0.699 −0.698 −0.697 −0.696 −0.695 values of g1(x) = x1 2 + x2 2 − 12 it g −1 −0.5 0 0.5 1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 progress of iterations due to g2(x) = x3 2 + x4 2 − 12 = 0 x3 x4 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −0.01 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 values of g2(x) = x3 2 + x4 2 − 12 it g Function value progress and progress of Lagrange function values: 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −52.005 −52 −51.995 −51.99 −51.985 −51.98 −51.975 −51.97 values of Lagrange funcion L(x,λ) it L 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −53.8 −53.6 −53.4 −53.2 −53 −52.8 −52.6 −52.4 −52.2 −52 −51.8 values of cost function f(x) it f 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 42. We can verify our solution - induct solution into KKT conditions: KKT1err = Ax −b + 2 diag(˜ λ)x = 10−13.     0 −0.0033 −0.0711 0.1421     g(x) = ( −0.7006 −0.0058 ) Example 8.5.3 (with large variability of radius)2 Let us consider input data A = fivediag(−1, −1, 4, −1, −1) ∈R12×12 b = Ay y = (2, 1, 0.5, 0, 0, 11, 10−5, −1, √ 2, −0.1, 4.1 ∗10−4, 143)T r = (2, 1, 0.5, 2, 10−3, 154)T eps = 10−4 Algorithm is over in 2 outer iterations. Iteration progress and progress of constraint functions values: −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 −1.5 −1 −0.5 0 0.5 1 1.5 progress of iterations due to g1(x) = x1 2 + x2 2 − 22 = 0 x1 x2 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −2.8 −2.6 −2.4 −2.2 −2 −1.8 −1.6 −1.4 values of g1(x) = x1 2 + x2 2 − 22 it g 2This example was introduced and solved in 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 43. −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5 progress of iterations due to g2(x) = x3 2 + x4 2 − 12 = 0 x3 x4 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −2 0 2 4 6 8 10 12 values of g2(x) = x3 2 + x4 2 − 12 it g −0.6 −0.4 −0.2 0 0.2 0.4 0.6 −0.4 −0.3 −0.2 −0.1 0 0.1 0.2 0.3 0.4 progress of iterations due to g3(x) = x5 2 + x6 2 − 0.52 = 0 x5 x6 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −0.025 −0.02 −0.015 −0.01 −0.005 0 0.005 values of g3(x) = x5 2 + x6 2 − 0.52 it g −2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 2.5 −2 −1.5 −1 −0.5 0 0.5 1 1.5 progress of iterations due to g4(x) = x7 2 + x8 2 − 22 = 0 x7 x8 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 values of g4(x) = x7 2 + x8 2 − 22 it g 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 44. −1 −0.5 0 0.5 1 x 10 −3 −8 −6 −4 −2 0 2 4 6 8 x 10 −4 progress of iterations due to g5(x) = x9 2 + x10 2 − 0.0012 = 0 x9 x10 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 x 10 −7 values of g5(x) = x9 2 + x10 2 − 0.0012 it g −150 −100 −50 0 50 100 150 −150 −100 −50 0 50 100 150 progress of iterations due to g6(x) = x11 2 + x12 2 − 1542 = 0 x11 x12 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −3284.38 −3284.375 −3284.37 −3284.365 −3284.36 −3284.355 −3284.35 values of g6(x) = x11 2 + x12 2 − 1542 it g Function value progress and progress of Lagrange function values: 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −4.0996 −4.0994 −4.0992 −4.099 −4.0988 −4.0986 −4.0984 x 10 4 values of Lagrange funcion L(x,λ) it L 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2 −4.1002 −4.1 −4.0998 −4.0996 −4.0994 −4.0992 −4.099 −4.0988 −4.0986 −4.0984 −4.0982 x 10 4 values of cost function f(x) it f 8 PROJECTED DUAL PROBLEM (PDP) ALGORITHM 45. We can verify our solution - induct solution into KKT conditions: KKT1err = Ax −b + 2 diag(˜ λ)x = 10−12.                      −0.0013 0.0007 0.0066 −0.0036 −0.4334 0.4405 0.0027 −0.0089 −0.0120 −0.0089 0 0                      g(x) = 103.         −0.0017 −0.0002 0.0000 −0.0000 0.0000 −3.2844         Inner iterations: • Initialization - number of CG iterations: 13 • 1. iteration - number of MPRGP iterations: 22 - number of CG iterations: 18 • 2. iteration - number of MPRGP iterations: 9 - number of CG iterations: 11 9 CONCLUSION 46. 9 Conclusion In this thesis, we used observations from simple algorithms to construct new very effective algorithm for solving problem of minimizing of quadratic function with separated quadratic constraints. We call it PDP. It represent a new way how to use Dual problem and projection to boundary of a set - it uses Inverse Dual problem to find corresponding update of Lagrange multipliers. It is probably the best update of Lagrange multiplier of previous iteration. First numerical tests imply good convergence, but proof was not constructed yet. Also precondition can improve number of inner CG and MPRGP iterations. In fact, contact problems imply minimizing problems with quadratic constraints, more-over linear equalities and inequalities. That is the reason, why PDP algorithm is useless in these cases. It has to be modified, probably using classic MPRGP algorithm. 10 REFERENCES 47. 10 References V. Vondr´ ak Numerical analysis I Syllabus VSB-TU Ostrava Z. Dost´ al Optimal Quadratic Programming Algorithms, with Applications to Variational In-equalities Springer, 2009. Z. Dost´ al, R. Kuˇ cera An optimal algorithm for minimization of quadratic functions with bounded spectrum subject to separable convex inequality and linear equality constraints MSM6198910027. R. Kuˇ cera Minimizing quadratic functions with separable quadratic constraints Optimization methods and software. 2007, vol. 22, issue 3, p. 453-467. R. Kuˇ cera Convergence rate of an optimization algorithm for minimizing quadratic functions with separable convex constraints SIAM J. Optim., Vol. 19, No. 2, pp. 846-862 J. Haslinger, R. Kuˇ cera, and Z. Dost´ al An algorithm for the numerical realization of 3D con-tact problems with Coulomb friction J. Comput. Appl. Math., 164-165 (2004), pp. 387–408. 10 REFERENCES 48. Appended CD includes these folders with matlab functions: • Chapter 3 – algorithm for generating figures in Chapter 3 • Chapter 4 – chapter4/cg - implementation of CG algorithm – chapter4/mprgp - implementation of MPRGP algorithm • Chapter 5 – chapter5/dual problem - figures in Section 5.2 – chapter5/invert dual problem - usage of Invert Dual problem • Chapter 6 – chapter6/constant update lagrange - constant update of Lagrange mul-tipliers – chapter6/sequence - sequence of Lagrange multipliers • Chapter 7 – chapter7/linear update - Linear update of Lagrange multipliers – chapter7/adaptive linear update - Adaptive linear update of Lagrange multipliers – chapter7/bisection - Bisection algorithm • Chapter 8 – chapter8/pdp_eq - implementation of PDP algorithm for Equality problem – chapter8/pdp_ineq - implementation of PDP algorithm for Inequality prob-lem
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https://ouhsc.edu/bserdac/dthompso/web/CDM/power/hypoth.htm
Hypothesis testing and power Hypothesis testing and statistical power All power and sample size calculations depend on the nature of the null hypothesis and on the assumptions associated with the statistical test of the null hypothesis. This discussion illustrates the core concepts by exploring the t-test on a single sample of independent observations. The null hypothesis Properties of the sample mean Sampling distributions Type 1 error and alpha (a) Rejection regions Factors that determine the location of the rejection regions Type 2 error and b b and power Power curves Download .txt files with SAS programs that create the graphs that appear on this page: sampling distributions power curves Online resources Russ Lenth's power and sample size page, including Java applets that explore influences of sample and effect sizes on power Power and sample size programs, University of California - San Francisco, including links to free programs Online calculator for Bonferroni adjustments (of alpha or z) for multiple comparisons. The null hypothesis A research hypothesis drives and motivates statistical testing. However, test statistics are designed to evaluate not the research hypothesis, but a specific null hypothesis. Therefore, researchers must begin by: specifying a null hypothesis (H 0) that relates to a population parameter. This requires knowing whether the outcome of interest can be summarized as, for instance, a mean, a count, or a proportion. For example, when we can measure the outcome variable at the interval or ratio scale, we can formulate a null hypothesis in terms of the population mean, which is designated by the greek symbol m. H 0: m=6 identifying a test statistic that relates to the hypothesized and unknown population parameter. In our example, which states a null hypothesis in terms of the population mean, a relevant test statistic is the t. calculating the test statistic (in this case, a t statistic) using sample data. Properties of the sample mean We calculate test statistics from information that we obtain from the sample. For example, we can calculate a t-statistic using the sample mean and sample variance. Although we collect just one sample, and therefore calculate a single sample mean, we understand that the sample that we have drawn is one of many that we might have drawn. In that respect, the sample mean is a continuous variable that could take on many values. Depending on the sample that we draw by chance, the mean's value could be anywhere on the illustrated number line. Somewhere on the number line is the true but unknown population mean m. To illustrate the relationship between the sample mean and the hypothetical but unknown population mean m, we add a second dimension to the "number line." This graph's vertical axis is a "second dimension" that illustrates the results we might obtain were we to draw many samples from a population. The vertical axis summarizes the frequencies with which we might obtain particular values for the sample mean. Common sense suggests that, if we collect a sample not once but many times, the samples' means would typically be close to, and often identical to, the population mean that forms the basis of the null hypothesis. However, we'll also collect samples whose means are smaller (like that of X 1) or larger (like that of X 2) than the true parameter. We'll occasionally collect a sample whose mean is quite different from the true value. We can be very specific about the relationship between the sample mean and the unknown population mean m if we can justify certain assumptions. In particular, if we can assume that we are measuring an outcome variable whose values are normally distributed, then statistical theory lets us state that the many samples that we might draw have means that are also normally distributed. To generate the graph below, we drew 10,000 samples, each with 10 observations, from a normal population of values with a known mean (m=6) and variance (s 2=2.5). The graph's vertical axis shows how often we randomly chose samples whose means equalled the values listed on the horizontal axis. The graph illustrates how, when this particular null hypothesis (H 0: m=6) is true, we will very often draw samples whose means are close to 6. In fact, statistical theory assures us that all these sample means will have a collective mean that exactly equals the population mean m. (This is true regardless of the population's distribution; it doesn't have to be normally distributed.) We expect a sample mean to equal, on average, the unknown population mean. E(xbar) = m where E refers to the statistic's "expected value." The graph illustrates that we might, by chance, collect samples whose means differ greatly from the true population mean of 6 (even though the probabilities of doing so are low.) Statistical theory predicts how much sample means will vary from their expected value. Var (xbar) = s 2/n In other words, the "sampling variance" of the sample mean variance depends on the population variance s 2 and on the number n of observations in the sample. The larger the sample, the smaller the variance, that is, the more precise our estimate of the population mean. Sampling distributions We can construct a graph of the sample means' distribution, like the one above, for any null hypothesis as long as we specify a population mean m 0 and variance s 2, and are confident in assuming that the variable of interest is normally distributed. Under these assumptions, every distribution looks vaguely alike; its shape and the location of its peak differ slightly depending on the hypothesized mean and variance. To eliminate this variability, we transform the sample means to a standard distribution like the t. Transforming sample information to a t value permits quick and consistent comparisons of samples from populations with different means and variances. Researchers are interested in sampling distributions, but not because they collect multiple samples. In practice, they generally collect a single sample for each combination of a study's independent variables. However, they understand that the they draw one sample out of many different samples that they might have drawn. Knowing the properties of sample means lets us relate any sample mean to the population's unknown mean and variance by using the t distribution. t = (xbar - m 0) / sqrt(S 2/n) (Because we don't know the population variance, we use the the sample variance S 2 to calculate the t-statistic. The graph below repeats the previous simulation, in which we drew 10,000 samples, each with ten observations, from a population of measurements that is normally distributed with a mean of 6 and a variance of 2.5. The horizontal axis represents, instead of sample means as in the previous simulation, the t-statistic calculated for each sample on the basis of its specific mean and variance. The graph illustrates that these t-statistics do, in fact, follow a t distribution, certain of whose values are tabulated in many statistics textbooks and online sources, including: The NIST/SEMATECH Engineering Statistics Internet Handbook.. distribution tables from StatSoft, Inc. Reviewing the equation that calculates t-values reveals that they are a ratio of two quantities: the difference between the sample and population means (xbar - m 0) the sample mean's standard deviation (sqrt(S 2/n)), also called the standard error of the mean | #### Definition Standard error: The standard deviation of a statistic that estimates a population parameter. Example Because the variance of a collection of sample means [Var(xbar)] estimates the population variance (s 2) through the relationship Var(xbar) = s 2/n and because the sample variance S 2 estimates s 2 then, Var(xbar) = S 2/n . The square root of this variance, is the standard deviation of the sample mean, also called the standard error of the mean (SEM), SEM = sqrt (S 2/n) | Examine the t statistic and you'll see that expresses the difference between the sample mean and the hypothesized population mean as a number of standard errors of the sample mean. The graph reveals that most sample means are close to the true population mean, within one standard error above or below the sample mean. Rarely does a sample mean differ from the true mean by two standard errors or more. Knowledge of the t distribution has produced tables that specify the probabilities of drawing samples whose means differ by various amounts from the true mean. Knowing these probabilities helps researchers decide whether the one sample they draw is consistent with the truth of the null hypothesis. This approach points up an important principle in hypothesis testing: WE BEGIN WITH THE ASSUMPTION THAT THE NULL HYPOTHESIS IS TRUE! Then, we draw a sample from the population. Next, we calculate a test statistic (like the t), from which we can calculate the probability that we obtained this sample if the null hypothesis is true. A corollary approach is to ask the question: How different must the sample mean be from the hypothesized mean m 0 before we suspect that the null hypothesis is not true, and decide to reject it? Type 1 error and alpha (a) The question is difficult to answer. Even sample means that are very different from the hypothesized mean are possible, just not very probable, when the null hypothesis is true. We must, therefore, accept the possibility that we could mistakenly reject the null hypothesis even when it's true. This type of mistake, a "type 1 error," is unavoidable. Researchers accept that they will occasionally commit type 1 errors when they examine the test statistics that they calculate from sample data. In practice, they "control type 1 error," that is, they specify the risk they are willing to take. Researchers customarily accept probabilities of committing type 1 errors of 0.05 or 0.01, designating whatever probability they elect with the symbol a. No rule exists, other than custom, to ordain the choice of a. Rejection regions We visualize the probability a as a portion or a region on a graph that illustrates the sampling distribution of the mean when the null hypothesis is true. The solid curve depicted below represents a particular t distribution, the one where df=n-1=9. The area under the curve represents the total probability that we might produce a given t-statistic. The area under the curve, by definition, is equal to one. That is because the graphs's horizontal axis illustrates every possible value for the t statistic that we might calculate for a given sample. The vertical axis shows the probability of obtaining any particular t-value. Every possible t statistic is accounted for, so the total probability is 1. Because the area under the t distribution's curve represents a probability of 1, regions under the curve represent probabilities that are proportional to the region's size. Two symmetrical (mirror-image) regions, one at the distribution's lower extreme and one at its upper extreme, together account for a=0.05 of the distribution's total probability. How do we interpret these "rejection regions," whose area equals the probability a? The regions identify t values (on the horizontal axis) that are relatively far from t's expected value of zero. Were we to draw many samples from a population, a (in this case, 5 percent) of the samples would have t-values that are this far from the expected value of zero. According to our knowledge of t distributions, t values that are equal to or less t=-2.26 or equal to or greater than t=2.26 are highly unusual; they occur in only five percent of samples of size 10 drawn from a population whose mean is 6. To illustrate the rejection regions' influence on our decisions, we "retransform" the graph's horizontal axis so that it shows sample means (xbar) instead of t values. Recall that we transformed the sample means' distribution to a t distribution through the equation: t = (xbar - m 0) / sqrt(S 2/n) Therefore, xbar = m 0 + t sqrt[S 2/n]To achieve this transformation, we have to assume that all of the sample variances are approximately equal to one another and, further, to the population variance s 2. xbar = m 0 + t sqrt[s 2/n] The critical values for t, -2.26 and 2.26, which define the borders of the rejection regions, are transformed to values for the sample mean of 4.87 and 7.13, respectively. xbar = m 0 +- t sqrt[S 2/n] xbar= 6 +- 2.26sqrt[2.5/10] xbar = 6 +- 1.13 = 4.87 , 7.13 If the null hypothesis (H 0: m=6) is true, we are unlikely (the chance is no more than than one in twenty) to draw a sample of n=10 whose mean is less than 4.87 or greater than 7.13. If we draw a sample whose mean is that large or that small, the sample is probably not part of a population with a mean of 6. Therefore, drawing a sample with such a mean justifies rejecting the null hypothesis. Alternatively, samples whose means fall between the critical values are more likely -- the probability is at least 0.05 -- to be part of a population whose true mean equals the one we've hypothesized. These samples do not justify rejecting the null hypothesis (H 0); when we draw such samples, we "fail to reject" H 0. Factors that affect the location of the rejection regions The equation that transforms t-values into values for "xbar," the sample mean: xbar = m 0 + t sqrt[S 2/n] suggests a way to specify the "critical" values for xbar (xbar c) that mark the rejection regions' boundaries: xbar c = m 0 +- t (a/number of tails), n-1 sqrt[S 2/n] The equation illustrates how the critical values for xbar depend on: m 0, the population mean that we specify under the null hypothesis t, a quantity determined by three a priori choices: (1) the value for a, (2) whether the hypothesis is one-tailed (directional) or two-tailed (non-directional), and (3) the choice of sample size n. S 2, the sample variance, which we must assume equals s 2, the population variance under the null hypothesis n, the sample size Effect of the choice of a The researcher chooses a value for a, the probability of making a Type 1 error. The risk of making such an error is part of the cost of making a decision. The larger the type 1 error that we accept, the larger the rejection region. | #### In our current example (H o: m 0=6, s 2=2.5), we accepted a type 1 error of 0.05 and calculated two sample means, 4.87 and 7.13, that formed boundaries for the rejection regions. | | #### We could, before drawing a sample, accept a larger risk of committing a type 1 error. We could set a at 0.20, a one in five chance of making a type 1 error. (Admittedly, 0.20 may be a larger probability than we'd accept comfortably in practice.) | | #### Increasing a enlarges the rejection region. Inserting this value for a into the equation yields critical values of 5.308 and 6.692 for the sample means that bound the rejection regions. | | | #### When we accept a larger a, we are more likely to draw a sample mean that leads us to reject the null hypothesis. Of course, we are also more likely to reject the null mistakenly. | Effect of the size of the population variance, and of the sample variance that estimates it The more diverse a population, the more likely that it contains members whose measures on some variable are relatively far from the population mean. Similarly, means of samples drawn from a highly variable population may lie relatively far from the true mean. When we draw a sample from such a diverse population, we can reject a null hypothesis only if the sample's mean (xbar) differs considerably from the hypothesized mean m 0. | #### Assuming the truth of our "model" null hypothesis (H 0: m=6, s 2=2.5), and accepting a type 1 error probability (a) of 0.05, specifies rejection regions bounded by sample means of 4.87 and 7.13. If we draw a sample of 10 observations to test a slightly different null hypothesis, one in which we estimate the population variance to be 16, ... | | #### the equation yields values of 3.139 and 8.861 for sample means that bound the rejection regions. | | | #### Thus, the larger our estimate of a populations' variance, the more a sample mean must differ from the hypothetical mean m 0=6 before we can reject the null hypothesis. | Effect of n, the size of the sample that we use to make a statistical decision | #### Up to now, we have tested a specific null hypothesis (H 0: m 0=6; s 2=2.5) by setting a at 0.05 and drawing a sample of 10 observations. Drawing a larger sample, say one with 50 observations, estimates more precisely the mean in the population from which we've drawn the sample. The larger the sample, the smaller the sample mean's standard error. When we estimate the population mean m 0 more precisely, a sample mean need not be as distant from the hypothesized mean m 0 to cause us to reject the null hypothesis. | | #### When we draw a sample of 50, our equation yields critical values of 5.551 and 6.449 for the sample means that bound the rejection regions. | | | #### These values are not as distant from the m 0 of 6 as the values for xbar (4.87 and 7.13) that we calculated for a sample of 10. | Type 2 error and b We have examined how researchers establish a null hypothesis, and identified a test statistic t whose value they evaluate to determine whether it inhabits a rejection region. To calculate the rejection regions' endpoints: xbar = m 0 +- t(a/number of tails), n-1 sqrt[S 2/n] we must specify a priori: the population mean m 0 under the null hypothesis a, an acceptable risk of committing a type 1 error the sample variance S 2, which we assume equals the population variance s 2 under the null hypothesis the size n of the sample that we can collect to calculate the test statistic We can also consider the probability b that we'll fail to reject the null hypothesis even though it's untrue. This second variety of mistaken statistical decision is a "type 2 error." The table below employs two columns to illustrate two "states of truth," one in which the null hypothesis is true, and another in which a specific alternative hypothesis (H a) is true. Against the columns, the table lists two rows that describe the decision, based upon the test statistic, to reject or "fail to reject" the null hypothesis. | #### Decision (made on basis of test statistic): | #### H 0 is true | #### H a is true | --- | #### reject H 0 | #### Type 1 error | #### correct decision | | #### fail to reject H 0 | #### correct decision | #### Type 2 error | We can calculate b, the probability of making a Type 2 error, if we can specify an alternative hypothesis H a in terms of a population mean and variance, and if we can assume that the population follows a normal distribution. We understand the logic in two steps. After drawing a sample, we first decide whether its mean is consistent with the truth of the null hypothesis (H 0: m 0=6; s 2=2.5). The figure illustrates that we retain (fail to reject) the null hypothesis if xbar is equal to or between the critical values we have calculated. If xbar is outside that region, we reject the null hypothesis. Second, we consider the probability that we will incorrectly fail to reject H 0 when a specific alternative hypothesis H a is true. Note that we must specify an alternative hypothesis. The figure below illustrates (shaded in red) the probability of incorrectly failing to reject the null (thereby committing a type 2 error) when a specific alternate hypothesis (H a: m a=8; s 2=2.5) is true. The curve immediately above represents the distribution of sample means (also called the "sampling distribution of the mean") under the specific alternative hypothesis. The red-shaded region accounts for 5.8 percent of the area under the curve. In other words, if the specific alternative hypothesis is true, the probability b of making a type 2 error is 0.058. b and power If b is the probability of making a type 2 error, then (1-b) is the probability of correctly rejecting the null hypothesis (H 0) when a specific alternative hypothesis (H a) is true. A test's ability to correctly reject the null, when an alternate hypothesis is true, is its power. When the alternate hypothesis is specified, power equals (1-b). Under the alternative hypothesis illustrated in the graph above (H a: m a=8), given a sample of ten observations whose variance is 2.5, the test's power is 1-0.058=0.942. A test's power depends on the specific alternative hypothesis. If we specify H a such that m a=8.5 and s 2=2.5, and draw a sample of ten, the sample mean's power to correctly reject the null is: (1-b) = 1 - 0.0114 = 0.98856 We can specify a different alternative hypothesis whose mean is closer to that of the null: H a: m a = 4.8; s 2=2.5. In this case, when we draw a sample of ten, the sample mean's power to correctly reject the null hypothesis is smaller: (1-b)= 1 - 0.446 = 0.554 Power curves The graphs illustrate that a test's power depends on the relative locations of: the "fail to reject region." The location of the rejection regions and, therefore, of the "fail to reject region," depends on the specific null hypothesis and on factors previously discussed. m 0, the population mean that we specify under the null hypothesis t, a quantity determined by three a priori choices: (1) the value for a, (2) whether the hypothesis is one-tailed (directional) or two-tailed (non-directional), and (3) the choice of sample size n. S 2, the sample variance, which we have to assume equals s 2, the population variance under the null hypothesis n, the sample size the sampling distribution of means under the alternative hypothesis. Power curves illustrate the effect on power of varying the alternate hypothesis. The curve illustrates how a sample of ten observations with a variance of 2.5 is quite powerful in correctly rejecting the null hypothesis (in this example, m 0=8) when the true mean is less than 6 or greater than 10. The curve also illustrates that the test is not powerful -- it may not reject the null hypothesis even when the true mean differs from m 0 -- when the difference is small. Increasing the sample size improves a test's power, as the graph below illustrates. Larger samples have the power to reject the null hypothesis correctly even when the true and hypothesized means differ very little. Last updated 7-1-2009 © Dave Thompson
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https://www.youtube.com/watch?v=FBKklFfRAC4
Translating Word Problems into Systems of Equations Professor Monte 1640 subscribers 3 likes Description 80 views Posted: 19 Mar 2024 This video uses an example to teach how to translate a word problem application into a system of equations. The problem is not solved in this video, it is merely an example of setting the problem up. The example used is a classic money (coin) problem using quarters and dimes. The basic steps are as follow: Name your variables. Write your two equations in words. Translate your word equations into algebraic equations. This process can be fairly simple and quick, following these steps to make finding the equations as easy and straightforward as possible. 1 comments Transcript: Hi there, it's Professor Monte, and we're going to translate word problems into systems of equations. Okay, so the steps, first, we're going to name our variables. Since we have a system of equations, we're going to have at least two variables. Oftentimes, they're just going to do two variables. Assuming there's two variables, we're going to write two word equations, and then we're going to translate the word equations into algebraic equations. We're not solving at this point. We're just translating from words into algebraic equations and a system of equations. All right, so here's our problem. "We have a bag of dimes and quarters. There's 30 coins, and the total value is $5.70. How many of each type of coin are there?" Okay, so we're looking for the number of dimes and the number of quarters. So I'm going to let x equal the number of dimes. I'm going to let y equal the number of quarters. All right, so that's what I'm getting is those are my two variables. Name your variables. Write two word equations. So let's figure out how to get these to make sense. Well, first of all, they say there's 30 coins. So my first word equation is going to be number of dimes (marker squeaking) plus number of quarters (marker squeaking) equals total coins. So we write equals, equals total coins. (marker squeaking) Okay, that's one equation. The other equation is going to be the value is 5.70. So the value of dimes (marker squeaking) plus the value of the quarters (marker squeaking) is going to equal the total value. (marker squeaking) All right, so let's go from there. Now let's look at the dimes plus quarters. Number of dimes plus number of quarters equals total coins. I'm just going to slip it right in here in between. Because the number of dimes, we know that's just x plus number of quarters, y, equals total coins. Well, there were 30 coins. That's one of my equations. I found if I write this out in words first, it's a lot easier to get the algebraic equation. Okay, so I've got x plus y equals 30, and that's for number of coins. (marker squeaking) But they also tell me the value, so now I need an equation for the value. Well, the value of dimes, remember a dime's 10 cents, or in dollars, it's 0.10. And a quarter's 25 cents, (marker knocking) or in dollars 0.25. Okay, I'm going to leave it in dollars since the answer's in dollars. If I wanted to just change it to cents, I'd call this 10 cents and 25 cents. But I'd have to call that 570 cents. And that's one thing you could do, but a lot of instructors want you to actually go in terms of money in dollars. So the value of the dime, well, remember, they're each worth 10 cents. But the total value of the dimes isn't 10 cents. It's 10 cents times the number of dimes. Think about if you were doing this yourself. You would count the number of dimes, you'd multiply by 10 cents. Oh, the number of dimes, that was x, times 10 cents. That's going to give you the total number you have in dimes. Plus, you're going to do the quarters, 25 cents, times however many of those there are, and that's going to give you the total, $5.70. And that's your other equation. 0.10x (marker squeaking) plus 0.25y equals 5.70. And that's the value of the coins. And so now we've got the two equations. We'll talk about in later videos how to solve the system of equations. There's different ways to do it, but once we get it set up, then we're ready to solve, and hopefully that helps you get it set up. My big advice to you, once you've named your variables, is write these two equations just in words. That's a lot easier for us to do, and then translate them into the algebraic equations.
8487
https://www.theropes.nyc/posts/three-pipes/
Math Ticker: Three pipes and a tank Math Ticker HomeAboutTutorialsAuthor☰ Three pipes and a tank problem solvingratiowork problemGMAT How to avoid fractions in work problems. Authors Affiliations Published Feb. 9, 2021 DOI Two pipes A and B can fill a tank in 60 min and 90 min respectively. A leak is present at 3 4 3 4 of the height. The leak alone takes 36 min to empty till 3 4 3 4 height of the tank. Find the time taken to fill the tank when all of the taps and the leak are opened simultaneously. 39 min 27 min 37 min 33 min 45 min If you calculate per-minute rates, you get fractions you’d like to add. There are two ways to avoid them. Introduce a probably fictitious unit for the work to be done so that the rates become integer-valued. Here we want to divide by 60, 90, 36, and 4. A common multiple of these numbers will do the trick. Therefore, you can assume that the tank contains 180 units of water (gallons, fl oz, pints, pips, whatever). You don’t have to calculate rates to compare or add productivities. It’s only important to let the pipes work the same amount of time. So let’s solve the problem using the latter strategy. Pipes A and B work together. To combine the two, we let them operate 180 minutes, a multiple of 60 and 90, and get A+B−→−−−180 min(3+2)tanks=5 tanks A+B→180 min(3+2)tanks=5 tanks The combination AB of the two pipes fills the tank in 36 minutes. The leak C empties 1/4 of the tank in 36 minutes, or the equivalent of 1 tank in 144 minutes. We combine the pipes C and AB by letting them work 144 minutes each, a multiple of 36: A B−C−→−−−144 min(4−1)tanks=3 tanks A B−C→144 min(4−1)tanks=3 tanks The combination ABC of the three pipes fills (the equivalent) of 1 tank in 48 minutes. Since AB supplies 3/4 of the tank and ABC the remaining 1/4, it takes 3 4×36 min+1 4×48 min=39 min 3 4×36 min+1 4×48 min=39 min to fill the tank. Footnotes
8488
https://mathoverflow.net/questions/497596/truth-of-a-conjecture-about-the-largest-incircle-of-convex-polygons
convexity - Truth of a conjecture about the largest incircle of convex polygons - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community MathOverflow helpchat MathOverflow Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Truth of a conjecture about the largest incircle of convex polygons Ask Question Asked 2 months ago Modified2 months ago Viewed 157 times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. The problem of determining the largest incircle of a convex polygon can be solved by constructing the Voronoi diagram of the polygon's edges and selecting the vertex of the diagram's graph with maximal distance to the boundary; Voronoi diagrams of convex polygons can be calculated in O(n)O(n) time. I was however looking for more "direct" methods and came up with the following: We can easily construct a tangential polygon T T with the same set of side-normals as the given convex polygon C C. Assume now that in both cases the edge lengths sum to 1 and side-normals point away from the polygon, I conjecture that: if side-normals of c i∈C c i∈C and of t i∈T t i∈T are parallel but the length of c i c i is smaller than that of t i t i then c i c i can't be tangent to C C's incircle. Questions: is the above conjecture true, resp., what are counter examples if the conjecture is true does that imply that no side of C C that is shorter than its counterpart in T T can't be tangent fo C C's incircle does the conjecture generalize to higher dimensions if we replace 'length' with the volume of facets Edit: looking at non-tangent edges is apparently the wrong thing to look at; it should rather be the edges that are tangent and adjacent to a non-tangent edge. Let now, as before, assume that the edge-lengths of the tangential polygon and of the non-tangential polygon sum to 1 and, that c i∈C c i∈C and t i∈T t i∈T have parallel normals for all sides of the polygons. I now conjecture: c i t i≥c j t j∀(c t,t j)⟹c i∩D¯≠∅c i t i≥c j t j∀(c t,t j)⟹c i∩D¯≠∅ when D¯D¯ is the closed disk of the largest incircle of C C Which reads that the side with the highest relative "growth" is tangent to the incircle and, as a consequence, that further tangents can be identified by removing c i c i and t i t i from the edgesets, rescaling the weight sums to 1 and applying the maximal "growth" criterion again. convexity euclidean-geometry polygons Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited Jul 13 at 15:51 Manfred WeisManfred Weis asked Jul 12 at 14:10 Manfred WeisManfred Weis 13.8k 4 4 gold badges 38 38 silver badges 80 80 bronze badges 3 I would restate the conjecture as "Suppose C C and D D are convex polygons with the same perimeter and the same set of outward-facing side-normals. Then any side of C C tangent to its largest incircle is at least as long as any side of D D with the same normal."Matt F. –Matt F. 2025-07-12 15:45:36 +00:00 Commented Jul 12 at 15:45 @MattF. please feel free to edit accordingly; your restatement seems superior from a mathematical point of view Manfred Weis –Manfred Weis 2025-07-12 17:00:15 +00:00 Commented Jul 12 at 17:00 @MattF. The OP added that D D is tangential, and I think it is essential.Ilya Bogdanov –Ilya Bogdanov 2025-07-13 12:18:18 +00:00 Commented Jul 13 at 12:18 Add a comment| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. I think that even if T T is tangential, this does not hold. Let T T be a tangential heptagon, number its sides clockwise. Its bottom (first) side is horizontal, and four opposite sides (with numbers 3,4,5,6) touch the incircle at points close to the topmost point (two tangency points are a bit to the left, and two to the right); the other two sides (2nd and 7th) are, say, vertical. Shrink T T to half its size, and then move the vertical sides to the left and to the right in order to preserve the perimeter; you obtain C C, with the largest incircle being the shrinked incircle of T T. Now, the 4th and 5th sides of C C are tangent to its largest contained circle, though they are half as long as the corresponding sides of T T. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Jul 13 at 15:43 Matt F. 659 3 3 silver badges 14 14 bronze badges answered Jul 13 at 12:29 Ilya BogdanovIlya Bogdanov 24.7k 56 56 silver badges 96 96 bronze badges 4 Yes, you can construct a hexagon based on the same idea.Ilya Bogdanov –Ilya Bogdanov 2025-07-13 12:30:08 +00:00 Commented Jul 13 at 12:30 could you please provide point coordinates for both heptagons, one set for the case when all sides are tangent to the largest incircle and another set for the case when at least one side is not tangent to the largest incircle?Manfred Weis –Manfred Weis 2025-07-13 13:20:40 +00:00 Commented Jul 13 at 13:20 Sorry, I do not want to compute such coordinates; I hope the idea is clear from the figure I’ve just added. Notice that you can make C C as wide as you wish…Ilya Bogdanov –Ilya Bogdanov 2025-07-13 14:21:00 +00:00 Commented Jul 13 at 14:21 1 ok thanks; the image suffices to check whether the conjecture fails in that case...Manfred Weis –Manfred Weis 2025-07-13 14:26:15 +00:00 Commented Jul 13 at 14:26 Add a comment| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions convexity euclidean-geometry polygons See similar questions with these tags. 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8489
https://sites.math.rutgers.edu/~zeilberg/EM18/TitchmarshZeta.pdf
THE THEORY OF THE RIEMANN ZETA-FUNCTION BY E. C. TITCHMARSH F.R.S. FORMERLY SAV!LlAN PROFESSOR OF GEOMETRY IN THE UNIVERSITY OF OXFORD SECOND EDITION REVISED BY D. R. HEATH-BROWN FELLOW OF MAGDALEN COLLEGE, UNlVERSlTY OF OXFORD CLARENDON PRESS· OXFORD 1986 Oxford University Press, Walton Street, Oxford OX2 6DP Oxford New York Toronto Delhi Bombay CalcutW Madra$ Karachi Petaling Jaya Singapore Hong Kong Tokyo Nairobi Dar es Salaam Cape Town Melbourne Auekland and associated companies in Beirut Berlin Ibadan Nicosia Oxford is a trade mark of Oxford Umversity Press Published in the United SWtes by Oxford University Press, New York ©The E:ucutors of the late Mrs K. Titchmarsh and D. R. Heath-Brown, 1986 All rights reserved. No part of this publication may be reproduced, stored in a retrieual system, or transmitted, in any form or by any means, electronic, ';;;';~~~f'm':sW:OO:~jns~f';;'t~~~;~t~~~:e, without This book is sold subject to the conditwn that 11 shall not, by way of trade or otherwise, be lent, re-sold, hired out or otherwise circulated without the publisher's prior consent in any form of binding or cover other than that in which it rs published and without a similar cond1tion mcludmg this condition being impoud on the subaequent purchaser British Library Cataloguing in Publication Data Titchmarsh, E. C. The theory of the Riemann zeta-function. 1. Calcul/Ul 2. Functions, Zeta 3. Riemann-Hilbert problems I. Title 515.9'82 QA320 ISBN0-19-853369-1 Library of Congress Cataloging in Publication Data Titchmarsh, E. C. (Edward Charles), 1899-The theory of the Riemann zeta-functwn. Bibliography: p. 1. Functions, Zeta. I. Heath-Brown, D. R. II. Title. QA246.T44 1986 512'.73 86-12520 ISBN 0-19-853369·1 (pbk.) Set by Macmillan India Ltd., Bangalore 560 025 Printed in Great Britain by Biddies Ltd. Guildford and King's Lynn PREFACE TO THE SECOND EDITION SINCE the first edition was written, a vast amount of further work has been done. This has been covered by the end-of-chapter notes. In most instances, restrictions on space have prohibited the inclusion of full proofs, but I have tried to give an indication of the methods used wherever possible. (Proofs of quite a few of the recent results described in the end of chapter notes maybe found in the book by I vic .) I have also corrected a number of minor errors, and made a few other small improvements to the text. A considerable number of recent references have been added. In preparing this work I have had help from Professors J. B. Conrey, P. D. T. A. Elliott, A. Ghosh, S.M. Gonek, H. L. Montgomery, and S. J. Patterson. It is a pleasure to record my thanks to them. OXFORD 19116 D.R.H.-B. PREFACE TO FIRST EDITION THIS book is a successor to my Cambridge Tract The Zeta. Function of Riemann, 1930, which is now out of print and out of date. It seems no longer practicable to give an account of the subject in such a small space as a Cambridge Tract, so that the present work, though on exactly the same lines as the previous one, is on a much larger scale. AB before, I do not discuss general prime-number theory, though it has been convenient to include some theorems on primes. Most of this book was compiled in the 1930's, when I was still researching on the subject. It has been brought partly up to date by including some of the work of A. Selberg and of Vinogradov, though a great deal of recent work is scantily represented. The manuscript has been read by Dr. S. H. Min and by Prof. D. B. Sears, and my best thanks are due to them for correcting a large number of mistakes. I must also thank Prof. F. V. Atkinson and Dr. T. M. Fleet for their kind assistance in reading the proof-sheets. OXFORD 1951 E.C.T. I CONTENTS I. THE FUNCTION ((s) AND THE DIRICHLET SERIES RELATED TO IT Definition of '(s). Dirichlet series connected with ((s). Sums involving CJ 0 (n). Ramanujan's sums. II. THE ANALYTIC CHARACTER OF ((s), AND THE FUNCTIONAL EQUATION AnalytiC COntinuation. Functional equation. The functiOilliJ:(B), e(s), and E(s). Values at negative integers. Siegel's proof of the functional equation. Zeros and the Hadamard product. Hamburger's theorem. Selberg's method, Eisenstein series and the Maass-Selberg formula. Tate's thesis. The p-adic zeta functions. III. THE THEOREM OF HADAMARD AND DE LA VALLEE POUSSIN, AND ITS CONSEQUENCES Methods of Hadamard, de la Vallee Poussin, and Ingham. The prime number theorem. Quantitative zero-free regions and bounds for ((s)-', C'(s)JC(s). Elementary estimates. Sieve methods. Proof via the Maa86--Seibel'g formula. IV. APPROXIMATE FORMULAE Estimates for exponential integrals. The approximate functional equation. The Riemann-Siegel formula. Weighted approximate func tiona! equations. The approximate functional equation for C(s)2. 13 45 71 V. THE ORDER OF ((s) IN THE CRITICAL STRIP 95 Thefunctionp(u). Weyl'smethod. Vander Corput'smethod. An improved rero-free region. Other methods for bounding p(i). The method of exponent pairs. Multiple sums and further work on /l( i>· VI. VINOGRADOV'S METHOD 119 Vinogradov's mean-value theorem. Vinogradov's zero-free region. Application to exponent pairs. The Vinogradov-Korobov estimate. VII. MEAN-VALUE THEOREMS The mean square. The fourth-power moment. Higher moments. Convexity bounds. Fractional moments. Weighted mean-values. The error term in the fonnula for the mean square. Atkinson's formula. The twelfth-power moment. More on fractional moments. Applications of Kloosterman sums. The mean-value theorems of lwaniec, and of Deshouillers and lwaniec. VIII. 0-THEOREMS Results for ((l+it) and 1/W+it). The range!,;; u < 1. Sharper bounds. 138 184 CONTENTS IX. THE GENERAL DISTRIBUTION OF ZEROS The Riemann-von Mangoldt formula. The functiollll S(t) and S 1(t). Gaps between ordinates of zeros. Estimates for N(a, T). Selberg's bound for f ~ N(u, T)du. Mean-valuesofS(t) andS 1(t). More on the distribution of the ordinates of zeros. Sign changes and distribution of S(l). Further work on N(a, T). Halilsz' lemma and the Large Values Conjecture. 210 X. THE ZEROS ON THE CRITICAL UNE 254 Riemann's memoir. The existence of an infinity of zeros on a = f. The function N0(T). The Hardy-Littlewood bound. Selberg's theorem. Functions for which the Riemann hypothesis fails. Levinson's method. Simple zeros. Zeros of derivatives e(m)(s). XI. THE GENERAL DISTRIBUTION OF VALUES OF ((s) 292 Values taken by ((a+ it). The case a> 1. The case f <a< 1. Voronin's universality theorem. The distribution of log((!+ it). XII. DIVISOR PROBLEMS 312 Basic results. Estimates for"~· The casek = 2. Mean-value estimates and 0-results. Large values of k. More on the case k = 2. More 0-results. Additional mean-value theorems. XIII. THE LINDELOF HYPOTHESIS 328 Necessary and sufficient conditions. Mean-value theorems. Divisor problems. The functions S(t), S 1(t), and the distribution ofzeros. XIV. CONSEQUENCES OF THE RIEMANN HYPOTHESIS 336 Deduction of the Linde)(){ hypothesis. The function v(u). Sharper bounds for C(s). The case <1 = 1. The functiona S(t) and S 1(t). Bounds for{(s) with" near f· Mean-value theorems for S(t) and S,(t). The function M(K). The Mertens hypothesis. Necessary and sufficient con-ditions for the Riemann hypothesis. More on S(t), S 1(t), and gaps between zeros. Montgomery's pair-correlation conjecture. Disproof of the Mertens conjecture, and ofTuran's hypothesis. XV. CALCULATIONS RELATING TO THE ZEROS 388 Location of the smallest non-trivial zero. Results of computer calculations. ORIGINAL PAPERS 392 FURTHER REFERENCES 406 e, THE FUNCTION (s) AND THE DIRICHLET SERIES RELATED TO IT 1.1. Definition of b(.s). The Riemann zeta-function {(.s) has its origin in the identity expressed by the two formulae ~I {(.s)=f:JI."' where n runs through all integers, and (1.1.1) (1.1.2) where p runs through all primes. Either of these may be taken as the definition of b(s); sis a complex variable, s = o+it. The Dirichlet series (l.l.l) is convergent for u > l, and uniformly convergent in any finite region in which a;;-;::: 1+8, 8 > 0. It therefore defines an analytic func-tion b(.s), regular for a > I. The infinite product is also absolutely convergent for a > 1; for so is this being merely a selection of terms from the series I n-o-. If we expand the factor involving pin powers of p-•, we obtain lJ('+i-+p'..+·.J. On multiplying formally, we obtain the series (1.1.1), since each integer n can be expressed as a product of prime-powers pm in just one way. The identity of ( 1.1.1) and ( 1.1.2) is thus an analytic equivalent of the theorem that the expression of an integer in prime factors is unique. A rigorous proof is easily constructed by taking first a finite number of factors. Since we can multiply a finite number of absolutely con-vergent series, we have TI('+~+-'..+···) ~ '+-'.+-'.+···· Ptt:.P p' p nl nll where n1, n2, •• , are those integers none of whose prime factors exceed P. THE FUNCTION ((8') AND Chap. I Since all integers up to P are of this fonn, it follows that, if {(8) is defined by (1.1.1), 1"·)- JJ(~-~n = I"·H-~-~-···1 I I <;; (P+I)"+(P+2)"+···· This tends to 0 asP-+ oo, if u > I; and (1.1.2) follows. This fundamental identity is due to Euler, and (1.1.2) is known as Euler's product. But Euler considered it for particular values of 8 only, and it was Riemann who first considered {(s) as an analytic function of a complex variable. Since a convergent infinite product of non-zero factors is not zero, we deduce that {(s) has no zerw for u > I. This may be proved directly as follows. We have for u > I (~-~)(~-M-··(1--l"·) = l+d:+~+···· where m1 , ~ .... , are the integers all of whose prime factors exceed P. Hence 1(~-~)···(l--l((•)l;;, I-(P~I)"-(P~2)'-···> 0 if Pis large enough. Hence l{(s)l > 0. The importance of {(s) in the theory of prime numbers lies in the fact that. it combines two expressions, one of which contains the primes explicitly, while the other does not. The theory of primes is largely concerned with the function 11(x), the number of primes not exceeding x. We can transform (1.1.2) into a relation between {(8) and 11(x); for if u>l, log((•) =- Liog(I--) =-i {tr(n)-tr(n-I)}log(I-~) p p n~ll =-%. tr(n)(log(l-~)-log(l- (n~lr)) = i tr(n) •s+>-8 -dx = 8 s· __::&.dx. (1.1.3) ,. 2 ,. x(X"-1) 11 x{X"-1) The rearrangement of the series is justified since 11(n) ...:;:;; n and log( I-n-•) = O(n-•). j THE DIRICHLET SERIES Again' and on carrying out the multiplication we obtain I ~l'(n) «8) = 61 '7/i' (a> I), (1.1.4) where .u(l) = I, .u(n) = (-I)k if n is the product of k different primes, and .u(n) = 0 if n contains any factor to a power higher than the first. The process is easily justified as in the case of {(s). The function p{n) is known as the MObius function. It has the property Il'(d) =I (q =I), 0 (q > 1), (1.1.5) where d I q means that dis a. divisor of q. This follows from the identity I= i.1i"~~) = i.!. L"(d) m~l n-1 q-lrf' djq It also gives the 'MObius inversion formula' g(q) =];; f(d), (1.1.6) f(q) = .L "(~)g(d), .. (1.1.7) connecting two functions f(n), g(n) defined for integral n. Iff is given and g defined by (1.1.6), the right-hand side of (1.1.7) is The coefficientofj(q) is .u(l) =I. Ifr < q,then d = kr,where klqfr. Hence the coefficient of f(r) is .L "(lr) = .L "'k') = o kjqjr k'lqfr by (1.1.5). This proves (1.1.7). Conversely, if gis given, andfis defined by (1.1.7), then the right-hand side of (1.1.6) is .L.L"(~g(r). dlq rid and this is g(q), by a similar argument. The formula may also be THE FUNCTION '(s) AND Chap. I derived formally from the obviously equivalent relations F(s)((s) ~ ~ g(n), F(s) ~ 1_ ~ g(n) L na "s)L. n•' where n-1 F(s) = ~ f(n). n-l 6i n" Again, on taking logarithms and differentiating (1.1.2), we obtain, for cr>l, (1.1.8) where A(n) = logp if n is p or a power of p, and otherwise A(n) = 0. On integrating we obtain 1og{(s) ~ ~A,(n) (a> 1), (1.1.9) 6 n• where A1(n) = A(n)flogn, and the value of log "s) is that which tends to 0 as cr ~ oo, for any fixed t. 1.2. Various Dirichlet series connected with {(a). In the first place "' ('(s) ~ "d(n) (a> 1), (1.2.1) ,6_n• where d(n) denotes the number of divisors of n (including I and n itself). For "" "" "" 1 l'(s)~ 2.t.2~~ 2n- 2 1• ,.-11-l v~l n~l ,.,-.. and the number of terms in ~he last sum is d(n). And generally {k(s) = ~ d~:) (a > 1), (1.2.2) wherek = 2, 3, 4, ... , anddk(n)denotesthenumberofwa.ys of expressing n as a product of k factors, expressions with the same factors ~n a different order being counted as different. For l'(s)~ i~ ... i~~ i~ 2 I, .. ,~1 1 .. ~-1 II: n-1 .. , .. .I'Jc-• and the last sum is d.,(n). t l THE DIRICHLET SERIES Since ;we have also ,.(,) ~ IJ(~-f.r~ IJ(~+~+t~+ .. .J. (1.2.3) on comparing the coefficients in (1.2.1) and (1.2.3) we verify the elementary formula d(n) ~ (m,+l) .. (m,+l) for the number of divisors of (1.2.4) n = P1''PW'· .p~. (1.2.5) Similarly from (1.2.2) d ( ) ~ (k+m,-1)! (k+m,-1)1 ( 2 ) kn m1!{k-I)! .. 'm,!{k-I)T' 1.'.6 We next note the expansions M ~ ~ l~(n)l (a> 1), {(2s) ~1 n8 where p.(n) is the coefficient in (1.1.4); {2(s) = ~ 2 .. (ul (u > 1), {(2s) 6 1 n~ where v(n) is the number of different prime factors of 11; and l'(s) ~ ~d(n') ( l) {(2s) 61 na u > , l'(s) ~ ~ (d(n)}' (a> I). {(2s) 61 ns To prove (1.2.7), we have (1.2.7) (1.2.8) (1.2.9) (1.2.10) and this differs from the formula for I/{(s) only in the fact that the signs are all positive. 'l'he result is therefore clear. To prove (1.2.8), we have THE FUNCTION ((s) AND and the result follows. To prove (1.2.9), {'(•) TI 1-p-" TI l+p~ ~(2s) = P (l-p--3)8 = P (I-p~)11 ~ u: {(l+p-')(1+2p~+3p-"+···l} ~ Q: {1+3p-'+···+(2m+l)p--+ .. }. and the result follows, since, if n is (1.2.5), d(n') ~ (2m,+l) ... (2m,+l). Similarly ''(•) TI 1-p-" TI l+p~ {(2s) = P (1-p-s)' = P (1-p_,)a Chap. I ~ Q: (l+p-•)(1+3p~+-··H(m+l)(m+2)p-~+ ... } ~ Q: {1+4p~+ .. +(m+i)'p-~+ ... }. and (1.2.10) follows. Other formulae are {(2s) ~ ~ A(n) (a> !), {(s) ,;f-1 n" (1.2.11) where A(n) =(-It if n has r prime factors, a factor of degree k being counted k times; (1.2.12) where r/J(n) is the number of numbers less than n and prime to n; and 1 - 2'~ {(s-1) ~ ~ a(n) (a> 2) (1.2.13) l-2-,6_ n" ' where a(n) is the greatest odd divisor of n. Of these, (1.2.11) follows at once from {(2•)-TI(J-p-·)-TI(-~ )-TI -~ ,. {(s) -P 1-p-18 -P I+p-s -P (l P +P ... ). Also THE DIRICHLET SERIES and (1.2.)2) follows, since, ifn = PT'···P~. Finally = 1-2_, 1-31-•I-51_,"" and (1.2.13) follows. Many of these formulae are, of course, simply particular cases of the general formula wherej(n) is a multiplicative function, i.e. is such that, if n = p~•p;>• ... , then /(n) ~ /(p'{")/(p'l.") ... Again, Ietfk(n) denote the number of representations of nasa product of k factors, each greater than unity when n > I, the order of the factors being essential Then clearly ~ /,(n) ~ {{(s)-1}' (a> 1). ~ n• (1.2.14) Let j(n) be the number of representations of n as a product of factors greater than unity, representations with factors in a different order being considered as distinct; and letj(l) = l. Then f(n) ~ ~ /,(n). ,, Hence ~/(n) • , {(s)-1 ,6, -..~ l+,~,ms)-1} ~ 1+ 1{{(,)-l} I ~2-{(s)" (1.2.15) It is easily seen that ~(8) = 2 for 8 = a, where a is a real number greater than I; and jC(8)1 < 2 for u >a, so that (1.2.15) holds for u >a. THE FUNCTION C(~r) AND Chap. I 1.3. Sums involving, u,.(n). Let u0 (n) denote the sum of the ath powers of the divisors of n. Then i.e. ((s){(s-a) ~ ~.!, ~ ~ ~ ~ ~ L "'· p.-1p. v~l •-1 JI, u > R(a)+I). 6. n" Since the left-hand side is, if a ;;j::. 0, IJ ('+~+~+ .. -)('+~+~+ .. -) (1.3.1) ~ IJ(I+l;;'"+l+p~+P"' + .. -) ~ IJ(I+l2;~+ .. .J we have (1.3.2) if n is (1.2.5), as is also obvious from elementary considerations. The formulat {(s){(•-a){(s-b){(s-a-b) ~ ~ u.(n)a0(n) (!.3.3) b(2s a b) ,6 n~ is valid for u > max{1,R(a)+1,R(b)+I,R(a+b)+1}. The left-hand side is equal to n 1-p-2.1-+a+b p (l p~)(1 p~+a)(1 p B+b)(I p~+a+b)' Putting p-s = z, the partial-fraction formula gives 1-pa+bz2 (I ')(! ~)(! Jh)(I p•"'z) I(' p" P' p""') = (1-p")(I-p11) 1-z -1-paz -1-pllz +1-p<~+llz 1 ~ (1-p<m+l)a_p<m+l)ll_~.p))z"' (I p")(I P'J.f:t 0 (I p":(I p")~ (I-p<•+'~)(I-p<•+"")"". t Ramanujan (2), B. M. Wileon (l). 1.3 THE DIRICHLET SERIES Hence b(s)b(s-a)C(s-b){(s-a-b} = TI ~ 1-p(m+l)a 1-p<m+~)b !, b(2s a b) P f::o 1-pa 1-pll p"'6 and the result follows from (1.3.2). If a= b = 0, (1.3.3) reduces to (1.2.10). Similar formulae involving a~lll(n), the sum of the ath powers of those divisors of n which are qth powers of integers, have been given by Crum (1). 1.4. It is also easily seen that, ifj(n) is multiplicative, and ~~ is a product of zeta-functions such as occurs in the above formulae, and k is a given positive integer, then can also be summed. An example will illustrate this point. The function aa(n) is 'multiplicative', i.e. if m is prime ton u,.(mn) = aa(m)aa(n). ~ "·':) ~ n ~ "·'~:). n-1 n p m-o p Hence and, if k =IT rf, Hence Hence 10 THE FUNCTION ((•) AND Chap. I 1.5. Ramanujan's sums.t Let C,~;(n) = !e-2nlurifk = L: cos2n:'", (1.5.1) ' ' where h runs through all positive integers less than and prime to k. Many formulae involving these sums were proved by Ramanujan. We shall first prove that <,(n) ~ .Z: I'(~) d. (!.5.2) dlk.dln. The sum is equal to kif kin and 0 otherwise. Denoting by (r, d) the highest common factor of rand d, so that (r, d)= l means t.hat r is prime to d, ~ ca(n) = ~ (r,d)~,r<d e-2nnri/d = 1'/.~;(n). Hence by the inversion formula of MObius (1.1.7) <,(n) ~ .Z: l'(~),,(n), '" and (1.5.2) follows. In particular ,,(l) ~,(k). (!.5.3) The result can also be written c.~;(n) = dr-f.a 1 .. p.(r)d. Hence t-~) = L p.~) dl-s dr=k,dln Summing with respect to k, we remove the restriction on r, which now assumes all positive integral values. Hence! i ck(n) = L p.(r) dl-s = ut-s(n) (1.5.4) k-l Jc'l r,djn r" {(a) ' the series being absolutely convergent for u > I since lck(n)l .::;;;;- u1(n), by (!.5.2). We have also i'~:) ~ i ~ .Z: "(5)a n~l n-1 rllkdln ~ .Z: I'(~) d i (~)• ~ {(s) .Z: 1'(5) d' 4 (L5.5) djk m-1 dlk t RBIIl&!lujan (3), Hardy (6). + Two more proofs are given by Ho.rdy, Ramanujan, 137-41. THE DIRICHLET SERIES We CB.Dialso sum series of the formt ~c,(n)f(n), 61 ns where j(n) is a. multiplicative function. For example, i '!l"l~(n) ~ i d~:) .Z: 3"(~) n-1 .. -1 lllk,llln. ~ .z: .,m i ~~~ lllk m-1 ~ <'I•) .z: •'-·,(~) TI (l+l-lp4) llJk Pill ;ra ~Ill"· rr k ~ IIP' the sum is k'4 u (.l+l-.\p4)- 2 (~)' 4 {.-(.1-l)p-'} n (.l+l-.lp'4 )+ PI pJk P p·;#p + ") (l;)' 4 (,-(,-l)p4){.-(,-l)p'4} ):;! (.l+l-.lp'4 )-... ,;:rk pp p~p,p• ~ k'4U (1 •+l-,\p4)-p:4 (,-(,-l)p4J) ~ k'4 !J ('-~+"('-~)(!-p:4))· Hence For example, in the simplest case j(n) = l, the series is ~AJ;,;,til· For given 8, n runs through those multiples of 8/q which are integers. If 8Jq in its lowest terms is 31jq1, these are the numbers 81 , 231, Hence the sum is .z: 8"m i (r; )' ~ {(s) .z: 3"(~)•>· lljk r-1 1 llik t Crum (I). 12 THE FUNCTION C(s) AND Chap. I Since S 1 = Sj(q, S), the result is ~ '•C:1 ~ {(s) ,2; s•-·~(i)(q,o)". (15.7) n.-1 lllk 1.6. There is another class of identities involving infinite series of zeta-functions. The simplest of these isf ,2; I_~~ ~~)log(("'). (16.1) p ~ n-1 We have log{(s) =22m 1 mr = i P<;;), '" P p m~t where P(s) = Lp-8 • Hence i ~~log{(n8) = i /L~)! P(:n.s2 = i P~s) 2 JL(n), n~l fi~l m~l r~l nlr and the result follows from (1.1.5). A closely related formula is ~ "(n) ~ ((s) ~ ~(n)log((.,), 6 ns 6 n (1.6.2) where v(n) is defined under (1.2.8). This follows at once from (1.6.1) and the identity Denoting by b(n) the number of divisors of n which are primes or powers of primes, another identity of the same class is (1.6.3) where cfo(n) is defined under (1.2.12). For the left-hand side is equal to ~ .12: (i.+f,.+f,.+-} m-1 p and the series on the right is Since t See Landau and Walfiao; (I), EsWnnann (1), (2). II THE ANALYTIC CHARACTER OF <(s), AND THE FUNCTIONAL EQUATION 2.1. Analytic continuation and the functional equation, first method. Each of the formulae of Chapter I is proved on the supposi · tion that the series or product concerned is absolutely convergent. In each case this restricts the region where the formula is proved to be valid to a hn.If-plane. For {(s) itself, and in all the fundamental formulae of § I. I, this is the ha.lf·plane u > I. We have next to inquire whether the analytic function {(s) can be continued beyond this region. The result is THEOREM 2.1. The function {(s) is regular far all values of s except s = I, where there is a simple pole uith residue I. It satisfies the functional equation (2.1.1) This can be proved in a considerable variety of different ways, some of which will be given in later sections. We shall first give a proof depending on the following summation formula. Let ljl(x) be any function with a continuous derivative in the interval [a, b). Then, if [ x] denotes the greatest integer not exceeding x, I ~(n) ~ f f(x) dx + f (x-[x]-llf'(x) dx+ a< .. .;;b a a +la-[a]-J:),!(a)-(b-[b]-!if(b). (2.1.2) Since the formula is plainly additive with respect to the interval (a, b] it suffices to suppose that n ~ a < b ~ n + 1. One then has I (x-n-t)~'(x)dx ~ (b-•-tl~(b) -(a-n -l,)~(a)-I ~(x)dx, ANALYTIC CHARACTER OF C(s) AND Chap. II on integrating by parts. Thus the right hand side of (2.1.2) reduces to ([b]-n)ljl(b). This vanishes unless b=n+l. in which case it is ljl(n + 1), as required. In particular, let .f,(n) = n---s, where s -¥=- 1, and let a and b be positive integers. Then ' ' "'\:' ~ = bl-s~al-8 ~sf x~[xJ,~tdx+!(b---a~a---a). (2.1.3) ~ n8 1~8 X"+ n-a+l a First take a> l, a= I, and make b-+ oo. Adding I to each side, we obtain {(8) ~ 8J·[x]-xHdx+!+i· (2.1.4) xs+1 s~l 2 ' Since [x]~x+! is bounded, this integral is convergent for a> 0, and uniformly convergent in any finite region to the right of u = 0. It therefore defines an analytic function of s, regular for a> 0. The right-hand side therefore provides the analytic continuation of '(s) up to u = 0, and there is clearly a simple pole at 8 = 1 with residue 1. For0<a<1wehave . ' . f [x]~xdx= -Jx-8dx=_! sf dx I 0 :rs+t 0 s-1' 2 1 Xf+i = 2' and (2.1.4) may be written {(s) ~ s J"[x]-x dx (0 <a< 1). (2.1.5) x'+' ' Actually (2.1.4) gives the analytic continuation of {(s) for a> -1; for if j(x) ~ [x]-x+t, j,(x) ~ J j(y) dy, thenf1(x) is also bounded, since, as is easily seen, for any integer k. Hence 2.1 THE FUNCTIONAL EQUATION " which tends to 0 as x1 -+ oo, x2 -+ oo, if a > -1. Hence the integral in (2.1.4) is convergent for a> -1. Also it is easily verified that ' •J[x]-x+idx~!+~ (a<O). xs+t s-1 2 ' Hence {(s) ~ 8 J[•J;;;.:,+idx (-1 <a< 0). (2.1.6) Now we have the Fourier series [x]-x+i = isin2n1Tx, (2.1.7) n=l n1T where x is not an integer. Substituting in (2.1.6), and integrating term by term, we obtain "s) = ~ ~ .!J""sin2n1TXdx -rr~n x-+t n-1 0 = ~ ~ (2n-rr)' I"" siny d 1T6 n o ya+t y ~ ;;(2w)'{-r(-s)).mii-"'{(l-s), i.e. (2.1.1). This is valid primarily for -1 <a< 0. Here, however, the right-hand side is analytic for all values of 8 such that a < 0. It therefore provides the analytic continuation of {(s) over the remainder of the plane, and there are no singularities other than the pole already encountered at 8 = 1. We have still to justify the term-by-term integration. Since the series (2.1.7) is boundedly convergent, term-by-term integration over any finite range is permissible. It is therefore sufficient to prove that . ~ IJ""sin2n1TX i~~;;: ~dx=O (~1M~+.)+o(~ j x'!:.) ~ o(>M~+•)· ' and the desired result clearly follows. ANALYTIC CHARACTER OF {(s) AND Chap. II The functional equation (2.1.1) may be written in a number of different ways. Changing s into 1-s, it is {(I-s)= 21-n-cos!snr(8}{(8). (2.1.8) It may also be written where x(s) = 2Sn-'-1 sin-fs7Tr(l-s) = rr-lr~(!8~s). and x(s)x(I-s) = 1. Writing g(s) ~ !-l)n-l•r(!,)((8), it is at once verified from (2.1.8) and (2.1.9) that g(,)~<(l-s). Writing we obtain E(•J~m+'•l E(•)~E(-•). (2.1.9) (2.1.10) (2.1.11) (2.1.12) (2.1.13) (2.1.14) (2.1.15) The functional equation is therefore equivalent to the statement that E(z) is an even function of z. The approximation nears = I can be carried a stage farther; we have ((,)~'~I +r+O(J,-IJ), where y is Euler's constant. For by (2.1.4) lim(((s)-!__J ~ J"[x]-xHdxH ........ 1 s-1 xt ' ~lim J"[x]-x dx+! ~ x' ' =~(I m 7 1 ~-logn+I) m-1 m ~lim(~ !+1-logn) ~ y. n,..., .f= 1 m+I (2.1.16) l.l. A considerable number of variants of the above proof of the functional equation have been given. A similar argument was ap:r)lied by Hardy,t not to {(s) itself, but to the function ~ (-!)•-• ~ (l-2'4 )((a). (2.2.1) 6 n• t Hardy (6). THE FUNCTIONAL EQUATION 17 This Diriohlet series is convergent for all real positive values of 8, and so, by a general theorem on the convergence of Dirichlet series, for all values of 8 such that u > 0. Here, of course, the pole of {(8) at s = l is cancelled by the zero of the other factor. These facts enable us to simplify the discussion in some respects. Hardy's proof runs as follows. Let f(x) = ~ sin(2n+1)x. 6 2n+l This series is boundedly convergent and f(x) = (-l)"'f7T for m11 < x < (m+l)7T (m = 0, 1, ... ). Multiplying by x"-1 (0 < s < 1), and integrating over (O,oo), we obtain "' (m+1)1r 00 !7T L (-1)"" I xs-1 dx = r(s)sin-fS?T L: __ I--m~o """" n~o (2n+1)B+l ~ r(s)•int'"(l-2~-1)((a+I). The term-by-term integration may be justified as in the previous proof. The series on the left is ~[!+ 1(-l)"{(m+ll'-m')]. This series is convergent for 8 < 1, and, as a little consideration of the above argument shows, uniformly convergent for R(s) .:-::;;; 1-8 < I. Its sum is therefore an analytic function of 8, regular for R(s) < L Butfors < 0 it is 2(1'-28+38 -••• ) = 2(I-2S+1){(-s). Its sum is therefore the same analytic function of 8 for R(8) < I. Hence, for 0 < s < I, ,..., 2,(1-2'+1)((-s) ~ r(a)sin!..,.(J-2+1)((s+l), and the functional equation again follows. l.3. Still another proof is based on Poisson •s summation formula •-~.f(n) ~ ~ lf(u)oo•2=udu. (2.3.1) If we put f(x) = lx 1- and ignore all questions of convergence, we obtain the result formally at once. The proof may be established in various ways. If we integrate by parts to obtain integrals involving sin 2nnu, 18 ANALYTIC CHARACTER OF {(s) AND Chap. II we obtain a proof not fundamentally distinct from the first proof given here. t The formula can also be used to give a proof dependingt on (l-2'~)((s). Actually cases of Poisson's formula enter into several of the following proofs; (2.6.3) and (2.8.2) are both cases of Poisson's formula. 2.4. Second method. The whole theory can be developed in another way, which is one of Riemann's methods. Here the fundamental formula is I f• x"-' {(a)= rw 0 e"'=l dx (u > 1). (2.4.1) To prove this, we have for"> 0 Hence r(s)((s) ~ i J x"-'c-= dx ~ J x"-'! c-= dx ~ J ,~:.'1 dx n=lo (I n-1 0 if the inversion of the order of summation and integration can be justified; and this is so by absolute convergence if a> I, since f j x•-•,-= ax ~ r<•)(<•l n-1 0 is convergent for u > 1. Now consider the integral l(s)=f~dz, e'-1 a where the contour C starts at infinity on the positive real axis, encircles the origin once in the positive direction, excluding the points ±2i1T, ±4i1T, ... , and returns to positive infinity. Here z•-1 is defined as e(B-l)logz when the logarithm is real at the beginning of the contour; thus l(logz) varies from 0 to 21T round the contour. We can take 0 to consist of the real axis from oo top (0 < p <' 21T), the circle lzl = p, and the real axis from p to oo. On the circle, jz4'-lj = eAizl. tMordell{2). t Ingham, Prime Numbers, 46. THE FUNCTIONAL EQUATION " Hence tb,e integral round this circle tends to zero with p if u > I. On making p -+ 0 we therefore obtain Hence f "" x"-1 f""(xe27Ti)'-1 J(s) =-&!-ldx+ e"'-l dx ' ' ~ (e""-l)r(s)((s) 2i1Tei1fs ~ r{l-s) {(s). e-i1fsr(l-s)f zs-1 ~(s)= ~ ez 1dz. a (2.4.2) This formula has been proved for u > I. The integral J(s), however, is uniformly convergent in any finite region of the 8-plane, and so defines an integral function of 8. Hence the formula provides the analytic continuation of {(8) over the whole s-plane. The only possible singu-larities are the poles of r(I-s), viz. s = 1, 2, 3, .... We know already that "s) is regular at s = 2, 3, ... , and in fact it follows at once from Cauchy's theorem that 1(8) vanishes at these points. Hence the only possible singularity is a simple pole at 8 = I. Here and I(ll~f~~2m, &-I a Hence the residue at the pole is I. If s is any integer, the integrand in J{s) is one-valued, and i(.s) can be evaluated by the theorem of residues. Since ez.:.! = l-!z+Bt~-B2~+ .. ·• where B1, B 2, ... are Bernoulli's numbers, 'we find the following values of {(s)' ({0) ~ -!, {{-2m)~ 0, ({1-2m)~ (-~:Bm (m ~I, 2, ... ). (2.4.3) To deduce the functional equation from (2.4.2), take the integral along the contour c .. consisting of the positive real axis from infinity to (2n+l}1T, then round the square with corners (2n+l)1T(±I±i), and then back to infinity along the positive real axis. Between the contours ANALYTIC CHARACTER OF {(s) AND Chap. II 0 and 0,. the integrand has poles at the points ±2i11', ... , ±2in11'. The residues at 2mi7T and -Zmi1r are together (2m7Teii,.)s-t+ (2m7Te~i,.r-1 = (2m7T)•-1eimr-1>2 cos f1r(s-1) = -2(2m11')'-1ei"'sin f7TS. Hence by the theorem of residues l(s) = f ~dz+47Tiei"8 sinf-n's i (Zm1r)w-1. e"-1 m~1 c. Now let a< 0 and make n-+ oo. The function 1/(e"-1) is bounded on the contours C.,, and zs-1 = O(lzl"-1). Hence the integral round C., tends to zero, and we obtain J(s) = 47Tiei,.wsinf7Tsm~ 1 (2m7T)8- 1 = 411'iei"8 sin lrrs(27T)8- 1t{l-s). The functional equation now follows again. Two minor consequences of the functional equation may be noted here. The formula {(2m)= 22m-l7T!m (!;)! (m = I, 2, ... ) (2.4.4) follows from the functional equation (2.1.1), with s =1-2m, and the value obtained above for {(1-2m). Also r(O) = -~log21r. (2.4.5) For the functional equation gives -~g~:; ~ -log2ff-jfftant"'+ ~(~i+ ~(~i· In the neighbourhood of s = I !fftan!sff ~- 8~ 1 +0(js-lj). ~(~; ~ ~i:i+··· ~ -r+ and ('(s) -(1/(s-l)')+k+... I ((s) (1/(s-l)}+y+k(s 1)+ ... ~ - s-1 +r+ .... where k is a constant. Hence, makil;tg 8-+ 1, we obtain -~(~; ~ -log2ff, and (2.4.5) follows. 2.5. Validityof(l.l.l)forall8. The original series (1.1.1) is naturally vaJ.id for a > 1 only, on account of the pole at 8 = 1. The series (2.2.1) is convergent, and represents (1-21-){(s), for a> 0. This series ceases 2.5 THE FUNCTIONAL EQUATION 21 to conver~ on a= 0, but there is nothing in the nature of the function represented to account for this. In fact if we use summability instead of ordinary convergence the equation still holds to the left of a = 0. THEOREM 2.5. The series 11~ 1 (-1)"-1n_, is summable (A) to the sun~ (I-21_,){(s)for all values of s. LetO I, and the result by analytic continuation for a > 0. We can now replace this by a loop-integral in the same way as (2.4.2) was obtained from (2.4.1). We obtain ~ f=:-~t=~x" = e-i,.sr_(~~ f w-1 xe-'" dw 6 1 n8 21Ti c 1+xe-w ' when 0 encircles the origin as before, but excludes all zeros of 1+xe-'", i.e. the points w = logx+(2m+1)i1T. lt is clear that, as x -+ 1, the right-hand side tends to a limit, uniformly in any finite region of the 8-plane excluding positive integers; and, by the theory of analytic continuation, the limit must be (I-21-..g(s). This proves the theorem except if s is a positive integer, when the proof is elementary. Similar results hold for other methods of summation. 2.6. Third method. This is also one of Riemann's original proofs. We observe that if a > 0 Hence if a> 1 the inversion being justified by absolute convergence, as in § 2.4-. 22 Writing ANALYTIC CHARACTER OF C(s) AND we therefore have "" Chap. II 12.6.I) 11·•1 ~-·~I xl•-'~lx) dx I• > I). 12.6.2) r(2s) 0 Now it is known that, for x > 0, i e-n'1f.:r:=~ i e-~<'"1", n~-oo "\IX n=- 1, (2.4.1) may be written {(s)r(s) = I' (I__I-~);tB-1 dx+_!_I +I"' xs:t dxi , e""-x s-e-' ' and this holds by analytic continuation for a > 0. Also for 0 < a < l !_ = -Ioo ,xB-1 dx. s-1 x ' THE FUNCTIONAL EQUATION Hence {(s)r(s) =I"(! !)xs-t dx (0 <a< 1). e"'-1 x ' Now it is known that the function f(x) = eZ"<z!)]-x,/:211") is self-reciprocal for sine transfonns, i.e. that fix)~ JW [fly)sinxydy. Hence, putting X= e.J(271) in (2.7.1), (l,)r(s) ~ l2n)l•l fi<JG'-' dg ~ 12n)i'J(§) [ g•-•dg [tly)singydy. If we can invert the order of integration, this is 2v+!~¥-l J fly) dy J t'-'singydg " ' = 2is+hrts-! l j(y)y-8 dy I u8- 1sin u du 23 12.7.I) 12.7.2) 12.7.3) = 2fs+'hris-i(21T)i8-!f(l-s){(l-s) 2cos}1r;r(l s)' and the functional equation again follows. To justify the inversion, we observe that the integral l fly)sin<y dy converges uniformly over o < 0 ~ g ~ a. Hence the inversion of this part is valid, and it is sufficient to prove that lh! J fly) dy U + J)<•-'sin<1J dg ~ 0. ~ ....... ooo o & . ' Now [<"-'sin -l. But • l [!x'-1 dx= - 28 (-I <a<O). Hence (2.8.!) Now (2.8.2) Hence r(s)1(s) ~ J• 2x i --1-x'-' dx ~ 2 i J• __ x' dx o A-1 4n!!.n-2+x2 fi-t o 4n21T2+x2 ~ 1T 2B-y = 2 ,6t 1 (2n1T) 8 -1 2cos!s1T = cos!s1r '(I-s), the functional equation. The inversion is justified by absolute con-vergence if -1 < a < 0. 2.9. Sixth method. The formulat c+i<O '<s)=.::: J (r'(I+z)_logz)z-~dz (-l<c 1; and the integrand is O(izl-"-1), so that the integral is convergent. and the formula. holds by analytic continuation, if a > 0. t Klo08terman (I). 2.9 THE FUNCTIONAL EQUATION 25 We may next transform this into an integral along the positive real axis after the manner of§ 2.4. We obtain '(s) = _sin1TB J {r'(l+x)_logx}x-..dx (0 <a< 1). '" o r(I+x) (2.9.2) To deduce the functional equation, we observe thatt r'(x) =lo X-~-zJ"" tdt r(x) g 2x 0 (t2+x2)(e21fl I) Hence ~g::;-logx = ~(~j+~-logx l f. tdt f. ' ( l l) = 2X-2 o (t2+x2)(e2"'t I)= -2 o t2+x2 e2,.t_I-2;ri dt. Hence (2.9.2) gives {(s) ~"'inns I" x~dxf·-'·(-'---'-) dt '" o o t2+x2 e2"'1-I 21Tf _ 2sin'"sJ"(-'---'-)tdtJ"" ~dx -1r 0 e21T1-l 21rt 0 t2+x2 -sin1Ts J• (-'---'-)t-•dt -cos!1rs 0 e2'~~'~-I 21rt = 2sin-!1Ts(2'")"-1 J·(-1--~)u-.. du e"-1 u ' = 2sin!1Ts(21T)~-1r(l-s)"l-s) by (2.7 .1). The inversion is justified by absolute convergence. 2.10. Seventh method. Still another method of dealing with nB), due to Riemann, has been carried out in detail by Siegel.t It depends on the evaluation of the following infinite integral. J eiw•!(4,.)+aw Let lll(a) = -----ew=r- dw, (2.!0.!) L where L is a straight line inclined at an angle i1r to the real axis, and t Whittaker and Watson,§ 12.32, exo.mple. t Siegel (2). 26 ANALYTIC CHARACTER OF {(s) AND Chap. II intersecting the imaginary axis between 0 and 27Ti. The integral is plainly convergent for all values of a. We have f eiiw"/.,. l)(a+l)-f!l(a) = ew-l (efa+l)w-e""')dw L = 1 eliw'/w+a"' dw = ei1ra• f efiW'I" dW, where W = w-2imz. Here we may move the contour to the parallel line through the origin, so that the last integral is e-1-t.,.l e-lp'J" dp = 21Tei""· Hence (2.10.2) Next let L' be the line parallel to L and intersecting the imaginary axis at a distance 21T below its intersection with L. Then by the theorem of residues But f eliw'/"+(a) = 27Ti. Eliminating fll(a+ I), we have 4>(a)~ 21Ti+2?Jei1f(a"-2a+i>• I+e-21ria (2.10.3) (UM) (2.10.5) 2.10 THE FUNCTIONAL EQUATION 27 If a =,!izf1T+}, the result (2.10.4) takes the form ~ dw =--2m---, f "''·:.-'•:'/~·•:"-l<=:·/•:•1:."' 21ri .e-fiz'I>T+t.z ew-1 e'"-1 e~-1 L :Multiplying by z"-1 (a> I), and integrating from 0 to ooe-1•,., we obtain The inversion on the left-hand side is justified by absolute convergence; in fact where c > 0, so that Now R(izw) = -crf../2. <>:>e-!lor f e-lh'J,-+iz I f e-liz•J,.+}z ---za-l dz = -~ ~z"-1 dz e=-I I+e-is,-ez-I ' ' z and where Lis the reflection of Lin the real axis. Hence or , . feliw'J,-+!w n-..-•r(}s),(s} = d•,.(s-1)2w-t1T~-1 r(!s) ~w-8 dw+ L , , , f e-liz'J,-+~= +e!I7TS2-117T-"t"8 -2f'(!-!s) ez=r-z~-l dz. (2.}0.6) z This formula holds by the theory of analytic continuation for all values of s. If s = l+it, the two terms on the right are conjugates. Hence /(s) = 71-isr(!sg(s) is real on a = f. Hence f(s) ~J(o+it) ~J(I-o+it) ~f(!-o-it) ~/(1-s), the functional equation. ANALYTIC CHARACTER OF {(s) AND Chap. II 2.11. A general formula involving "8). It was observed by Mtintzt that several of the formulae for ,(8) which we have obtained are particular cases of a. formula containing an arbitrary function. We have formally j r-1,.t 1 F(nx) dx = .. ~1 [ x"-1 F(nx) dx ~ I~ J Y'-'F(y) dy n=l 0 ~ ((s) J Y'-'F(y) dy, where F(x) is arbitrary; and the process is justifiable if F(x) is hounded in any finite interval, and O(x-"), where a:> 1, as x ~ oo. For then Il~l J I?J'-'F(y)( dy n-1 0 exists if 1 < a < a:, and the inversion is justified. Suppose next that F'(x) is continuous, bounded in any finite interval, and O(x-.8), where {3 > 1, as x -;.. oo. Then as x---i'- 0 I F(=l- J F(ux) du ~ x J F'(ux)(u-[u]) du n-1 0 o 1/x a> ~ x J 0(!) du+x J O((ux)-ll) du = 0(1), 0 1/Z i.e. I F(nx) ~ ~f· F(v)dv+O(!) ~ '+0(1), n~1 X X say. Hence f• x'-' I F(nx) dx ·-· ' " ~ J' x'-'( I F(nx)-~)ax+-'-+ J·,., I F(nx)dx, o ,.1 x 8-1 t n=l and the right-hand side is regular for a> 0 (except at 8 = 1). Also for a< 1 ~ = -c s"" ...s-2 dx. 8-1 1 tMUntz(l). 2.11 THE FUNCTIONAL EQUATION 29 Hence we have MUntz's formula ((') J Y'-'F(y)dy ~ J x'-'( ~ F(nx)-~ J F(v) dvj dx, (2.!!.1) o o n 1 o valid for 0 < a < I if F(x) satisfies the above conditions. If F(x) = e-"' we obtain (2.7.1); if F(x) = e-,.x' we obtain a formula equivalent to those of§ 2.6; if F(x) = I/(l+x2) we obtain a. formula which is also obtained by combining (2.4.1) with the functional equation. If F(x) = x-1 sin7TX we obtain a formula equivalent to (2.1.6), though this F(x) does not satisfy our general conditions. If F(x) = I/(l+x)2, we have • I J, , 1 1 LF(=l--F(v)dv~ ,L----.. -t x o n=t (l+nx)2 x ~ ',[-dd:logf(f+Il] -~. X ~ {~t/x X Hence (::~,~((s) ~ r g•-•(fi,logr(t+Jl-~)ag, and on integrating by parts we obtain (2.9.2). 2.12. Zeros; factorization formulae. THEOREM 2.12. g(8) and E(z) are integral junctions of order l. It follows from (2.1.12) and what we have proved about "s) that t{8) is regular for u > 0, (s-1g{8) being regular at s = I. Since e(s) = t(l-.~), g(8) is a.lso regular for u < 1. Hence t(s) is an integral function. Also jr{}8)j = j I e-uuiH dul ~ l e-uuia-1 du= r(ta)= O(eAa!Oga) (a>O), (2.12.1) and (2.1.4) gives for u ~ l, js-lj >A, ((s) ~ 0(1'1! ~)+0(1) ~ O((sl). (2.12.2) Hence (2.1.12) gives t(s) = O(eAI~Iloglsl) (2.I2.3) for a~ i, ]8] >A. By (2.1.13) this holds for u ~!also. Hence g(s) is of order I at most. The order is exactly I since as s ---i'- oo by real values log,(s)"' 2""", logf(8) ,...,is logs. 30 Hence also ANALYTIC CHARACTER OF ,(11) AND S(z)=O(e"'l'tlogt.ot) (jzj >A), Chap. II and 3(z) is of order 1. But S(z) is an even function. Hence 3(·/z) is also an integral function, and is of order !- It therefore has an infinity of zeros, whose exponent of convergence is f. Hence S(z) has an infinity of zeros, whose exponent of convergence is 1. The same is therefore true of ~(s). Let p1, p2, ••• be the zeros of g(s). WehaveaJrea.dyseenthat,(s)hasnozerosfora > 1. It then follows from the functional equation (2.1.1) that "s) has no zeros for a< 0 except for simple zeros at s = -2, -4, -6, ... ; for, in (2.1.1), ,(1-s) has no zeros for u < 0, sin}m has simple zeros at s = -2, -4, ... only, and r(I-s) has no zeros. The zeros of '(s) at -2, -4, ... , are known as the 'trivial zeros'. They do not correspond to zeros of g(8), since in (2.1.12) they are cancelled by poles of r{!8). It therefore follows from (2.1.12) that g(8) has no zeros for a > I or for a < 0. Its zeros p1, p2, ... therefore all lie in the strip 0 ~ o ~ 1; and they are also zeros of {(8), since 8(8-1)r(}8) has no zeros in the strip except that at 8 = 1, which is cancelled by the pole of {(8). We have thus proved that {{8) has an infinity of zeros p1, p2, ... in the strip 0 ~a~ 1. Since (l-21-){(8) = 1-~+~---· ~ 0 (0 < 8 < 1) (2.12.4) and {(0) 0, {{8) has no zeros on the real axis between 0 and 1. The zeros p1, p2, ... are therefore all complex. The remainder of the theory is largely concerned with questions about the position of these zeros. At this point we shall merely observe that they are in conjugate pairs, since {(8) is real on the real axis; and that, if p is a zero, so is 1-p, by the functional equation, and hence so is 1-p. If p = f3+iy, then 1-p = 1-,B+iy. Hence the zeros either lie on a = f, or occur in pairs symmetrical about this line. Since g(8) is an integral function of order 1, and f(O) = -{(0) = }, Hadamard's factorization theorem gives, for all values of s, f(s) ~ t'~'I)(l-~)e'IP, (2.12.5) where b0 is a constant. Hence (2.12.6) THE FUNCTIONAL EQUATION where b = b0+!log11. Hence also <'l•l ~ b-1_-' r'l!•+1l "'(!_ ') ((•) s-1 2 r(!•+1) + 7 8-p + p . Making s-+ 0, this gives ~~~i ~ b+1-4 ~~W· Since {'(0)/{(0) = log211 and r'(I) = -y, it follows that b = log211-l-!r. 31 (2.12.7) (2.12.8) 2.13. In this sectiont we shall show that the only function which satisfies the functional equation (2.1.1 ), and has the same general charac-teristics as {(s), is {(8) itself. Let G(s) be an integral function of finite order, P(s) a polynomial, and j(s) ~ G(s)fP(s), and let /(•)~ ~ ... f::tns (2.13.1) be absolutely convergent for a > I. Let f(s)r(!s)11-~' = g(l-s)r{l-fs)11-iu->, (2.13.2) where theseriesbeingabsolutelyconvergentforrf< -a< 0. Thenf(s) = G{(s), where 0 is a constant. We have, for x > 0, ll+ioo tfo(x) = dn 21"' f(s)r(is)1t-~sz-i• ds 00 2+1oo = 2 ~ f r(ts)(n-n2x)-fw ds n-l 2-ioo = 2 .. ~ 1 a,.e-rrn'z. Also, by (2.13.2), ll+i"' 41•) ~ .k,L g(1-s)r(i-!•)•-~'4>x-l•ds. We move the line of integration from o = 2 to a= -1-a. We observe f Hamburger (1)-(4). Siegel (1). 32 ANALYTIC CHARACTER OF ((s) AND Chap. II thatj(s) is bounded on a= 2, and g(l~s) is bounded on u = -1-~; since r~~l,> ~ O(IW-'>· it follows that g(l-s) = O(ltli) on u = 2. We can therefore, by the Phragmen-Lindelof principle, apply Cauchy's theorem, and obtain -«-l+ioo ~(x) ~ ~ f g(l-,)r(i-l-1"-''x-1• ds+ J. R,. -«-1-i<>O where R1 , R2,. •• , are the residues at the poles, say s1, •• , sm. Thus J, R, ~ J. x-l•Q,(logx) ~ Q(x), where the Q~(logx) are polynomials in logx. Hence f(x) ~ ~ ~ -•1•"" r(!-!•H='/x)-1•1• ds+Q(x) n-1 -«-1-i 0), and integrate over (O,oo). We obtain ~ ~"- = ~ ~e-2"111+! J., Q(x)e-wt'x dx, 6t7T(t2+n') 6tt o and the last term is a sum of terms of the form [ x" loghxe-"'""' dx, where the b's are integers and R(a) > -1; i.e. it is a sum of terms of the form to' logPt. Hence .! a .. (,+ 1 . +-, 1 )-mH(t) = 2'1T f b,.e-2"'"1, n=t ~n ,n n-l where H(t) is a sum of terms of the form t"log.Bt. Now the series on the left is a meromorphic function, with poles at ±in. But the function on the right is periodic, with period i. Hence (by analytic continuation) so is the function on the left. Hence the residues at ki and (k+ l)i are equal, i.e. ak = ak+l (k = l, 2, ... ). Hence a" = a 1 for all k, and the result follows. THE FUNCTIONAL EQUATION 2.14. Some series involvin~ ((s). We havet ((sJ-,~ 1 ~ '-!'{(('+'l-1J-'(;";l)m,+2J-1J-.. r2.14.1J for all values of s. For the right-hand side is 1 _ 1 ~ l((s-l)s~ (s-1)+11~ ···) s-l,6-2n•-1 1.2 n2+ 1.2.3 na+ ~ 1-,~~~, •• 1_,((1-M'-·,,-:11 1 ~( 1 I s-1~ = 1- s-I{::-2 (n-1)"1 - n"-1-7 ~((•)-,~~-The inversion of the order of ~;ummation i~; ju~;tified for a > 0 by the convergence of ~ l ~I'-Ul1::1:~l~ ~ ~(('-~)-'"-1). ,6-2n"-16 (k+I)! nk+2 f:?.n" n The series obtained is, however, convergent for all values of s. Another formula! which can be proved in a similar way is (2.14.2) also valid for all values of s. Either of these formulae may be used to obtain the analytic con-tinuation of {(s) over the whole plane. 2;15. Some applications of Mellin's inversion formulae.§ Mellin's inversion formulae connecting the two functionsf{x) and )j(s) The simplest example is .. .., f(x) ~ 2, f \J(s)x-• ds. (2.15.1) 2m a-i 0). (2.15.2) From (2.4.1) we derive the pair f(x) ~ ,,~ 1 , \J(s) ~ r(s)((s) (a> 1), (2.15.3) t Landau, Handbuch, 272. t Rama.swami (1). § See E. C. TiWhma.rsh, /ntrodWJii tl· (2.15.4) The inverse formulae are thus and a+ioo ~I l'(s)((s)x-•ds~_2 (a> I) 2m e:.:-1 a-i.., a+ioo ~ I ~~l'(s)((2s)x~ ds ~ ~(x) (a> il· 2 .. ·-"" (2.15.5) (2.15.6) Each of these can easily be proved directly by inserting the series for {(s) and integrating term~by-term, using (2.15.2). As another example, (2.9.2), with s replaced by 1-s, gives the Mellin pair f(x) ~ ~g::;-logx, ~(8) ~ -~((1-s) (0 <a< 1). (2.15.7) Stn7T8 The inverse formula is thus a+i«> r'(l+x)Jo x = _2 I ,(1-s)x-Bds f(I+x) g 2ia-ico sin7TB · (2.15.8) Integrating with respect to x, and replacing s by 1-s, we obtain a+ioo logr(l+x)-xlogx+x = ..!., I '~s)x" ds (0 <a< 1). (2.15.9) 2t 8Sill7T8 a-ioo This fonnula is used by Whittaker and Watson to obtain the asymp-totic expansion of log r(l+x). Next, let f(x) and !J(s) be related by (2.15.1), and let g(x) and C!i(s) be similarly related. Then we have, subject to appropriate conditions, c+i«> .:o /,;; I \;ls)ffi(w-s) ds ~ I f(x)g(x)x•-> dx. (2.15.10) c-iro 0 Take for example ~{s) = t»{s) = r(s)~(s), so that f(x) ~ g(x) ~ 1/(e"-1). Then, if R(w) > 2, the right-hand side is ( I 2 3 ) ~ ::;w+a-+4"+ ... r(w) ~ r(w)g(w-1)-i(w)). 2.15 THE FUNCTIONAL EQUATION 35 Thusif1<c 1, j I }:e-<m'+•.,.,x•-•dx~ l'(w) ~ ~--1-. om-1a=1 /::1~1(m2+n2)w This may also be written l'(w)G~:~:) -Ww)), where 1'(n) is the number of ways of expressing n as the sum of two squares; or as !'(w)g(w)"(w)-{(2w)}, where 'l'}(w) = 1-w-3-w+S-to-.... Hencet if! < c < R(w)-! <+«<> /,;; I l'(s)l'(w-s)((2s),(2w-2s) ds ~ l'(w)i'(w)"(w)-{(2w)). c-i(x) ~ I /(1)!!(1) cosxl dt can be evaluated. Let/(!)~ l<l(il)l' ~ <l(il)<l(-il), where(x) ~ ! [ <I( it)</>( -il)!l(!)y" dl ~ l l<l(il)</>( -il)<{!+il)y" dt fH«> ~ 2i~y I <11•-ll<l(!-•l ~ 2i~ I <11•-ll Chap. II ~ ~ ~ ( 2 .r r(wl(n~J·-• dw-3 :J: rtwl(n~J·-• dw) ~ ~% (n~w) ~ r•'•l•' -~% (n~J' e-•'•l•'. Hence J E(t)cosxt dt = 21r2 f (21Trt4e-llz12-3n2e-&tl2)exp( -n~e-2:1:), 0 n~l (2.16.1) Again, putting f(s) ~ !/(•+!), we have l+i«> (x) ~ -~ J ~r(IH•pd•((•)Y'd• 2t-vy 8 !-•· f+ioo ~ -'y in the notation of§ 2.6. Hence j t~~i oosxt dt ~ b(el•-2.-l•f(e-"')). (2.16.2) The case c/>(8) = I'(fs-!) was also investigated by Ra.manujan, the result being expressed in terms of another integral. 2.17. The function ~(8, a). A function which is in a sense agenerali~ zation of ((s) is the Hurwitz zeta-function, defined by • I ~(8,a) = 6(n+a)B (0 1). This reduces to ~(s) when a= 1, and to (28-l){(s) when a= -f. We shall obtain here its analytic continuation and functional equation, which are required later. This function, however, ha.s no Euler pro-duct unless a= i or a= l, and so does not sha.re the most characteristic properties of t(s). THE FUNCTIONAL EQUATION 37 As in§~-· {(8,a) = ~ !... foo ~-le-2cos{!11(8-1)+2mmJ} = -2(2m1T)"-le1.,.•sin(i7Ts+2m7Ta). Hence, if a < 0, {( ) 21'(1-s)(. ! ~ cos2mmJ ! ~ sin2mmJ) 8,a = (21r)1_, sm 1T8,!;: 1 ----;n~+cos 1rB.f:: 1~. (2.17.3) If a= 1, this reduces to the functional equation for ((s). NOTES FOR CHAPTER 2 2.18. Selberg has given a very general method for obtaining the analytic continuation and functional equation of certain types of zeta-function which arise as the 'constant terms' of Eisenstein series. We sketch.aform of the argument in the classical case. Let .7£ = {z = x + iy: y > 0} be the upper half plane and define aud E(z,s)= ~ y•_ (ze . .',u>l) c•d"'-""lcz+dl2s (cod)"' I B(z, •) ~ ((2s)E(z, •) ~ I y• (zE.>t'", u > 1), c.d- -"' lcz+dl2• (cod)#(0.0) these series being absolutely and uniformly convergent in any compact subset of the region R(s) > 1. Here E(z, s) is an Eisenstein series, while B(z, 8) is, apart from the factor y 8, the Epstein zeta-function for the lattice generated by 1 and z. We shall find it convenient to work with B(z, 8) in preference to E(z, s). 38 ANALYTIC CHARACTER OF {(s) AND Chap. II We begin with two basic observations. Firstly one trivially has B(z + 1, s) = B( -1/z, s) = B(z, s). (2.18.1) (Thus, in fact, B(z, s) is invariant under the full modular group.) Secondly, if .6. is the Laplace-Beltrami operator A= -y2 -+-(a' o') 8x2 iJy2 ' then A--- -sl-s ---( Y') Y' fcz+df2s -( )lcz+df2s' (2.18.2) whence 8B(z, s) = s(l- s)B(z, s) (u > 1). (2.18.3) We proceed to obtain the Fourier expansion of B(z, s) with respect to x. We have where B(z, s) = L a11(y, s)e21rinx, ' f e-21finxdx a,(Y, s) =y• ~ o ~+d+icyf2s ' " • f e-'•'••dx =20nys"2s)+2ysc~l d~L."' lcx+d+icyf2•' 0 with 0, = 1 or 0 according as n = 0 or not. The d summation above is and the sum over k is cor 0 according as cfn or not. Moreover 2.18 THE FUNCTIONAL EQUATION and f oo e-2..tnyu 8 w-!Ks-f(2n:fnfy) (vz + 1)" dv = 2n: (lnly) r(s) (n ;'0), in the usual notation of Bessel functionst. We now have where and B(z, s) = ljl(s)y" + tjl(s)y 1 -• + B 0(z, s) (u > 1), ~(s) ~ 2((28), ~(s) ~ 2ni r(s- !l ((28 -1) r(s) 39 (2.18.4) "' Ks-t(2n:ny) B 0(z,s)=8n:8yi L n"-iu1 28 (n)cos(2nnx)~. (2.18.5) We observe at this point that K..,(t)<;€t-ie-t (t---+oo) for fixed u, whence the series (2.18.5) is convergent for all s, and so defines an entire function. Moreover we have B0(z, s) o:t e-Y (y---+ oo) (2.18.6) for fixed s. Similarly one finds OBo(z,s) o:te-:v (y---+ oo). ay (2.18.7) We proceed to derive the 'Maass--Selberg' formula. Let D = {ze.n": lzl:;;:: 1, IR(z)l ~!}be the standard fundamental region for the modular group, and let Dy= {zeD: l(z)~ Y}, where Y ~ 1. Let R(s), R(w) > 1 and write, for convenience, F = B(z, s), G = B(z, w). Then, according to (2.18.3), We have ff dxdy ff dxdy {s(i-s)-w(i-w)) FGY'~ (GAF-FAG)Y' Dy Dy = ff(FV 2G-GV2F)dxdy n, ~ f (FVG-GVF)·dn, '"• t see Wat!lon. Tlu!ory of &sse/ {u11ctioM §6.16. ANALYTIC CHARACTER OF {(s) AND Chap. II by Green's Theorem. The integrals along x = ±! cancel, since F(z + 1) = F(z), G(z + 1) = G(z) (see (2.18.1)). Similarly the integral for lzl = 1 vanishes, since F( -1/z) = F(z), G( -1/z) = G(z). Thus l {s(l-s)-w(1-w)} ffFad~:Y = f (F%i(x, Y)-G(x, Y))dx. Dy -f (2.18.8) The functions ys andy1 -s also satisfy the eigenfunction equation (2.18.3) (by (2.18.2) with c = 0, d = 1) and thus, by (2.18.4) so too does B0(z, s). Consequently, if Z ~ Y, an argument analogous to that above yields 'l {s(1-s)-w(1-w)} f f F0 G0 d:~y t y -t = f (Fo 0~ 0 (x, Z)-G0 a;o (x, Z) )ax -! l -f (F0 °~ 0 (x, Y)-G/(~ 0 (x, Y))dx, -! where F 0 = B 0(z, s), G0 = B 0(z, w). Here we have used F 0(z+ 1) = F 0(z) and G0(z+ 1) = G0(z). (Note that we no longer have the corresponding relations involving -1/z.) We may now take Z ..... oo, using (2.18.6) and (2.18.7), so that the first integral on the right above vanishes. On adding the result to (2.18.8) we obtain the Maass-Selberg formula ['(1-s)-w(1-w)] ~ii(z, ,)ii(z, w)d;~y l ~I (F~(x, Y)-G¥,'(x, Y))dx l -f(FoO~o(x,Y)-GoO(~o(x,Y))dx -! ~ (s-w){~(s)~(w) 1"-•-•-Q>(s)Q>(w)YH•-') + (1-s- w) {Q>(s) ~(w) Y•-w- ~(s)Q>(w) Y•-•), (2.18.9) where THE FUNCTIONAL EQUATION ii(z s) _ {B(z, s) (y ~ Y), ' -B 0(z, s) (y > Y). 2.19. In the general case there are now various ways in which one can proceed in order to get the analytic continuation of 1J and 1/J. However one point is immediate: once the analytic continuation has been established one may take w = 1 - s in (2.18.9) to obtain the relation Q>(s) Q>(1- ') ~ ~(s) ~(1- s), (2.19.1) which can be thought of as a weak form of the functional equation. The analysis we shall give takes advantage of certain special properties not available in the general case. We shall take Y = 1 in (2.18.9) and expand the integral on the left to obtain (s- w)a(s + w)~(s)~(w) + p(s, w)~(s) + y(s, w)~(w) + b(s, w) ~ 0, (2.19.2) where l cx(u)=(1-u) ffy-"dxdy-1= -2 J(1-x2)f(l-u)dx D, 0 and {J, y, b involve the functions¢ and B 0 , but not t/J. If we know that ((s) has a continuation to the half plane R(s) > a 0 then ljJ(s) has a continuation to R(s) > ta0 , so that ex, {1, y, b are meromorphic there. If (s- w)cx(s + w)t/J(w) + {J(s, w) = 0 identically for R(s), R(w) > 1, then p(s,w) 1/J(w)= (s-w)cx(s+w) (2.19.3) (2.19.4) which gives the analytic continuation oft/J(w) to R(w) > !a0 • Note that (s- w)cx(s + w) does not vanish identically. If(2.19.3) does not hold for all s and w then (2.19.2) yields ~(s) ~ (s- w)a(s+ w)~(w) + P(s, w)' (2.19.5) which gives the analytic continuation of 1/J(s) to R(s) > in 0 , on choosing a suitable win the region R(w) > 1. In either case ((s) may be continued to R(s) > a 0 -1. This process shows that {(s) has a meromorphic continuation to the whole complex plane. 42 ANALYTIC CHARACTER OF C(B) AND Chap. II Some information on possible poles comes from taking w = 8 in (2.18.9), so that B(z, w) = B(ZS). Then (2u-l) ~I.B(z,s)l 2 d;~y ={l¢(s)I2 Y 2"'- 1 -IY,(s)I 2 Yl 2"} ~(s)~Y'"-~(s)~Y-'" +(2u-1) 2it . If t =/= 0 we may choose Y ~ 1 so that the second term on the right vanishes. It follows that lr/f(s)jllYI-2" ~ llj.l(s)l2 ¥2"-1 for u ~ !-· Thus lj1 is regular for a ~! and t :f= 0, providing that ¢ is. Hence ((s) has no poles for R(s) > 0, except possibly on the real axis. If we take ! < R(s),R(w) < 1 in (2.19.5}, so that ¢(s) and (w) are regular, we see that 1/J(s) can only have a pole at a point s0 for which the denominator vanishes identically in w. For such an s0 , (2.19.4) must hold. However a:(u) is clearly non-zero for real u, whence ljt(w) can have at most a single, simplepoleforrealw > f, and thisisatw = s0 • Since it is clear that ((8) does in fact have a singularity at 8 = 1 we see that s0 =1. Much of the inelegance of the above analysis arises from the fact that, in the general case where one uses the Eisenstein series rather than the Epstein zeta-function, one has a single function p(8) = 1/J(s)/tP(s) rather than two separate ones. Here p(s) will indeed have poles to the left of R(s) =f. In our special case we can extract the functional equation for ((s) itself, rather than the weaker relation p(s) p(l-s)= 1 (see (2.19.1)), by using (2.18.4) and (2.18.5). We observe that ns-l12q1 _ 28(n) = nl12-sq28 _ 1 (n) and that K,.(z) = K_,.(z), whence x-sr(s)B0(z, s) is invariant under the transformation 8 ___,. 1 -8. It follows that n:-sr(s)B(z, 8) -n:8 - 1r(1-s)B(z, 1-8) ~ {A(s)-A(t-s))Y'+{A(s-j)-A(l-s))y' where we have written temporarily A(8) = 2n:-•r(s)((2s). The left-hand side is invariant under the transformation z ...... -1/z, by (2.18.1), and so, taking z = iy for example, we see that A(s) =A(! -8) and A(s -!-) = A(1-8). These produce the functional equation in the form (2.6.4) and 2.20 THE FUNCTIONAL EQUATION indeed yield n:-•r(s)B(z, s) = ns-tr(l-8)B(z,1-8). 2.20. An insight into the nature of the zeta-function and its func· tional equation may be obtained from the work of Tate [1 ]. He considers an algebraic number field k and a general zeta-function (({,c)~ f {(a)c(a)da, where the integral on the right is over the ideles J of k. Here{is one of a certain class of functions and cis any quasi-character of J, (that is to say, a continuous homomorphism from J to C ")which is trivial on k ". We may write c(a) in the form c0(a)lal•, where c0(a) is a character on J (i.e. lc0(a)l = 1 for aE J). Then c0(a) corresponds to J., a 'Heeke character' for k, and (( {, c) differs from ((s, x) ~ IJ {1- x(P)(NP)-•)-' (where P runs over prime ideals of k), in only a finite number of factors. In particular, if k = 0, then (({,c) is essentially a Dirichlet L-series L(8, x). Thus these are essentially the only functions which can be associated to the rational field in this manner. Tate goes on to prove a Poisson summation formula in this idelic setting, and deduces the elegant functional equation (({,c)~ ((/.c) where r is the 'Fourier transform' of{, and C(a) = Co(a)lall-•. The functional ~uation for ((s, x) may be extracted from this. In the case k = Q we may take c0 identically equal to 1, and make a particular choice f = {0 , such that {0 = { 0 and ((f,.l·i•) ~ "-!• r(is){(s). The functional equation (2.6.4) is then immediate. Moreover it is now apparent that the factor n: -!• r(fs) should be viewed as the natural term to be included in the Euler product, to correspond to the real valuation ofO. 2.21. ltisremarkablethatthe values of((s)for8 = 0, -1, -2, ... , are all rational, and this suggests the possibility of ap-adic analogue of ((s), interpolating these numbers. In fact it can be shown that for any prime p and any integer n there is a unique meromorphic function {P.,.(s) defined ANALYTIC CHARACTER OF ((s) Chap. II for sE ZP, (the p-adic integers) such that (P',.(k) = (1-p-k)C(k) for k ~ 0, k =n (mod.p-1). Indeed ifn ¢ 1 (modp -1)then(P.,.(s)will be analytic on ZP, andifn = 1 (modp- 1)then (P',.(s) will be analytic apart from a simple pole at s = 1, of residue 1-(1/p). These results are due to Leopoldt and Kubota [1J. While these p-adic zeta-functions seem to have little interest in the simple case above, their generalizations to Dirichlet L-functions yield important algebraic information about the corresponding cyclotomic fields. III THE THEOREM OF HADAMARD AND DE LA VALLEE POUSSIN, AND ITS CONSEQUENCES 3.1. As we have already observed, it follows from the formula ,(,) ~ IJ(~-~r (a> I) (31.1) that {(s) has no zeros for o > 1. For the purpose of prime-number theory, and indeed to determine the general nature of ((s), it is necessary to extend as far as possible this zero-free region. It was conjectured by Riemann that all the complex zeros of {(s) lie on the 'critical line' o = !- This conjecture, now known as the Riemann hypothesis, has never been either proved or disproved. The problem of the zero-free region appears to be a question of extending the spl!ere of influence of the Euler product (3.1.1) beyond its actual region of convergence; for examples are known of functions which are extremely like the zeta-function in their representation by Dirichlet series, functional equation, and so on, but which have no Euler product, and for which the analogue of the Riemann hypothesis is false. In fact the deepest theorems on the distribution of the zeros of ((s) are obtained in the way suggested . .But the problem of extending the sphere of influence of (3.1.1) to the left of o = 1 in any effective way appears to be of extreme difficulty. By (1.1.4) ..':__ ~ ~ ~(n) (a> 1), ((s) ,6 n~ where l,u(n)l ~ l. Hence foro near to I I I I . I A C(8) ~ 6no = {(o) < o-1' i.e. j((s)j > A(a-1). Hence if '(s) has a zero on o = I it must be a simple zero. But to prove that there cannot be even simple zeros, a. much more subtle argument is required. It was proved independently by Hadamard and de la Vallee Poussin in 1896 that ((s) hasnozeroson the line o = I. Their methods are similar in principle, and they form the main topic of this chapter. THE THEOREM OF HADAMARD Chap. III The main object of both these mathematicians was to prove the prime-number theorem, that as x-+ co 'l'T(x) ...... lo=x· This had previously been conjectured on empirical grounds. It was shown by arguments depending on the theory of functions of a complex variable that the prime-number theorem is a consequence of the Hadamard-de Ia Vallee Poussin theorem. The proof of the prime-number theorem so obtained was therefore not elementary. An elementary proof of the prime-number theorem, i.e. a proof not depending on the theory of {(s) and complex function theory, has recently been obtained by A. Selberg and Erd&. Since the prime-number theorem implies the Hadamard-de Ia Vallee Poussin theorem, this leads to a new proof of the latter. However, the Selberg-ErdOs method does not lead to such good estimations as the Hadamard-de Ia Vallee Poussin method, so that the latter is still of great interest. 3.2. Hadamard's argument is, roughly, as follows. We have for u > 1 log((s) ~ '<;'~_I~'<;' !+f(s), (3.2.1) 7 m~l mpms 7 r where j(s) is regular for u > !- Since {(s) has a simple pole at 8 = 1, it follows in particular that, as u-+ 1 ( u > 1 ), '<;'.!..~log__!. (3.2.2) fPa u-1 Suppose now that 8 = 1+it0 is a zero of {(8). Then if s = u+it0, as u-+ I (u> I) '<;' oos(t,logp) ~ Iogj{(s)j-Rf(s) ~ log(u-1). (3.2.3) 7 pa Comparing (8.2.2) and (3.2.3), we see that cos(t0 logp) must, in some sense, be approximately -I for most values of p. But then cos(2t0 logp) is approximately 1 for most values of p, and logiC(u+2itoll,......"" cos(2tologp),...... ""!,.....Jog! 7 pa 7Pa u-I' so that 1+2it0 is a pole of {(s). Since this is false, it follows that W+it,)'f'O. To put the argument in a rigorous form, let 8 = L: -ia• p = L: ~(t;~ogp), Q = 2: cos(2;!ogp). . . . 3.2 AND DE LA VALLEE POUSSIN Let 8', P', (j·be the parts of these sums for which (2k+I)1T-<X ~ t0 logp ~ (2k+I)7T+(}: 47 for any integer k, and a fixed, 0 < (}: < f'l'T. Let 8", etc., be the re-mainders. Let.\= 8'JS. If E is any positive number, it follows from (3.2.2) and (3.2.3) that P< -(1-<)8 if u-I is small enough. But P' ~ -8'= -.\8 and P" ~ -S"cos(}: = -(I-.)8cosa. Hence -{.+(1-.)cosa}8 < -(I-E)8, i.e. (I-.)(1-cosa) 1). (3.3.3) Now, keeping t fixed, let a~ I. Then ('(a) ~ O{(a-1)-'). a_nd, ifi+~tisazeroofC(s), C(u+it) = O(u-1). Also C(u+2it) = 0{1), smce C(s) lS regular at I+2it. Hence the left-hand side of (3.3.3) is O(u-1), giving a contradiction. This proves the theorem. There are other inequalities of the same type as (3.3.I), which can be used for the same purpose; e.g. from 5+8cos,P+4cos2,P+cos3,P = {l+cos,P)(I+2cos,P)2 ~ 0 (3.3.4) we deduce that ('(a)j((a+it)l'i((a+2it)l'l((a+3it)i ;;;, I. (3.3.5) This, however, has no particular advantage over (3.3.3). 3.4. Another alternative proof has been given by Ingham. f This depends on the identity C 2 (sg(st;:~C(s-ai) = ~ lu,.~:)l 2 (a> 1), (3.4.!) where a is any real number other than zero, and a.,(n) ~ l d•'. t Ingham (3). AND DE LA VALLEE POUSSIN This is the particular case of (1.3.3) obtained by putting ai for a and -ai for b. Let u0 be the abscissa of convergence of the series (3.4.I). Then a0 .::;;; I, and (3.4.1) is valid by analytic continuation for a> a0 , the function f(s) on the left-hand side being of necessity regular in this half-plane. Also, since all the coefficients in the Dirichlet series are positive, the real point of the line of convergence, viz. s = a0, is a singularity of the function. Suppose now that l+ai is a zero of C{s). Then 1-ai is also a zero, and these two zeros can~l the double pole of C 2(s) at s = I. Hencef(s) is regular on the real axis as far ass= -1, where C(2s) = 0; and so u0 = -I. This is easily seen in various ways to be impossible; for example (3.4.I) would then givef{!) ~I, whereas in fact/(!)= 0. 3.5. In the following sections we extend as far as we can the ideas suggested by § 3.1. Since C(s) has a finite number of zeros in the rectangle 0 .:::;; a .::;;; I, 0 .:::;; t .:::;; T and none of them lie on a = 1, it follows that there is a rectangle 1-8.::;;; a.::;;; I, 0.:::;; t.:::;; T, which is free from zeros. Here S = S(T) may, for all we can prove, tend to zero as T ~ oo; but we can obtain a positive lower bound for S(T) for each value ofT. Again, since 1/C{s) is regular for a = I, I .::;;; t ~ T, it has an upper bound in the interval, which is a function ofT. We also investigate the behaviour of this upper bound as t ~ oo. There is, of course, a similar problem for C{s), in which the distribution of the zeros is not imme-diately involved. It is convenient to consider all these problems together, and we begin with C(s). THEOREM 3.5. We have ((s) ~ O(logt) uniformly in the region 1--1A .:::;;-a.:::;;-2 (t>t0), ogt U'here A is any positive constant. In particular W+it) ~ O(logt). In (2.1.3), take a> I, a= N, and make b -+00. We obtain (3.5.1) (3.5.2) (3.5.3) 50 THE THEOREM OF HADAMARD Chap. Ill the result holding by analytic continuation for u > 0. Hence for u > 0, t>l, ~a(.~.) +O(N;-•)+O(.V-•) (3.5.4) In the region considered, if n ~ t, [n--$1 =n-o= e-alogn ~ exp( -(1-lo~t)logn} ~ n-1e..i, Hence, taking N = [t], {('l ~ ~,o(M+o(X,)+oaJ+o(M ~ O(log N)+ 0( I) ~ O(log !). This result will be improved later (Theorems 5.16, 6.1 .. ), but at the cost of far more difficult proofs. It is a1so easy to see that {'(') ~ O(Iog't) (3.5.5) in the above region. For, differentiating (3.5.3), N ro ''(s) =- 'logn + f [xl-x+f (1-slogx)dx-~ ns x"+l n-2 X - NI-sJogN-Nl-•)z+f.~.Y-•IogN, 8-l (8-1 and a similar argument holds, with an extra. factor log ton the right·hand side. Similarly for higher derivatives of {(s). We may note in passing that (3.5.3) shows the behaviour of the Dirichlet series (1.1.1) foro,;: I. If we take a= 1, t-=/=; 0, we obtain {(l+il)- f ____! = (L+it) f«>[x]-x_+! ck+ N_-' 1 -t~:V-1-il, L... nt+il x2+il tt > N which oscillates finitely as .N ~ oo. For u < 1 the series, of course, diverges (oscillates infinitely). 3.6. Inequalities for l/{(8), {'(8)/"8), and log "8). Inequalities of this type in the neighbourhood of u = I can now be obtained by a slight elaboration of the argument of§ 3.3. We have for u > I 1 •<·~it) I ~ g(a))lj{(·+2it)jl ~ a((~~:~,J· (3.6.1) 3.6 AND DE LA VALLEE POUSSIN Also {(1-fit)-{(a+il) ~ -! {'(u+il) du ~ O{(a-l)log't) (3.6.2) for u > 1-Aflogt. Hence l'(l+it)l > A1 (u-I)i -A 2(u-l)log2t. logit The two terms on the right are of the same order if u-1 = !og-Dt. Hence, taking u-I = A 3log-9t, where A 3 is sufficiently small, IW+i<)l > Aiog-'t. (3.6.3) Next (3.6.2) and (3.6.3) together give, for 1-Alogt < u Alog-'t-A(l-a)log't, (3.6.4) and the right-hand side is positive if I-u < Alog-8t. Hence ,(8) has no zeros in the region a> I-A log-"t, and in fact, by (3.6.4), ~ ~ O(log't) (3.6.5) in this region. Hence also, by (3.5.5), ~(~i ~ O(Iog't), (3.6.6) and log{(')~ I" {'(u+it) du+Iog((2+i<) ~ O(log't), (3.6.7) 1 {(u+tt) both for u > I-Alog-"t. We shall see later that all these results can be improved, but they are sufficient for some purposes. 3.7. The Prime·number Theorem. Let n(x) denote the number of primes not exceeding x. Then as x ~ oo n(x)"'lo;x. (3.7.1) The investigation of n(x) was, of course, the original. pu_rpose for which "8) was studied. It is not our purpose to pursue th~ stde of the theory farther than is necessary, but it is convenient to msert here a proof of the main theorem on 11(x). We have proved in (1.1.3) that, if u > 1, f ro ~(x) '-log"s) = 8 z x(XS-I) (W.;, We want an explicit formula. for 11(x), i.e. we want to invert the above 52 THE THEOREM OF HADAMARD Chap. m integral formula.-We can reduce this to a case of Mellin's inversion formula as follows. Let w(s) = f x"+l7~~1) dx. Then log{(s)w(s) =I"' 1T(X) dx. s xs+t (3.7.2) ' This is of the :Mellin form, and w(s) is a comparatively trivial function; in fact since n(x) ~ x the integral for w(s) converges uniformly for a ;;:::. t+O, by camparison with [xi<'(::..- I)' Hence w(s) is regular and bounded for a ~ l+S. Similarly so is w'(s), since w'(s) = J 'IT(X)Iogx iJ!+~;~l)2 dx. We could now use Mellin's inversion formula, but the resulting formula is not easily manageable. We therefore modify (3.7.2) as follows. Differentiating with respect to s, _ {'{8)+log{(s) '() ~ I"n(x)logxdx. s"s) ss +w 8 z x"+l Denote the left-hand side by cfo(s), and let ( ) _I' n(u)logud gx---- u, u 0 h(x) ~ j g~) du, 71(x), g(x), and h(x) being zero for x < 2. Then, integrating by parts, tfo(s) = J g'(x)x--s dx = s j g(x)x-s-1 dx 0 0 = 8 j h'(x)x-e dx = 82 r h(xp;-e-l dx (a> I), 0 0 3.7 AND DE LA VALLEE POUSSIN Now h(x) i.a continuous and of bounded variation in a.ny finite interval; and, since 'l'l'(x) ~ x, it follows that, for x > 1, g(x) ~ xlogx, and h(x) ~ xlogx. Hence h(x):rk-2 is absolutely integrable over (O,oo) if k < 0. Hence k+i< h(x)~_J I tj!-8)x~ds (k (I-8)a ' •+<-h(x)~-21 I +(:)x'ds (c>l). .. 8 c-i«> The integral on the right is absolutely convergent, since by (3.6.6) and (3.6. 7) rfo(8) is bounded for a ~ 1, except in the neighbourhood of s = 1. In the neighbourhood of s = I rfo(s) = s~1+log8~I+···· and we may write where !fo(s) is bounded for u ~ I, js-lj ~ 1, and if( a) has a. logarithmic infinity as 8-+ I. Now c+i h -I I x' I I ~(8) (x)- 2m (s- 1)81ds+2;n ~zB ds. c-iao e-i«> The first term is equal to the sum of the residues on the left of the line R(s) = c, and so is x-logx-1. In the other term we may put c = 1, i.e. apply Cauchy's theorem to the rectangle (1±iT, c±iT), with an indentation of radius r: round s = 1, and make T-+ oo, r:-+ 0. Hence h(x) = x-logx-1+~ I"'if1(1+it)xudt. 2n -• (!+it)' The last integral tends to zero as x-+ oo, by the extension to Fourier integrals of the Riemann-Lebesgue theorem.t Hence h(x)- x. (3.7.3) t See my lmrodteaion lo th$ Thwry of Fouriu lntegmlll, Theoram I. 54 THE THEOREM OF HADAMARD To get back to 1r(x) we now use the following lemma: Let j(x) be be positive non-decretUJing, and as x--+ oo let ft~)du-x Then f(x)-x. If S is a. given positive number, (1-S)x < J llfldt < (1+0)x (x > x0(S)). Hence for any positive " x(l+~) :t(l+f) x I ¥d·~ I ~~)du- V~)du " . . < (1+0)(l+•)x-(1-S)x ~ (2S+c+o<)X. But, sincej(x) is non-decreasing, Chap. III x(l+f) x{l+•) X{l+~l f f~) du ~f(x) f ~ > j(x) J x(~~") = 1 ~,/(x). Hence j(x) < x(I+c:-)(1+8+~)· Taking, for example, e = "1/8, it follows that lini~~l. Similarly, by considering j /~) du, z(l-•) we obtain and the lemma follows. Applying the lemma twice, we deduce from (3.7.3) that g(x)-x, and hence that 1T(x)logx,..., x. 3.8. THEOREM 3.8. There is a constant A such that "8) is not zero for u;;;::::1- 1~t (t>t0). 3.8 AND DE LA VALLEE POUSSIN We have for u >I -R{f(~:J =~I;~~ cos(mtlogp). Hence, for a > 1 and any real y, _ 3 {'(a) _ 4R {'(a+ir) -R {'(a+2iy) ((a) ((a+ir) ((a+2iy) "iogp =f.;:. pma {3+4cos(mylogp)+cos(2m.ylogp)} ~ 0. Now -f(~i<a~1+0( 1 ) Also, by (2.12.7), -f(~/ ~ O(iogt)-L (,~P +~)· p where p = fl+iy runs through complex zeros of {(s). Hence -R(f(~il ~ O(iogt)-L ((a fi;1t,, y)'+~,!A p Since every term in the last sum is positive, it follows that -R(f(~il < O(iogt), and also, if f3+iy is a particular zero of {(a), that -R({'(a+ir)) 01 1 ((a+ir) < ( ogy)- a-fi· From (3.8.2), (3.8.3), (3.8.5), (3.8.6) we obtain 3 4 17_ 1- a-f3+ O(logy) ~ 0, or say a~!- a~f3 ~ -Allogy. Solving for {3, we obtain l-{3 >- l-(u-l)A 1 logy ::--- 3/(a I)+A 1logy' The right-hand side is positive if a-1 = fA 1flogy, and then the required result. 1-f3~1~zy' (3.8.1) (3.8.2) (3.8.3) (3.8.4) (3.8.5) (3.8.6) THE THEOREM OF HADAMARD Chap. Ill 3.9. There is an alternative method, due to Landau,t of obtaining results of this kind, in which the analytic character of {(s) for cr.::;;; 0 need not be known. It depends on the following lemmas. LEMMA a:. If f(s) is regular, and lf(s)l < ,M (M >I) f(s,) in the circle js-s0 ] :::;;; r, then l f'(s) 2; I I< AM (ls-s,l,;; ir), f(s)-P s-p T where p runs through the zeros of f(s) such that ]p-s0 j :::;;; !r-The function g(s) = f(s) II (s-p)-1 is regular for js-s0 ] :::;;; r, and not p zero for js-s0 \ :::;;; tr. On js-s0 ] = r, ]s-p] ~ !r ~ js0-p], so that 1:~~:)1 ~If~~:) TI (':~:ll,;; If~~:>~ < ,M, This inequality therefore holds inside the circle also. Hence the function h(s) ~ log(:1~!))' where the logarithm is zero at s = s0, is regular for ]s-s0 j :::;;; }r, and h(s,) ~ 0, R{h(s)} < M. Hence by the Borel~Caratheodory theorem+ I hi•) I <AM 11•-•ol ,;; f'), and so, for js-so\ :::;;; fr, lh'(s)l ~ ~~ J ."J:l....d,l <AM_ 2m (z-s)2 r 1a~s1~1~ This gives the result stated. (3.9.1) LEMMA f3. If f(s) satisfies the conditioM of the previous lemma, and has no zeros in the right-hand half of the circle js-s0 j :::;;; r, then -R(f'ls,)) <AM; f(s,) r while if J(s) has a zero Po betw~?.en s0--!r and s0, then R(f'(s,)) AM I -j(so) < r- So-Po. t Landau (14). t Titchn:uush, Theory of Functi<ms, § 5.5. 3.9 AND DE LA VALLEE POUSSIN 57 Lemma a:·gives -R(f'(s,)) IMA y. Let f(s) satisfy the conditions of Lemma a:, and let 1 !'1•,)1 <!J!. f(so) r' Suppose also tha.tf(s) #- 0 in the part a~ a0-2r' of the circle js-s0 j:::;;; r, u•here 0 < r' < fr. Then ~~~:1 <A'¥ 11•-•,1 ,;; r'). Lemma. a: now gives -R(f'l•lj < A !J!- 'R_I < A !J!. f(s) r L.. s-p r for. ~11 s_ in l_s-s0_1 .:::;;; fr, cr ~ cr0-2r', each term of the sum being positive m this regwn. The result then follows on applying the Borel~ Caratheodory theorem to the function -j'(s)jf(s) and the circles js-s0 j = 2r', js-s0 j = r'. 3.10. We can now prove the following general theorem, which we shall apply later with special forms of the functions 8(t) and cfo(t). THEOREM 3.10. Let as t ~ oo in the region 1-tl(t),;;;;; a,;;;;; 2 (t ~ 0), where cfo(t) and 1j8(t) are positive non-decreasing functions oft for t ~ o, such that O(t),;;;;; I, cfo(t) ~ oo, and (3.10.1) Then thtre is a constant A 1 such that {(s) has no zeros in the region 0 >-!-A 012'+!) (3.10.2) ::or 1cfo{2t+I)' Let P+iy be a zero of {(s) in the upper half-plane. Let 1+e-4<2y+l).:::;;; ao.:::;;; 2, So= cr0+iy, s~ = a0+2iy, r = 8(2y+I). 58 THE THEOREM OF HADAMARD Then the circles j.s-80 j ~ r, js-s~j ~ r both lie in the region u~ 1-0(t). Now 1-'-1 < ~ < Aeof><2i'+ll, {(s0 ) u0-l and similarly for s~. Hence there is a constant A 2 such that ((•,) ((s,) Chap. ill I {(s) I< eA,(2y+tl, I b(~) I< eA,(2y+ll, in the circles js-s0j ~ r, js-s~j ~ r respectively. We can therefore apply Lemma~ with M = A 2 ,P(2y+l). We obtain R(('(a,+2iy)) < A,~( 2r+l), (3.10.3) -((a0+2iy) 8(2y+l) and, if Hence f3 > uo-fr· -R(('(a,+iy)) < A,f(2y+l) __ 1_. b(u0+iy) 8(2y+l) u0-/3 b'(ao) 1 -C(un) ,.., uo-1. - ~(~:: < u 0: l' where a can be made as near l as we please by choice of uo. Now (3.8.2), (3.10.3), (3.10.5), and (3.10.6) give -""--+ 5A,f(2y+l) __ 4_;;;, 0, u 0-l 8(2y+l) aG-f3 ( 3a 5A, f(2y+l))-' Uo-fJ ~ 4(u0-I) +4 0(2y+l) ' 1-~:;, (3<>__+"6f(2y+l))-' -(a,-1) 9' 4(u0-l) 4 8(2y+l) (3.10.4) (3.10.5) (3.10.6) (I-""- 5A, f(2y+l)(a,-1))/(---""-+ 5A,f(2y+ll). -4 4 8(2y+l) 4(a0-1) 4 8(2y+l) To make the numerator positive, take a = ~. and a,-l ~ 40~,!~~~!:~. this being consistent with the previous conditions, by (3.10.1), if Y is large enough. It follows that I-P 124:~:~~2;+1) 3.10 AND DE LA VALLEE POUSSIN 59 as required. If (3.10.4) is not satisfied, ~ <:;, a,-!r ~I+! 8(2y+l)_t8(2y+l) 40A,~(2y+l) ' which also leads to (3.10.2). This proves the theorem In particular, we can take 8(t) = f, t/J(t) = log(t+.2). This gives a. new proof of Theorem 3.8. 3.11. THEOREM 3.11. UnderthehypotheseaojTheorem 3.10wehave {'(s) ~ a(f(2t+3l). _I _ o(f(2t+3l) ((s) 8(2!+3) ((s)-8(2!+3) (3.11.1), (3.11.2) uniformly for In particular W+it) _ o(f(2t+3)) ('(!+it)-8(2!+3) ' c-1- ~ o(f(2t+3)) ((i+it) 8(2!+3). (3.11.3) We apply Lemma y, with (3.11.4), (3.11.5) 8 _ 1 +A1 8(2t0+3) . o-2 t/J(2to+3) +lto, r = 8(2to+3). In the circle Js-s0 [ ~ r and ((s) ( ei<O) (f(2! +3) ) {(so)= 0 O"o-1 = 0 8(2t:+3) e<W•+l) = O{e4.~2to+3l}, ns,) ~a(-~-)~ a(f(2t,+3l) _ o(~(2t,+3l) {(so) o-0 -I 8(2t0+3) ·---,- · v;'( e) chan therefore take M = A¢(2to + 3). Also, by the previous theorem ., s as no zeros for ' t ~ t0+I, a-;;:-:: l-A1 8{2(to+I)+I} I-A 8(2t0+3) f{2(t,+l)+l}-'~(2t,+3)' Hence we can take 2r' = ~ 8(2t0+3) 2 f(2t,+3)' ('(s) ~ o(f(2t,+3)) ((s) 8(2t,+3) Hence l·•-s, I <:;, 3A, 8(2!0+3) 4 ~(2!,+3)' for and in particular for t = t0 , u ;;:-:: 1-~ 8(2to+3) This is (3.11.1), with to instead oft. 4 t/J{2to+3J' 60 Also, if THE THEOREM OF HADAMARD l- ~ 8(2t+3) ,:.::: 17 s;:: I+ 8(2t+3) 4 ¢(2!+3) ~ ~ ¢(2!+3)' 1+~ '( 8(2!+31) J o(¢(2!+3)) du <;;Jog, I+ ¢(2!+3) + " 8(2!+3) < JogA8i~:~";~) +0(1). Chap. Ill (3.11.6) Hence (3.11.2) follows if a is in the range (3.11.6); and for larger a it is trivial. Since we may take O(t) = {, "'(t) = log(t+2), it follows that ('(•) ~ O(Jogt). ,---( 1 ) ~ O(logt) (3.11.7), (3.11.8) ((•) ,, in a region u ~ 1-A/logt; and in particular ~g:::; ~ O(Jogt), ((!~it)~ O(Jogt). (3.11.9), (3.1LIO) 3.12. For the next theorem we require the following lemma. LEMMA 3.12. Let where an = O{~(n)}, t/J(n) being non-decreasing, and ~~~o(-~-) ;S-1 nu (u-1)"' a8 u-'1--1. Then if c > 0, u+c > l, xis not an integer, .and N is the integer nearest to x, (3.12.1) 3.12 AND DE LA VALLEE POUSSIN 61 If xis an integer, the corresponding resuU is .,1 c+iT ~ ~+~ = ~ J f(s+w)~dw+O{T(u+:-l)"}+ n-1 c-iT ( ~(2x)x'-"logx) (~(x)x-") +0 --T-- +0 -T-. (3.12.2) Suppose first that xis not an integer. If n < x, the calculus of residues zk( T + T + T'')(;)"~~L -oo-iT c-iT c+iT gives Now _ o( (xfn)" ) +O( (x/nY s· du ) -Tlogxfn logxfn"" u2+T2 - o( (xfn)' ) -Tlogxfn ' and similarly for the integral over (-oo-iT,c-iT). Hence !, x we argue similarly with -oo replaced by +oo, and there is no residue term. We therefore obtain a similar result without the term 1. Multiplying by ann-sand summing, If n < !x or n > 2x, llogxfnl >A, and these parts of the sum are o(~ Ia. I) o( 1 ) :{= 1n"tc -(u+c-l)o: • If N < n:::;; 2x, letn = N+r. Then log;~ log:t;> ~> ~-62 THE THEOREM OF HADAMARD Chap. Ill Hence this part of the sum is o{t~s(2x)xl-a--c L ~} = O{tf(2x)x1-0'-Il}ogx}. l<(r<;;z A similar argument applies to the terms with !x ~ n < N. Finally _l_aN_I _ _ of .f(N) ) _ o(.f(N)x'-•~) N·~llogx(NI -\N·~Iog(1+(x-N)(N) -Tx=Ni . Hence (3.12.1) follows. If x is an integer, all goes as before except for the term ~c+JiT~=~lo c+~T =~(i1f+o(.!.)). 211tx' w 2mx" g c-~T 21Ttx" T c-iT Hence (3.12.2) follows. 3.13. THEOREM 3.13. We have I ~~(a) ,(8)= .. ~ n;-at all points of the line a = 1. Take a .. = p.(n), a:= 1, u = I, in the lemma, and let x be half an odd integer. We obtain 2 ~(a)~~ 'J+<T_1_x" dw+o(c&')+o(logx). n<>:: nw 21ft c-iT '(s+w) w Tc T The theorem of residues gives ~ T {('~w) ~ dw ~~+zk cr + 7T + r) c-iT c-iT -8-iT -li+iT if 3 is so small that {(s+w) has no zeros for R(w) ;;> -o, ll(s+w)l <;:; lti+T. By§ 3.6 we can take 3 =A log-9T .. Then -li+iT T J '(s:w) ~ dw = o( x-11 log 7T J ~(O~~v2)) -ll-iT -T T" =0{x---lllog7T f ..J( 1~v2)} = O(x---lllogl'T), -Till AND DE LA VALLEE POUSSIN and 'J+iT 1 x" dw ~ o(log'T I' . d ) - o(x'log''l') -ll+iT {(s+w) w T -ll x u --T-}' and similarly for the other integral. Hence .~ ~~~~-~ ~ o(fo)+o0T)+o("'I~'T)+o0o~T)· Take c = lflogx, so that X'= e; and take T = exp{(Iogx)l/10}, so that logT = (logx)1110, 8 = A(logx)-91 10, xll = TA., Then the right-hand side tends to zero, and the result follows. In particular ~ 1 p.~n) = 0, 3.14. The series for ~'(s)j{(s) and log {(s) on a= I. Takingt a,.= A(n) = O(logn), o: = 1, a= I, in the lemma, we obtain 2 A(n) ~ -~ 'J+iT{'(,+w)x"dw+O(c&') oflog'x) n<:c nB 2mc-iT {(s+w) w Tc + r T . In this case there is a pole a.t w = 1-s, giving a residue term ~~~~)- ;~ (s ;/:: 1), a-logx (s = 1), where a is a constant. Hence if s =1=- 1 we obtain -.;;' A(n) + {'(') _ x'-• ~ o(c&') + 0 (log'x) oflog'"'l') (''log'T) ,.~ ns {(s) 1-s Tc r T + r ~-,+O -T-7 Taking c = 1/logx, T = exp{(logx)l/10}, we obtain as before -.;;' A(n) {'(') x'~ 6 --;;-+ {(s) -1_ 8 = o(l). (3.14.1) The term x1-'/(l-s) oscillates finitely, so that if R(s) = 1, s =1=- 1, the series ~A(n)n-8 is not convergent, but its partial sums are bounded. If s = I, we obtain 6 A~n) = logx+0(1), (3.14.2) or, since -.;;' A(n) ~ -.;;' logp + ~ -.;;' Jogp ~ -.;;' Iogp 6 n f<;c P f,:2 p~:c pm f<;c P +0(1), ,L logp = logx+O(l). (3.14.3) P<:c p t See (l.l.S). THE THEOREM OF HADAMARD Cbap. III SinceA1(n) = A(n)flogn, and lflogn tends steadily to zero, it follows that is convergent on u = I, except for t = 0. Hence, by the continuity theorem for Dirichlet series, the equation log C(s) = ~ AI(n) £=2 ns holds for a = l, t oF 0. To determine the behaviour of this series for s = I we have, as in the case of 1/C(s), where c = !flogx, and Tis chosen as before. Now b T1og{(w+1)~dw~bC'r + -r + T)+b f. c-iT c-iT -/l-iT -ll+iT C where 0 is a. loop starting and finishing at s = -0, and encircling the origin in the positive direction. Defining 0 as before, the integral along a= -0 is O(x-llJog10T), and the integrals along the horizontal sides are O(xcT-llog9 T), by (3.6. 7). Since ~{loge(w+l)-log~} is regular at the origin, the last term is equal to ------; log--dw. 1 I 1 xw Zn-t ww Since a I I l dw I 2 ------; log--= -"L:'A0 log w 2m ww ..-;n a this term is also equal to I I lx"'-1 2;i logw----w- dw-log8. a Take 0 to be a circle with centre w = 0 and radius p (p < 8), together 3.14 AND DE LA VALLEE POUSSIN 65 with the fiegment ( -0, - p) of the real axis described twice. The integrals along the real segments together give l I' ( l )·-·-! l I' ( l )·-·-I ---: log --,--.-,-- -- du-------; log --- du 2m ue-•17 -u 2m uei17 -u . ' =- j x-:-1 du =-llfue-v;l dv p plog:~: = Jr~e-t> dv-11 T"'~ dv+log(Ologx) p]OKZ 1 ~ r+1og(81ogx)+o(l) if plogx -Jo. 0 and 3logx ~co. Also flog~ x"';;l dw = o(plog~logx). )u:)-p Taking p = lflog2x, say, it follows that ~ A~n) = loglogx+r+o(I). The left-hand side can also be written in the form 2:~+ 2: 2: m 1 •· 1><:~: p m;;.z P"'<z p As x -+ oo, the second term clearly tends to the limit Hence """.!. = loglogx+r- ~ """~+o(l). P~P f=z 7mpm (3.14.4) (3.14.5) 3.15. Euler's product on a = I. The above analysis shows that for u = I, t -=;6 0, where p runs through primes and q through powers of primes. In fact the second series is absolutely convergent on a = 1, since it is merely a rearrangement of 66 THE THEOREM OF HADAMARD which is absolutely convergent by comparison with Hence also Taking exponentials, = ""'log~ 1 -(o = 1, t:? 0). 7 1-p-<' •(•) = n ,~p~· p Chap. III (3.15.1) i.e. Euler's product holds on u = I, except a.t t = 0. At s = 1 the product is, of course, not convergent, but we can obtain an asymptotic formula for its partial products, viz. ( ') ,,. TI l-p '"'""Iogx" ... To prove this, we have to prove that f(x) = -log n ('-~) = loglogx+r+o (I). P<• Now we have proved that Also g(x) = ,£,A 1 7 ~n) = loglogx+r+o(I). oo I <~6,mpm' P"'>.:ll (3.15.2) which tends to zero as x-+ co, since the double series is absolutely convergent. This proves (3.15.2). AND DE LA VALLEE POUSSIN It will sJso be useful later to note that n(1+~)- 6eY~~s·. p.;;z p >r (3.15,3) For the left-hand side is n l-l/p 2 ,,eY}o xn(~-.!) _ eY!ogx = 6eY}ogx p.;;x 1-l/p g P pZ -"2) 17"1 • Note also that (3.14.3), (3.14.5) with error term 0(1), and (3.15.2) can be proved in an elementary way, i.e. without the theory of the Riemann zeta-function; see Hardy and Wright, ThE Theory of Numbers (5th edn), Theorems 425 and 427-429. Indeed the proof of Theorem 427 yields (3.14.5) with the error term o(l). logx NOTES FOR CHAPTER 3 3.16. The original elementary proofs of the prime number theorem may be found in Selberg {2] and ErdOs [1 ], and a thorough survey of the ideas involved is given by Diamond (1]. The sharpest error term obtained by elementary methods to date is x(x) = Li(x)+ O[x exp{ -(log x)l-e} ], (3.16.1) for any e > 0, due to Lavrik and Sobirov . Pintz has obtained a very precise relationship between zero-free regions of ( (s) and the error term in the prime-number theorem. Specifically, if we define R(x) = max{IJt(t)-Li(t)l: 2 ~ t ~ x}. then log R;x)""' m!n { (1- p)logx +log lyl}. (x ..... oo), the minimum being over non-trivial zeros p of ((s). Thus (3.16.1) yields (1-fJ)log x+log jyj p (log x)l~t for any p and any X. Now, on taking log X= (1- f1)- 1 loglyl we deduce that !-p,.(loglyl)5 , •• for any e' > 0. This should be compared with Theorem 3.8. 3.17. It may be observed in the proof of Theorem 3.10 that the bound ((s) = O(e.P(t)) is only required in the immediate vicinity of s 0 and sO. It would be nice to eliminate consideration of sO and so to have a result of 68 THE THEOREM OF HADAMARD Chap. Ill the strength of Theorem 3.10, giving a zero-free region around 1 +it solely in terms of an estimate for { (s) in a neighbourhood of 1 +it. Ingham's method in §3.4 is of special interest because it avoids any reference to the behaviour of ((s) near 1+2iy. It is possible to get quantitative zero-free regions in this way, by incorporating simple sieve estimates (Balasubramanian and Ramachandra ). Thus, for exam-ple, the analysis of § 3.8 yields L logp{l+cos(mylogp)}.:::; ..2.-~~p+O(logy). P•m p"'a u-1 (1-However one can show that L {l+cos(ylogp)}" 1 X X X<p<;;.2X og for X:;.: y2, by using a lower bound ofChebychev type for the number of primes X !. y > 0, and that b satisfies 1 - p ~ {J ~ (log y) - i. Then there is some r with b ~ r ~ 1 for which ,, n(y, r, r) + n(2y, r, r) 'P {}2 (1 - fJ). (3.18.1) Roughly speaking, this says that if 1 - fJ is small, there must be many other zeros near either 1+iy or 1+2iy. Montgomery gives a more precise version of this principle, as do Ramachandra (1] and Balasubramanian and Ramachandra [3 ]. To obtain a zero-free region AND DE LA VALLEE POUSSIN 69 one couples hypotheses of the type used in Theorem 3.10 with Jensen's Theorem, to obtain an upper bound for n(t, r, r). For example, the bound ((s)«(l+T'-")logT, T~ 1tl+2, which follows from Theorem 4.11, leads to n(t, r, r) ~ rlog T+loglog T+log~. (3.18.2) On choosing b = (loglog y)/(log y), a comparison of (3.18.1) and (3.18.2) produces Theorem 3.8 again. One can also use the Epstein zeta-function of §2.18 and the Maass-Selberg formula (2.18.9) to prove the non-vanishing of ((8 ) for u = 1. For, if 8 =!+it and ¢(8) = 2((28) = 0, then l~(l:+it)l' ~ ,P(s)~(l-s) ~ ¢(s)¢(1-s) ~ l¢(l:+ i~l' ~ 0, by the functional equation (2.19.1). Thus (2.18.9) yields If- -dxdy D B(z, 8)B(z, w) ~ = 0 for any w f 8, 1-8. This, of course, may be extended to w = 8 or w = 1-8 by continuity. Taking w =!-it= 8 we obtain ~fiB<•. •ll' d~~y ~ o so that B(z, s) must be identically zero. This however is impossible since the fourier coefficient for n = 1 is according to (2.18.5), and this does not vanish identically. The above contradiction shows that ((28) =F 0. One can get quantitative estimates by such methods, but only rather weak ones. It seems that the proof given here has its origins in unpublished work of Selberg. 3.19. Lemma 3.12 is a version of Perron's formula. It is sometimes useful to have a form of this in which the error is bounded as x---+ N. 70 THE THEOREM OF HADAMARD Chap. Ill LEMMA 3.19. Under the hypotheses of Lemma 3.12 one has '~=~ crTf(s+w):::_dw+O{T(a::-1)~} '--n• 2ttt J w n ,;;x c-iT +0{~(2x)x~•logx }+o{~(N)x-•min(Tix"-Ni' 1) }· Th" f llows at once from Lemma 3.12 unless x- N = O(xjT). In the latte;sca':e one merely estimates the contribution from the term n = N ~ 0{ ~(N)N-•), and the result follows. IV APPROXIMATE FORMULAE 4.1. IN this chapter we shall prove a number of approximate formulae for ((B) and for various sums related to it. We shall begin by proving some general results on integrals and series of a certain type. 4.2. LEMMA 4.2. Let F(x) be a real differentiable function 81.Wh that F'(x) is monotonic, and F'(x);;::: m > 0, or F'(x) ~ ~m < 0, throughout the interval [a, b ]. Then (4.2.!) Suppose, for example, that F'(x) is positive increasing. Then by the second mean-value theorem b ' f ( F( )} dx ~ f F'(x)cos(F(x)) dx a COS X a F'(x) ~ F':a) f F'(x)cos(F(x)) dx sin{F(fl}-sin(F(al) F'(a) ' and the modulus of this does not exceed 2/m. A similar argument applies to the imaginary part, and the result follows. 4.3. More generally, we have LEMMA 4.3. Let F(x) and G(x) be real functions, G(x)JF'(x) monotonic, and F'(x)JG(x) ~ m > 0, or ~ -m < 0. Then 11 G(x)ei1'1x) dx I~~-The proof is similar to that of the previous lemma.. The values of the constants in these lemmas are usually not of any importance. 4.4. LEMMA 4.4, Let F(x) be a real function, tun:ce differentiable, and let FH(x) ~ r > 0, or F"(x) ~ -r < 0, throughout the interval [a, b]. Then (4.4.1) APPROXIMATE FORMULAE Chap. IV Consider, for example, the first alternative. Then F'(x) is steadily increasing, and so vanishes at most once in the interval (a, b), say at c. Let b c-8 c+8 b I= I eiFf;c)dx =I+ I+ I= I 1+I2+I 3 , a a c-8 c+8 where 0 is a positive number to be chosen later, and it is assumed that a+O ~ c ~ b~O. In I3 F'(x) = f F~(t) dt > r(x~c) ;;:::. rS. Hence, by Lemma. 4.2, I 1 satisfies the same inequality, and 11 21 ~ 28. Hence j/j <;;~+28. TakingS= 2r-l, we obtain the result. If c < a+S, or c > b~S, the argument is similar. 4.5. LE.MMA 4.5. Let F(x) satisfy the cowiitions of the previous lemma, ami let G(x)JF'(x) b~ monotonic, ami IG{x)l ~ M. Then /j G(x)eiFdxl < ~· The proof is similar to the previous one, but uses Lemma 4.3 instead of Lemma 4.2. 4.6. LEMMA4.6. Let F(x) be real, with derivatives up to the third order. Let and 0 < ..\2 < -F"(x) < A..\2, jF"(x)j <;;AA,, throughout the interval (a, b). Let F'(c) = 0, where a ~c ~b. TMn in the case (4.6.1) (4.6.1) (4.6.2) (4.6.3) (4.6.4) In the case (4.6.2) the factor eti" is replaced by e-li"'. If F'(x) does not vanish on [a, b] then (4.6.5) holds without the leading term. APPROXIMATE FORMULAE If F'(x) dOes not vanish on [a, b] the result follows from Lemmas 4.2 and 4.4. Otherwise either (4.6.1) cr (4.6.2) shows that F(x) is monotonic, and so vanishes at only one point c. We put assuming that a+S < c < b-S. By (4.2.1) AJ.o c+8 c+8 J ~ J exp[i{F(c)+(x-c)F'(c)+!(x-c)'F"(c)+ c-8 c-8 +1(x-c)'F"(c+8(x-c))] dx o+a = eiFfe) I e·}l(x·-<">'F"(c)[I+O{(x-cf..\3}]dx ,, o+8 = eiF(c) I eii(x--c)"F 6(c)dx+O(S"'t). ,, Supposing F"(c) > 0, and putting i(x-c)'F.(c) ~ u, the integral becomes i8'.F'(c) «> (F~!)}t I ~du~(F~!))l(f~du+0(8~~) ' ' ~ (2tr)fe.¥• a(!\ {F'(c)jl + SA;}' Taking S = (-\A3)---i, the result follows. If b-8 < c < b, there is also an error o+8 eiF(c) I ef-i(x--cl'F"(r)dx = o(-1-l = o(-1 -) and also O(A-t)• 0 (b-c)AJ iF'(b)j ' ' and similarly if a < c ~ a+S. 74 APPROXIMATE FORMULAE Chap. IV 4. 7. We now turn to the consideration of exponential sums, i.e. sums of the form I eZ...U(n.J, wheref(n) is a real function. If the numbersf(n) are the values taken by a functionj(x) of a. simple kind, we can approximate to such a sum by an integral, or by a sum of integrals. LEMMA 4.7.t Letf(x) be a real function witA a cantin1UJU8 and steadily decreasing derivativef'(x) in (a, b), amf,letf'(b) = o:,j'(a) = [3. Then ' a~.o;;b et1r(f(n) = .. -.,2:<fi+'IJ e 2"i{f(a:J-v~J dx+O{log(,B-o:+2)}, (4.7.1) wMre '1'J is any positive WMtant less than I. We may suppose without loss of generality that 1)-1 < o: ~ "'• so that v ;<: 0; for if k is the integer such that TJ-l < a.-k ~ "'• and h(x) ~ /(x)-kx, then (4.7.1) is ' a<~.;:b et:rrih(n> = "''-'l)<v~k<.ot+'lj etriiA<zHv-k);zJ dx+O{log(.,8' -o:' +2)}, where o:' = o:-k, {3' = /3-k, i.e. the same formula for h(x). In (2.1.2), let fjJ(x) = eh-!.f(.o;:J. Then ' ' a<?;.;behif = j e2"if(zldx+ 1 (x-[x]-f)2nij'(x)e21rifdx+0(1). Also if x is not an integer; and the series is boundedly convergent, so that we may multiply by an integrable function and integrate term-by-term. Hence the second term on the right is equal to . ' -2iL fsin~I"'TXe'l.trif(zlj'(x)dx v-la . ' = L ~I (e-2rriv:r:_e2trivz)e2'1rlf(z)j'(x) dx. v-l a The integral may be written ' ' !.. J f'(x) d(e2'1rl{/{2:)-v:r:l) _ ! J !.J3:)__d(el'lrl{f+vzJ) .,; • /'(z)-> -2m • J'(x)+• . t van W!r Coxput (1). APPROXIMATE FORMULAE Since 1 {~\~v is steadily decreasing, the second term is o(l+-;). by applying the second mean-value theorem to the real and imaginary parts. Hence this term contributes o(~,r/+,J ~ o(~~)+o(~~) ~ O{Jog(~+Z)}+O(l). Similarly the first term is O{ftf(v-{3)} for v ~ {3+1), and this contributes Finally and the integrated terms are O{log(f3+ 2)}. The result therefore follows. 4.8. As a. particular case, we have LEMMA 4.8. Let f(x) be a real differentiable function in the interval [a, b ], letf'(x) be monotonic, and let 1/'(x)l ~ 8 < 1. Then ' a<<;;be2"i/(n) = J e'lfri/(zldx+O(l). (4.8.1) Taking 'I < 1-8, the sum on the right of (4.7.1) either reduces to the single term v = 0, or, ifj'(x) ~'I or ~ -'I) throughout [a,bJ, it is null, and ' 1 e'lfrif(z) dx = 0(1) by Lemma 4.2. 4-9. THEOREM 4.9. t Let [(x) be a real function with ckriuatives up to the third order. Let {'(x) be steadily decreasing in a~ x ~ b, and ['(b)= a, ['(a)= {3. Let x. be ckfined by /'(•.) ~ ' (• < ' ,;;; ~). t van der Corput (2). 76 APPROXIMATE FORMULAE Chap. IV L<t A,,;; lf"(x)l < A.l,, IJ"'(x)l <A.I,. Then I e21rif(n) = e--1-m L: e2"i~(xvl-~;'l + O(Azl)+ a<.n4:b a<v.;;fJ jj {x.)jo +O[Iog{2+(b-a).I,)]+O{(b-a).lf.I!J. We use Lemma 4. 7, where now ~-a~ O{(b-a).\,). Also we can replace the limits of summation on the right·hand side by (a:+I,,B-1), with error 0(.\;ai). Lemma 4.6. then gives ' "" f "" •'""'"''~·~) L... e~·b•i{/(x)-vxJ dx = cii" L ---+ «+l<v<fil-1 a «+l<v<,B-1 Jj"(x.)ji + L O("•"~l+ L; (o(I)+o(,i_)J· <>+I<v<,B-1 o:+l<v<,B-l V-~ ,.-v The second term on the right is 0{(~-a).l;!,\f) ~ O{(b-a)At ,\f), and the last term is O{log(2+P-a)} ~ O[log{2+(b-a).I,)J. Finally we can replace the limits {rx+I,,B-1) by (o:,,B] with error O(~t). 4.10. LEMMA 4.10. Letj(x) satisfytheaameconditiansasinLemma 4.7, and let g(x) be a real positive decreasing function, with a continuous derivative g'(x), and let jg'(x)j be steadily decreasing. Then ' a<.~;;;;bg(n)e21Ti/(n) = a-TJ<~ 0, jtj < 2nxf0, when 0 is a given constant greater than l. We have, by (3.5.3), ((s) ~ ~ .!, N'~• +•f"[uJ-u+!du-tN~ ~ ns l-s us+l n-1 N ~ ~ .!. N'~• +o(1.1)+ow~•). .6i n' 1-s Nu (4.11.2) The sum is of the form considered in the above lemma, with g(u) = u- 0 , and Thus Hence Hence j(u) ~ _tl~!"• f'(u) ~ -2',. If'(•) I ,;; 2:. < ~· = Nl-_xl-+ O(x-o). 1-s Making N ~ oo, the result follows. t Hardy and Littlewood {3). 78 APPROXIMATE FORMULAE Chap. IV 4.12. For many purposes the sum involved in Theorem 4.ll contains too many terms (at least Altl) to be of use. We therefore consider the result of taking smaller values of x in the above formulae. The form of the result is given by Theorem 4.9, with an extra factor g(n) in the sum. If we ignore error terms for the moment, this gives Taking ~ g(n)e2'1Ti/(:n),....., e-l<ri ,2: e2'11"i~(:J:.l-v:•l g(x.). a(Zn)i{ 1 +o(n }· (4.12.2) xi•) ~ (~f"-te«+l•>(I+O(nl· (4.12.3) Hence the above relation is equivalent to 2; ~ ~ xi•) 2; ;h· x<n.;;N l/2•N<v"-l/2m: The formulae therefore suggest that, with some suitable error terms, {(s)-- 2: ~+x(s) 2: ~· .... ., """ where 2""11 ~ 111. 4.12 APPROXIMATE FORMULAE 79 Actually the result is that {(s) = L ~+x(s) L n:_8 +0(x-")+O(It1~-"y"-1) (4.12.4) n;;;x n;;;y for 0 < u < 1. This is known as the approximate functional equation. t 4.13. THEOREM 4.13. If his a positive constant, 0 < u < I, 27TXY = t, x > h > 0, y > h > 0, th 0. By Lemma. 4.10 2 ~ ~ 2 [''::• du+O(x~•Jog(~-~+2)). x<n.;;;N i/2TrN-'1 t, the first term is v = 0, i.e. N J ~ = -!'!:-s-xl--8' us 1-8 Hence by ( 4, 11.2) {(s) = "".!..+ "" JN e2n-ivu du+O(x-"logt)+O(tN-"), L. ns L. us n<.;:r 1;;;v;;;v+q., since x1-•j(l-8) = O(:r") = O(x-"logt). Now [ •'::·· du ~ r(l-,)("J'"('. and by Lemma 4.3 J • u~",-'"""''"'~·~••> du ~ a(~) ~ o(N~•). "' N v--.,(tj27TN) v J u-8ebivudu = [u 1 -s e~!.>rivu]"' _ 2-nivf u1-Be2,.ivudu 0 l-8 0 l-80 ~ o(x',-")+o(i.-7:;-;,x))-t Hardy and Littlewood (3), (4), (6), Siegel (2). 80 APPROXIMATE FORMULAE Chap. IV Hence There is still a possible term corresponding to y-7] < v .::;;; y+ 7J; for this, by Lemma 4.5, giving a term Finally we can replace v .::>;; y-"1 by v ~ y with error a([("y'y-'r(l-,)[y"-') ~ O(tl-•1f'-'). AlsO fort> Q x(s) = 2S~-lsint81Tr(l-8) = 2sn-"-1{ _ e-:"" + O(e-t""t)}r(I-s) ~ ("rt'P(1-,)(1+0(e-•')). Hence the result follows on taking N large enough. It is possible to prove the full result by a refinement of the above methods. We shall not give the details here, since the result will be obtained by another method, depending on contour integration. 4.14. Complex-variable methods. An extremely powerful method of obtaining approximate formulae for {(s) is to express b(s) as a contour integral, and then move the contour into a position where it can be suitably dealt with. The following is a simple example. Alternative proof of Theorem 4.11. We may suppose without loss of generality that xis half an odd integer, since the last term in the sum, which might be affected by the restriction, is O(x-a), and so is the possible variation in xt-w/(1-s). APPROXIMATE FORMULAE Suppose finrt. that 0 > 1. Then a simple application of the theorem of residues shows that x+i"" {(s)- I n-w = In-s= - ~ f z-8 cot1TZ dz n:r 2t :r-i"" 1 J' 1 •s+i<» . xl-8 = _2l (cot 17z-i)z-•dz-2i (cot?TZ+l)z-sdz- 1 8. :r-i"" X The final formula holds, by the theory of analytic continuation, for all values of s, since the last two integrals are uniformly convergent in any finite region. In the second integral we put z = x+ir, so that [cot?Tz+il = I+2e2"• < 2e-2"r, and [z-sl = [z[-aelargz < z-aellll'IN:i:an(r/x) < z-aelllr/z. Hence the modulus of this term does not exceed x-•J e-2"r+llr/xdr= ~. 0 27T-[t[/x A similar result holds for the other integral, and the theorem follows. It is possible to prove the approximate functional equation by an extension of this argument; we may write -cot1TZ-i = 2i ~ e2vA, i.e. for comparatively small values of x, if n is large. However, the rest of the argument suggested is not particularly simple, and we prefer another proof, which will be more useful for further developments. 4.15. THEOREM 4.15. The approximate functional equation (4.12.4) holds for 0 h > 0, y > h > 0. It is possible to extend the result to any strip -k <::: a< k by slight changes in the argument. l•'or a> l b(s) = ~ !+! s""x"-le-m" dx. n~ln" r(s)o e""-1 82 APPROXIMATE FORMULAE Chap. IV Tra.naforming the integral into & loop-integral as in§ 2.4, we obtain ~ 1 e-.:"8r(l-s) J w-te-mw '(a)= 61 ns+~ a ew-1 dw, where 0 excludes the zeros of e 10-1 other than w = 0. This holds for all values of s except positive integers. Let t > 0 and x .:!( y, so that x :::;;: ..j(tj27r). Let a~ I, m ~ [x], y ~ tj('kx), q ~ [y], "~ 2"1J. We deform the contour 0 into the straight lines q, 0 2, 03, 04 joining oo, cTJ+i7J(l+c), -CTJ+i1J(l-c), -cTJ-(2q+l}ni, co, where cis an absolute constant, 0 < c ,;; f. If y is an integer, a •mall indentation is made ~ above the pole at w = iTJ. We have then <(s) ~%.. ~+x(•) ~ n' 1~+ ,-, .. ;~-•)( J + J + J + J )· c. c, c. c. Let w = u+iv = pei (0 < tfo < 21T), Then /W-1) =pa-le--It/>, On 04, rp;;:, in", p > A1), and /e"'-11 >A. Hence I f I ~ a( "•->d•• r ,-m• du) ~ O(e""1-l•') ~ O(e"•-!»l). c, -"'1 On Ca, rp ~ in+ arctan 1 ~c > !1r+c+A where A > 0, since ' ' arctano=J~>J~-8 I+,u2 (l+p.)2 -1+8' ' ' Hence w-1e-mw = 0("17"-Ie--t(j-n-+<'+4l+"""') = 0(7J"-te-t(V+-4l) and jew-11 >A. Hence j = 0(7J"e--t Aeu. Hence w'-'e-mw [ f (I + ) ) ~ = 0 7J"-1expl-tarctan~-(m+I)u ]· Since m+I ~ x = tf7J, and !!farctan(I+c)7J+~)= -~+.! 0 du\ u 7J u2+(I+c)27J2 7J > • 4.15 APPROXIMATE FORMULAE we have arctan (I+c)TJ+~ ~ arcta.nl+c +c u " ' = !'"+c-arctan 1 ~c = i'"+A, sincefor0<8<1 8 f d~ e arctan6 < 0 (l-,u)2 = 1_ 6. Hence = 0(7J"e-<f,.+-4lt)+0(7J"-1e-"'P') = 0(7J"e-<Vr+4 lt). Finally consider 0 2• Here w = i7J+.:%.e-li", where~ is real, j.:\j < ../2C7J· Hence w'-' ~ exp((s-l){!i~+Iog("+Arl;•))] = exp((s-l){!i'"+logTJ+~e-li"-~ ~e-fi"+0(~)}] Also e:::7 = oe7:-e-::) (u ~ 0), = o(:::~ which is bounded for u < -P,. and u > !'"; and je-"""1 = e-M/1!-Jz. Hence the part with juj > }1r is (u < 0), = o{7Ja-Ie-!"~l e-A>.•'I-•tt'L} = 0(7Jat--e-llrl). The argument also applies to the part juj <!'"if je"'-11 >A on this part. If not, suppose, for example, that the contour goes too near to the pole atw = 2tpi. Take it round an arc of the circle jw-2q1Tij = P,.. On this circle, w = 2q1Ti+-fnet0 APPROXIMATE FORMULAE Chap. IV and log(w"-le-m"') = -!m:rre'~ll+(s-1){fi7T+log(2qn+f"7re111ji)} teiO = -!m'ITe10-}7Tt+(s-l)log(2q7T)+4q+0(1). Since this is Hence 1117T-~ = 2~-t = 0(1), -!•t+(s-I)log(2qo)+O(l). [ws-te-m"'l = O(qo--te-i"l). The contribution of this part is therefore 0(7]<1-le-l<rl). Since we have now proved that C(s) = ~ ~+x(s) ~ nL.+O{ti--o-(e-AI+1"/o-t-i+ 71o--l)}. The 0-terms are O(c")+OIW"•-")+o(tt-o(~)"-') = O(e-A1 )+0(x-o-)+0(t-ix1 -o-) = O(x-o-). This proves the theorem in the case considered. To deduce the case x ~ y, change 8 into l-8 in the result already obtained. Then {(1-s) = L nL,+x(1-8) _L ~+O(xo--1). .... , ""'" Multiplying by x(s), and using the functional equation and x(s)x(l-s)~ I, we obtain Interchanging x andy, this gives the theorem with x ~ y. 4.16. Further approximations.t A closer examination of the above analysis, together with a knowledge of the formulae of§ 2.10, shows that the 0-terms in the approximate functional equation can be replaced by a.n asymptotic aeries, each term of which contains trigono-metrical functions and powers oft only. t Siegel (2). 4.16 APPROXIMATE FORMULAE 85 We shall sconsider only the simplest case in which x = y = ,j(tj21T), 1"/ = .j(2rrt). In the neighbourhood of w = iTj we have (s-I)log7 ~ (s-l)log I+-.-w ( w-i•) i7J ITj ~ (a+it-l)(w-:i" !(w-:i•)' + ... J ''1 2 l7J = f,;(w-i1"/)+~(w-i7J)'+···· Hence we write e(l-l)log(w/i1J) = e(1J/21rXw-i'1)-t(i/4,.XW-i1J)'.P(4-;:1)• where say. Now ~ = (-8-~ -i-Vt+iz)tfo(z) = a-+l~izZ,P(z). dz z+-.t z -.t Hence (z+-.'t) I na,.zn-l = (a-I+iz2) i a .. zn, n=l n-o and the coefficients an are detennined in succession by the recurrence formula (n+Ib't.an+l = (a-n-l)a .. +ian-2 (n = 2, 3, ... ), this being true for n = 0, n = I also if we write a_2 = a_1 = 0. Thus (a-l}(a-2) a2 = --2-, ,--, It follows that (4.16.1) (not uniformly inn); for if this is true upton, then a,.+t = OWln+linl-i)+O(t-l<"-tJ+Ii(n-2Jl-}) = O(t-i rfo(z) = .. ~ 0 a,.z"+rN(z). rN(z) = ..!. f tfo(w)zN dw, 2ml' ur"(w-z) APPROXIMATE FORMULAE where r is a contour including the points 0 and z. Now log<f(w) = (s-l)log(l+~)+!iw2-iw.Vt = (a-l)log(l +~)+iw2 ~ {~~:-~(~f· Hence for jwj ~ ~,.,It we have R{log,l(w)) <:; la-lllog~ + lwl'· ~ Lf,. Chap. IV Let jzj < f'\i't, and let r be a circle with centre w = 0, radius PN> where Mjzj ~ P.v ~ ~,lt. Then t:v(z) = O(jzj·Yp_yNe5P116>1). The function p-N&J"/6VIhas the minimum (5ej2N'\i't)~N for p = (2N..Jtf5)l; PN can have this value if 2lii«(2N'\I't)t 3.y 20 z,~ ~ 5 t. Hence ( ( 5, )IN) r.v{z) = 0 jzjN 2N'\i't ( N :'2, 1 ,1 ,;:: ~(2Nvt)l)· ~50, '""'21 5 ~ o("·-•d·'('¥TJ for N < At. The case where the contour goes near a pole gives a similar result, as in the previous section. 4.16 APPROXIMATE FORMULAE 87 In the firSt N terms we now replace 0 2 by the infinite straight line of which it is a part, 0' 2 say. The integral multiplying a,. changes by o{ '7"-le-~11"1 I e-(,\'fb)+(').!211",'2) -(m+IX.\1~2)(J(k) n d>..}. b Since m+I "";;;:- t/'7 = '7/(277), this is ok•e-l·• J"e-''J•"CI,;nr d+ h We can write the integrand as e-A'/8" X e-A•fe"(~)" .j(2n) ' and the second factor is steadily decreasing for >.. > 2.j(n7T), and so throughout the interval of integration if n < N < At with A small enough. The whole term is then o("l'j"-le-i"t-<'l'/3271"')(-"-)") = 0{'7"-le-i"f-(l/167r)(!.Vt)n} 2.j(2n) · Also a .. = (r~~-rn+l)z-n = o{(2~~~Y"}· Hence the total error is o{'7"-te-i-"t--(f/ts">~ (}Vt)n(2~ 1 )~~~} = o{'i"-te-l"t--(t/ta"> ~: (1 5;!)1"}· Now (tfn)l-"' increases steadily up ton= tje, and so if n <At, where A< 1/e,itis O(eft.Alogl/A). Hence if N < At, with A small enough, the whole term is O(e-<f,.+All). We have finally the sum . _1 N~l a., I e(i/47rXW-i'))'+{')/27rXw-i'l)-mw • (t"l'j)" L.... in(21T)in ew l (W-t"l'j)n dw. n-o Ci The integral may be expressed as -I exp(i (w+2m1Ti-i1j)2+.!L(w+2m1Ti-i1j)-mw) x L 47T 27T X (w+~::~;i7J)" dw, where Lis aline in the direction argw = f1T, passing between 0 a.nd 27Ti. 88 APPROXIMATE FORMULAE Chap. IV This is n! times the coefficient of go~ in _ f exp{i;(w+2m1Ti-i1))2+ " .. I .. 1) dw +-(w+2mm-t1J -mw+e(w+2m1Tt-HJ .,-------1 2'" e-~ -exp(i(2mn-"1(~-tm+<)) f exp(~+w(;-2mH)),::l L where 'Y(a) = COS1T~ 8 2 ::-aa-kl, = 21T( -1 )m-Ie-ttt-<si,.,./Sl'f"(~ _ 2m+e)ei'"~· = 217(-l)m-Ie-lil-<&i,.,./8) i 'YV'>(;!-2m)~! (}i:f'lv, ,..~o IL P-o Hence we obtain N-1 1 'v-n di1l'(s-ll(2n-t)is-i21T(-l)m-Ie-tll-<5i,./8l L L v!{:~2v)!2"X n~o v.;;fn ( 2)'·-· ( ) X ; ~ a,.'J"(n-2vl ;-2m. Denoting the last sum by SN, we have the following result. THEOREM 4.16. If 0:::,;; a::;;;; 1, m = [.J(t/27T)], and N t+O(t-1) no;;z n,;;"' = 2 L n-icos(!9--tlogn)+O(t-l). n("'c)ho,{t-(2m+I),I(2nt)-i•} O(d. (4.17_ 5) t cos...j(2wt) + ) 90 APPROXIMATE FORMULAE Chap. IV 4.18. A different type of approximate formula has been obtained by Meulenbeld. t Instead of using finite partial sums of the original Dirichlet series, we can approximate to C(s) by sums of the form ~c/>(:~z), where rfo(u) decreases from I to Oas uincreases from 0 to 1. Thisreduoes considerably the <D.'der of the error terms. The simplest result of this type is C(s) = 2 L ~~~/z +x(s} L n:,-"""' n<¥ -x(s) L n:,+ 2x<:-l) L nL.+o(Ja+k+xa~lt)• ¥<n.<:.l!l ¥<n<2u valid for 27TXY = Jtj, Jtl ~ (x+I)t, -2 < u < 2. There is also an approximate functional equation:j: for {C(s)}1. This is {((•)}' ~ L d~~) +x'(s) L :\"!.+O(xl-•Iogt), (4.18.1) "'"'"' .... v where 0 ~ u ~ 1, xy = {t/2w)2, z ~ h > 0, y ~ h > 0. The proofs of this a.re rather elaborate. NOTES FOR CHAPTER 4 4.19. Lemmas 4.2 and 4.4 can be generalized by taking F to be k times differentiable, and satisfying IF'<11>(x)l ~ .1. > 0 throughout [a, b]. By using induction, in the same way that Lemma 4.4 was deduced from Lemma 4.2, one finds that ' f eiF(x)dx4 11 A-IIk, The error term O(A 2-!.1. 3-!) in Lemma 4.6 may be replaced by 0(..1.:J 1 Aj), which is usually sharper in applications. To do this one chooses 6 = A3-t in the proof. It then suffices to show that . f. e""'(eif(x) -1) dx «(H)- '• (4.19.1) iff has a continuous first derivative and satisfies /(x) 4 x 30- 3 , t Menlenbeld ( 1 ). t Hardy and Littlewood (6), Titchmareh (21). APPROXIMATE FORMULAE {'(x) 4 x 2'0- 3. Here we have written A= !F'(c) and f(x) ~ F(xH)- F(c)-!x'F"(c). H 0 ~ (AO) 1 then (4.19.1) is immediate. Otherwise we have (l0)-1 d + f + f . -(A0)-1 (A0)-1 " The second integral on the right is trivially O{(A0)-1 }, while the third, for example, is, on integrating by parts, . f . 2 eif(xl_I (2i..tzeilx )-.-dx = 2tA.x ().0)-1 [ ., ···"·'-'J' s· ., . d (·'"''-') e'"x ---e'"x --- dx 2i.b (~<>)-• dx 2i..tx ().0)-1 l f(x)l s· lxi{'(x)e'"''-<·'""-1) I ~ max -+ dx x = (l<>)-1,0 Ax 2iAx2 ().0)-1 «(H)-'+ j,1;:1~:jdx «(H)-' as required. Similarly the error term 0{ (b- a)AtAf} in Theorem 4.9 may be replaced by O{(b-a)Ai}. For further estimates along these lines see Vinogradov [2; pp. 86-91] and Heath-Brown [11; Lemmas 6 and 10]. These papers show that the error term O((b-a)AiAb can be dropped entirely, under suitable conditions. Lemmas 4.2 and 4.8 have the following corollary, which is sometimes useful. LEMMA 4.19. Let {(x) be a real differentiable function on the interval [a,b],let{'(x) be monotonic, and letO<A:%; lf'(x)l:%; 9 < 1. Then L e2><if{n) 4sA-t. a<n<>l> 92 APPROXIMATE FORMULAE Chap. IV 4.20. Weighted approximate functional equations related to those mentioned in §4.18 have been given by Lavrik and Heath-Brown [3; Lemma 1], [4; Lemma 1]. As a typical example one has ns)k = ~d,.(n)n-sws(~)+x(s)" ~d,.(n)ns-lwl-8(;) + O(x 1 -"log"(2 + x)e -~'1 4) (4.20.1) uniformly fort~ 1, Ia I,:::;; !t, xy = (t/2rr.)", x, y ~ 1, for any fixed positive integer k. Here W (U) = ~ Cf+i"'( max(O, -a)). The advantage of (4.20.1) is the very small error term. Although the weight W 8(u) is a little awkward, it is easy to see, by moving the line of integration to c = ± 1, for example, that W8 (U)= { O(u-') (u ~ 1), l+O(u)+ o{u"(log~)-lt'} (O)" x"e"'~ 1 • x 21n r(!s) z (c > max(O, 1-a)), and moves the line of integration to R(z) = - d, d > max(O, a), giving 2xi +((s),.+Res(z = 1-s). The residue term is easily seen to be O{x1 -a log"(2+x)e-t'/4}. In the integral we substitute z = -w, x = (t/2x)"y- 1, and we apply the 4.20 APPROXIMATE FORMULAE 93 functional· equation (2.6.4). This yields -d+•"'' f ( (tt)-"12 r{l(s+z)} ((s+z>)" x"e"'~ 2xi -d-•"" r(~s) z as required. Another result of the same general nature is l((!+it)12" = m.t= 1 d,.(m)d,.(n)m -t-itn -!+itWt(mn)+ O(e-t'f2) (4.20.2) for t ;<: 1, and any fixed positive integer k, where 1+1"" W(u)~l f (•,r(l:<!+it+z))r{!<!-it+z)) )' , ,,dz ' rri ,,. r(!<!+it)) r{!<t-it)) u e -;-· This type of formula has the advantage that the cross terms which would arise on multipling (4.20.1) by its complex conjugate are absent. By moving the line of integration to R(z) = ±tone finds that W1(u)=2+0{ul}og"(~)} (O<u.::;:I), and W1(U) = O(u -!)for u ;3 1. Again better estimates are possible. The proof of (4.20.2) is similar to that of (4.20.1), and starts from the formula t mJ= 1 d,.(m)d,.(n)m -!-itn-!+itW 1(mn) ~ --'-o 'f··· (·-· r{l:(t+ it+z)) rwl:-it+z)) 2rro ,,. r{l:(t+it))r{!<!-it)) C<t+it+z)C(!-it+z) r e"'~. 94 APPROXIMATE FORMULAE Chap. IV 4.21. We may write the approximate functional equation (4.18.1) in the form ((s)2 = S(s, x) + x(s)2S(l-s,y)+R(s,x). The estimate R(s, x) ~ x!-<>Iogt has been shown by Jutila (see I vic [3; § 4.2]) to be best possible for t!~ lx-~1 ~ tf. Outside this range however, one can do better. Thus Jutila (in work to appear) has proved that R(s, x) ~ t!x-"(log t)log ( 1 +-7-)+ t- 1x 1-"(y"+log t) for 0 ~ u ~ 1 and x p t p 1. (The corresponding result for x ~ t may be deduced from this, via the functional equation.) For the special case x = y = tj2n one may also improve on(4.18.1). Motohashi , , and in work in the course of publication, has established some very precise results in this direction. In particular he has shown that where .&(x) is the remainder term in the Dirichlet divisor problem (see § 12.1). Jutila, in the work to appear, cited above, gives another proof of this. In fact, for the special case u = -!, the result was obtained 40 years earlier by Taylor (1). v THE ORDER OF {(s) IN THE CRITICAL STRIP 5.1. THE main object of this chapter is to discuss the order of ~(a) as t-+-oo in the 'critical strip' 0.:::;;; a.:::;;; l. We begin with a general dis· cussion of the order problem. It is clear from the original Dirichlet series (1.1.1} that {(a) is bounded in any half-plane a~ 1+8 > I; and we have proved in (2.12.2) that '(•) ~ 0(111) (a;;,!). For a < f, corresponding results follow from the functional equation '(•)~x(s)W-s). In any fixed strip lX.:::;;; a:::;;; (3, as t-+ oo by (4.12.3). Hence and lx(•)l ~ (~t· '(•) ~ 0(11-•) (a ,;; -S < 0), '(•) ~ 0(11") (a;;, -S). Thus in any half-plane a ~ a0 '(•) ~ 0(111'), k ~ k(a0 ), (5.l.I) i.e. C(a) is a function of finite order in the sense of the theory of Dirichlet series.t For each a we define a number fL(a) as the lower bound of numbers e such that '(a+il) ~ 0(1111). It follows from the general theory of Dirichlet seriest that, as a function of a, p;(a) is continuous, non-increasing, and convex downwards in the sense that no arc of the curve y = p;(a) has any point above its chord; also p;(a) is never negative. Since ((a) is bounded for a~ 1+8 (8 > 0), it follows that !'(a)~ 0 (a> 1), (5.1.2) and then from the functional equation that J'(a) ~ !-a (a < 0), (5.1.3) These equations also hold by continuity for a = 1 and a = 0 respec-tively. t See Titchmarsh, Theory of FunelionB, §§ 9,4, 9.41. t Ibid.,§§5.65,9.41. 96 THE ORDER OF {(s) C~p.V The chord joining the points (0, i) and (1, 0) on the curve y = p.(u) is y = l-tu. It therefore follows from the convexity property that ~(a) ~ l-!a (0 <a< 1). (5.1.4) In particular, fL(i):,;;;;; -!,i.e. ((!+it)~O(t!+<) (5.1.5) for every positive E. The exact value of p,(a) is not known for any value of a between 0 and I. It will be shown later that p.( l) < t, and the simplest possible hypothesis is that the graph of p.(a) consists of two straight lines ~(a)~ !-a (a~ !), 0 (a> tl· (5.1.6) This is known as LindelOf's hypothesis. It is equivalent to the state-ment that ((!+it)~OW) (5.1.7) for every positive "· The approximate functional equation gives a slight refinement on the above results. For example, taking a= f, x = y = ,J(tj21T) in (4.12.4), we obtain ((Hit)~ L nJ,.+O(l) L j:.+O(t-l) n<;;.J(t/2,.) n<;;4(1/21t) ~ o( L ~)+ort-11 n.;;>'(t/i11') ~O(tl). (5.1.8) 5.2. To improve upon this we have to show that a certain amount of cancelling occurs between the tenns of such a sum. We have ' ' L n-s = L n-oe-Ulog,. n~a+l n-a+l and we apply the familiar lemma of 'parti!tl summation'. Let bl ~ b2 ~ ••• ~ b.. ~ 0, and Bm = a1+a2+ ... +am where the a's are any real or complex numbers. Then if Js,..j.:;;;;M (m=l,2, ... ), ja1 b1+a2 b2+ ... +anbnl.:;;;; Mb 1• For a1b1+ .. +a,.b11 = b1 s1+b2(B:a-s1)+ ... +b.,(s..-s.,1) (6.2.1) = s1(b1-b:a)+s11(b2-b3)+ ... +s,._1(b,._1-b,.)+s.._b,.. 5.2 IN THE CRITICAL STRIP Hence latbt+ ··+a,.bnl ~ M(b1-b2+ ... +b..__1-b,+b11 ) = Mb 1• If 0 .:;;;; b 1 ~ b2 ~ ••• ~ b., we obtain similarly ialbl+···+anbnl ~ 2Mb 11• If a,.= e-Uloa-n, b .. = n-a, where u ~ 0, it follows that i n-8 = o(a-a max I I e-itlognj). (5.2.2) n~a+l a<c.;;b n-a+l This raises the general question of the order of sums of the form ' :E = n~~+l e2rri/(nl, (5.2.3) whenj(n) is a real function of n. In the above case, j(n) = -t~:gn. The earliest method of dealing with such sums is that of Weyl,t largely developed by Hardy and Littlewood.+ This is roughly as follows. We can reduce the problem of :E to that of ' S = n-f+I e2rriO(nl, ;::e g(n) is a. polynomial of sufficiently high degree, say of degree k. jSj 2 = LLe21ri(q(m)-u(n)) = ,L,Le21fi{u(n+~)-u(n)) m ,. ~ n (5.2.4) with suita~le limits for the suiDB; and g(n+v)-g(n) is of degree k-1. By repeatmg the process we ultimately obtain a sum of the form ' Sk= L e2wi(An+~<l. n=a+t We can now actually carry out the summation. We obtain S -~1-e2wi<b-a)AI 1 I kl-l-e2,.u- ~ jsin?T,\j' (5.2.5) lfjcosec?T.\j is small compared with b-a, this is a favourable result, and can be used t~ give a non-trivial result for the original sum S. An alternative method is due to van der Corput.§ In this method we approximate to the sum :E by :he corresponding integral J e21fif~ldx, t Weyl (1), (2). u + Littlewood (2), Landau (15). §van dt!r Corput (1}-(7), van der Corput and Kokama (1), Titeluna.nili (8HI2). 98 THE ORDER OF ((s) Chap. V and then estimate the integral by the principle of stationary phase, or some such method. Actually the original sum is usually not suitable for this process, and intermediate steps of the form (5.2.4) have to be used. Still another method has been introduced by Vinogradov. This is in some ways very complicated; but it avoids the k-fold repetition used in the Weyl-Hardy-Littlewood method, which for large k is very 'uneconomical'. An account of this method will be given in the next chapter. 5.3. The Weyl-Hardy-Littlewood method. The relation of the general sum to the sum involving polynomials is as follows: LEMMA 5.3. Let k be a poaitive integer, t ~ 1, b-:a ~ tt-lJ(k+ll, and ~~ 1 exp{-it(~-~~+···+(-l~:: 1 mk)}\ ~ M (~<:; b-a). Then l .. ~t+l e-ttlognl <AM. For ~I'~ exp(-it("':a-(-1)'-'m')-. (1-I)'mk+' )JI 6: 1 ··+ kak d (k+l)ak+l +··· <:;2M~ le.(t)l e:")' [ ( (b-a)'+' )] ~ 2Mexp t (k+I)ak+t+··· ( (b-a)'+'/( b-a)) ~ 2Mexp t~ 1-----a ,:::;:; 2Mez. IN THE CRITICAL STRIP 99 ~.4. Tile simplest case is that of {(!+it), and we begin by working folio~ We require the case k = 2 of the above lemma, and also the LEMMA. Let Then Putting m' = m-r, this takes the form I ! e2:rri(2amr---<»""+,8r) ,;:: ~~~ I ~ 4:rrfu:nvl m,. ""' .. ~~~t+l ii'e ' t ~he~,tecorrespoHnding to each value of r, m runs over at most"· consecu-Ive m gers. ence, by (5.2.5}, ' e-• = P-+2 ,.~ 1 min(,u, jcosec21ro:rj). 5.5. THEOREM 5.5. m+it) ~ O(tl!oglt). Let 2ti .:::; a < At, b ~ 2a, and let By§ 5·3• Lv = O(M), where M is the maximum of S ~ ! exp(- ·1 ( m I m' )) v m-l ~ a+v,u =2 (a+yp.)2 for p.' ~ ,u. By§ 5.4 this is o[(~+~'min(~·leosee-'-'" f))Jl. r-1 2(a+vp.)2 100 THE ORDER OF ((s) Chap. V Hence L ~ O{(N+I)~l)+O[('f I Nf 'f min(~, \oo,ec 21.:,~)•1Wl v~l v~l r~l ~ O{IN+I)~:}+o[(N+I)l(~%min(~·l•o•ec 210:,~)'1Wl Now tr tr trp,{2a+(2v+I)p:} 2(a+vJL)2. 2{a+(v+l)p}2· 2(a+vp.) 2{a+(v+l)~-t} 2 ' which as v varies lies between constant multiples of tr~-t/a 3 , or, by (5.5.1), ofrffl-2· H:nce for the values ofv for which !trj(a+vp.) 2 1ies in a certain interval {l-rr,(l±!)1T}, the least value but one of [sin2(a~·vfL)2 1 is greater than Arfp,2, the least but two is greater than 2Ar/f-1-2,,the least but three is greater than 3ArfJL2 , and so on to O(N) = O(ta) terms. Hence these values of v contribute fL+o(~+~+ ... ) = JL+o(~togt) = o(~logt). The number of such intervals {l1T, (l±il1r} is o{(N+I)~+l}. Hence the v-sum is Hence O{(N +I)logt}+O(~logt). [ ,,( . )]' L = O{{N+l)JLl}+O(N+I)l ~ (N+l)logt+t;:-Iogt ~ O{(N +l)~l)+O{(N+I)~tloglt}+O{(N +l)l~Iogt} = O(a!t1-Iogit)+O(at-flogt). If a = O(ti), the second term can be omitted. Then by partial summa· tion b L n:+il = O(tilogit) (b ~ 2a). n~a+l By adding O(log t) sums of the above form, we get L ntl+it = O(t! lo~t). 21h;n<;(t/21r)i IN THE CRITICAL STRIP 2: ;;} .. ~ o( 2: ;!,) ~ O(tl). n<2ti n<21i Also The result therefore follows from the approximate functional equation. 5.6. We now proceed to the general case. We require the following lemmas. LEMMA 5.6. Let f(x) = ~+ .. be a polytWmial of degree k with real coefficients. Let B=Ie2-rrif(m) where m ranges over at most p. consecutive integers. Let K = Zk-1. Then for k;;;-.2 [S[K ~ zzx,u.K-t+zKp.K-k,.,,.~!-•min(p., [cosec(1Takl r1 ... rk_1)1) where each r varies from 1 to JL-1. Fork = l the sum is replaced by the single term min(J.', [cosec ?Tal). We have [8[2 = ~e2Ri{J(m)-/(m1l = ~f:e2Ri[/(m)-/(m-riJ} (m' = m-r1 ) ~ "i_ 1 !Btl• r,~-1'+1 where 81 = ~ e1hri{/(m)-/(m-~iJ) = ~ e21ri(O£k1-,mt-•+···> and, for each r1, m ranges over at most 1-' consecutive integers. Hence by HOlder's inequality [8[2,;;; C.~"tl+lly-2/K(,.,:~,.~~ISII}Kr'K ~ (2JL)l-2/K(p.fK+ 'I} [81[~K)21K, r,=-~<+1 where the dash denotes that the term r 1 = 0 is omitted. Hence IBIK ~ (2t-t)iK-t(JLiK+ ,.,:~:+IIStf}K). If the theorem is true for k-1, then j81j}K ~ 2Kp.fK-t+ +2}KJL!K-k+lr,,.~.-•min(JL,Icosec{7T(akr1)(k-l)!r1 •• rk_1}[). 102 THE ORDER OF ((B) Chap. V Hence ISJK :S; 2-!K-I,uK-1+ + 2{KJ.tK-I+2Kp.K-k ..... ~k-'min(,u,j cosec(m:l':k! r 1 ... rk_1) 1), and the result fork follows. Since by§ 5.4 the result is true fork= 2, it holds generally. 5.7. LEMMA 5.7. For a< b ~ 2a, k ~ 2, K = 2k-I, a= O(t),t > l0 , L = n~+ln-il = O(ai-t/Ktt/((k+tlKliogl/Kt)+O(at-1/((k+IJKlJogkiKt). If a ,;;; 4tli(k+IJ, then I; = O(a) = O(ai-tJKtlf((k+IlKl) as required. Otherwise, let JL = [fat-1/(k+IJ], a.nd write Then I;v = O(M), where M is the maximum, for p,' ,;;;;; ll'· of Sv = •~t exp{-it(a.:v~-~(a;:JL)2+···+(-I)k-tk(a~:JL)k)}· By Lemma 5.6 S = O(,ul-IIK)+O[t.ti-k/K( "" min(JL.Icosec~£k-1)!rl ... rk-ti)J"Kl· v ,.,,:Sk-• I 2(a+vfl)k Hence ~ ~ O{(N+i)~>'-""J+ +0[,.1->/K I ( L minHcosect(k~~~:~);• 'llrKJ v~l ,.,, •• ,rl-I [ ( NH ( I t(k-I)'' ' lll'IK] +0 JLI-k/K(N+I)l-I!K ~ _L min JJ-, cosec~-f_v~)·"k-l v-lr, ... ,rk-1 by HOlder's inequality. 5.7 Now as V varies, IN THE CRITICAL STRIP t(k-l)lr1 ... rk-l t(k-1)lr1 ... rk 1 2(a+v,u)" 2{a+(v-1)p.}" 103 lies between constant multiples of t(k-I)!r1 ... rk_1 p.a-k-t, i.e. of (k-I)!r1 ... rk-tf'-k. The number of intervals of the form {l11, (l±f)7T} containing values of ft(k-I)!r1 ... rk_1(a+vp.)-k is therefore O{(N+I)(k-1)!rr .. rk_1 p.-k+I}. The part of the v-sum corresponding to each of these intervals is, as in the previous case, p.+O((k-1)~;1 .. rk_J+o(2.(k-~;r1 ... rk_J+ .. = p.+O( p.klogt . \ = o(~-'k logl\ (k-l)!r1 ... rk_J rr .. rk_J Hence the v-sum is O{(N +I)logt}+O(~~.~~=-tJ Summing with respect to r 1, ... , rk-t• we obtain O{(N +I ),.'-1 logt}+O(,.'Iog't). Hence :!: ~ O{(N+I),.1-•K)+O{(N+I),.1-UK]og"KtJ+O{(N+I)1-•K,.Jog'IKt). The first term on the right can be omitted, and since N+I = o(b:a +I)= O(t1/(k+l)) the result stated follows. 5.8. THEOREM 5.8. If l ia a :fixed integer greater than 2, and L = 2l-1, then ~(B)= O(tl/W+IJL)logl+I/Lt) (u = 1-1/L). The second term in Lemma 5. 7 can be omitted if Taking k =land applying the result O(logt), times we obtain I n-•1 = O(Nl-1/Ltlf{(l+IJL}logl/Lt), n,;;N for N;,;;; t21Cl+ ll}ogi-It. Similarly, fork< l, we find I n-•1 = O(Nl-1/Kt1/l(k+l)K)logl!Kt) t•~••2llog-kt <11,;; N (5.8.1) (5.8.2) 104 THE ORDER OF ((8') Chap, V for t21(11+2)}0 g-Rt < N,:::.;; t21{k+lllog1- 11 t. The error term here is at most O(Nl-1/Lf"}og' t) with ( 1 1) 2 1 ( 1 1) 1 "~ L-K k+2+(k+1)K' p~- L-K k+K-Thus p ~ 1/L. When k = l-1 we have ( 1 2) 2 2 2 1 tl= L-L i+l+IT=l(l+l)L <(l+t)L' and for 2:::.; k ~ l-2 we have ( 1 1) 2 1 k-1 1 a:::.; 4K.-K k+2+(k+1)K= 2(k+l)(k+2)K:::.;O<(l+l)L' It therefore follows, on summing over k, that (5.8.2) holds for N:::.; tilog- 1t. Hence, by partial summation, we have L n-s = O(tl/{(l+l)LJiogl+l/Lt), n.;(t/211)! L n•-I = O(t2u-l+l/({l+l)L)Jogl/Lt), n,;;(t/2n)l and the theorem follows from the approximate functional equation. 5.9. van der Corput's method. In this method we approximate to sums by integrals as in Chapter IV. THEOREM 5.9. If f(x) is real and twice differentiable, and o < >., ,;;; J"(x) ,;;; 1M, ("" ~. ,;;; -r<•) ,;;; IM,) throughout the interval [a, b], and b ;:-.a+ 1, then •<f'<, '"''"'' ~ O{h(b-a)~tJ+ O(.l;t). If ,\~ ;;;.:: 1 the result is trivial, since the sum is O(b-a). Lemmas 4.7 and 4,4 give 0{(~-•+1).\,-lJ+O{Iog(~-•+2)}, where ~-• ~ f'(a)-f'(b) ~ O{(b-a)IM,}. Since log(~-•+2) ~ 0(~-•+2) ~ O{(b-a)IM,}+0(1) ~ O((b-a)hAl)+0(1), the result follows. Otherwise 5.10 IN THE CRITICAL STRIP 105 5.10. )'uEmu. 5.10. Let f(n) be a real functiun, a< n ~ b, and q a positive integer not exceeding b-a. Then I L e'"'""'l <A~~+A(b-a:fl L e"""""'-"""IJI. a<n.«b q q r-1 a b. Then the inner sum vanishing if n ~ a-q or n > b-1. Hence I L e"if<•JI,;;; I L:l! e'•;J!m+•ll n q n m-1 ,;;; H~ 1 ~l~:""'m+·'I'J'-Since there are at most b-a+q ~ 2(b-a) values of n for which the inner sum does not vanish, this does not exceed Now = q+ !!e2.,.i{f(nHn)-/(l'+nll+ !!et:rri{f(m+,.J-/(f.'+")}· J<<m m<JJ-Hence ~ ,, ~ lm.;l e21Ti/(m+n)l ,;; 2(b-a)q+ 21 ~ ~' e2"i{/(m+n)-/(f.'+"))l· In the last sum, j(m+n)-f(p,+n) =f(v+r)-j(v), for given values ofv and r, 1 ~ r ~ q-1, just q-rtimes, namely p, = 1, m = r+I, up top, = q-r, m = q, with a consequent value of n in each case. Hence the modulus of this sum is equal to 1:~:(q-r) f e217M~+~J-/(v))l ~ q:~:~ ~ e2ni{J(v+r)-Mlj. (5.10.1) Hence I~ e"""''l,;;; H<(b-a)'q+<(b-a)q ~I~ e••<U<•+•H<•"Il'· and the result stated follows. 106 THE ORDER OF ((s) Chap. V 5.11. THEOREM 5.11. Letj(x) be real and have continwus derivatives up to the third order, and let .\3 ~ f'"(x) ~ hA3, or .\3 ~ -J"'(x) ~1M3 , and b-a ~ I. Then a<'..;l> e2"il<nl = O{ht(b-a)At}+O{(b-a)lA;l}. Let g(x) ~ f(x+,)-f(x). Then where x < ( < x+r. Hence r..\3 :::;;; g"(x) :::;;; hrA3 , or the same for -g"(x). Hence by Theorem 5.9 a q~ q ;f1 3 3 ~ o(b;;")+O{h(b-a)'ql,\t+(b-a)q·l,\01)! ~ O{(b-a)q·l)+O(hl(b-a)ql,\[)+O((b-a)tq·!,\0!). The first two terms are of the same order in .\3 if q = fA;lJ provided that .\3 :::;;; l. This gives O(ht(b-a )Al )+ O{(b-a)!,\;i) as stated. The theorem is plainly trivial if ,\3 > I. The proof also requires that q :::;;; b-a. If this is not satisfied, then b-a = 0(..\;t), b-a ~ O((b-a)!,\;i), and the result again follows. 5.12. THEOREM 5.12. m+it) ~ O(tllogt). Takingf{x) = -(277)-1tlogx, we have f"(x)~-~. Hence if b ::::; 2a the above theorem gives ~ O(altt)+O(ad), 5.12 IN THE CRITICAL STRIP 107 and the seoo"nd tenn can be omitted if a ::::; tf. Then by partial summa-tion (5.12.1) Also, by Theorem 5.9, I n·" ~ O(ti)+O(ad), a<n.-;b and hence by partial summation L n:,.~ o((WJ+o((~)!j. a<n.-;b Hence (5.I2.1) is also true if tt <a< t. Hence, applying (5.I2.I) O(log t) times, we obtain .L nAu= O(t-1-Iogt), •« and the result follows. 5.13. THEOREM 5.13. Letf(x) be real and have continuous derivatives up to the k-th order, where k ~ 4. Let -'k ~ J.k (or the same for -j<kl(x)). Let b-a ~I, K = 2k~t. Then I elhrif(nl = O{h21K(b-a)i\~{2K-2l}+O{(b-a)1~2/Ki\,t0K~2J}, a<noO:b where the constants implied are independent of k. If i\k ~ I the theorem is trivial, as before. Otherwise, suppose the theorem true for all integers up to k-1. Let g(x) ~ j(x+,)-f(x). Then g(k-l)(x) =J<k~l'!(x+r)-j<kl(x) = rj<kl(f), where X < e < x+r. Hence ri\k ~ g(k-l)(x) ~ hri\k. Hence the theorem with k-l for k gives la<ntb-r eZlriQ'(n)j < Atk'fK(b-a)(ri\k)tJ<K-2J+A2(b-ap~41K(ri\k)~tJ(K~2) (writing constants A 1, A 2 instead of the O's). Hence :~:~a<~b-relhrig(n)/ < Alk'IK(b-a)q1+1/(K-2)i\kl{(K~2)+ +2Az(b-a)l~'fKq1~1/(K-2)i\klf(K~2) since ' l r~l/(K~2) < r~l/(K-2) dr _ ---- ::;::: 2ql~l/(K~2) o-> f ql~l/(K-2) r~t 0 -l-1/(K-2)""' 108 THE ORDER OF ns) Chap. V forK~ 4. Hence, by Lemma 5.10, I ell"il<nl~Aa(b-a)q-l+A,(b-a)-l-q-i{Ath4JK(b-a)q1+1/(K-2J,\,V<K-2)+ a<n.;;b +2A11(b-a)l-41Kq1-1/<K-2l,\il/(K-2)}t :::;;; A3(b-a)q-i+A,A[h21K(b-a)ql/l2K---4.l,\l~2K--'l+ +A,(2A2)i(b-a)t-1/Kq-li<2K-4.l,_kl/(2K-4.l. To make the first two terms of the same order in Ak, let q = [A;;ll<K-l)J+I. Then Aj;ll<K-1) ~ q.::;;; 2.._;;lf(K-1), ql/(2K-4l,\fU2K-4) :( 2l/(2K-4l,'\,~/(2K-4.){1-l/(K-l)}.::;; 2..\l/(2K-2), q-I/(2K-4))._i;l/(2K-4) ~ .._;;1/(2K-2l, and we obtain I a<~<;be2"i/(n)l ~ (A 3+2A4Af)h21K(b-a)AlJ@tK-2l+ +A,(2A 2)f(b-a)t-2/K,\,;;I/(2K-2l. This gives the result fork; the constants are the same fork as for k-I if A 3+2A4 Af ~ A1, A,.(2A2)t ~ A2, which are satisfied if A1 and A 2 are large enough. We have assumed in the proof that q :::;;; b-a, which is true if 2.\,;-ll<K-1) ~ b-a. Otherwise 1a<~..;;be2"'i/(n)l ~ b-a ~ (b-a)l(2A,i;lf(K-ll)i ~ 2i(b-a)1-2/K,\k1/(l!K-2J, and the result again holds. 5.14. THEOREM 5.14. If l ~ 3, L = 21-1, a= 1-l/(2£-2), ~(s) = O(t1/{llL-2 llogt). We apply the above theorem with f(x) ~ _tiogx, 2rr If a< n ~ b ~ 2a, then (-l)'(k-l)!t 27Txk • (k-l)!t"' If"'( II"' (k-l)!t 211(2a)k """' x """' 2111lk ' and we may apply the theorem with Ak = ~1T(;,})!kt' h = 2k. (5.14.1) 5.14 IN THE CRITICAL STRIP 109 Hence ! n-il= o[22k/Ka((k-l)!t)l/(2K-2)]+0[ai-2/K((k-l)!t)-l/(2K-ll)l a<n..;;b 211(2a)k 21T(2a)k = O(al-k/(2K-2JtiJ<2K-2l)+O(ai-2/K+k/(2K-ll}t-IJI2K-lll). (5.14.2) The second term can be omitted if a< AtKJ<kK-2K+2J. (5.14.3) Hence by partial summation ,L n-.. = O(al-a-k/(2K-2Jtl/<2K-2l) a<n<;;b subject to (5.14.3). Taking a= l-l/(2L-2), _L n-.. = O(al/<2L-2J-k/12K-llJtik2K-2l). a<n.;;;b First take k = l. We obtain Hence L n-.. = O(tli<2L-2l) (a < AOLJ(IL-llL+lll). a<n.;;;b n.<;;tLI-2LH)n-8 =ijL/(IL-~LH)<~,;;tL/(IL-2L+2l +•• = O(tl/(2L-2))+0(t1/(2L-2l)+ .. = O(t1 1<2L-llllogt). Next L ~~ L + L +--·• tLIUL-1L+O)<n.<;;t it<n<;;t ll<n,;;}t and to each term ! corresponds a k < l such that 2-"'t<no;;;l!l-'"j tKJ{(k+l}K-2K+I) < 2-mt ~ tK/(kK-2K+2). Then ,2 ~ = O{t11/<llL-2J-k/<2K-2llK/IIk+l)K-2K+Il+l/12K-llJ}. 2-"'l<n..;;:l-'"t The index does not exceed that in (5.14.6) if ( l k ) K I I 2L-2-2K-2 (k+I)K-2K+I +2K-2 <;; 2L-2' which reduces to L-K ~ (l-k)K, i.e. 21-k_I ~ l-k which is true, Since there are again O(log t) terms, L ~ = O(tl/12L-2 >Iogt). jL/(IL-2L+"l}<n..:;t (5.14.4) (5.14.5) (5.14.6) The result therefore follows. Theorem 5.12 is the particular case l = 3, L = 4. no THE ORDER OF {(s) Chap. V 5.15. Comparison between the Hardy-Littlewood result and the van der Corput result. The Hardy-Littlewood method shows that the function p.(a) satisfies P-(1 - 2L~) ~ (k+:)zk-t' and the van der Corput method that P-(1 -21~2) ~ 21~2· For a. given k, determine l so that l-zlz-;~2 < l-2L1 ~ l-21~2· Then (5.15,2) and the convexity of p.(a) give 1 l-1 1 (5.15.1) (5.15.2) ( 1 ) 2f=i-2f=2 I 2f=L2- 2f<=i I p.I-zk-1:::;;; z 1 l 211 2+ l-1 l 21-2 zl-t_z- zl_z 21-1=2- zi_z 2}-k-1 1 = l2l-1_2f+2:::;;; (k+l)2k-l if (k+I)(2f-t_zk-1) .:s;;; (l-2)2l-1+2. Since zk-t > (21-1-2)/(l-1), this is true if (k+I)(zl-t_ zl-t_z):::;;; (l-2)2'-1+2, l-1 i.e. if k+l ~l-1. Now zk-t ~ zl72 <~:::;;; 2f-3 if l ~ 8. Hence the Hardy-Littlewood result follows from the van der Corput result if l ~ 8. For 4 ::s;; l :::;;; 7 the relevant values of I- a are H.-I~. v. d. c. The values of k and lin these cases are 3, 4, 5 and 5, 6, 7 respectively. Hence k,;;::; l~2 in all cases. IN THE CRITICAL STRIP 5.16. THEOREM 5.16. W+i<)~o(~). loglogt We have to apply the above results with k variable; in fact it will be seen from the analysis of§ 5.13 and§ 5.14 that the constants implied in the O's are independent of k. In particular, taking a= I in (5.14.4), we have ,2: nll+it = O(a-k/(2K-2Jtl/(2K-2l) (a < b ~ 2a) a<n-<;;b uniformly with respect to k, subject to (5.14.3). If tK/{(k+l)K-2K+l) <a,;;;;; tK/(kK-2~+2) it follows that ,L: nll+tt = O(t11<2K-2l-kK/I\2K-2)\kK-K+l)l) = O(t-l/{2\k-UK+2l). a-: ( logt ) { '} A \ exp (logt)~ > logt. Hence the above sum is bounded. Also 2 Lu = O(logtR/{(r-l)R+ll) =of~) n,.;IBI{!•-llB+In (r-l)R+l ~ a(Iog~ ~ a(.~). r } loglogt This proves the theorem. The same result can also be deduced from the Weyl-Hardy-Little-wood analysis. 112 THE ORDER OF {(s) 5.17. THEOREM 5.17. Fort> A {(s) = O{logat), u ~ 1-(lof~:~w·, {(s) of=- 0, a :;3: 1-Allol~;~~ (with some A 1), and I ~ o(~). {(l+it) loglogt ('ll+i•l~o(~) {(l+it) loglogt · Chap. V (5.17.1) (5.17.2) (5.17.3) (5.17.4) We observe that (5.14.1) holds with a constant independent of l, and also, by the PhragmCn-LindelOf theorem, uniformly for a;;, 1-!/(2£-2). Lett be given (sufficiently large), and let Then and similarly Hence l = [lo~2log(lol;.!~~)l L::;::: 2(Iil0&'2)log(logtjloglogtJ-1- 1 logt """ - 2Ioglogt' L~!~. 4loglogt l >- i >-loglogt-logloglogt-log2loglogt >- (loglogt)2 2L-2 <>- 2L:? log2 logt ,.... logt for t >A (large enough). Hence if then u ;:<: 1_ (loglog t)2 logt u;?: l-2Ll_2. Hence (5.14.1) is applicable, and gives {(s) = O(t11<2L-2>Iogt) = O(t1 1LJogt) = O(t<'loglogtflogt>logt) = O(logst). This proves (5.17.1). The remaining results then follow from Theorems 3.10 and 3.11, taking (fort> A) 8(1) ~ (lof!;~'l', f(l) ~ 51oglogt. IN THE CRITICAL STRIP 113 5.18. Jn,this section we reconsider the problem of the order of '(!+it). Small improvements on Theorem 5.12 have been obtained by various different methods. Results of the form ((!+it)~ O(t"loglt) 163 27 229 19 15 ct = 988' 164' 1392' il6' 92 with were proved by Walfisz (1), Titchmarsh (9), Phillips (1), Titchmarsh (24), and Min (l) respectively.t We shall give here the argument which leads to the index fj4• The main idea of the proof is that we combine Theorem 5.13 with Theorem 4.9, which enables us to transforyt a given exponential sum into another, which may be easier to deal w1th. THEOREM 5.18. (!Hit)~O(t'""'l· Consider the sum l:l = Z n-il= Z e-illogn, a<n..;;b a<n"'b where a< b ~ 2a, a< A"/t. By§ 5.10 1:,~ o(~)+o((~~I:E,I)l). (5.18.1) where q ~ b-a, and 1:2= z e-if{log(n+r)-logn). a<n,;;b-r We now apply Theorem 4.9. to 1:2. We have f(x) = -~{log(x+r)-logx}, j'(x) = 2-rrx::+r)' f'"(x) = ~ 3x~ix3;risr2. We can therefore apply Theorem 4.9 with .:\2 = tra-3 , As= tra-4 • Thus :E, ~ e-l•' 2: 1J:;:•~:1 +o(~)+o(log(2+~))+o(~). a.<v.:.{J ,. (5.18.2) where f(') ~ f(x,)-,.., " ~ j'(b-,), p ~ f'(a). Actually the log-term can be omitted, since it is O(t!ria-~). Consider next the sum L e2m</><~) (rx < r ~ {3). a.<v..;:y t Note that the proof of the lemma. in Titchmarsh (24) is incorrect. The !l!lllDl& ahould be replaced by the corresponding theorem in Titchrnarsh (16). THE ORDER OF ((6) Chap. V The numbers xp are given by tr _ . I( 2 2tr)l 21Txp(x~+r)- v, I.e. x~ = 2 r +;; -fr. Hence tf,'(v) = {f'(x~)-v}~-x~ = -x, = !r-Mr2+~)l, since rv ,;; rf'(a) = 21Ta~: 2 +r) ,;; 2::2 ,;; ~;. It follows that where K1, K 2, ... , and tk depend on k only. We may therefore apply Theorem 5.13, with h = 0(1), and Hence Ak = K 3(tr)i(trfa2)~-k = K3(tr)1-ka2k-1. ~ e2rr;~(y) = of~( ~2k~l )I/12K"-2)) o((~)l-2/K( a2k-l )-1/(2K-2)) n<~,;;y la2 tk-¥-1 + a2 tk-Irk-1 · Also IJ"(x,)l-i is monotonic a.nd of the form O(t-lr-iaf). Hence by partial summation L e:rri<j.{")1 = O{(tr)f-.k-t)/(2K-2J-l-}+ "<•<~ 1/ (x,) I • + O{(tr)}-2/K +{k-IJI(2K-2la4/K-}-<2k-IJI(2K-2>}. Hence I q-t -,L II:21 = O{(tq)f--i}+ q r-1 +O{(tq)l-2/K+{k-l)/(2K-2Ja4/K-f--{2k-1J/(2K-2l}+0{(tq)-fa-l}+0{(tq)fa-f}. Inserting this in (5.18.1), and using the inequality we obtain (X+Y+ ... )t,;; Xt+YI+··· I:1 = O(aq-l)+O{(tq)1--<k-tJI(4K-4Ja<2k-l.W<tK-4l+l}+ +O{(tq)f-l/K+(k-t)/(4K-4la2/K+:l-<2k-tJ/(4K---4l}+O{(tq)-lafJ+0{(tq)fafoJ. The :first two terms on the right are of the same order if q = [ai3K-2k-2)/(3K-k-2>t--{K-k)/(3K-k-2)], and they are then of the form O(acaK-2)JI~K-k-lllt(K-.tlii~K-k-SJl) = O(t<'K-Zk-t)'{'<SK-.1:-Z)}) (a< Av't). 5.18 IN THE CRITICAL STRIP Fork= 2~.3, 4, 5, 6, .. , the index has the values 1 3 5 17 73 2' 7' 12' 41' ffi•••• and of these H is the smallest. We therefore take k = 5, q ~ [aHt-ll] (a> tH), and obtain I:1 = O(aH·tH)+O(aU~tlh)+O(aHti--H)+O(a-i-k!c). This also holds if q ~ b-a, since then 116 L, ~ O(b-a) ~ O(q) ~ O(a!lt--ll), I which is of smaller order than the third term in the above right-hand side. It is easily seen that the last two tenns are negligible compared with the first if a = 0( -./t). Hence by partial summation 2: n:+U = O(aloti~)+O(a-ill-tJ.'oir) (a> tU). a<n"b Applying this with a= N, b = 2N-1; a= 2N, b = 4N-1, .. until b ~ [A"J, we obtain L ;;k, ~ O(tlt•)+O(N--h~tlh) N<n<;;.1<11 ~ O(th'<) (N > th). We require a subsidiary argumentfor n,;; t;."¥-, and infaot (5.14.2) with k = 4 gives ! n-" ~ O(altl<) a<n.;;:b 2: n~\u = O(a-.'dl'.r), a<.ll<::b and by adding terms of this type as before (a<At1), L n!1 +il= O(tl£17 1"+-l.-) = O(tN..-) = O(tN,). ,...;;11/17 The result therefore follows from the approximate functional equation. NOTES FOR CHAPTER 5 5.19. Two more completely different arguments have been given, leading to the estimate (5.19.1) 116 THE ORDER OF {(s) Cbsp. V Firstly Bombieri, in unpublished work, has used a method related to that of §6.12, together with the bound r r1 L. exp{2:n:i(a.x+,Bx2)}1 6da.d{3.:o;:P3JogP, J J lO<>P 0 0 to prove (5.19.1). Secondly, (5.19.1) follows from the mean-value bound (7.24.4) oflwaniec . (This deep result is described in §7.24.) Heath-Brown [9) has shown that the weaker estimate ~J(f) ~ ft follows from an argument analogous to Burgess's treatment of character sums. Moreover the bound ,u(!) ,;: fr,;, which is weaker still, but none the less non-trivial, follows from Heath-Brown's fourth-power moment (7.21.1), based on Weil's estimate for the Kloosterman sum. Thus there are some extremely diverse arguments leading to non-trivial bounds for .u(!). 5.20. The argument given in §5.18 is generalized by the 'method of exponent pairs' of van der Corput (1), (2) and Phillips (1). Let s, c be positive constants, and let.'F(s,c) be the set of quadruples (N, I, f,y) as follows: (i) Nand yare positive and satisfy yN-s;:: 1, (ii) I is a subinterval of (N, 2N], (iii) f is a real valued function on I, with derivatives of all orders, satisfying (5.20.1) for n ;:.0. We then say that(p, q)isan 'exponentpair'ifO ~p ~! ~ q ~ 1andif foreachs > Othereexists asufficientlysmallc = c(p, q, s) >Osuch that (5.20.2) uniformly for (N, I,{, y)e!F(s, c). We may observe that yN-s is the order of magnitude of f'(x). It is immediate that (0, 1) is an exponent pair. Moreover Theorems 5.9, 5.11, and 5.13 correspond to the exponent pairs(!, f), (!,f), and ( 1 2'-k-1)· 2"-2' 2"-2 5.20 IN THE CRITICAL STRIP 117 By using Lemma 5.10 one may prove that A(p ) = (-p- p+q+1) 'q 2p+2' 2p+2 is an exponent pair whenever (p, q) is. Similarly from Theorem 4.9, as sharpened in § 4.19, one may show that B(p, q) ~ (q-J:,p+J:) is an exponent pair whenever (p, q) is, providing that p + 2q ;:: f. Thus one may build up a range of pairs by repeated applications of these A and B processes. In doing this one should note that B2(p, q) = (p, q). Examples of exponent pairs are: B(O, 1) ~ (J:, J:), AB(O, 1) ~(#,f), A'B(O, 1) ~ ), BA'B(O, I)~ (f, f), A•B(O, I)~<.,;,.. H). BA'B(O, 1) ~(it,~), ABA'B(I, O) ~ (-/lr. Jt), BA•B(O, I)~ (!j-, if), ABA'B(O, 1) ~ (!J-, it), A'BA'B(O, 1) ~ (Jlr, H), BABA'B(O, 1) ~ (,',-, ft). To estimate the sum~~ of §5.18 we may take f(x) =/;log x, ' Y=2;• 8= 1, so that (5.20.1) holds for any c ;:: 0. The exponent pair (it, §t) then yields L, ~tl!a!t whence L n -f-it~ t-M-a-& 4: rM-a<n<>:t> for a 4: ti. We therefore recover Theorem 5.18. The estimate p(!) ~ ~of Phillips (1) comes from a better choice of exponent pair. In general we will have "(J:) "J:(p+q-J:), providing that q ;:: p + !- Rankin has shown that the infimum of frklzzwxh:0142J. 5.21. The list of bounds for p(!) may be extended as follows. m ~ 0·164979 .. Walfisz(l), & ~ 0·164634 .. Titchmarsh (9), ,w,- ~ 0·164511 ... Phillips (I) 0·164510 .. Rankin & ~ 0·163793 .. Titchmarsh (24) <t ~ 0·163043 ... Min(1) :1klzzwxh:0027 ~ 0·162162 .. Haneke ,\/,/,- ~ 0·162136 .. Kolesnik ;',',- ~ 0·162037 .. Kolesnik [ 4] m~ (}162004 ... Kolesnik [5 ]. The value J.r was obtained by Chen [1 ], independently of Haneke, but a little later. The estimates from Titchmarsh (24) onwards depend on bounds for multiple sums. In proving Lemma 5.10 the sum over r on the left of (5.10.1) is estimated trivially. However, there is scope for further savings by considering the sum over rand 1' as a two-dimensional sum, and using two dimensional analogues of the A and B processes given by Lemma 5.10 and Theorem 4.9. Indeed since further variables are introduced each time an A process is used, higher-dimensional sums will occur. Srinivasan has given a treatment of double sums, but it is not clear whether it is sufficiently flexible to give, for example, new exponent pairs for one-dimensional sums. VI VINOGRADOV'S METHOD 6.1. STILL another method of dealing with exponential sums is due to Vinogradov.t This has passed through a number of different forms of which the one given here is the most successful. In the theory of the zeta-function, the method gives new results in the neighbourhood of the line U=l. Let be ~ polynomial of degree k ;;:: 2 with real coefficients, and let a and q be mtegers, S(q)= L e2rri/(nl, a<n.;;a+q ' ' J(q, l) = J ... J I S(q) 1 21 da1 .. dak, " ' ;:ee !:::.~~~of the order of J(q,l) as a function of q is important in Since S(q) = O(q) we have trivially J(q, l) = O(qtl). Less trivially, we could argue as follows. We have {S(q)}k = L e2"i'9nl+-·+np+ ... n,, ... ,n~; On integrating over the k-dimensional unit cube, we obtain a zero factor if any of the numbers mt+ ... +m:-nt-... -ni (h = 1, ... , k) is different from zero, Hence J(q, k) is equal to the number of solutions of the system of equations mt+ ... +mZ = nt+ ... +ni (h =I, ... , k), where a < mp ~ a+q, a < n~ ~ a+q. But it fol1ows from these equations that the numbers n~ are equal (in some order) to the numbers mp. Hence only the mp can be chosen t Vinogradov (1}-{4). Tchudakoff (1H5), Titchmarsh (20), Hua (1). 120 VINOGRADOV'S METHOD Chap. VI arbitrarily, and so the total number of solutions is O(qk), Hence J(q, k) ~ O(q') and J(q, !) ~ O{q"-"J(q, k)) ~ O(q"-'). This however is not sufficient for the application (see Lemma 6.8). F~r any in~ger l, J(q, l) is equal to the number of solutions of the equations m~+ ... +mf = n~+ ... +nf (h = 1, 2, ... , k), where a < m,. ~ a+q, a < n,. ~ a+q. Actually J~q, l) is independent of a; for putting M,. = m.,-a, N,. = n,.-a, we obtam ± (M,+a)' ~ ± (N,+a)' (h ~ I, .. , k), v=l ,.~1 which is equivalent to ±Mz~INZ (h~I, ... ,k), v=l ,.-1 and 0 < M,.:::;;; q, 0 < N,.:::;;; q. Clearly J(q, l) is a non-decreasing function of q. Le n n be two sets of integers, let 6.2. LEMMA 6.2. t ml•'~' mk, t•··:• k k ll sh = .. ~ 1 m!, s11 = .. ~1 n,., and let uh, u',.. be the h-th elementary symmetric Ju.rwtions of the m,. and n,. respectively. If [m,.[ :::;;; q, ln .. I :::;;; q, and th£n Clearly and Now Hence [s~~-s~l::::.;; qA-t (h =I, ... , k), (6.2.1} la,-a~l <;; f(2kq)'-1 (h ~ 2, ... , k). (6.2.2) [sAl :::;;; lcqh, [s~[ ~ kq", ioAI ~ (~)q":::;;; lchqll. o1 = !(s~-s1 ). [oa-o;[ = !l(s~-Ba)-(s~~-s~)[ ~ ![(s1 -s~)(s1 +s~)l+!lsa-8;[ <;; kq+!q <;; jkq, the result stated for h = 2. Now suppose that (6.2.2) holds with h = 2, .. ,j-1, where 3 :s;;;j ~ 1:, so that loA-o}.[ ::::.;,; (21cq)A-t (h = 1, .. ,j-1). By a. well-known theorem on symmetric functions 8;-o1s1_1+oa8J-I--"'+(-1)ijoJ = 0· VINOGRADOV'S METHOD Hence [o1-uj[::::.;; ~[s1-s)J+~I [ohsi-h-a~sj_hj J J A=t js1-sj[ 1 1~ , , , ~ -1 -. -+J f:t )(oh-oh)s;-A+uh(s;-h-s;-h)i :s;;; qi~l +~ ~ {(2kq)h-lkqi-h+(kq)hqi-h-1} J J h-1 ~ qH(I+ ~ (2'-'+I)k') J h-t qi-t~ qi-1(2k)i-1 :s;;; T 6o 2hk'' = T 2k-z :s;;; (2kqJi-tj( 2:~ 1 l :s;;; H2kq)i-t::::.;; f(2kq)i-t since 2kf(2k-l) :s;;; 2 andj ~ 3. This proves the lemma. 121 6.3. LEMMA 6.3. Let 1 < f} < q, and let g1, ... , gk be integers satisfying 1 < Ut < g2 < ... < gk :s;;; G, g,.-g,._1 > I. (6.3.1) For each value ofv (1.:::;;; v :s;;; k) let m., be an integer lying in the interval -a+(g,-l)q/G < m, <;; -a+g,qJG, where 0.:::;;; a :s;;; q. Then the number of sets of suck integers m1, ... , mkfor which the values of 8 11 (h = 1, ... , k) lie in given intervals of lengths not exceeding qk-t, is .:::;;; ( 4kG)ik(k-ll. If xis any number such that )xi .:::;;; q, the above lemma gives ' i(x-m1) ... (x-mk)-(x-n1) ... (x-nk)l :s;;; ... ~ 1 )a11 -a~Jix)k-A :s;;; qk-t[ l+fht(2k)h-t} ( 3 (2k)'-2k) = qk-1 1 + 42k=l ~ (2kq)k-l since k ~ 2. If n1, ... , nk satisfy the same conditions as m1 , ... , mk, then /mk-n,.l ~ qfG for v = 1, 2, ... , k-1. Hence, putting x = nk, (q/G)k-tlmk-nkl .:::;;; (2kq)k-t, i.e. Thus the number of numbers m, satisfying the requirements of the theorem does not exceed (2kG)k-t+1.:::.;; (4kG)k-t. 122 VINOGRADOV'S METHOD Chap. VI Next, for a given value of mk, the numbers m1, •.. , mk-t satisfy similar conditions with k-1 instead of lc, and hence the number of values of mk-t is at most {4(k-I)G}k-11 < (4kG).t-1• Proceeding in this way, we :find that the total number of sets does not exceed (4kQ)(k-l)+(.t-2)+-M = (4kG)ik(.t-t), 6.4. LEMMA 6.4. Under the aame conditions aiJ in Lemma 6.3, the number of sets of integers m 1, ... , m.~:for which the numbers 8,. (h = 1, ... , k) lie in given intervals of lengths not excwling cqh(1-llk>, where c > 1, does not exceed (2c)k(4kG)i~>-Oq!<>-o. We divide the hth interval into 1 + [~~:.1:] ~ 2cql-h/k parts, and apply Lemma 6.3. Since llt (2cqt-hlk) = (2c)kq-i<.t-1) we have at most (2c)"'q-i<k-1) sets of sub-intervals, each sa.tisfying the conditions of Lemma. 6.3. For each set there are at most (4kG)ik<k-t) solutions, so that the result follows. 6.5. LEMMA 6.5. Let k < l, let f(n) be as in §6.1, and let ' ' I= [ ···f1Zm.r~, ... Zm,uI2JS(qt-ll.t)lt(l-.t> do:t ... do:.~~:, where zm.tJ~ = (u~-1)1-"'q~n.;;u..a-"'Q e2wif(n) and the g. satisfy (6.3.1) with 1 < G = 2"' < q. Then I~ 23k+<m+2J!li.t-U-mk(l-k)kkilik-1Jqf.t-}J(q1-l/k,l-k). Wehave 1 1 I = N,~,Na"'Y(N 1 , ••• , Nk) f···l ebiCNta.t+.-+N•«•lj S(ql-1/k)jW-kJ drx1 ••• drx.k ' ' ~N.~,N&'J!(Nl, ... ,Nk) [ ,,, f IS(ql-1/k)Jt(l-kJ drx.l ... drx.t, where 'Y(N 1, ... , N.~~:) is the number of solutions of the equations m~+ ... +ml-n~-... -ni = N,. (h = 1, ... , k) form. and n. in the interval (g.-1) 2-"'q < x ~ g.2-"'q. Moreover N,. runs over those integers for which one can solve N11 = n;"+ ... +n.J~.~~:-m?-... -m' 1 ~.t• 6.5 VINOGRADOV'S METHOD where m:' and n~ lie in an interval (a,a+qi-tlk]. As in §6.1 we can shift each range through -a, i.e. replace a by 0. Then Nh ranges over at most 2(1-k)qhO-tlk> values. Hence by Lemma 6.4, for given values of n 1, ... , nk, the number of sets of (mw .. • m..,) does not exceed {4(l-k)}k(2"'+2k)fk(k-llq}(k-1). Also (n 1, ... , n..,) takes not more than (1 +2-"'q)k,;,;;,; (21-mq)k values. Hence N,,!:,N&'Y(~, ... , Nk) ~ {4(l-k)}kk-lk(k-1)2(m+2Jfk(k-1)-mk+kq!k-!, and the result follows. 6.6. LEMMA 6.6. The result of Lemma 6.51wlds whether the g's satisfy, (6.3.1) or not, if m has the value Since M~ [log1_J· klog2 j ZM,u, ... ZM,uaj2 ~ (21-Mqf~k, it is sufficient to prove that (21-Mq)2k ~ 23k-t{.li+2Jik(k-l)-Mk(l-k)kkfk(k-t)qlk-t, or that or that (ik+i)logq ~ ik(k+I)Miog2+!k(k~I)Iog4k, or that logq ~ kMlog2+~~; 1 1 )Iog4k. Since this is true if which is true fork ;;::: 2. M ~ kl~!gq2-l, klog2 ~ k~k; 1 1 )Iog4k, log2 ~ !~~log4k, (6.6.1) 6.7. LEMMA 6.7. The sd of integers (g1, ... , g 1), where k < l, and each g., ranges over (I, G], is Baid to be weU-spaced if there are at least k of them, say gh, ... , g11, satisfying g1. -gi•-• > I (v = 2, ... , k). The number of sets which are not well-spaced is at moat, lt 3lGJ<-1. Let g;, ... , gl denote g1, ... , g1 arranged in increasing order, and let f. = g~- g~ 1 • If the set is not well-spaced, there are at most k ~ 2 of the numbers f., for which f.,> 1. 124 VINOGRADOV'S METHOD Chap. VI Consider those sets in which exactly h (0,;; h,;;; k-2) of the numbers J., are greater than 1. The number of ways in which these hf.'s can be chosen from the totall-1 is r~ 1 ). Also each of the h//s can take at most G values, and each of the rest at most 2 values. Since g~ takes at most G values, the total number of sets of ifv arising in this way is at most (1, 1) G'<+tzi-k-t. The total number of not well-spaced sets g~ is therefore ~ 'f e~l) 0"+121-h-1 ~ Qk-1 'f (--,:1 )z1-h-1 h-o h~o '(; Qk-1(1-f--2)1-t < 31Gk-t. Since the number of sets g, corresponding to each set g', is at most l!, the result follows. 6.8. LEMMA 6.8. If l;;;::: !P+~k and M is defined by (6.6.1), then J(q,l) ~ max(l,M)482l(l)lllkk{k<k-tlq'li.l-kl/k+:k-}J(ql-l/k,l-k). Suppose first that M is not less than 2, i.e. that q ;;?: 22k. Let p. be a positive integer not greater than JJ<I -1. Then P. ~ kl~~gq2 -1, i.e. 21-'+1 ~ qtlk. 2~ p Let S(q) ~ I I ''"'""' ~ I z", o~1 (0-1)2-'''t:;:Ul!-Pq o~t say. Then {S(q)}' = L zf'.Ul ... ZNJr' where each Uv runs from 1 to 21-', and the sum contains 21-'1 terms. We denote those products Z~".o' ... Zf',11, with well-spaced g's by Z~. The number of these, Mf' say, does not exceed 21'-i. In the remaining terms we divide each factor into two pn.rts, so that we obtain products of the type Zf'+l,o, ... Zf'+t,o,, each g lying in (I, 21-'+1). The number of such terms, M 11+1 say, does not exceed l! 31 2f' f'). Consider, for example,! 1 Z~) 2 • The general z~ can be written z~<.o• ... z~<>hz.u.oH• ... z~<.O•' where g1, ••• , gk satisfy (6.3.1) with G = 2~-<. Now, since the geometric mean does not exceed the arithmetic mean, )Zf'.Oh• ... z,...OI)t ~ l~k .i )Z,u.o.l2(l-k>, p~k+l We divide these Zf'.o• into parts of length qt-1/k_I (or less). The number of such parts does not exceed [ 2-P.q J 2-~<q 2-JJq ift-1/k_I +1 ~ qi-1/k_I +2-~<-lqtfk < fql-I/k+2-~L-lqtfk ~ 21-f'+lk-2lJ+2l+k'+k .Jl,.(l-k)kkik(k -1lq2(l-kJik+ik-{J(q1-tJk, l- k). A similar argument applies to z:n, with p. replaced by m. Hence -" J(q,l) ~ M ,l: 2m(fk'+!k-2l)M~x m-p X 22l+k'+k(l-k)kk¥<rk-1lq2(l-kllk+\k-1J(ql-1/k, l-k). A],o M m"'%:p. 2"'ffk'+lk-2lJM~ ~ 2P(-I-k'+fk-2l)+2J'l+ f 2m(lk"+fk-2l)(l!)262l22(m-l)(k-l) m-p.+l = 2fl'<k"+kl+(l!)262l f 2m(fk'+fk-2l-2)-2{k-1) m~p.+l ~ 22 JJI+(l!)262l ~ 2(l!)262l, 126 VINOGRADOV'S METHOD Chap. VI smce we can start with an integer p. such that 2"1 < l!. (Indeed we may tfike Jl = 1.) Hence J(q, I>.:%; Mz21+k•+k+l(l!)2B2'Ikktk<k-llq2<l-kJ!k+!k-!JJ(q'-ltk, 1 ~k), and since the result follows. If M < 2, i.e. q < 22k, divide S(q) into four parts, each of the form S(q'), where q' :( iq::;;; qt-tik. By HOlder's inequality IS(q)l21::;;; 421-1 L IB(q')l2l::;;; 42l-lq2k(l-l/k) 218{q')l2(1-kl_ Integrating over the unit hypercube, J(q,l) ~ 42z-tq2kklzzwxh:0145 L J(q',l-k) ~ 42lq2k(l-l/k)J(ql-I/k,[-J.:), and the result again follows. 6.9. L~MMA 6.9. lfr is any non-negative integer, and l ~ !k2+ !k +kr, then J(q,l) ~ K'1ogrq.q21-~k<k+tJ+Ii, where S, = ~k(k+Il(l-~r. 1klzzwxh:0029 = 482l(l!) 2lkk~"'<k-1)klzzwxh:0421+1) <;; 3kq. k+l k+l Since this is independent of n', it follows that ::~:: J. .. JIS0(q)l21dcx1 .. da:,. :«; ~kq J .. J jS0(q)l 21 dcx1 ... dcx11 since " ' :«; 3kq2 21J(q,l), (6.12.8) S0(q)21 :«; 221-t(jS(q)j21 +~ J IS(p)j2Ldp). 0 Defining l as in Lemma 6.10, we obtain from (6.12.4), (6.12.6), (6.12.8) and Lemma 10 ~ 2k+5knQt-"h{3ke64tklog"kqf}"hlogq+q. Now q ~ 2A. tl<k+ n ~ 2Q41<k+H. Hence lSI~ Ae33klog"kQt-ir-+3fl<k+t)IJtogQ+ZQ4/(k+1l 130 VINOGRADOV'S METHOD Chap. VI and the result follows, since !r-3/{(k + l)l} -;:. Jr and l < 3k2 logk. 6.13. LEMMA 6.13. /f/(x) satisfies the ccmditionsof Lemma 6.12 in an. interval [P+l, P+N], where N ~ Q, and >.-1::;;; Q ~ ).-1, (6.13.1) then \ .. ~t: 1 e2•if<nl\ < Ae33klogtkQ1-Plog Q. (6.13.2) If >.-i ,;;;;; N, the conditions of the previous theorem are satisfied when Q is replaced by N, and (6.13.2) follows at once from (6.12.3). On the other hand, if >.-l > N, then f..~t: 1 e2•if(nll ~ N < >.-i ~ Qi:::;;; Ql-P, and (6.13.1) again follows. 6.14. THEOREM 6.14. ((I+it) ~ O{(logtloglogt)l). Let j(x) = klzzwxh:0422~glogt)i = 1-uo logtt ' with a sufficiently small A. Hence in this region 'lsi~ o(~;;k)+Oi11 ~ o( 01-·')+0(1) l-u0 [ log't l = 0 exp{Aloglt(loglogt)l}~, (loglogt)t We can now apply Theorem 3.10, with 8(!) ~ A(loglogt)l "It)~ A logl!(loglog!)l. ~·'I' Hence there is a region A a~l loglt(loglog t)i (6.15.1) 132 VINOGRADOV'S METHOD Chap. VI which is free from zeros of {(s); and by Theorem 3.11 we have also ((l~il) = O(loglt(loglogt)l), rg:::; = O(loglt(loglogt)l). NOTES FOR CHAPTER 6 • (6.15.2), (6.15.3) 6.16. Further improvements have been made in the estimation of J(q,l). The most important of these is due to Karatsuba who used a p-adic analogue of the argument given here, thereby producing a considerable simplication of the proof. Moreover, as was shown by Steckin , one is then able to sharpen Lemma 6.9 to yield the bound J(q, !) ~ ck'logkq21-fk(k+l) +" .. for l ~ kr, where k ~ 2, r is a positive integer, Cis an absolute constant, and Jr = fk 2(1 - l/kY. Here one has a smaller value for br than formerly, but more significantly, the condition l ;::. fk 2 + l-k + kr has been relaxed. 6.17. One can use Lemma 6.13 to obtain exponent pairs. To avoid confusion of notation, we take [to be defined on (a, b], with a< b ~ 2a andA.-!~a~A.- 1 .Then I L e2" if(n) I ~ Ae33k log"ka 1- P log a. aklzzwxh:0150 Now suppose that (N, I, [, y) is in the set fF (s, t) of § 5.20, whence ftx,.x-~-k ~ l/(~: 1 ~~7)l ~ ftx,.x-s-k with s(s+1) ... (s+k-1) (lk =y (k+1)! We may therefore break up I into O(s+k) subintervals (a, b] with b ~ (!)il(s+~<>a, on each of which one has 1/(IHll(x)l A. ~ (k+ij! ~ 2A., with A= §-tx,.a-~-k. Wenowchooseksothatl -!-~ N ~ 2N ~ ,t-Iforall a in the range N ~ a ~ 2N. To do this we take k ~ 7 such that N~<-t N~< --<!N1-~~-. rx,._1 rxk (6.17.1) 6.17 VINOGRADOV'S METHOD 133 Not~ that N"!rxk tends to infinity with k, if N ~ 2, so this is always posstble, providing that (6.17.2) The estimate (6.17.1) ensures that 2N ~A. t, and hence, incidentally, that A. < 1. Moreover we also have if N ~2s+l<+ 2 , and so ..t-+~ N. It follows that L e2~•f(n)~ske33k'\ogkN1-PlogN •<I for N ~ 28 +1<+2, subject to (6.17.2). We shall now show that (p,q) ~ ( I ,1---'-) 25(m-2)m21ogm 25m2logm (6.17.3) (6.17.4) is an exponent pair whenever m ~ 3. If yN2-s-m ~ 1 then (yN-s)PN" ~N, and the required bound (5.20.2) is trivial. If (6.17.2) fails, then yN-s ~sN" and, using the exponent pair (!h-, HB-> = ASB(O, 1) (in the notation of §5.20) we have L e 2~i[(n)~8(yN-s)rh-Ntit~8Nill-~ N<J ~ (yN-s)PN<J •<I as required. We may therefore assume that yN2-s-m < 1, and that (6.17.2) holds. Let us suppose that N ~ max(28 +m+2, 2(ts+ 1)m). Then (6.17.1) yields N" 1 <~·i·s;1.s:2 .... s+:-2yN1 ~!(max(i,1)y-1 yN1-s <2(ts+1)k lNm-1, whence (ts: 1y-m <2(!s+1)m-l. Since N ~ 2(~-s + l)m we deduce that k ~ m. Moreover we then have N ~ 2B+m+ 2 ~ 2B+k+ 2, so that (6.17.3) applies. Since k is bounded in 134 VINOGRADOV'S METHOD terms of p, q and s, it follows that L e2"<f(n) ~P•<~·• Nt-Plog N 4p.q•s N<~ "'' if N y.P'<l•BI, and the required estimate (5.20.2) follows. Chap. VI 6.18. We now show that the exponentpair(6.17.4) is better than any pair (o:, fJ) obtainable by the A and B processes from (0, 1), if m ~ 106• By this we mean that there is no pair (IX, fJ) with bothp ~ 1:1. and q ~ (3. To do this we shall show that (6.18.1) Then, since 5425m2 log m < (m- 2)3 for m ~ 106 , we have q + 5pi- < 1, and the result will follow. Certainly (6.18.1) holds for (0, 1). Thus it suffices to prove (6.18.1) by induction on the number of A and B processes needed to obtain (o:, {J). Since B 2(o:, (3) = (oc, {3) and A(O, 1) = (0, 1), we may suppose that either (o:, {J) = A(y, b) or (IX, {J) = BA (y, 0), where (y, 0) satisfies (6.18.1). In the former case we have l y+b+1 ( ' )t }'+2~5yl ( ' )! fJ+OO. = 2y+2 + 5 2y+2 :;::-2y+2 + 5 2y+2 :;::: 1 for 0 ~ y ~ t. and in the latter case 2y+1 ( b )! 2y+1 ( • )! fJ+ 5a:i=2)'+2+ 5 2y+2 :;::2y+2+ 5 2y+2 :;::: 1 for 0 :::;; y :::;; !- This completes the proof of our assertion. The exponent pairs (6.17.4) are not likely to be useful in practice. The purpose of the above analysis is to show that Lemma 6.13 is sufficiently general to apply to any function for which the exponent pairs method can be used, and that there do exist exponent pairs not obtainable by the A and B processes. 6.19. Different ways of using J(q, l) to estimate exponential sums have been given by Korobov {1] and Vinogradov (see Walfisz {1; Chapter 2) for an alternative exposition). These methods require more information aboutfthan a bound (6.12.1) for a single derivative, and so we shall give the result for partial sums of the zeta-function only. The two methods give qualitatively similar estimates, but Vinogradov's is slightly simpler, and is quantitatively better. Vinogradov's result, as given by Walfisz [1), is L n-fr 4':at-P a<n.,b (6.19.1) 6.19 VINOGRADOV'S METHOD 135 for a < b :::;; 2a, t :;;::: 1, where k:;;::: 19, and 1 p = 60000k"2· The implied constant is absolute. Richert has used this to show that ((a+ it)~ (1 + t 1 00{ 1 -" 1~(log t)i, (6.19.2) uniformly for 0 :::;; a :::;; 2, t :;;::: 2. The choices O(t) = ( loglog t )i, ¢(t) = loglog t 100log t in Theorems 3.10 and 3.11 therefore give a region a:;;::: 1 ~A (log t)-i(Ioglog t)-l free of zeros, and in which ('(s) ((s) -<(log t)l(loglog t)l, ((~) .. (log t)l(loglog ol. The superiority of (6.19.1) over Lemma 6.13 lies mainly in the elimi-nation of the term exp(33k 2 log k), rather than in the improvement in the exponent p. Various authors have reduced the constant 100 in (6.19.2), and the best result to date appears to be one in which 100is replaced by 18.8(Heath-Brown, unpublished). 6.20. We shall sketch the proof ofVinogradov's bound. The starting point is an estimate of the form (6.12.4), but with ' L ehi{f(uv+~•J-tklzzwxh:01512,.iF(uvJI 1 ~ ql-1 ~~~e2"iF(uv)l' = ql-1 ~'1(v)( ~e2"iF(uv))' =ql-1 L n(al' ... ,a,)L'1(v)e2"iG(,.,, ... ,,..;.,), ,.,, ....... Chap. VI where l11(v)l = 1, n(ap ... ,ak) denotes the number of solutions of u 1h+ ... + u1 h = ah (1 ~ h ~ k), and Now, by HOlder's inequality again, one has Here L n(ap .. ,ak) = q1 , ...... and .. , ....... . Moreover where and n(tp ... , •~o> is the sum of 11(v1) ... '7(v21) subject to vth+ ... +v,h-v,+th_ ... -v2,h = 'h (1 ~ h ~ k). VINOGRADOV'S METHOD Since ln(-r1 , .. ,Tk)l;:;; J(q, l), it follows that I L;e2"iF(uvJI 212 ~ q41'-41J(q, [)2 (I (IILexp(2ttiAhah-rh)l) h- 1 ~ ..... ~ q4l"-41J(q,l)2 fJ (L min (lqh, lcscttAh-rnl)). h~t tn !37 At this point one estimates the sum over -rh, getting a non-trivial bound whenever q- 2h ..:(!A hi..:( 1. This leads to an appropriate result for the original sum (6.20.1), on taking l = [ ck2 ] with a suitable constant c. If we use Lemma 6.9, for example, to estimate J(q, l), then (K2r)(21•)-'<(1. One therefore sees that the implied constant in (6.19.1) is indeed independent of k. VII MEAN-VALUE THEOREMS 7.1. The problem of the order of ((s) in the critical strip is, as we have seen, unsolved. The problem of the average order, or mean-value, is much easier, and, in its simplest form, has been solved completely. The form which it takes is that of determining the behaviour of ~,![((a+it)['d' as T --lo- oo, for any given value of u, We also consider mean values of other powers of '(s). Results of this kind have applications in the problem of the zeros, and also in problems in the theory of numbers. They could also be used to prove 0-results if we could push them far enough; and they are closely connected with the 0-results which are the subject of the next chapter. We begin by recalling a general mean-value theorem for Dirichlet series. • b g(s)=k~ be absolutely convergent for a> a1, u >a~ respectively. Then for a:> a1, {3 >u2, T w lim!T Jf(a:-1--it)g(fi-it)dt= "anbi:"· T.,2 Ln~X+~' -T n~I (7.1.1) For the series being absolutely convergent, and uniformly convergent in any finite t~range. Hence we may integrate term~by~term, and obtain T ! f f(a:+it)g({j-it) dt = ~ a.,b,.+"" "ambn 2sin{Tlogn/m). 2T_T f=1 n~X+ft Li;,.,~~ 2Tlogn/m The factor involving Tis bounded for all T, m, and n, so that the double series converges uniformly with respect to T; and each term tends to zero as T ~ oo. Hence the sum also tends to zero, and the result follows. 7.1 MEAN·VALUE THEOREMS In particular, taking b,. = ii,. and a: = f1 = a, we obtain I IT "' 2 J.~2T lf(u+it)j 2 dt = 2: ~:~ (u > u1). -T n~l (7.1.2) These theorems have immediate applications to {(s) in the half~ plane a > 1. We deduce at once T ~.~~-I l"u+it)j2 dt = "2u) (a> I), (7.1.3) and generally ,. J;i!!,~~~ ~~l(a+it){(vl({3-it) dt = ~+•l(a:-j-{1) (a:> l, f1 >I). (7.1.4) Taking a .. = dk(n), we obtain T lim.!--f [((a+it)["' dt ~ .:;;- dl(n) (a> 1). (7.1.5) T....,.,.,2T L., n2a ~T n~I By (1.2.IO), the case 1.: = 2 is T lim~ f [((a+it)[' dt ~ ('(Za) (a> 1). (7.1.6) '1'--..,2'1 -'1' {(4a) The following sections are mainly concerned with the attempt to extend these formulae to values of a less than or equal to 1. The attempt is successful for k ~ 2, only partially successful for k > 2. 7.2. We require the following lemmas. LEMMA. We have L 2: tnUnal~gnfm = O(T2-2aJogT) (7.2.I) O<m<n<T for!~ a< I, and uniformly for!~ u ~ u 0 < 1. Let~~ denote the sum of the terms for which m < .fn, ~ 2 the remain~ der. In ~1 , lognfm >A, so that ~~ < Amtn~T m-an-<>< ACtT n-ay< AT2-2o, In ~2 we write m = n-r, where 1 ~ r ~ tn, and then Hence lognfm = -log(1-rfn) > rjn. ~! < A 2 2 (n-;}:an-a <A 2 nl-2a 2 ~ < AT2-2a}og T. n<T r<>f>• n<T •·<>!" 140 MEAN-VALUE THEOREMS Chap. VII LEMMA. (7.2.2) Dividing up as before, we obtain :E1 = o[(~n-ue-hrJ = O(S2u-2), ( • '" I) ( I) :E2 = 0 L nl-2ae-Sn L T = 0 Q2a-2log 8: • n~l r~t and THEOREM 7,2. T lim -T 1 J l((a+it)l' dt ~ ((2u) (a> j). T-• ' We have already accounted for the case u > I, so that we now sup-pose that!< u ~ 1. Since t ~ I, Theorem 4.11, with x = t, gives ((s) ~ 2 n-•+O(t-•) ~ Z+O(t-•), 11<1 say. Now J T IZI2 dt = J [~m-a-il L n-a+il] dt ' ~ '~"~ m-:n°0 J (~)" dt (T 1 = max(m, n)) m<T n<T 7., = L n-2a(T-n)+"" '\;m-an-" (nfm?'1'-{n/m)iTo n<T -4.:.-7 dognjm = T .. t1'n-2u nfTnl-2o+O( L L m"n"l~gnjm) O<tn<n<T ~ T{((2a)+O(T'-'"))+O(T'-'")+O(T'-'" log T), provided that a< 1. If a= I, we can replace the a of the last two terms byj,say. Ineithercase -1• Hence f IZI' dt ~ T((2a). i f l((s)l'dt ~ f IZI'dt+O(JIZ!t-•dt)+o(lt-'" dt) 1 1 l l = {tZI'dt+o([IZi2dt {e- 2"dtY+O(log1') ~ r!ZI'dt+0((7'logT)l)+O(Iog7'), and the result follows. 7.2 MEAN-VALUE THEOREMS It will be useful later to have a result of this type which holds uniformly in the strip. It ist THEOREM 7.2 (A). [l{(u+it}l2dt < ATmin(logT, u~i) uniformly for t :s;;;; u :s;;;; 2. Suppose first that l ,:::;;; u ,:::;;; !- Then we have, as before, T r IZ12dt < T L n-2o+O(T2-2UlogT) i n<T uniformly in u. Now 2 n~2u,:::;;; 2 n-1 < AlogT n<T n<T • A and also ,:::;;; I+ f u-?.u du < a-f Similarly 1'2-2uJog T,:::;;; Tlog T, and also, putting x = (2u-l)log T, p2-2ulog T = !Txcl'j(u-!),:::;;; fT/(a-f). This gives the result for u :s;;;; f, the term O(t-") being dealt with as before. If 1 :s;;;; u :s;;;; 2, we obtain fiZI'dt < T 1;n-l+O(TllogT), i n 2, and x = t/(21T-v'logt), y = .Vlogt. Then, since x<!+it) = 0(1), ((!+it)~ J. n+"+O( .~, n-l)+O(dloglt)+O(log-lt)· = "~"' n-i-«+O{logl:t) ~ Z+O(loglt), T say. Since I (logh)' dt ~ O(TlogiT) ~ o(TlogT), it is, as in the proof of Theorem 7.2, sufficient to prove that T [1Zi 2 dt--TlogT. T T Now J [Z[2dt = f m~,m-f-itn~xn-~+ildt. In inverting the order of integration and summation, it must be remembered that x is a function oft. The term in (m, n) occurs if x > max(m,n) = T 1/(21T-v'logT 1) say, where T1 = T1(m,n). Hence, writing X= Tj(2n..JJogT), (tZ[ 2 dt = L~ r m-{-iln-i+ildt 0 m,n<"X T, -T' 1 o('T,(n,n))+o(''-1-)· -.. -:S. n + .~ n f<,.{x_"')(mn)lognfm The first term is TlogX+O(T) ~ TlogT+o(TlogT). The second term is Olfx ,llogn) ~·O(X,IlogX) ~ O(T), and, by the first lemma of§ 7 .2, the last term is O(XlogX) ~ O(NlogT). This proves the theorem. MEAN-VALUE THEOREMS 143 7.4. We shall next obtain a more precise form of the above mean-value formu1a. t THEOREM 7.4. ,. [ l(l!+it)l'dt ~ TlogT+(2y-1-log2~)T+O(Tl+<). (7.4.1) ·we first prove the following lemma, LEMMA. If n < Tj27T, '+iT ~ ~J x(t-s)n~ 8 da = 2+0(nHog(~/2nn))+oC 0!t~· (7.4.2) !-•T If n > Tj27r, c > !. .k}: xl1--' d, ~ o(,,1;~;nfT))+o(T~; 1 ). We have This has poles at s = -2v (v = 0, 1, ... ) with residues (-1)~21+2~71"2~ ~ Also, by Stirling's formula, for -7r+S < arg(-s) < 7T-S xl1-s) ~ (~t'2 ,~~'~ (1+0(~))· The calculus of residues therefore gives .k ( T+ !r+ 7")xl 1 -~d, -oo~tT, !-iT, !+iT, (7.4.3) = ~ (-ljV2H·2P7r2Pnb ;S (2v)! = 2cos2trn = 2. Also, since elnrlf(--a) = O(e!"l), t Ingham (I) Obtained the errQr term O(T11og T); the method given here is due to Atkinson(!). MEAN-VALUE THEOREMS Chap. VII 7'(1-s)n-• ds ~ o[ j ( 1 a~:T,t• ,-•n-• dal !+iT 1 -oo ! ~ o[n-l J (2~;l' dal ~ o(nf!og(;,12.-n))• and similarly for the integral over ( -c:o-iT 1, !-iT 1). Again, for a fixed a, Hence where x(1-s) ~ ('ft•-• ,-•<-!••(1+0(~)) (t ~ 1). f+iT 1 T, I x(1-s)n-!ds = n-ie-li11" I eiF(lldt+O(n-llogTl), !+iT T F(t) ~ tlogt-t(log2~+1+logn), F'(t) = logt-log217'n. Hence by Lemma 4.2, the last integral is of the form 0(tog(;/2~)) uniformly with respect to T 1. Taking, for example, T1 = 2eT > 411en, we obtain (7.4.2). Again TTx(1-s)n~& ~ n~d•• [(.J;t'•"""'dt+O(n~ J ~-ldt), c+i 1 1 and (7.4.3) follows from Lemma 4.3. In proving (7.4.1) we may suppose that Tf211 is half an odd integer; for a change of 0(1) in T alters the left-hand side by O(Tl), since {(f+it) = O(tl), and the leading terms on the right-hand side by O(log T). Now the left-hand side is 1 [iciHi•ll'dt ~ t [m+i•lm-it)dt -T -T !+iT jHT ~ 1 I ,(,)((!-•) d.~~, I x(1-•lC'I•) ds 2t 2l -l-iT f-lT i-+iT i+i1' ~ I xl1-s) " d(n)ds+~ I x(1-s)(c'l•)- " d(n)) & 2' Lns 21-L..ns 4-iT n<;;TJ2tr l-iT n.;;T/2,. = 11+1 1, say. 7.4 MEAN-VALUE THEOREMS By (7.4.2), 1 1 ~ 2~ I d(n)+O( 2: ntlo dl(~/'~))+o(tog T 2: d~7))· n<>T/2,. n<;;T/2,. g n<;;T/2,. The first term ist 2~('!_tog'!+(2y-I)'!+ O(Tl)) 211 211 271 ~Tlog T +(2y-1-log 2~)T+O(Tl). Sincet d(n) = O(n~), the second term is 0(..«~4,.n1-:.)+0{Tth T/47r~TI2•(Tj2-!)-n} = O(TfH). The last term is also clearly of this form. Hence 1 1 = TlogT+(2y-l-log271)T+O(TiH). Next, if c > 1, I,~f.( T + r)x(l-s)(c'(s)- 2: d~~))ds+ }-iT c·HT .... T12" c+iT +~ L d(n) I x(I-sJn-Bds-A, n>T/2,. c-iT A being the residue of 11x(l-sg2(s) at s = l. 146 Since x{l-s) = OW-l), and {2(u+iT) and L d(n)n-s are both of the form n<;;T/2• O(T'-"'') (a,;;; 1), O(T•) (a> 1), the firstt.enn is By (7.4.3), the second term is o( T•-t 2: d~~) (tog(2:,/T) -j-1)) n>T/21f ~ o(Tt•• 2: n- ~~2~)) + o(T•-t 2: nL.) T/211"TI• =O(T{H). Since c may be as near to I as we please, this proves the theorem. A more precise form of the above argument shows that the error-term in (7.4.1) is O{Ttlog2T). But a more complicated argument,§ t See§ 12.1,or Hardy and Wright, An Introduction to the Theory of Numbers, Theorem320. 1 Ibid.Theorem315. § Titchmarah (12). , .. MEAN. VALUE THEOREMS Chap. VII depending on van der Corput's method, shows that it is O(T1-Iog2T); and presumably further slight improvements could be made by the methods of the later sections of Chapter V. 7.5. We now pass to the more difficult, but still manageable, case of l'(s)j4 • We first provet THEOREM 7.5. ,. lim! I i ((u+it)j 4 dt = {"(2a) (a>!). T---+ro T 1 ((4u) Take x = y = .j(tj21T) and a > ! in the approximate functional equa-tion. We obtain {(s) = _L ~+x(s) _L -;t~+O{t~l) = Z1+Z2+0(t-i), say. Now 7V<I(I/21t) n<<l(ll2:rr) ( 7.5.1) = 2: (mn~v)u (~r where each variable runs over {l,.J(t/21T)}. Hence I T IZ,i'dt ~IT '-1 -(e)" dt 1 1 L.- (mnfLv)"' mn '" ",.f:!T"•l (mn~v)' f (~)" dt, where T1 = 21Tmax(m2,n2,,u.2,"2) The number of solutions of the equations mn = p.v = r is {d(r)}2 if r < ..j(Tj21T), and in any case does not exceed {d(r)}2• Hence T 2: m! ,, ~ T 2: (d;;!l' + o(r 2: (d;~.r) mn~~~~( ) f'<-'(T/2-Ir) ,t(T/21r);;;r<T!27T ~ T ~ {d(r))' ~ T('(2a). (7.6.2) 6 r2u {(4a) t Hardy and LittlewQod (4). MEAN-VALUE THEOREMS 1<7 Next 2 (m~\2a < 2 27T(m2~::~~2+v2) • mn~p.~ mn-p.~ and the right-hand side, by considerations of symmetry, is 87T L (m::v)a ~ 87T L m(:::~;:> = O(T€ L m2-2a L n-2<>) mn-JW ~ O(T•(Ti<>~'"'+I)logT) ~ O(TI~•+•)+O(T•). The remaining sum is o( ' 'ilqj<l(rL)- o(r· 2: ~-1 -) ~ O(T•~••+•) O<o<~T/'" (qr)" log(r/q) -(qr)"log (r/q) ' by the lemma of§ 7 .2. Hence T Now let I [Z ['dt ~ T ('(2a) 1 1 {(4a) j(T) ~ fj I n•~'j' dt. 1 n<,(l/21r) The calculations go as before, but with a replaced by 1-a. The term corresponding to (7 .5.2) is T 2 ~ 2(~;; = O(T2a+€), r<AT and the other two terms are 0( Ti+ a+€) and O(Jl2a+€) respectively. Hence j(T) ~ O(T"'+•), and, since x(s) = O(tf-a), T T f 1 Z2i4 dt < A f t2-4aj'(t) dt T =-o A[t2-4aj(t)]~ +A(4a-2) [ tl-4aj{t) dt = O(T2-2a+€)+0( l tt-20" f-€ dt) = O(T2-2a+E). The theorem now follows as in previous cases. 7.6. The problem of the mean value of j((f+it)j 4 is a little more difficult. If we follow out the above argument, with a = }, as accurately as possible, we obtain fl((!+it)['dl ~ O(Tiog'T), i (7.6.1) MEAN-VALUE THEOREMS Chap. VII but fail to obtain an asymptotic equality. It was proved by Inghamt by means of the functional equation for {{(s)}2 that T f !((!+it)i' dt ~ Tiog'T +O(Tiog'T). (7.6.2) 2~' ' The relation (7.6.3) is a consequence of a result obtained later in this chapter (Theorem 7.16). 7.7. We now pass to still higher powers of {(s). In the general case our knowledge is very incomplete, and we can state a mean-value formula in a certain restricted range of values of u only. THEOREM 7. 7. For every p08itive integer k > 2 T J.~ ~ J J'(u+it)j 2kdt = ~ d!~) (a> 1-1)· (7.7.1) 1 n~l This can be proved by a straightforward extension of the argument of § 7.5. Starting again from (7.5.1), we have 1Zti2k ="" 1 (nt···nk)". ~ (m1 ... mkn1 ... nda m1 ... mk where each variable runs over{1,,.j{t/2Tr)}. The leading tenn goes in the same way as before, d(r) being replaced by dk(r). The main 0-tenn is of the form o(T• 2: 2: -1-) ~ O(T~'-•1+•). O<'l<r<ATt! (qr)u logrjq The corresponding term in j(T) ~ J I '5' n•-'1" dt 1 n<~ti:'2>t) is O(Tka+•), and since lx12k = O(tk-2ka), we obtain O(Tk(t-a}+') again. These terms are o(T) if u > 1-1/k, and the theorem follows as before. 7.8. It is convenient to introduce at this point the following notation. For each positive integer k and each u, let P.k(u) be the lower bound of positive numbers ' such that T ~ r i((a+it)l"dt ~ O(TI). t Ingham (I). 7.8 MEAN-VALUE THEOREMS , .. Each P.k(u) has the same general properties as the function p.(u) defined in§ 5.1. By (7.1.5), P.k(u) = 0 for a> I. Further, as a function of u, ftk(a) is continuous, non-increasing, and convex downwards. We shall deduce this from a. general theorem on mean-values of analytic functions.t Let f(s) be an analytic function of s, real for real s, regular for a ~ a except pos8ibly for a pole at s = s0, and O(e•l1 1) asltl ~ oofor every positive o: and a ~ a. Let ex < {3, and suppose that for all T > 0 T f 1/(o+it)j'dt,;; C(T"+l), " (7.8.1) !1/(~+it)i' dt,;; C'(T'+l), (7.8.2) where a ~ 0, b ~ 0, and G, C' depend on j(s). Then for ex < u < {1, T ~ 2, T I IJ(a+il)l2 dt:::;; K(CT"1fJ-oXI{3-C<l(C'Tb)(a-e<)}(/1-e<l, (7.8.3) ~T where K depends on a, b, o-:, fJ only, and is bounded if these are bounded. We may suppose in the proof that o-: ~ f, since otherwise we could apply the argument toj(s+i-ex). Suppose first thatj(s) is regular for a~ ex. Let u+i» ~ f r(s)f(s)z-s ds = <fo(z) (a~ o-:, Jargzl < ~7T). o-i"' Putting z = 'ixe-1 8 (0 < 8 < l1r), we find that r(a+it)j(o+it)e-i<a+ilXf,.-81, <fo(ixe-iS) are Mellin transforms. Let Then, using Parseval's formula and HOlder's inequality, we obtain f(a) = 21T ll<~-~( • )'•-·J!~-·· ::;; 2TT I l¢>[2x~-I dx [ l¢>12x2fJ-t dx ~ {J(o)}$-•li$~>{I(~))'"-"'$-•J. t Hardy, Ingham, Bnd POlya (1), Titchma.rsh {23). MEAN-VALUE THEOREMS T Writing F(T) ~I lf(•+;t)l'dt,:;; C(T"+l) we have by Stirling's theorem (with various values of K) J((X) < K j (t2«-1+l)jf((X+it)l2r2S1dt = K I F(t){23(t2a:-1+ l)-{2(X-l)t2"'-2}e-2& dt <KG I (t<~+I)23(t2<>:-l+l)e-2&dt <KG I (t<1+2"'-1+1)3e-2&dt < KG{S-<~- 2"'+ 1 +1) < KGS-1•-2<>:11. Similarly for i{f3). Hence Also J{u) < K(GS-a-2"'+1)(~-oJ~~-"'l(G'S---t>-2~+lyo-"')/(fl-"'l = KS-2o+I(GS-a)(~-a)f(~--"')(G'S-bj(a-"')f(~-"'l. Chap. VII 1/5 t/5 J(u) > K J" J/(u+it)j2J2<7-l dt > KS-2o+1 J jf(u+it)l! dt. 1/25 1 128 Putting 3 = 1fT, the result follows. Ifj(s) has a pole of order kat s0, we argue similarly with (s-s0 )kj(s); this merely introduces a factor T2k on each side of the result, so that (7.8.3) again follows. Replacing Tin (7.8.3) by fT, !-T, .. , and adding, we obtain the result: If T T I lfl•+;t)l'dt ~ O(T"), f lfl~+;t)l'dt ~ O(T'), T f jf(u+it)j2dt = O{Tla(,B-a)+b(a-O<))!<,B-n)}. Takingf(s) = {k(s), the convexity of J.t~;(u) follows. 7,9 MEAN-VALUE THEOREMS 151 7.9. An alternative method of dealing with these problems is due to Carlson, t His main result is THEOREM 7.9. Let uk be the lower bound of numbers u such that Then forO 0, c > 1, c > u. For the right-hand side is 1 'J+im ro "' c+i, ;;-;;; r(w-') 2: O.,.a•-wdw~ 2: <>,I J r(w-s)(on)•-wdw .. c-ioo n~l n"' n~l ns 21Tic-ioo c-u+i<» = ~~ ~-bc_.,.L, f(w')(On)-w'dw' = ~~e~n. ,6n' The inversion is justified by the convergence of jwrc-a+;lv-t))l i l:;la·~av. -oo n~l Taking a,.= dk(n),f(s) = {k(s), c = 2, we obtain ~ dk(n) ~ I 2 J+f"" L. ----:;;;-e- "=------; f(w-s){k(w)Ss-t~ dw (a< 2). n=l 21Tt2-ia> Moving the contourtoR(w) =a, where u-1 <a:< u, we pass the pole of r(w-s) at w = s, with residue {k(s), and the pole of {k(w) at w = 1, where the residue is a finite sum of terms of the form Km,n['<"'l(I-s)lognO.Ss-1. t Ca.rlson (2), (3). !52 MEAN-VALUE THEOREMS Chap. VII This residue is therefore of the form 0(8a-lHe-..t111), and, if 8 > It[-""', it is of the form 0(e-A111). Hence «+i {k(s) = ~~-~e~n-~ f r(w-s){k(w}88-Wdw+0(e-AIII). L... n• 2:rn n-1 «-i Let us call the first two terms on the right Z 1 and Z2• Then, as in previous proofs, if a > !, J T ( "' d%(n) -38n) (""" """dk(m)dk(n)e-(m+n>8) 1Ztf2dt = 0 T 61 ~e +0 ~ .. ~ mana[Iogmjn[ ,,. ~ O(T)-t- 0(~.~ m•-·~;'~~l:gffl/nl) ~ O(T)-t-0(8'•-'-•) by (7.2.2). Also, putting w = o:+iv, IZ,I,;; S•-• I" lr(w-s)('(w)l dv 2rr ,;; 3~([ lr(w-s)l dvllr(w-s)("(w)l dvy. The first integral is 0(1), while for [t[ ~ T ( T + f)lr(w-s)("(w)l dv ~ cr + J)cAIHiviA'dv ~ O(e-AT). -ao 2T -ao 2T Hence T 2T T ) J [Z2 [2dt = o{o2a-2« J [~(w)[2k dv J [r(w-s)[ dt +0(02 a-2«) iT -2T IT Hence = o{o2a-2) = 0(82a-2«ptwt<«lH). J [~(s)[U: dt = O(T)+0(82a-2-~)+0(82a-2c>p1+f't<«lH). Let 8 = ~-lll+f'.t<«ll/ll--l-l+p.~o:(a:)" 7.9 MEAN-VALUE THEOREMS !53 For such values of a, replacing T by !T, !T, ... , and adding, it follows that (7.9.1) holds. Hence ak is less than any such a, and the theorem follows. A similar argument shows that, if we define ak to be the lower bound of numbers a such that T ~ f [~(a+it)[ 2kdt = O(T€), (7.9.2) then actually a~ = ak. For clearly a~ ~ ak; and the above argument shows that, if o: > aA,, and a < o:, then T J "(a+it)[ 2kdt = O(T)+0(02<r-2-€)+0(020"-to:T1+<). IT Taking 8 = T-ll, where 0 <A< lj(2-2a), the right-hand side is O(T). Hence ak ~ o:, and so ak ~ a~. It is also easily seen that ~JTI{(a+it)l 2kdt......, ~~~ak). T L n2a 1 n~1 For the term 0( T) of the above argument is actually i-T i d!~~) e-2Sn = JT i: d!~~) +o (T), n=1 n~l and the result follows by obvious modifications of the argument. This is a case of a general theorem on Dirichlet series. t THEOREM 7.9 (A). If p.(a) is the p.1unction defined in§ 5.1, 1-ak ~ ~:~7::~1 ) fork= I, 2, .... Since ~(a+it) = O(tf'{O")+£), T T [ [~(a+it)[2kdt = o{T2 fL(O")H f "(a+it)jllk-2dt}, and hence Since P.k-1(ak_1) = 0, this gives P.k(ak_1) ~ 2p.(ak_1), and the result follows on taking o: = ak_1 in the previous theorem. t See E. C. Titchmarsh, Tluwry of Function8, § 9.51. MEAN-VALUE THEOREMS Chap. VII These formulae may be used to give alternative proofs of Theorems 7.2, 7.5, and 7.7. It follows from the functional equation that P.k(1-a) = P.k(a)+2k(a-!). Since P.k(ak) = 0, P.k(1-ak) ~ 0, it follows that ak ~f. Hence, putting a:= 1-ak in Theorem 7.9, we obtain either uk =for i.e. Hence a~.;= !. or ak ~ 1-1+2k~:k-!l' 2ak-1 ~ 2k(ak-!)(l-ak). (7.9.3) For k = 2 we obtain a2 = f, but for k > 2 we must take the weaker alternative (7.9.2}. 7.10. The following refinementt on the above results uses the theorems of Chapter Von ,u{a). THEOREM 7 .10. Let k be an inteyer greater than 1, and let vbe determined by (7.10.1) Then (7.10.2) The theorem is true fork= 2 (v = 1). We then suppose it true for alll with 1 < l < k, and deduce it for k. Take l = (v-1)2v-2+1, where vis determined by (7.10.1). Then p.1(a:) = 0, provided that a:> I- 2l+2~-l-2 = l-2V~l' Taking a:= 1-2-v+1+t:, we have, since T T -T 1 f l~(a:+it)l 2kdt ~max l~(a:+it)1 2k-:>.I_T 1 f l{(a:+it)l:>.ldt, 1 I.;;t.;;T 1 P.k(x) ~ 2(k-l),u(a:)+,u 1(a:) ~ 2(k-!)!'(o) ~ 2{k-(~;1t,~:~~ 1 -1} t Davenport (l),Ha$elgrove (1). 7.10 MEAN-VALUE THEOREMS by Theorem 5.8. Hence, by Theorem 7 .9, a ~ l-2-v+l(2k+2"-2)-l = 1--~ k -.c::; (v+l)2v-I 2k+2"-2' The theorem therefore follows by induction. For example, if k = 3, then v = 2, and we obtain a3 ~ i instead of the result a3 ~ i given by Theorem 7.7. !55 7.11. For integral k, dk(n) denotes the number of decompositions of n into k factors. If k is not an integer, we can define dk(n) as the coefficient of n-w in the Dirichlet series for ~k(B), which converges for a > 1. We can now extend Theorem 7.7 to certain non-integral values of k. THEOREMt 7.11. For 0 < k ~ 2 (7.11.1) This is the formula already proved for k = 1, k = 2; we now take 0<k<2. Let (N(s) ~ [I 1 _~~· "N(•) ~ ((•)/(N(•). P 0, {(N(s))' ~IT (1-p-•)-> ~ ~d>{n), P 0, and d~(n) = d.(n} if n < N, and 0 ~ d~(n) ~ d.(n) for all n. Hence T 1im-T 1 f I(N(a+it)j-'dt ~ ~ {d>{n)}' (a> 0), (7.11.2) T, ~ nZo 1 n-1 and (7.11.3) t Ingham (4); proof by Davenport (I). 156 MEAN· VALUE THEOREMS We shall next prove that T Chap. VII lim lim.!,Ji~(u+it)-~,v(a+itJJ 2kdt=0 (a>t). (7.11.4) N->'1>T_..rol 1 By HOlder's inequality ~ J r ,,Nr" dt ~ (~ J r"rll' d·)"(~ j'rc.,r""'-" d,r-·'. 1 l 1 (7.11.5) Now {7J,y(s)-I}2 is regular everywhere except for a pole at s = 1, and is of finite order in t. Also, for q > .f, T T J I"N(a+it)-!['dt ~ J {l+2"[((a+i<)i)'dt ~ O(T). ' 1 Hence, by a theorem of Carlson, t T oo ' r !c.Jr ( +it)-If'dt~ ''"(n) :7~ '1' 1 7b a f:-t n2" for a > !, where PN is the coefficient of n -s in the Dirichlet series of {1JN{s)-I}2. Now PN(n) = 0 forn < N, and 0 :( PN(n) ~ rl(n) foralln. Since! d2(n)n-2" converges, it follows that T lim lim.\ J 11Js(a+it)-Ij4 dt -=.c 0; (7.11.6) .v-oo'-',.""1 1 (7.11.4) now follows from (7.11.5), (7.11.6), and (7.11.3). We can now deduce (7.11.1) from (7.11.3) and (7.11.4). We have+ ([r<r"d•r ~([r<v+HNi"d<J" ,;: (Jrc,r"def +(TrH,I"dtJ". 1 ' where R = 1 if 0 < 2k ~ 1, R = lj2k if 2k > 1. Similarly (TrcNi"df ~ (!r(i"df +Uic-c":"de)". ' ' ' and (7 .11.1) clearly follows. t See Titchmarsh, Tl11my of F'uncliona, § 9.51. t Hardy, Littlewood, and P6lya, Inequaliliu, 'fhoorem 28. 7.12 MEAN-VALUE THEOREMS 7.12. An alternative set of mean-valne theorems.t Instead of considering integrals of the form T I(T) ~ f il(a+it)j"dt where T is large, we shall now consider integrals of the form J(3) ~ ( j((a+it) l"e-• dt ' where S is small. The behaviour of these two integrals is very similar. If J(S) = 0(1/S), then T I(T) < e f )~(a+it)) 2ke-tl 1' dt < eJ(ljT) = O(T). Conversely, if I(T) = O{T), then J(3) ~ ll'(t)e-•dt ~ [I(t)e-&J~+Ol I(t)c&dt ~ o(3 J•·-•dt) ~ O(I/3). Similar results plainly hold with other powers of T, and with other functions, such as powers of T multiplied by powers of log T. We have also more precise results; for example, if l(T) ,, CT, then J(S) ,... CjS, and conversely . If l(T) ,... CT, let )I(T)-CT) ~ f.T forT~ T0• Then ~ 00 00 J(3) ~ 3 J I(t)e~•dl+3 J {I(t)-Gt)e-•dt+G3 J te-"dt. 0 ~ ~ The last term is Ce...ST•(T 0+ 1/S), and the modulus of the previous term does not exceed £(T 0+l/8). That J(8) ,, CJS plainly follows on choosing first T0 and then 8, The converse deduction is the analogue for integrals of the well~known Tauberian theorem of Hardy and Littlewood,:j: viz. that if an ~ 0, and then n~oanx" .- l~x (x ....,..J) N n~oan'""N• t Titchmarsh(l),(19). t Soo Titchm&nlh, Theory of Functions,§§ 7.51-7.53. !58 MEAN-VALUE THEOREMS The theorem for integrals is as follows: lff(t) ;;>Of.,. all t, and as S-+ 0, then as T-+ oo. l f(t),-· dl -~ [f(t)dt"' T We first show that, if P(x) is any polynomial, . ' l f(t)e-& P{e-&) dt,....., ~ [ P(x) dx. Chap. VII (7.12.1) (7.12.2) It is sufficient to prove this for P(x) = x". In this case the left-hand side is . ' f f(t)e--{k+llBt dt ,..,_I =! f x" dx. 0 (k+I)O 0 0 Next, we deduce that [j(t)r''g(e-") dt- ~! g(x) dx (7.12.3) if g(x) is continuous, or has a discontinuity of the first kind. For, given E, we cant construct polynomials p(x), P(x), such that p(x) ,; g(x) ,; P(x) and j {g(x)-p(x)} dx ~ «, l {P(x)-g(x)} dx :os;; £. Then rrms J j(t)e-&g(e-&) dt,:;;;; Hills Tj(t)e---&P(e-&) dt s-o 0 S-->-o 0 1 1 = [ P(x) dx < [ g(x) dx+£, and making ~~:-+ 0 we obtain . ' ~S [ j(t)e---&g(e-&) dt :( J g(x) dx. Similarly, arguing withp(x), we obtain ~8 JJ(t)e---&g(e---&) dt;;;:::. J g(x) dx, ..." " and (7.12.3) follows. t See Titchmarsh, Thwry of FunctWns, § 7 .53. 7.12 MEAN-VALUE THEOREMS Now let g(x) ~ 0 (0::::;;; x < e-1), = 1/x (e-1::::;;; x::::;;; 1). Then "' 1/li J f(t)e-"g(,-") dl ~ J f(t) dl " " and j g(x)dx~ j'!i~ 1. 0 1/'-"' Hence [fit) dt- ~· which is equivalent to (7.12.2). If f(t) ;:. 0 for all t, and, for a given positive m, J j(t)e-" dt- ~logm~, (7.12.4) T then j f(t) dt- TlogmT. (7.12.5) The proof is substantially the same. We have f j(t)e- o), (7.12.6) then ! t-lj(t),-• dt- c r~~-:t aft-• (0 < ~ < o). (7.12.7) ob~~~tiplying (7.12.6) by (8-'I]).B-t and integrating over (?],CO), we 160 MEAN-VALUE THEOREMS Ch.:tp. VII Now r e-&(8-1)).8-1 d8 = e-'l)t I e-21x.B-1 dx = e--'l)tt-.Br(f3), ' ' f • f• x~-1 P(fi)P(•-fi) 3-•(o-nW-1 do~ --dx ~ ,~-· _r()' -, (1J+x)n: (1. ' ' and the remaining term is plainly o (1111-«) as 7J-+ 0. Hence the result. 7.13. We can approximate to integrals of the formJ(3) by means of Parseval's formula. If R(z) > 0, we have 2+i 2+ioo "' _I__ f P(s)('(s),~ ds ~ 2 d,(~) J P(s)(nz)-• ds ~ I d,(n)e-=, 2-ni :1-ioo n~l 21T~ 2-i<» n~l the inversion being justified by absolute convergence. Now move the contour to u = (1. (0 < o: < 1). Let Rk(z) be the residue at 8 = I, so that Bk(z) is of the form Let Then ~(a~kl+a~klJogz+ .. +akk~llogk-lz). tf,k(z) = n~tdk(n)e-'1111-Rk(z). a+i<o _I__ J f(s)('(-• ds ~ ~,(,). 2wl a-i<>O Putting z = ixe-i&, where 0 < 8 < !-'"' we see that q,k(ixe-i8), r(sgk(s)e-ilf.r-3~ are Mellin transforms. Hence the Parseval formula gives (7.13.1) (7.13.2) .;;. j [f(a+it)Ck(a+it)[2e(,.-2S;~ dt = j lrfok(ixe-i8)1 2x 20'-l dx. --' (7.13.3) Now as It I-+ oo lf(a+it)l ~ .-1<1 "1W-l"(2w){1+0(t-')}. Hence the part of the t-integral over ( -oo, 0) is bounded as 8-+ 0, and we obtain, for ! < u < 1, j tll0'-l{I+O(t-l)}l{(u+it)12ke-281 dt = j lrfok(ixe-i8)1 2x20'-l dx+O(l). 0 ° (7.13.4) 7.13 MEAN-VALUE THEOREMS 161 In the case u = f, we have lr(f+it)l2 = r.sech11t = 27Te-.,ili+O(e-3"'111). The integral over ( -oo, 0), and the contribution of the 0-tenn to the whole integral, are now bounded, and in fact are analytic functions of 8, regular for sufficiently small 181. Hence we have 7.14. We now apply the above fonnulae to prove THEOREM 7.14. As 0-+ 0 ! l((!+i<)i'e-" dt ~~log~. In this case R 1(z) = 1/z, and rfot(z) = .. ~~e-~~~-~ = e~~~-~· Hence (7.13.5) gives (7.13.5) (7.14.1) f •l((!+it)l'e-"'dt~J·r (" 1 ~) 1 ~~'dx+0(1). exp txe txe-' ' ~~~ The x-integrand is bounded unifonnly in 8 over (O,r.), so that this part of the integral is 0(1). The remainder is dx ~ f• __ d_x _ +iei8 f• I .. {exp(ixe-18)-l}{exp(-ixei8)-1} .. exp( ixei8) 1 x . "I"' 1 dx +I"' dx (7.14.3) -te-.. exp(ixe-i8)-l X .. Xi' The last term is a constant. In the second tenn, turn the line of integra-tion round to (7T, ?r+ioo). The integrand is then regular on the contour for sufficiently smalliOI, and is O{x-1exp(-xcos8)} as X-+00. This integral is therefore bounded; and similarly so is the third tenn. 162 MEAN-VALUE THEOREMS Chap. VII The first term is Jmtln~l exp(-imxe-1 1l+inxe 111) dx _ ~ ~ exp{ (tn+n)1Tsin8-i(m-n)7TCOsS} - ,fb 1 .6 (m+n)sinS+i(m n)cosO _ ~e-2mr6lnll 2 ~ "~e(m+n)sinScos{(m-n)l'fCosS}e-<"<+"matnll_ -L.. 2nsin0 + L.. .L.. (m+n)2 sin28+(m-n)2 cos28 #~I m~2 n~l ., ~ '~ 1 (m-n)co~sin{(m-nl':~~~ e-(m+>~Prsiu& -~ ;£;:2 {= 1 (m+n)2sin2 8+(m-n)2cos2 0 =:Et+:E2+:E3, the series of imaginary parts vanishing identically. Now :El = 2s:nslog-l-e-12,.stnll,...,~Jog~, I~ I< 2 ~ m~ 2m sinS __ e-"'"''"' ~ o(s I me-•"''"') ~ o(~). 2 11f':;2 (;:1 (m-n)2cos23 m~2 8 and, since jsin{(m-n)7Tcos8}1 = jsin{2(m-n)7Tsin2 l8}j = O{(m-n)82}, :!:, ~ o(s• I mf ,-"'"''"') ~ o(s• I me-"'"'"')~ 0(1). m-2 n-1 m~z This proves the theorem. The case!< a< 1 can be dealt with in a similar way. The leading term is f • • I"" x2"-l L c-21l.tSinllx2a-ldx = e2.t:~in3_J dx '" n~l '" I I"' y2a-l 1 s"' y211-l = (2sin8)2" elf-1 dy,...., (28)2" eY-1 dy 21rsinli 0 = (~)'w f(2u)"2a). Also (turning the line of integration through -}1r) 1 e-{(nHn)alu li+i{m-n)cos lilzx2a-l dx = o(e--<m+n)1rslnli [ e--<m-n)ucosli(1T2a-l+y2a-1) dy} ~0---, ( e-rsluli) m-n (7.14.4) 7.14 MEAN-VALUE THEOREMS 163 and the terms with m =F n give o(~ e-"'"'"'~m~.)~ o(~log~)· Hence J"' t2"-1j {(u+it)j2e-2& dt"' r(2a){(2a) 22"820" . (7.14.5) ' Hence by (7.12.6), (7.12.7) Jl((a+it) l'e-" dt ~ ((:•) . (7.14.6) 7.15. We shall now show that we can approximate to the integral (7.14.1) by an asymptotic series in positive powers of 8. We first requiret THEOREM 7.15. As z-+ 0 in any angle jargzj ~A, wher-e A< !n", (7.15.1) where the b,. are conatanta. Nears= 1 r(s){'(•)•~ ~ {1-y(s-1)+ .. J(_I+r+···)' ~{1-(s-1)1og•+···l s-1 z = z(s~ip+y-~ogz s~1 +· Hence by (7.13.1), with k = 2, f d(n)e-n~ = y-logz +!. a.f+i f{8){2(s)z-s da (0 < Cl < 1). n=l Z 27Tl <>.-i Here we can move the line of integration to u = -2N, since f{8) = O(jtjKe-t,.~l), {2 (8) = O(jtjK) and z-2 = O(r-ae>.l). The residue at 8 = 0 is {2(0) = !- The poles of r(s) at a= -2n are cancelled by zeros of {2(s). The poles of f(a) at 8 = -2n-1 give residues ---=-!: {2(-2n-1)z2n+l = ---~z2•HI (2n+1)! (2n+1)!(2n+2)2 · The remaining integral is O(jzj2 ·v), and the result follows. The constant implied in the 0, of course, depends on N, and the series taken to infinity is divergent, since the function I d(n) e-= cannot be continued analytically across the imaginary axis. tWigert(l). 164 MEAN-VALUE THEOREMS Chap. VII We can now provet THEOREM 7.15 (A). Aa 3-+ 0, for every positive N, f "' j{(!+it)j2e-2&dt = y-l~g 47T8 + f cn8"+0(1:W+1) 0 2sm8 n=O the constant of the 0 depending on N, and the c .. being con&tants. We observe that the term 0(1) in (7.14.2) is " ' ! J j{(!+it)[ 2e-(w+'1l~sech7Tt dt-! [ j{(!+it)j 2e(.,.-2~;llsech7Tt dt, and is thus an analytic function of 8, regular for j8j < 1r. Also f • (-1 ___ 1 )( I +!___) dx 0 exp(ixe-ia)-1 ixe-W exp(-ixe'~.S)-1 ixei1l is analytic for sufficiently small j8j. We dissect the remainder of the integral on the right of (7.14.2) as in (7.14.3). As before "' 7f+i<» f I dx f I d' .,. exp(-ixei.S)-1 x =.,. exp(-izei.S)-1 '-' and the integrand is regular on the new line of integration for sufficiently smalljS[, and, if 8 = f+i7J, z = '"+iy, it is O{y-1 exp(-ycosfe-'1)} as y-+ oo. The integral is therefore regular for sufficiently small )OJ. Similarly for the third term on the right of (7.14.3); and the fourth term is a constant. By the calculus of residues, the first term is equal to . {; 1 1 2 m 6 irifl exp{ -2in1Te2i0)-1 + f . dy + ' [exp((i~-y)e-;'}-1][ exp(( -i~+y)e")-1 r As before, they-integral is an analytic function of 8, regular for 101 small enough. Expressing the series as a power series in exp(2i11'eli&), we therefore obta.in J i{(f+it)l2e-2& dt = 21reUl i; d(n)exp(2in1Te2ifl)+ ~anOn (7.15.2) o n-1 n-o for llil small enough and R(O) > 0. t Kober (4), Atkinaon {1). 7.15 MEAN-VALUE THEOREMS Let z = 2i1T{l-c21&) in (7.HU). Multiplying by 21rea, we obtain 21rea n~ 1 d(n)exp(2in11'e2i&) = y-lo~~=~ 0 sin0) +~+ N-< + n?;O bn{2i1T(1-e21&)}2n+I+0(02-2(z) = ! d(n)e-n:_y-logz ·~=1 z (7.16.1) In this case the contribution of small x is not negligible, but is sub-stantially the same as that of the other part. We have "'+ioo 1>2(~) = ~"'1"' r(s}{2 (s)z8 d<~ (0 < o: < I) ! I-f"'+ia:~ r(l-s){2(1-s)z1-•ds 21Ti 1-<>:-ia:~ Now r(I-s)/x2 (s) = 22- 287T-28 COs2!s1Tr2(s)r(l-s) = 21- 2111T1- 28 cotts1Tr(s) = 2 1- 28111- 28(-i+O( ~-t~rl )) r(s) (t ~ ±oo). [sm!...,.l If z = ixe-i& (x > x0, 0 < li < !1r), the 0 term is ( 1 -f<>:+ia:~ e-tm ) 0 ",,;., [sin!s~l[r(s)[esG) = ;~z 1 7Ha:~ 21- 281T1- 28P(s){2 (s)z-. ds+O(x"') 1-.. -i,(4n',)+O(x"), where o: may be as near zero as we please. (7.16.2) 166 MEAN-VALUE THEOREMS Chap. VII We also use the results • (I I) L d2(n)e-'"'l = 0 ~log3-, n-1 'T/ TJ (7.16.3) i n2d2(n)e-"'~ = a(~ logs!) n-1 7J TJ (7.16.4) as 7J ~ 0, By (1.2.10) 2+i"' 2+i ~ f r(•) ('(') "-· d• ~ -~ ~ d'(n) f r(•)(n")-• a, 2m 2 _ 1 , {(2s) 2m ,.=1 2_;"' = .. ~ 1 d2(n)e-n'l. (7.16,5) Hence c+ioo ~ d'(n)c''~R+!, J r(•) f('2 ('))"-•d• (! <c 8 c-1"' ~ R+O("~), where R is the residue at s = 1; and R = ~ (alog3 !+blog2!+clog!+a), 1J TJ TJ 7J where a, b, c, d are constants, and in fact I I a = 3!,(2) = ~-This proves (7 .16.3); and (7 .16,4) can be proved similarly by first differentiating (7 .16.5) twice with respect to "'· We can now provet THEOREM 7,16. As 8--)- 0 Jtm+it)['c-«dt~~~log'~· Using (7.13,5), we have f [((!+it)['e-"'dt ~ f!Uixe-")i'dx+O(l), ' ' and it is sufficient to prove that J[~,(ixe-")i'dx ~,b ~log'~· t Titchrn&nh (1). 7.16 MEAN-VALUE THEOREMS For then, by (7.16.2), T~~.(ixe-'')l'dx~ JHi·~"ll'~~ Ji~·(ox?ll'~ 0 1/2,. 1/21f ~ J"' I27Txe-11i,P2(47T2ixe-i8)+0(x"')[2~ (0 <a<!) x' l/21r = Jlr$2(ixe-a)+O(x"'-1)[2dx = I lr$2 {ixe-i8)12dx+O( I lr$2(ixe-t.'l)[2dx I x 2"'-2 dx)l+ 21r lt1t 21T -1 11 ' 1+0( 1 1 • 1) 0() -g;;>)log)l :J!og)l+ I, and the result clearly follows. It is then sufficient to prove that ,[ 1%/(n)exp(-inxe-<>) I' dx ~ ,b ~log'~, for the remainder of (7.16.1) will then contribute 0(8-ilog11/ll). As in the previous proof, the left-hand side is equal to ~ 2 e-h,.sin/1 ~~d(n) 2nsinS+ + 2 ~ ~ 1 d{m)d(n) (m+n)sinllcos{2(m-n)1TCOS8} e-!(m+n)Jrsinll_ ,6 2 £=1 {m+n)'sin2ll+(m n)2 cos28 _ 2 ~ m~ld(m)d(n) (m-n)cosllsin{2(m-n)1TCOsS} e-2(m+n)lrslnll 167 ~2 f='1 (m+n)1sin2 ll+(m-n)2cosZll =l:l+I:z+I:s. 168 MEAN-VALUE THEOREMS Chap. VII j:E2j ~ 2 i "i 1 d(m)d(n} m~:~~!~se-2mnalnll m=lln~l ( = 4sinS ~ d( ) -llmnslnllm~ld(m~r) cos'S .6a m m e r~ rll = 4sinli~ .!: ~ md(m)d(m-r)e-a,.mslnll. cos23 6 rll m6+1 The square of the inner sum does not exceed m.%+1 m2d2(m)e-2,.msln3 m-~+l d2(m-r)e-2:rrmsi.D.II :::,;; m~l m2d2(m)e-ll,.mslnll m~l d2(m)e-2mnslnll ~ o(~log'~)o(~Iog'~) ~ o(~Iog'~) by (7.16.3) and (7.16.4). Hence l:, ~ o(~Iog'~)· Finally (as in the previous proof) l:3 = o(sz m~Z m1+~e-2m1rslnll) = 0(3-~). This proves the theorem. It has been proved by Atkinson (2) that fit(Hilll'<-"dt ' ~ HA!og'~+Blog'~ +Olog'~+Dlog~+E)+o((~)+'). where A=~· B = -~{2log271-6y+ 24:;2>). A method is also indicated by which the index H could be reduced to a. 7.17. The method of reRidues used in§ 7.15 for j~(f+it)j 2 suggests still another method of dealing with j{(!+it)i4. This is primarily a question of approximating to j 1 .. ~~ d(n)exp( -inxe-'11) ~~ dx = J I ~ exp(in:e-ta)-lls dx 2'11' 2'11' n~l = i i j {exp(imxe i8) l~xp( inxe'~3) 1r m~l n~1 1 ,. MEAN-VALUE THEOREMS In the terms with n ;;::. m, put x = g{m. We get ~1~s· d< ~tmn~ {exp(ige i3)-l}{exp(-inm tge'~3)-lf -·~ Approximating to the integral by a. sum obtained from the residues of the first factor, as in§ 7.15, we obtain as an approximation to this "'I~"" l 211 eiB ~~ m n~ r~ exp{ -2i(nrfm)JTe2~=-i = oGv~d 2 ;v)e-2•ralu z3 • l-1 v-l 170 MEAN-VALUE THEOREMS Chap. VII Using Schwarz's inequality and (7.16.3) we obtain o(~ e-1~; 1 n 23 ~ log3 ~} = o(~ log'~)· Actually it follows from a theorem of Ingham (1) that this term is oalog3~)· 7.18. There are formulae similar to those of§ 7.16 for larger values of k, though in the higher cases they fail to give the desired mean-value formul&.t We have Now = 2k-ks:rri-.tRcosk}s11cosec1Tsrk-1(s). For large s rk-t(s) ,.... a(k-1X1-{le--<.t-t1s(217)ilk-tJ. Now r{(k-l)s-fk+l} , {(k-l)s)<k-tls-Jk+{e--<k-Il•(211)L Hence we may expect to be able to replace rk-l(s) by (k-1)-<>->»!>-1(2~JIIk-Or((k-1)8-!k+1}. Also, in the upper half-plane, cos":ts?Tcosecs71 ,...., (!e-i1•")k ~.:: = -2'-kie-ism~k-Il, We should thus replace r(I-s)x-k(s) by -i. 21-ks11t-kBe-ismik-ll(k-l )--<k-l)s+i-t-~(211Wt-2lr{(k-l )s-!k+ 1 }. Hence an approximation to ~,~;(1/z) should be ifik(~) = -i(211)ik! dk(n)inx 1-a+ioo n=l X l-ii<o r{(k-l)s-fk+l}(k-I)-<k-06+lk-~e-il1r(ik-1)(2k.,..tnz)-• d8. t Soo also BeUman (3). 7.18 MEAN-VALUE THEOREMS 171 Putting 8 = (w+!k-1)/(k-I), the integral is -i(211)lk~J r(w)(k-l)-w-te-i17\4-k-tXw+}k-ll!lk-1)(2k1Tknz)-<w+~k-1W<-t>dw = -i(21T)fkz(k-1 )-ie-i17}k-I)"f(2k1Tknz)- x x exp{ -(k-1 )eimik-t)!(k-l>zkllk-t)~f(k-I)-·bf~112<k-tlln--<ik-tl/lk-tJx X ck exp{-(k-1 )el--l,.2k{(k-I)11k/(k-I)(nx)lllk-I)e-i81<k-1)}, where jCki = 1. We have, by (7 .13.5), . . j I ((f+it)["e-"' dt ~ J l~,(ix.-''JI' dx+a(1) ' ' ' . =I l AO(m+n)(m-n), ( " m-> 1 ) ~.~a 'm•e-''"''"''- ~a( f m•e-''"''"') ~a(-!) ,f;:2 ~ m-n m~l (,\O)IH ' ~.~a(~ m•e-'••1 •' ~'-1 ) ~ a(-1 )· m-2 n~t m-n (AO)IH Hence, for ,\ < A, J -1~ (ixe-")l'dx ~ a(-1-). ,\ k (..\0)1+< 172 MEAN-VALUE THEOREMS Chap. VII ' . f lf,(ix.-")1' dx ~ f I +•(ixe',.) I'~· o t/A Also and by the above formula this should be approximately (27T)kf(k-l) ~X X jl.! <!~~~:~11 exp{-(k-1)i(27T)kf(k-IJ(nx)l/\k-IJe-ilit(k-Il}l 2 __!,!?_. 11 ,~. n=l n xll-k/(k-tJ Putting x = fk-1, this is (27T)klx f . I ~ d,(n) I' X A-•t•~-u ,6 n<fk-IJI<k-IJ exp{-(k-1)i(27T)kl(k-tlntttk-1lfe-illl<k-1l} df, and we can integrate as before. We obtain K ~ ~ d,(m)d,(n) :f:t f=t (mnj(!k-1}1\k-IJ X exp[(k-1 )(27T)kl-mtl\k-tl)i cos 8J(k-1 )-X (nl/(k 11 mtl<k ll)icos~~;t/\:~ 1 J~:~:~:~)!:~~~(kll)s:~};~~<~~~· where K depends on k only. The terms with m = n are ( l • d'l ) 0- '-~exp(-K8ntAA-1l,-lAA-lJ)) = 0~~ !___) 8.ft n S(AO)•. The rest are a("" __ I __ exp(-K0ml/(k-1J).-l/(k-1l)) ~>...;:. (mn)l-~k-1JI<k-IJ mtl<k 1J_nt~k-1J · Now m-• .L n<ik IJ/\k-ll(m!Ak-tJ_nl}(k-11) ·-· !m I m-1 = o{ L n(tk-1)/(k l)ml/(k l)+ L m<fk I)J(k I)+~ 1) 1( )) n-1 fm m-n = O(mt--<-lk-tJtH). MEAN-VALUE THEOREMS Hence we obtain oL~~~ m•cxp(-KSmtl\.\-ll\)} = o{[ x•exp(-KSxW·-1}..\-l/{k-u)dx} Altogether J lei Hit) I"•-"' dt ~ a(1 " 8 ;,,.)+a((8 :_,)"'). and taking,\ = Mk-t, we obtain [ l'(f+it);2ke-281 dt = O(O-ik-•) (k ~ 2). This index is what we should obtain from the approximate functional equation. 7.19. The attempt to obtain a non-trivial upper bound for [ rm+itW'e-• at for k > 2 fails. But we can obtain a lower boundt for it which may be somewhere near the truth; for in this problem we can ignore .f,k(ixe-ili) for small x, since by (7.13.5) (7.!9.1) and we can approximate to the right-hand side by the method already used. If k is any positive integer, and a > 1, ('(a)~ n ('--'-)-' ~ n ~ ij,+m-I)' _I~~ d,(n) P P 8 P 6a (k-l)!m! pms ;S-1 ns If we replace the coefficient of each term p-nu by its square, the coefficient of each n-s is replaced by its square. Hence if Fk(.s) = ~ df~:), then F. 8 ~ n ~ (lk+m-l)!)' l -~ k() P ~' {k-l)!m! pmi -l}ik-}, and so represents an analytic function, g(s) say, regular for u > L and bounded in any half-plane u ~ !+S; and· Fk(s) = SA.'(.s)!J(s). Now i a:(n)c-2" 81n8 = ~ 2 J+i"'r(s)Fk(.s)(2sinS)---a ds. n=I 21Tt 2-ir.c Moving the line of integration just to the left of u = I, and evaluating the residue at 8 = I, we obtain in the usual way since here there is a pole of order k2+ 1 at 8 = 0. We can now prove THEOREM 7.19. For any fixed integer k, and 0 < 8 ~ 80 = 80(k), ! j((J+it)j"'-"dt;:, ~log'' i· an!he integral on the right of (7.19.1) is equal to (7.18.1) with A= l; 2::1 "'j~Iogk' ~· while kz+2::s = aGlog"'-1~)· The result therefore follows. 7.20 MEAN-VALUE THEOREMS 175 NOTES FOR CHAPTER 7 7 .20. When applied (with care) to a general Dirichlet polynomial, the proof of the first lemma of § 7.2 leads to ' f ~~a.,n-•tl 2 dt = ~ la.,I2 {T+O(nlog2n)}. 0 However Montgomery and Vaughan [I] have given a superior result, namely ' f ~~a.n-'f dt ~ ~ I•.I'{T+O(n)) (7.20.1) Ramachandra has given an alternative proof of this result. Both proofs are more complicated than the argument leading to (7.2.1). However (7.20.1) has the advantage of dealing with the mean value of ((s) uniformly for u ~ !· Suppose for example that r1 = !· One takes x = 2T in Theorem 4.11, whence ((!+it)~ L n-1-"+0(T-1) ~ Z+O(T-~, .,,;2T say, for T:::;: t :<::;; 2T. Then " f IZI'dt~ L n-'{T+O(n)) ~ TlogT+O(T). T .,,;2T Moreover Z <:g T~. whence Then, since [ O(T-I)'dt ~ 0(1), we conclude that " J K(f+it)l2 dt = TlogT+O(T), MEAN-VALUE THEOREMS Chap. VII and Theorem 7.3 follows (with error term O(T)) on summing over t T, t T, ... . In particular we see that Theorem 4.11 is sufficient for this purpose, contrary to Titchmarsh's remark at the beginning of §7.3. We now write JI((J:+it)l'dt ~ Tlog(~)+(2y-1) T+E(T). 0 Much further work has been done concerning the error term E( T). It has been shown by Balasubramanian that E( T) ~ T t + '. A different proof was given by Heath-Brown . The estimate may be improved slightly by using exponential sums, and lvic [3; Corollary 15.4) has sketched the argument leading to the exponent -fo- + s, using a lemma due to Kolesnik [ 4 ). It is no coincidence that this is twice the exponent occurring in Kolesnik's estimate for Jl(f), since one has the following result. LEMMA 7.20. Let k be a fixed positive integer and lett )'c 2. Then log 2t ((f+it)k~(logt)(1+ f IC(f+it+iu)l"e-l"ldu )-(7.20.2) -log' I This is a trivial generalization of Lemma 3 of Heath-Brown [2 ], which is the case k = 2. It follows that C(!+it)2 4: (logt)4+(logt)maxE{t±(logt) 2 }. (7.20.3) Thus~ if ll is the infimum ofthoseo: for which E(T) 4: r~, thenp(f)::;; ill· On the other hand, an examination of the initial stages of the process-for estimating ((f+it) by van der Corput's method shows that one is, in effect, bounding the mean square of C (f +it) over a short range (t- ~. t + ~). Thus it appears that one can hope for nothing better for Jl(f), by this method, than is given by (7 .20.3). The connection between estimates for C(! +it) and those for E( T) should not be pushed too far however, for Good has shown that E(T) = Q(T!). Indeed Heath-Brown later gave the asymptotic formula f ' E(t)' dt ~ f(2n) -I((!)' Tl + O(Thog'T) (7.20.4) ((3) 0 from which the above Q..result is immediate. It is perhaps of interest to 7.20 MEAN-VALUE THEOREMS note that the error term of (7 .20.4) must be 0{ T f(log T)- I}, since any estimate O{F(T)} readily yields E(T)~{F(T)logT}t, by an argu-mentanalogous to that used in the proof of Lemma o: in 14.13. It would be nice to reduce the error term in (7.20.4) to O(T l+') so as to include Balasubramanian's bound E(T) 4: rt+•. Higher mean-values of E(T) have been investigated by I vic [I] who showed, for example, that ' f E(t)Bdt4: TH•. (7.20.5) 0 This readily implies the estimate E(T) ~ Ti+•. The mean-value theorems of Heath-Brown and Ivic depend on a remarkable formula for E(T) due to Atkinson . Let 0 <A !-(NT +!'!')'. 2Jt 2 2Jt 4 Then E(T) = !: 1 +I:2+0(log2T), where !:1 = 2-:1 "~N (-1)"d(n)(~+~ )-t {sinh 1 (~ y}-1 sin{(n) with (7.20.6) (••)' {(n) = tn+2Tsinh- 1 2T +(n2n 2 +2nnT)l, (7.20.7) and !:2 = 2 L d(n)n -t log-sing(n) ( T )-' ,.,N 2nn where g(n) = Tlog 2~n- T-tn. Atkinson loses a minus sign on {1; p 375 ]. This is corrected above. In applications of the above formula one can usually show that !:2 may be ignored. On the LindelOf hypothesis, for example, one has n~x d(n)n -J-•T ~ T• 178 MEAN-VALUE THEOREMS Chap. VII for x ~ T, so that :E2 ~ T' by partial summation; and in general one finds I:2 4 T2s<(!}+•. The sum :E 1 is closely analogous to that occuring in the explicit formula (12.4.4)for .&(x) in Dirichlet's divisor problem. Indeed, if n = o(Ti) then the summands of (7.20.6) are (-l)nc~;yd~~) cos( 2j(2xnT)-~)+o( r.d~;>). 7.21. Ingham's result has been improved by Heath-Brown to give T Il((t+it)l'dt~ J, c0 T(IogT)"+0(TI+<) (7.21.1) where c4 = (2n:2) t and c3 =2{4y-l-log(2x)-12('(2)n: 2}x-2. The proof requires an asymptotic formula for n~N d(n)d(n+r) with a good error term, uniform in r. Such estimates are obtained in Heath-Brown (4] by applying Weil's bound for the Kloosterman sum (see §7.24). 7 .22. Better estimates for a k are now available. In particular we have u 3 ~-&and u4 :::; §-.The result on u4 is due to Heath-Brown . To deduce the estimate for u 3 one merely uses Gabriel's convexity theorem (see § 9.19), taking a = i• f3 = i . ..l = t. p. = t. and u = -i2-· The key ingredient required to obtain u 4 ::;; i is the estimate [l((t+it)l"dt-< T'(IogT)" (7.22.1) of Heath-Brown . According to (7.20.2) this implies the bound p.(!)::;; l In fact, in establishing (7.22.1) it is shown that, if IC(!+ itr)l ~ V(> O)for 1::;; r::;; R, whereO <lr::;; Tand tr+l-tr ";:31, then R <t T 2 V- 12 (log T)lG, and, if V ";:3 T h-(log T)2, then R<t TV- 6 (logT)B. Thus one sees not only that C <! + it) <t tl (log t)!, but also that the number 7.22 MEAN-VALUE THEOREMS 179 of points at which this bound is close to being attained is very small. Moreover, for V "d3 T h-(log T)2, the behaviour corresponds to the, as yet unproven, estimate T f K<!+it)IGdt I IS(x)ldx )e-G'Ktr, (7.22.2) where K runs over powers of2 in the range Ti::;; K::;; TG-2logaT, and S(x) = S(x, K, T) = K\: (-1)nd(n)etHn> withf(n) as in (7.20. 7). The bound (7.22.2) holds uniformly for log2 T::;; G ::;; Th. In order to obtain the estimate (7.22.1) one proceeds to estimate how often the sum S(x, K, T) can be large, for varying T. This is done by using a variant of Halitsz's method, as described in §9.28. By following similar ideas, Graham, in work in the process of publication, has obtained T I I((~+ it)l"'dt-< T" (log T)"'. (7.22.3) Of course there is no analogue of Atkinson's formula available here, and so the proof is considerably more involved. The result (7 .22.3) contains the estimate p.(t) ::;; h (which is the case l = 4 of Theorem 5.14) in the same way that (7.22.1) implies p(!)::;; !· 7.23. As in §7.9, one may define u,., for all positive real k, as the infimum of those u for which (7.9.1) holds, and 11',. similarly, for (7.9.2). "" MEAN-VALUE THEOREMS Then it is still true that a k = a~, and that T f l((a+it)l 2kdt = T~d,.(n)2n 2"+0(TI-~) Chap. VII for a> a,., where b =(a, k) > 0 may be explicitly determined. This may be proved by the method of Haselgrove [1 ]; see also Turganaliev [1 ]. In particular one may take O(a, f) = t(a -!)for l < u < 1 (I vic [3; (8.111)] or Turganaliev [1 ]). For some quite general approaches to these fractional moments the reader should consult Ingham (4) and Bohr and Jessen (4). Mean values for u = ! are far more difficult, and in no case other than k = 1 or 2 is an asymptotic formula for T f I((J:+it)l"dt ~ I,(T), 0 say, known, even assuming the Riemann hypothesis. However Heath-Brown [7) has shown that T(logT)""~Ik(T)4T(IogT)k" (k=~). Ramachandra [3), having previously dealt with the case k = !· Jutila [ 4] observed that the implied constants may be taken to be independent of k. We also have l"(T)p T(logT)"" for any positive rational k. This is due to Ramachandra [ 4] when k is half an integer, and to Heath-Brown in the remaining cases. (Titchmarsh (I; Theorem 29] states such a result for positive integral k, but the reference given there seems to yield only Theorem 7.19, which is weaker.) When k is irrational the best result known is Ramachandra's estimate l"(T) p T(log T)k"(loglog T)-k'. If one assumes the Riemann hypothesis one can obtain the better results and 111(T)-tT(logT)k' (O~k~2) (7.23.1) 7.23 MEAN· VALUE THEOREMS 181 for which see Ramachandra [ 4] or Heath-Brown [7 ]. Conrey and Ghosh have given a particularly simple proof of (7.23.1) in the form with I,(T);, {C,+o(1)} T(logT)'', c, ~ {r(k'+1)}-' ~ {( 1-~ r .t (r~~;(~))' p-• }· They suggest that this relation may even hold with equality (as it does when k = 1 or 2). 7.24. The work of Atkinson (2) alluded to at the end of §7.16 is of special historical interest, since it contains the first occurence of Kloosterman sums in the subject. These sums are defined by S(q;a,b)= ± exp(~(an+bii)), n.~l q (7.24.1) {n..q)=l where nft = 1(mod q). Such sums have been of great importance in recent work, notably that of Heath-Brown [ 4] mentioned in §7 .21, and of Iwaniec and Deshouillers and lwaniec , (3] referred to later in this section. The key fact about these sums is the estimate IS(q; a, b)l <; d(q)ql(q, a, b)l, (7.24.2) which indicates a very considerable amount of cancellation in (7.24.1). This result is due to Weil when q is prime (the most important case) and to Estermann in general. Weil's proof uses deep methods from algebraic geometry. It is possible to obtain further cancellations by averaging S(q; a, b) over q, a and b. In order to do this one employs the theory of non-holomorphic modular forms, as in the work of Deshouillers and lwaniec [1 ]. This is perhaps the most profound area of current research in the subject. One way to see how Kloosterman sums arise is to use (7 .15.2). Suppose for example one considers J I((J:+it)l' l.~u u-•f e-•!Tdt. (7.24.3) 0 Applying (7.15.2) with 2b = 1/T+ilog(v/u) one is led to examine • (2ninu ) I d(n) exp -u e•IT . ""' 182 MEAN-VALUE THEOREMS Chap. VII One may now replace e'1Tby 1 +(i/T) with negligible error, producing J. d(n) •• pc·~·")··P(-2 :;") ~ zk .'f n·>(2~:)· nH)d· where D(s.~)= n~t d(n)exp( 2 tt~nu)n-s. This Dirichlet series was investigated by Estermann [1 ], using the function {(s, a) of §2.17. It has an analytic continuation to the whole complex plane, and satisfies the functional equation n(e ") ~ 2v',,r(l-•)' {n(l-e, ii)-coe(ns)D(l-•, ii)} 'V (2n)2-2s U V providing that (u, v) = 1. To evaluate our original integral (7.24.3) it is necessary to average over u and v, so that one is led to consider for example. In order to get a sharp bound for the innermost sum on the right one introduces the Kloosterman sum: ( 2 · -) " (2ninm) Lexp..!!!:!!!!.=Lexp--L 1 u«V V 111 , 1 V 11 .;U (..,u)-1 (m•u)-1 """'(modu) =I exp __ t... -L..exp---" (2ninm) '<' {' .(', (2nia(m-u))} m"'t V .. ,.u Va=t V (m.u)=I =! f S(u;a,n) L exp(- 2n:iau). U n= 1 u.;:U V and one can now get a significant saving by using (7 .24.2). Notice also that S(u; a, n) is averaged over u, a and n, so that estimates for averages of Kloosterman sums are potentially applicable. By pursuing such ideas and exploiting the connection with non-holomorphic modular forms, Iwaniec showed that ~. 's·· "' l((i+i41'dt .. (RA+TRIA-I)T' '· 7.24 MEAN-VALUE THEOREMS 183 for 0 :s:; tr,.;; T, tr+ 1 -tr ~ l\ "d3 Ti. In particular, taking R = 1, one has T+Ti f l((j+it)l'dt .. T!., (7.24.4) T which again implies Jt(!) :s:; t. by (7.20.2). Moreover, by a suitable choice of the points tr one can deduce (7.22.1), with T 2 +< on the right. Mean-value theorems involving general Dirichlet polynomials and partial sums of the zeta function are of intP.rest, particularly in connection with the problems considered in Chapters 9 and 10. Such results may be proved by the methods of this chapter, but sharper estimates can be obtained by using Kloosterman sums and their connection with modular forms. Thus Deshouillersand lwaniec , [3) established the bounds T fl((t+it)l 4 1 L ann111 2 dt4;T'(T+TiNZ+TfNt) L ja1112 n<:;N no>;N 0 (7.24.5) and J rl((j+i41'1 L a.m"l'l L b.n"l 7 dl mo>;M n<:;N 0 .. T'(T+T!MiN+T!MNl+MINi)( L la.l')( L lb.l')(7.24.6) mo>;M n<:;N for N :s:; M. In a similar vein Balasubramanian, Conrey, and Heath-Brown showed that fl((j+il)l'l.t•(m)F(m)m-!-•'1' d! ~ CT+O,(T(log T)-'), 0 (7.24.7) JJ(m)tt(n) -( T(m, n)2 ) C= .... ~M~F(m)F(n)(m,n) log 2n:mn +2y-1 forM :s:; Tilr-•, where A is any positive constant, and the function F satisfies F(x) 4; 1, F'(x) 4; x- 1• The proof requires Weil's estimate for the Kloosterman sum, ifT!:s:; M:s; T11r-•. VIII !l-THEOREMS 8.1. Introduction. The previous chapters have been largely con-cerned with what we may call 0-theorems, i.e. results of the form <(•) ~ O(j(t)}, 1/((s) ~ O{g(t)) for certa.in values of a. In this chapter we prove a corresponding set of 0-theorems, i.e. results of the form <(•) ~!l{~(t)}, lf'(s) ~ !l{~(t)), the 0 symbol being defined as the negation of o, so that F(t) = O{,P(t)} means that the inequality iF(t)l > A,P(t) is satisfied for some arbitrarily large values of t. If, for a given function F(t), we have both F(t) ~ O(j(t)}, F(t) ~ !l(j(tl}, we may say that the order of F(t) is determined, and the only remaining question is that of the actual constants involved. For a> I, the problems of {(a+it) and 1/{(a+it) are both solved. For ! ~ u ~ l there remains a considerable gap between the 0-results of Chapters V-VI and the 0-results of the present chapter. We shall see later that, on the Riemann hypothesis, it is the D:-results whtch represent the real truth, and the 0-results which fall short of it. We are always more successful with D:-theorems. This is perhaps not surprising, since an 0-result is a statement about all large values oft, an D:-result about some indefinitely large values only. 8.2. The first D: results were obtained by means of Diophantine approximation, i.e. the approximate solution in integers of given equa-tions. The following two theorems are used. DIRIOHLET's THEOREM. Given N real numbers a1, ~ •••• , aN, a positive integer q, and a positive number t0, we can find a number tin the range t,,;;; t,;;; t,qN, (8.2.1) aM integers x1, .x2, ••• , xN, 8UCh tOOt )tan-xnJ ~ Ifq (n = I, 2, ... , N). (8.2.2) The proof is based on an argument which was introduced and employed extensively by Dirichlet. This argument, in its simplest form,.is that, if there are m+ I points in m regions, there must be at least one region which contains at least two points. D-THEOREMS Consider the N -dimensional unit cube with a vertex at the origin and edges along the coordinate axes. Divide each edge into q equal parts, and thus the cube into qN equal compartments. Consider the qN+I points, in the cube, congruent (modi) to the points (ua1,ua2, ... ,uaN), where u = 0, t0 , 2t0 , ••• , qNt0• At least two of these points must lie in the same compartment. If these two points correspond to u = u1 , u = u2 (u1 < u2), then t = u 2-n1 clearly satisfies the requirements of the theorem. The theorem may be extended as follows. Suppose that we give u the values 0, t0, 2t0, ••• , mqNt0• We obtain mqN+I points, of which one compartment must contain at lea.st m+ I. Let these points correspond to u = u1, ••• , um+1· Then t = u 3-u1, ••• , um-u1, all satisfy the require-ments of the theorem. We conclude that the interval (t0, mqNto) contains at least m solutions of the inequalities (8.2.2), any two solutions differing by at least t0• 8.3. KRONECKER'S THEOREM. Let a1, a2, ... , aN be linearly independent real numbers, i.e. numbers such that there is no linear relation •\a1+---+ANaN = 0 in which the coeificients ,\1>··· are integers rwt aU zero. Let b1, ... , bN be any real numbers, and q a given positive number. Then we can find a number t and integers x 1, ••• , xN, such that ltan-b ... -x,.) ~ Ifq (n = 1, 2, ... , N). (8.3.I) If all the numbers b ... are 0, the result is included in Dirichlet's theorem. In the general case, we have to suppose the a,. linearly independent~ for example, if the a ... are all zero, and the b ... are not all integers, there is in general no t satisfying (8.3.I). Also the theorem assigns no upper bound for the number t such as the qN of Dirichlet's theorem. This makes a considerable difference to the results which can be deduced from the two theorems. Many proofs of Kronecker's theorem are known. t The following is due to Bohr (15). We require the following lemma LEMMA. If cfo(x) is pruitive and contimwus for a ~ x <:;;; b, then lim ( J {cfo(x)}n dx)t/n = max cfo(x). n->«> a a,;;z.;;b A similar result holds for an integral in any number of dimensions. t Bohr (lfi), (16), Bohr and Jossen (3), Eatermann (3), Lettenmeyer (I). 186 0-THEOREMS Chap. VIII Let M = max¢(x). Then ( 1 {~(x))• dx)"" <;; {(b-a)M•)I/• ~ (b-aJ'''M. Also, given o:, there is a.n interval, (a,fJ) say, throughout which cfo(x) ~M-o:. Hence ( 1 f~(•lJ' dx)"';;;, f(~-·HM -•l"l''" ~ (~-·l'''(M -•). and the result is clear. A similar proof holds in the general case. Proof of Kronecker'B theorem. It is sufficient to prove that we can find a number t such that each of the numbers e2"i(a.l-b,.) (n= 1,2, ... ,N) differs from 1 by less than a given E; or, if N F(t) =I+ .. ··~:}2"t<a..t-b..l, that the upper bound of JF(t)l for real va.luesoftisN +I. Let us denote this upper bound by L. Clearly L .:::;;;; N +I. Let G(cfot,r/J2, ... ,¢N) = 1+ n~1e27Ti</>., where the numbers ¢v r/>2, ... , 4>N are independent real variables, each lying in the interval (0, 1). Then the upper bound of JGJ is N+l, this being the value of jGj when rfo1 = ¢1 = ... = ¢N = 0. We consider the polynomial expansions of {F(t)}k and {G(,P1 , ... ,,PN)}I:, where k is an arbitrary positive integer; and we observe that each of these expansions contains the same number of terms. For, the numbers a1, a 2, ••• , aN being linearly independent, no two terms in the expansion of {F(t)}k fall together. Also the moduli of corresponding terms are equal. Thus if then {G(.Pt•···• .PN)}k = I+ L Cqe2,.i{.\o.•~•+···+>.. . .v~.vl, {F(t)}k = I+ I Cqe2,.i{.\0•1(a 11-bJ+···+-\ .• 'I'(a.,l-b.v)) = 1+ICqe2n«.c..t-fl.l, say. Now the mean values Fk= lirn!:TfTiF(t)j2kdt T-2 -T 8.3 ll-THEOREMS 187 and Gk = j j ··) IG(¢1, ••• ,¢v)j2kd.pl .. df.v are equal, each being equal to I+}.;C:. This is easily seen in each case on expressing the squared modulus as a product of conjugates and integrating term by term. Since N + 1 is the upper bound of I G I, the lemma gives !,~G!I2k=N+J. Hence also ¥~FJf2k = N+I. But plainly forallva.luesofk. HenceL~N+I,andsoinfactL=N+I. This proves the theorem. 8.4. THEOREM 8.4. If u > l, t!ten j{(s)j <;;((a) for all values of t, while i((s)l;;;, (1-•)((a) for some indefinitely large values oft. We have l((s)j ~I~ n-•1 <;;I n-• ~((a), n=l n~l (8.4.1) (8.4.2) so that the whole difficulty lies in the second part. To prove this we use Dirichlet's theorem. For all values of N ~(s) = f n-"e-illogn + f n-<>-if, tt~l n~N+l and hence (the modulus of the first sum being not less than its real part) N • j{(s)l ~ n~ 1 n-"cos(tlogn)-n~~+In-". (8.4.3) By Dirichlet's theorem there is a number t (t0 .:::;;; t .:<( t0 q"") and integers X1, .•• , xN, such that, for given Nand q (q ~ 4), l tl~gn -xnj-"(! (n = 1, 2, ... , N). -n q Hence cos(tlogn) ~ cos(2n/q) for these values of n, and so 0-THEOREMS Chap. VIII Hence by (8.4.3) I((•) I ;;, oos(2nfq)((a)-2 }; n-o. NH Now . I , {(a)= In-"> u-"du = -, n=l a-l ' and ! n-" < Ju-"du = Nl-a. N+l v a-l Hence l{(s)l ~ {cos(27T/q)-2N1-"}{(a), (8.4.4) and the result follows if q and N are large enough. THEOREM 8.4 (A). The function {(s) is unbounded in the open region a> I, t >8 > 0. This follows at once from the previous theorem, since the upper bound {(a) of {(s) itself tends to infinity as a-+ I. THEOREM 8.4 (B). The function {(l+it) is unbounded as t-+00. This follows from the previous theorem and the theorem ofPhragmCn and LindelOf. Since {(2+it) is bounded, if {(l+it) were also bounded {(s) would be bounded throughout the half-strip I ~ a ::;; 2, t > S; and this is false, by the previous theorem. 8.5. Dirichlet's theorem also gives the following more precise result.t THEOREM 8.5. However large 1-t may be, there are values of s in the region a > I, t > 1-t, for which l((s)l > Aloglogt. (8.5.1) Also ((!+it) ~ l2(loglog t). (8o5.2) Take t0 = I and q = 6in the proof of Theorem 8.4. Then (8.4,4) gives l((s)l;;, (!-2N'-")/(a-l) (8.5.3) for a value oft between I and 6N. We choose N to be the integer next above 81/(a-Il, Then I((•) I;;, _ _!;;, log(N-1) > AlogN (8.5.4) 4(a-I) 4log8 for a value oft such that N > Alogt. The required inequality (8.5.1) then follows from {8.5.4). It remains only to observe that the value oft in question must be greater than any assigned t1, if a-1 is sufficiently small; otherwise it would follow from {8.5.3) that {(s) was unbounded f Bohr and Landau (1). 8.5 0-THEOREMS 189 in the region a> I, I < t ~ t1; and we know that {(s) is bounded in any such region. The second part of the theorem now follows from the first by the PhragmCn-LindelOf method. Consider the function /(•) ~ lo!~=~•' the branch ofloglogs which isreal fora> I, and is restricted to lsi > I, a> 0, t > 0 being taken. Then f(s) is regular for I ~a~ 2, t > S. Also I log log sl ,...... log log t as t --+ oo, unifonnly with respect to a in the strip. Hence f(2+it)-+ 0 as t-+ oo, and so, if f(l+it)-+ 0, f(s)--+ 0 unifonnly in the strip.t This contradicts (8.5,1), and so (8.5.2) follows. It is plain that arguments similar to the above may be applied to all Dirichlet series, with coefficients of fixed sign, which are not absolutely convergent on their line of convergence. For example, the series for log {(s) and its differential coefficients are of this type. The result for log {(s) is, however, a corollary of that for t(s), which gives at once llog{{s)l > Iogloglogt-A for some indefinitely large values oft in a > l. For the nth differential coefficient oflog{(s) the result is that I(~)" log((•)! > A,.(loglogt)" for some indefinitely large values oft in a> l. 8.6. We now turn to the corresponding problems! for 1/{(s). We cannot apply the argument depending on Dirichlet's theorem to this function, since the coefficients in the series _ != ~~-t{n) {(s) n~ ns are not all of the same sign; nor can we argue similarly with Kronecker's theorem, since the numbers (logn)/27T are not linearly independent. Actually we consider log {{s), which depends on the series L p-8, to which Kronecker's theorem can be applied, THEOREM 8.6. The function lj{(s) iB unbounded in the open region a> I, t > 8 > 0. We have for a~ 1 I II I . l I . I I jlog{(s)-~p; = .f~2mpm• o::;.f,~2pm=.fp(p-l)=A. t See e.g. my Theory of Functions,§ 5.63, "ith tho angle transformed into a strip. ~ Bohr and Landau (7). 0-THEOREMS Chap. VIII Now R( L _I) ~ ! oo•(tl~gp,) ~ I co•(tl~Jll'>J + ! _1,. p p' n~t Pn :n~l Pn n~X+lp,. Also the numberslogp,. are linearly independent. For it follows from the theorem that an integer can be expressed as a product of prime factors in one way only, that there can be no relation of the form P~'p~• ... p;'" = I, where the A's are int<>gers, and therefore no relation of the form A 1 logp1+-·-+"-.vlogp.v = 0. Hence also the numbers (logp .. )/27T are linearly independent. It follows therefore from Kronecker's theorem that we can find a number t and integers x1, •• , X,v such that !t 10;:"-l-x,! ~ t (n =I, 2, ... ,N), 1tlogp,.-1T-27TX,\,;;; 17T (n =I, 2, ... , "K). Hence for these values of n cos(tlogp,.) = -cos(tlogp,.-1T-2?Tx,),;;; -cos!11 = -!, andhenoe R(L ~)~-~I~+ ! -~. p p n~t1n n~.\"+1 p,. Since !Pn-1 is divergent, we can, if His any assigned positive number, choose u so near to I that! pna > H. Having fixed u, we can choose N so large that Then Since H may be as large as we please, it follows that R(L p-8 ), and so logl{(s)l, takes arbitrarily large negative values. This proves the theorem. THEOREM 8.6 (A). The function tg(l+it) is unbounded as t-+ oo. This follows from the previous theorem in the same way as Theorem 8.4 (B) from Theorem 8.4 (A). We cannot, however, proceed to deduce an analogue of Theorem 8.5 for 1/{(s). In proving Theorem 8.5, each of the numbers cos(tlog n) has to be made as near as possible to I, and this can be done by Dirichlet's theorem. In Theorem 8.6, each of the numbers cos(tlogp,.) has to be made as near as possible to -I, and this requires Kronecker's theorem. 8.6 0-THEOREMS 191 Now Theorem 8.5 depends on the fact that we can assign an upper limit to the number t which satisfies the conditions of Dirichlet's theorem. Since there is no such upper limit in Kronecker's theorem, the corre-sponding argument for 1/{(s) fails. We shall see later that the analogue of Theorem 8.5 is in fact true, but it requires a much more elaborate proof. 8. 7. Before proceeding to these deeper theorems, we shall give another method of proving some of the above results. t This method deals directly with integrals of high powers of the functions in question, and so might be described as a short cut which avoids explicit use of Diophantine approximation. T We write M{lf(s)l 2} =lim!__ J lf(a+it)! 2 dt, T-+oo2T -T and prove the following lemma. LEMMA. Let w b g(s)=~~~· be absolutely wnvergentfor a given value of u, and let every m with bm 'I= 0 be prime to every n with c,. :f=. 0. Then for such u M{jg(s)h(s)l'} ~ M{lg(sii'}M{jh(s)l'}. By Theorem 7.1 M{jg(s)l'} ~ .~, !!:;;', ~f{!h(s)l'} ~! Now ro d g(s)h(s)= 6-;=;· where each term dr r-1< is the product of two terms bm m -s and c .. n -s. Hence N{lg(s)h(s)l'} ~ ~ r;;;' ~ L L li:::)",;· ~ M{lg(s)j')N{Ih(s)l'). We can now prove the analogue for 1/{(s) of Theorem 8.4. THEOREM 8.7, fju > l, then for all values of t, while I I I,;: ((a) [(8) """{(2a) I I I>- (I ) ((a) ~ ,.--€ {(2a) for some iru:Uifinirely large values of t. f Bohr and Landau (7). (8.7.1) (8.7.2) 192 0-THEOREMS We have, for u > 1, Since we have also ~ l~l•ll ~ Il (1+!) ~ Il (1-p-'") ~ lM, ;S n° P ]P P I-p-0 ~(2a) and the first part follows. To prove the second part, write dsJ ~ IT ( 1-~l,,,,). {((!)}'~IT (1-~)'hv(s)}'. By repeated application of the lemma it follows that M(l((:)l") ~IT M([(1-Pi.)[")M{I"v(s)l"). Now, for every p, M1 r1-~ 1 .. 1 ~ ~lTr1-~ 1 .. d •• since the integrand is periodic with period 21Tflogp; and M{l,v(s)l"} ;;.1, Cha;. VIII since the Dirichlet series for {7JN(a)}k begins with I+ .... Hence N 2,./]ogp. M(-1-)>-Illogp. f 11-l_l"dt l{(a)j2k "'"' 27T p~ . »=1 0 ( 2,./logp )'I" Now lim f [1-.2_1 2 k dt = max 11-J:.[ =I+!, k---+cto 0 P 8 o;;;:!o:;;2or/logp p 8 pu Hence ~[ M(111:l1 ,.Jt';;. IT (t+Jkl· Since the left-hand side is independent of N, we can make N -+ ao on the right, and obtain . [ ( 1 )]"" {Ia) ~ M l{(s)j2k ~ {(2o)" 8.7 0-THEOREMS 193 Hence to any E corresponds a k such that [M(I((:)I',Jr > (!-•)(~~~)' and (8.7.2) now follows. Since {(a)/{(2o-)-+ co as a-+ I, this also gives an alternative proof of Theorem 8.6 It is easy to see that a similar method can be used to prove Theorem 8.4 (A). It is also possible to prove Theorems 8.4 (B) and 8.6 (A) directly by this method without using the Phra.gmt'in-LindelOf theorem. This, however, requires an extension of the general mean-value theorem for Dirichlet series. 8.8. THEOREM 8.8. t However large t0 may be, there are values of s in the region a> I, t > t0 for which I dsJI > A loglogl. Also W~il) ~ !l(loglogl). As in the case of Theorem 8.5, it is enough to prove the first part. We first prove some lemmas. The object of these lemmas is to supply, for the particular case in hand, what Kronecker's theorem lacks in the general case, viz. an upper bound for the number t which satisfies the conditions (8.3.1). LEMMA a. If m aM n are different positive integers, Forifm [ > -1 -. n max(m,n) log~~logn~I =~+~~+ .. >~· LEMMA {1. Ifp1, ... ,pN are thefirstN primes, andJl1, ... , P.N are integers, not all 0 (not necessarily positive), then N lloghl~·I>PN~N (p.=maxjp.,.j). For IT p~ = uJv, where ·-• u ~ II Pi:'. v ~ II Pi:'. ~ .. >o ~ .. <o t Bohr and Landl!.u (7). "' !!-THEOREMS and u and v, being mutually prime, are different. Also max(u,v) <-QP:. ~~. and the result follows from Lemma a::, Chap. vm LEMMA y. The numbe1- of aolutiDn8 in positive or zero integers of the ~ For N = 1 the number of solutions is k+I, so that the theorem holds. Suppose that it holds for any given N. Then for given vN+l the number of solutions of does not exceed (k-vN+l+I)N ~ (k+I)N; and vN+l can take k+I values. Hence the total number of solutions is~ (k+I)N+l, whence the result. LEMMA 0. For N >A, there exits a t satisfying 0 ~ t:::;; exp(N6) for which cos(tlogp,.) < -1 + ~ (n ~ N). Let N > 1, k > I. Then where l~0 x,.)k = };c(v0, ... ,vN)~ ... x"z:, c(p0, ... ,v,.,.) = ;o!.~.~N!' ! v,. = k. The number of distinct terms in the expansion is at most (k+ 1 ).v < k2N, by Lemma y. Hence so that Let so that c~: c)'.; :~:c· I I < k"' I c', l;c2 > k-2.V(,!c)l = k-L'V(N+I)2k. N F(t) = 1 - .. f:}itlOilP~, {F(t)}"" =I c(v0, •.. ,11N)(-l)~•+···+~aexp(it ~ 11,)ogp .. ), jF(I)"' ~ ~Jcc'(-l)~•.+''"'•xp(it~ (•.-V.Jlogp,) =l::l+:I:., where I:1is taken over values of (11,11') for which 111 = v;, v1 = v;, ... , and 8.8 !!-THEOREMS 1:2 over the rema.inder. Now T ~ f et<V dt = I (a = 0), I I JT I let"'T-11 2 T' •'"' dt ~ i•T .; !•IT (• # 0). T TI JiF(t)l"'dt;;,~,c'-~,II( •<;<;! IT. o v,. "n ogpw Hence By Lemma fJ, since the numbers~~ .. -~~~ are not all 0, Hence T >N ~ f IF(t)l"' dt;;, I c'- 2PT 2; 2; cc' ' ;;,k-"'(IcJ•- 21/'(IcJ' = (k-2N_ 2~V")(N+I)2k. In this we take k = N4, T =eN', and obtain, for N >A, Hence (~ !IF(I)I"'dt)''"' ?(N+I)e-1110NINH>>>N+I-fz.,. Hence there is a tin (0, eN") such that !F(t)l >N+I-4. 195 Suppose, however, that cos(tlogp11) ~ -I+lfN for some value of n. Then IF(t)l =::;;; N-I+!I-eiHogp~l = N-1+"'2(1-costlogp .. )t ( l)l N I :::;;; N-1+"'2 2-N .:::;;; +I-2N, a contradiction. Hence the result. 0:-THEOREMS Chap. VIII We can now prove Theorem 8.8. As in§ 8.6, for u > I log+O(l). Let now N be large, t = t(N) the number of Lemma 8, 8 = 1/logN, and u = 1+8. Then log.:!>-- :z:~(~~~ >- (1-.!.)i !! ~ j{(s)j ~ p~ ~ N 1 Pi. N+tp'J. ( ') l w l ( ') • l > 1-N 2 -;;-2 2 1-N Vog((c)-A)-2 2 (Anlogn)• p N+lp.,. .V+l AS 11 ANt-u logmsJT > -A-Nlog3-logN u-1 >-A, I(~)[>~~ AlogN > Aloglogt. The number t = t(N) evidently tends to infinity with N, since 1ms) is bounded in jtj ~A, u ~ 1, and the proof is completed. 8.9. fu Theorems 8.5 and 8.8 we have proved that each of the inequalities IW+it)[ > Aloglogt, 1/IW+it)[ > Aloglogt is satisfied for some arbitrarily large values of t, if A is a su~table constant. We now consider the question how large the constant can be in the two cases. Since neither I {( l +it) [floglog t nor I {(I +it) j-1jloglog t is known to be bounded, the question of the constants might not seem to be of much interest. But we shall see later that on the Riemann hypothesis they are both bounded; in fact if A~ jjffiiW+it)l, ~ ~ linll/l((l+it)i, (8.9_ 1) t--><» loglog t 1-->oo loglog t then, on the Riemann hypothesis, .\ ~ 2eY, JL ~ .geY, (8.9.2) where y is Euler's constant. 8.9 ll-THEOREMS There i~J therefore a. certain interest in proving the following results. t THEOREM8.9(A). E!lf~~~~t~j ~eY. THEOREM 8.9 (B). E!!! 1 /~~~~~i:)j ~ ~eY, Thus on the Riemann hypothesis it is only a factor 2 which remains in doubt in each case. We first prove some identities and inequalities. AB in§ 7.19, if F,(,) ~ ~ l) (8.9.3) ... ~ ns and (8.9.4) then (8.9.5) Now for real x -'f. d~ 'f. d1 -; 0 ll-x!ei.f>j 21.:"""""; 0 (l-2Vxcos4+x)k' < 8·9·6) Using the familiar formula }',,)~!f. ~~<11 ___ _ n 71' 0 {z-.J(z2-1)cosP}"+l (8.9.7) for the Legendre polynomial of degree n, we see that (8.9.8) Naturally this identity holds also for complex x; it gives F,(s) ~ IJ (I-~-•)A'(:~;:) ~ ''(') IJ P,,(:~;:). A similar "'t of formulae holds for lfns). We have ( 8 . 9 " 9 ) g(~))'~ IJ(·-~)' ~ Q(•-?+"-'~:-;')p',.-·+'~~)')· t Littlewood (5}, (6}, Titclunanili (4}, (14). 0-THEOREMS Chap. VIII Hence (8.9.10) where the coefficients b~;(n) are determined in an obvious way from the above product. They are integers, but are not all positive. The form of these coefficients shows that g,(x) ~ ~ f•l i __ k!_xlme<m;l'd~ 21r ,. ... ~o m!(k-m)! ~ ~ fll+xle'0i"df ~ ~ J(t+2xlws~+x)'~· ' ' Comparing this with the formula we see thatt Hence P,(z) ~ ~ J {z+,(z'-l)eos~)·d~ gk(x) = (1-x)kPk(l+x). 1-x G,(a) ~ n (1-p-')'P,(l+p-•) ~I n p,(l+p-'). P 1-p__, ~"(s) P k 1-p"""" We have also the identity (8.9.14) Fk+I(s) = ~llk+ 1 (s)Gds). (8.9.15) t This formula is, essentially, Murphy's well-known fonnula Jl(oos8) = eos .. i8F(-k, -k; 1; -tan'i8) with :r: = -tan1i6; cf. Hobson, SphericalandEUipMlidal Harmcmics, pp. 22, 31. 8.9 0-THEOREMS AgainforO<x<! "'"' l f d~ f.t(x) >;(I u-=-zvxcos,P+x)"' •I' = n(l.!~x)2k f { 1 -1~v;~~~:s 0 ;~xr !Up •lk = n(l_!v'x?"' f {1+0(~)r d,P > 2k(1~~x)2k (8.9.16) if k is large enough. Hence also ( ) - (l )" " ( ) (l +Jx)"•' (l+Jx)" (8.9.17) gk X--X + fk+l X>~> -3k--for k large enough; and g,(x) .;;~! (l+'•J"~ ~ (l+'xJ" (8.9.18) for all values of x and k. 8.10. Proof of Theorem 8.9 (A). Let a> I. Then f T(t-l'!)11<•+it)j"dt~ JT(1-l'!) ~d,(m) ~d,(n)dt T T L. m"+it L.., na-if -T -T m~l n-1 ~ !~\~) J(~-~)dH- LLd''(:!~~n) J(~-~)(~)"dt n=l -T ,.,..,. -T ~ T ~ dl(n) + "" d,(m)d,(n) 4sin'{!Tlog(n/m)) ,.-'S n2a £;'F-f (mn)a Tlog2(n/m) ~ di(n) ~ T 6 ~ = TF~.:(2a). (8.10.1) Since (from its original definition) f.t(p-k) ;;;::. 1 for all values of p, ( 1 ( 1 )-") F,(Zu) ;;, IT f,(p-'') ;;, n 2k 1---. p,;;z p<;;z p (8.10.2) for any positive x and k large enough. Here the number of factors is n(x) < Axflogx. Hence if x > Vk n ~ (')'''"' _ ( Axlog2k) . p.;;;:r: Zk;;;::. 2k - exp -logx > e . (8.10.3) 0-THEOREMS Chap. VIII Also n 1-p-• " l-p-• " (I I) log ~= L.log~= L0 17--p<;x p p.;;x p p.;;x P P ~ L o(Jogp f ~) ~ o(cu-l) L logp) ~ O((u-I)logx). p.;;x 1 p p.;;x p (8.I0.4) Fd2a) > e-.dX-.1k(a-I)Iogx Il (1-!)-2 "', po>;x p Hence and (~ J (~-~)l({u+it)l"dt)"" > e-A<I'-A<•-''0'" n (~-~r -T po>;x 1 as x-+ co, by (3.15.2). > {eY+o (l)}e-A:rlk-A(a-t)louJogx Letx = fJk, wherek-t < 8 < I,a.ndu = I+TJflogk,whereO < TJ ;lllo>;T (~ f (~-~)("~r af" +(~ [(~-~)m::Tdf" ' ' Hence (2)'"' 2 < T a-I+2112kma,T· ma,T > 2-li2k{eY+o (I)}e .Ja-A'l(]og k-log~)- 2Jog k S TII2kTJ. LetT= 7]-41<, so that Then loglog T = logk+log(4log~)· ma,T > 2-112k{eY-j-o (I)}e-Aa-A11{loglog T-log(4log~)-log~}­ -21]{loglog T-log( 4log~) }· Giving 8 and 'I arbitrarily small values, and then making k-+ co, i.e. T -+ co, we obtain lini~L>-eY loglogT"'"" ' where, of course, a is a function of T. 8.10 0-THEOREMS The result now follows by the PhragmCn-LindelOf method. Let f(s) ~ loglo~\~+hi) where h > 4, and let A~ Tim I((I+it)l. loglogt 201 We may suppose.\ finite, or there is nothing to prove. On a = I, t ?:::: 0, we have Also, on a = 2, If(•) I ,;; ll((ls)l < A+• (t > t,). og ogt If(•) I~ o(l) < A+• (t > t,). We can choose h so that lf(s)l < "-+~' also on the remainder of the boundary of the strip bounded by a = I, a = 2, and t = I. Then, by the PhragmCn-LindelOf theorem, lf(s)j < A+E in the interior, and so . Tim JM! ~ Tim jf(s)L__ <;; A. loglogt loglog(t+h) Hence A ;;::: eY, the required result. 8.11. Proof of Theorem 8.9 (B). The above method depends on the fact that dk(n) is positive. Since bk(n) is not always positive, a different method is required in this case. Let a> I, and let N be any positive number. Then ~IT I ~b,(n) I' dt ~ ~IT " b,(m) " b,(n) dt T n 8 T L ma+iJ L..... na-iJ 0 n.;;. 0 mo>;N n.:;N Now ~ " bl(n) +~" "b,(m)b,(n) IT(")" dt L...., n2a TL.....L..... mana m n.:;N m+N o >- "'b~(n)! "'~lbk{m)bk(n)j_2 :::--- n..:fJv n2a T Li;:+~ mana llognfmj' I n I n+I I I log-?:::: log- ?:::: -;;::: -N, m n 2n 2 so that the last sum does not exceed Since {{a),..... I/(a-I) as a-+ I, and ,(2) >I, we have, if a is sufficiently near to I, 202 D·THEOREMS Hence the above last sum is less than 4N T(u-I)2k" Also 1 1; ~ b,(n)l,;:: ~ )b,(n)) <I" )b,(n)) {k(s) .. 7N ns """ .. ~· n" Niu-t .. .tfN nF+f Chap. vm I (m•Hl)' I ( 2 )' < NTa=t {(a+ I) < N"Ta=I C1-l for a sufficiently near to 1. Since for a > 2 G,(a),;; IJ (I+pF)" ~ IJ u~:-~:r ~ (~'tf)". we have simil&rly G (2u)- ' bi(n) = ..;;;;:' bi(n) < !_ ~b:(n) k L., n2o L.. n2u Na-1 n"+l n.;;.N n>N "> < G,(a+Il < l_(m•Hl)" < -'-(~)'"'· Na-1 Na-1 {(a+I) Na-1 u-1 These two differences are therefore both bounded if N---( 2 )"'"-" a-1 With this value of N we ha.ve by (8.9.17). Now log TI :!~~: ~ O((a-l)logx) •<• as in (8.10.4). Hence, as in (8.10.3) and (3.15.3), T1 {~(l+~t'} > e-.A.x-Ak(a-l)lolll'"'{b+o(l)}2klog2kx p<x where b = 6erj-rr2, B.ll n.THEOREMS 203 Choosing x and u as in the last proof, N (2logk)2klogkJTJ+2k (u-1)2k< -"-• and we obtain T I J 1!+0(1)1 2 dt > e-A3k-ATJk{b+o(J)}2klog2k8k-T 0 ~?(s) 4 (2logk)2klogk/TJ+2k -T -"-+O(I}. Finally, let Then ( 2logk ) 0 logk loglogT =-= logk+log -"-+2 +Ioglog~ < (l+£)logk fork> k1 = k1(£, 1)). Hence T -~ J I_!+ O{l) 12 dt > e-ASk-A11k{b+o(l)}2kf!oglog T -log~)2k +0{1). T 0 'k(s) \ I+t 0 Let Since the first term on the right of the above inequality tends to infinity with k {for fixed 0, 1), and t) it is cleartha.tM~.T tends to infinity. Hence j,"~sl+O(Ill < 2M!,T if k is large enough, and we deduce that 4~T > 1e-ASk-A11K{b+oBll2ke o J l+t" og0 Giving 0, £, and 1J arbitrarily small values, and then varying T, we obtain The theorem now follows a.s in the previous case. 0-THEOREMS Chap. VIII 8.12. The above theorems are mainly concerned with the neighbour-hood of the line a= I. We now penetrate further into the critical strip, and provet THEOREM 8.12. Let a be a .fixed number in the range l ~a< I. Then the inequality l{{a+it)l > exp(log•t) iB satisfied for some irulefinitely large vq,lues oft, provided that a< I;-a. Throughout the proof k is supposed large enough, and 8 small enough, for any purpose that may be required. We take l < a < I, and the constants C1 , C 2, ••• , and those implied by the symbol 0, are independent of k and S, but may depend on a, and onE when it occurs. The case a = lis deduced finally from the case a > f. We first prove some lemmas. LEMMA a. Let k-1( l)m ! (k) rt•l''(•) ~ ' -m.a;;; + 6o (s=I)"Hl in the neighbourhood of s = I. Then jaJ:JI < e0 •k (I ~ m:::;;; k). The~) are the same as those of§ 7.13. We have r(s) =,J 0c,.(s-l)", ~k(s) = (s-1)-kn~oe~kJ(s-1)", where jc,.j :s;;,; c;, je~lj :s;;,; C; (C2 > I, C 3 > I). Hence e;kJ is less than the coefficient of (s-I)" in L~o D;(s-IJ"t = {l-(7a(s-t)}-k = ~o fk!:;~!r G;(s-l~". Hence ml(a(kll ~~k-~-lc e(kJ/ •-t~+n-l)!m · m n~o k-m-n-1 n 6 2 (k-l)!n! 3 LEMMA/3. < kC[c~d!~~~l} 1 2 < ea,k, ~ llr{a+it)('(a+it)ol•-"1' dt > [I,J 1ddn)exp(-inxe-ia)/ 2 x2u-1 dx-exp(C,klogk). t Titchm1U'&h(4). 8.12 0-THEOREMS By (7.13.3} the left-hand side is greater than 2 J j<f.k(ixe-ifJ)I2x2u-1 dx;:;, Jl,~ 1 dk(n)exp(-inxe-,.8 )j 2 x2u-t dx-' ' Since jlog(ixe-ifJ)j:::;;; logx+!1T, I Rk(ixe-i8)j .:::;;; ~{la~klJ+ ja~klj(logx+!7T)+ .. +a~k!l(logx+f7T)k-t} :::;;; keC,k(log:+i1T)k-I, and J (logx+!7T)2k-2x2<>-3 dx < r (2logx)2k-2x2u-3 dx+ .r 1T2k-2x2<>-3 dx 1 i 1 -r(2k-l) + 1T2k-2. -2(1-u)2k-I 2-2a The result now clearly follows. LEMMA y. l!n~ldk(n)exp(-inxe-ill)j2x2u-t dx t "'dl:() I"" > ~~ L -~~e-2nalnll_C6 Jog"8 L di(n)e-nsln8. n'"l u~1 The left-hand side is equal to "~ 1 n~ 1 dk(m)dk(n) J exp(imxe"8-inxe-;8)x2u-l dx ' ~ 2 + 2 ~ l:, + l:,. m~n m;£n ~OW Je-2nxsinfJx2<>-1dX = (2nsin0)-2U r e-Yy2<>-ld!J, 1 2n;inS and for 2nsin0:::;;; 1 ro ro f e-Yy2u-l dy ~ J e-Vy2<>-l dy = C, > ('7 e-2n sin 8, 2nsln8 1 while for 2nsin0 > 1 I e-lly2u-ldy> J e-Vdy=e-2nsin8. 2nslnll 2nSLnll 0-THEOREMS Chap. vm Hence :tl = ! dl(n) I"' e-2n.:cslnllx2<>-l dx > !l! ~ di,(n) e-2nslnll n~I 1 ,32<>6-t n2<> -Also, using (7.I4.4), ~ e-i~~rnll "' = 0 8 ~ ~,-. -,L dk(m)e-:"uinll dk(m-r)e-{<m-~)•lnll ~~1 m~r+1 ~e-1rf<ln81 • • ~ Gs L-' ,~ L d~(m)e-mslnll 2: di{m-r)e--<m-r)llfn/l)l r~l m~r+l m~r+l "' e-t~slnll "' "' <Gs L -,-2: di(m)e-m•ln8 < 0 8 log~ ""di(m)e-miln8, ,.~1 m~t .S rf=t This proves the lemma. LEMMA .S. For q > I exp(os(lo:kr"} < Fk(a) < exp(0 10 k21")-lt is clear from (8.9.6) that fk(x) ~ (I-Yx)-2k (0 < x <I). Also it is easily verified that ((k+m-1)1)' (k'+m-1)! (k-l)!ml ~ (k2-I)!m!-Hence, for 0 < x < I, , ~ (k'+m-1)' Jk(x) ~ L. (k2-I)! Txtn = (1-x)-k' m-o m. Hence logFk(a) = P"~k'logik(p-")+ rh.k"logfJ,p-<>) ::;;;;: 2k ~k'log(I-p-l<>)-I+k2~k•log(I-p-<>)-1 ~ O(k,J,rl•)+O(k',.~.,p-•) ~ O{k(k••)•-l•)+O{k'(k••)•-•) ~ O(k"•). 8.12 a-THEOREMS On the other hand, (8.10.2) gives logFk(a) > 2k! log(l-p-~")-1-! log2k P<"' p 2k !p-l"-011 ~log2k pq: logx w-t" x > 012 klogx -Oulogxlog2k. Taking X-12 (c k )'" -2011 logk the other result follows. Proof of Theorem 8.12 for ! < u < 1. It follows from Lemmas f3 and y and Stirling's theorem that f "' )~(a+it))2ke-2&t2<>-l dt > 5.:! ~ di,(n) e-2naJn8 Oll" L.. n2" 0 n-1 I~ -Ou)ogb 6 di(n)e-nsln8_Q15eC•klogk, Now, if 0 <.: < 2a-1, ~~(n) > Fk(2a)-016 .fS ?;2a(?M)~ = Fk(2a)-q6 8(Fk(2a-.:) > exp{ 09(lo: k) 11 "} -0 16 8~ exp{0 10 k21(2<>-€l}, and I di(n)e-nBinll < 0 11 ! fil(n)(n0)~--2<> = 011 0€-2 <>Fk(2a-f) n~l n-1 < Ol18€-2<>exp{qok2/(2a-~)}. Let .S = exp{ -~k2J<2u-(>}. Then J fb{a+it)f"e-"'t'•-• dt > f,.[c,exp(c.(1 o:k)"")-c,c,.--Ouq,~k21<2<>-€)]-o~5ea.klogk > ~exp{o,Co:kt"}· 208 0-THEOREMS Suppose now that [C(a+it)[ <;; exp(log"t) (t ;:> t,) where 0 < o: < 1. Then ! I C(a+it)f",-'•t'•-' dt ,; cg + J ,,.,,,.,,.,,,, dt. If t > k2JS2, k > k0 , then ~ < Vt < !!____ 8 21og"-t Hence Chap. VIII ~ ~~ ~ J e2klol('l-2&t2"-l dt .:s;; e2ki<>I('(AN8'> J e-2&t2o--I dt+ J e-&t2a-l dt 1 1 k•JlJ• <e2kii)ga(k't)'J~. Hence (--"--)l/u = o(klog"-~) = O(kl+(2<t)f(2"-~l), Iogk S Hence !~I+~• a 2a-E and since E may be as small as we please ~~~+;. o:;?;l-cr. The case u = f. Suppose that m+it) ~ O(exp(logPt)) (0 < fl < tl· Then the function /(s) ~ C(s)exp(-log"a) is bounded on the lines u = f, a = 2, t > t0 , and it is O(t) uniformly in this strip. Hence by the Phragmen-Linde!Oftheoremf(s) is bounded in the strip, i.e. for! < u < 2. Since this is not true for!< a< 1-fJ, it follows that fl;:;, •. NOTES FOR CHAPTER 8 8.13. Levinson (1] has sharpened Theorems 8.9(A) and 8.9(B) to show that the inequalities f((l+it)l ~e 1 loglogt+0(1) 8.13 and 0-THEOREMS --1-. -~ ~ (loglogt -logloglog t) + 0(1) 1((1+tt)/ n:2 209 each hold for arbitrarily large t. Theorem 8.12 has also been improved, by Montgomery [3 ]. He showed that for any a in the range ! < a < 1, and for any real 9, there are arbitrarily large t such that R{e'.9log((a+it)} ~ !'n-(a-!)- 1(log t) 1 -a(Joglog t)-a. Here log ((s) is, as usual, defined by continuous variation along lines parallel to the real axis, using the Dirichlet series (1.1.9) for a > 1. It follows in particular that ((a+ it)= n{exp(_isr (logt)l-a )} <!<a< 1), a-t (loglogW and the same for ((a+it)- 1• For a=t the best result is due to Balasubramanian and Ramachandra , who showed that ( (logH)l ) r.,~:~+HI((!+it)/ ~ exp f (loglogH)i if(log T'f ~ H ~ Tand T ~ T(r5), whereJisanypositiveconstant. Their method is akin to that of§ 8.12, in that it depends on a lower bound for a mean value of /((!+it)/ 2k, uniform ink. By constrast the method of Montgomery uses the formula (logl)2 ~ f e-i8Jog((a+it+iy)( 8in/yY {1+cos(9+ylogx)}dy -(logl)2 ~ L A(n) .-·-"(l:-flog"f)+o{x(logt)-'). (8.13.1) liogn/xl><flogn X This is valid for any real x and 9, providing that ((s) -=f. 0 for R(s) ~a and /I (s) - t 1 ~ 2(log t )2_ After choosing x suitably one may use the extended version of Dirichlet's theorem given in § 8.2 to show that the real part of the sum on therightof(8.13.l)is large atpointst1 < ... < tN~ T,spaced at least 4(1og T)2 apart. One can arrange that N exceeds N(a, T), whence at least one such tn will satisfy the condition that ((s) 0 in the corresponding rectangle. IX THE GENERAL DISTRIBUTION OF THE ZEROS 9.1. In § 2.12 we deduced from the general theory of integral functions that '(s) has an infinity of complex zeros. This may be proved directly as follows. We have I l l 1 I 1 (I I) (I I) 3 v+32+··· < 22+2.3+u+··· = :;;+ 2=a + a-4 + ... = 4· Hence for u ~ 2 I((•) I,;;!+.,!.+~+···,;;!+~+···<~' (9.1.1) and l{(s)[ ~ 1-~- ... ~ 1-~- ... > ~-(9.1.2) Also Rg(s)} =I+ cos(t;~g2) +··· ~ 1-~- ... > ~· (9.1.3) Hence for a > 2 we may write log{(s) ~ log[{(s)[+iarg{(s), where arg ~(s) is that value of arcta.n{IC(s)/RC(s)} which lies between -!7T and !11. It is clear that [log{(•)l <A (u;, 2). (9.1.4) For u < 2, t -=1=- 0, we define log '(s) as the analytic continuation of the above function along the straight line (o+it, 2+it), provided that ((s) # 0 on this segment of line. Now consider a system of four concentric circles 010 02, 03,. 04, with centre 3+iT and radii 1, 4, 5, and 6 respectively. Suppose that {(a) # 0 in or on Of. Then log {(a), defined as above, is regular in Of. Let M1 , M2, M3 be its maximum modulus on 011 02, and 03 respectively, Since {(a) = O(t""'), R{log {(a)} <A log T in Of, and the Borel-" Caratheodory theorem gives M3 ~ 6 2 ~~A1ogT+ :~:logl{(3+iT)I < AlogT. Also M1 <A, by (9.1.4). Hence Hadamard's three·circles theorem, applied to the circles 01, 02, 03, gives M2 ~ M~MC <AlogPT, where 1-cr: = p = log4jlog5 < 1. 9.1 GENERAL DISTRIBUTION OF ZEROS 211 Henoe {(-!+iT)~ O(exp(logiT)) ~ O(T'). But by (9.1.2), and the functional equation (2.1.1) with u = 2, 1{(-l+iT)[ >AT!. We have thus obtained a. contradiction. Hence every such circle Of contains at least one zero of {(a), and so there are an infinity of zeros. The argument also shows that the gaps between the ordinates of successive zeros are bounded. 9.2. The function N(T). Let T > 0, and let N(T) denote the number ofzeros ofthe function {(a) in the region 0 ~ a ~ 1, 0 < t ~ T. The distribution of the ordinates of the zeros can then be studied by means of formulae involving N(T). The most easily proved result is THEOREM 9.2. As T --+ co N(T+I)-N(T) ~ O(logT). (9.2.1) For it is easily seen that N(T+I)-N(T),;; n(,l5), where n(r) is the number of zeros of {(a) in the circle with centre 2+iT and radius r. Now, by Jensen's theorem, ' " f ) d' ~ 2. J logi{(2+iT+3e")l dO-logl{(2+iT)[. ' 2~ ' ' Since l{(s) 1 < t""' for -1 ~ u ~ 5, we have log[{(Z+iT+3e")l <AlogT. ' Hence f n~)dr <AiogT+A < AlogT. ' ' ' J n~) dr ~ J n~) dr ~ n(~5) J ~ = An(~5), 0 -.'6 •'5 Since the result (9,2.1) follows. Naturally it also follows that N(T+h)-N(T) ~ O(log T) for any fixed value of h. In pa.rticula.r, the multiplicity of a multiple zero of {(a) in the region considered is at most O(log T). 212 GENERAL DISTRIBUTION OF ZEROS Chap. IX 9.3. The closer study of N(T) depends on the following theorem.t If Tis not the ordinate of a zero, let S(T) denote the value of 'IT-Iarg((!+iT) obtained by continuous variation along the straight lines joining 2, 2+iT, i+iT, starting with the value 0, If Tis the ordinate of a. zero, let S(T) ~ S(T+O). Let L(T) =]; TlogT-1 +~:; 27TT+~· THEOREM 9.3. As T ~ rx; N(T) ~ L(T)+S(T)+O(I/T). (9.3.1) (9.3.2) The number of zeros of the function E:(z) (see § 2.1) in the rectangle with vertices at z = ±T±~i is 21V(T), so that 2N(T) ~ _l f ?:'(')a, 21T'i E(z) taken round the rectangle. Since E:(z) is even and real for real z, this is equal to 'l'+fi ~' ?.+iT !+iT ~iu + n~·(~:J,~~u + I )~(~:d· T T+ii :a 2+iT = ~dargg(s), where d denotes the variation from 2 to 2+iT, and thence to !+iT, along straight lines. Recalling that we obtain Now and by (4.12.1) •-!T+O(IfT). Adding these results, we obtain the theorem, provided that Tis not the ordinate of a zero. If T is the ordinate of a. zero, the result follows from t Backlund (2), (3). 9.3 GENERAL DISTRIBUTION OF ZEROS 213 the definitions and what has already been proved, the term 0( 1/T) being continuous. The problem of the behaviour of N(T) is thus reduced to that of S(T}. 9.4. We shall now prove the following lemma. LEMMA. Let 0:::::;; a< fJ < 2. Letf(s) be an analytic function, real for reals, regular for a~ a: except ats = 1; let iR/(2+i!)l ;;, m > o and 1/(a'+it')I:::,;;MaJ (a'~a, I:::,;;t':::,;;t). Then if T is not the ordinate of a zero of f(s) largf(a+iT)I..::;; fOg{(2_:)/( 2-fJ)}(logM"'T+2+log~)+~ (9.4.1) fora ~fJ. Since argf(2) = 0, and I If(•)) argf(s) = arctanlRf(s) , where Rf(s) does not vanish on a = 2, we have l-'gf(2+iT)I < !•· Now if Rf(s) vanishes q times between 2+i7' and fJ+iT, this interval is divided into q+ I parts, throughout each of which R{f(s)} ~ 0 or R{f(s)} ..::;; 0. Hence in each part the variation of argf(s) does not exceed 7T. Hence largf(s)l ..::;; (q+i)1T (a~ fJ). Now q is the number of zeros of the function g(') ~ Hf(,+iT)+f(,-iT)) for I(z) = 0, {3:::::;; R(z):::::;; 2; hence q:::::;; n(2-fJ), where n(r) denotes the number of zeros of g(z) for lz-21 ..::;; r. Also 2-a< 2-a< f n~)dr~ J n~)dr ~n(2-fJ)log!=~· 0 2-{3 and by Jensen's theorem 1-a< 2rr f n(r) dr ~I f loglg(2+(2-a)•''JI d8-loglg(2)1 r 2• ' ' < logM,.,T+2+log 1/m. This proves the lemma. GENERAL DISTRIBUTION OF ZEROS We deduce THEOREM 9.4. As T -+ oo S(T) ~ O(log T), i.e. N(T) = ! Tlog T- I+log 2" T+O(logT). 277' 277' Chap. IX (9.4.2) (9.4.3) We apply the lemma with j(s) = {(s), ox = 0, p = !, and (9.4.2) follows, since {(s) = O(tA). Then {9.4.3) follows from (9.3.2). Theorem 9.4 has a number of interesting consequences. It gives another proof of Theorem 9.2, since (0 <8 0 are arranged in a sequence Pn = fln+irn. so that Yn+1 ;<: Yn• then as n-+ 00 IPnl......, Yn"' 1~":. · (9.4.4) Wehave N(T),....., 2~TlogT. Hence 27TN(y,±I)......, {r,.:.!:::l)log(y,.±l),....., y,.logy,.. Also Hence Hence and so N(y,.-1) ~ n ~ N(rn+I). 211n,....., y,.logy,. logn......, Iogyn, 2.-Yn"'logn" Also IPnl,....., Yn• since {1,. = 0(1). . We can also deduce the result of§ 9.1, that the gaps between the ordinates of successive zeros are bounded. For if jS(t)l ~ Clogt (t;;:: 2), N(T+H)-N(T) ~ 1 TJ+Hlog.'...dt+S(T+H)-S(T)+O(I_) 21T T 21T T If T (') ;;:: 2;log2;-0{log(T+H)+IogT}+O T, which is ultimately positive if H is a constant greater than 4tr0. The behaviour of the function S(T) appears to be very complicated. It must have a discontinuity k where T passes through the ordinate of a zero of ~(s) of order k (since the term 0(1/T) in the above theorem is in fact continuous). Between the zeros, N(T) is constant, so that the 9.4 GENERAL DISTRIBUTION OF ZEROS 215 variation of S(T) must just neutralize that of the other terms. In the formula (9.3.1), the term tis presumably overwhelmed by the variations of S( T). On the other hand, in the integrated formula T T 7' f N(t) dt ~ f L(t) dt+ f S(t) dt+O(Iog T) ' ' ' the term in S(T) certainly plays a much smaller part, since, as we shall presently prove, the integral of S(t) over (0, T) is still only O(log T). Presumably this is due to frequent variations in tl>e sign of S(t). Actually we shall show that S(t) changes sign an infinity of times. 9.5. A problem of analytic continuation. The above theorems on the zeros of {(s) lead to the solution of a curious subsidiary problem of analytic continuation.t Consider the function P(s) ~ L 1_, (9.5.1) p P' This is an analytic function of s, regular for a > 1. Now by (1.6.1) P(s) ~ ~ ~~)log((ns). (9.5.2) n~l As n-+ oo, log {(m) ,.., 2-ns. Hence the right-hand side represents an analytic function of s, regular for a > 0, except at the singularities of individual terms. These are branch-points arising from the poles and zeros of the functions {(ns); there are an infinity of such points, but they have no limit-point in the region a > 0. Hence P(s) is regular for a > 0, except at certain branch-points. Similarly, the function ' ~ ('Ins) Q(s) ·~ -P (s) ~- ~.~(n) ((ns) (9.5.3) is regular for a > 0, except at certain simple poles. We shall now prove that the line a= 0 is a natural boundary of the functions P(s) and Q(s). We shall in fact prove that every point of a = 0 is a limit-point of poles of Q(s). By symmetry, it is sufficient to consider the upper half-line. Thus it is sufficient to prove that for every u > 0, 8 > 0, the square 0 <a < 8, u < t ~ u+ll (9.5.4) contains at least one pole of Q(s). t La.ndauaodWalfisz(l). 216 GENERAL DISTRIBUTION OF ZEROS Chap. IX Asp -+ oo through primes, N{p(u+8)} ......-I; (u+8)plogp, N(pu),...., -kuplogp, by Theorem 9.4. Hence for all p ~ p0(8, u) N{p(u+O)}-N(pu) > 0. (9.5.5) Also, by Theorem 9.2, the multiplicity v(p} of each zero p = f3+iy with ordinate y ~ 2 is less than A logy, where A is an absolute constant. Now choose p = p(15, u) satisfying the conditions P>i. p;;,~, p;;,p,(o,u), p>Aiog{p(u+o)). There is then, by (9.5.5), a zero p of ,(8) in the rectangle i ~ u < I, pu < t ~ p(u+l3). (9.5.6) Since y > pu ~ 2, its multiplicity v(p) satisfies v(p) <A logy<;; Alog(p(u+o)} 0. t Eatermann (I). 9.6 GENERAL DISTRIBUTION OF ZEROS 217 9.6. An approximate formula for t'(s)fb(s). The following approxi-mate formula for rf8)g(8) in terms of the zeros near to 8 is often useful. THEOREM 9.6 (A). If p = f:J+iy runs throw;h zeros of {(8), {'(s) ~ L I__+O(logt), {(8) )l-yl08-p (9.6.1) uniformly for -1 :>;;; u :>;;; 2. Take f(8) = {(8), 8u = 2+iT, r = 12 in Lemma a: of § 3.9. Then M = A log T, and we obtain f'(s) ~ L I__+O(logT) (9.6.2) {(s) lp-s.I<>:s 8-p for [8-8ul ~ 3, and so in particular for -I :>;;; u ~ 2, t = T. Replacing T by tin the particular case, we obtain (9.6.2) with error O(Iogt), and -I ~ u ~ 2. Finally any term occurring in (9.6.2) but not in (9.6.1) is bounded, and the number of such terms does not exceed N(t+6)-N(t-6) ~ O(logt) by Theorem 9.2. This proves (9.6.1). Another proof depends on (2.12.7), which, by a known property of the r-function, gives {'(a)~ L (-'-+~)+O(logt). {(8) P s-p p Replacing 8 by 2+it and subtracting, {'(s) ~ '-" (_I__ __ I)+O(logt), {(s) "7 s-p 2+tt-p sinoe {'(2+it)f{(2+it) ~ 0(1). Now L 2+~t-p = ll-~nO(l) = O(logt) )1-y)<>:I by Theorem 9.2. Also 2: (,~ -2 :, ) ~ 2: , 2;;;i,-l+n<y"l+n+l P + p t+n<y'(t+n+l ( p)( p) ~ '-" of-'-) ~ L o(..!.) ~ oflog(t+nl) t+n<y~+n+l (y-t)2 t+n<y.;:t+n+l nZ \ n2 ' again by Theorem 9.2. Since ! 1~;-n) < 2 Io!22t + L: lo!;n = O(Iogt), n=l n<;:t n>l 218 GENERAL DISTRIBUTION OF ZEROS Chap. IX it follows that 2: (-'---1 -) ~ O(logl). y>l+l 8-p 2+~t-p Similarly 2: (-'---1 -) ~ O(logl) yd-1 8-p 2+tt-p and the result follows again. The corresponding formula for log t(s) is given by THEOREM 9.6 (B). We have logt(s) = 11 ~,,}og(s-p)+O(logt) (9.6.3) uniformly for -I ,;;; q ::;;; 2, where log {(s) has its usual meaning, and -1r < Ilog(s-p) ::;;; '"· Integrating (9.6.1) from s to 2+it, and supposing that tis not equal to the ordinate of any zero, we obtain log{(s)-log{(2+il) ~,.-~.,{log(s-p)-log(2+il-p)}+O(logl). Now logt(2+it) is bounded; also log(2+it-p) is bounded, and there are O(log t) such terms. Their sum is therefore O(log t). The result therefore follows for such values of t, and then by continuity for all values of s in the strip other than the zeros. 9.7. AsanapplicationofTheorem 9.6 (B) weshallprovethefollowing theorem on the minimum value of {(s) in certain parts of the critica.l strip. We know from Theorem 8.12 that i{(s)l is sometimes large in the critical strip, but we can prove little about the distribution of the values of t for which it is large. The following resultt states a much weaker inequality, but states it for many more values oft. THEOREM 9.7. There is a coMtant A suck that each interval-{T, T+I) contains a value oft for whick i{(s)l > 1-A (-1 <;;a<;; 2). (9.7.1) Further, if His any number greater than unity, then 1{(•)1 > T-•H (9.7.2) for -1.::;;;: u ~ 2, T.::;;;: t ~ T+1, except possibly for a set of values oft of measure If H. Taking real parts in (9.6.3), logl{(s)l = 11 ~,./ogls-pi+O(logt) ;;;::: 11 ~0 Ioglt-yi+O(logt). (9.7.3) t VWiron (I), La.nda.u (8), (18), Hobeiael (3). 9.7 GENERAL DISTRIBUTION OF ZEROS "' Now T+l IPln(y+t,T+l) J ~ logl<-rl d< ~ ~ J logl<-rl dl T 11-Yf'.-;t T-t,;;y.-;T+t ma:s:(y-l,T) , .. ;;;, ~ J Iogl<-rl dl T-t..;;y.-;T+2 y-l ~ (-2) > -A!ogT. T-t..;;y<;;T+i Hence ll-~•.a loglt-yl > -A log T for some tin (T, T+1). Hence logl{(s)l > -A log T for some t in (T, T+1) and all u in -1 ~ u ~ 2; and logi{(•)l > -AH!ogT except in a. set of measure 1/H. This proves the theorem. The exceptional values oft are, of course, those in the neighbourhood of ordinates of zeros of {(s). 9.8. Application to a formula of Ramanujan.t Let a and b be positive numbers such that ab = 1r, and consider the integral I J a-28 ~ ds = __! f ~ r(}-s) ds 21ri {(1-2s) 21ri ..;1T {(2s) taken round the rectangle (l±iT, -!±iT). The two forms are equiva-lent on account of the functional equation. Let T~oo through values such that IT-yl > exp(-A1 yflogy) for every ordinate y of a zero of {(s). Then by (9. 7.3) Iogl{(a+iT)I;;;,-~ A,yf!ogy+O(!ogT) > -A,T IT-yl<t where A 2 < }1r if A1 is small enough, and T > T0• It now follows from the asymptotic formula for the r -function that the integrals along the horizontal sides of the contour tend to zero as T ~ oo through the above values. Hence by the theorem of residuest -f+ia> l+i') L.n2wt a L...n n-1 -}-ioo n~l = n~~~~e--ialn>', Evaluating the other integral in the same way, and multiplying through by ._Ia, we obtain Ramanujan's result \ 1 a ~ ~-t(n) e-(a/nl',,b ~ (:~ c--i.l>.'nl' = __ !_ 'bP !1l-.:::-te2 ,tS n "~ n z,'b L... ~'(p) · (9.8.1) We have, of course, not proved that the series on the right is con-vergent in the ordinary sense. We have merely proved that it is conver-gent if the terms are bracketed in such a way that two terms for which ]y-y'l < exp(-A 1y/logy)+exp(-A 1 y'jlogy') are included in the same bracket. Of course the zeros are, on the average, much farther apart than this, and it is quite possible that the series may converge without any bracketing. But we are unable to prove this, even on the Riemann hypothesis. 9.9. We next prove a general formula concerning the zeros of an analytic function in a. rectangle.t Suppose that cfo{s) is meromorphic in and upon the boundary of a rectangle bounded by the lines t = 0, t = T, u = o:, u = {3 ({3 > o:), and regular and not zero on a= {3. The function logcfo{s) is regular in the neighbourhood of a= {3, and here, starting with any one value of the logarithm, we define F(s) = logcfo(s). For other points s of the rectangle, we define F(s) to b~ the value obtained from logcfo(,B+it) by continuous variation along t =constant from ,B+it to a+it, provided that the path does not cross a zero or pole of cfo(s); if it does, we put F(s) = }~~ 0 F(a+it+i£). Let v(a', T) denote the excess of the number of zeros over the number of poles in the part of the rectangle for which a > a', including zeros or poles on t = T, but not those on t = 0. ~ Then J F(s) ds = -27ri J v(a, T) da, (9.9.1) the integral on the left being taken round the rectangle in the positive direction, t Littlewood(4). 9.9 GENERAL DISTRIBUTION OF ZEROS We may suppose t = 0 and t = T to be free from zeros and poles of cfo(s); it is easily verified that our conventions then ensure the truth of the theorem in the general case. We have ~ ~ T J F(s)ds ~ J F(a)da- J F(a+iT)da+ J {F(~+it)-F(a+itl)idt. " " () (9.9.2) The last term is equal to T {J fJ o+iT f i dt f f(a+it) da ~ f da f f.0) ds o(a+it) ,S(s) ' () " " 0 and by the theorem of residues o-+iT fJ fJ+•T fJ+iT f !;~;a, ~ (f + J -,L ) !;~; M- z.i,(a, T) = F(a+iT)-F(a)-21Tiv(a, T). Substituting this in (9.9.2), we obtain (9.9.1). \Ve deduce T THEOREM 9.9. If S1~T) = J S(t) dt, then S1(T) = ~ J log"(a+iT}i da+O(I}. {9.9.3} ! Take cfo(s} = {(s), a:= f, in the above formula, and take the real part. \Vc obtain ~ T ~ J log:((a)l do- J a,g((~+it) dt- J logl((a+iT)I da+ l 0 ~ T + 1 .,g((}+it) dt ~ 0, (9.9.4) the term in v(a, T}, being purely imaginary, disappearing. Now make f3 ~ oo. We have log((s) ~ log(t+},+ .. .) ~ 0(2-•) as a~ oo, uniformly with respect tot. Hence arg {(s) = 0(2-""}, so that the second integral tends to 0 as ,B ~ oo. Also the first integral is a constant, and 1 log! ((a+iT)i ' T Since S1(T) = O(log T), the middle tennis O(T-IJog 'I'), and the last termis "" ~ o(j 1 j~'a•) ~ o( _ ["'~{+ J ~i) ~ 0(Io~1'). Hence the result follows. A similar result clearly holds for T f S~t) dt (0 <IX< l). It has recently been proved by A. Selberg (5) that S(t) ~ !l±{(logt)l(Ioglogt)-l} (9.9.7) with a similar result for S1(t); and that S,(t) ~ n.{(logt)l(Ioglogt)-'}. (9.9.8) 9.10. THEOREM 9.IO.t S(t) ha8 an infinity of changes of sign. Consider the interval (y,., Yn+l) in which N(t) = n, Let l(t) be the t Tirehmo.rsh{l7). 9.10 GENERAL DISTRIBUTION OF ZEROS 223 linear function oft such that l(y .. ) = S(y,J, l(Yn+l) = S(y,.+1-0). Then for y,. < l < Yn+l l(t)-S(t) ~ {S(y.H-0)-S(y.)} t-y. -{S(t)-S(y,)} Yn+l-Yn ~ -{L(y.H)-L(y,)} l=:r, + { L(t)-L(y,)J+O(~). Yn+l-Yn y,. using (9.3.2) and the fact that N(t) is constant in the interval. The first two terms on the right give -L'(<)(t-y.)+L'(o)(t-y.) (y, < 0 < t, y" < g < y,H) = L~(f1 )(7]-t)(t~y,.) (t1 between g and 7]) ~ 0(1/y.) since y,.+I~y, = 0(1). Hence y.,, ,.,, J S(t)dt~ J Z(t)dt+o(rn:~~r") Y• ~ !{y.H-y,){S(y.)+S(y.H-O)}+O(Y"H=J'-")· y, Suppose that S(t) ~ 0 for t > t0• Then gives Hence Hence N(y,);;, N(y,-0)+1 B(y11 ) ~ S(y,.-O)+I ~I. ,., f S(t) dt;;, }(y,H-y.)+O(b<"~~y") Y• r S(t) dt;;, !lr.v-r.,), '·· contrary to Theorem 9,9 {A). Similarly the hypothesis S(t) ,;;;; 0 for t > t0 can be shown to lead to a contradiction. It has been proved by A. Selberg (5) that S(t) changes sign at least T(log T)~e-.1'\og\og T times in the interval (0, T). 9.11. At the present time no improvement on the result S(T) ~ O(log T) is known. But it is possible to prove directly some of the results which would follow from such an improvement. We shall first provet THEOREM 9.11. The gaps between the ordinates of successive zeros of C(s) tend to 0. t Littlewood (3). 224 GENERAL DISTRIBUTION OF ZEROS CLap. IX This would follow at once from {9.3.2) if it were possible to prove that S(t) ~ o (log t). The argument given in§ 9.1 shows that the gaps are bounded. Here we have to apply a similar argument to the strip T-0 ~ t ~ T+O, where 8 is arbitrarily small, and it is clear that we cannot use four concentric circles. But the ideas of the theorems of Borel-CarathCodory and Hadamard are in no way essentially bound up with sets of concentric circles, and the difficulty can be surmounted by using suitable elongated curves instead. Let D, be the rectangle with centre 3+iT a.ndacornerat -3+i(T+S), the sides being parallel to the axes. We represent D4 conformally on the unit circle D~ in the z-plane, so that its centre 3+iT corresponds to z = 0. By this representation a set of concentric circles lzl = r inside D~ will correspond to a set of convex curves inside D4 , such that as r-+ 0 the curve shrinks upon the point 3+iT, while as r-+ 1 it tends to coincidence with D4• Let D~, D' 2, Ifa be circles (independent, of course, of T) for which the corresponding curves D1, D2, D3 in the s-pla.ne pass through the points 2+iT, ~I+iT, -2+iT respectively. The proof now proceeds as before. We consider the function /(')~log({)}, where s = s(z) is the analytic function corresponding to the conformal representation; and we apply the theorems of Borel-Caratheodory and Hadamard in the same way as before. 9.12. We shall now obtain a more precise result of the same kind.t THEOREM 9.12. For every large positive T, "s) has a zero P+iy satisfying A lr-TI <logloglogif'" This was first proved by Littlewood by a detailed study of the con-formal representation used in the previous proof. This involves rather complicated calculations with elliptic functions. We shall give here two proofs which avoid these calculations. In the first, we replace the rectangles by a so.ccession of circles. Let T be a. large positive number, and suppose that {(s) has no zero P+ir such that T-8 ~ y ~ T+8, where 8 < !. Then the function j(s) ~ log ((s). where the logarithm has its principal value for a > 2, is regular in the rectangle _ 2 ~a~ 3, T-8 ~ t ~ T+S. t Littlewood {3}; proof8 given here by TiWhma.rah {13), Kra!ru18Chke (1). 9.12 GENERAL DISTRIBUTION OF ZEROS 225 Let c~, ell, C", r~ be four concentric circles, with centre 2-fvS+iT, and radii lS, 1$. f8, and 8 respectively. Consider these sets of circles for v = 0, I, ... , n, where n = (12j8J+I, so that 2-lM ~-I, i.e. the centre of the last circle lies on, or to the left of, a = -1. Let mv, Mv, and M~ denote the maxima of ]f(s) 1 on ell, e", and C" respectively. Let A 1, A2, ••• denote absolute constants (it is convenient to preserve their identity throughout the proof). We have R{J(s)} < A1 log T on all the circles, and 1/(2+iT)] < A 2 • Hence the Borel-Caratheodory theorem for the circles c(l and ro gives 3+18 Mn <g_~8 (A 1 logT+A 2 ) = 7(A 1 logT+A2), and in particular 1/(2-)o+iT)i < 1(A,logT+A,). Hence, applying the Borel-Carath€odory theorem to C1 and r1, M 1 < 7{A 1 logT+i/(2-!S+iT)I} < (7+72)A 1 logT+72A 2• So generally M~ < (7-j- ... +7"+1)A1 IogT+7v+tA2, or, say, Mll < 1vA3 1ogT. (9.12.1) Now by Hadamard's three-circles theorem .Mv ~ 1n~M~, where a and b are positive constants such that a+b = 1; in fact a = log i/log 3, b = log 2/log 3. Also, since the circle 0~ 1 includes the circle cv, mv ~ M~_1 • Hence M~ ~ M~ 1 Mt (v = 1, 2, ... ,n). Thus lf/ 1 ~ MgMt, 11f 2 ~ M~M~ ~ Mg'MfM~, and so on, giving finally Hence, by (9.12.1), Mn ~ Mff'Mf-'l>Mr-•~.> ... M~. Now Mn ~ Mg"1a"-'b+2a"-'b+ ... +nb(Aalog T)«"-'b+a"-'b+ ... +b. an-Ib+2a"-2b+ ... +nb < n 2, an-1b+an-2b+ ... +b = b(J-an)/(1-a) = 1-an. Hence Mn ~ Mff'7n'(A3 log T)l-a" < A 4 7n'(log T)l-a", since Mn is bounded as T <--+ oo. But l'(s)l > tA• for a~ -1, t > t0, so that M" > A 5 logT. Hence 45 <A,. 7n'(log T)-"", loglogT < {~f(n 2 log7-log~)· logloglogT < nlog~+A 6 logn, 226 GENERAL DISTRIBUTION OF ZEROS Chap. IX so that s <!_! < __ A __ • n-1 logloglog T a.nd the result follows. 9.13. Second Proof. Consider the a.ngula.r region in the a-plane with vertex at a= -3+iT, bounded by straight lines making angles ± !IX(O < o: < tt) with the real axis. Let w = (s+3-iT)wfcx. Then the angular region is mapped on the ha.lf-pla.ne R(w) ~ 0. The point 8 = 2+iT corresponds to w=5'"'"'· Let Then the angular region corresponds to the unit circle in the z-pla.ne, and s = 2+iT corresponds to its centre z = 0. If s = a+iT corre-sponds to z = -r, then (a+3)•• = w = 5"/ 01 ~~;, i.e. Suppose that t(s) has no zeros in the angular region, so that log t(s) is regular in it. Let a= i+iT, -l+iT, -2+iT correspond to z = -r1, -r1 v -r8 respectively. Let M10 M:!, M3 be the :rna.xima. of llog C(a) I on the a-curves corresponding to lzl = r1 , r1, r3. Then Hadamard's throo-circles theorem gives lo M. ~ lografrslo M + logr2/r1Jo M.. g 2 """ logr3{r1 g 1 logr8{r1 g 8 It is easily verified that, on the curve corresponding to jzj = r1, u > J. For if w = f+i1J, then u = -3+(f2+1J2 )o:<l2"'cos(;arctan~)· which is a. minimum at 1J = 0, for given g, if 0 < re < -fn'; and the minimum is -3+f"'"', which, as a function of e. is a minimum when f is a minimum, i.e. whenz = -r1• It therefore follows tha.t logM 1 <A. Since R{log {(s)} <A log T in the angle, it follows from the Borel-Ca.ra.tbeodory theorem tha.t Ma < l~ra (AlogT+A) <All~gr~· 9.13 GENERAL DISTRIBUTION OF ZEROS 227 Hence logMs ~A+ logr2/~log(Alog T)· logr3/r1 l-r3 Now if r1, r2, and r8 are sufficiently near to I, i.e. if re is sufficiently small, log(l + r2-rt) logr2/r1 = r1 ,::: (r2-r1)! logr3/r1 log( 1+ ra r 1 r1) """""' ra-r1 ' l-r1 l-r2 and r2-r1 = I+r1 -l+r2 I+r2 < (/o)"'"'-(i)"l« ra-r1 1-r 1 _ 1-ra I+r3 (/u)"''"'-·W,.'"' I+r1 I+ra < 1-A(U"''"'· Hence Also 1/(l-r3) < A5 ... '"'· Hence logM 2 <A+{I-A(8)"'«}{loglogT+~log5+A}. Let a = 11/(clogloglog T). Then logM 2 < A+{I-A(loglog T)-<' 1<m!}{loglog T+clog5logloglog T+A} < loglog T -(loglog T)t if clog! < l and T is large enough. Hence M2 < logTe-(l<mlogT)I <£ logT (T > T0(..:)). In particular logl((-l+iT)) < ,]ogT, )((-!+iT))< T•. But 1((-l+iT)) ~ )x(-!+iT){(2-iT)i > KTi. We thus obtain a contradiction, and the result follows. 9.14. Another resultt in the same order of ideas is THEOREM 9.14. For any fixed h, however small, N(T+h)-N(T) > K!og T fo' K ~ K(h), T > T,. This result is not a consequence of Theorem 9.4 if h is less than a certain value. Consider the same angular region as before, with a. new a: such that t Not previoU8ly published. 228 GENERAL DISTRIBUTION OF ZEROS Chap. IX tan o: ~ f, and suppose now that "s) has zeros p1, p2, •• , Pn in the angular region. Let F(') ~ (•-p-;~•-p,). Let C be the circle with centre i+iT and radius 3. Then ls-pvl ~ I on C. Hence IF(s)l ~ l'(s)l < TA on C, and so also inside 0. Let f(s) = log F(s). Then f(s) is regular in the angle, and Rf(•) <A log T. Also j(2+iT) ~ log((2+iT)-J,log(2+iT-p,) ~ 0(1)+ f 0(1) ~ O(n). v~l Let M10 M2, and M3 now denote the maxima of 1/(s)l on the three s-curves. Then A M3 <-(log 7'+n). l-r3 Also M1 < An, as for /(2+iT). Hence loglf(-l+iT)I,;;; logM, < logr~(A+logn)+~r2frt 10g(~~±~~L~) logr3fr1 logr3/r1 l-r3 < A+logn+~-'-1 log-+log --log';, ( I (log T)) logr3fr1 l-r3 n < A+logn+{l-A(!)'i•)(~log5+log( 10~ T)) as before. But 1/(-l+iT)I ~ [log((-l+iT)-J,log(-l+iT-p,)l ;;, logl((-!+iT)I- I 0(1) v~l > A 1 log T-A2 n, say. If n > }(A 1fA 2) log T the theorem follows at once. Otherwise 1/(-l+iT)I > }A,logT, and we obtain toge 0 : r) < A+{l-A(~)"I"'}{~~og s+IogC 0 : ~}· A(l)'i•log('o~T) < A+{l-A(I)'i•);;log5, 9.14 GENERAL DISTRIBUTION OF ZEROS and hence loglog- <-log-+log-+A <-, ( log~ '" 9 I A n <X 4 e~"'~Iog T. This proves the theorem. 9.15. The function N(a, T). We define N(a, T) to be the number of zeros fJ+iy of the zeta-function such that fJ > u, 0 < t :::;;,; T. For each T, N(u, T) is a non-increasing function of u, and is 0 for u ~ l. On !he Riemann hypothesis, N(u, T) = 0 for u > f. Without any hypothesis, all that we can say so far is that for!<u<l. N(a, T) ,;;; N(T) <AT log T The object of the next few sections is to improve upon this inequality for values of u between l and I. We return to the formula (9.9.1). Let ~(s) = {(s), <X= u0, p = 2, and this time take the imaginary part. We have v(u, T) = N(u, T) (u T T 21T J N(u, T) du = J logl{(u0+it)1 dt- J logl{(2+it)j dt+ "• 0 0 + j arg,(u+iT) du+K(u0), where K(u0) is independent ofT. We deducef THEOR.EM9.I5. ljf:::;;;u0 :::;;; I,andT-+oo, ' T 2TT 1 N(u, T) du = J logl{(u0+it)1 dt+O(logT). We have T J logl((2+it)ldt~R iA,~n)n-~~-i ~ 0(1). o n~2 n -1- ogn Also, by§ 9.4, arg "u+iT) = O(log T) uniformly for u ~ f, if. T is not the ordinate of a zero. Hence the integral involving arg {(u+iT) is O(log T). The result follows if T is not the ordinate of a zero, and this restriction can then be removed from considerations of continuity. t Littlewood (4). GENERAL DISTRIBUTION OF ZEROS THEOREM 9.15(A).t For any fixed a greater than!, N(a, T) ~ O(T). For any non-negative continuousf(t) ' ' b~a j logf(t) dt <;; log(b~a j j(t) dt). Thus, for! <a< l, T T j logl{(a+it)l dt ~! pogl{(a+it)l'dt <;; !Tlog(~ !l{(a+it)l'dt) ~ O(T) by Theorem 7.2. Hence, by Theorem 9.15, J N(a,T)da ~ O(T) for a0 >f. Hence, if al = l+!(u0-!), Chap. IX N(a,, T) ,;;; _I s·· N(a, T) da,;;; ~I' N(a, T) da ~ O(T), ao-ul o, ao-"! a, the required result. From this theorem, and the fact that N(T) ,...,AT log T, it follows tha.t all but an infinitesimal pr<>portit:m of the zeros of {(s) lie in the strip l-3 < a < !+8, however small 8 may be. 9.16. We shall next prove a number of theorems in which the O(T) of Theorem 9.15(A) is replaced by 0(1'8), where 8 < Lt We do this by applying the abO\'e methods, not to '(s) itself, but to the function {(s)Mx(s) ~ {(s) 2: ~~~)· n I, Mx(s)-+ 1/{(s) as X-+ oo, so that t(s)Mx(s)--+ I. On the Riemann hypothesis this is also true for 'i < a ~ 1. Of course we cannot prove this without any hypothesis; but we can choose X so that the additional factor neutralizes to a certain extent the peculiarities of {(a), even for values of a less than I. Let fx(a) ~ {(s)M,(a)-1. t Bohr and Landau (4), Littlewood (4). ~ Bohr and Landau (5), Carlson (I), Landau (12). Titchma1'8h (5), Ingham (5). GENERAL DISTRIBUTION OF ZEROS We shall' first prove THEOREM 9.16. If for some X= X( a, T), pt-l(q):::;;; X< TA, T f 1/x(•)l' dt ~ O(T"•'log"T) jT 231 as T -+CC, uniformly for a ;;;;:= o:, where l(a) is a positive non-increasing function with a bounded derivative, and m ia a comtant ;;;;:= 0, then N(a, T) = O(Tl(qllog"'+IT) uniformly for a ;;;;:= o:+ 1/log T. We have fx(s) = {(s) L: f.!~~)-1 = L:an~-:l, n<X where a1{X) = 0, and a,(X) ~ "2: ~(d) ~ 0 (n <X), 'I• la,(X)I ~ lL ~(d)l <;; d(n) "" X 0 , lj" .(a)''< ( "'.;' d(n))' ~ O(X"-') <_I< 1, -' I ""-" L.., n2 2X 2 n)<X so that h(s) o:j=- 0. Applying (9.9.1) to h(s), and writing we obtain ' T 21T I v(u,fT,T)du= J {logjh(u0-f-it)j-logllt(2-f-·it)l}dt+ "• lT -1-j {argh(u·i-iT) -argh(u-1-~iT)} du. Now loglh(a)l <;; log{I+:fx(')l'},:; 1/x(s)l', so that, if a0 ;;:=a, T T I log !h(u0-f-it)l dt:::;;; I ifx(u0+it)j2 dt = O(TfC<>ollogmT). lT fT Next -logjh(2+it)l:::;;; -log{l-lfx(2-l-if)[ 2} ~ 2lfx(2-Ht): 2 < X- 1 232 so that GENERAL DISTRIBUTION OF ZEROS -Jloglh(2+it)l dt < £ ~ O(T"••). jT Chap. IX Also we can apply the lemma. of§ 9.4 to h(s), with a = o, {3 ;;::: f, m;;::: }, and M,.,~ = O(XAT4 ). We obtain for a ~ f. Hence ! a J N(a,fT, T) du;;:, (u1-a0)N(u1,}T, T) "• "• ifu0 < a1 ,.,;;; 2. Taking u1 = a0+I/logT, we have pll..ao) = '['l(a>)+O(), Hence N(a1 , fT, T) = O('['l(aVIogm+lT). Replacing T by fT, lT, ... and adding, the result follows. 9.17. The simplest application is THEOREM 9.17. Forany:fixedain f <a< 1, N(u, T) = O(T4tr(1-alH). We use Theorem 4.11 with x = T, and obtain fx(') ~ L ;;l. L ~(~)-I+O(T-'IMx(s)l) m<T n<X n ~ Lb"~~)+O(T-•X•-•), (9.I7.I) where, if X < T, bn(X) = 0 for n < X and for n > XT; and, as for a.,, )b11(X)j ~ d(n) = O(n•). Hence fi"L b.~~Td• ~iT 2: !b.~~) I'+ 2:2: t::!;. [(~)" dt ~ ~ ~ o( T .~ .L.) + o(.f.;<~T (mn)• .\ogmjn) = O(TXt-2aH)+O{(XT)2~ItuH} 9.17 GENERAL DISTRIBUTION OF ZEROS 233 by (7.2.1). These terms are of the same order (apart from E's) if X = p2a~l, and then J 12: b·~:r dt ~ O(T •• ,.,,.). jT The 0-term in (9.17.1) gives O(T1 - 217X 2- 217 ) = O(T1- 217X) = 0(1). The result therefore follows from Theorem 9.16 • 9.18. The main instrument used in obtaining still better results for N(u, T) is the convexity theorem for mean values of analytic functions proved in § 7.8. We require, however, some slight extensions of the theorem. If the right-hand sides of (7.8.1) and (7.8.2) are replaced by finite sums I C(T"+I), I C'(T'+I), then the right-hand side of (7 .8.3) is clearly to be replaced by K ~ ~ (CTayP--al/(fl-~l(C'Tb)<"-~)/<fJ-~l. In one of the applications a term pa}og'T occurs in the data instead of the above Ta. This produces the same change in the result. The only change in the proof is that, instead of the term f ( u)"""-' K o 8 e-2u du = (5a+2~-l' we obtain a term f (it+t~-1 log'ie-2u du J "(u)•+••-'( I I ) -••d K l ,I = 0 8 log's+4log33logu+ .. e u < sa+2~-l og s· THEOREM 9.18. If ~(!+it)= O{flogc't), where c' ~ J, then N(a, T) = O(T1<1+2cXt-alJog5T) uniformly for ! :::;;; a :::;;; I. IfO < 8 < 1, T T f If (I+S+it)i'dt ~ ' 'ax(m)ax(n) J ("')"dt x L..., L..., mt+8nl+B n 0 m;;>Xn;;>X 0 = T 2 :~l~l +2 2 2 a~;j::+\n) sin\~:r::/~jn) n;;>X Xo;;;m."\: Xo;;;m<n 234 GENERAL DISTRIBUTION OF ZEROS Chap. IX (putting x = Xyllf) Hence 2 ::~~ < Al~~f < X~3 .,x since Xall = e2Slogx > i(28logX)3. Also, since 1 < log.+A-1 < logA+A-i forA> 1, "" d(m)d(n) "" d(m)d(n) "" d(m)d(n) ft.m~ (mn)1+flognfm < ftm-f.. (mn)1+f + ftm4::. mfnH.t'(mn)flognjm T ( ~ d(n))' +"" d(m)d(n) J"1+f dx < .:f1 n 1+f ~f;: (mn)ilognfm .. ~ < ('(l+fl+ J•l+f L" d(m)d(n) dx 1 X 2-t{ m<n~ (mn)•lognfm < ,.11 +')+ J"(l+f)Aiog'x d :! o, ~ 1 ~X<~t' Hence f lfx(l+8+it)1 2 dt < A(f+ 1 )o-'. (9.18.1) For u = i we use the inequalities 1/xl' <;; 2(1(1'1Mxl'+1), T J IM lt+it)('dt« T L l''(n)+2 LL ll'(m)!'(n)( u x ....,. •<X n m<n<X(mn)llognjm <;; T" ~+2""-~1-.. .fxn f<nfg(mn)llognfm by (7.2.1). < A(T+X)IogX, t The first result follow~ eaeily from (7.16.3); for the second, see Ingham (1); the argument of§ 7.21, and the first result, give an W~:trn log :I'. 9.18 GENERAL DISTRIBUTION OF ZEROS 235 T Hence J l/x(i+it)l 2 dt < AT2c(T+X)log2c'(T+2)logX. (9.18.2) T The convexity theo:em therefore gives J 1/x(a+it)(' dt jT = o{ (f+ 1 )a-4r--lll(j+llJ{T2c(T+X)log2c'(T+2)log X}(1+3-oll(i+8> ~ o(~+X pfc(t-al (XTZc)((2a-I)Illf(f-tal(8"log3(T+2)logX)(l-t3-aJICi-t3>). 84 X2a-1 Taking 8 = 1jlog(T+X), we obtain O{(T+X)T''"-•'X'-~Iog'(T+X)). If X = T, the result follows from Theorem 9.16. For example, by Theorem 5.5 we may take c = !, c' = f. Hence N(a, T) ~ O(Tl<•-•>log'T). (9.18.3) This is an improvement on Theorem 9.17 if a > i-On the unproved LindelOf hypothesis that ~(l+it) = O{tf), Theorem 9.18 gives N(a, T) = O(T2<1-alH). 9.19. An improvement on Theorem 9.17 for all values of u in ! < u < 1 is effected by combining (9.18.3) with THEOREM 9.19(A). N(u, T) = O(Tf-oJogST). We have T T f lfxiHit)('dt <A f lm+it)I'(Jfxit+it)('dt+AT ' ' T T l <A (I I'IHit)('dt f iMxiHit)('dt l+AT. ' ' Now Mi:(s) = 2 ~· len I~ d(n). n<X' Hence T f IM (!+it)(' at<;; T L d'(n)+2 LL d(m)d(n) 0 x n<X' n m<n<X'(mn)ilognfm < ATlog'X+AX2Jog3X. T Hence f lfx(l+it)l 2 dt < ATl(T+X2)ilog2(T+2)log2X. (9.19.1) 236 GENERAL DISTRIBUTION OF ZEROS Chap. IX From (9.18.1), (9.19.1), and the convexity theorem, we obtain T J l/x(a+it)l 2 dt jT ((T ) )<•-l>l<1> = 0 :x+l 8-4 {Ti(T+X2)tlog2(T+2)log2X}M~-a>llt+~>. If X= T}, 8 = lflog(T+2), the result follows as before. This is an improvement on Theorem 9.17 if l < u < £. Various results of this type have been obtained, t the most successful+ being THEOREM 9.19(B). N(u, T) = 0(TJ(I-a)l{2-o:ol}og5T). This depends on a two-variable convexity theorem;§ if J(a,A) ~ ([IJ(a+i!)l'"dt)'. then J(a,pA+q~) ~ O{JP(o,A)J•(p,~)) (o< a< p), where P=~~· q=~=-:· We have [l!xl!+i!)ll dt <A J'lcil+i!)lliMxll+it)lldt+AT " " < A(J'Icl!+it)l' dt)1(/Wxll+it)1' dt); +AT < A{Tlog'IT+2IP{IT+X)logX)l+AT < A(T+X)log'(T+X). (9.19.2) In the two-variable convexity theorem, take o: = t, f1 = l-1-8, A= !, JL =!,and use (9.18.1) and (9.19.2). We obtain T f lfx(a+it)I 11K dt < A{(T+X)log2(T+X)}!1<t-~a-+jlll{ (f+ 1 )s-'t-{l/(t-}a+!lll, where K = pA+q,u lies between! and~. Taking X= T, S = I flog T, we obtain T [ Jfx(u+it)jl/K dt < AT3<1-a)/(2-allog'T. t Titchiil3I'!Ih (5), Ingham (5), {6). t Ingham (6). §Gabriel {1). 9.19 GENERAL DISTRIBUTION OF ZEROS 237 The result now follows from a. modified form of Theorem 9.16, since logll-.f11,;; log(l-1'1/xl') a,OJ,ol,O<y<;;T Hence an equivalent problem is that of the sum (9.20.1) I IP-!1· (9.20.2) O<yo(T There are some immediate results.t If we apply the above argument, but use Theorem 7.2(A) instead of Theorem 7.2, we obtain at once ' J N(a, T) da < ATlog(min(logT,log 00 ~!)) (9.20.3) for f ~ u 0 ~ 1; and in particular ' f N(a, T) du ~ O(TloglogT). (9.20.4) l These, however, are superseded by the following analysis, due to A. Selberg (2), the principal result of which is that f N(a, T) du ~ O(T). l We consider the integral Julm+it)~l!+il)l'dt, t Selberg {5). t Littlewood {4). (9.20.5) 238 GENERAL DISTRIBUTION OF ZEROS Chap. IX where 0 < U ~ T and t/1 is a function to be specified later. We use the formulae of §4.17. Since •'' ~ {x(!+it))-t ~ (~t·-•·'(t+oGJ)· we have Z(t) ~ z(t)+i(t)+O(d), (9.20.6) where z(t) = (!)1"e-l'" I n-l-il 21Te n.;;:c and x = (t/27T)f. LetT~ t.:::;;; T+U, T = (T/21T)!, -r' = {(T+U)j211}t. Let zl(t) = (~tu e-f:rri 6 n-~-it. Proceeding as in§ 7 .3, we have T+U f [z(t)-z,(t)l'dt~ o(u 2: ~)+O(TI!ogT) '1' 1'<n...;1" ~ o(uT';:-T)+O(Tt!ogT) ~ O(U'fT)+O(TI!og.T). (9.20.7) 9.21. LEMMA 9.21. Let m and n be positive integera, (m, n) = 1, M = max(m, n). Then The terms with miL = nv contribute u2:2: I _ U 2: I _ U 2: 1 m,..~1w ~-......... rm,:;1'(rn.rm)l- (mn)!,. .... ,Mr The remaining terms are 0(~ .. ~ (~·)l[lo:(n•/m~)l) ~ o(~ .. ~ (m~n·)l[l~(n•/m~)l) ~ o(M <<~.JM• (<.l)ljl!g.I/J ~ O(MO,.log(MT)}, and the result follows. 9.22 GENERAL DISTRIBUTION hF ZEROS 9.22. LEMMA 9.22. Defining m, n, M M before, and supposing Tl;;T, T+U 239 f zl(t)(~r dt ~ ~ L ~+O(MTI)+O(U'JT)+O(Tk) T </mo<r,;tfn (9.22.1) if n ~ m. If m < n, the first term on the right-hand side is to be omitted. The left-hand side is The integral is of the form considered in§ 4.6, with F(t) = tlog~, 27Tp..vm c~-n-. Hence by (4.6.5), with~= (T+U)-1, As= (T+U)-2, it is equal to (2nc)!el•'~ '" + 0( Tl) + o{min(--1-, Tt)} llogc/TI + o{ minCogi(T~ U)/cl' T!) }· (9.22.2) with the leading term present only when T ~ c ~ T + U. We therefore obtain a main term (9.22.3) where p.. and v also satisfy T2njm ~ p..v ~ T'2njm. The double sum is clearly zero unless n .::.:; m, as we now suppose. The v-summation runs over the range v1 ~ v.::.:; v1 l' where v1 = <2njmp and v2 = min(<'2njmp, <),and p runs over •nfm ~ p ~ •· The inner sum is therefore v2 - v1 + O(n) if nip., and O(n) otherwise. The error term O(n) contributes O{(mn)h} = O(MTi) in (9.22.1). On writing p = nrwe are left with 2•("')t L (v2-vl). n •fm.;;r.;;r;fn Let v3 = •'2jmr. Then v2 = v3 unless r < <' 2/m-c. Hence the error on 240 GENERAL DISTRIBUTION OF ZEROS Chap. IX replacing v2 by v3 is o{("')' I (~~,)} ~ o{("')!(~~~+1)("'~,)} n •fm.;;r<r'"/mt mr n mr m t { - (''2~'2)2} {(m)!('''~'2)} = 0 (mn) ! --+0 ---' n ' ~ O(U'T-')+O(MlUT-l). Finally there remains ~~ I 1 (mn) tfm<;;r.;;tfnr Now consider the 0-terms arising from (9.22.2). The tenn O(Ti) gives Next ""'·( 1 rl) L.., L, ~ mm jlog(2trp.vmfnT) I' ~ .... ~.;; .. ~o(r·" ',min(-1 -,rl))· L,r~" jlog(rmjnT2)j "'' Suppose, for example, that n < m, Then the terms with r· < !nT2jm or r > 2nT2jm are o(r• L ~) ~ O(T'r) ~ O(Tl+<). r.;;T• In the other terms, let r = [nT2jm]-r'. We obtain o(r• ~ (nr';m)lj,' 01/~m'Jm)) (I OJ < 1) ~ o( r·('if)ltog r) ~ O(Tl+•), omitting the term.s r' = -1, 0, 1; and these are O(TfH), A similar argument applies in the other cases. 9.23 GENERAL DISTRIBUTION OF ZEROS 241 9.23. LEMMA9.23. Let (m,n)=l with m,n-:o:;;X-:o:;; r!.I[TH--:<.::; u~ T, then T+U f Z 2(t)(~rdt= (m~)!{log 2!n +2y)+ O(UiT-HogT). Let Z(t) = z 1 (t) +z1 (t) +e(t). Then ,.u [ {z 1 (t)+z 1 (t))'(~)"dt T+U T+U T+U ~I Z(t)'(~)"dt+O( I IZ(t)e(t)idt)+o( I Je(t)i'dt} We have ' ' ' f~e(t)1 2 dt~ O(U'/T)+O(Tllog T) ~ O(ll'/T) ' by (9.20.7), and "" I IZ(t)l'dt ~ O(Ulog T)+O(T!+•) ~ O(Ulog T), ' by Theorem 7.4. Hence ]"IZ(t)e(t)jdt ~ O{(U'/T)l(Ulog T)l) ' by Cauchy's inequality. It follows that "" ~ [ (z,(t)2 +z,(t) 2 +2z 1 (t)z,(t)}(~)"dt+O(UIT-!loglT). By Lemmas 9.21 and 9.22 the main integral on the right is ~ m~) ( I 'c+ I ")+O{T!X'tog(XT))+O(XTI)+ r!!<T/nr r!!<T/mr +0(U2JT)+O(Tfo) 242 GENERAL DISTRIBUTION OF ZEROS Chap. IX whether n ~ m or not. The result then follows, since I -+ I -~log-+2r+o- . 1 1 ,, (X) r,;;~fnr r,;.•fmr mn 't and since the error terms O{T!X2 log(XT) }, O(XT!), 0(U2jT), O(Tfo) and O(UXT-!) are all O(UiT-Hog T). 9.24. THEOREM 9.24. f N(a, T) da ~ O(T). • Consider the integral where Clearly T+U T+U I~ J l((!+it)~(!+it)['dt ~ J Z2(t)[~(l+it)['dt, ifJ(s) =,6-8,r1-s fxl'(f")p.(p}N(f") p.(,),.<.J,,,, ,l''(p)/</>(p) ,1"'(p)/(p) "''') ,ix"'(p)N(p) [S,[<;;.p:,) for all values of r. Now (9.24.1) where m = qj(q, r), n = rj(q, t). Using Lemma 9.23, the main term contributes to this For a fixed q <X, Now 9.24 GENERAL DISTRIBUTION OF ZEROS Hence the. second factor on the right is 2:2: p.(f")p.(p) 2: </>(•)~ 2:<~>(•) 2:2: p.(p,)p.(p)' r<X,pr<X t/J(pr} ~lq,vlr vlll f'<X,('<X tfo{pr) ,,, Put pr = l. Then pvjpr, pv[l, i.e. p[(l/v). Hence we get 2: </>(•) 2: ~ 2: p.(p). viii l~lf pl(lfv) The p·sum is 0 unless l = v, when it is 1. Hence we get Hence ( "p.'(p))-' ( " p.'(p))-' ~ I s,o,(p) ~ L. :;:--' l s, ~ L. :;:--' l 11<-.Xr<X p<X 'f' p p<X 't' p and 2 2 OqO,(q,r)logq = ( 2 j 2 (p))-1 01 1og 1 = 0. q<Xr<X pa(n) be defined by ~ .(•)- ~-a-!) 6 n• -'(a) ' so that "'(n) ~ •""" p.(m) ~ .... n (1-1)· a ~ ml+a p[n pl+a Let if(n) be defined by ~~(·)~ ('(•-_l_l, Then and hence Hence and {= 1 n' '(s) -('(s-1) ~ ((s)%, ~~~). nlogn ~I ~(d). .,. q~ ,.fx.OqO,(q,r)log(q,r) =d"{.x.f(d)ql~'1tx0q0r =db if.(d)clq.~X Oqt 243 Now Hence Since we have GENERAL DISTRIBUTION OF ZEROS ~(n) ~ [~ <1.1•1] ~ <ll•)(logn+ L lo~~)· Ba amO pin P 1 " ~'(p))-' z '5'8,S,(p)log(q,,),;; 2logX L.., -;,:---() . q<Xr~ p<X 't' p •~ ~~~2. ~ IJ (I+~;;~.)~ IJ (I+(p-ll)po) ~ ((,+!) IJ (I-PL.)(l+(p__Il)p')' L p.-2(p) ,.AlogX. p<X c/J(p) The contribution of aJl the above terms to I is therefore o (u log T\ + O(U) ~ O(U) log X} on taking, say, X = Trh-. The 0-term in Lemma 9.23 gives O(UlT-IIog T) L L ~ q<:X ,(q)4J(r) ~ O(UlT-IJog T)O(X) ~ O(UIT-&!og T). Taking say U = T-H, this is O(U). Hence I= O(U). By an argument similar to that of§ 9.16, it follows that . f (N(a, T+U)-N(a, T)}da ~ O(U). ! Chap. IX 9.24 GENERAL DISTRIBUTION OF ZEROS 24i> Replacing T by T+U T+2U, ... and adding, O(TJU) terms, we obtain f {N(a, 2T)-N(a, T)) da ~ O(T). I Replacing T by fT, fT, ... and adding, the theorem follows. It also follows that, if } < a :( I, N(a, T) ~ ;;-~ J N(a', T) do' ~<rt-} ,;; 2-, j'· N(a, T) da ~ a('!)· a-"2" a--! I (9.24.2) Lastly, if f(t) is positive and increase.g to infinity with t, all but an infinitesimal proportion of the zeros of '(s) in the upper half-plane lie in the region (o-!1 <~~)( The curved boundary of the region a ~ i+ <lit), Tl < t < T logt <i(Tl) lies to the right of a= a1 =!+log T' and ( T ) (TlogT) N(a,,T)~O a,-! ~0 <i(Tl) ~o(TlogT). Hence the number of zeros odside the region specified is o(Tlog T), and the result follows. NOTES FOR CHAPTER 9 9.25. The mean value of S(t) has been investigated by Selberg (5). One ha. T IIS(t)l"dt- k!!~ 1 1 ,. T(loglogT)' (9.25.1) for every positive integer k. Selberg's earlier conditional treatment (4) is discussed in§§ 14.20---24, the key feature used in (5) to deal with zeros off the critical line being the estimate given in Theorem 9.19(C). Selberg (5) also gave an unconditional proof of Theorem 14.19, which had pre-viously been established ou the Riemann hypothesis by Littlewood. 246 GENERAL DISTRIBUTION OF ZEROS Chap. IX These results have been investigated further by Fujii [1 ], and Ghosh , , who give results which are uniform ink. It follows in particular from Fujii [1} that ' fiS(t+h)-S(t)l2dt = n 2 Tlog(3+hlog T) + 0[T{log(3+hlog T)}l] " (9.25.2) and ' IIS(t+h)-S(t)l"d<« T{Ak'log(3+hlogT))' (9.25.3) uniformly for 0 .::;; h ::;; ! T. One may readily deduce that NJ(T) ~ N(T)e-AJ 1, where N/T) denotes the number of zeros {3 + iy of multiplicity exactly j, in the range 0 < y .::;; T. Moreover one finds that # {n: 0 < }'11 .::;; T, Yn+ 1 -y,. >-: A/logT} 4: N(T)exp { -AA.i(IogA.)-!}, uniformly for A:;::: 2, whence, in particular, (9.25.4) for any fixed k;;::. 0. Fujii [2) also states that there exist constants).> 1 and J.l < 1 such that and Yn+t-Yn >-A 2njlogy,.""" Yn±t-Yn:;,;;p 2:njlogy,. (9.25.5) . (9.25.6) each hold for a positive proportion of n (i.e. the number of n for which 0 < Yn ~Tis at least AN(T) if T ~ T0). Note that 2x/log Yn is the average spacing between zeros. The possibility of results such as (9.25.5) and (9.25.6) was first observed by Selberg . 9.26. Since the deduction of the results (9.25.5} and (9.25.6) is not obvious, we give a sketch. If M is a sufficiently large integer constant, 9.26 GENERAL DISTRIBUTION OF ZEROS then (9.25.2) and (9.25.3) yield " I IS(t+h)-S(!)I'dtl> T and " I IS(t+h)-S(!)I'dt .. T uniformly for By HOlder's inequality we have 2T 2T I IS(t+h)-S(!)I'd<" ([ IS(I+h)-S(t)ldty X (JIS(t+h)-S(t)l'd T. We now observe that hlogT ( 1 ) S(t+h)-S(t)~N(t+h)-N(t)-~+0 logT, for T::;.;; t ::;.;; 2T, whence " I IN(t+h)-N(t)- hi;:Tidtl> T. We proceed to write h = 2nMlflog T and b(t,l) = N(t+ 1::~ )-N(t)-l, 247 248 so that Thu• and hence GENERAL DISTRIBUTION OF ZEROS hlogT M-l ( 2xmA ) N(t+h)-N(t)---~ L b t+-1 T'l. 2n m=O og 2T+21lmA/logT T<t =~: f lt5(t,A)Idt T+2wml/1ogT 2T ~ M I lb(t, l)ldt + 0(1), 2T I lb(t,l)ldt,. T uniformly for 1 ~ A :E;; 2, since M is constant. Chap. (9.26.1) Now, iflisthe subset of [T,2T] on whichN(t+ 1!:~) = N(t), { b(t,l)+2l (tel), lb(t,l)l<; b(t,l)+2l-2 (te[T,2T]-[J, so that (9.26.1) yields 2T T 1 is chosen sufficiently close to 1. Thus, if { 2nl } 8= n:T~y .. ~2T,}',.+ 1 -y01 "li'logT, then T<Om(l)<O L<> •• , -y.,)+0(1), ..• ,,. GENERAL DISTRIBUTION OF ZEROS so that T 4: 8 IogT' by (9.25.4) with k = 2. It follows that #S,.N(T), proving that (9.25.5) holds for a positive proportion of n. 249 (9.26.2) Now suppose that p is a constant in the range 0 < p < 1, and put U = {n: T ~ Y, ~ 2T}, and V= {ne U: i'n+l -y,.,;: l::Jtr}• whence #U = {;togT+ O(T). Then " L <> •• ,->..)+0(1) nEU-V 2n:p 2n:A ;.logT(#U-#V-#S)+logT 8+0(1) 2n" (T ) 2n(l-") ( T) =logT 2,;"logT-#V + logT #S+O logT . If the implied constant in (9.26.2) is q, it follows that # V )> N(T), on taking Jl = 1- v, with 0 < v < q(..l-1)/(1- 'f). Thus (9.25.6) also holds for a positive proportion of n. 9.27. Ghosh was able to sharpen the result of Selberg mentioned at the end of §9.10, to show that S(t) has at least TOogT)exp( AloglogT ) (logloglog T)! <~ sign changes in the range 0 ~ t ~ T, for any positive 6, and A = A(O), T ~ T(O). He also proved (Ghosh ) that the asymptotic formula (9.25.1) holds for any positive real k, with the constant on the right hand 250 GENERAL DISTRIBUTION OF ZEROS Chap. IX side replaced by r(2k + 1)/r(k + 1)(2:n:)2". Moreover he showed (Ghosh ) that iS(t)l _ 1 J (loglog I) - f(), say, has a limiting distribution in the sense that, for any u > 0, the measure of the set of t e {0, T] for which f(t) ~ u, is asymptotically TP(u). (A minor error in Ghosh's statement of the result has been corrected here.) 9.28. A great deal of work has been done on the 'zero-density estimates' of §§9.15-19, using an idea which originates with Hal8.sz . However it is not possible to combine this with the method of §9.16, based on Littlewood's formula (9.9.1). Instead one argues as follows (Montgomery [1; Chapter 12 ]). Let M Js)((s) ~ ~a.n-•, so that a,.= Ofor2 ~ n ~X. If((p) = 0, wherep = fJ+iyandfJ > t. then we have e-1/Y+ L a,.n-Pe-n!Y= L ann-Pe-nfY n>X n-1 2+ioo ~ -'-c f M x<s+ p) (s + p)r(s) Y• ds, 2n< 2-i~ by the lemma of §7.9. On moving the line of integration to R(s) = !-fJ this yields Mx<t)r(l-p)YI-P+ +-'-c f MJ!+il)((!+iQr(t-P+i(t-y))YI-•+<<•-'>dt, 2n< since the pole of r(s) at s = 0 is cancelled by the zero of {(s + p). If we now aBSume that logZT ~ y ~ T, and that log T <Clog X, logY <Clog T, 9.28 GENERAL DISTRIBUTION OF ZEROS 251 then e-l~Y~ 1 and M x(l)r(l-p)Y' -P = o(l), whence either I L a.n-'e-•1'1? I "" In the latter case one has for some tP in the range ltP -yl,;:; log2 T. The problem therefore reduces to that of counting discrete points at which one of the Dirichlet series :Ea,. n- se -nJ Y, M /B), and ((s) is large. In practice it is more convenient to take finite Dirichlet polynomials approximating to these. The methods given in §§9.17-19correspond to the use of a mean-value bound. Thus Montgomery [1; Chapter 7] showed that rtlln t I a.,n-s, 12 4 (T+N)(logN)2 n~lla,.l2n-2o­ for any points s, satisfying (9.28.1) (9.28.2) and any complex a,.. Theorems 9.17, 9.18, 9.19(A), and 9.19(B) may all be recovered from this (except possibly for worse powers of log T). However one may also use Hal8.sz's lemma. One simple form of this (Montgomery [1; Theorem 8.2]) gives r~~~ .. ~~ a,.n-··r 4(N+RTi)(1ogT) .. tlla,.l2n 2" (9.28.3) for any points s, satisfying (9.28.2). Under suitable circumstances this implies a sharper bound for R than does (9.28.1). Under the LindelOf hypothesis one may replace the term RT! in (9.28.3) by RT• N i, which is superior, since one invariably takes N.::;; T in applying the Halitsz lemma. (If N ;;?:- T it would be better to use (9.28.1).) Moreover Montgomery [1; Chapter 9] makes the conjecture (the Large Values 252 Conjecture): GENERAL DISTRIBUTION OF ZEROS rtlln~l a,.n-•·f ~(N+RT') n~lla.,l2n-2« for points sr satisfying (9.28.2). Using the HalBsz lemma with LindelOf hypothesis one obtains N(u,T)<t.T", i+e~a~l. (Halasz and Turim {1 ], Montgomery [1; Theorem 12.3 ]). If the Large Values Conjecture is true then the LindelOfhypothesis gives the wider range ! + E ~ u ~ 1 for (9.28.4) 9.29. The picture for unconditional estimates is more complex. At present it seems that the Haltisz method is only useful for u ~ i-· Thus Ingham's result, Theorem 9.19(B), is still the best known for ! < a ~ l· ' Using (9.28.3), Montgomery [1; Theorem 12.1] showed that N(u,T)4T 2n-<rJI.,.(logT) 14 (!~u~l), which is superior to Theorem 9.19(B). This was improved by Huxley [I]' to give N(a, T) 4i T31<3a-n(tog T)44 (!,;;: u ~ 1). (9.29.1)' Huxley used the Halilsz lemma in the form R<{Nv-2 n~l \a.,l2n-:k+TNV-6( .. ~lla.,l2n-2a Y}(logT)2, for points sr satisfying (9.28.2) and the condition I I a.n-•r\;;. V. "~ 1 In conjunction with Theorem 9.19(8), Huxley's result yields N(u, T) <t Tl215(t-aJ(logT)44 (f,;;: u,;;: 1), (c. f. (9.18.3)). A considerable number of other estimates have been given, for which the interested reader is referred to lvic [3; Chapter 11 ]. We mention only a few of the most significant. Ivic showed that Mu T <{T(3-3a)/(7a-4)H (f,;;:u,;;:fl) ( ' ) T<9-9a)/{8a-2lH (fl,;;: (I,;;: 1), which supersede Huxley's result (9.29.1) throughout the range ! < q < 1.Jutila gaveamorepowerful,butmorecomplicated,result, '·" GENERAL DISTRIBUTION OF ZEROS 253 which has a similar effect. His bounds also imply the 'Density hypothesis' N(CI,T)-4 T 2 - 2a+•, for ft.:;;;q.:;-;;1. Heath-Brown im-proved this by giving N(C1,T)<T<9-9a>l<7a-t)+• (ft.:;;;q.:;-;;1). When C1 is very close to 1 one can use the Vinogradov-Korobov exponential sum estimates, as described in Chapter 6. These lead to N(CI, T) ..:g TA(l-a)I(Jog T)A' 0 for suitable numerical constants A and A', (see Montgomery [1; Corollary 12.5 ], who gives A = 1334, after correction of a numerical error). Selberg's estimate given in Theorem 9.19(C) has been improved by Jutila (2] to give N(a, T) 0. 9.30. Of course Theorem 19.24 is an immediate consequence of Theorem 19.9(C), but the proof is a little easier. The coefficients (j used in § 9.24 are essentially r Jl(r)r- 1 logXjr log X' and indeed a more careful analysis yields ' f l((f+it)\21 L Jl(r)logXjr r-!-it\2 dt,..., T(1+ logT)· 0 r.;;,x logX logX Here one can take X.:;;; r!-£ using fairly standard techniques, or X.:;;; Tl-r-• by employing estimates for Kloosterman sums (see Balasubramanian, Conrey and Heath-Brown [1 ]). The latter result Yields (9.24.1) with the implied constant 01}845. X THE ZEROS ON THE CRITICAL LINE 10.1. General discussion. The memoir in which Riemann first con~ sidered the zeta-function has become fa.mous for the number of ideas it contains which have since proved fruitful, and it is by no means certain that these are even now exhausted. The analysis which precedes his observations on the zeros is particularly interesting. He obtains, as in§ 2.6, the formula , where !'( .. )w-!•,(s) ~ -1-+ s~(x)(xl•-•+x-1-l•) dx, s(s-1) 1 Multiplying by !s(s-1), and putting s = !+it, we obtain E(t) = }-(t2+l) l ,P(x)x-l-cos(!tlogx) dx. Integrating by parts, and using the relation 41'(1)+~(1) ~ -·· which follows at once from (2.6.3), we obtain 8(t) ~ 4 [ ~{xi~'(x)}x-lcos(!tlogx) dx, Riemann then observes: (10.1.1) (10.1.2) 'Diese Function ist f"Ur aile endlichen Werthe von t endlich, und laast sich nach Poten:z:en von U in eine sehr schnell convergirende Reihe entwickeln. Da fUreinen Werth vons, dessen reeller Bestandtheil grOSSer aJs 1 ist, log ~(s) = - I log( 1-p-) endlich bleibt, und von den Logarithmen der i.ibrigen Fa.cWren von E(t) dasselbe · gilt, so kann die Function E(t) nur verschwinden, wenn der imaginii.re Theil von t zwischen li und - ii liegt. Die Anzahl der Wurzeln von E(t) = 0, deren reeller Theil zwischen 0 und T liegt, ist etwa = f;logf;-f;; denn dass Integral J dlogE(t) positive um den ln~grifi der Werthe von_' erstreckt, deren imaginii.re Theil zwischen ii und - il, Wld deren reeller Theil zwischen 0 und T liegt, ist (bis auf einen Bruchtheil von dar Ordnung der GrOsse I/T) gleich {Tlog(Tj2?r)-T}i; dieses Integral aber i.st _g~eich _der ~~ der in diesem Gebiet liegenden Wurzeln von E(t) = 0, mult1phcirt m1t 2m. Man findet nun in der That etwa so vie! reelle Wurzeln innerhalb dieser Grenzen, und es ist sehr wahrscheinlich, dass aile Wurzeln reelle sind.' 10.1 ZEROS ON THE CRITICAL LINE 255 This statement, that all the zeros of E(t) are real, is the famous 'Riemann hypothesis', which re:ma.ins unproved to this day. The memoir goes on: 'Hiervon wii.re &llerdings ein strenger Bewei..s zu wiinschen; ich babe indess die Aufsuchung desselben na.ch einigen fliiehtigen vergeblichen Versuchen vorUiufig bei Seite gelassen, da. er fur den Iilichsten Zweck meiner Untersuchung [i.e. the explicit formula for 1r(.x)] entbehrlich schien.' In the approximate formula for N(T), Riemann's l(T may be a mistake for log T; for, since N(T) has an infinity of discontinuities at least equal to l, the remainder cannot tend to zero. With this correction, Riemann's first statement is Theorem 9.4, which was proved by von Mangoldt many years later. Riemann's second statement, on the real zeros ofE{t), is more obscure, and his exact meaning cannot now be known. It is, however, possible that anyone encountering the subject for the first time might argue as follows. We can write {10.1.2) in the form .E(t) = 2[ {u)cos ut du, (10.1.3) where (10.1.4) This series converges very rapidly, and one might suppose that an approximation to the truth could be obtained by replacing it by its first term; or perhaps better by {u) = 2n2cosh~ue-2"ooab2u, since this, like (u), is an even function of u, which is asymptotically equivalent to (u). We should thus replace .E{t) by .E(t) = 4n2 [ cosh~ue- 2""0"h 2"cosutdu. The asymptotic behaviour of E(t) can be found by the method of ateepest descents. To avoid the calculation we shall quote known Bessel-function formulae. We havef Kja) = [e-acoah"coshzudu, and hence .E{t) = n2{K!+-!-u{2n")+K:-!-u{2n)}. For fixed z, as v -+ oo I,(•) ~ (}•)"/r{v+l), t Watson, Theory of Buul Funcli<>1W, 6.22 (5). 256 ZEROS ON THE CRITICAL LINE Hence LL}it(21T) .-.--n--l-l-~ 1 -,---- ,...., __!____ d"" (-'-) 1 (--'--)~i 1 e-iirr, "' r{-1-~tt) 1T.J2 211 211e .W.+}it l-t-+}u(21T) ,.._,!'(~~ .. +!it)= O(ei"1 t-¥), Kf+!~1 (27T) = }ncosec?t(t+!it){L-:--!u(21T)-I£+!u(21T)} --~e-tm(~t(~tue~i ... Hence .E(t) ,.., 1!'~2-ftie-l""cos(!tlog!+i7T). 21re The right-hand side has zeros a.t ltlog~+l7T = (n+f)1T, and the number of these in the interval (0, T) is flogf-f+O(l). The similarity to the formula for N(T) is indeed striking. However, if we try to work on this suggestion, difficulties at appear. We can write E(t)-E(t) ~ _[ {(u)-(u)}e'"' du. To show that this is small compared with :E:(t) we should want to the line of integration into the upper h~If-pla.ne, at least as l(u) = !?T; and this is just where the series for $(u) ceases to Actually IE(!) I> Arte+"l((!+i!)l. and IC(!+it)l is unbounded, so that the suggestion that E(t) is approximation to E(t) is false, at any rate if it is taken in the obvious sense. 10.2. Although every attempt to prove the Riemann hypothesis, all the complex zeros of '(s) lie on a = f, has failed, it is '(s) has an infinity of zeros on a = }. This was first proved in 1914. We shall give here a number of different proofs of this First method. t We have E(t) ~ -W'+ll~+l•r(!Hi!l((Hi!). where E(t) is an even integral function of t, and is real for real t. A zero t HIU'dy(I). 10.2 ZEROS ON THE CRITICAL LINE 257 of '(s) 011 a = l therefore corresponds to a real zero of E(t), and it is a. question of proving that E(t) has an infinity of real zeros. Putting x = -io: in (2.16.2), we have ~I"' E(t) cosh ext dt = e-li"'-2eli"'f(e2i"') 11'0 t2+i: = 2coslo:-2eii"'{!+rf(e2i"')}. (10.2.1) Since {(!+it) = O(t""), E(t) = O(tAe-i:rrt), and the above integral may be differentiated with respect to ex any number of times provided that o: < f7T. Thus ~I"' E(t) t2"coshcxtdt = (- 1) .. cos-fo: -2(d) 2"eli"'{'+·'·(e"")) 11' 0 t2+! 2211-1 do: y 'f' • We next prove that the last term tends to 0 as ex-+ f11', for every fixed n, The equation (2.6.3) gives at once the functional equation x-l-2xkf(x) = xl-2x-t,PG)· ~(x) ~ x-!~(~)Hx-1-t. Hence f(i+S) = ~e-n''l!'(i+S) = ~ (-l)"e-n•,.& ~2,P(41l)-,P(8) ~ ~&"'(~)-~,(~)-~· It is easily seen from this that l+f(x) and all its derivatives tend to zero as x .--;.- i along any route in an angle ]arg(x-i)l < tn'· We have thus proved that lim I"' E(t) t2"cosho:tdt = (-l)"trcos-1-n'. ·-l·' t'+t 2" (10.2.2) Suppose now that S(t) were ultimately of one sign, say, for example, positive for t ~ T. Then say. Hence lim I"' :(t~t2"coshcxtdt='L, "',.l,.T t +.j: T" I ~(t~ t 2"coshcxtdt :s;;; L 7' t+-\ 258 ZEROS ON THE CRITICAL LINE for all a: < }1r and T' > T. Hence, making a:-+ .p, Hence the integral T' f E(t) gzncoshfn""tdt ~ L. T <'+t f "" E(t) t2"cosh!-ntdt 0 tll+l Chap. X is convergent. The integral on the left of (10.2.2) is therefore uniformly convergent with respect to or: for 0 ~a::;;; p, and it follows that f "" 8(t) 2 ,. h, d _ (-l)"'n'cos}77 0 tll+l t cos :J1Tt t -22" for every n. This, however, is impossible; for, taking n odd, the right-hand side is negative, and hence • T f S(t) tll"coshf?Ttdt <- J E(t) 1 t2 ncoshf1Ttdt T tll+l & t2+4 ;;:<: m for 2T::;;; t::;;; 2T+ l. "" 2T+l f E(t) tZ"cosh 1-t dt A J mt2" dt >- m(2T)2 ... T t2+f 4" 9' liT -:?' Hence m22"<K, which is false for sufficiently large n. This proves the theorem. 10.3. A variant of the above proof depends on the following ofFejer:t Let n be any positive integer. TMn the number of changes in sign the interval (O,a) of a continuouafunctianf(x) is not kaa than the of changes in aign of the aequence /(0). 1 f(t) dt. .... j f(t)t· dt. We deduce this from the following theorem of Fekete:t t Fejer(l). t Fekete(!). }0.3 ZEROS ON THE CRITICAL LINE 259 The member of changes in sign in the interval (O,a) of a continUOU8 functicmf(x) ia not le.98 than tM number of changea in sign of the aequence where f(a), /,(a). ... , /,(a), (10,3.2) (v= 1,2, ... ,n), f,(x) ~ f(x). To prove Fekete's theorem, suppose first that n = 1. Consider the curve y = f 1(x). Now / 1(0) = 0, and, if f(a) and f 1(a) have opposite signs, y is positive decreasing or negative increasing at x = a. Hence J(x) has at least one zero. Now assume the theorem for n-1. Suppose that there are k changes of sign in the sequence / 1(x), ... , /,.(x). Thenf 1(x) has at least k changes of sign. We have then to prove tha.t (i) ~/(a) and/ 1(a) ha.ve the same sign, j(x) has at least lc changes of stgn, (ii) ifj(a) and/ 1(a) have opposite signs,f(x) baa at least k+ 1 changes of sign. Each of these cases is easily verified by considering the curve y =J 1(x). This proves Fekete's theorem. To deduce Fej6r's theorem, we have /,(x) ~ (•~I)'! (x-t)'-'f(t) dt, and hence a a /,(a)~ (•~I)! J (a-t)'-'f(t)dt ~ (•~I)! J f(a-t)t'-' dt. ' ' We may therefore replace the sequence (10.3.2) by the sequence /(a), J f(a-t) dt, ... , J j(a-t)t•-• dt. (10,3.3) ' ' Since the number of changes of sign ofj(t) is the same as the number of changes of sign off( a-t), we can replaoef(t) by f(a-t). This proves Fej6r's theorem. To prove that there are an infinity of zeros of{(a) on the critical line, we prove as before that Renee lim J"" E(t) t2 "coshr::ttdt= (-l)"1TCOBi1T <><-+:{,. 0 t2+! 22n • f" E(t) t2"coshr::ttdt 0 t2+t 260 ZEROS ON THE CRITICAL LINE h&s the same sign as (-1)11 for n = 0, 1, ... , N, if a= a(N) is large enough and "' = "'(N) is near enough to !'If. Hence E(t) has at least N changes of sign in (O,a), and the :result follows.t 10.4. Another method! is based on Riemann's formula. (10.1.2). Putting x = e11" in (10.1.2), we have E(t) = 4 J«> ~{e3"if/(eh)}e-l"cosut du du 0 = 21 fll(u)cosut du, sa.y. Then, by Fourier's integral theorem, (u) ~ ~ J E(t)co.ut dt, a.nd hence also ''"'(u) ~ (-:)• [E(t)t'"cosutdt. Since f(x) is regular for R(x) > 0, ([)(u) is regular for -}1r < l(u) < Let Then (2n)!c, ~ (-l)"''"'(O) ~ ~ J E(t)t'"dt. Suppose now that .S(t) is of one sign, say E(t) > 0, for t > T. en > 0 for n > nn, since «> T+ll T f E(t)t'" dt > f E(t)t'" dt- J jE(t)lt'" dt o T+l 0 > (T+I)'" TrE(t)dt-T'" J jE(t)l dt. T+l o It follows that ~nl(iu) increases steadily with u if n > 2n0• But in fll(u) and all its derivatives tend to 0 as u-+ }i11 along the axis, by the properties of ifl(x) obtained in§ 10.2. The theorem follows again. 10.5. The above proofs of Hardy's theorem are all similar in they depend on the consideration of 'moments' f f(tW'' dt. The t Fekete{2). t P0lya{3). 10.5 ZEROS ON THE CRITICAL LINE 261 methodt·depends on a contrast between the asymptotic behaviour of the integrals r Z(t)dt. rjZ(t)jdt, T T where Z(t) is the function defined in§ 4.17. If Z(t) were ultimately of one sign, these integrals would be ultimately eqnal, apart possibly from sign. But we shall see that in fact they behave quite differently. Consider the integral J (x(•))-le(a)da, where the integrand is the function which reduces to Z(t) on u = }, taken round the rectangle with sides a=}, a= t, t = T, t = 2T. This integral is zero, by Cauchy's theorem. Now J+2iT 2T f (x(•))-le(s) d8 ~ i J Z(t) dt. !+iT T Alw by (4.12.3) tx(•))-1 ~ (z',;)'"-:+J•,-!•<-~•·(t+oaJ)· Hence, by (5.1.2) and (5.1.4), (x(a))-le(a) ~ O(tl•-l.tH•+<) ~ O(tl.,.) (i,;;; a,;;; 1), = O(tta-}+•) = O(t~+<) (I< a~~). The integrals along the sides t = T, t = 2T are therefore O(Ti+<). The integral along the right-hand side is r(M+!•,-~.,,,(t+o(f))•H•tJidt. The contribution of the 0-term is " f O(d) dt ~ O(Tl). T The other term is a. constant multiple of ~ n-! jT(~)~+}ite-iit-Ulogndt. n-1 7, Now d'( t ) I ([i2 }tlog2,;- !t-tlogn = 21 . H~nce, by Lemma 4.5, the integral in the above sum is O(Ti), uniformly Wtth respect to n, so that the whole sum is also O(T~). t Seo Landau, Vorl"-"Un!fen, ii. 78--85. ZEROS ON THE CRITICAL LINE Combining all these results, we obtain y Z(t) dt = O(Tl). On the other hand, But 2T l-+2iT 2+iT 2+21T ~HiT i J t(t+it) dt ~ J {(') ds ~ J + J + J 7' }+iT i+t7' 2-riT 2+2iT Chap. X (10.5.1) [ oc I ]2+2iT I' ~ s- L --+ O(Tl) da ~ iT+O(Tl), n-2 n•Iog n HiT !-Hence r IZ(t)l dt >A~. Hardy's theorem now follows from (10.5.1) and (10.5.2). Another variant of this method is obtained by starting again from (10.2.1). Putting a= !?T-S, we obtain I , E(t) cosh{(!1r-S)t}dt = 0(1)+0( I exp(-n21rie-2i8J} t2+! n=I 0 ~ 0(1)+0(.~, e-•''"'"") ~ O(i)+Oq e-•'"'""dx) ~ as S- 0. If, for example, E(t) > 0 fort > t0, it follows that for T > r]Z(t)] dt ~ 1 rZ(t)dt 1 <A r.:!i•lel"'dt T T T or • < ATl I §~etnl-!trr dt < AT-1-I .E(t) coshf(t"IT-..!._T)t) t2+l t2+! \ 2 T • ~ O(Tt.Tl) ~ O(T!). This is inconsistent with (10.5.2), so that the theorem again follows. 10.6. Still another method.t depends on the formula (4.17.4), viz. Z(t) = 2 ~ cos(D-~~log~+O(t-l), t Tit.clllllQrsh(ll). 10.6 ZEROS ON THE CRITICAL LINE where x = -.j(tj21T). Here tJ. = D-(t) is defined by x(l+it) = e-2i{Jq), so that O'(t) ~ -~ x'IHit) ~ -~ho ~-~ r'(!-tit)~ J.:'liHit)l 2 xlt+•tl 2\ g 2 rtl-!it) 2 rt!+Ftl 11 q 2 1 I"' udu ~ -, og~+, og(,\r+!t )-l+<t'-R' {u'+l!+!it)')(e""-1)' and we have O'(t) ~ llogt-flog2~+0(1/t), O(t)- !tlogt, O'(t)- ~· The function tJ.(t) is stea.di.ly increasing for t ~ t0• If v is any positive integer ( ~ v0), the equa.tion tJ.(t) = V1T therefore has just one solution, say t~, and t~,.... 21TVjlogv. Now Z(t,) ~ 2(-1)" L cos(tJ:gn) +O(t;l). •<• The sum (t) = """cos(t~logn) _ 1 +cos(t~log2) g ~ ~ .Jn --./2 +··· consists of the constant term unity and oscillatory terms; and lihe formula suggests that g(t~) will usually be positive, and hence that Z(t) will usually change sign in the interval (tp, t~+l>· We shall prove THEOREM 10.6. As N-+ 00 .~. Z(t,,)- 2N, It follows at once that Z(t2v) is positive for an infinity of values of v, and that Z(t2v+1l is negative for an infinity of values of v; and the existence of an infinity of real zeros of Z(t), and so ofE(t), again follows. We have N N ,2 g(f2vl = L L cos(t~~ogn) ~=M+I ~=M+l n<;;.,'(loo/21>") =N-M+ ,L ~ ,L cos(~~logn), lo;;n;;;;-.'(fo.v/2,.) .,..;;t0yo;;l,!l where r = max(t2M+2• 27m2). The inner sum is of the form I cos(2•¢(")), 264 ZEROS ON THE CRITICAL LINE Chap. X where We may define t~ for all v;;;::: v0 (not necessarily integral) by !9-{t~) = vtr. Then c/''(v)=!;~~' so that Hence if,'(v) is positive and steadily decreasing, and, if vis large enough, ,P"(v) = -2?Tiogn 8-"(t2vl ,..,-~log~ <-A~ {&'(t,)}3 t 2,log3t2, la.ylog3ta.,,: Hence, by Theorem 5.9, L cos(t2vlogn) = o(t., llogglln )+o(tlNlog~) ,.;;r..,.-;t,,. t Nlo t2N login = O(tivloglt2.vl· Hence 2: J"n ,2 cos(t2,logn) = 0{4 . ..,.log~fa.vl 2<n<;;~(l,,./211') ;;;::l.,,;;t., = O(NlloglN). Hence and a similar argument applies to the other sum. 10.7. We denote by N0(T) the number of zeros of {(s) of the form !+it (0 < t::::;; T). The theorem already proved shows thatN 0(T) tends to infinity with T. We can, however, prove much more than this. THEOREM 10.7.t N0(T) >AT. Any of the above proofs can be put in a more precise form so as to give results in this direction. The most successful method is similar in principle to that of§ 10.5, but is more elaborate. We contrast the behaviour of the integrals <+H I = f E(u) :::"! e-u/T du, where T ~ t ~ 2Tand T-oo. t Hardy and Littlewood (3). 10.7 ZEROS ON THE CRITICAL LINE 265 We use the theory of Fourier transforms. Let F(u),j(y) be functions related by the Fourier formulae F(u) ~ "r!~,J f(y)e'~ dy, f(y) ~ ,2- I" F(u)e-''" du. -v(21T);X) Integrating over (t,t+H), we obtain t+H oo I I I e>uH_J F(u) du ~ ,-f(y) -.- •"' dy, I -v(21T) -oo ty I+II I eiull_l sothat t F(u)du, j(y)~ a.re Fourier transforms. Hence the Pa.rseval formula. gives Ii'J"Fr•l d·l'd, ~ 11/rvll'"'i";,!.!_lt~ dy. If F(u) is real, 1/(Y)I is even, a.nd we have [IT'F(u)du,,dt~ 2 J if(y)('4'in;,!l!ydy <;; 2H' rlf(y)('dy+B f~Bf!Ji'ay. (10.7.1) o 1/H Now (2.16.2) may be written __!_I" S(t) e«• dt ~ !elLe-ll~(e--<). 21T.., t 2+! Putting t = -i(l?T-!8)-y, it is seen that we may take F(t) =I .E(t) e(l,.-l8>+lulf(ei(l,.-8)+2Y). LetH ~I. The contribution of the firstterminf(y) to (10.7.1) is clearly O(H). Putting y = logx, G = e11H, we therefore obtain "I'+H I' G ) 1 f F(u) du dt = o{H2 [ llf(e1 (l,.-8lx2)12 dx + +0( I·l~(e'x')I'~)+O(H). (10.7.2) u log2x 266 ZEROS ON THE CRITICAL LINE Chap. X Now I!J.(ei(i,.-3>xz)lz = ~n~t e-n.,...:'(sJo.hiC083)1 2 = 2 e-lln~m::•sma+ 2 2 e-4m'w:z:'sJn3+><m'-n')orztco.sa. n~l ,. .. ,. As in§ 10,5, the first sum is O(x-1S-l-), and its contribution to (10. 7.2) is therefore o( H' J x-'3-l dx)+o( J !~!:,:) ' G ~ O{H'(G-I)S-i}+O(S-l/IogG) ~ O(HH). The sum with m -=j:. n contributes to the second term in (10. 7.2) terms of the form by Lemma. 4.3. Hence the sum is o(H' ~ ~' e-<m'+•'""'"') ~ o(H' ~ e-m'"'"'~' I ) L., L., m2-n2 L.. m L.. m-n -•-m~ -~ o(H' i lo~m ,-m'"'"') ~ o(H•( L lo~m + L ,-m'"'"')) m~ -~ ~~ ~ o(H'Iog'~) ~ O(Hs-1) for 8 < 80(H). The first integral in (10.7.2) may be dealt with in the same way. Hence "II+H 12 [ [ F(u) dul d! ~ O(Hs-1). Taking;;= 1/Tand T> T0 (ll), it follows that OT j ill' d! ~ O(HTI). 10.8. We next prove that where J > (AH+'Y)T-1, ri'Yi'd! ~ O(T) (0 < H < T). T (10.7.3) (10.8.1) (10.8.2) 10.8 ZEROS ON THE CRITICAL LINE We have,·if s =!+it, 7',;;; t ~ 2T, eint TljS(!)I 1,+! > Aj((J+il)l. Hence TlJ > A7H l~(~+iu)l du > AI7H ~U+iu) duj ~AjTI L ni~'"+O(T-I))dul ' n<AT ~AH+o(!T L ~d•/)+OiHT-1) t ll<;;n<.AT 267 ~AH+o(j L (nl+"'~'logn -;:.r.}logn)I)+O(HT-I). 2o;;n<AT It is now sufficient to prove that J OT I L nl+"Ilo .I'd!~ O(T), T 2.;;;n<.4T g and the calculations are similar to those of§ 7.3, but with an extra factor logmlogn in the denominator. To prove Theorem 10.7, let S be the sub-set of the interval (T, 2T) where I= J. Then J jl[ dt = J J dt. s s OT ( OT )' Now jlljdt,;:;;;jjljdt~ Tjjlj 2 dt T-tj (All+"'¥) dt OT > AT-iHm(S)- T-i j j'Yj dt > AT-lHm(S)-T-i(T rj'l'j2dtt > AT-!Hm(S)-AT!, where m(S) is the measure of S. Hence, for H ~ 1 and T > T0 (H), m(S) < ATH-l. 268 ZEROS ON THE CRITICAL LINE Chap. X Now divide the interval (T, 2T) into [Tf2H] pairs of abutting intervals j 1, j 2, each, except the last j 2, of length H, and each j 2 lying to the right of the corresponding j 1• Then either j 1 or j 2 contains a zero of E(t) unless j 1 consists entirely of points of 8. Suppose that the la.tter occurs for v j 1's. Then vH,;;; m(S) < ATH-l. Hence there a.re, in (T, 2T), at least T(1 A) T [Tf2HJ-• > n 3--m > 4ii zeros if H is large enough. This proves the theorem. 10.9. For many years the above theorem of Hardy and Littlewood, that N0(T) >AT, was the best that was known in this direction. Recently it has been proved by A. Selberg (2) that N0 (T) >AT log T. This is a remarkable improvement, since it shows that a finite propor-tion of the zeros of {(s) lie on the critical line. On the Riemann hypo-thesis, of course, N,(T) ~ N(T)--/;TiogT. The numerical value of the constant A in Selberg's theorem is very small.t The essential idea of Selberg's proof is to modify the series for '(s) by multiplying it by the square of a partial sum of the series for g(s)}-l. To this extent, it is similar to the proofs givezi in Chapter IX of theorems about the general distribution of the zeros. We define~ by ,.j'~s) = ~ ~ (u > 1), o:1 = 1. It is seen from the Euler product that a:~' o:~ = ~v if (/-'. v} = 1. Since the series for (1-z)l is majorized by that for (1-z)-l, we see that, if then Jo:~J ,;;; c4 ,;;; 1. -./C(s) = ~ ~· o:~ = 1, Let Pv=a:v(I-1 ::;) (I,;;v<X), Then IP.I,; 1. t It. was ca.lcula.ted in an Oxford diuertation by S. H. Min. 10.9 ZEROS ON THE CRITICAL LINE 269 All sums involving Pv run over [1, X] (or we may suppose P~ = 0 for v~X). Let 10.10. Lett c+i<» <!>(•) ~ .k,J. r(!B)w-l•{(a)1(B)1(1-B),.d8 where c > I. Moving the line of integration to a = f, and evaluating the residue at 8 = 1, we obtain i+i<» (z) ~ !z,l(l)>(0)+!, J r(!-l•{(•h)•,l(i-s)"ds . .. !-i<» ~ jz,(1)1(0)-~ f. ~(·; 11(!+;1)1'•" dt. 21T t +.-On the other hand, (z) ~ .k :! L L P,P. r rt!sl~-1· n•~~:,~ ds n-1 ~ v c~w ~ :! L L P•!•exp(-:.:.')· n~l 11-v Putting z = e-«t•-l3l--11', it follows that the functions F(t) ~ ~ E(t) 11(Hilll'~l.-l", ~(lmlt'H f(y) ~ !.11(1)1(0)-z-1:! L L ¥ exp (- ~;.~') n-1 I" ~ are Fourier transforms. Hence, as in§ 10.7, "('+h )2 1/h "' l j F(u) du dt,; 2k' [ 1/(Y)I' dy+B J,lf(y)l'y-' dy (10.10.1) where h ,;;; 1 is to be chosen la.ter. Putting y = logx, G = ellh, the first integral on the right is equal to t Titclunarsh(26). 270 ZEROS ON THE CRITICAL LINE Chap. X Calling the triple sum g(x), this is not greater than G G G 2 J lif>(l;:~O)I' dx+2 J lu(x)l' dx < !lif>(I),(0)1'+2 J lg(x)l' dx. . . . Similarly the second integral in (10.10.1) does not exceed lif>(I)if>(O)I' +2 J•lg(x)l' dx 2G log2G 0 log2x · 10.11. We have to obtain upper bounds for these integrals as 8-+ 0, but it is more convenient to consider directly the integral J(x, 0) ~ j lg(u)l'u-9 du (0 < 0,; !, x;;, I). This is equal to +i1r(m~: 2 - n~ 2 )u2coss}?· Let :E1 denote the sum of those terms in which mKjJ.. = np.f11, and :E11 the remainder. Let (KV,,\p.) = q, so that 101 = aq, ~ = bq, (a, -b)= I. Then, in :E1, ma = nb, so that n = ra, m = rb (r = 1, 2, .. ). Henoo 10.11 ZEROS ON THE CRITICAL LINE 271 where K(8), and later K1{8}, are bounded functions of 9. Hence we obtain o~,t (j e->' dy+O(x,l,)) -~ [f e-•'y-9 dy+O{(x,I")'-'J] + ' ' +TJfB-!K(B) [ [ e-u'y-0 dy+O{(x'-''1)1-0J-] +O{x1- 0log(2 + '1- 1 )} -..JTT + Kt(8)7)!8-! +O(xt-e I (2+ -I)} -~ e e og 'I . Putting 7J = 21TKtp.2q-2ain8, it follows that ~. ~ 2 ( 2 ,~~~\te:?+ K,JB) (2~'ino)l'-!s(O)+ +O{x 1 -6 log(2+'1-1 ) L IP"P-.PPP.I}• e ><'-!" lv where S(O) ~ 2 ('L)'-' ~.~,p.~,. ,.~v Kf-l-Av Defining r/la(n) as in § 9.24, we have qt-6 = ~ q,_fJ(p) = pl~.JL 4>-e(p). Hence S(O) ~ 2 ;,,(p)( 2 ~;!.;,)'-p<X' PI"• (10.11.1) Let d and d1 denote positive integers whose prime factors divide p. Let K = dK', v = d1v', where (K',p) = 1, (v',p) =I. Then L: ~ = L: {it-1 od L: ~~:~ L: ~Jf. piKV pldd1 1 t<" v" Now, for (K',p) = 1, Hence the above sum is equal to I ""'IXdiXd ""' IX,.• I X ""' IXv'J X log2X p"fi;J, dl-9d~ ,..fgld K'l-6 og ""(f2 v·fxtd• 7 og dl v'. 10.12. LEMMA 10,12, We have L ;f,'t.log£-~o((~)'Iogl~TI(I+~)I) (10.12.1) ,.·<,x/d Pip uniformly wit-h respect to e. ZEROS ON THE CRITICAL LINE Chap. X We may suppose that X;::: 2d, since otherwise the lemma. is trivial. l+ia:> Now ~I ;d8=0 (Ol). ,. 8 ,.., Also L: ;;;&~ I1 (~-~)· ~ L:(~-~)-l "((1~9+•1' (tt',pl-1 (p,pl~l PIP Hence the left-hand side of (10.12.1) is equal to ~ ']'" ~ ('ID' I1 (I-p'~ .. t· "111~9+•1" (10.12.21 1-ioo Pip There are singularities at 8 = 0 and s = 8. If 8 ;::: {log(Xjd)}-t, we can take the line of integration through 8 = 8, the integral round a small indentation tending to zero. Now ~((l~itll < Altl for all t (large or small). Also I1 (I-p'l'"r ~ o(l I1 (l+p>k)ll ~ o( I1 (I+~))· Pip Pip PIP Hence (10.12.2) is o((~)'IJ (~+~)ll~;~~:) ~ o((~)'IJ (~+~)l M· and the result stated follows. If 8 < {log(Xjd)}-1, we take the sa.me contour as before modified by a. detour round the right-hand side of the circle lsi = 2{log(Xfd)}-1• On this circle I (X/d)•l ~ e2, the p-product goes as before, and 1((1-9+•11 > A!og(Xfdl. Hence the integral round the circle is o(Iog-l~TI ('+~)l I~~~~~ o(Iogl~ TI (1+~)1). PIP PIP p The integral along the part of the line a = 8 above the circle is o((~)'TI(~+M 1 f ~)~o((~)'Iogl~Il(~+~)l). Pip .A(IOIJX/d)-t Pip p The lemma. is thus proved in all cases. 10.13 ZEROS ON THE CRITICAL LINE 10.13. LEMMA 10.13. ') 1'1,""·1 ~a(~ I1 (1 +~))· pt;t, l p PIP p Defining tvd as in § 10.9, we have 2 la;;d,l ~~tad,= L~• plddt l pldd, 1 pJD where Dis a. number of the same class a.s d or d1, ~ ~ I1 (~-~)-'~a(~ TI(I+~))· pp]p p pplp 10.14. LEMMA 10.14. 8191 ~ o(1!'~) uniformly with respect to 9. In particular 8(01 ~ o(1o:x)· By the formulae of§ 10.11, e.nd the above lemmas, "P.P, ~ o(-1 - " 1"'""'1 (:!)'Iogl:!Iogl! I1 (~+~)) :2. Kt-811 log2X ~. dl-8~ d d d1 PIP p ~a(~ I1 (' +~) L 1••••.1) logX pip' p pldd, dd1 Hence 8191 ~ o(/; L: ~-o~PI TI (~+~)') ogX p.;;x• P Pip p (X" " I ( I)') = 0 log2X L.. pt+8 n 1+p p,;,x• PIP 273 Henoo ZEROS ON THE CRITICAL LINE S(8)- o( X" ')' " I ) -log2X nfi-• p.fx.,,. (npl)l-Kln! = o(t!:~ ~ nf+o p.h.,,. pt+o} ~ 0h~::X ~ a L ;J n-1 p..;;x• ~o(,!~)· Chap. X 10.15. Estimation of :E1• From (10.11.1), Lemma. 10.14, and the inequality 1,8,1 ~ 1, we obtain L ~ o( 1~)+o( (~lxX')' )+o(x'-'Iog(X/~)X'l •x) ' BI8x'logX ~fBx'IogX 8 og · We shall ultimately take X= o--c and h = (alogX)-I, where a and c are suitable positive constants. Then G = xa = J-nc. If :c :<.:;; G, the last two terms can be omitted in comparison with the first ifG}(2 = O(d-l-), i.e. if (a+ 2)c ~ t- We then have :El = o(ot&xO~ogx)· (10.15.1) 10.16. Estimation of :E2. If P and Q are positive, and x ~ 1, f "" -Pu'+iQu' du - ! I"" e-Pv iQv d . - o(~ "' e 1i,b -2.,. VF+I e v -xDQ}' e.g. by applying the second mean-value theorem to the real and imaginary parts. Hence :E2 = o[~ L hL' jm;:2- n~2~-~exp(-1T{m;:2+n~2)sin0}]. K~V 11l11. The terms with mx/A > np.fv contribute to the m, n sum o( ~ ,-•m'·''-'"'' L (m;:·-•:,T)· m-1 n<m..ouol-¥< Now mx(mxv-nAp.) A'v ' 10.16 ZEROS ON THE CRITICAL LINE 275 Hence the m, n sum is o(~ m%t (~+lo~X))e-mn'K'~-•sina} ~ o{~(l+ logX)log~+~log 2 ~} K J.p (j KJ.l (j ~ o(""Iog!)+o(~log'!J. K 0 K/L 0 since, as in §10.15, we have X= ;;-c, with 0 < c ~ l· The remaining terms may be treated similarly. Hence :E2 = o(~ ~ (~log~+~log2 ~)} = o(~log 2 ~)· (10.16.1) 10.17. LEMMA 10.17. Undertheassumptionsof§10.15 l/TF(u) duj' dt ~ o(~li!gx)· (10.17.1) By (10.15.1) and (10.16.1), J(x,O) ~ o(~;,gx) (10.17.2) uniformly with respect to 8. Hence G G G fig(x)j'dx~- J x'~dx~[-x0J]~+0 f x'-•Jdx . . . ~ o(~lei~gx)+o(e l~lex~gx) ~ 0(~i~!gGx)· taking, for example, 8 = f. Also ! 8J(G,8)d8 ~ Jjg(x)j'dx! Ox-0 d8 ~I jg(x)i'(Io~'x -2Xt:ogx -xli~g'x)dx >- J•lg(x)j' dx -~ J•lg(x)l' dx ~ log2x 2 ---zr-a a 276 ZEROS ON THE CRITICAL LINE Chap. X since G = elih ;;:: e. Hence • l J 1 i~;!~' dx,;; f 8J(G,8) d8+fJ(G,i) l ~a( [ ~t(/:Ogx)+a(~IGf~ogX) ~ a(~liog~logx)· Also ~(0) ~ a(X), ~(!) ~ a(logX). The result therefore follows from the formulae of§ 10.10. 10.18. So far the integrals considered have involved F(t). We now tum to the integrals involving iF(t)l. The results about such integrals are expressed in the following lemmas. LEMMA 10.18. f • iF (I) I' dt ~ a( log I/~ )· -• ~tJogX By the Fourier transform formulae, the left-he.nd side is equal to f • f•l e-iil~l• I' 2 lf(Y)I' dy ~ 2 ----...---~(I),(0)-g(x) dx ' ' ,;; 4! ig(x)l' dx+a(X'log'X). Taking x ~I, 8 ~ {log(!/~))-• in (10.17.2), we have Hence f "" lg(u)j2e-iogu/I.IQII'l/3J du = o( log I/8 )· 1 aliogX ·-· f l()l'd -a(logll~) 1 gu U-aliogX. We can estimate the integral over (8-11, oo) in a. comparatively trivial manner. As in§ 10.11, this is less than 10.18 ZEROS ON THE CRITICAL LINE 277 Using, for example, K2..l.-2sin() > AX-2() >AlP (since X= {J-c with c < t), and IP .. I :s;; 1, this is ~ a(X2log2X j e-Ail'u' du) = O(X2log2X e-.4./ll'), ·-· which is of the required form. 10.19. LEMMA 10.19. [ (J'IF(u)l au)' at~ a(~:~~~~)· For the left-hand side does not exceed n~riF(u)i'du)dt~h fiF(u)i'du Jat~h' [IF(u)i'du, -co t -«> u.-A -«> and the result follows from the previous lemma. 10.20. LEMMA 10.20. If~ ~ 1fT, T f iF(t)i dt > ATI. We have ( 'J'+T + ~r + 1'J·(·)~·(·) ds ~ o. f+i ll+i ll+iT f+iT Since rfo(s) = O(X-1-) for u ~ }, the first term is O(X), and the third is a(XTI). Also ((s)p'(s) ~I+~';;;· where la ... l :::;;;: da(n). Hence ll+iT ., H-iT f ((s)p'(s) ds ~ i(T-I)+ 2: a. f ~ 2+i n=l 2+i ~ i(T-I)+a(i d,(n)) n= 2 n2 logn ~iT+a(l). T It follows that f ((i+it),l'(t+it) dt ~ T. 278 Hence ZEROS ON THE CRITICAL LINE l)F(t)) dt >A l d)((!+it),\'(i+it)) dt T >AT-! f l((!+it),S'i!+it)) dt ~1' > AT-iiJ m+it),S'(Hit) dtl >AT-1. 10.21. LEMMA 10.21. T l+h J dt J )F(u) I du > AhTt. ' ' The left-hand side is equal to T+h mln(T;u) T u T Chap. X J )F(u)) du J dt;;, J )F(u)) du J dt ~ h J )F(u)) du, 0 ma.x(O,u~h) h u-h h and the result follows from the previous lemma.. 10.22. THEOREM 10.22. N,(T) >AT log T. Let E be the sub-set of (0, T) where <+h l'+h f )F(u)) du > [ F(u) dul. For such values oft, F(u) must change sign .in (t,t+h), and hence so must E(u), and hence tCi+iu) must have a zero in this interval. Since the two sides are equal except in E, 1 dt'J~F(u)j au;;, 1 ('['IF(•) I du-ITF(u) aul) dt ~ [('f~F(u)) du-ITF(u) dul) dt > AhTL [I TF(u) dul dt. The left-hand side is not greater than 10.22 ZEROS ON THE CRITICAL LINE "' by Lemma. 10.19 with 3 = IfT. The second term on the right is not greater than ( T Tl<+h I' )l AhiTi f dt f I F(u) du dt < loglX by Lemma 10.17. Hence { E)' A '(logX\l Tl m( ) "> 1 T• logT} -A 2hi-loglT' where A1 and A11 denote the particular constants which occur. Since X= pc and h = (alogX)-1 = (aclogT)-1, {m(E)}i > A 1 ciT-l-A2(ac)!Tt. Taking a small enough, it follows that m(E) >A3 T. Hence, of the intervals (O,h), (h,2h), ... contained in (O,T), at least [A 3 Tfh] must contain points of E. If (nh, (n+l)h) contains a point t of E, there must be a zero of ~(!+iu) in (t,t+h), and so in (nh,(n+2)h). Allowing for the fact that each zero might be counted twice in this way, there must be at least ![A,Tfh] > ATlogT zeros in (0, T). 10.23. In this section we return to the function E(t) mentioned in § 10.1. In spite of its deficiencies as an approximation to E(t), it is of some interest to note that aU the zeroo of E(t) are real.t A still better approximation to ll>(u) is (u)cosut du, and we shall also prove that all the zeroo of E(t) are real. The function K.ll'(a) is, for any value of a, an even integral function of z. We begin by proving that if a is real all ita zeroo are purely imaginary. It is known th&t w = ~(a) satisfies the differential equation £.(•~) ~ (•+~)w. This is equivalent to the two equations ~ = Jf ~ = (a+~)w. t POly• (1),(2), (4). 280 ZEROS ON THE CRITICAL LINE Chap. X These give It is also easily verified that w and W tend to 0 as a -+ oo. It follows that, if w vanishes for a certain z and a = a 0 > 0, then I {IWI'+I•'+•')(w('}~ ~ 0. Taking imaginary pa.rt:.a, 2ixy J':l' da ~ 0. Here the integral is not 0, and R;,(a) plainly does not vanish for z real, i.e, y = 0. Hence x = 0, the required result. We also require the following lemma. Let c be a positive constant, F(z) an integral junction of genus 0 or 1, which takes real values /M real z, and has no complex zeros and at least one real zero. Then all the zeros of are alilo real. F(z+ic)+F(z-ic) (10.23.1) We have F(z) = C&e=fr (1-;-)ezlrx.., n=l n where C, a:, o:1, ••• are real constants, IXn 0 for n = 1, 2, ... , I a.,:;-11 is convergent, q anon-negative integer. Letz be a zero of (10.23.1). Then so that 1 ~jF(z-ic)j' ~ (x2+(y-c)2)'IT"" (x-a:,.)2+(y-c)2 F(z+ic) x2+(y+c)2 n=t (x-a .. )2+(y+c)2' If y > 0, every factor on the right is < l; if y < 0, every factor is > 1. Hence in fact y = 0. The theorem tha.t the zeros of E(t) a.re a.U real now follows on taking F(z) = Rtu(27T), c = ;. 10.24. For the discussion of E"'(t) we require the following lemma.. Let lf(t) I < Ke-ltlo+3 for some positive 8, ao that F(•) ~ ~~!~) [ f(t)<'" dt 10.24 ZEROS ON THE CRITICAL LINE 281 is an integral function of z. Let all the zeros of F(z) be real. Let f(t) be an integral function oft of genus 0 or 1, real for real t. Then the zeros of G(•) ~ ~ s· f(t)4>(it)•"' dt vl2~) -• are also all real. Weha.ve where the constants a.re all real, and! a;n11 is convergent. Let ~.(t) ~ Ct"•"'IJ(1-~)<''"-Then f,.(t)-+ f(t) uniformly in any finite interval, and (as in my Theory af FundianB, § 8.25) l~.(t)( < K•"'" uniformly with respect to n, Hence G(•) ~ lim ~ s· f(t)~.(it)<"' dt ~ lim G,(•), n-.oo..,(27T) -oo ......,."" say. It is therefore sufficient to prove that, for every n, the zeros of Gn(z) are real. Now it is easily verified that F(z) is an integral function of order less than 2. Hence, if its zeros are real, so are those of (D-a)F(•) ~ ,@~{<-~F(•)} for any real eo:. Applying this principle repeatedly, we see that all the zeros of H(•) • ~ ll'(D-a1) •.• (D-.,)F(z) ~ ~(!,)_i f(t)(it)'(it-a1) ••• (it-.,) 1 by L s = ~ Xo(n) = ! ! ! !__ ! o() ,6 n' l"+2,+ag+ 4"+&+ ... , ~ x 1(n) 1 i i 1 1 L 1(s) = 6_ ---na = p+2i-3i-4s+&+···· ~ xz(n) 1 i i 1 1 La(s) = 6 --ne = p-28+aa-4'+s-+···· L 3(8 ) = ~ x::> =~-}a-~+~+~+···· Each x(n) has the period 5. It is easily verified that in each case x(m)x(n) ~ x(mn) if m is prime to n; and hence that L(•> ~ IJ(I-x:,T' (a> I). It is also easily seen that L,(•>~(I-M((•). so that L0(s) is regul&r except for a simple pole at s = 1. The other three series are convergent for any real positive a, and henoe for a > 0. Hence L1(s), L 2(s), and L3(s) are regular for u > 0. Now consider the function /(•) ~ !sec8(e-"L,(•He"L,(•)} = i+ ta;,8-ta.;8 -~+~+··· ~ ~(((•.!)+tan8((•,!)-tan8((a,!)-((8,1)}, where '{a, a) is defined as in § 2.17. 10.25 ZEROS ON THE CRITICAL LINE 283 By (2.17)J(s) is an integral function of s, and for a < 0 it is equal to 2r(I-a)(. , , 5,(27T)l-B 8Jil T"B X ~ ( 2m7T 4m7T 6m1r 8m11) I x £:: 1 cos----s-+tan8cos----s--ta.n9cos----s--cos----s- m1,+ ~( 2m~ 4m~ . 6m~ . Sm~) I) +cos}11s ~ sin----s-+tan9sin5 -tan9sm-y-sm----s- m 1_, m~l 4r(l-•)cos!~• ~ (. 2m~+t 8 . 4m~) I = &(27T)l-s ,£:;1 sm5 an sm5 mt-•" If 471 . 871 (. 27T ll . 47T) sin-+tan9sm- = tan9 sm-+tanvsm-=-, 5 5 5 D (10.25.1) this is equal to 4r(I-s)cosi71s(sin!:::+tan9sin~)/(I-s). 5"(27T)l-11 5 5 The equation {10.25.1) reduces to sin29 = 2cos¥ = ..;5;- 1, and we take 9 to be the root of this between 0 and !11. We obtain tan8 ~ ,'(IO~::'~>-•, sin¥+tan0sin~ = ~· and f(s) satisfies the functional equation f(s) = ~1~s)cosl~j(1-s). 5B-a(27T)l-11 There is now no difficulty in extending the theorems of this chapter tof(s). We can write the above equation as (~tr(H¥lf(•) ~ (~)H'r(l-!s)f(l-s), and putting s = !+it we obtain an even integral function oft analogous to E(t). We conclude that f{s) has an infinity of zeros on the line a = }, and that the number of such zeros between 0 and Tis greater than AT. On the other hand, we shall now prove that f(s) has an infinity of zeros in the half-plane a > 1. 284 ZEROS ON THE CRITICAL LINE If p is a prime, we define a(p) by o(p) ~ W+i)x,(pl+!(l-i)x,(P). so that <1:(p) = ±l or ±i. For composite n, we define a(n) by the equation o:(n,.n2) = o:(n1)o:(n2). Thus ja(n)j is always 0 or I. Let M(s,x) ~ ~ o(n~~(n) ~ fJ (1- o(p~~(p)r. n~l P where x denotes either Xt or x2• Let Now N(s) ~ !{M(s,x,)+M(s,x,)). o:(Plxt(P) = t(l+i)x~+t{l-ilxt X2• o:(p)x2(P) = }{l+ilxt Xa+!{I-i)xi. Chap. X and these are conjugate since x~ = xi and x~ and Xt x2 are real. Hence M(s, x1 ) and M(s, x2) are conjugate for real s, and N{s) is real. Let s be real, greater than I, and -+ l. Then logM(s,x,) ~ ~o(p;;:,(p)+0(1) ~ W+i) "xl(p)+!(l-i) "x,(p)x,i1'.)+ 0 (1). 7 pB f ps Now x~ = x 3 and Xt x 2 = Xn· Hence Hence "xltP) ~ "x,(p) ~log L,(s)+0(1) ~ 0(1), f p8 f pB logM(s,x1) = !(l-i)log 8~ 1 +0(1), N(s) = RM(s,x1 ) = ~(s~l)cos(flog 8~ 1)e0U>. It is clear from this formula that N{s) ka& a zero at each of the points 8 = I+e-{2m+l}>r (m = 1, 2, ... ). 10.25 ZEROS ON THE CRITICAL LINE Now for a~ 1+3, and x = x1 or X:z• log L(s+i-r, x)~logM(s, x) ~ .~(log(1- a(p;(p))-log('-p;•x(Pl))+a(~) ~a(.~ ~(Pl;,:r''l) +a(F,). .... 285 Let a:(p) = e:zwifi 0 and a > 0, there is a "T such that lf(s+i,)-N(s)l < • (a? 1+8). Let 31 > 1 be a zero of N(8). For any 'rJ > 0 there exists an 7}1 with 0 < 7}1 < 7J, 7Jt < 81-1, such that N(s) =f=. 0 for fs-s1f = 7Jl· Let E= min fN(s)f IS-8,1-')t and 3 < s1-7]1-l. Then, by Rouche's theorem, N(s) and N(•)-{N(s)-j(s+iT)) have the same number of zeros inside fs-s1 1 = TJ1, and so at least one. Hencef(s) has at least one zero inside the circle fs-81-i-rl = 1]1• A slight extension of the argument shows that the number of zeros of /(s) in u > I, 0 < t ~ T, exceeds AT as T ~ oo. For by the extension of Dirichlet's theorem (§ 8.2) the interval (t11, mqPt0) contains at least m values of t, differing by at least to. such that l tlogp -x' I ~! (p ~ P). 2w P q The above argument then shows the existence of a zero in the neigh-bourhood of each point s1 +i(-r+t). 286 ZEROS ON THE CRITICAL LINE Chap. X The method is due to Davenport and Heilbronn (1), (2); they proved that a class of functions, of which an example is L 2 (m~:5nll)s' m,n, I. It has been shown by calculationt that this particular function has a zero in the critical strip, not on the critical line. The method throws no light on the general question of the occurrence of zeros of such functions in the critical strip, but not on the critical line. NOTES FOR CHAPTER 10 10.26. In§ 10.1 Titchmarsh's comment on Riemann'sstatement about the approximate formula for N(T) is erroneous. It is clear that Riemann meant that the relative error {N(T) -L(T)}/N(T) is O(T-1). 10.27. Further work has been done on the problem mentioned at the end of§ 10.25. Davenport and Heilbronn (1), (2) showed in general that if Q is any positive definite integral quadratic form of discriminant d, such that the class number h(d) is greater than 1, then the Epstein Zeta-function (o(•) ~ L Q(x,y)-• (u >I) "'Y--"'l (x•y)f(0.0) has zeros to the right of u = 1. In fact they showed that the number of such zeros up to height Tis at least of order T (and hence of exact order T). This result has been extended to the critical strip by Voronin , who proved that, for such functions ("'(s), the number of zeros up to height T,for! < a 1 ~ l(s)::.:; u2 < 1, is alsooforderatleast T(andhence of exact order T). This answers the question raised by Titchmarsh at the end of § 10.25. 10.28. Much the most significant result on N 0(T) is due to Levinson [2 ], who showed that N 0(T) ;;, aN(T) (10.28.1) for large enough T, with 1): = 0·342. The underlying idea is to relate the distribution of zeros of ((s) to that of the zeros of C (s). To put matters in t Potter and Titchmarsh (I). 10.28 ZEROS ON THE CRITICAL LINE 287 their proper perspective we first note that Berndt has shown that . , T( T ) # {s = a+lt: 0 < t ~ T, ( (s) = 0} = 2; log4n'-1 +O(log T), and that Speiser (1) has proved that the Riemann Hypothesis is equivalent to the non-vanishing of('(s) forO< a<!· This latter result is related to the unconditional estimate # {s= a+it:-1 <u <f, T 1 <t~ T 2,('(s)=O} = #{s= u+it:O <a<!, T1 < t::.:; T2,{(s) = 0} +O(log T2 ), (10.28.2) zeros being counted according to multiplicity. This is due to Levinson and Montgomery , who also gave a number of other interesting results on the distribution of the zeros of (' (s). We sketch the proof of (10.28.2). We shall make frequent reference to the logarithmic derivative of the functional equation (2.6.4), which we write in the form ('(,)+ ('(1-s) ~ logn-!(r'(!s) + r'(!- !•!) ((s) ((1- s) r(j-s) r(!- !-s) ~ -F(s), say. We note that F(! +it) is always real, and that F(s) ~ log(l/2n) + 0(1/1) (10.28.3) (10.28.4) uniformly fort:;::, 1 and lui::.:; 2. To prove (10.28.2) it suffices to consider the case in which the numbers T1 are chosen so that ((s) and {'(s) do not vanish fort= Ti, -1 ~a::.:;!- We examine the change in argument in C'(s)/{(s) around the rectangle with vertices !-O+iTI> !-0+iT2, -l+iT2, and -l+iTt> where 0 is a small positive number. Along the horizontal sides we apply the ideas of §9.4 to ((s) and {'(s) separately. We note that ((s) and {'(s) are each O(tA) for -3::.:; u::.:; 1. Moreover we also have I(( -1 +iT) I~ Tj, by the functional equation, and hence also 1{'(-1 +iT 1 )1 ~ Tfl~i.=-:::~;j ~ T1,logT 1 , by (10.28.3) and (10.28.4). The method of§ 9.4 therefore shows that arg ((s) and arg C'(s) both vary by OOog T.) on the horizontal sides of the 288 ZEROS ON THE CRITICAL LINE Chap. X rectangle. On the vertical side u = - 1 we have ~~; = log(~)+O(l), by (10.28.3) and (10.28.4), so that the contribution to the total change in argument is 0(1). For the vertical side u = !-b we first observe from (10.28.3) and (10.28.4) that a(- n!+it)) "1 (10.28.5) ((!+it) ~ if t ~ T 1 with T 1 sufficiently large. It follows that a(-('(!- 0+ it))"! ((!-O+•t) (10.28.6) for T 1 ,;; t ~ T 2, if 0 = J(T2) is small enough. To see this, it suffices to examine a neighbourhood of a zero p =! + iy of {(s). Then ('(s) ~ ---"'--+m'+O(Is-pl), ((s) s-p where m ~ 1 is the multiplicity of p. The choice s = ! +it with t ,. y therefore yields R(m') ;;?:- 1, by (10.28.5). Hence, on taking s = ! - J +it, we find that a(-('(s)) ~-1 mb 1 ,+a(m')+0(1s-pl) ;,! ((s) s-p for Is-pi small enough. The inequality (10.28.6) now follows. We therefore see that arg ('(s)/((s) varies by 0(1) on the vertical side R(s) = t- 0 of our rectangle, which completes the proof of (10.28.2). If we write N for the quantity on the left of (10.28.2) it follows that N 0(T,)-N0(T,) ~ {N(T,)-N(T,)} -2N+O(log T,), (10.28.7) so that we now require an upper bound for N. This is achieved .by applying the 'mollifier method' of §§9.20--24 to ('(1-s). Let v(u, T 1, T2) denote the number of zeros of ('(1-s) in the rectangle u ~ R(s) ~ 2, T 1 < l(s) < T 2• The method produces an upper bound for , f v(u,T1,T2)du, (10.28.8) which in turnyieldsanestimateN ~ c{N(T2)-N(T1 )} for large T 2• The constant c in this latter bound has to be calculated explicitly, and must 10.28 ZEROS ON THE CRITICAL LINE 289 be less than!- for (10.28.7) to be of use. This is in contrast to (9.20.5), in which the implied constant was not calculated explicitly, and would have been relatively large. It is difficult to have much feel in advance for bow large the constant c produced by the method will be. The following very loose argument gives one some hope that c will turn out to be reasonably small, and so it transpires in practice. In using (10.28.8) to obtain a bound for N we shall take u = t-ajlog T 2 , where a is a positive constant to be chosen later. The zeros p' = {J' + iy' of ('(1- s) have an asymmetrical distribution about the critical line. Indeed Levinson and Montgomery showed that L (!- {J') - ~ 2 T loglog T, O<y',;T 1t whence fJ' is t- (log log y')/log y' on average. Thus one might reasonably hope that a fair proportion of such zeros have {J' < u, thereby making the integral (10.28.8) rather smalL We now look in more detail at the method. In the first place, it is convenient to replace ('(1- s) by ('(s) ((s) + F(s) ~ G(s), say. If we write h(s) = tt-!s r(!s) then (10.28.3), together with the functional equation (2.6.4), yields ((I-s) ~ _ F(s) h(s) G (s) h(J-s) so that G(s) and C(l- s) have the same zeros fort large enough. Now let ~(s) ~ L b,n-• (10.28.9) .. , be a suitable 'mollifier' for G(s), and apply Littlewood'sformula(9.9.1)to the function G(s)l/f(s) and the rectangle with vertices u +iTt> 2+iT2 , 2+iTl' u+iT2• Then, as in §9.16, we find that , logT2J N<;-a-v(u,TI>T2)du '· logT 2J ~~ logiG(u+it)l/f(u+it)ldt+O(logT2). '· 290 ZEROS ON THE CRITICAL LINE Cbap. X Moreover, as in §9.16 we have J' logiG(u+ i~·(u+ it) ldt T, T, ~!(T2 -T1)1og(Tz~T 1 f IG(u+it)Y.,(u+it)l2dt). T, Hence, if we can show that I' IG(u +it).(u+ it) I' dt- c(a) (T,- T,) (10.28.10) T, for suitable T 1, T 2, we will have (10.28.11) whence N 0(T2)-N0(T1 ) ~ (1-log:(a) +o(l)){N(T2)-N(T1)} by (10.28. 7). The computation of the mean value (10.28.10) is the most awkward part of Levinson's argument. In he takes'y = T2!-• and b ( ) u-tlogy/n n=Jln n logy . This leads eventually to (10.28.10) with c(a) = e2a(!__+!) _ !2~+!-~. 2a3 24a 2a 3 a 2 24a 12 12 The optimal choice of a is roughly a = 1·3, which produces (10.28.1) with = 0·342. The method has been improved slightly by Levinson , , Lou [I] and Conrey and the best constant thus far is ct = 0·3658 (Conrey [1 ]). The principal restriction on the method is that on the size of y in (10.28.9). The above authors all takey = T2l-•, but there is some scope for improvement via the ideas used in the mean-value theorems (7 .24.5), (7.24.6), and (7.24.7). 10.29 ZEROS ON THE CRITICAL LINE ,., 10.29. An examination of the argument just given reveals that the right hand side of (10.28.11) gives an upper bound for N + N, where N = # {s =!+it: T1 < t ~ T2, C'(s) = 0}, (zeros being counted according to multiplicities). However it is clear from (10.28.3) and (10.28.4) that C'<l+ it) can only vanish if((! +it) does. Consequently, if we write N<rl for the number of zeros of ((s) of multiplicity r, on the line segment s = l + it, T 1 < t ~ T 2, we will have N = L (r-l)N<rl. r=2 Thus (10.28. 7) may be replaced by N"'- '~' ('-2)N''' ~ {N(T,)-N(T,)}-2(N+N')+O(log TJ. If we now define N<rl(T) in analogy to N, but counting zeros l +it with 0 < t :s;; T, we may deduce that N"'(T)-I ('-2)N'"(T) ~ •N(T), (10.29.1) for large enoughT, and ll = 0·342. In particular at least a third of the non-trivial zeros of ((s) not only lie on the critical line, but are simple. This observation is due independently to Heath-Brown and Selberg (unpublished). The improved constants ll mentioned above do not all allow this refinement. However it has been shown by Anderson that (10.29.1) holds with ll = 0·3532. 10.30. Levinson's method can be applied equally to the derivatives .;(s) of the function ~(s) given by (2.1.12). One can show that the zeros of these functions lie in the critical strip, and that the number of them, Nm(T) say, for 0 < t :s;; T, is N(T)+Om(logT). If the Riemann hypo-thesis holds then all these zeros must lie on the critical line. Thus it is of some interest to give unconditional estimates for ~~~nfNm(T)-l#{t:O < t ~ T, ,;(i+it) = 0} =llm, say. Levinson , showed that a: 1 ~ 0·71, and Conrey improved and extended the method to give ll1 ~ 0·8137, ~X 2 :;?; G-9584 and in general llm = l+O(m- 2). XI THE GENERAL DISTRIBUTION OF THE VALUES OF {(s) 11.1. IN the previous chapters we have been concerned almost entirely with the modulus of ~(s), and the various values, particularly zero, which it takes. We now consider the problem of '(s) itself, and the values of s for which it takes any given value a. t One method of dealing with this problem is to connect it with the famous theorem of Picard on functions which do not take certain values. We use the following theorom:t If f(s) is regular and never 0 or I in js-s0 j ~ r, and lf(s0)j :::;;; a, then jf(s)j:::;;; A(a,8)for js-s0 j:::;;; Or, where 0 < () < 1. From this we deduce THEOREM 11.1. '(s) takes every value, with one possible exception, an infinityojtimesinanystrip l-0 t0+I, and consider the functionf(s) = g(s)-a}/(b-a) in the circles C, C', of radii !8 and !8 (0 < 8 < I), and common centre s0 = I+!-8+iT. Then )f(•,ll,;;" ~ {W+JSI+ lal)/lb-a), andf(s) is never 0 or I in 0. Hence lf(s)l <A(") in 0', and so ]~(u+iT)] < A(a,b,a) for l,:;:; u,:;:; 1+!8, T > t0+l. Hence ~(s) is bounded for a > l, which is false, by Theorem 8.4 (A). This proves the theorem. We should, of course, expect the exceptional value to be 0. If we assume the Riemann hypOthesis, we can use a similar method inside the critical strip; but more detailed results independent of the Riemann hypothesis can be obtained by the method of Diophantine approximation. We devote the rest of the chapter to developments of this method. t Soo Bohr (1)-(14), Bohr and Courant (1), Bohr and Jesaen (1), (2), (5), Bohr and Landau (3), Boroluleni.us a.nd Jessen (1), JeiiE!f!n (1), va.n Kampen (1), va.n Kampen ll.lld Wintner {1), Kershner (1), Kershner and Wintner (I), (2), Wintner (l)-{4). t See Landau's Ergdmiallf! der Funktiowmtheorie, § 24, or Valiron's Integral Funetiono, Ch. VI,§3. 11.2 DISTRIBUTION OF THE VALUES OF ((s) 293 11.2. We restrict ourselves in the first place to the half-plane a > 1; and we consider, not ~(s) itself, but log"s), viz. the function defined for a > l by the series log((•) ~- t (p-'+!r"+ ... J. We consider at the same time the function fr~l ~ t Jogp(p~+r"+ ... J. We observe that both functions are represented by Dirichlet series, absolutely convergent for a > I, and capable of being written in the form where fn(z) is a power-series in z whose coefficients do not depend on s. In fact fn(z) = zlogpn/(l-z) in the above two cases. In what follows F(s) denotes either of the two functions. 11.3. We consider first the values which F(s) takes on the line u = a 0, where a0 is an arbitrary number greater than I. On this line F(s) = n~l/,(p;o•e-illogp~), and, as t varies, the arguments -tlogp11 are, of course, all related. But we shall see that there is an intimate connexion between the set U of values assumed by F(s) on a = u0 and the set V of values assumed by the function of a.n infinite number of independent real variables ()1, 82, .. We shall in fact show that the set U, which is obviously contained in V, is everywhere denBe in V, i.e. that corresponding to every value v in V (i.e. to every given set of values 81, 82, ... ) and every positive e, there exists at such that jF(a0+it)-vj <e. Since the Dirichlet series from which we start is absolutely convergent for a= a0, it is obvious that we can find N = N(a0 , e) such that L-~+/n(p;ooeh'ip~)j < fe (11.3.1) for any values of the P.n• and in particular for J.ln = ()n, or for J.t .. = -(tlogp.,)/211', 294 THE GENERAL DISTRIBUTION OF Chap. XI Now since the numbers logp" are linearly independent, we can, by Kronecker's theorem, find a number t and integers g1, g2, ••• , UN such that J-tlogp ... -21T8 .. -21fUnl < '1 (n = 1, 2, •.. , N), '1 being an assigned positive number. Since j,..(p;v•e21ri8) is, for each n, & continuous function of 8, we can suppose '1 so small that I J, {f.(p;"••'""·)-/.(p;••e~••"•·)}l < l•· (II.3.2) The result now follows from (11.3.1) and (11.3.2). 11.4. We next consider the set W of values which F(s) takes 'in the immediate neighbourhood' of the line u = u0, i.e. the set of all values of w such that the equation F(s) = w baa, for every positive 0, a root in the strip )a-u0 ) < 8. In the first pla.oo, it is evident that U is contained in W. Further, it is easy to see that U is everywhere dense in W. For, for sufficiently small3 (e.g. for 3 < f(u0-l)), IF'(s) I < K(u0 ) for all values of 8 in the strip )o-ao) <a, so that )F(a0+it)-F(a1+it)) < K(a0))o1-a0 ) ()a1-a0 ) < 3). (11.4.1) Now each value w in W is assumed by F(s) either on the line a = a0, in which case it is a u, or at points a 1 +it arbitrarily near the line, in which case, in virtue of (11.4.1}, we can find au such that )w-u) < K(a0))ol-ao] < ti, We now proceed to prove that W is identical with V. Since U is con-tained in and is everywhere dense in both V and W, it follows that each of V and W is everywhere dense in the other. It is therefore obvious that W is contained in V, if Vis closed. We shall see presently that much more than this is true, viz. that V consists of all points of an area, including the boundary. The following direct proof that Vis closed is, however, very instructive. Let v be a limit-point of V, and let vv (v = 1, 2, ... ) be a sequence of v's tending to v. To each vv corresponds a point Pv(81.v, 92r .. ) in the space of an infinite number of dimensions defined by 0 ~ B ... v < 1 (n = 1, 2, ... ), such that lb(o0, 81..,, ... ) = Vv. Now since (P,.) is a bounded set of points (i.e. all the coordinates are bounded), it has a limit-point P (fJt, e:', ... ), i.e. a point such that from (P,.) we can choose a sequence (P,,) such that each coordinate B..,.,. of P..~ tends to the limit 0: as r-+ oo. 11.4 THE VALUES OF ((s) It is now easy to prove that P corresponds to v, i.e. that ql(a0 , fJ!, ... ) = v, so that v is a point of V. For the series for v..,., viz. .. ~lf .. (p;;a•e2.,.;0....,.), 295 is uniformly convergent with respect to r, since (by Weierstrass's M-test) it is uniformly convergent with respect to all the 8's; further, the nth term tends to j.,(p;;a•e 2'fri0~) as r-+ oo. Hence v = ~vv, = !~~n~1/.,(p;;aoe21ri0.,,) = ql(oo, 8f, ... ), which proves our result. To establish the identity of V and W it remains to prove that V is contained in W. It is obviously sufficient (and also necessary) for this that W should be closed. But that W is closed does not follow, as might perhaps be supposed, from the mere fact that W is the set of values taken by a bounded analytic function in the immediate neighbourhood of a line. Thus e-z' is bounded and arbitrarily near to 0 in every strip including the real axis, but never actually assumes the value 0. The fact that W is closed (which we shall not prove directly) depends on the special nature of the function F(s). Let v = Cl'(oo, 8 1, 82, ... ) be an arbitrary value contained in V. We have to show that vis a member of W, i.e. that, in every strip lu-u0 ) <3, F(s) assumes the value v. Let G(a) = .. ~ 1 J .. (p;;se2'fri0~), so that G(u0) = v. We choose a small circle C with centre u0 and radius less than fJ such that G(s) =I= v on the circumference. Let m be the minimum of IG(s)- vi on C. . Kronecker's theorem enables us to choose t 0 such that, for every s m C, JF(s+it0)-G(s)/ < m. The proof is almost exactly the same as that used to show that U is everywhere dense in V. The series for F(s) and 0(8) are uniformly convergent in the strip, and, for each fixed N, f.J .. (p;oe2.,.,,..) is a continuous function of u, p.1, ... , P.N· It is therefor: sufficient to show that we can choose t0 so that the difference between the arguments of Pfi8 at 8 = ao+ito and p.;•e27110• at 8 = u0, and consequently that 296 THE GENERAL DISTRIBUTION OF Chap. XI between the respective arguments at every pair of corresponding points of the two circles is (mod 21T) arbitrarily small for n = 1, 2, ... , N. The possibility of this choice follows at once from Kronecker's theorem. We now have F(s+i!,)-v ~ {G(s)-v)+{F(s+ii,)-G(s)), and on the circumference of G IF(s+it0)-G(s)l < m.;;; jG(s)-vj. Hence, by Rouche 's theorem, F(s + it0)- v has in C the same number of zeros as G(s)- v, and so at least one. This proves the theorem. 11.5. We now proceed to the study of the set V. Let Yn be the set of values taken by j.,.(p;;8) for u = a0, i.e. the set taken by j,.(z) for [zj = p:;;(J•. Then Vis the 'sum' of the sets of points J-;,, Vz, ... , i.e. it is the set of all values v1+v2+ ... , where v1 is any point ofl{, v2 any point of Vz, and so on. For the function log {(s), V... consists of the points of the curve described by -log(I-z) as z describes the circle lzl = p;o•; for ''(s)/,(8) it consists of the points of the curve described by -(dogp,)/(1-z). We begin by considering the function ''(s)/,(8). In this case we can find the set V explicitly. Let z,,)ogp,. w,=- 1-z, · As zn describes the circle 1z .. 1 = p;; 0 •, w, describesthecirclewithcentre C -_p;;2uologp" n-l-p;;2u0 and radius Let and let p;; 0•logp" Pn= l-p;;2u, • w .. = c .. +w~ = c .. +p .. ei<f~, Then V is the set of all the values of c+ n~lp,.ei<f~ for independent </J1, tfo2, •••• The set V' of the values of I Pn ei<fo~ is the 'sum' of an infinite number of circles with centre at the origin, whose radii p1, p2 , ... form, as it is easy to see, a. decreasing sequence. Let v;. denote the nth circle. 11.5 THE VALUES OF {(s) Then V~ + V~ is the area swept out by the circle of radius Pa a.s its centre describes the circle with centre the origin and radius p1• Hence, since p 2 < p 1 , V; + V~ is the annulus with radii p 1 - p 2 and p 1 + P2· The argument clearly extends to any finite number of terms. Thus V~ + ... + V.N consists of all points of the annulus N N Pt- L Pn ~ lwl ~ L Pn> n~2 n~l or, if the left-hand side is negative, of the circle N lwl .;;;•~•'"' It is now easy to see that (i) if p1 > Pa+Pa+···· the set V' con.sists of all paints w of the annulus Pt- f Pn ~ lwl ~ f p .. ; n~2 n~t (ii) if p1 < p2+Pa+ .. , V' consists of all points w for which lwl ~ .. ~1 p,. For example, in case (ii), let w0 be an interior point of the circle. Then we can choose N so large that • N Nt1 Pn < .. ~/ .. -lwol· Hence lies within the circle v; + ... + VN for any values of the .Pn• e.g. for .PN+t = ... = 0. Hence N WI= n~/"eio/>.,. for some values of cfov· ., tfo,, and so Wo = n~tp .. ei• as required. That V' also includes the boundary in each case is clear on taking all the cfo, equal. The complete result is that there is an absolute constant D = 2·57 determined as the root of the equation 2-Dlog2 = ~ p;;Dlogp, 1-2-W 6 l-p,;-2D ' 298 THE GENERAL DISTRIBUTION OF Chap. XI such that for a0 > D we are in case (i), and for 1 < a0 ~ D we are in case (ii). 'l'he radius of the outer boundary of V' is R ~ ('(2o0) _ ('(o0) {(2a0) {(u0) in each case; the radius of the inner boundary in case (i) is r = 2p1-B = zt-aolog2/(I-2-Za•)-R. Summing up, we have the following results for {'(s)/{(s). THEOREM 11.5 (A). The values whick ns)/{(s) takes on the. line q = cr0 > 1 form a set everywhere dense in a region R(a0 ). If u0 > D, R(u0 ) is the annulus (boundary included) with centre c and radii rand R; if u0 ~ D, R(u0) is the circular area (boundary included) with centre c and radiU8 R; c, r, and Rare continuous functions of a 0 defined by c ~ ('(2o0)/((2o0), R ~ c-('(o0)/((o0), r ~ 2'-••log2((J-2-''•)-R. Further, a8 O"o -+ co, lime= limr =limB= 0, limcfR = Iim(R-r)JR = 0; aa a0 -'J-D, limr = 0; and a8 a0 -+l, IimR =co, lime= {'(2)/((2). THEOREM 11.5 (B). The set of values which {'(s)/{(s) takes in the immedia:U neighbourhood of a= a0 is identical with R(a0 ). In particular, since c tend8 to a finite limit and R to infinity as (10--+ l, nB)/,(B) takea all value& infinitely often in the strip l < a < l +8, for an arbitrary positive 8. The above results evidently enable us to study the set of points at which r(B)/,(B) takes the assigned value a, We confine ourselves to giving the result for a = 0; this is the most interesting case, since the zeros of r(s)f'(s) are identical with those of r(s). THEOREM 11.5 (C). There iB an absolute constant E, between 2 and 3, Buck that r(B) # 0 for 17 > E, while ''(B) kaa an infinity of zerOB in every strip between 17 =I and 17 =E. In fact it is easily verified that the annulus R(170) includes the origin if 170 = 2, but not if a0 = 3. 11.6. We proceed now to the study of log '(s). In this case the set V consists of the 'sum' of the curves V,. described by the points w .. = -log(I-z,.) as z,. descdbes the circle jz,.j = p;;a•. In the first place, V,. is a convex curve. For if u+iv = w =f(z) =f(x+iy), 11.6 THE VALUES OF {(s) 299 and z describes the circle izl = r, then ~+i~ =J'(z){l+i~) =J'(z)xtiy. Hence arctan~= arg{zf'(z)}-p. A sufficient condition that w should describe a convex curve as z describes izl = r is that the tangent to the path of w should rotate steadily through 211 as z describes the circle, i.e. that arg{zf'(z)} should increase steadily through 211. This condition is satisfied in the case f(z) = -log(l-z); for zj'(z) = z/(1-z) describes a circle enclosing the origin as z describes jz I = r < I. If z = reiB, and w = -log(I-z), then u = -llog(I-2rcos8+r2 ), v =arctan rsinB . l-rcos8 The second equation leads to rcos8 = sin41±cosv(r2-sin2v)i. Hence, for real rand 8, lvl ~ arcsinr. If cos81 and cos82 are the two values of cos8 corresponding to a given v, (I-2rcos81+r2)(1-2rcos82+r2) = (I-r2)2. Hence if u1 and u 2 are the corresponding values of u, u1+u1 = -log(I-r2). The curve V,. is therefore convex and symmetrical about the lines u= -!log(l-r2) and v=O. Its diameters in the u and v directions are !log{(l+r)/(1-r)} and arcsinr. Let and Wn = cn+w~, c = ,.~/" = flogC(2a0). Then the points w~ describe symmetrical convex figures with centre the origin. Let V' be the 'sum' of these figures. It is now easy, by analogy with the previous case, to imagine the result. The set V', which iB plainly symmetrical about both axes, is either (i) the region bounded by two convex curves, one of which is entirely interior to the other, or (ii) the region bounded by a Bingle convex curve. In each case the boundary is included as part of the region. This follows from a general theorem of Bohr on the 'summation' of a. series of convex curves. 300 THE GENERAL DISTRIBUTION OF Chap. XI For our present purpose the following weaker but more obvious results will be sufficient. The set V' is included in the circle with centre the origin and radius R = ~ ~logl+p,;-"• = -flog' 2(ao). ,6 2 1-p,;-"• ~(2a0 ) If a0 is sufficiently large, V' lies entirely outside the circle of radius arcsin2-"•~ ~ flogl+p;;"• =arcsin2-"•+!log 1+2-""-R. n~ 1-p;;"• 1-z-ao If and so if u0 is sufficiently near to 1, V' includes all points inside the circle of radius n~t arcsinp;"•. In particular V' includes any given area, however large, if u0 is suffi.~ ciently near to 1. We cannot, as in the case of circles, determine in all circumstances whether we are in case (i) or case (ii). It is not obvious, for example, whether there exists an absolute constant D' such that we are in case (i) or (ii) according as u0 > D' or 1 < u0 ::::;;; D'. The discussion of this point demands a. closer investigation of the geometry of the special curves with which we are dealing, and the qUestion would appear to be one of considerable intricacy. The relations between U, V, and W now give us the following analogues for log {(s) of the results for ''(s)j,(s). THEOREM 11.6 (A). On each line a= u0 > 1 the values of log '(s) are everywhere dense in a region R(u0) which is either {i) the ring-shaped area. bounded by two convex curves, or (ii) the area bounded by one convex curve. For sufficiently large values of a0 we are in case (i), and for values Of u0 sufficiently near t.o 1 we are in case (ii). THEOREM 11.6 (B). The set of values whieh log '(s) tam in the immediate neighbourhood of a = u0 is identical with R(u0). In particular, since R(u0 ) includes any given finite area when u0 is sufficiently near 1, log {(s) takes every value an infinity of times in 1 < a < 1+8. As a consequence of the last result, we have THEOREM 11.6 (C). the function '(s) tam every value except 0 an i?tfi.nity of times in the strip 1 < a < 1 +~l. This is a more precise form of Theorem 11.1. 11.7 THE VALUES OF ((s) 301 11.7. We have seen above that log {(s) takes any assigned value a an infinity of times in a > 1. It is natural to raise the question how often the value a is taken, i.e. the question of the behaviour for large T of the number Ma(T) of roots of log{(s) =a in a> 1, 0 < t < T. This question is evidently closely related to the question as to how often, as t-+oo, the point (a1 t,a2 t, ... ,aNt) of Kronecker's theorem, which, in virtue of the theorem, comes (mod 1) arbitrarily near every point in the N -dimensional unit cube, comes within a. given distance of a.n assigned point (bvb2, ... ,bN). The answer to this last question is given by the following theorem, which asserts that, roughly speaking, the point (a1 t, ... , aNt) comes near every point of the unit cube equally often, i.e. it does not give a preference to any particular region of the unit cube. Let ~ .... , aN be linearly independent, and let y be a region of tk N-dimensional un~t cube with volume r (in the Jordan sense). Let Iy(T) be the sum of the tntervals between t = 0 dnd t = T for which the point P (a1 t, ... ,aNt) is (mod1) inside y. Then }.~ Iy(T)JT = r. The region yissaid to have the volume r in the Jordan sense if given "• we can find two sets of cubes with sides parallel to the axes, of v~lumes rl and r2, included in and including y respectively, such that rl~(:::;;; r::::;;; r2+e. If we call a point with coordinates of the form (a1t, ... ,aNt), modi, an 'accessible' point, Kronecker's theorem states that the accessible points are everywhere dense in the unit cube 0. If now Yv y2 are two equal cubes with sides parallel to the axes, and with centres at accessible points P1 a.nd P2, corresponding to t1 and t2, it is easily seen that lim~(T)/-\,,(T) ~ I. ~or (a1_t, ... ,aNt) will lie inside y2 when and only when {a1(t+t2~t1 ), .. } hes mstde y1• Consider now a set of p non-overlapping cubes c, inside O, of side f, ~ach of which has its centre at an accessible point, and q of which lie Inside y; and a set of Poverlapping cubes c', also centred on accessible Points, whose union includes C and such that y is included in a union of Q o~ them. Since the accessible points are everywhere dense, it is POsstble to choose the cubes such that q/P and Qjp are arbitrarily near to r. Now, denoting by ~lc( T) the sum of t-intervals in (0, T) corresponding to the cubes c which lie in y, and so on, t l,(T)/fi I,(T).; I,v:).; t I,(T)jfi J,(T). 302 THE GENERAL DISTRIBUTION OF Chap. XI Making T -+ oo we obtain ~ ~ ~ l~c:) ~ ~· a.nd the result follows. 11.8. We can now prove THEOREM ll.S (A). If u = u0 > 1 is a line on which log ~(s) comes arbitrarily near to a given number a, then in every strip u0-0 < u < u 0 +8 the valm! a is taken more than K(a, u 0, O)T times ,for large T, in 0 < t < T. To prove this we have to reconsider the argument of the previous sections, used to establish the existence of a root of log ((s) = a in the strip, and use Kronecker's theorem in its generalized form. We saw that a sufficient condition that log ((s) = a may have a root inside a circle with centre u0 + it0 and radius 28 is that, for a certain N and corresponding numbers Op···· ()N, and a certain 'I= q(n0 , J, Op···· ON) 1-tologp.,-2w0.,-2?rg11 1 < 71 (n = 1, 2, ... , N). From the generalized Kronecker's theorem it follows that the sum of the intervals between 0 and T in which to satisfies this condition is asymptotically equal to (7J/21t)NT, and it is therefore greater than H7J/21T)NT for large T. Hence we can select more than M7Jf21T)NTf3 numbers t~ in them, no two of which differ by less than 43. If now we describe circles with the points o0+it;. as centres and radius 23, these circles will not overlap, and each of them will containazerooflog t(s)-a. This gives the desired result. We can also prove THEOREM 11.8 (B). There are p08itive constants K1(a) and K2(a) such that the number Ma(T) of zer08 of logt(s)-a in u >I satisfies the inequalities Kl(a)T < Ma(T) < K2(a)T. The lower bound follows a.t once from the above theorem. The upper bound follows from the more general result that if b is any given cr:mBtani, the number of zeroa of ((s)-b in a> !+3 (3 > 0), 0 < t < T, is O(T) as T-+co. The proof ofthis is substantially the same as that of Theorem 9.15 (A), the function ((s)-b playing the same part as t(s) did t}J.ere. Finally the number of zeros of log((s)-a is not greater than the number of zeros of ((s)-ea, and so is O(T). 11.9. We now turn to the more difficult question of the behaviour of {(s) in the critical strip. The difficulty, of course, is that C(s) is no 11.9 THE VALUES OF '(s) 303 longer represented by an absolutely convergent Dirichlet series. But by a. device like that used in the proof of Theorem 9.I7, we are able to obtain in the critical strip result,s analogous to those already obtained in the region of absolute convergence. As before we consider log {(s). For a ~ I, log t(s) is defined, on each line t = constant which does not pass through a singularity, by con· tinuation along this line from u > I. We require the following lemma. LEMMA. If f(z) is regular for )z-z01 ~ R, awl f f I/(•) I' dxdy ~ H, 1£-.ftl<i;R then I/(•) I,; f(~u~~ (lz-•01.;; R' < R). For if lz' -z01 ~ R', {W)}' ~ !. I {/(•)}' dz ~-'..I'' {f(z' +n"))' dO. 2m z-z' 211 le-£'1~1' o Hence R-R' R-R' 2,. lf(z')l 2 I rdr ~!_I Ilf(z'+reiO)I 2rdrd8 ~ !!, 271 21t ' ' ' and the result follows. THEOREM II.9. Let u0 beafixednumberin the range l 0, are everywhere dense in the whole plane. Let This function is similar to the function {(s)Mx(s) of Chapter IX, but it happens to be more convenient here. Let 3 be a positive number less than }(u0-}). Then it is easily seen as in§ 9.19 that for N ~ N0(u0, E), T ~ T0 = T0(N), T f I{N(a+it)-11 2 dt < ET uniformly for u0-3 ~ u ~ o1+3 (o1 >I). Hence T 0"!.+8 Hence I I i{N(a+it)-1)2dudt < (u1-o0+23)ET. 1 a0 -ll v+l- a1+1l J J J{N(o+it)-II2 dodt < (o1-a0+23)""E Y-l- ao-ll 304 THE GENERAL DISTRIBUTION OF Chap. XI for more than (1-.je)T integer values of v. Since this rectangle contains the circle with centre s = u+it, where o-0 :::;,; a:::;: n 1 , v-!+0 ~ t ~ v+!-0, and radius 0, it is easily seen from the lemma that we can choose fJ and € so that given 0 <'I < 1, 0 < q' < 1, we have I~N(u+it)-1] < 7J (u0 ~a~ u1) (11.9.1) for a set of values oft of measure greater than (l-7J')T, and for N ~ N0(a, 11• '1'), T ~ T0(N). Let RN(') ~-I Log(I-p,;-•) (u > !), N+' where Log denotes the principal value of the logarithm. Then (N(s) ~ exp{RN(sl). We want to show that RN(s) = Log{N(s), i.e. that JIRN(s)J < ftr, for u ~ 110 and the values oft for which (11.9.1) holds. This is true for u = a1 if u1 is sufficiently large, since JRN(s)]-+ 0 as a1 -+ oo. Also, by (11.9.1), R{N(s) > 0 for a0 .::::;;: a::::;;; a1, so that IRN(s) must remain between -}11 and -fn' for all values of u in this interval. This gives the desired result. We have therefore IRNI•II ~ jLog[I+{{NI•)-1})1 < 2 i1NI•I-ll < 2~ for u0 z::;;: u z::;;: u10 N ~ .N'o(a0, 'tJ, '1'), T ~ T0(N), in a set of values of t of measure greater than (1-'tJ')T. Now consider the function N FN(u0+it) = -... ~/og(I-p,;-ao-il), and in conjunction with it the function of N independent variables N 41N(610 ... ,8N) =-I log(l-p,;-a•ez'lfiB.). ,, Since I p;a• is divergent, it is easily seen from our previous discusSion of the values taken by log{(s) that the set of values oflllN includes any given finite region of the complex plane if N is large enough. In particular, if a is any given number, we can find a. number Nand values of the 8's such that We can then, by Kronecker's theorem, find a number t such that JFN(u0+it)-aJ is arbitrarily sma.ll. But this in itself is not sufficient to prove the theorem, since this va]ue of t does not necessarily make JRN(s)J amaJl. An additiona.l argument is therefore required. 11.9 THE VALUES OF {(s) 305 Let 4tM.N = -~ log(l-p;aoe2'1Ti9,.) = ~ ~ p;;mrroe'twim9 •• .. ~]?'+~ L L m n~M+l m~l Then, expressing the squared modulus of this &a the product of con-jugates, and integrating term by term, we obtain ! I . . ! I~M.NI' dOMH·· dON ~ .~t J;, P:";," N "' 1 "' < L p;'"' L ;;;><A L p;'"', n~M+l m~l n-M+l which can be made arbitrarily small, by choice of M, for all N. It therefore follows from the theory of Riemann integration of a con-tinuous function that, given o:, we can divide up the (N-M)-dimensional unit cube into sub-cubes q~, each of volume.\, in such a way that .\ .f m 11 ~x J41M.Ni2 < !o:2. Hence forM~ M0(o:) and any N > M, wecanjind cubes of total volume greater than lin which j4tM.NI < o:. We now choose our value oft as follows. (i) Choose M so large, and give B;,, ••. , trM such values, that 4tM(B;., ... ,O'M)=a. It then follows from considerations of continuity that, given t:, we can find an M-d.imensional cube with centre e;, ... , O'M and side d > 0 throughout which (ii) We may also suppose that M has been chosen so large that, for any value of N, 14tM.Nl To(N), the inequality JRN(s)j -ldMT if Tis large enough. There are 306 THE GENERAL DISTRIBUTION OF Chap. XI therefore values of t for which the point lies in one of these cubes, and for which at the same time JRN(s)j < lE. For such a value oft llog((•)-al <;; iFN(s)-ai+IRN(•)I and the result follows. <;; lM(8,, ... , BM)-•I+IM,NI+IRN(•)I < lE+!E+!E = E, 11.10. THEOREM 11.10. Let l <a< p < 1, and let a be any complex number. Let ~a,,e(T) be the number of zeros of log "s)-a (defined as bef01'e) in the rectangle a: <a< {J, 0 < t < T. Then there are positive constants K1(a,cx.,{J), K 1(a,a,f3) suck that K,(a,a,fi)T < M., •. p(T) < K,(a,a,fi)T (T > T,). We first observe that, for suitable values of the (J's, the series - ... ~~log(I-p;-•e 2'""') is uniformly convergent in any finite region to the right of a = }. This is true, for example, if (}n = in for sufficiently large values of n; for then I p;;'e"'"· ~ I (-I)"p;;', .. >n. n>n1 which is convergent for real 8 > 0, and hence uniformly convergent in any finite region to the right of the imaginary axis; and for any O's ~ )p;•e27ri8•) 2 = IP;;2a is uniformly convergent in any finite region to the right of u = f. H a is any given number, and the 8's have this property, we can choose n1 so large that 1-.~t,Iog(I-p;;'e'""·)l < < (u ~ !(a+fi)), and a.t the same time so that the set of values of --~ 1 log(l~p;;icx--i,6e21rlll.) includes the circle with centre the origin and radius )a)+ )E). Hence by choosing first 8,.,+1, ... , and then 810 ••• , 81 ", we can find values of the 8's, sa.y e;,, 8',., ... , such that the series G(s) = - .. ~ 1 log(I-p;;•elori6~) is uniformly convergent in any finite region to the right of a = l, and G(!a+tfi) ~ a. 11.10 THE VALUES OF ((s) 307 We can then choose a. circle 0 of centre !o+l,8 and radius p < l(,B-a} on which G(s) of= a. Let m =min )G(s)-a). BODO Now let Then, as in the previous proof, f ... j I I it11M.N(s))2 d8M+1 .. dON dadt <A ll~ 1p;;za. 0 0 ls-}cx-l,61.;;:l(,6-<>l Hence for M ;;;::, M0(E) and any N > M we ca.n find cubes of total volume greater than ! in which I I )41M.N(s))2 dadt < E 1~-ia:-l,Bio:;i(ft-ot) and so in which (by the lemma of§ 11.9) l.,,N(•)I < 2(</n)l(fi-a)-1 (1•-!a-!fil <;; !(fi-a)). We also want a little more information about RN(s), viz. that RN(s) is regular, and )RN(s)) < 'IJ, throughout the rectangle for a set of values of t0 of measure greater than (l-7J')T. As before it is sufficient to prove this for {N(s)-1, and by the lemma it is sufficient to prove that jj 10+1 ~(t,) ~ I du I I{N(•)-11' dt < ' « 10-1 for such fo, by choice of N. Now T fJ T lo+l I f(t,) dt ~ I du I dt, I I{N(•)-11' dt 1 .. 1 to-t ,6 T+l t+l fJ T+l :<.:;;Ida [ i{N(s)-1)2 dt 1L dt0 = 2 Ida j ){N(s)~ 1)2 dt < ET by choice of N as before. Hence the measure of the set where "'(to) > "'!E is less than .JET, and the desired result follows. 308 THE GENERAL DISTRIBUTION OF Chap. XI It now follows as before that there is a set of values oft0 in (0, T), of measure greater than KT, such that for IB-ia::-1,81 ~ }(,8-o:) [,.,~ 1 Iog(I-p,;-gezmfr..)- .. ~ 1 log(I-p;~-u·)/ < !m, io!>M.N(s)j < !m, a.nda.lso IRN(s+it0)1 < lm. At the same time we can suppose that M has been taken so large that !G(s)+ n~llog(I-p;se2"i8;.)1 < !m (a;;;::: o:). Then jlog((s)-G(s)l < m on the circle with centre }a+l.B+it0 and radius p. Hence, as before, log {(a)-a has at least one zero in such a circle. The number of such circles for 0 < t0 < T which do not overlap is plainly greater than KT. The lower bound for Ma,a,fl(T) therefore follows; the upper bound holds by the same argument as in the case a > l. It has been proved by Bohr and Jensen, by a more detailed study of the situation, that there is a K(a,tX,/3) such that M.,.1(T) ~ K(a, ,,p)T. An immediate corollary of Theorem 11.10 is that, if Na,<:<,f>( T) is the number of points in the rectangle l < a < u < {j < 1, 0 < t < T where ((a) ~ a (a # 0), then N.,,.1(T) > K(a,,,p)T "(T > T0 ). For {(s) =a if Iog{(s) =log a, any one value of the right-hand side being taken. This result, in conjunction with Theorem 9.17, shows that the value 0 of {(s), if it occurs at all in a > !, is at any rate quite exceptional, zeros being infinitely rarer than a-values for any value of a other than zero. NOTES FOR CHAPTER!! 11.11. Theorem 11.9 has been generalized by Voronin , , who obtained the following 'universal' property for ((s). Let D, be the closed disc of radius r < t. centred at s = f, and let {(s) be any function continuous and non-vanishing on Dr, and holomorphic on the interior of Dr. Then for any e > 0 there is a real number t such that maxl((s+it)-f(s)l <e. (11.11.1) BED, 11.11 THE VALUES OF Wt) 309 It follows that the curve y(t)=(C(a+it),C'(a+it), ... ,(<n-ll(a+it)) is dense inC", for any fixed a in the range!< a< 1. (In fact Voronin establishes this for a = 1 also.) To see this we choose a point z = (z0 , Z 1 , ••• , z,. 1)withz0 1-0, and takef(s)to be a polynomial for which pm>(a) = zm for 0 ~ m < n. We then fix an R such that 0 < R < t -Ia- fl, and such that f(s) is nomranishing on the closed disc ls-ai~R. Thus, if r=R+Ia--fl, the disc D~ contains the circle Is -a I = R, and hence (11.11.1) in conjunction with Cauchy's inequality m' lg(zo)l ~ ~ lz~~I~R lg(z)l, yields l((a+it)-zml ~ ~e (0 ~ m < n). Hence y(t) comes arbitrarily close to z. The required result then follows, since the available z are dense inC". Voronin's work has been extended by Bagchi [I] (see also Gonek [I]) so that D, may be replaced by any compact subset D of the strip! < R(s) < 1, whose complement inC is connected. The condition on f is then that it should be continuous and non-vanishing on D, and holomorphic on the interior (if any) of D. From this it follows that if I) is any continuous function, and h 1 < h 2 < ... < hm are real constants, then ((s) cannot satisfy the differential-difference equation l){((s+hl), C'(s +hl), ... , c<n,)(s+hl), C(s+h2). C'(s+h2), ... , ( '"''(s+ h,) •... } ~ 0 unless I) vanishes identically. This improves earlier results of Ostrowski [I] and Reich [I]. ll.I2. Levinson has investigated further the distribution of the solutions Pa = Pa + iya of C(s) =a. The principal results are that # {pa:O :<.>; Ya ~ T} =-[;log T+O(T) and #{Po' 0" '" <; T.IP, -l:l ;>b)~ O,(T) (b > 0). Thus (c.f. § 9.15) all but an infinitesimal proportion of the zeros of ns)- a lie in the strip f- t5 < a < f + 0, however small b may be. 310 THE GENERAL DISTRIBUTION OF Chap. XI In reviewing this work Montgomery (Math. Reviews 53# 10737) quotes an unpublished result of Selberg, namely I <P. -v-1 1 T(loglogT)l. O,;;:J· • .;;T 47t (11.12.1) il.:<l This leads to a stronger version of the above principle, in which the infinite strip is replaced by the region l•-!l < ~(t)(loglogt)l, logt where rjl(t) is any positive function which tends to infinity with t. It should be noted for comparison with (11.12.1) that the estimate L <P. -j) ~ O(logT) 00:: ~ • .;; T is implicit in Levinson's work. It need hardly be emphasized that despite this result the numbers p a. are far from being symmetrically distributed about the critical line. 11.13. The problem of the distribution of values of((!+ it) is rather different from that of ((u +it) with!< a< 1. In the first place it is not known whether the values of ((!+it) are everywhere dense, though one would conjecture so. Secondly there is a difference in the rates of growth with respect tot. Thus, for a fixed u >},Bohr and Jessen (1), (2) have shown that there is a continuous function F(z; u) such that ~m{tE [-T, rpog((a+it)ER)- UF(x+iy;a)dxdy (T-oo) for any rectangle R c {; whose sides are parallel to the real and imaginary axes. Here, as usual, m denotes Lebesgue measure, and. log((s) is defined by continuous variation along lines parallel to the real axis, using (1.1.9) for a > 1. By contrast, the corresponding result for u = ! states that -'-m{te[-T T]' log((}+it) eR}--'-ffe-Osuch that m{te[O,T):i((}+it)I~V}~Texp(-A log2V ). loglogT uniformly for 1 ~ V ~ log T . XII DIVISOR PROBLEMS 12.1. Tn divisor problem of Dirichlet is that of determining the asymptotic behaviour as x-+ oo of the sum D(x) ~ .~.d(n), where d(n) denotes, as usual, the number of divisors of n. Dirichlet proved in an elementary way that D(x) ~ xlogx+(2y-I)x+O(xl). (12.1.1) In fact ~ [ ,lx]'+2 L ([~]-[ -'xJ) m.;;V:r; ~ 2 L [~]-[-'xJ' m.;;V,z ~ 2 L ~~+0(1))-{-'x+O(I)}' m.;;"x ~ 2x{Iog-'x+y+O(x-l))+O(,Ix)-{x+O(,Ix)), and (12.1.1) follows. Writing D(x) ~ xlogx+(2y-l)x+t.(x) we thus have 6.(x) = O(xl-). (12.1.2) Later researches have improved this result, but the exact order of A.(x) is still undetermined. The problem is closely related to that of the Riemann zeta.~function. By (3.12.1) with a11 = d(n), s = 0, T-+ oo, we have I <I+<• x• D(x) ~ ~ ('(w)- dw (o > 1), 2m w c-iao provided that xis not an integer. On moving the line of integration to the left, we encounter a double pole at w = 1, the residue being xlogx+(2y-l)x, by (2.1.16). Thus A.(x) = __!., c'I+i<:<> '2(w)~dw (0 < c' < 1). 2m c'-ioc W 12.1 DIVISOR PROBLEMS 313 The more general problem of D,(x) ~ I d,(n), •<• where dk:{n) is the number of ways of expressing n as a. product of k factors, wa.s also considered by Dirichlet. We have I "I+<• x• D,(x) ~ 2,;i ('(w)wdw (o > 1). <-<-Here there is a. pole of order k a.t w = 1, and the residue is of the form xPk(logx), where P.1: is a. polynomial of degree k-1. We write D,(x) ~ xP,(logx)+t.,(x), (12.1.3) so tha.t aa(x) = a(x). The classical elementary theoremt of the subject is ak(x) = O(x1-llkiogk-2x) (k = 2, 3, ... ,). (12.1.4) We have already proved this in the case k = 2. Now suppose tha.t it is true in the case k-1. We have D,(x) ~ I I ~ I d,,(n) n,,.. .•• nl.;;;:c mn,;;x ~ mb'" nbim d,,(n) + x'"l,« n<~m d,,(n) = mbllln "lxtm dk-l(n)+ n.;;;~-tfl dk-l(n) :ctll<o;;;:c/n 1 ~ L 4-.(~)+ L ~~-x"'+O(I))a,,(n) m,;;:cill n.;;;:c•-•ll = L Dk-t(~)+x L dk-~(n)_xllkJ1,_t(x1-1fk)+ m.;;;x•J& no;;x•-•Jl +O{D,,(x•->~')}. Let us denote by P.t::(z) a. polynomial in z, of degree k-1 at moat, not always the same one, Then " log'-'m (I ~)+ o(log'-•fl ~ ----;n- = Pk og -~--j" Hence L ~pk-t(~) = xpk(Iogx)+O(xl-I/klogk-2x). m<;;:clll Also ';:' t. (~) ~ o(x'-"'-''Io-'-'x ~ -1 -) L.. k-t m o mt-1/(k:-I) m..:zl/1 m.;;:c/l = O{xi-1/(k-t)}ogk-sx.xlf{k(.l:-l)J} = 0(x1-lfklogk-3x). t See e.g. Landlr.u (6). 314 DIVISOR PROBLEMS Chap. xn The next term is x """ 4-1(n)-J_1(n-l) = x 2: Dk_1(n)+ xJ_1(N), n<;;7.,1a n n.;;"'•-•ran(n+I) N+l where N = [x1-l./1c], Now x "';;;;:' P,.1(logn)+xNPk-t(logN) = xp (logx)+O(xliklog"'-2x) ,.,..~, 1~ n+l N+l k = Cx-x 0 -- +O(xN-l/lk-llJogk-3N) L (log'-'n) nl+l/(k-tJ n>:t•->1! = Ox+O(x1-ll"log"'-ax). Finally xllk Dk_1(xt-I/k) = x11k{xt-t!kp,._1(1og xt-tlk) + O(,xlt-t/k}{t-t,~k--tl) Iogk-Sx)} = xp,._t(logx)+O(xt-1/k}og"'-sx). This proves (12.1.4). We may define the order ak of llk(x) as the least number such that .6.k(x) = O(xcr..t+<) for every positive E. Thus it follows from (12.1.4) that C(k ~k-;;1 (k = 2, 3, ... ). (12.1.5) The exact value of r:xk has not been determined for any value of k. 12.2. The simplest theorem which goes beyond this elementary result is THEOREM 12.2, t (Xk ~ !~~ (k = 2, 3, 4, ... ). Take an= dk(n), ljJ(n) = n~, a:= k, 8 = 0, and let x be half an odd integer, in Lemma. 3.12. Replacing w by s, this gives D,(x) ~~IT {'(s)~ds+O(T(c,.1)')+o(x;') (c > 1). t Voronol (1), Landau {5). 12.2 DIVISOR PROBLEMS 315 Now take the integral round the rectangle -a-iT, c-iT, c+iT, -a+iT, where a > 0. We have, by (5.1.1) and the Phragmen-Lindelof principle, {(8) = O(tca+lx~-o-m .. +el) in the rectangle. Hence ~+iT ~ ,.LT {k(s)~ds = o(l pk< .. +txc-o-){(a-t<>)-lxoaa) = O(Tk f xk(s) ::':as n~l -a-iT nl-B 8 T = ix-a .! dk(n~ f xk( -atQ (nx)~ dt. n~l nH"" -T -a+~t x( -a+ it)= Ce-itlogt+Uiog2.,.+itta+i+O(t"-i) and I -~ + o(~) -a+it-it t2 · The corresponding part of the integral is therefore T -iOk f eikt(-logt+lotr2w+ll(nx)itt<a+llk-1 dt+O(T 1. This integral is of the form considered in Lemma. 4.5, with F(t) = kt(-logt+log27T+l)+tlognx. Since F"(t) = -~.:::;;; -~, the integral is 316 DIVISOR PROBLEMS Chap. XII uniformly with respect ton and x. A similar result holds for the integral over (-T, -1), while the integral over (-1, I) is bounded. Hence <'>,(x) ~ o(T(c":l)')+o(x;·)+or·:•-')+ +x-a% ~~~ O(T<a+flk-l) _ (-"' ) (x'") (T<••l~-l) - 0 T(c-1)' +0 T +0 x" • Taking c = 1+£, a= E, the terms are of the same order, a.pa.rt from E's,if Hence The restriction that x should be half an odd integer is clearly unnecessary to the result. 12.3. By using some of the deeper results on {(s) we can obtain a. still better result for k ? 4. THEOREM 12.3.t cxk ~ ~~! (k = 4, 5, ... ). We start as in the previous theorem, but now take the rectangle as far as a = ! only. Let us suppose that m+it)~o(t'). Then {(s) = O(t~(c-u)J(e-fl) uniformly in the rectangle. The horizontal sides therefore give o(f pk,(c-all-l:t"" da) = O(Tk>.-Ixi)+O(T-lx"). l f+iT T Also J ''(s)~ds ~ O(xl)+O(xl J lm+it)l'~)· i-iT 1 Now T T f lm+it)l'dt,;; max lm+it)l'-•f lm+it)l'dt 1 t l.;;t.;;T 1 t ~ o(T<'-<>' {lm+it)l'~)· t Hardy and Littlewood (4). 12.3 DIVISOR PROBLEMS Also ~(T) ~! I"Hit)i' dt ~ O(T'"), by (7.6.1), so that f T I m+it)l'dt ~ fT ~'(t) dt ~ r~(')]T + fT TJ!)_dt t t t 1 t2 ' ' ' ~ O(T')+O(f ,bdt) ~ O(T•). !+iT Hence J Ck(s)~ds = O(xi)+O(xi-Tlk-4);+•). i-iT Altogether we obtain Ak(x) = O(T-1x")+O(xlTk;-t)+O(xiTlk-~J).H). 317 The middle term is of smaller order than the last if i\ ~ l· Taking c = I +e, the other two terms are of the same order, apart from E's, if T=xl/{2(k---').+2). This gives Ak(x) = O(x!12(k-4J).+l)/{2(k-4J).+2HH). ~aking A= l+E (Theorems 5.5, 5.I2) the result follows. Further slight Improvements for k ~ 5 are obtained by using the results stated in § 5.18. 12.4. The above method does not give any new result for k = 2 or k = 3. For these values slight improvements on Theorem 12.2 have been made by special methods. THEOREM 12.4. t a:2 ::s;; ~· The argument of§ 12.2 shows that t.(x) ~ ~ ~ d(n) -•J•iT x'(•) x'ds+o('I"<')+o(x') 2m~ nt, s xa T (I2.4.I) n-l -a-iT where a> 0, c > I. Let T 2/(41T2x) = N +f, where N is an integer and consider the terms with n > N. As before, the integral over I ~ / ~ T is of the form T x~l+« f eiF(ll{t2a+O(t2a-I)} dt, 02.4.2) t van der Corput (4). 318 where DIVISOR PROBLEMS F(t) = 2t(-logt+log27T+l)+tlognx, F'(t) = log 471~. Hence F'(t) ~log N:·r and (12.4.2) is "":.~(0(tog{n~;H)})+O(T'"l)· For n ;;<: 2N this contributes to (12.4.1) o(~ ~ d(n)) ~ O(N') xa ~ n1 +~~ ' fi.~2N and for N < n < 2N it contributes Chap. xn o(T'" 'N d(n) ) - o(N• ~ ~) ~ O(N•). --;;a n=~+l nt+alog{n/(N+!-)} -£=1m Similarly for the integral over - T :::;. t ~ -1; and the integral over -1 < t < 1 is clearly O(x-a). Ifn~N,wewrite -oHT io - ( if• + -f<T + -of-iT + ifT ). f ~ f -a-iT -ioo iT -i = _ ! J cos2 !w7Tr(w)r(w-1){27T"(nx)}2-2wdw ••' 1-ioo ~ -4i J(~)[K,{4•-'(nx))+ ... Y,{4n"(nx))] in the usual notation of Bessel functions. t The first integral in the bracket is I •"'''( A+f+O(t-'))dt ~ o{log{(~+!)/n) }· whioh gives ~ d(n) O(N') ,6 nlog{(N+!l/•} t See, e.g., Titchmanih, Fouriul~. (7.9.8), (7.9.11). DIVISOR PROBLEMS 319 as befote; and similarly for the second integral. The Ia.st two give ~~d~lf(~)"da)~o(~d~l((S)")~o((~)")-~ Altogether we have now proved that ~ t.(x) ~ _ 2 :• ~ d~:)[K,{4n"(nx))HnY,{4n~(nx))]+O(~)+o(~)· ' (12.4.3) By the usual asymptotic formula.et for Bessel functions, this may be replaced by d(x) = wx: 2 ~ d~~) cos{4w"(nx)-f1r}+O(x-i)+O(;)+o(;). (12.4.4) Now Consider the sum 2 e'.,.'"'(mn"'), !Nim<n<;',N/m We apply Theorem 5.13, with k = 5, and f(n) ~ 2~(mnx), f'''(n) ~ A(mx)ln-1. Hence the sum is o(l!( (mx)l )+,) o((l!)'((N/m)l)io) m (NJm)f + m (mx)f ~ O{(Nfm)H(mx)n)+O((Nfm)H(mx)--.T). Replacing N by iN, !N, ... , and adding, the same result holds for the sum over 1 ~ n ~ Nfm. Hence the first term on the right of(12.4.5) is O(NHxi< mt,N m--t)+O(Nth:--.T m];N m--H) ~ O(Ntlxi<)+O(Nt!ri>). Similarly the second inner sum is 0{(-'N)H(mx).h}+O{(-'N)tl(mx)-h), and the whole sum is 0( Nftx-h- m~N m-h) + 0( Nfx--n m~IN m-,-) ~O(Nttxi<)+O(Nffirio). t Watson, Theory of BUHI. Funoeions, H 7.21, 7.23. 320 DIVISOR PROBLEMS Hence, multiplying by e-li .. and taking the real part, .td(n)co•{4-rr,/(nx)-v} ~ O(NHxn)+O(NI!,.,). Using this and partia.l summation, (12.4.4) gives Chap. XII t.(x) ~ O(NH-ixl+h)+O(Nl!-lxl-h)+O(N")+O(N-!,....) ~ O(Niln'<)+O(Ntlxh)+O(N")+O(N-lx'-l). Taking a= E, c = l+t:, the first and last terms are of the same order, apart from t:'s, if N = [xH]. Hence .6.(x) = O(xHH), the result stated. A similar argument may be applied to .6.3(x). We obtain xl "<;' d (n) , ("'") .6.3(x) = 7 L. -Tcos{&rr(nx)a}+O T , '" 3 nz) n (12.4.6) and deduce that The detailed argument is given by Atkinson (3). H the series in (12.4.4) were absolutely convergent, or if the terms more or less cancelled each other, we should deduce that a 2 ,:;;; !; and it may reasonably be conjectured that this is the real truth. We shall see later that a2 ;;;:. l, so that it would follow that a:1 = l· Similarly from (12.4.6) we should obtain a:3 = !-; and so generally it may be conjectured that k-1 CX.k=~· 12.5. The average order of6.k(x). We may define flk, the average order of 6.k(x), to be the least number such that ~ j t.l(y) dy ~ O(x•fh+<) for every positive E. Since ~ f 6l(y) dy ~ ~ f O(y'"'+<) dy ~ O(x'••+<), we have flk ::;;;;. rxrc: for each k. In particular we obtain a. set of upper bounds for the fl~o from the above theorems. As usual, the problem of average order is easier than that of order, and we can prove more about the fl~o than about the ""k· We shall first prove the following theorem. t t Titchmarsh (22). 12.5 , DIVISOR PROBLEMS 321 THEOBEM 12.5. Let Y~o be the lower bound of positive numbers a for which Then flrc: = 'Yk; and J I"' ){(a+it))llk dt =I""( ) ~2a-1 d 21T_"' Ja+itj2 o ukxx x provided that u > flk· c+iT We have Dk(x) = -2 1 . Jim J (k(a) x" d8 (c > I). mT ...... oo a c-iT (12.5.1) (12.5.2) Applying Cauchy's theorem to the rectangle y-iT, c-iT, c+iT, y+iT, where y is less than, but sufficiently near to, 1, and allowing for the residue at a = 1, we obtain ")'+iT 6.k(x) =~lim J (k(a) x" da. (12,5.3) 2mp_,."' s y-iT Actually (12.5.3) holds for Y~o < y < 1. Fort (k(a)js-+ 0 uniformly as t--+ ±co in the strip. Hence if we integrate the integrand of (12.5.3) round the rectangle y' -iT, y-iT, r+iT, y' +iT, where 'Yk <y' <y < 1, and make T-+ oo, we obtain the same result withy' instead of y. If we replace x by ljx, (12.5.3) expresses the relation between the Mellin transforms /(•) ~ 6,(1/•), il'(•) ~ ''(•)/•, the relevant integrals holding also in the mean-square sense. Hence Parseval's formula for Mellin transforms+ gives 1 I• l"r+it)l" I• (I) J• 2;;: lr+it)ll dt = di X xlly-1 dx = 6.f(x)x-2y~l dx provided ;:at Yk < y < l. o o (12.5.4) It follows that, if 'Yk < y < 1, X I 6.f(x)x-2 Y-1 dx < K = K(k, y), !X X I 6.t(x) dx < KXlly+1, !X t Byana.pplica.tionofthelemma.of§ 11.9. t Soo Tit.:Jhma.rsh, Theory of Fourier Integrol8, Theorem 71. 322 DIVISOR PROBLEMS and, replacing X by fX, !X, ... , and adding, X [ ai(x) dx < KX!r+l. Hence 13.~: ~ y, and so flk :::;;; Yk· The inverse Mellin formula is 'k;a) = j ak(~)xs-1 dx = j t.k(x)x---8-l dx. ' ' Cho.p.XD (12.5.5) The right-hand side exists prima.rily in the mean-square sense, for Y.t <a < 1. But a.otua.lly the right-hand side is uniformly convergent in any region interior to the strip fl.t < a < 1; for X X X )l f I A.~;(x)lx-a--1 dx ~ { J a:(x) dx I x-Zo-Z dx tx !X IX = {O(X2P!+tH)O(X-2<1-t)}i = O(Xfh-V+€), and on putting X = 2, 4, 8, ... , and adding we obtain J IA~c(x)lx-a-t dx < K. It follows tha.t the right-hand side of (12.5.5) represents an analytic function, regular for f1.~: < u < 1. The formula therefore holds by analytic continuation throughout this strip. Also (by the argument just given) the right-hand side of (12.5.4) is finite for f3k < r < 1. Hence so is the left-hand side, and the formula. holds. Hence 'i'k ::::;;; f1", and so, in fact, 'i'k = p". This proves the theorem. 12.6. THEOREM 12.6 (A).t ~.-;, k-;/ (k ~ 2, 3, ... ). Iff< a< 1, by Theorem 7.2 Hence c.T< fl((a+it)i'dt.;;( fi((a+it)i"dt)"'( fdot"'. iT iT i-T T J )t(a+it))2kdt ~ 2k-lQ~T. iT t Titohmanh (22). 12.6 DIVISOR PROBLEMS Hence, if 0 < a < f, T > 1, • T T f IJ(a+it)l"at> f l((a+it)l"dt ()' f I(( +'t)l"d _,, )a+it)ll !T Ja+itJll > T\7' a t t T 323 > G"Tk{t-2 a)-s J )t(I-u-it))2k dt (by the functional equation) iT ~ 0"2k-t01-o-Tk<t-2o)-t. This can be made as large as we please by choice ofT ifu<!(k-1)/k. Hence k-1 'i'k~2k and the theorem follows. THEOREM 12.6(B).t o:k ~ k~1 (k = 2, 3, ... ). Foro:" ~Pk· Much more precise theorems of the same type are known. Hardy proved first that both d(x) > Kxi, d(x) < -Kxl hold for some arbitrarily large values of x, and then that xi may in each case be replaced by (xlogx)iloglogx. 12.7. We recall that(§ 7.9) the numbers ak are defined as the lower bound!! of a such that T ~ f l((a+it)i"'dt ~ 0(1). We shall next prove THEOREM 12.7. For each integer k ~ 2, a necessary and sufficient condition that is that ~. ~ k-;/ uk::::;,;;k-:2· (12.7.1) (12.7.2) Suppose first that (12.7.2) holds. Then by the functional equation T T f!{(a+it))lkdt = o{pk(t-20')! J'(1-a-it)j2kdt} = O(Tk(t-20")+1) t Hardy {2). 324 DIVISOR PROBLEMS Chap. xn for a < f(k-1)/k. It follows from the convexity of mean values that for T f ]{(a+it)]2k dt = O(TI+<i+l/V<HI2k-a)k) k-1-E: k+I+E: -~<a<~-The index of T is less than 2 if T k-l+E: a>zr· Then f (((u+it)l" dt ~ O(T-') (8 > 0). ]a+it] 2 jT Hence (12.5.1) holds. Hence Yk ~ !(k-1)/k. Hence Pk~f(k-1)/k, and so, by Theorem 12.6(A), (12.7.1) holds. On the other hand, if (12.7.1) holds, it follows from (12.5.2) that T jl{(a+it)]2kdt = O(T2) for a> f(k-1)/k. Hence by the functional equation T f ]{(a+it)]'kdt = O(Tk(l-2<1>+2 ) for a< f(k+l)fk. Hence, by the convexity theorem, the left-hand side is O(TI+~) for a= f(k+l)fk; hence, in the notation of§ 7.9, u; <;; !(k+l)fk, and so (12.7.2) holds. 12.8. THEOREM 12.8. f p, ~ •. p, ~ !, p, ,;;; '· By Theorem 7.7, ak ~ 1-lfk. Since 1-~~k~1 (k~3) it follows that Pa = l, Pa = !-The available material is not quite sufficient to determine p,. Theorem 12.6 (A) gives p, ~ l To obtain an upper bound for it, we observe that, by Theorem 5.5. and (7.6.1), SI((Hit)l'dt~ o(TI+< SI((Hit)l'dt) ~ O(Tt+<), ' ' t The value of Jls i8 due to Hardy (3), and that of fJ• to ~ (4); for fJ, -Titobma:8h (22). 12.8 DIVISOR PROBLEMS and, since u 4 ~ i'!; by Theorem 7.10, !1((/o+itll' dt ~ o(Tt rl(({o-it)('dt) ~ O(T'I+<). Hence by the convexity theorem T jl ((u+itAo dt ~ O(T<-\'o+<) for fo < <1 < !- It easily follows that y4 ~ ,, i.e. /J4 ~ ~-NOTES FOR CHAPTER 12 325 12.9. For large k the best available estimates for ak are of the shape ak ~ 1- Ck -f, where Cis a positive constant. The first such result is due to Richert . (See also Karatsuba [1 ], I vic [3; Theorem 13.3] and Fujii .) These results depend on bounds of the form (6.19.2). Fortherange4 ~ k.:::; Bone hasak ~ f-1/k(Heath-Brown )while for intermediate values of k a number of estimates are possible (see I vic [3; Theorem 13.2]).1nparticularonehasa9 .:::; ,a10 .:::; U.,a11 .:::; Jlo-, and a:I2 ~ t-12.10. The following bounds for a2 have been obtained. fo3o- = 0·330000 . . van der Corput (2), H= 0·329268. van der Corput (4), it ~ 0·326086 . Chih , Richert . if = 0·324324 . Kolesnik [1 ], -fo1f.r- = 0·324273 . Kolesnik , M- = 0·324074.. Kolesnik , !H- = 0·324009 . Kolesnik [5 ]. In general the methods used to estimate a2 and ,u(!) are very closely related. Suppose one has a bound I I exp [2n:i{x(mn)l +ex- 1 (mn)~}] ~ (MN)~x 2 .9-f, M<m.t;;M1 N<n-"Nt (12.10.1) for any constant c, uniformly for M < M 1 ~ 2M, N < N1 .:::; 2N, and MN ~ x2--ta.Itthenfollowsthat,u(!) ~ !9,a2 .:::; 9, andE(T) ~ T·9+• (for E( T) as in § 7 .20). In practice those versions of the van der Corput 326 DIVISOR PROBLEMS Chap. XII method used to tackle ,u(i) and tx 2 also apply to (12.10.1), which explains the similarity between the table of estimates given above and that presented in §5.21 for ,u(!)- This is just one manifestation of the close similarity exhibited by the functions E(T) and i(x), which has its origin in the formulae (7 .20.6) and (12.4.4). The classical lattice-point problem for the circle falls within the same area of ideas. Thus, if the bound (12.10.1) holds, along with its analogue in which the summation condition m = 1 (mod 4) is imposed, then one has #{(m, n)e.Z2: m2+n2 ~ x} = n:x+O(x-9+•). Jutila has taken these ideas further by demonstrating a direct connection between the size of A(x) and that of ((f +it) and E(T). In particular he has shown that ifa:2 =! thenp(f) ~ 'lfir and E(T) 4: Th+•. Further work has also been done on the problem of estimating a:3 • The best result at present is a:3 :::;; ~.due to Kolesnik [S). For a:4 , however, no sharpening of the bound a:4 :::;; ! given by Theorem 12.3 has yet been found. This result, dating from 1922, seems very resistant to any attempt at improvement. 12.11. The 0-results attributed to Hardy in §12.6 may be found in Hardy . However Hardy's argument appears to yield only d(x) = n+ ((x logx)hoglogx), (12.11.1) and not the corresponding n _ result. The reason for this is that Dirichlet's Theorem is applicable for n+, while Kronecker's Theorem is needed for the n result. By using a quantitative form of Kronecker's Theorem, Corradi and Kiltai showed that A(x)~n-{xlexp(c (loglogx)l ,)}. (logloglog xF for a certain positive constant c. This improved earlier work of Ingham and Gangadharan . Hardy's result (12.11.1) has also been sharpened by Hafner [1 1 who obtained d(x) = n+ [(x logx)~(loglogx}t(a+Zlog2)exp { -c(logloglogx)i} 1 for a certain positive constant c. Fork ;:: 3 he also showed that, for a suitable positive constant c, one has d 11(x) = 0. [(x log x)<ll-tl1211(loglog x)o exp { -c(logloglog x)i} ], 12.11 where DIVISOR PROBLEMS k-1 a=~(klogk+k+1) and n. is n+ fork =3 and n± fork ;<:4. 327 12.12. As mentioned in §7.21 we now have a 4 ~ j., whence /14 = j-, (Heath-Brown [8 )). Fork = 2 and 3 one can give asymptotic formulae for I A,(y)'dy. 0 Thus Tong [1 1 showed that f :tf."lh-1)/h '"' A (y)'dy~---~ d (n)2n-(~<+Ili~<+R (x) o h (4k-2)n2 n~l h h with R2(x) 4 x(logx)5 and (k ;>3). Taking a 3 :::;; ft (see §7.22) yields c3 :::;; lj. However the available information concerning a,. is as yet insufficient to give c11 < (2k-1)/k foranyk;:: 4.1tisperhapsofinteresttonotethatHardy's result (12.11.1) implies R 2(x) = O{xi(logx)-!}, since any estimate R 2(x)-« F(x) easily leads to a bound .1.2(x) 4 {F(x) logx}!, by an argu-ment analogous to that given for the proof of Lemma a: in § 14.13. I vic [3; Theorems 13.9 and 13.10] has estimated the higher moments of ti.2(x) and .1.3(x). In particular his results imply that I dz(y)8dy 4: x3+•. 0 For d 3(x) his argument may be modified slightly to yield I l..1.3(y)l 3 dy 4 x 2+•. 0 These results are readily seen to contain the estimates a:2 :::;; !. {12 ~ ! and a:3 :::;; t. /13 :::;;! respectively. XIII THE LINDELOF HYPOTHESIS 13.1. THE LindelOf hypothesis is that ((Hit)~O(t') for every positive E; or, what comes to the same thing, that '(•+it) ~ O(t') for every positive E and every u ~ !; for either statement is, by the theory of the function JL(rr), equivalent to the statement that ,u(u) = 0 for a ~ !. The hypothesis is suggested by various theorems in Chapters V and VII. It is also the simplest possible hypothesis on p.(a), for on it the graph of y = p.(u) consists simply of the two straight lines y ~ !-• (•.;;;!). y ~ 0 (•;;, !). We shall see later that the Linde!Of hypothesis is true if the Riemann hypothesis is true. The converse deduction, however, cannot be made -in fact (Theorem 13,5) the LindelOf hypothesis is equivalent to a much less drastic, but still unproved, hypothesis about the distribution of the zeros. In this chapter we investigate the consequences of the LindelOf hypothesis. Most of our arguments are reversible, so that we obtain necessary and sufficient conditions for the truth of the hypothesis. 13.2. THEOREM 13.2.t Alternative necea80117J and 8'Ujficient conditions for the truth of the LindelOf hypotkeais are T ~ f fm+i!)[" dt ~ O(T') (k ~ I, 2, ... ); (I3.2.I) T ~ f ['(•+it)[" dt ~ O(T•) (u > !. k ~I. 2, ... ); (I3.2.2) T ~Jr•(•+it)f"dt~ ~dl(n) (•>!, k~ I,2 •... ). (I3.2.3) T L n2v 1 n~l The equivalence of the first two conditions follows from the convexity theorem(§ 7.8), while that of the last two follows from the analysis of § 7.9. It is therefore sufficient to consider (13.2.1). t Hardy and Littlewood (5). 13.2 THE LINDELOF HYPOTHESIS 329 The necessity of the condition is obvious. To prove that it is sufficient, suppose that ((f+it) is not OW). Then there is a positive number.\, and a sequence of numbers !+it~, such that t~-)'- oo with v, and rm+i<,lr >c.: (C > o). On the other hand, on diflerentiatJ:tg (2.1.4) we obtain, fort~ 1, rn!+i<ll < Et. E being a positive absolute constant. Hence rm+itl-m+it,)[ ~I 1 n!+iu)dul < 2E[t-t,[t, < jC~ ,, if [t-t~l ~ t; 1 and vis sufficiently large. Hence rm+it)f > w,; (ft-t,f.;;; t;'). Take T = it~, so that the interval {t~-t; 1 , t~+t; 1 ) is included in {T, 2T) if v is sufficiently large. Then 'l.T t,+t;' J ]{(-f+it)j2k dt > J (!C~.)2k dt = 2(!C)2ktek,-t, T t"-t;• which is contrary t-o hypothesis if k is large enough. This proves the theorem. We could plainly replace the right-hand side of (13.2.1) by O(T4 ) without altering the theorem or the proof. 13.3. THEOREM 13.3. A necessary and sufficient condition for the truth of the LindelOf hypothesis is that, for every positive integer k and q > f, ''(') ~ 2: d,~:) + O(t->) (t > 0), (I3.3.I) n.;;ta where 8 is any given positive number leas than I, and,\= .(k,8,a) > 0. We may express this roughly by saying that, on the LindelOf hypo-thesis, the behaviour of {{s), or of any of its positive integral powers, is dominated, throughout the right-hand half of the critical strip, by a section of the associated Dirichlet series whose length is less than any positive power oft, however small. The result may be contrasted with what we can deduce, without unproved hypothesis, from the approxi-mate functional equation. Taking an= dk(n) in Lemma 3.12, we have {ifx is half an odd integer) c+iT .Z:d~:) = ~ J {k(s+w)~dw+O(T(a+~-I)k) n 1-a+E. Now let 0 < t < T-1, and integrate round the recta.ngle !-a-iT, c-iT, e+iT, f-u+iT. We have ~ I 'k(s+w)~dw = 'k(s)+ zl-B P(~. logx) 2m w 1-8 1-s rectangle = '"(s)+O(xJ.-oHt-1+~), P being a polynomial in its arguments. Also ( ·r + tTiT) ''(•+w>~dw ~ O(x'T-'"> f-0'-;T c+iT by the LindelOf hypothesis; and i-a+iT T I 'k(s+w>~ dw = a(xt-0' I l'k), where 8 is arbitrarily small. The result may be used to prove the equivalence of the conditions of the previous section, without using the general theorems quoted. 13.4. Another set of conditions may be stated in terms of the numbers O:.t and P.t of the previous chapter. THEOREM 13.4. Alternative mWJsary and su.ffi.cient canditiC»UJ for the truth of the LindelOf hypothesis are rxk ~ l (k ~ 2, 3, ... ), (13.4.1) p,,;; t (k ~ 2, 3, .. ), (13.4.2) Pk=k~l (k ~ 2, 3, ... ). (13.4.3) As regards sufficiency, we need only consider (13.4.2), since the other 13.4 THE LINDELOF HYPOTHESIS 331 conditions &re formally more stringent. Now (13.4.2) gives "Yk <!,and T I l{(a+i!)l" d! ~ 0(1) (a> 1 ) lu+iW· !"• !~ J J l{(a+i!)l"' d! ~ O(T') (a>!). IT The truth of the LindelOf hypothesis follows from this, as in§ 13.2. Now suppose that the LindelOf hypothesis is true. We have, as in § 12.2, Now integrate round the rectangle with vertices at f-iT, 2-iT, 2+iT, f+iT. We have 2±iT J 'k(s)~ds = O(x2T•-1), f±iT 7T{'(s)~ds ~ o(xl JIHi!l'-1 d!) ~ O(xlT•). l-iT -T The residue at s = 1 accounts for the difference between D.t(x) and ~k(x). Hence ~k(x) = O(xiT•)+O(x2T€-l). Taking T = x2, it follows that rxk < l· Hence also f3.t f, N(a, T+I)-N(a, T) ~ o(log T). The necessity of the condition is easily proved. We apply Jensen's formula. ,. 1 I"' log--~-loglf(n")l dO-loglf(O)I, r1 ••• rn 21T 0 where r1, ..• are the moduli of the zeros of f(s) in [sf~ r, to the circle with centre 2+it a.nd ra.dius i-lS, f(s) being {(s). On the LindelOf t Backlund (4). 332 THE LINDEL6F HYPOTHESIS Chap. Xlli hypothesis the right· hand side is less than o (log t); and, if there are N zeros in the concentric circle of radius 1~}8, the left-hand side is greater than Nlog{(!-!S)/(!-!8)). Hence the number of zeros in the circle of radius 1-!3 is o(logt); and the result stated, with a = !+8, clearly follows by superposing a. number (depending on 8 only) of such circles. To prove the converse,t let 01 be the circle with centre 2+iT and radius 1-8 (8 > 0), and let :E1 denote a summation over zeros of b{B) in 01• Let 02 be the concentric circle ofradius i-28. Then for 8 in 01 ~(8) ~ {'(s)_' I_~ o(log~. b(s) L..1s-p 8 -, This follows from Theorem 9.6 (A), since for each term which is in one of the sums L~s~p· L s~p. 11-yl<l but not in the other, Is-pi ;;;::: 8; and the number of such terms is' O(logT). Let 03 be the concentric circle of radius J-M, G the con,cen.tri<> ci•·c]e of radius !- Then ifJ(s) = o (log T) for s in G, since each term is 0(1), and by hypothesis the number of terms is o (log T). Hence Hadamard's three~circlea theorem gives, for s in 03, 1~(8)1 <{o(logT))•{O(S-'logT))~ where o:+,B = I, 0 < ,8 < l, a and ,8 depending on a only. Thus in 01 ~(8) ~ o(logT), for any given a. Now j ~(s) do~ log((2+it)-log((t+3Hit)-i+" -I, Qog(2+it-p)-log(!+3S+it-pl) ~ 0(1)-Iogm+•S+itJ+o(logT)+ +I, log(!+3S+it-p), since~~ has o (log T) terms. Also, if t = T, the left-hand side is o (log T). Hence, putting t = T and taking real parts, logi{(H3S+iT)I ~ o(logTJ+ I, logi!+3S+iT-pl. Since I!+M+iT-pl <A inc;, it follows that logl{(t+3S+iT)I < o(logT), i.e. the LindelOf hypothesis is true. t Littlewood (4-). 13.6 THE LINDEL6F HYPOTHESIS 333 13.6. THEOREM 13.6(A).t On f1le Liwlewj hypothesis S(t) ~ ~logt). The proof is the same as Backlund's proof(§ 9.4) that, without any hypothesis, S(t) = O(logt), except that we now use t(s) = O(t•) where we previously used '(s) = O(t.A.). THEOREM 13.6 (B).t On the LindeWj hypo(hesis S,(t) ~ o(logt). Integrating the real part of (9.6.3) from ! to !+3a, j+S3 i+s3 J logl{(s)l do~ > [ logls-pl da+O(Slogt). i ly=n J loglo-fil do;> J loglo-!-!81 do~ 3S(log!S-l). ! l Hence I+" J logl{(8)l do~ > o(slog~)+O(Slogt) ly~iJ"'<I o l ~ O(Slogl/S.logt). Also, as in the proof of Theorem 13.5, log((8) ~ I,Iog(s-p)+o(logt) (t+3S <;;;a<;;; 2). Hence ' ' J logl{(8)l do~ I, J logls-pl da+o(Jogt) !+33 i+s3 ~I, O(l)+o(logt) ~ o(logt). Hence, by Theorem 9.9, ' S,(t) ~ ~ J logl{(s)l da+O(i) i ~ O(SloglfS.logt)+o(logt)+O(l), and the result follows on choosing first a and then t. t Cram6r (I), Littlewood {4). t Littlewood (4). 334 THE LINDELOF HYPOTHESIS NOTES FOR CHAPTER 13 Chap. xm 13.7. Since the proof of Theorem 13.6(A) is not quite straightforward we give the details. Let g(z) ~ t{((z+2+iT)+((z+2-iT)) and define n(r) to be the numberofzerosofg(z) in the disc )zl ~ r. As in § 9.4 one finds that S(T) < n(f) + 1. Moreover, by Jensen's Thorem, one ha. R '" f n~) dr = ~ f log)g(Reill))d9-loglg(O)I. (13.7.1) 0 0 With our choice ofg(z) we have log lg(O)) = log)R((2 +iT) I= 0(1). We shall takeR= i + 0. Then, on the LindelOf Hypothesis, one finds that I((Rei~+2±iT)J::::,;; T• for cos 8 ;;:: - 3/(2R) and T sufficiently large. The remaining range for ,9 is an interval of length 0(0!). Here we write R(Reis+ 2) = 11, 80 that ! -:-0 .::.:;: a ~ !- Then, using the convexity of the J1 function, together Wtth the facts that p(O) =! and, on the LindelOf Hypothesis, that J.t(!) = 0, we have J.t(u) ::.;;: b. It follows that I((Rei9+2±iT)I ~ T~+< for cos 9 ~ - 3j2R, and large enough T. We now see that the right-hand side of (13.7.1) is at most Since we conclude that O(e logT)+ O{Oi(o +e) logT}. R o f n(r) Rn(j)~ ----,:-dr 0 n(j) ~ o{ GH-I(o+<) )logT}. and on taking 0 = e1 we obtain n(i) = O(ek log T), from which the result follows. · 13.8. It has been observed by Ghosh and Goldston (in unpublished 13.8 THE LINDELOF HYPOTHESIS work) that the converse of Theorem 13.6(B) follows from Lemma 21 of Selberg (5). THEOREM 13.8. lf S 1(t) = ~logt), then the Lind£16{ hypothesis holds. We reproduce the arguments used by Selberg and by Ghosh and Goldston here. Let ! ::.;;: u ::.;;: 2, and consider the integral 5+1<Xl 1 f log((s+iT)ds 2tti 4-(s-u)2 · 5-ioo Since log((s+iT)~2-B(•J the integral is easily seen to vanish, by moving the line of integration to the right. We now move the line of integration to the left, to R(s) = a, passing a pole at s = 2 + u, with residue -!log((2+u+iT)=O(l). We must make detours around s =l-iT, if u < 1, and around s =p-iT, if u < fJ. The former, if present, will produce an integral contributing O(T- 2 ), and the latter, if present, will be It follows that -'f-• du 4-{u+i(y-T))' · 0 "" /J-" 1 f log((n+it+iT) dt- L f ~u = O(l), 2tt 4+t2 IJ>., 4-{u+t(y-T)}2 0 for T 3 1. We now take real parts and integrate for i ::.;;: u ::.;;: 2. Then by Theorem 9.9 we have "" P-t ! f S~~~;) dt ~ P~l f O) ifi,-TI.;1, R 4-{u+i(y-T)}2 3 0, otherwise. If u>f is given, then each zero counted by N(n, T+l)-N(a, T) contributes at least !(n- !)2 A to the sum on the right of (13.8.1), whence N(u, T + 1)-N(n, T) = o(log T). Theorem 13.8 therefore follows from Theorem 13.5. XIV CONSEQUENCES OF THE RIEMANN HYPOTHESIS 14.1. In this chapter we assume the truth of the unproved Riemann hypothesis, that all the complex zeros of {(a) lie on the line u = f. It will be seen that a perfectly coherent theory can be constructed on this basis, which perhaps gives some support to the view that the hypothesis is true. A proof of the hypothesis would make the 'theorems' of this chapter essential parts of the theory, and would make unnecessary much of the tentative analysis· of the previous chapters. The Riemann hypothesis, of course, leaves nothing more to be said about the 'horizontal' distribution of the zeros. From it we can also deduce interesting consequences both about the 'vertical' distribution of the zeros and about the order problems. In most cases we obtain much more precise results with the hypothesis than without it. But even a proof of the Riemann hypothesis would not by any means com· plete the theory. The finer shades in the behaviour of {(s) would still not be completely determined. On the Riemann hypothesis, the function log {{s), as well as {{s), is regular for a>! (except at s = I). This is the basis of most of the analysis of this chapter. We shall not repeat the words 'on the Riemann hypothesis', which apply throughout the chapter. 14.2. THEOREM I4.2.t We have log((')~ O{(logt)'-'•+<) (14.2.1) uniformly for l < u 0 .::;;: u .::;;: I. Apply the Borel-Caratheodory theorem to the function log {(z) and the circles with centre 2+it and radii ~--!3 and 1-3 (0 < 3 < !J. On the larger circle R{log((z)} ~ logl{(z)l < Alogt. Hence, on the smaller circle, 3-28 3-!8 llog((z)lo;; tBAlogt+Tilogl{(2+it)ll < A3-1 logt. (14.2.2) t Littlewood (1). 14.2 CONSEQUENCES OF RIEMANN HYPOTHESIS 337 Now ap!ly Hadamard's three-circles theorem to the circles Ct. c,, 03 with centre u1+it (I < u 1 .::;;: t), passing through the points I+TJ+it, u+it, l+3+it. The radii are thus r1 = u1-I-1), r, = a1-u, r, = a1-f-3. If the maxima of llog((z)l on the circles are M 1 , M 2 , M 3 , we obtain Hence llog{(a+itll < (~r-a(Al~gt)" < 1J~:sa(IogtJ2-2u+0(~)+0<•1)+0(1/u.l. The result stated follows on taking 8 and 1J small enough and a1 large enough. More precisely, we can take u1 = ~ = ~ = loglogt; since {logt)O<~> = eO(~l<>glogt) = eOt0(e)), i.e. we have both ((,)~O(t'). ~~O(t') (14.2.5) (14.2.6) for every a > f. In particular, the truth of the LindeWf hypothesis follows from that of the Riemann hypothe&is. 338 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV It also follows that for every fixed a > f, as T -+ oo I T dt ((2a)T , j((a+it)j'- ((4a) · Fora> l this follows from (7.1.2) and (1.2.7). For l < u,;; 1 it follows from (14.2.6)and the analysis of§ 7.9, applied to 1/{(s) instead of to {k(s). 14.3. The functiont v(u). For each a> l we define v(u) as the lower bound of numbers a such that It is clear from (14.2.3) that v(u).:::;;,; 0 for a> 1; and from (14.2.2) that v(u) ~ 1 for l < u ~ 1; and in fact from (14.2.1) that v(u) ~ 2-2u for l - !:__ ~A,(n) ::--2" 6 na' and hence v(u} ~ 0 if a is so large that the right-hand side is positive. Since this is certainly true for a~ 3. Hence v(u) = 0 for u ~ 3, Now let l < u1 < a < u 2 ~ 4, and suppose that !og((a,+it) ~ O(!og"t), !og((a,+it) ~ O(!og't). Let g(a) ~log ((s){log(-ia))-~•l, where k(s) is the linear function of 8 such that k(a1) = a, k(all) = b, viz, k(s) = (s-a1)b+(all-8)a aa-al Here {log( -is)}-k e) denote the branches which are real for a = 0. Thus Iog(-is) = logt+Iog(1-~) = logt+oO)· loglog(-is) = loglogt+log{1+0(tl:gt)} ~ loglogt+O(I/t). t Bohr and Landau {3), Littlewood {5). 14.3,. Hence CONSEQUENCES OF RIEMANN HYPOTHESIS l{log(-i8)}-.l:(lll = e-R~)loeloa<-UJl = e-k -l· Hence v(a} = 0 for a > l. Since v(a) is finite for every a > !. we can take a = v(a1)+t:, b = v(a1}+t: in (14.3.1). Making t:-+ 0, we obtain v(a) ~ (a-adv(a1J+(a1-a)v(a1), all a1 i.e. v(a) i8 a convex function of a. Hence it is continuous, and it is non-increasing since it is ultimately zero. We ca.n also show that ''(s)f"s) has the same v-function as log '(.s). Let v1(8) be the v-function of {'(s)f'(s). Since ('(a)~.!., f Iogb(') d' ~ o(~(logt)"'"-"+<), "8) 21r1 (8-z)2 8 ls-.,1=11 we have v1(a) ~ v(a-8) for every positive 8; and since v(a) is continuous it follows that v1(a)~11(a). We can show, as in the case of 11(a), tha.t 111(a) is non-increasing, and is zero for a ;a: 3. Hence for a < 3 i.e. ' Iogb(•) ~ -f('(x+it)dx-!ogb(3+it) a b(x+tt) ~ o(J (logt)••<•l+•dx)+O(I) ~ O{(logt)'•<•l+<), v(a) ~111 (a). CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV The exact value of v(a) is not known for any value of a less than l. All we know is THEOREM 14.3. For}< a< 1, 1-a ~ v(a) ~ 2(1-a). The upper bound follows from Theorem 14.2 and the lower bound from Theorem 8.12. The same lower bound can, however, be obtained in another and in some respects simpler way, though this proof, unlike the former, depends essentially on the Riemann hypothesis. For the proof we require some new formulae. 14.4. THEOREM 14.4.t As t-+ 00, - ('(•) ~ ~ A(n) ,~•·+ I s•~Pf(p-s)+a(s~tlogt), (14.4.1) t(s) 6 n• p uniformly for 1 ~ a ,;;; J, e-"1 ,;;; 0 ~ I. Taking a,.= A(n), j(s) = -b'(s)/t(s) in the lemma of§ 7.9, we have ro 2+ico ""'A(n)e-Sn = -~ f r(z-s)r(z)O"-"dz. 6 nR 2"1Tt2-iro b(z) (14.4.2) Now, by Theorem 9.6(A), ~((• 1 1~ 2; :. +a(logt), .., s 11-yl<ls "2" ty and there are O(Iog t} terms in the sum. Hence f,~; ~ a(Iogt) on any line a of:. J. Also ('(s) ~ a(~)+a(Io t) ((s) mmft-yf g uniformly for -1 ~a~ 2. Since each interval (n,n+1) contains values of t whose distance from the ordinate of any zero exceeds Ajlogn, there is at,. in any such interval for which fg: = O(log2t) (-1 ~ u ~ 2, t = t,.). t Littlewood (<5), to the end of§ 14.8. 14." CONSEQUENCES OF RIEMANN HYPOTHESIS 341 By the theorem of residues, .k( T + T· + 't + 'f)rr,-·~r,~;,·~·d, 1!-it,. 2+it,. !+it. !-it,. ~ ('(•) + I f(p-s)S'~P-r(l-s)S•~•. b(s) -t.<y<t. The integrals along the horizontal sides tend to zero as n -+ oo, so that ro !+iro L A(n) e-8n = - __!__, f P(z-s) r(z) O•-z dz-n-1 n" 2"1T~ !-iro b(z) _ 1'(•1- I r(p-s)s•~,+r(l-•JS'~'. ((•) p Since P(z-s) = O(e-...:lv-tl), the integral is at[ ,~•"-''log( IY 1 + 2)s•~i dy) " = o{[ e-...:lv-ll}og(l2ti+2)0a-1-dy}+ +a(({+ J)ct""log(fyf+2)S•~!dy) ~ a(s•~lJogt)+a(s•~l) ~ a(S'~iJogt). Also r(l-s)Ss-1 = O(e-...:tsa-1) = O(e-AtO-i) = O(e-At+i"t) = O(e-A') = O(S<>-1-Iogt). This proves the theorem. 14.5. We can now prove more precise results about t'(s)JC(s) and log C(s} than those expressed by the inequality v(a) ~ 2-2u. THEOREM 14.5. We have f,~; ~ a{(logt)'~"'), lo ((') ~ a((log t)'~"'). g loglogt uniformly for 1 < a0 ::;;; u ~ a 1 < 1. We have lf,~il.;; ,;t A!:),,+,.._l t )r(p-•ll+a(S•~!Jogt). (!4.5.1) (14.5.2) 342 Now CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV 00 l+ico _LA(n)e~n = -~ I r(z-a)~'(z)Sa-zdz = 0(8°-1), n-1 na 21Tt. 2-i<» '(z) since we may move the line of integration to R(z) = f, and the leading term is the residue at z = 1. Also jr(p-s)l < Ae-Aiy-11 uniformly for a in the above range. Hence ~ !P(p-s)j <At e-.d.ll-yl =A n~t n-1~1-yl<ne-Aif-yl. The number of terms in the inner sum is O{log(t+n)) ~ O(log!J+O{Iog(n+l)}. Hence we obtain o(.te-"{Iogt+log(n+l)}j ~ O(logl). Hence f(~~ ~ 0(3"-'J+O(O"-llogi)+O(O"-ilogl). and taking 8 = (log t)-11 we obtain the first result. Again for u0 ~a :::;;; u1 log ((•) ~ log ((a, +il)- I"' ~((x+it) dx o ~ x+t-t) ~ 0{ (log t)•-••·+•}+ o( r (log!)'-"' dxl ~ O{(logt)'-"'•+<}+Of(logl)'-'"). \Ioglogt If u ~ a2 < a1 and E < 2(u1-u2), this is of the required form; l!nd since u1 and so a 1 may be as near to 1 as we please, the second result {with u11 for a1 ) follows. 14.6. To obtain the alternative proof of the inequality v(a) ~ 1-a we require an approximate formula for log "8). THEOREM 14.6. For fixed a: and o such that } < a: < o :::;;;: 1, and e-"":::;;;: 8:::;;;: 1, log((•) ~ ~A,(n),-a•+0{8•-•(log')""'+<}+O(l). 6 n' 14.6 "' CONSEQUENCES OF RIEMANN HYPOTHESIS 343 Moving the line of integration in (14.4.2) to R(w) = a:, we have "' «+i«> LA(n)e-&n.= -''(s)_r(l-a)8B-l-~ I r(z-a)r(z)O..'dz. ,.~ 1 n• ,(8) 2m «-i«> {(z) Since {'(a)f'(8) has the v-function v(u), the integral is of the form o{oo-e< I e-Aiv-tl{log(IYI+2)}-<«)H dy} = 0{8°-"(logt)-<«)+€}; and r(I-s)S•-1 is also of this form, as in§ 14.4, Hence - ('(•) ~ ~ ~le-"'+0{3"-•(logl)''"'"}. "8) 6 ns This result holds uniformly in the range [ u, J ), and so we may integrate over this interval, We obtain log "8)- ~A~:) e-3n.+O{S0-"(log t)I'(«>H} = log{(H+it)- ~~e-ll"= 0(1), 6n•+ .. as required. 14.7. Proof that v(u);;;:::. 1-o. Theorem 14.6 enables us to extend the method of Diophantine approximation, already used for u > I, to values of u between i and I. It gives log I((•) I ~ ~ A,(n) oo•(tlogn)e-'"+0{8"-•(logt)""'+<}+O(l), 6 no ~A(n) ( • ) = L.. ....!..cos(tlogn)e-8n+O '5' e-3n +0{8°-"(logt)-<(1>+§}+0(1) n~l no n-1:'+1 for all values of N. Now by Dirichlet's theorem(§ 8.2) there is a number tin the range 211 ~ t ~ 21rg_N, and integers .x1, ... , .xN, such that l tlogn -.x,.l ~! (n = 1, 2, ... , N). 2w q Let us assume for the moment that this number t satisfies the condition 344 Now CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV as in§ 14.5. Hence log I{(•) I> K(a)8"·' +O('-""~+o(N'-")+o(S"-•(logt}'<~+<}+O(I). logN T} q Take q = N = [8-], where a > 1. The second and third terms on the right are then bounded. Also logt .::>;; Nlogq+log21T,;; falog~+log21T, so that 8.,;;;; K(logt)-1/a+€. Hence logj{(s)j > K(logt)l-"-'l+O{(logt)"-"+v(o:l+ll'}, where 11 and r/ are functions of a which tend to zero as a -+ 1. If the first term on the right is of larger order than the second, it follows at once that v(a) ~ 1-a. Otherwise a:-a+v(o:) ~ 1-a, and making o: -+ a the result again follows. We have still to show that the t of the above argument satisfies r"1 ~ 8. Suppose on the contrary that 8 < e--'~1 for some arbitrarily small values of 3. Now, by (8.4.4), I{(•) I;;;, (cos~-2N•-•){(a) > ~(!-2N'-•) q a-1 for a> 1, q ;;;<= 6. Taking a= l+logSjlogN, I{(•) I> a~ I~ AlogN >A log~> Att. Since lb(s)j -+00 and t;;::: 21T, t """ oo, and the above result contradicts Theorem 3.5. This completes the proof. 14.8. The function ~(l+it). We are now in a. position to obtain fairly precise information about this function. We shall first prove THEOREM 14.8. We have jlogC(1+it)j ~ logloglogt+A. (14.8.1) 14.8 CONSEQUENCES OF RIEMANN HYPOTHESIS ... In particular W+it) ~ O(loglogt), (14.8.2) {(!~it) ~ O(loglog t). (14.8.3) Taking u = 1, o: = i in Theorem 14.6, we have jlogC(1+it)j ~ n~A~n)e-8n+0(31"logt)+O(I) ~ ~ A~n) + n=:+I e-8n+0(3ilogt)+O(I) ~ loglogN+O(e-8Nj3)+0(Sl:logt)+O(I) by (3.14.4). Taking 3 = log-•t, N = 1+[1og5t], the result follows. Comparing this result with Theorems 8.5 and 8.8, we see that, as far as the order of the functions C{I+it) and 1/,(l+it) is concerned, the result is final. It remains to consider the values of the constants involved in the inequalities. 14.9. We define a function fi(a) as fi(a) ~ 2:"L· By the convexity of v(a) we have, for ! < a <a' < I, v(u') ~ (l-a')v(u1~~u'-a)v(l) = ~=: v(a), i.e. fi(a') <;;fi(a). Thus fi(a) is non-increasing in(!, 1). We write fi(!) ~ lim fi(a), fi(l) ~ lim fi(a). u-+!+O 0"-->1-0 Then by Theorem 14.3, for ! <a < I, ! <;;; fi(l) <;;; fi(a) <;;; fi(!) <;;;I. We shall now provet THEOREM 14.9. As t """ oo I{(I+it)l <;;; 2fi(I)•Y{I+o(l)}loglogt, l{(l~it)l.;;; 2fi(l)~{l+o(l)}loglogt. t Littlewood (6). (14.9.1) (14.9.2) 346 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV We observe that the 0(1) in Theorem 14.6 is actually o(l) i£8-+0. Also, taking a = 1, Sl-o.(Iogt)Pfc.l~ = o(I) if a= {logt)-2.8<«)-'1 (1) > 0). ,2: L e-8P .. e-3mP. p m>l 'JP' This is evidently uniformly convergent for 8 ~ 0, the summand being less thanp-"'. Since each term tends to zero with 8 the sum is o(l). Hence "" e-&nP log~(I+it) = £::. mpm<l+11J+o(l) ~- "log(l-'""'')+o(l) f p1+1J The second term is O(e~/8) = o(l) if ur = [8-1 -~]. Also 1-- ~ 1-- ::;:: I+-. I I ,-•PI I p pl+it-...::; p Hence, by (3.15.2), Now logJ{(l+it)j <;;- L log(l-~)+o(l) P<w = loglogur+y+o(I), j~(l+it)j.::,;; eY+o{lllogw. logw ~ (l+~r)log~ = (l+E){2,8(a)+11}loglogt, 14.9 CONSEQUENCES OF RIEMANN HYPOTHESIS 347 and taking a: arbitrarily near to 1, we obtain (14.9.1). Simila.rly, by (3.15.3), 1 ( 1) logj{(l+it)j ~ P~ log l+p +o(l) = loglog=+log~+o(l), and (14.9.2) follows from this. Comparing Theorem 14.9 with Theorems 8.9 (A) and (B), we see that, since we know only that ,8(1),;;; 1, in each problem a factor 2 remains in doubt. It is possible that ,8(1) =},and if this were so each constant would be determined exactly. 14.10. The function S(t). We shall next discuss the behaviour of this function on the Riemann hypothesis. If l <a< a< ,8, T < t < T', we have {JHT' <:<+iT' <:<+iT {J+iT. log((•) ~ .k( f + f + f + f )lo:~~) d,. fJ+iT fJ+iT' <:<+iT' <:<+iT Let~ > 2. By (14.2.2), ll+iT z f log((•) d' ~ a(l__ f jlog((x+iT)j dx) ~ a(log'l'. ~• ~T ~~ <:<+iT <:< fJ+iT ,., fJ+iT Also f lo:!~) dz = L A1(n) f z~: dz. 2+iT n-l Z+iT Now ~f+iT n-~ [ -n~ lfl+iT 1 jjf+iT n--z dz= -------d• z-s (z-s)logn 2 .T logn (z-s)z 2+iT +• Z+iT ( I ) (I s• dx ) ( I ) = O 1i2{t-T) +O ?ii -«> (x-u)2+(t-T)2 = O n 2(t-T) . Hence and hence fJ+iT f log ((•) d, ~ a(_l__), Z-8 t-T Z+iT fJ+iT f log1t•l dz ~ oPog T\ Z-8 rt-TJ <:<+iT uniformly with respect to {J. Similarly for the integral over (JJ+iT', a+iT'). 348 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV /l+iT' f ~'<'l a,~ o(T'-'1'. fl+iT Z-8 {J-;; 1 Also Making f1 ~ oo, it follows that log{(s) = ~ "'+JiT'log~(z) dz+O(log~+O(lo~T')· (14.10.1) 2m s-z t-Tj T -t a+iT A similar argument shows that, if R(s') < }, Takings' = 2a:-o+it, so that s'-z = 2a:-a+it-(a:+iy) = cx-iy-(a-it), and replacing (14.10.2) by its conjugate, we have From (14.10.1) and (14.10.3) it follows that log'(•) ~ !, •+JiT"log[{('ll dz+O(log'l'+oflog, T') ~ 1Tl s-z t-T} \T -t (l4.l0.4-) «+iT and Iog~(s) = ! «+JiT'arg~(z) dz+O(~~+Of!o~T')· (14.10.5) 7T s-z t-T} \T -t <~<+iT 14.11. We can now show that each of the functions max{log l{(s)[,O}, max{-log [{(a)[,O}, max{arg {(s), 0}, max{-arg {(s), 0} has the same ~·-function as log {(s). Consider, for example, max{arg {(s), 0}, and let its v-function be v1(u). Since [arg{(s)[,;; [log{(s)[ we have at onoe CONSEQUENCES OF RIEMANN HYPOTHESIS Also ('14.10.5) gives T" arg{(s) = ~ j (a o:~~~t-y)2 arg{(o:+iy) dy+G(~':~)+o(~~-=;) (14.11.1) T" < A(log T')••<•>+• f o-o dy+ o(log T) + oflog T') T (u-o:)2+(t-y)2 t-T \T'-t < A(logW•(cx)H+O(t-llogt), taking, for example, T = !t. T' = 2t. It is clear from this that v1(u) is non-increasing. Also the Borel-Caratheodory inequality, applied to circles with centre 2+it and radii 2-o:-3, 2-o:-23, gives [log{(oH+it)[ < ~ ((logt)•,(>)H+ lo;l) +~flog[{(2+it)[[. If o:+O < l, so that v(o:+3) > 0, it follows that v(o:+O):s;;;v1(o:)+£". Since E and 3 may be as small as we please, and v(u) is continuous, it follows that v(o:)::::;;; v1{o:). Hence v1(u) = v(u) (! <a< I). Simil&rly all the v-functions are equal. 14.12. 0-resultst for S(t) and 81(t). THEOREM 14.12 (A). Each of the inequalities S(l) > (logt)l~•, S(t) < -(logt)l~• has solutions for arbitrarily large values of t. Making ex-+! in (14.11.1), by bounded convergence .. (14.12.1) (14.12.2) arg{(s) ~ f (a !;;:;:;, y)'S(y)dY+Oco~t) (a>!). !• (14.12.3) If S(t) < Iogat for all large t, this gives .. arg{(s) < Alogat f (u-!t~tt-y)2 dy+OC 0 ;t) l• < Alog"t+O(t-1 logt). f Landau (I), Bohr IUld Landau (3), Littlewood (5). 1 . J.J ~ ~ 1 1~ , '· CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV . , .c 1 ~he abi;e :n~ilsis shows that this is false if a < v(u), which is satisfied ~ ~·'':"'if a < land u is near enough to f. This proves the first result, and the ' other may be proved similarly. THEOREM 14.12 (B). S,(t) ~ !l{(logt)l-•). From (14.10.5) with a:-+! we have log{(•) ~ i fti 8)Yl. dy+O(l) s-'l-tY -tt ~ i [- S,(y) lu + Iu ~- dy+O(l) s-l-1-Y ft it (s-f-ty)2 ~ I"~ dy+O(l) (14.12.4) (s-!-ty)2 l• since 81(y) = O(logy). The result now follows as before. In view of the result of Selberg stated in§ 9.9, this theorem is true independently of the Riemann hypothesis. In the case of S(t), Selberg's method gives only an index ! instead of the index l obtained on the Riemann hypothesis. 14.13. We now turn to results of the opposite kind.t We know that without any hypothesis S(t) ~ O(logt), S,(t) ~ O(logt), and that on the LindelOf hypothesis, and a fortiori on the Riemann hypothesis, each 0 can be replaced by o. On the Riemann hypothesis we should expect something more precise. The result actually obtained is ·, THEOREM 14.13. S(t)~o(~). loglogt S(t)-o( logt ) 1 -(loglogt)2 • We first prove three lemmas. LEMMA a:. Let ~(t) ~ max !S,(u)i, l>;;;tu;;t so that rf.(t) is non-decreasing, and rf.(t) = O(logt). Then S(t) ~ 0[(~(2t)logt)l]. t Landau (11), Cnun6r (I), Littlewood (4), Titchmarllh (3). (14.13.1) (14.13.2) CONSEQUENCES OF RIEMANN HYPOTHESIS This' is independent of the Riemann hypothesis. We have N(t) ~ L(t)+R(t), where L(t) is defined by (9.3.1), and R(t) ~ S(t)+O(l/t). Now Hence Hence Hence N(T+x)-N(T) ;;, 0 (0 < x < T). R(T+x)-R(T);;, -{L(T+x)-L(T)) > -AxlogT. 'f R(t)dt ~ xR(T)+ J {R(T+u)-R(T))du T 0 > xR(T)-A l ulog T du > xR(T)-Ax2 log T. T+< R(T) <~j R(t)dt+AxlogT BI(T+x~-S~(T) +O(~)+A:r log T ~ o(~(~T))+o(~)+Ax1og7' 351 Taking x = {rf.(2T)flog T}{, the upper bound for S(T) follows. Similarly by considering integrals over (T-x, T) we obtain the lower bound. LEMMA {3. Let u ~ 1, and let F(T) ~ max!log{(•)!+loglT (a-!;;, -1 -, 4 <:; t <:; T). loglogT Then log,(s) = O{F(T+I)e-.A(a-!J1oglogT} ( l l -::<' 2+1oglog T ~a""" 2' We apply Hadamard's three-circles theorem as in§ 14.2, but now take We obtain where 3 l I 111 = 2+loglogT' 'I=!. J = loglogT' 11 ~f. M2 < AMg = AJJ 3(I/M 3 )1-", M 3 ~F(T+I), and 1-a ~ log':>/iog':> ~log(!+ a-!-8)/log("•-l-8) r2 r1 u1-a u1-I-7J >A(a-!-3). 352 CONSEQUENCES OF RIEMANN HYPOTHESIS Ch&p. XIV Hence M 2 ~ AF(T+ l)(logtT)-A(a-!--6l. This gives the required result if a ~ ~. and for i ~ a ~ 2 it is trivial, if the A is small enough. LEMMA y. For a > t, 0 < ~ < ft, <+I log((s) ~ i J S(y). dy+o(P(21))+0(1). (14.13.3) H •-l-•Y g We have J 2t~dy=[ St(Y) lzt -iftt~dy l+f s-f-iy s-f-iy t+f Hf (s-f-iy)2 ~ o(~(21)) + o(~l"l J~ dy ) ~ o(fl"l) E (a ll'+(y t)' E ' 1+1 and similarly for the integral over (it, t-f). The result therefore follows from (14.12.4). Proof of Theorem 14.13. By Lemmas a andy, •+I log((s) ~ 0(~(41)logl)l J ((a-l)'-;(y-t)')l +0~~~~))+0(1) H ~ o[(~(41)logl)l .~1 ] +0(~~~))+0(1) for a-l ~ lfloglog T, 4 ~ t ~ T. Taking g _ A~~(41))1 I -log t (log log T)t' we obtain log((s) ~ O[(logT)l(loglogT)l(~(4T))I]. Hence by Lemma~. for a-! ~ ljloglog T, log{(s) = O((log T)i(loglog T)l{!)b(4T+4)}le-.A:(a-t>to~loeTJ. Hence j logl((s)l da ~ O[(logT)l(loglogT)-l(~(4T+4))1]. -l-+1/l<>trl<>aT (14.13.4) 14.13 CONSEQUENCES OF RIEMANN HYPOTHESIS Again., the real part of (14.13.3) may be written i logl((s)l ~ f (a-l~'+x'(S(I-x)-S(I+x)}dx+O(~~~))+O(I). 353 (14.13.5) Hence l+i< t I log I((•) Ida~ I arctan~ (8(1-x)-S(t+x)) dx+ j 0 X +0(1'~(21)/E)+O(!') ~ 0[{(~(41)logl)l]+0(!'~(2t)/E}+O(!')· Taking p. = lfloglog T, and e as before, !+1/loglogT j log 1((•)1 da ~ O[(log T)l(loglog T)-l(~(4T))!]. (14.13.6) Now (14.13.4), (14.13.6), and Theorem 9.9 give 8,(1) ~ O[(logT)I(loglogT)-1(+(5T))!] (4,;;; t,;;; T). (14.13.7) V e.rying t and taking the maximum, +(T) ~ O[(log T)l(loglog T)-t(~(5T))l]. Let ~(T) ~max (loglogl)'q(1), 4.o:;;to:;;T logt so that rfo(T) is non-decreasing and logT ~(T),;;; (loglog T)'~(T). Then (14.13.7) gives ~(T) ~ o[___lo[2' (~(5T))l] (loglogT)'I. ~(T)(loglog T)' 0[(~(5T)}l] ~ 0[(~(5T,))I] (T ,;;; T,). logT Varying T and taking the maximum, ~(T,) ~ 0[(~(5T,))I]. But f(5~) < 51#(11) for some arbitrarily large 11; for otherwise ~(5"1,) ;;;, 5"~(1,), i.e. rfo(T) >AT for some arbitrarily large T, which is not so, since in fact +(T) ~ O(log T), ~(T) ~ O((loglog T)'). Hence ~(T,) < A(~(T,))!, ~(T,) l+1 • (14.13.8) 14.14. Theorem 14.13 also enables us to prove inequalities for {(s) in the immediate neighbourhood of u = t, a region not touched by previous arguments. We obtain first THEOREM 14.14 (A). "!+it)= o{exp(A 10~1!~t)}· (14.14.1) We have 8(t+x)-8(!) ~ {N(t+x)-N(!)}-{L(!+x)-L(!)}-{f(!+x)-/(!)}, wherej(t) is the 0(1/t) of(9.3.2), and arises from the asymptotic formula for logr(s). Thusj'(t) = O(lft2), and since N(t+x) ~ N(t) 8(!+x)-8(!) > -Axlogt+a(xf!') > -Axlog!. Hence, by (14.13.5), ~ f x21ogt ( logt \ logl((s)l <A 0 (•-ll'+x' dx+a t(loglog!)'J +a( I) < Aglogt+a(t(l:;~~!)')+a(J) uniformly for u > -f, and so by continuity for u = -f. Taking < ~ lfloglog! the result follows. THEOREM 14.14(B). We have Alogt 1 ( 2 ) 1 I{( )I Alogt -loglogt og (a-!)loglogt < og 8 <loglogt (! <",;;! +Afloglogt), (14.14.2) ••g((s) ~ a(~ 1 10 1 g! ) (j,;;",;; t+Afloglog!). (14.14.3) og ogt By (14.!3.1) and (14.13.3), I ( - (12_ J dx l a(___l<£) a( 1 I log (s)- 0 loglogt 0 .J{(f.1-l)2+x2} + ~(loglogt) 2 + · 14.14 CONSEQUENCES OF RIEMANN HYPOTHESIS Now i tl<o-j-) I du Jdx' 0 ~{(• W+x') ~ 0 ~(l+x'')' which is less than I if g ~ u-f, and otherwise is less than tl<a-f) l+ J .t;: ~ l+log_L, 1 x u-l 355 Taking g = 1/loglogt, the lower bound in (14.14.2) follows. The upper bound follows from the argument of the previous section. Lastly, taking imaginary parts in (14.13.3), I a.g((s) ~ [ x';;:-~il'{8(!+x)-8(!-x))dx+ a( log! ) I + t(loglog!)' +a(l) = o{lo~!~t f x2;(~~!)' dx}+o{"l:;~~t)t}+O(l). I • Now J .z2_;(~~~)2 dx < f x2:(~~l)2 dx = f7T, ' ' Hence, taking f = 1, (14.14.3) follows uniformly for u > -f, and so by continuity for u = f. In particula.r log,(•) ~a(~) (• ~ ~+.4) (14.14.4) loglog t 2 loglog t · From (14.14.4), (14.5.2), and e. Phragmen-LindelOf argument it follows that ((lo t)2-2a) log '(•) ~ a lo~log! (14.14.5) uniformly for 1 A 2+ loglogt ~a~ 1-S. 14.15. Another result in the same order of ideas is an approximate formula. for log {(s), which should be compared with Theorem 9.6 (B). THEOREM 14.15, Fort~ a~ 2, log{(s) = '5' log(s-p)+O(logtlogloglogt), lt-yki7toglogl loglogt (1 •. 15.1) where p = l+ir "'"" tlwough zerD8 of C(s). CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV In Lemma. a: of§ 3.9, let j(s) = {(s), s0 = ~+~li+iT, r =-faa, where 8 = 1/loglog T. By (14.14.4) I I I (AlogT) {(s0) ~ exp loglog T · The upper boWld in (14.14.2) gives I {(s) I < exp{l:gll:~ ~} for ]s-s0 ] ~ r, a;;::: 1; and for ]s-s0 1 ~ r, a<}, the functional equa-tion gives ]{(s)] < Atl-a]{(1-s)] < At"~exp(1:;l:~~) < exp(1::l:~~)· It therefore follows from (3.9.1) tha.t log {(a)-log {(s0)= 1 ... -p~t8i<~Slog(s-p)+ + '5" log(s0-p) ~ o( log T ) ls.-pf'Ztaf<IS loglog T for ]s-s0 ] ~fi-r, and so in particular for l ~ u ~ l+S, t = T. Now Also Hence ( logT) log {(so) = 0 loglog T . So=P = :}ali+i(T-y}. ::Jaa ~ ]s0-p] <A, and so, if the logarithm has its principal value, log(s0-p) ~ o(Iog~) ~ O(logloglog T). Also the number of values of p in the above sums does not exceed by Theorem 14.13. Hence '5" I ( ) _ o(log Tlogloglog T\ 1,.-pf.i;IB/4 og 8o=P -loglogT -r 14.15 CONSEQUENCES OF RIEMANN HYPOTHESIS 357 Since tT-y] ~ 3 if ]s0-p] ~ 2li/.J3, the result follows, with T fort and l ~ u ~ f+S. It is also true for 1+3 < u ~ 2, since in this region log{(s) = o(1o~!;T}' and, as in the case of the other sum, '5" log(s-p) ~ 0 {l<>gTiogloglog '1'. l8o-p(;;;:23/<13 -loglogT -, This proves the theorem. For C'(s)/{{s) we obtain similarly from Lemma. a: of§ 3.9 ~(~i ~ }' 8~ +O(logt). ]1-yl.;;:'ttoglogl p (14.15.2) 14.16. THEOREM 14.16. Each interval [T, T + 1] contains a value oft such that ( 1 t ) J((s)] > exp -A 10;l~gt (! ~ u ~ 2). (14.16.1) Let b = 1/loglog T. Then the lower bound (14.16.1) holds automati· cally for u ;:. ! +b, by (14.14.4). We therefore assume that t ~ u ~ t+ b. If s = u +it and s0 =! +b +it then, on integrating (14.15.2), we find log ((s) ~ I log('-=.E..)+o(~) ((so) lt-yl.:;;.l s0 -p loglog T · M~~over log((s0) = o(to~l!;T) by (14.14.4) so that, on taking real .. ,~ logl((•)l~ I logi'-=.E..I+O(~) lt-yl.;;<~ S0- P loglog T ~ L logl!=!!+o( logT) 11 _ 11 ,. 6 2b loglog T ' since js-pl ~ ]t-yl and js0 -p] ~ 215. We now observe that T+l min(y+.I,T+l) ! L log~dt= I f log~dt 11-ylo:;.l 26 T-,),;;y.;;T+l+.l 2b IIllli(y-;;~ 2b and the result follows. In pa.rticul&r, if E is any positive number, each (T, T+l) contains a t such that (14.16.2) 14.17. Mean-value the01·emst for S(t) and81(t). We consider first S1(t). We begin by proving THEoREM 14.17. For tT ~ t ~ T, 8 = T-i, nS (t) = 0 _ ~A 1(n)cos(tlogn) e-.Sn+0(-1-) (14 17.1) 1 6 nllog n loglog '1' where 0 ~ Jlogj((a)j da. Making fJ ..-r co in (9.9.4), we have wS,(t) ~ 0- flogj((a+it)j da. t Now, integrating (14.4.1) from 8 to J+it, log~(s)-log~(J+it) =~A~:) c.Sn_ ~~~e-8n+ (14.17.2) +I j 88•-Pr(p-81) da1+0(8°-llogt) {!--•O+ L n-•) n ~ 2 2o;;n>;; 1(~ n> 1/~ ~ O{(b•-'+2-•b)logt). (14.17.3) Hence, for ! ~ a ~ ~. . log~(s) = I~~Sn)e-Bn+ 2:j 88•-Pr(p-s1)da1 +0(bl)+0(8o~1Jogt), n~2 p 0 t Littlewood (6), Titchmarsh (2}. 14.17 CONSEQUENCES OF RIEMANN HYPOTHl"SIS and integrating over ! ~ a ~ J, .1. .1. .~. j log((•) da ~ j (~A~~) ,-lln)da+ 'L: j (a,-!)8'•-Pr(p-•,) da,+ i i n 1 P t +O(b~)+0(8ilogt). Also, by (14.17.3), Jlog((•)da~ J (!A~:),-••)da+O(ol), -l -l n~2 and the inversion being justified by absolute convergence. Hence -nS 1(t) = C- ~At(n)cos(tlogn)e_8,.+ ,6 nllogn Now l'(p-s1) = O(e-.Aiy-,~1) (jy-tj ~ I), r(p-•,) ~ o(IP~,) ~ 0({(a,-tl'~lr-ti')l) (ly-tl <I). Hence f = ly-ti<"fuoglogt + l{loglogt~y-tl<l + ly-~;..1 t = a()y-ll<~oglog!J ()0>-ida1)+ I t +0( L - 1 -f(a,-!)8••-lda,)+ 1/loglogt..;;ly-tl<l],'-tj! l +0( ~ e-.41-y-11 J (a1-f)80t-i da1). ly- ;;ol i 360 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV Now t "" I s•n-! dul <I e-a:logl/ll dx = log-l(ljS), ! ' t • J (a1-})Sa•-i du1 < J xe-zlogl/& dx = log-2(1/S). I ' As in§ 14.5, '5' e-Aiy---tl = O(log t). ly-1f';.t Also, by (14.13.1), for t-1 ~ t' ~ t+I N(t'+-1 -)-N(t'}~o(~)-loglog t loglog t '5' 1- o( 1ogt) l:r-tJ.;;ti\oglogt -loglogt ' Hence and .L ,~.~ ) .L ,~. Hl/loglogt<y.;;t+l m<~ogt l+m/loglogt<;;y<t+(m+l)jloglogt Hence = L o(-/1 11 - I l~g t ) = O(log tlogloglog t). m<loglogt m og ogt og ogt I~o( __ Io_g_!)+oflogtlogloglogt) o(~) -o( 1 ) p loglj8loglogt r log2 1JS + log2 IJ8 -loglogt for the given S and t. This proves the theorem. 14.18. LEMMA 14.18. If a.~ 0(1), a < .. ~ o(ohlog~)· This proves the first part. In the second part we have a pair o£ logarithms running throughout the remainder terms, and this is easily seen to produce the extra log1 1ja in the result. 14.19. THEOREM 14.19. Aa T-+ oo, _I_JT{S(t))'dt~~+_l_ ~ Al(n). T 0 1 112 21T2,6nlog2n Let f(t) = Ce-8- ~~J5n)cos(tlog11:) e-8 ... 6 nllogn Then, as in the lemma, _2_ fT {f(t)}2dt = 02e-28+~ ~ Af{n) e-28n+O~og I/8)• TIT 2 n~ll nlogln r TO and we can replace a by 0 in the first two terms on the right with error 0(~)+0( i~~:g-:::) ~ O(~l+ 0( L Io!'n)+o( L,ni:g'n) n~l n.;;;t/8 n>ll ~ 0{1flog(1/o)). Hence, taking a = T-1, ~ J{f(t)}'dt ~ C'+~in~~~~ +O(Io~T)· !T n-Il Hence T T ' ~ f {S,(t))' dt ~ ~ f (f(t)+O(logl~g T)J dt IT iT T T ~ ~ J {f(t))'dt+O(Tiog~ogT J If(!) I dt)+o((logl:gT)')' jT iT 362 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV and, since J 1/(!)l dt <;; [ J dt J {f(t))' dt]l ~ O(T), }T JT fT it follows that ! JT{S,(t)}'dt~~+_l i A!(n) +0(-1 -). TIT 11' 11 2.,._ 1 nlog2n loglogT Replacing T by !T, lT, ... and adding, we obtain the result. 14.20. The corresponding problem involving S(t) is naturally more difficult, but it has recently been solved by A. Selberg (4). The solution depends on the following formula. for r(a)/{(8) THEOREM 14.20. Without any hypotheai8 ~'(s) __ 'A....(n) + x2. -~· those at z = -2q and z =pare x-2<!-a_xZ<-2<!-a) xP-•-z2{p-a) (-2q-a)2 ' ~· respectively. The result now easily follows. 14.21. THEOREM 14.21. Fort > 2, 4 :::;;,; X :::;;,; t2, I I u1 = 2+1ogx' we have S(t) ~I" A.(n)sin(tlogn)+0(-'-1 "A.(n)l+o(logt)· Tr ... ~. n"' logn logx ... ~. n"'+" f logx By the previous theorem, ('(u+it) A,(n) {xlU-al+x1-"} 2wxf-cr 1 ((u+it)=-n~,2 na+it+O ~ + logx ~(u 1 -!) 2 +(t-}') 2 (14.21.1) for u ;;o: u1, where lwl < I. Now x2(1-a)+xl-a.::.;,: x1-2a+x-" <2xf-u. t2 1ogx logx Hence Now by (2.12.7) {'(•) "( I I) ,- ~ L., -+- +O(Iogt) . .,(a) P a-p p Hence R {'(a+it)- " ( a-! +-'-)+O(lo t) {(a+it) - f (a-!)'+(t y)' t+r' g ' a-! 0(1 t) = f(u !JI+(t y)l+ og , CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV Taking real parts in (14.21.2), substituting this on the left, and taking f((Tt u~~t y)2+ O(logt) = -R L ~:,<:J+¥2:(<;1-=.-;;~~tt=y-}2 (lw'l < 1). n<~ 1' Hence ( 1 -¥)L(a1 -hl~~tt-y)2 = -R L ~~J+O(logt). 1' n1-:>!, Hence (14.2!.3) Inserting this in (14.21.2), we get ('(a+it) ~- "'AA~I+o(xl-ol "'A.(n)I)+O(xl-oiogt). ~(a+~t) .. ~. na+it .. ~, na:t+it (14.2!.4) Now By (14.2!.4) J,~ I] 1 ~:~Jda+o(J1 ~:~:!,J J x~-'da)+o(logt Jxl-'da) a 1 a, u 1 ~I "' A.(n) +o(_l_J"' "'-·~~~ o(Iogt) ""~' n"•+11logn logx .. ~. na:t+it + logx · 14.21 CONSEQUENCES OF RIEMANN HYPOTHESIS Also, by (14.21.4) with (T = (11, IJ.)&( -')l('(a,+it)l 2 """ (Tl !" {(at+it) ~ o((a,-1>1 L ~:,(:2J)+O((a,-!)logt} n<x• ~ 0(Io~xl.~. ~~(:2J)+oG::~J· It remains to estimate .fs. For l < (T ::;;;; a 1, I(("(a,+it)('(a+it))~ LI(-I ___ I_)+O(logt) ~(a1 +it) {(a+it) P a1+it-p (T+it-p L (1-y){(a-!)'-(a,-!)'} ~ ' ((a !i'+(t y)'){(a, !l'+(t ri'}+O(logt). Hence I I(('(a,+it) _ ('(a+it))l ~((T1 +it) {(u+it) "" lt-yi((Tt-il 2 0 I ,;;f{(a-ll'+(t-y)'j{(a,-!)'+(1 rl')+ (ogt). Hence "' (a,-!J' I• )t-y) da 0{( -')lo t) iJal::;;;; ~ (at-fl2+(t-y)2 (a f)2+(t y)2+ (Tt T g ' I <;; !w(a,-!) L (a -~',~~:-y)'+O((a,-!)logt) ' ' ~ o(I J "' A,(n)J)+o(~'-)· logx n~• na:t+it logx by (14.21.3). The theorem follows from these results. Theorem 14.21 leads to an alternative proof of Theorem 14.13; for taking x = .J(Iogt) we obtain 8(t) ~ {~. ~) + 0(lo~x I'"£') +o(:::~J = O(x)+O(x)+OG~:~) -o(~) -loglogt · 366 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV 14.22. THEOREM 14.22. FO'I' pa~x:o;;; Ti (O<a~f) We have S(t)+! 2 'in(t!ogp) ~! 2 A(p)-A.(p)pl-oo ,in(tlo )+ 7T p<x• Vp 7T pq:• pllogp gp +OLo~x/1 ~:~!';/) +O(/ P~,P''~;~f]~gp/) + +a(!/ ' A,(p')/)+o('' 1) o(logt) logx P6,p'!i..•n+ill ~fPfr + logx · The last term is bounded if fT ::;;;; t ~ T, x ;;;::,. Ta, where a is a. fixed positive constant. The last term but one is o(2 f -{-.) ~ o(2 J"~h) ~ O(l). p ~p p p Now consider the first term on the right. If p < x, A(p)-A.(p)pl-0 • ~ (1-pl-"•)Jogp = (I-p-lJlog:r)togp = (1-e-logp/log:r:)logp = of!og2p)• \logx and, ifx <p ~ x2, it is O(A,(p)} ~ o(!ogp!ogx'fp) ~ o(log'p)· logx logx Hence the first term is the imaginary part of where Now 2p~' p<:r:' A(p)--;A.(p)p!-oo ~ O(!ogpj!ogx). ogp 14.22 Since CONSEQUENCES OF RIEMANN HYPOTHESIS ' ~ o(log'p) - o(logp) ~ log2x -logx' the first term is o(T lol x 2 ~l') ~ O(T), g p<.-.:' p by (3.14.3). The second term is o( 2 2 ~~l)+o( 2 2 i"•""l ) p p ql-P<:t:'fp<q<pp.(p q)jp ~ o( L: ~o~~P x)' +0( 2 ~~-logp) p<x'p g p<:>:' g ~ O(x')+O(x') ~ O(T) ifx~ ..JT. 367 A similar argument clearly applies to the second term. In the third term, the sum is of the form where a;,= 0(1); and ~ O(T)+O(!og'x)+O(log'x) ~ O(T). Similarly for the fourth term, and the result follows. 14.23. THEOREM 14.23. If Ta ,;;; x ,;;; Ti, 368 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV This is T L L -f.I J sin(tlogp)sin(tlogq) dt P Anlogn) Hence the first term is .L pl:gp ~ 0(1). Alw !Tloglog T+O(T). llogpfql > Ajp > Ajx'. Hence the remainder is o(x'( L J,)') ~ O(x'.x') ~ O(x'), p<o:'p and the result follows if x ~ T-1-. ]4.24. THEOREM 14.24. T f {S(t)pdt"'hTloglogT. T T J (S(t))'dt~ J (sct)+~ L sin(tJogp)~ L sin(~ogp))'at iT iT P<z' p p))'dt--1-T p<x' p T -~ J (s(!)+~ L sin(tJogp)) L sin(~ogp) dt+ iT p<;~:' p P I, holds also for u = I. On the Riemann hypothesis we can go much farther than this.t THEOREM I4.25 (A). The aeries .;:';~(n) n~ n~ ia convergent, and its aum i8 IJ,(a), for every s with a > i· (14.25.1) In the lemma of§ 3.I2, take an= p.(n), f(s) = 1/,(s), c = 2, and x half an odd integer. We obtain 2+tT "\:' ~(n)- I J 1 x"'dw+o(x') L. n8 -21ri '(s+w) w T nT 'J+iT )-1 -x"'dw+-l_+o(x'), ~,;;; + + ((•+w) w ((•) T 2-iT }-<1+&-iT f-u+&+•T where 0 < 3 < a-1. By (14.2.5), the fu.st and third integrals are and the second integral is o{x~-u!-& l (l+ltl)-l+E dt} = O(xi-u-t&T~). t Littlewood (I). 370 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV Hence 'p.(n) = !+0(T-tHx2)+0(T~:d·-u+8). (« n" ~(s) Taking, for example, T = x3, the 0-terms tend to zero as x-+oo, and the result follows. Conversely, if (14.25.1) is convergent for u > f, it is uniformly con· vergent for a ;;:: u0 > !, and so in this region represents an analytic function, which is 1/{(s) for u > 1 and so throughout the region, Hence the Riemann hypothesis is true. We have in fact THEOREM 14.25 (B). The convergence of (14.25.1) for a>~ is a necessary and 81Jjficient condition for the truth of the Riemann hypothesis. We shall write Then we also have THEOREM 14.25 (C). A necessary and sufficient condition for the Riemann hypothesis is M(x) = O(x!H). {l4.2s.2) The lemma of§ 3.12 with s = 0, x half an odd integer, gives }+5-iT i+5+iT 2+iT ~.k(,L .L +)J(~)'Edw+o(~) (14.25.3) ~ o(l (l+lvl)-'+ f, and the Riemann hypothesis follows. 14.26. The finer theory of .M(x) is extremely obscure, and the results are not nearly so precise as the corresponding ones in the prime-number problem. The best 0-result known is CONSEQUENCES OF RIEMANN HYPOTHESIS 371 TliEORIW 14.26 (A).t M(x) = o{xiexp(A 10~~;x)}· (14.26.1) To prove this, take 8 = l+logl!g T' T = x2, in the formula (14.25.3). By (14.14.2), ~"~)~ ~ exp(Al:;~;T) on the horizontal sides of the contour. The contribution of these is therefore o{x 1~exp{A 10~~;T)} = o{exp(A 1:;~;x)}. On the vertical side, (14.14.2) gives 1 __ 1 __ 1 & ex (A logv lo 2loglog !'\ {(i+8+iv) '"""' p loglogv g loglogv-} for v0 ~ v ~ T. Now it is easily seen that the right-hand side is a. steadily increasing function of v in this interval. Hence ){(~)~ ~ exp{A 10~ 1~;T) (v0 ~ v ~ T). Henoe the integral &long the vertical side is of the form O(xi•')+O(xl••ex (A log T ) JT l 1 ~ ~ ~(n) ~ ~ M n (1 __ 1_) ~ 8 f• M(x)dx '(s) ,6 n" ,6. ( ) n8 (n+l)" 1 r+1 · Suppose tha.t (14.26.3) IM(x)l ~ llfo (1 ::;;;; X< Xo), ::;;;; axt (x;;;::. Xo)-Then J oo dx f• dx < JsJM() 1 ri+ jsJS 1 ;;;+i ~21•1M,+~. ·-· (14.26.4) But if p = l+ir is a simple zero of "s}, and s = u+iy, u--+ !, then I I I [('l ~ ((•)-((p) ~ (u-!)('(p)' We therefore obtain a contradiction if ~ < IP(~(p)l' This proves the theorem. 14.27. Formulae connecting the functions of prime-number theory with series of the form etc., are well known, and are discussed in the books of Landau and Ingham. Here we prove a similar formula for the function M(x). THEOREM 14.27. There is a sequence T,, v ::;;;; T, :::;;; v+ 1, BUCh that M(x) ~lim " --"'---2+ ~ (-l)•-'( 2~/x)"' (14.27.1) ,"' lym..p{'(p) .&:1 (2n)!n{(2n+I) if xis not an integer. If xis an integer, M(x) is to be replaced, by M(x)-!~(x). In writing the series we have supposed for simplicity that all the zeros of ,(8) are simple; obvious modificationsarerequiredifthisisnot so. 14.27 CONSEQUENCES OF RIEMANN HYPOTHESIS 373 Foia. fixed non-integral x, (3.12.1), with an = p.(n), 8 = 0, c = 2, and w replaced by 8, gives M(x) ~ __!, 'J+<T ~ l_da+o(!)· 271-iT 8 "8) T If xis an integer, fp.(x) is to be subtracted from the left-hand side. By the calculus of residues, the first tenn on the right is equal to ' xP f (-1)n-1(27Tjx)2n ,,fu, p('(p) - 2 + ,6 (2n)!n((2n+l) + +:bCvr··+ -·NrT+ T ).~.~da. 2-iT -:lN-1-iT -2N-1+iT where Tis not the ordinate of a zero. Now Here -ONf-l+iT x" diJ :lN+fll+iT xl-l •1<•1 ~ (l-e)((I-e)d 8 -W-1-iT 2N+2-iT U.'+l+iT f xt~ ~-111" 1 1-s cos!s'ITr(s) «S)ds. 2.V+2-iT r~s) = O(jeHs-fnogsl) = O(ea-{a-flloglsl+i,.ltl) = O(eo-{o-~)logo+~,.~l). Hence the integral is T 0{ f ~(~):aN+Ze2N+2-{2N+illog(2N+2)dt}, -T which tends to zero as N --+ oo, for a fixed T. Hence we obtain " xP 2 ~ (-1)•-'(2~/•1" I (-oof-;T 'J+iT) x" d L p('(p)-+ L (2n)'n((2n+l)+2,;l + a((•l '· lyi+iT f x' f x'~ f x'~ 2'->,o I 8'(s) d8 = (1-8)"1-8)d8 = 1-s cos!s?TI'(8) «S}diJ -«>+iT 2+iT 2+iT 374 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV Also by (14.16.2) we can choose T = T .. (v,;;; T,.:::;;; v+l) such that ~ ~ O(t') (t <;;a<;; 2, t ~ T,). Hence for -I ..:;;; u:::;;; f, t =Tv 1 ( IW-l) WJ ~ O 1((1-i}l ~ O(t'). II+ iT" Hence f -"'-a.~ O(T'-'). s{(s) • -l+iTv Similarly for the integral over (2-iT, -oo-iT), and the result stated follows. It follows from the above theorem that :2: lp('~p)l is divergent; if it were convergent, L:p,~p) would be uniformly convergent over any finite interval, and M(x) would be continuous. 14.28. The Mertens hypothesis.t It was conjectured by Mertens, from numerical evidence, that IM(n)l < .Jn (n > 1). (14.28.1) This has not been proved or disproved. It implies the Riemann hypo-thesis, but is not apparently a consequence of it. A slightly less precise hypothesis would be M(x) ~ O(xl). (14.28.2) The problem has a certain similarity to that of the function !f(x)-x in prime-number theory, where ~(•) ~ I A(n) • ... On the Rieme.nn hypothesis, t/J(x)-x = O(xfH), but it is not of the form O(xi), and in fact ~(x)-x ~ ll(xllogloglogx). (14.28.3) The influence of the factor logloglog x is quite inappreciable as far as , t See reference!! in Landau's Htmdbtu;h, and v011 Sterneck (1). 14.28 CONSEQUENCES OF RIEMANN HYPOTHESIS 375 the calculations go, and it might be conjectured that ( 14.28.2} could be disproved similarly. We she.llshow, however, that there is an essential difference between the two problems, and that the proof of (14.28.3} cannot be extended to the other case, at any rate in any obvious way. The proof of (14.28.3) depends on the fact that the real part of L:~ y>O p is unbounded in the neighbourhood of z = 0. To deal with M(x) in the same way, we should have to prove that the real part of f(z) ~ L ~:,.( ) (R(z) > 0) y>O p~ p is unbounded in the neighbourhood of z = 0. This, however, is not the case. For consider the integral 1 f , ... 2,;l s{(s) a. taken round the rectangle (-1, 2, 2+iT .. , -1+iT .. ), where the T .. are those of the previous section, and an indentation is made above 8 = 0. The integral along the upper side of the contour tends to 0 as n-+ co, and we calculate that 2+ioo -I+ioo 2 1 f'"' 1 f'"" 1 f'"' /(z) = 2;i II 8'(8)ds-2;i -1 s'(s)ds+2;i -1 s'(s)ds. The last tenn tends to e. finite limit as z -+ 0, Also leiszl = e!l-zl ~ell (s = -I+it, z = x+it, x > 0) and 1"(-1+it) = O(t-l). The second term is therefore bounded for R(z) >0. The first term IS equal to 00 I Hoc 1 :2: f '"' Now, tfn >I, Hence --: f'(n) -ds 211'$ sn• a~t 2 O+Jiooell(iz-loga) _ ( 1 ) , -.-ds- 0 n 11 logn 376 CONSEQUENCES OF RIEMANN HYPOTHESIS Cha.p. XIV uniformly in the neighbourhood of z = 0. Hence ~ p.(n) 2 +Ji<» ~ ds. = 0(1). L sn8 n=2 2 If z = rei9, we have Hence ~0(1)+ J~dA ~ O(IJ+ J~ax =0(1)+ f~+ fex-;Idx+ J~dx ' ' ' ~log~+ 0(1). /(')~~log~+ 0(1), and consequently RJ(z) is bounded. 14.29. In this section we shall investigate the consequences of the hypothesis that x ! (M;xl)'ax ~ O(logX). (14.29.1) This is less drastic than the Mertens hypothesis, since it clearly follows from (14.28.2). The corresponding formula with M(x) replaced by ifl(x)-x is a consequence of the Riemann hypothesis. t THEOREM 14.29 (A). If (14.29.1) is true, all the zeros of {(s) on t1u critical line are simple. By (14.26.3), I I I J• )M(x)) _ f•)M(x)) _I dx «8} ~jsjl ~dx-[sjl ~xi-a+!-,;: , (J.M'<•IaxJ• ~)l ~ j.'L(J.~ax)l· """' I I x"+l x"+i-(a-j)i-X"+i ' ' ' t cramer (5). 14.29 CONSEQUENCES OF RIEMANN HYPOTHESIS 377 Let f(X) ~! (M;xl)' dx Then Let p be a. zero and a = p+h, where h > 0. Then u = l+h, and hence I~ (lp+hl). '(p+h)- 0 h (14.29.2) This would be false for h -+ 0 if p were a. zero of order higher than the first, so that the result follows. Multiplying each side of (14.29.2) by h, and making h-+ 0, we obtain t~p) ~ O()p)). (14.29.3) We can, however, prove more than this. THEOREM 14.29 (B). If (14.29.1) i8 true, L:)p.-:p))' (14.29.4) ia convergent. This follows from an argument of the 'Bessel's inequality' type. We have fx(M(x) " xP-' )' O~l --x-ly~Tp{'(p) dx X X ~J(M(x))'ax+ L L __ I -JxP+P'-'dx-1 X lyi<T )y'J<TppT(p){'(p') 1 -2 L -:-Jx M(x)xP-'dx. lyi+p~ldx= p+p'-1 = 0 lr+r'l · Hence the sum of these terms is less than K1 = K1(T). In the last sum we write X X X J M(x)xP-ldx= J M(x)xP-2(1-i)dx+~f M(x)xP-1 dx. ' ' ' The last term is o(~ !IM(x)lddx) ~ o[~(~ M~~x)dx! xdx)'] by (14.29.1). Also ~ o(f M;~x))t ~ O(log!X), ! M(x)xP-'(1-x)dx ~ .kJ ((w)w~::~::!, l)dw. (14.29.5) To prove this, insert the Dirichlet series for lf"w) on the right-hand side and integrate term by term. This is justified by absolute con-vergence. We obtain co l+ioo L ~(n) f I x••P-'-1 n~t 21Ti iWw(w+p)(w+p l)dw. 2-ioo Evaluating the integral in the usual way by the calculus of residues, we obtain L ~(n)(X'-':::;"-'- X';"")~ L ~(n) Jx(xP-'-"';')dx n.;;x P P .. _,.x ,. ~ f? ~(n)xP-'(1-:f)dx ~ J .M(x)xP-'(1-:f) dx, 1 ...... x l and (14.29.5) follows. 14.29 CONSEQUENCES OF RIEMANN HYPOTHESIS 379 Let ·U be not the ordinate of a zero. Then the right-hand side of ( 14.29.5) is equal to .k(I <r +J: \I +JJ+ +sum of residues in -U < l(w) < U. Let p" run through zeros of {(a) with imaginary parts between - U and U. Let U > T. Then there is a pole at w = 1-p, with residue log X (1-p)('(l-p). At the other p" the residues are ('(p")ef;.·:~~~~<e+p) ~ o(l(p"+p ll)(p"+p)l) ~ o(ly"!yl')' by (14.29.3), and L I·! I'..; L ~<K,, -~~~~u Y Y Y -y Y +y where K 1 depends on T, if JyJ < T, but not on U. Again l+ioo f xw+p-1_1 ((w)w(w+p)(w+p dw ~ o(xi J" ~) I) u v(v+y)2 ll+iU ~ 0(u<;:y)) ~ 0(u<ff~T))' and similarly for the integral over (2-ioo, 2-iU). Also by (14.2.6) and the functional equation I ( ltl-l ) , ((±+it)~ O Wf-it)l ~ O(Jtl•-•). Hence, since Jw+pl ~ f, [w+p-l/ ~ f, }+iU U f X•"-'- I dw ~ o[ f ..1!1'2. dv) ~ 0(1). l-•u ((w)w(w+p)(w+p-1) -u (n+•')l Finally, by Theorem 14.16, we can choose a sequence of values of U such tha.t {(~) = O([wl) (t = U, } ~ u ~ 2). 380 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV By the functional equation the same result then holds for l ~ u ~ l also. Hence HJw x••P-'-1 ( xl ) ( xl ) ~(w)w(w+p)(w+p~l)dw=O {fl-~(U+rl' = 0 ijl-~(U-T) 2 ' -!-+iU and similarly for the integral over (2-iU, !-iU). Making U-+ oo, it follows that X I M(x)xP-2(1--xx) dx = ~ +R, ' (1-pl((l-p) where! Ri < 1\ 3 = K 3(T) if lrl < T. Hence we obtain o,;;AlogX+logX L ~I ''(I ll'-2logX L ~I ''(I )I'+ )ykT P':> p lyi<..T pr, p +AlogLY+K4(T), ' _1.:;:::A _:!____ K4(T~ 1 .f+"! I~ lnHirl+ 1 fn,)a..l !+iy > :!-At!t-yl y A A A >e-tf.>t· t Cramer and Landau (I). 14.30 CONSEQUENCES OF RIEMANN HYPOTHESIS 381 Suppose on the contrary that (t-t-3,t+t-3) is free from ordinates of zeros. Theorem 14.15 gives logl((i+illl ~ '5' logl!-ri+Oflog!logloglog!)· tt-yt,.i11.oglogt -loglogt There are O(log tjloglog t) terms in the sum, each being now O(log t). Hence loglm+il)l ~ a(llogl'! )• 1 ~ a(ex (A log''))· og ogt '(f+it) p loglogt Now 11'-t•P(fs)"s) is real on a= l- Hence -!logn+~ f'(t'l + ('(s) 2 f(!sl ((s) is purely imaginary on a= -f. Hence, on a= f, l ('(s)l ('(') f'(t') TI8f ;;, -R ((') ~ -ilogn+!R f(t'l ~ !log!+O(ll ~ oo. Hence (without any hypothesis) I('(Hilii;;,I((Hilll (<><,). This proves the theorem. 14.31. Let -f+iy, t-Hy' be consecutive complex zeros of '(s). If (14.29.1) is true We have y'-y>~exp(-Allolgy )· y og ogy o ~ fni+i!)d< ~ (r'-ri1'(Hirl+ f (r'-<W(Hi!ld<. y y Hence by (14.29.3) r'~y <Ali (y'-!)("(Hi!id<[ ' y• <A max l("(!+illl J (y'-!ld!~A(y'-y)' max !("(!+ill I· y,.t.;;y' y y.;;t.;y· Now ("(!+ill ~ ~ l mt::t,:•"l i"" dO ~ a(~ li((Hi!+""l I do) = o{~exp(A 10~!~t)(I+t~r)} 382 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV by (14.14.1) and the functional equation. Taking r = lflogt, ~"(!+it)= o{exp(A 10~ 1~~t)} and the result follows. 14.32. N eceasary and sujficient condition& for the Riemann hypofhesis. Two such conditions have been given in § 14.25. Other similar con~ ditions occur in the prime·number problem. t A different kind of condition was stated by M. Riesz.t Let "': (-1)'"""' ) F(x) ~ 6, (k-1)! {(2k)" (14.32.1 Then a simple application of the calculus of residues gives a+ioo a+ioo Fx -~ f __ x' __ ci.<J=_i f r(l-s)x-ds, ( ) -2 P(s){(2s).un~s 2~ {(2s) a-ioo a-ioo where ! < a < l. Taking a just greater tha.n !, it clearly follows that F(x) ~ O(xt••). On the Riemann hypothesis we could move the line of integration to a= f+E (using (14.2.5)) and obtain similarly F(x) ~ O(xl+•). (14.32.2) Conversely, by Mellin's inversion formula, I'(I-s) =-I"' F(x)x-1-scl. !; the analytic function represented is therefore regular for a > !, and the truth of the Riemann hypothesis follows. Hence (14.32.2) is a necessary and sufficient condition for the Riemann hypothesis. A similar condition stated by Hardy and Littlewood§ is "': ( -x)' -O(x-l) (14.32.3) 6,k!{(2k+l)-. These conditions have a superficial attractiveness since they depend explicitly only on values taken by t(s) at points in a> 1 ~but actually no use has ever been made of them. t Landau, Vork.mngen, ii. 10~56. tM.Ri682:(l). § Hardy and Littlewood (2). 14.32 CONSEQUENCES OF RIEMANN HYPOTHESIS 383 Conditions for the Riemann hypothesis also occur in the theory of Farey series. Let the fractions hjk with 0 < h ~ k, (h, k) = 1, k ~ N, arranged in order of magnitude, be denoted by r, (v = I, 2, ... , II>(N), where (N) ~ ~(l)+ ... +~(N)). Let ~. ~ r,-vj(N) be the distance between r, and the corresponding fraction obtained by dividing up the interval (0, I) into II>(N) equal parts. Then a necessary and sufficient condition for the Riemann hypothesis is:f: •l!)'l ( 1 ) ,?t 8~ = 0 Nt-~ . (14.32.4) An alternative necessary and sufficient condition is§ :~:10,1 = O(NiH). (14.32.5) Details are given in Landau's Vorle.&ungen, ii. 167-77. Still another conditionll can be expressed in terms of the formulae of § 10.1. IfE(t) and(fl(u) are related by (10.1.3), a necessary and sufficient condition that all the zeros of E(t) should be real is that (14.32.6) for all real values of x andy. But no method has been suggested of showing whether such criteria are satisfied or not. A sufficient conditiontt for the Riemann hypothesis is that the partial sums "~ 1 v... of the series for '(s) should have no zeros in a > 1. NOTES FOR CHAPTER 14 14.33. The argument of §14.5maybeextended to the strip!,;; a 0 .::;; a .::;; ~.giving ('(s) = o(0"-l- 1)+0(J"-ilogt). ((s) 1-a The choice 0 = (log t) 2 then yields ~~; ~(logi2-:"-l t Franel (I). § Landau (16). II See P61ya (3), § 7. tt Turan (3). 384 CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV uniformly for u0 ~ u ~ ! and t :;:: 2, and hence log {(s) ~ { ::: ",~, 1 ,, _ 1 if l+log:ogt ~ u~ j-, 1 (l g a) loglog t + logloglog t if «o ~a~ l+loglogt · These results, together with those of§ 14.14 are the sharpest conditional order-estimates available at present. 14.34. Then-result given by Theorem 14.12(A) has been sharpened by Montgomery [3 ), to give S(t) ~ o,( (logt)l ) (loglogt)t on the Riemann hypothesis. A minor modification of his method also yields It may be conjectured that these are best possible. Mueller has shown, on the Riemann hypothesis, that if c is a suitable constant, then S(t) changes sign in any interval [T, T+cloglogT]. Further results and conjectures on the vertical distribution of the zeros are given by Montgomery , who investigated the pair corre-lation function F(a.,T)=-1-L Tt~I1-Y'>w(y-y'), N(T) O<p'.;;T where w(u) = 4/(4 + u2). This is a real-valued, even, non-negative function of a:, and satisfies F(a., T) =a.+ T-241ogT+O(lo~T )+0(a.T~- 1 )+0(T-i~) (14.34.1) fora.;;?: 0, whenceF(a., T) ..... a. as T-+oo, uniformly forO< 0 ~a.~ 1-0. Montgomery conjectured that in general F(a:, T) ..... min (a., 1) (14.34.2) 14.34 CONSEQUENCES OF RIEMANN HYPOTHESIS uniformly for 0 ~ 0 ~ a. ~ A. This is related to a number of conjectures on the distribution of prime numbers. (See Gallagher and Mueller [I], Heath-Brown , and joint work of Goldston and Montgomery in the course of publication.) From (14.34.2) one may deduce that { , 2•• 2•P} # y,ye[O,T]:logT~y-y'~logT { I ' (sin••)' } -N(T) o(a,P)+ 1- -.-du for fixed a., {J, as T-+ oo. Here fJ(a., {J) = 1 or 0 according as a. ~ 0 ~ fJ or not. Using (14.34.1), Montgomery showed that L' m(p)',;; {!+o(1)}N(T), O(T);, {i+o(1)}N(T), (14.34.3) on the Riemann hypothesis. The conjecture (14.34.2) would indeed yield Nm(T)"""' N(T), i.e. 'almost all' the zeros would be simple. Montgomery also used (14.34.1) to show that li.minf Yn±t-Yn :<0·68· ••• (2(T) L IM(J:+iy)('(J:+iy)l'. O<y.;:T O ogy where the polynomial P(x) is chosen optimally as j-x-tx2. One may CONSEQUENCES OF RIEMANN HYPOTHESIS Chap. XIV write the sums occurring in (14.34.5) as integrals and 1 f \'(s) M ' 2;[ P ((s) (s)( (s)ds --"--, f ~((s)) M(s)M(1-s)('(s)('(1-s)ds, 2m p~S taken around an appropriate rectangular path P. The estimation of these is long and complicated, but leads ultimately to the lower bound N<'>(T);, {!~+o(1)}N(T). The estimate (14.34.4) has also been improved, firstly by Montgomery and Odlyzko {1], and then by Conrey, Ghosh and Gonek . The latter work produces the constant 0·5172. The corresponding lower bound limsupy"+t-Yn ;3A.>l n~<X' 2tt/logy,. (14.34.6) has been considered by Mueller {1], as well as in the two papers just cited. Here the best result known is that of Conrey, Ghosh, and Gonek {1], which has A. = 2·337. Indeed, further work by Conrey, Ghosh, and Gonek, which is in the course of publication at the time of writing, yields A. = 2·68 subject to the generalized Riemann hypothesis (i.e. a Riemann hypothesis for ((s) and all Dirichlet L·functions L(s, x).) Moreover it seems likely that this condition may be relaxed to the ordinary Riemann hypothesis with further work. Hone asks for bounds of the form (14.34.4) and (14.34.6) which are satisfied by a positive proportion of zeros (as in § 9.25) then one may take constants 0·77 and 1·33 (Conrey, Ghosh, Goldston, Gonek, and Heath-Brown ). 14.35. It should be remarked in connection with § 14.24 that Selberg (4) proved Theorem 14.24 with error term O(T), while the method here yields only 0{ T(loglog T)!}. Moreover he obtained the error term O{TOoglog T)"- 1 } for (14.24.1). 14.36. The argument of the final paragraph of § 14.27 may be quantified, and then yields L l('(l:+iy)I·',.T. lri"'T uniformly forT ;3 T0 , assuming the Riemann hypothesis and that all the zeros are simple. However a slightly better result comes from combining 14.36 CONSEQUENCES OF RIEMANN HYPOTHESIS 387 the asymptotic formula L 1('(!:+ iy)l'- f.N(TXlog T)' O<y .. T of Gonek {2] with the bound (14.34.3). Using HOlder's inequality one may then derive the estimate L. ' -1-.- .. T, CT<y 0 is a suitable numerical constant. 14.37. The Mertens hypothesis has been disproved by Odlyzko and te Riele , who showed that limsup~>1·06 x-"-' yx and lim inf ~ < - HMJ9. x-+oo yx Their treatment is indirect, and produces no specific x for which IM(x)l_> xi. The method used is computational, and depends on solving numencally the inequalities occurring in Kronecker's theorem, so as to make the first few terms of(14.27.1) pull in the same direction. To this extent Odlyzko and te Riele follow the earlier work of Jurkat and Peyerimho:ff [1 ], but they use a much more efficient algorithm for solving the Diophantine approximation problem. 14.38. Turan (3) conjectured that '" ,l(n) L..-;30 """ n (14.38.1) for all x > 0, where A.(n) is the Liouville function, given by (1.2.11). He showed that his condition, given in § 14.32, implies the above conjecture, which in turn implies the Riemann hypothesis. However Haselgrove proved that (14.38.1) is false in general, thereby showing that Turiln's condition does not hold. Later Spira {1] found by calculation that has a zero in the region a > 1. XV CALCULATIONS RELATING TO THE ZEROS 15.1. It is possible to verify by means of calculation that all the complex zeros of {(s) up to a certain point lie exactly (not merely approximately) on the critica.lline. As a simple example we shall find roughly the position of the first complex zero in the upper half-plane, and show that it lies on the critical line. We consider the function Z(t) = e10o{(f+it) defined in§ 4.17. This is real for real values oft, so that, if Z(t1) and Z(ta) have opposite signs, Z(t) vanishes between t1 and t2, and so {(s) has a zero on the critical line between !+it. and !+i~. It follows from (2.2.1) that{(!) < 0, then from (2.1.12) that t(f) > O, i.e. that E(O) > 0; and then from (4.17.3) that Z(O) < 0, We shall next consider the value t = &rr. Now the argument of§ 4.14 shows that, if x is half an odd integer, Hence, taking t > 0, I Z(t)- ""cos(tlogn-8-)1 ~ ~+~· L nl '"""t 21TX-t •<• For x = ~. t = 6?T, the right-hand side is about 0·6. We next require an approximation to &. We have so that ,.,- (' . ) -u r(i-!it) e -X 1-+~t -1T P(!+!it)' & ~ -!tlog~+llog r(t+!it) ~ }tlog-'--!•-l•+o(!). 21T t (15.1.1) (15.1.2) It may be verified that the term 0(1/t) is negligible in the calculations. Writing{}= 27tK, and using the values log 2 = 0·6931, log 3 = 1·0986, it is found that K = 0·1166, 3log3-K = 3·179 3log2-K = 1·963, 3log4-K = 4·042, 15.1 CALCULATIONS RELATING TO THE ZEROS 389 approximately. Hence the cosines in {15.1.2) are all positive, and cos 2-rrK = 0·74 .... Hence Z(&u) > 0. There is therefore one zero at least on the critical line between t = 0 andt= &u. Again, the formulae of§ 9.3 give N(T) ~ 1+2K+~t.arg{(s), where ~ denotes variation along (2, 2+iT, !+iT). Now R ns) > 0 on a = 2, and an argument similar to that already used, but depending on (15.1.1), shows that R((s) > 0 on (2+iT, !+iT), if T = &u. Hence lf>arg{(•)l <b. and N(IJ,) < f+2K < 2. Hence there is at most one complex zero with imaginary part less than &u, and so in fact just one, namely the one on the critical line. 15.2. It is plain that the above process can be continued as long as the appropriate changes of sign of the function Z(t) occur. Defining K = K(t), as before, lett~ be such that Then (15.1.2) gives K(t,) ~ !•-I (• ~I, 2, ... ). (15.2.1) Z(t,) (-I)'~ oos(t~~ogn) If the sum is dominated by its first term, it is positive, and so Z(t~) has the sign of (-1)~. If this is true for v and v+l, Z(t) has a zero in the interval (tv, t~+1 ). The va.lue t = 61T in the above argument is a rough approximation to t2• The ordinates of the first six zeros are 14·13, 21·02, 25·01, 30·42, 32·93, 37·58 to two decimal places.t Some of these have been calculated with great accuracy. 15.3. The calculations which the above process requires are very laborious if tis at all large. A much better method is to use the formula ( 4.17 .5) arising from the approximate functional equation. Let us write t= 21T'U, and a,.= a,.(u) = n-fcos2?T(K-ulogn), h([) ~ 00, 2•(['-f-llr). COS27T~ t See the refamncoo Gram (6), Linda!()£ (3), in Landau's Hmadbw:>h. 390 CALCULATIONS RELATING TO THE ZEROS Chap. XV Then (4.17.5) gives Z(2nu) ~ 2J,"•(u)+(-l)m-'u-lh(,lu-m)+R(u), where m = [ Yu], and R{u) = O(u.-1). The a,.(u) can be found, for given values of u, from a table of the function cos 21l'X. In the interval 0 ~ f ~ !, hU) decreases steadily from 0·92388 to 0·38268, and h(I-<J ~ h(<J. For the purpose of calculation we require a numerical upper bound for R(u). A rather complicated formula. of this kind is obtained in Titchma.rsh (17), Theorem 2. For values of u which are not too small it can be much simplified, and in fact it is easy to deduce that IR(u)l < ~ (u > 125). This inequality is sufficient for most purposes. Occasionally, when Z(27rU) is too small, a second term of the Riemann-Siegel asymptotic formula has to be used. The values of u for which the calculations are performed are the solutions of (15.2.1), since they make a:1 alternately 1 and -I. In the calculations described in Titchmarsh (17), I began with u = 1·6, K = -0·04865 and went as far as u = 62·785, K = 98·5010. The values of u were obtained in succession, and are rather rough approximations to the uv, so that the K's are not quite integers or integers and a half. It was shown in this way that the first 198 zeros of ~(8) above the real axis all lie on the line u = f. The calculations were carried a great deal farther by Dr. Comrie. t Proceeding on the same lines, it was shown that the first 1,041 zeros of ~(8) above the real axis all lie on the critical line, in the interval 0 < t < 1,468. One interesting point which emerges from these calculations is that Z(tvl does not always have the same sign as (-1)v. A considerable number of exceptional cases were found; but in each of these cases there is a neighbouring point(. such that Z((.) has the sign of (-1)v, and the succession of changes of sign of Z(t) is therefore not interrupted. 15.4. As far as they go, these calculations are all in favour of the truth of the Riemann hypothesis. Nevertheless, it may be that they do t See Titchmarsh (18). 15.4 CALCULATIONS RELATING TO THE ZEROS 391 not go far enough to reveal the real state of affairs. At the end of the table constructed by Dr. Comrie there are only fifteen terms in the series for Z(t), and this is a very small number when we are dealing with oscillating series of this kind. Indeed there is one feature of the table which may suggest a change in its character farther on. In the main, the result is dominated by the first term at> and later terms more or less cancel out. Occasionally (e.g. at K = 435) all, or nearly all, the numbers a:" have the same sign, and Z(t) has a large maximum or minimum. As we pass from this to neighbouring values oft, the first few~ undergo violent changes, while the later ones vary comparatively slowly. The term a,. appears when u = n1, and here cos211(K-ulogn) = cos11{ulog(ujn2)-u-}+ ... } = cos11(n2+l+ ... ) = (-l)"cos}n+. and !uaf'ldU Mathematik. Mathe~ Annalen. MathematillCheZeit.rehrijt. Pn>eeedings of the Cambridge Phikuluphical Society. Proceeding" of the Lond l, GOttitl{/er Nachrichten (l9ll), 409-28. (3) Sur !'existence de valeurs arbitrairement petites de lafonction ((a) = C I, 0Vf!t"sigt Videnak. Selsk. Kr.lbenhavn (I911), 201-8. (4) Sur Ia fonction {(s) dana le demi-pla.n u > 1, O.R. 154 (1912), I078-81. (5) V"ber die Funktion r(8)/{(s), J.M. 141 {1912), 217-34. (6) En nyt Bevis for, at den Riemann'ske Zetafunktion {(a) = {(cr+it) bar uendelij mange NuJpunkten indenfor Parallel-strimlen 0 ,-;;;; u,;;;,; I, Nyt. Tidsa.jor Math. (B) 23 (1912), 81-5. (7) Om de Vaerdier, den Riemann'ske Funk.tion {(u+it) a.ntager i HaJvplanen u > I, 2. Skand. Math. Kongr. (1912), 113--21. (8) Note sur Ia fonction zeta. de Riemann {(8) = {(u+U) sur Ia droite a= 1, Over.tigt Vidensk. Selsk. Kr.lbenhatm (l!H3), 3-11. (9) Losung des absoluten Konvergenzproblems einer allgemeinen Kle.sse Dirichletscher Reihen, A.M. 36 (1913), 197-240. (10) Sur Ia fonction {(a) de Riemann, C.R. 158 (1914), 1986-8. (II) Zur Theorie der Riemannschen Zetafunktion im kritischen Streifen, A.M. 40 (1915), 67-100. (12) Die Riemannsche Zetafunktion, Deutsche Math. Ver. 24 (1915), 1-17. (13) Ober eine que.si-periodiache Eigenschaft Dirichletacher Reihen mit An-wendung auf die Dirichletschen L-Funktionen, M.A. 85 (1922), ll5-22, (14) Ober diopba.ntische Approximationen und ihre Anwendungen auf Dirich· Ietache Reihen, besonders auf die Riemannsche Zetafunktion, 5. Skand. Math. Kongr. (1923), 131-54. (15) Another proof of Kronecker's theorem, P.L.M.S. (2), 21 (1922), 315-16. (16) Again the Kronecker theorem, J.L.M.S. 9 (1934), 5-6. BoHR, H., and Cou:&ANT, R. (1) Neue Anwendungen der Theorie der Diophantischen Approxima.tionen auf die RiemaiUlsche Zeta.funktion, J.M. 144 (1914), 249-74. BoHR, H., a.nd JESSEN, B. (1), (2) tlber die Werteverteihmg der Riemannschen Zeta.funktion, A.M. 54 (1930), 1-35 and ibid. 58 (1932), 1-55. (3) One more proof of Kronecker's theorem, J.L.M.S. 7 (1932), 274-5. (4) Mean-value theorems for the Riemannzeta.-function, Q.J.O. 5 (1934), 43-7. (5) On the distribution of the va.lues of the Riemann zeta·ftmetion, Amer. J. Math. 58 (1936), 35-44. BoHR, H., a.nd LANDAu, E. (I) Ober da.s Verhalten von {(s) und {il(a) in der Ni!.he dar Oeraden u = 1, Giittinger Nachrichten (1910), 303-30. {2) tlber die Zetafunktion, Rend. di Palermo, 32 (1911), 278-85. (3) Beitrage zur Theorie der Riemannschen Zeta.fun.ktion, M.A. 74 (1913), 3-30. (4) Ein Satz iiber Dirichletache Reihen mit Anwendung auf die {-Funktion und die L-Funktionen, Rend. di Palermo, 37 (1914), 269-72. (5) Sur lea zero& de Ia. fonction {(a) de Riemann, C.R. 158 (1914), 106-10. (6) tJber das Verha.lten von 1/{(8) auf der Oeraden a= 1, &Ottinger Nach-rlchten (1923), 7I-80. (7) Na.chtrag zu unseren Abha.ndlungen aua den Jahrg§ngen 1910 und 1923, Glittinger Nachrichten (1924), 168-72. 394 ORIGINAL PAPERS BOHR, H., LANDAU, E., and LITTLEWOOD, J. E. (1) Sur Ia fonction t(s) dans le voisinago de Ia. droite u = l, BuU. Acad. Belgique, 15 (1913), 1144-75. 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Astr. och Fysik, 16 (1922), No. 18, and 19 (1926), No. 25. CHOWLA, S. D. (1) On some identities involving zeta-functions, Journal Indian Math. Soc. 17 (1928), 153-63. CoRPUT, J. G. VAN DEll. (1) Zahlentheoretische Abschti.tzungen, M.A. 84 (1921), 53-79. (2) VerschilrfWlg der Abschatzung beim Teilerproblem, M.A. 87 (1922), 39-65. (3) Neue zahlenthcoretische Abschii.tzungen, erste Mitteilung, M.A. 89 (1923), 215-54. (4) Zum Teilerproblem, M.A. 98 (1928), 697-716. (5) Zahlentheoretische Abschii.tzungen, mit AnwendWlg auf Gitterpunkt. probleme, M.Z. 28 (1928), 301-10. (6) Neue za.hlentheoretische Abschatzungen, zweite Mitteiltmg, M.Z. 29 (1929), 397-426. (7) ttber Weylsche Summen, Mathe!matica B (1936-7), 1-30. CORPUT, J. G. VANDER, and KOKSMA. J. F. (1) Sur l'ordre de grandeur de Ia fonction '(s) de Riemann dans Ia ba.nde critique, Annale& d6 Toulous6 (3), 22 (1930), l-39. CRAIG, C. F. (I) On the Riemann ,.Function, BuU. A mer. Math. Soc. 29 (1923), 337-40. CaAME:a, H. (1) i.Jber die Nullstellen der Zetafunktion, M.Z. 2 (1918), 237-41. (2) Studien tiber die Nullstellender Riema.nnschenZetafunktion, M.Z. 4 (1919), 104-30. ORIGINAL PAPERS 395 (3) Bemerkung zu der voratehenden Arbeit des Herrn E. Landau. M .z. 6 (1920), 155-7. (4) V"berda.s Teilerproblem von Piltz, ArkivfiJr Mat. Astr. och.Fysik,I6 (1922), No.2!. (5) Ein Mittelwertsa.tz in der Primzahltheorie, M.Z. 12 (1922), 147-53. CRAMER, H., and LANDAu, E. (1) V"ber die Zeta.fWlktion auf der Mittellinie des kritischen Streifens, Arkiv /IJr Mat. Astr. och Fysik, 15 (1920), No. 28. CRUM,M.M, (I) On some Dirichlet series, J.L.M.S. 15 (1940), 10-15. DAVENPORT, H. (1) Note on mean-value theorems for the Riemann zeta-ftmction, J.L.M.S. 10 (1935), 136-8. DAVENPORT, H., and HEILBRONN, H. (1), (l) On the zeros of certain Dirichlet series I, II, J.L.M.S. ll (1936), 181-5 and 307-12. !>ENJOY, A. (1) L'hypothMe de Riemann sur Ia distribution des zeros de "s), reliee A Ia. th6orie des probabilites, O.R. 192 (1931), 656-8. DEURING, M. {1) Ima.ginll.re qua.dratische Zahlk6rper mit der Kla.ssenzahl1, M.Z. 37 (1933), 405-15. (2) On Epstein's zeta-function, Annals of Math. (2) 38 (1937), 584-93, ESTERMANN, T. {1) On certain functions represented by Dirichlet series, P.L.M.S. (2), 27 (1928), 435-48. {2) On a problem of analytic continuation, P.L.M.S. 27 (1928), 471-82. (3) A proof of Kronecker's theorem by induction, J.L.M.S. 8 (1933), 18-20. FAVARD, J. (I) Sur la repartition des points oU une fonction prtll!que petiodique prend tme valeur donnOO, O.R. 194 (1932), 1714-16. FEJER, L. (I) Nombre de changements de signe d'une fonction dans un intervalle et sea moments, C.R. 158 (1914), 1328-31. FEKETE,M. (1) Sur tme limite inf6rieure des changements de signa d'une fonction dans un intervalle, C.R. 158 (1914), 1256-S. (2) The zeros of Riemann's zeta·function on the critical line, J.L.M .S. 1 (1926), 15-19. FLE'l"'', T. M. (1) On the function ~~sin~,J.L.M.S. 25 (1950), 5-19. (2) On a coefficient problem of Littlewood and some trigonometricaJ. sums, Q.J.O. (2) 2 (1951), 26-52. FRANEL, J, (1) Les suites de Farey et le probl8me des nombres premiers, GOtting6r Ncwh-richten (1924), 198-201. 396 ORIGINAL PAPERS GABRIEL, R. :M. (1) Some results concerning the integrals of moduli of regular functions a.long certain curves, J.L.M.S. 2 (1927), 112-17. GRAM, J.P. (1) Tafeln fUr die Riemannsche Zetafunktion, Skriften Kobenhavn (8), 9 (I92S), 311-25. GRONWALL, T. H. (1) Sur Ia fonction {(s) de Riemann au voisinage de u = 1, Rend. di Palermo, 35 (1913), 95-102. (:Z) Ober da.s VerhaJten der Riemannschen Zeta-funktion auf der Geraden u = l, Areh. der Math. u. Phys. (3) 21 (1913), 231-8. GROSSMAN, J. (I) i)ber die Nullstellen der Riema.nnschen Zeta-funktion und der Dirich-Jetachen L-Funktionen, Di88erlation, GQttingen (1913). G~~~Dfo~~~ for {(s) in the critical strip, J.L.M.S. 14 (1939), 97-IC.O. (l) Some Fourier transforms in prime-nUTilher theory, Q.J.O. 18 (1947), 53-64. (3) Some formulae for the Riemann zeta-function, J.L.M.S. 22 (1947), H-18. (4) Fourier reciprocities and the Riemann zeta-function, P.L.M.S. 51 (1949), 4-01-14. H.U)AMARD, J. (1) Une application d'une fonnule integrale relative aux series de Dirichlet, Btdl. Soc. Math. de France, 56, ii (1927), 43-4. HAMBUBGER, H. (1 ), (2), (3) tlber die Riemannsche Funktionalgleichung der ,.Funktion, M.Z. lO (1921), 240-54; u (1922), 224-45; 13 (1922), 283-3ll. (4) tiber einige Beziehungen, die mit der Funktionalgleichung der Riema.nn-schen ,.Funktion a.quivalent sind, M.A. 85 (1922), 129--40. HARDY, G. H. (1) Surles zerOs de Ia fonction ,(8) de Riemann, C.R. 158 (1914), 1012-14. (l) On Dirichlet's divisor problem, P.L.M.S. (2), 15 (1915), 1-25. (3) On the average order of the arithmetical functions P(n) and Ll(n), P.L.M.S. (2), 15 (1915), 192-213. (4) On some definite integrals considered by Mellin, Mes8enger of Math. 49 (1919), 85-91. (5) Ramanujan's trigonometrical function e~(n), P.C.P.S. 20 (1920), 263-71. {6) On the integration of Fourier series, Messenger of Math. 51 {1922), 186--:92. (7) A new proof of the functional equation for the zeta-function, Mal. Tid8-skrift, B (1922), 71-3. (8) Note on a theorem of Mertens, J.L.M.S. 2 (1926), 70-2. HARDY, G. H., INGHAM, A. E., and P6LYA, G. (1) Theorems concerning mean values of analytic functions, Proc. Royal Soc. (A), 113 (1936), 542-69. HARDY, G. H., and LITI'LEWOOD, J. E. (1) Some problems of Diophantine approximation, Internal. Congress of Math., Cambridge {1912), 1, 223-9. (2) Contributions to the theory of the Riemann zeta-function and the theory of the distribution of primes, A.M. 41 (1918), 119-96. ORIGINAL PAPERS (3), The zeros of Riemann's zeta-function on the critical line, M.Z, 10 (1921), 283-317. (4) The approximate functional equation in th~ theory of the zeta.-function, with applications to the divisor problems of Dirichlet and Piltz, P.L.M.S. (2), 21 (1922), 39--74. (5) On Linde10f's hypothesis concerning the Riema.nn zeta-function, Proc. Royal Soc. (A), 103 {1923), 403-12. (6) The approximate functional equations for ns) and ' 2(s), P.L.M.S. (2), 29 (1929), 81-97, liARTliiAN, P. (1) Mean motions and almost periodic functions, Trans. Amer. 1l:fath. Soc. 46 (1939), 66-81. liASELGROVE, c. B. (1) A connexion between the zeros and the mean values of "s), J.L.M.S. 24 (1949), 215-22. HASSE, H. (I) Beweis des Analogons der Riemannschen Vermutung flir die Artinschen undF. K. SchmidtschenKongruenzzeta.funktioneningewissenelliptischen Fillen, Giittinger Nacht"idden, 42 (1933), 253-62. HAVILAND, E. K. (1) On the asymptotic behavior of the Riemann g.function, Amer. J. Math. 67 (1945), 411-16. HEeKE, E. (1) fiber die L6sungen der Riemannschen Funktionalgleiclumg, M.Z. 16 (1923), 301-7. (2) 'Ober die Bestimmtmg Dirichletscher Reihen durch ihre Funktional· gleichung, M.A. ll2 (1936), 664-99. (3) Her1eitung des Euler-Produktes der Zetafunktion und einiger L·Reihen aus ihrer FunktiollB,lgleichung, M.A. U9 (1944), 266-87. HBILBRONN, H. {1) itber den Primzahlsatz von Herm Hoheisel, M.Z, 36 {1933), 394-423. HILLE, E. (1) A problem in 'F8(ltorisatio Numerorwn', Acta ArUh. 2 (1936), 134-44. HOHEISEL, G. (1) NormaJ:folgen und Zetafunktion, Jahresber. Bchles. GeuU. 100 (1927), l-7. (2) Eine Illustration zur Riemannschen Vermutung, M.A. 99 (1928), 150-61. {3) tlber das Verha.lten des reziproken Wertes der Riemannschen Zeta· Funktion, Sitzungsber. PTeUS~J. Akad. Wisa. (1929), 219-23. (4) Nullstellenanzahl und Mittelwerte der Zetafunktion, Sitzungsber. Pr(fUII8, Akad. Wi88. (1930), 72-82. {5) Primzahlproblerne in der Analysis, Sil~ungsber. Preuas. Akad. Wis8. (1930), 580--8. HOLDER, o. (1) 'Ober gewisse der MObiusschen Funktion p.(n) verwandte zahlentheoretische Funktionen, der Diriehletschen Multiplikation und eine Verallgemeine-rung der Umkehrungsfonneln, Ber. Verh. sii.eh8. Akad. Leipzig, 85 (1933), 3-28. HtrA,L.K. (1) An improvement of Vinogmdov's mea.n-value theorem and several applica· tiona, Q.J.O. 20 (1949), 4S-6l. ORIGINAL PAPERS HUTCHINSON, J. I. (I) On the roots of the Riemann zeta-function, Trans. Amer. Math. Soc. 27 (1925), 49-60. !NGHAlll, A. E. (1) .Mean-value theorems in the theory of the Riemann zeta-function, P.L.M.S. (2), 27 (1926), 273-300. (2) Some asymptotic formuloo in the theory of numbers, J.L.M.S. 2 (1927), 202-8. (3) Notes on Riemann's C-function and Dirichlet's £-functions, J.L.M.S. 5 (1930), 107-12. (4) Mean-valuetbeoremsandtheRiemannzeta-function,Q.J.0.4(1933),278-90, (5) On the difference between consecutive primes, Q.J.O. 8 (1937), 255-66. (6) On the estimation of ~V(ri, T), Q.J.O. 11 (1940), 291-2. (7) On two conjectures in the theory of numbers, A mer. J. Math. 64 (1942), 313-19. JARNix, V., and L.o;nAu, E. (I) Untersuchungen iiber einen van der Corputschen Satz, M.Z. 39 (1935), 745-67. JESSEN, B. (I) Eine Integrationstheorie fiir FWlktionen unendlich vieler Veriinderlichen, mit Anwendung auf da.s Werteverteilungsproblem f(ir fastperiodische Funktionen, insbesondere fiir die Riemannsche Zetafunktion, Mat. Tidsskrijt, B (1932), 59-65. (2) Mouvement moyen et distribution des valeurs des fonctions presque· periodiques, 10. Skand. Math. Kangr. (1946), 301-12. JESSEN, B., and WINTNER, A. (1) Distribution functions and the Riemann zeta-function, Trans. A mer. Math. Soc. 38 (1935), 48--88. KA_c, M., and STEINHAUS, H. (I) Surles fonctions independantes (N), Stwlia Math. 7 (1938), 1-15. KAMPEN, E. R. VaN (1) On the addition of convex curves and the densities of certain infinite convolutions, A mer. J. Math. 59 (1937), 679-95. KAMPEN, E. R. VAN, and WINTNER, A. (1) Convolutions of distributions on convex curves and the Riemann zeta-function, Amer. J. Mmh. 59 (193i), 175--204. KERSHNER, R. (I) On the values of the Riemallll C-function on fixed lines u >I, Amer. J. Math. 59 (1937), 167-74. KERSHNER, R., and 'VINTNER, A. (I) On the boundary of the range of values of C(s), Amer. J. Math. 58 (1936), 421-5. (2) On the asymptotic distribution of C'/C(s) in the critical strip, Amer. J. Math. 59 (1937), 673-8. KmNAST,A. (1) Uber die Dirichletschen Reihen fiir CP(s), LP(s), Comment. Math. Helv. 8 (1936), 359-70. KLoosTERMAN, H. D. (1) Een integraal voor de C-functie van Riemann, Ohriatiaan Huygens J.l!ath. Tiid&chrift, 2 (1922), 172-7. I • ORIGINAL PAPERS 399 KLUYVER, J. C. (I) On certain series of Mr. Hardy, QtWrt. J. of Math. 50 (1924), 185-92. KOBER, H. (I) Tra.nsformationen einer bestimmten Besselschen Reibe sowie von Potenzen der Riemannschen C·Funktion und von verwa.ndten Funktionen, J.M. 173 (1935), 65-78. (2) Eine der Riemannschen verwandte Funktionslgleichung, M.Z. 39 (1935), 630-3 . {3) Funktionen. die den Potenzen der Riemannschen Zetafunktion verwandt sind, und Potenzreiben, die iiber den EinheitskrelS nicht fortsetzbar sind, J.M. 174 (1936), 206-25. (4) Eine Mittelwertformel der Riema.nnschen Zetafunktion, Compositio Math. 3 (1936), 174-89. KOCH, H, VON (1) Contribution A Ia theorie des nombres premiers, A.M. 33 (1910), 293-320. KOSLIAKOV, N. (1) Some integral representations of the square of Riemann's function E(t), C.R. Acad. Sci. U.R.S.S. 2 (1934), 401-4. (2) Integral for the square of Riemann's function, G.R. Acad. Sci. U.R.S.S. N.S. 2 (1936), 87-90. (3) Some formulae for the functions C(s) and C,(s), C.R. Acad. Sci. U.R.S.S. (2), 25 (1939}, 567-9. KRAMASCHKE, L. (1) Nullstellen der Zetafunktion, Deutsche Math. 2 (1937), 107-10. KuSJIIIN,R. {1) Surles zero de Ia fonction '(s) de RiemaWl, C.R. Acad. Sci. U.R.S.S. 2 (1934), 398-400. LANDAU, E. (1) Zur Theorie der Riemannschen Zetafunktion, Viertel,jahrssehr. Naturf. Gea. Zilrich. 56 (1911), 125-48. (l) tlber die Nullstellen der Zetafunktion, M.A. 71 (1911), 548-64. (3) Ein Satz Uber die ,.Funktion, Nyt. Tidss. 22 (B), (1911), 1-7. (4) "'Ober einige Summen, die von den Nullstellen der Riemannschen Zeta-funktion abhangen, A.M. 35 (1912), 271-94. (5) Ober die Anzahl der Gitterpunkte in gewissen Bereichen, GOUinger Nach· rWhten (1912), 687-771. (6) Geli:iste and ungelOSte Probleme aus der Theorie der Primzahlverteilung 1md der Riemannschen Zetafunktion, Jahresbe:r. der Deutsehen Math. Ver. 21 (1912), 208-28. (7) Geltiste and ungeltiste Problems aus der Theorie der Primzahlverteilung und der Riemannschen Zetafunktion, Proc. f) Internal. Math. Crmgr. (1913), 1, 93-108. (8) "'Ober die Hardysche Entdeckung unendlich vieler Nullstellen der Zeta-fWlktion mit reellem Teill, M.A. 76 (1915), 212-43. (9) Ober die Wigertsche asymptotische Funktionalgleichung fiir die Lam-bertsche Reibe, Arch. d. Math. u. Phys, (3), 27 (1916), 144--6. (10) Neuer Beweis eines Satzes von HelTil Valiron, Jahreaber, der IkutscAen Math. Ver. 29 (1920), 239. (11) 1l'ber die Nullstellen der Zetafunktion, M.Z. 6 (1920), 151-4. 400 ORIGINAL PAPERS (12) 'Ober die Nullstallen der Dirichletschen Reihen und der RiernB.ll.n8Chen '·Funktion, Arkivfiir Mal. Astr. och Fyaik, 16 (1921), No.7. (13) tlber die MObiussche Funktion, Rend. di Palermo, 48 (1924), 277-80. (14) 'Ober die Wurzeln der Zetafunktion, M.Z, 20 (1924), 98--104. (15) Uber die {-Funktion und die L-Funktioncn, M.Z. 20 (1924), 105--25. (16) Bemerkung zu der vorstehenden Arbeit von Herm Franel, GiiUinger Nachrichten (1924), 202-6. (17) "Ober die Riema.nnsche Zetafunktion in der Nilhe von u = I, Rend. di PaleT'TTW, 50 (1926-), 423--7. (18) ltber die Zetafunktion und die Ha.da.mardsche Thoorie der ganzen Funkti-onen, M.Z. 26 (1927), 170--.5. (19) "Vber daa Konvergenzgebiet einer mit der RieilliiJlii.SCben Zetafunktion zusa.mmenhii.ngenden Reihe, M.A. 97 (1927), 251-90. (20) Bemerkung zu einer Arbeit von Hm. Hoheisel iiber die Zetafunktion, Silzungsber. Pr 1, GOtlinger Nach· richten (1933), 81-91. LANDAU, E., and WALFISZ, A, (I) tlber die Nichtfortsetzba.rkeit einiger durch Dirichletsche Reihen definierter Funk.tionen, Rend. di Palermo 44 (1919), 82-6. LERCH, M. (I) Vber die Bestimmung der Koeffizienten in der Potenzreihe fUr die Funktion S J, O.R. 154 (1912), 263-6. (2) Researches in the theory of the Riemann {-function, P.L.M.S. (2) 20 (1922), Records xxii-xxviii. (3) Two notes on the Riemann zeta-function, P.O.P.S. 22 (1924), 234-42. (4) On the zeros of the Riemann zeta-function, P.O.P.S. 22 (1924), 295-318. (5) On the Riemann zeta-function, P.L.M.S. (2), 24 (1925), 175-201. (6) On the function lj{{l +ti), P.L.M.S. (2), 27 (1928), 349-57. MAIER, W. (1) Gitterfunktionen der Zahlebene, M.A. ll3 (1936), 363-79, MALURKAR, S. L. (1) On the application of Herr Mellin's integrals W some aeries, Journal Ifldian Math. Soe. 16 (1925), 130-8. MATTSON, R. (1) Eine neue Darst.ellung der Riemann'schen Zetafunktion, Arkiv fOr Mal. Alllr. ooh Fyllik, 19 (1926), No. 26. ORIGINAL PAPERS 401 MELLIN, H. (I) tlber die Nullstellen der Zetafunktion, Annales Aead. Scienliarium Fen. nicae (A), 10 (1917), No. 11. MEULENBELD, B. (I) Een approxllnatieve Functionaa.lbetrekking van de Zeta.functie van Riemann, Dis8ertalion, Groningen, 1936. MIKOL.ls,M. (1) Sur !'hypothese de Riemann, O.R. 228 (1949), 633-6. MIN, S. H. (1) On the order of {(j+il), Trans. Amer. Math. Soc. 65 (1949), 448-72. MIYATAKE, Q. (I) On Riemann's e-function, T6hoku Math. JourtUJl, 46 (1939), 160-72. MoRDELL, L. J. (I) Some applications of Fourier series in the analytic theory of numbel"S, P.O.P.S. 34 (1928), 585-96. (2) Poisson's summation formula and the Riemann zeta-function, J.L.M.S. 4 (1929), 285-91. (3) On the Riemann hypothesis and imaginary quadratic fields with a given class number, J.L.M.S. 9 (1934), 289-98. MUNTz, C. H. (I) Beziehungen der Riem8.Illll!chen {-Funktion zu willkiirlichen reellen Funktionen, Mat. Tidllllkrijl, B (1922), 39-47. MUTATKER, V. L. (1) On some formulae in the theory of the zeta-function, Journal Indian Math. Soc. 19 (1932), 220-4. NEVANLINNA, F. and R. (I), (2) tJber die NullsWilen der Riemann.schen Zetafunktion, M.Z. 20 (1924), 253-63, and 23 (1925), 159-60. OsTROWSKI, A. (I) Notiz tiber den Wertevorrat der Riema.nnschen {-Funktion am Rande des kritischen StreifenB, JahresberWlU Deut8(;h, Math. Verein. 43 (1933), 58-64. PALEY, R. E. A. C., and WIENER, N. (1) Notes on the theory and application of Fourier transforms V, TrantJ. AfiWT. Math. Soc. 35 (1933), 768-81. PHILLIPS, ERIC (1) The zeta-function of Riemann; further developments of van der Corput's method, Q.J.O. 4 (1933), 209-25. (2) A note on the zeros of {(s), Q.J.O. 6 (1935), 137-45. POL, B. VANDER, (1) An electro-mechanical investigation of the Riemann zeta-function in the critical strip, Bull. A mer. Math, Soc. 53 (1947), 976--81. POLYA, G. (1) Bemerkung tiber die lnWgraldarstellung der Riema.nnschen e-Funk.tion, A.M. 48 (1926), 305-17. (2) On the zeros of certain trigonometric integrals, J.L.M.S. I (1926), 98-9. (3) tlberdie algebraisch-funktiontheoretischen Untei'8Uchungen von J, L. W. V. Jensen, Kgl. Danllke Videnskabernes Sel8kab. 7 (1927), No. 17, (4) tJber trigonometriache Integrale mit nur reelle-n Nullstellen, J.M. 158 (1927), 6-18. 402 ORIGINAL PAPERS PoPOv, A. I. (1) Several series containing primes and roots of ((s), C.R. Acad. Sci. U.R.S.S., N.S, 41 (1943), 362-3. POTTER, H. S. A., and TrrcHMARSa, E. C. (1) The zeros of Epstein's rt:eta-functions, P.L.M.S. (2) 39 (1935), 372-84. RADEMACHER, H. (1) Ein neuer Beweis fill' die Funktionalgleichung der (-Funktion. M.Z. 31 (1930), 39-44. RAMANUJAN, 8. (1) New expressions for Riemann's functiona t'(s) and E(t), Quart. J. of Math. 46 (1915), 253-61. (2) Some formulae in the analytic theory of numbers, Musenger of Math. 45 (1915), 81-4. (3) On certain trigonometrica.l sums and their applications in the theory of numbers, TraTUJ. Camb. Phil. Soc. 22 (1918), 259-76. RAMASWAMI, V. (I) Notes on Riemann's (-function, J.L.M.S. 9 (1934), 165-9. RIESZ,M. (I} Sur l'hypoth6se de Riemann, A.M. 40 (1916), 185--90. SCHNEE, W. (1) Die Funktionalgleichung der Zetafunktion und der Dirichletschen Reihen mit periodiachen Koeffizienten, M.Z. 31 (1930), 378--90. SELBERG, A. (I) On the zeros of Riemann's zeta-function on the critical line, Arch. for Math. og Ncuurv. B, 45 (1942), 101-14. (2) On the zeros of Riemann's zeta-function, Skr. Norske Vid. Akad. Oslo (1942), no. 10. (3) On the normal density of primes in small intervals, and the difference between consecutive primes, Arch. for Math. og NCUurt•. B, 4 7 (1943), No.6. (4) On the remainder in the formula for N(T), the number of zeros of {(s) in the strip 0 < t < T, Avha:ndlinger No1"8ke Vid. Akad. Oslo (1944), no. 1. (5) Contributions to the theory of the Riemann zeta-function, Arch. for Math. og Naturv. B, 48 (1946), no. 5. (6) The zeta-function and the Riemann hypothesis, 10. Skand. Math. Kongr. (1946), 187-200. (7) An elementary proof of the prime-number theorem, Ann. of Math. (2) 50 (1949), 305-13. SELBERG, S. (1) Bemerkung zu einer Arbeit von Viggo Brun uber die Riemannsche Zeta-funktion, Norske Vid. Sekk. Forh. 13 (1940), 17-19. SIEGEL, C. L. (1) Bemerkung zu einem Satz von Hamburger ii.ber die Funktionalgleichung der Riemannschen Zetafunktion, M.A. 86 (1922), 276--9. (2) "Ober Riemanns Nachlass zur analytischen Zahlentheorie, QueUen und Btudien zur Geschichte der Mcuh. Astr. und Physik, Abt. B; Studien, 2 (1932), 45-80. ' (3) Contributions to the theory of the Dirichlet L-series and the Epstein Zeta-functions, Antu~/.8 of Math. 44 (1943), 143-72. SPEISER, A. (I) Geometrisches Z\U' Riema.nnschen Zetafunktion, M.A. IIO (1934), 514-21. ORIGINAL PAPERS STEEN; E:. w. P. (I) A linear transformation connected with the Riemann zeta-function, P.L.M.S. (2), 41 (1936), 151-75. STERNECK, R. VON (I} Neue empirische Daten tiber die zahlentheoretischc Funktionu(n), Interrnzt. Ccmgress of ~uath., Cambridge (1912), I, 341-3. SzAsz,O. (I) Introduction to the theory of divergent series, Cincinnati, 1944 (Math. Rev. 6, 45). TAYLOR, P. R. (I) On the Riemann zeta-function, Q.J.O. 16 (1945), 1-21. TCHUDAKOFF, N. G. (I) Sur Ies zeros de la fonction {(s), C.R. 202 (1936), 191-3. (2) On zeros of tho function {(s), C.R. A cad. Sci. U.R.S.S. 1 (x) (1936), 201-4. (3) On zeros of Dirichlet's £-functions, MaJ. Sbornik (I) 43 (1936), 591-602. (4) On Weyl's sums, ~ttat. Sbornik (2) 44 (1937), 17-35. (5) On the functions {(s) and 1T(x), C.R. Ac.ad. Sci. U.R.S.S. 21 (1938), 421-2. THIRUVENR;ATACIIABYA, V. (I) On some properties of the zeta-function, Journal Indian Math. Soc. 19 (1931), 92-6. TITCHXARSH, E. C. (I) The mean-value of the zeta-function on the critiealline, P.L.M.S. (2) 27 (1928), 137-50. (2) On the remainder in the formula for N(T), the number of zeros of {(s) in the strip 0 < t < T, P.L.M.S. (2), 27 (1928), 449-58. (3) A consequence of the Riemann hypothosis, J.L.M.S. 2 {1927), 247-54. (4) On an inequality satisfied by the zeta-function of Riemann, P.L.M.B. (2), 28 (1928), 70-80. (5) On the zeros of the Riemann zeta-function, P.L.M.S. (2) 30 (1929), 319-21. (6) Mean value theorems in the theory of the Riemann zeta-function, Messenger of Mcuh. 58 (1929), 125-9. (7) The zeros of Dirichlet's L-functions, P.L.M.S. (2) 32 (1931), 48s-500. (8)-(12) On van der Corput's method and the zeta-function of Riemann, Q.J.O. 2 (1931), 161-73; 2 (1931), 313-20; 3 (1932), 133-41; 5 (1934), 98-105; 5 (1934), 195-210. (13) On the Riemann zeta-function, P.C.P.S. 28 (1932), 273-4. {14) On the function 1/W+it), Q.J.O. 4 (1933), 64.-70. {15) On Epstein's zeta-function, P.L.M.S. (2) 36 (1934), 485-500. (16) The lattice-pointa in a circle P.L.M.S. (2) 38 (1935), 96--115. (17), (18) The zeros of the Riemann zeta-function, Proo. Royal Soc. (A), 151 (1935), 234-55, and ibid. 157 (1936), 261-3. (19) The mean value of IW-+itll'. Q.J.O. 8 (1937), 107-12. (20) On t(s) and 77(x), Q.J.O. 9 (1938), 97-108. (21) The approximate functional equation for ,~(s), Q.J.O. 9 (1938), 109-14. (22) On divisor problems, Q.J.O. 9 (1938), 21&-20. (23) A convexity theorem,J.L.M.S. 13 (1938), 196--7. (24) On the order of {(~-+it), Q.J.O. 13 (1942), 11-17. (25) Some properties of the Riemann zeta-function, Q.J.O. 14 (1943), 16-26. (26) On the zeros of the Riemann zeta (Unction, Q.J.O. 18 (1947), 4-16. 404 ORIGINAL PAPERS TORELLI, G. (1) Studio sulla fWlZione {(s) di Riemann, Napoli Rend. (3) 19 (1913), 212-16. TsuJr,M. (I) On the zeros of the Riemann zeta.-fnnction, Proc, Imp. Acad. Tokyo, 18 (1942), 631-4. Tm!.AN, P, (1) tlber die Verteilung der Primza.hlen I, Acta Szeged, lO (1941), 81-104. (l) On Riemann's hypothesis, Bull. Acad. Sci. U.R.S.S. 11 (1947), 197-262. (3) On some approximative Dirichlet-polynomials in the theOry of the zeta-function of Riemann, Danske VidenBk. Selakab, 24 (1948), Nr. 17, TuruNo, A. M. (1) A method for the caJculation of the zeta-function, P .L.M.S. (2), 48 (1943), 180-97. UI.'ZINGER, A. A. (1) Die reellen Ztige der Riemannschen Zeta.funktion, Di8sertation, ZUrich, 1934 (see Zentralblattfiir Math. 10, 163), VALIRON, G. (I) Sur lea fonctions entierea d'ordre nul et d'ordre fini, Annates de Toulouse (3), 5 (1914), 117-257. VALLJ!!E POUSSIN, C. DE LA {1), {l) Surles zeros de {(s) de Riemann, O.R. 163 (1916), 418-21 and 471-3. VINOGRAoov, I. M. (1) On Weyl's sums, Mat. SbQTnik, 42 (1935), 521-30. (2) A new method of resolving of certain general questions of the theory of numbers, Mat. Sbomik (1), 43 (1936), 9-19, (3) A new method of estimation of trigonometricalsUIIl8, Mat. Sbornik (1), 43 (1936), 175-188. (4) The method of trigonometrica.l sums in the theory of numbers, Trav. I1U1t. Math. Steklo.ff 23 (1947). VORONo:t, G. (1) Sur un problema du ca.lcul des fonctions a.sympt<ttiques, J.M. 126 (1903), 241-82. (2) Sur unefonction transcendante et ses applications A lasommationde quelques series, Amiale8 de l'Ecole N01male (3) 21 (1904), 207-68 and 459-534. WALFisz, A. (1) Zur Abechii.tzung von {(!+it). GOttinger Nachrichlen (1924), 155-8. (2) "Ober Gitterpunkte in mehrd.imensionalen Ellipsoiden VIII, Trovaux de l'Institut Math. de Tbiliasi, 5 (1938), 181-96. WALTHER, A. 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WEYL,H, (1) "Ober die Gleichverteilung von Zahlen mod. Eins, M.A. 77 (1916), 313-52. (2) Zur Abecha.tzung von {(I+ti), M.Z. 10 (1921), 88-101. Wurrr.AXER, J. M. (I) AninequalityfortheRiemannzeta.-function, P.L.M .S. (2), 41 ( 1936), 544-52. (2) Amean-value theorem for analytic functions, P.L.M.S. (2), 42 (1936), 186-911. WIGERT, S. (1) Sur Ia s6rie de Lambert et son application A Ia thOOrie des nombres, A.M. 41 (1916), 197-218. (2) Sur Ia. thOOrie de Ia fonction {(a) de Riemann, Arkiv f01" Mae. Aatr. och Fyllik (1919), No. 12. (3) On a problem concerning the Riemann {-function, P .G.P.S. 21 (1921),17-21. Wn.soN,B,M. (1) Proofs of some formulae enunciated by Ramanujan, P.L.M.S. (2) 21 (1922), 23fH55. WILTON, J. R. (I) Note on the zeros of Riemann's {·function, Meat~enger of Math. 45 (1915), 180-3. (2) A proof of Burnside's formula for log f(x+ 1) and certain allied properties of Riemann's {-function, MellSffi{Jer of Math.. 52 (1922), 90-3. (3) A note on the coefficients in the exp8.11Sion of {(8, x) in powers of 8-1, Quart. J. of Math. 50 (1927), 329-32. (4) An approximate functional equation for the product of two {-functions, P.L.M.S. (2), 31 (1930), 11-17. {5) The mean value of the zeta-function on the critical line, J.L.M.S. 5 (1930), 28-32. WINTNER,A. (1) A note on the distribution of the zeros of the zeta-function, A mer. J. MoJ,h. 57 (1935), 101-2. (2) A note on the Riemann ~-function, J.L.M.S. 10 (1935), 82-3. (3) The almost periodic behavior of the function 1/{(l+it), Duke Math. J. 2 (1936), 443-6. (4) Riemann's hypothesis and almost periodic behavior, Reviata Oi. Lima, 41 (1939), 575-85. (5) On the asymptotic behavior of the Riemann zeta-function on the line u = l, A mer. J, Math. 63 (1941), 575-80. (6) Riemann's hypothesis and harmonic analysis, Duke Math. J. 10 (1943), 99-105. (7) The behavior of Euler's product on the boundary of convergence, Duke Math.J. 10 (1943), 429-40. 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On the zeros of the Riemann Zeta function and L-series -II, Hardy-Ramanujan J. 5 (1982), 1-30. BERNDT, B. C. The number of zeros for (<~J (s), J. London Math. Soc. (2), 2 (1970), 577-580. BURGESS, D. A. On character sums and L-series. II, Proc. London Math. Soc. (3), 13 (1963), 524-536. CHEN, J.-R. On the order of((!+ it), Chinese Math. Acta, 6 (1965), 463-478. CHIH, T.-T. l1] A divisor problem, Acad. Sinica &i. Record, 3 (1950), 177-182. CONREY, J. B. Zeros of derivatives of Riemann's xi-function on the critical line, J. Number Theory, 16 (1983), 49--74. CoNREY, J. B., and GHOSH, A. [I] On mean values of the zeta-function, Mathematika, 31 (1984), 159--161. CoNREY, J. B., GHOSH, A., GOLDSTON, D., GoNEK, S. M., and HEATH-BROWN, D. R. On the distribution of gaps between zeros of the zeta-function, Quart. J. Math. Oxford (2), 36 (1985), 43-51. CONREY, J. B., GHoSH, A., and GONEK, S. M. A note on gaps between zeros of the zeta function, Bull. London Math. Soc. 16 (1984), 421--424. CoRRADI, K., and KATAI, l. [1 J A comment on K. S. Gangadharan's paper entitled "Two classical lattice point problems", Magyar Tud. Akad. Mat. Fiz. Oszt. Kiizl. 17 (1967), 89--97. FURTHER REFERENCES 0ESHOUILLERS, J.-M., and IWANIEC, H. (1] Kloosterman sums and Fourier coefficients of cusp forms, Invent. Math. 70 (1982), 219-288. Power mean values of the Riemann zeta-function, Mathematika, 29 (1982), 202-212. Power mean values of the Riemann zeta-function II, Acta Arith. 48 (1984), 305-·312. DIAMOND, H. Elementary methods in the study of the distribution of prime numbers, Bull. Amer. Math. Soc. 7 (1982), 553-589. ERDOs, P. On a new method in elementary number theory which leads to an elementary proof of the prime number theorem, Proc. Nat. Acad. &i. USA, 35 (1949), 374-384. ESTERMANN, T. On the representation of a number as the sum of two products, Proc. London Math. Soc. (2), 31 (1930), 123-133. On Kloosterman's sum, Mathematika, 8 (1961), 83--86. FuJII, A. On the distribution of the zeros of the Riemann Zeta function in short intervals, Bull. Amer. Math. Soc. 81 (1975), 139--142. On the difference between r consecutive ordinates of the Riemann Zeta function, Proc. Japan Acad. 51 (1975), 741-743. On the problem of divisors, Acta Arith. 31 (1976), 355--360. GALLAGHER, P. X., and MUELLER, J. H. Primes and zeros in short intervals, J. Reine Angew. Math. 303/304 (1978) 205--220. GANGADHARAN, K. 8. Two classical lattice point problems, Proc. Camb. Phil. Soc. 57 (1961), 699-721. G~TIHO~- Riemann's zeta function-sign changes of S(T), Recent progress in analytic number theory, Vol I, 2&--46, (Academic Press, London, 1981). On the Riemann Zeta function- Mean value theorems and the distri-bution of IS(T)j, J. Number Theory, 17 (1983), 93-102. G~~;K~a~iic properties of zeta and L-functions, Thesis, Univ. Michigan, Ann Arbor, 1979. [2J Mean values of the Riemann zeta-function and its derivatives, Invent. Math. 75 (1984), 123-141. G~~· :in n-Resultat fur das quadratische Mittel der Riemannschen Zetafunktion auf der kritische Linie, Invent. Math. 41 (1977), 233-251. Hi~~E~=~ ~mega theorems for two classical lattice point problems, Invent. Math. 63 (1981), 181-186. On the average order of a class of arithmetic functions, J. Number Theory, 15 (1982), 36--76. FURTHER REFERENCES HALAsz, G. [1) Uber die Mittelwerte multiplikativer zahlentheoretischer Funktionen, Acta Math. Acad. &i. Hungar. 19 (HISS), 36&-403. HALAsz, G., and Tu&AN, P. [I] On the distribution of Roots of Riemann zeta and allied functions, I, J. Number Theory, 1 (1969), 121-137. HANEKE, W. [1) Verschiirfung der Abscbiitzung von ((f+it), Acta Arith. 8 (1962-63), 357-430. HARDY, G. H. On Dirichlet's divisor problem, Proc. London Math. Soc. (2), 15 (1916), 1-25. HABELGKOVE, C. B. A connection between the zeros and the mean values of {(s), J. London Math. Soc. 24 (1949), 215-222. [2) A disproof of a conjecture of POlya, Mathematika, 5 (1958), 141-145. HEATH-BROWN, D. R. The mean square of the Riemann Zeta-function, Mathematika, 25 (1978), 177-184. The twelfth power moment of the Riemann Zeta-function, Q!Ulrt. J. Math. Oxford (2), 29 (1978), 443--462. Hybrid bounds for Dirichlet L-functions, Invent. Math. 4 7 (1978), 149-170. The fourth power moment of the Riemann Zeta-function, Proc. London Math. Soc. (3), 38 (1979), 385--422. [5) Simple zeros of the Riemann Zeta-function on the critical line, Bull. London Math. Soc. 11 (1979), 17-18. Zero density estimates for the Riemann Zeta-function and Dirichlet L-functions, J. London Math. Soc. (2), 20 (1979), 221-232. Fractional moments of the Riemann Zeta-function, J. London Math. Soc. (2), 24 (1981), 65-78. Mean values of the Zeta-function and divisor problems, Recent progress in analytic number theory, Vol I, 115-119, (Academic Press, London, 1981). Hybrid bounds for L-functions: a q-analogue of van der Corput's method and a t-analogue of Burgess's method, Recent progress in analytic number theory, Vol I, 121-126, (Academic Press, London, 1981). Gaps between primes, and the pair correlation of zeros of the zeta- Th~~~;:~k~~§~~:!~ri~e (!~~~e~;:rem, J. Number Theory, 16 (1983), 242-266. HUXLEY, M. N., On the difference between consecutive primes, Invent. Math. 15 (1972), 155-164. INGHAM, A. E. [I] On two classical lattice point problems, Proc. Camb. Phil. Soc. 36 (1940), 131-138. IVIC,A. (1) Large values of the error term in the divisor problem, Invent. Math. 71 (1983), 513-520. [2) A zero-density theorem for the Riemann zeta-function, Trudy Mat. Irnd. Steklov. 163 (1984), 85--89. FURTHER REFERENCES 409 [3) .The Riemann zeta-function, (Wiley-lnterscience, New York, 1985). IWANIEC, H. Fourier coefficients of cusp forms and the Riemann Zeta-function, Expose No. 18, Semin. Theor. Nombres, Universite Bordeaux, (1979/80). JURKAT, W., and PEYERIMHOFF, A. [1) A constructive approach to Kronecker approximations and its appli-cations to the Mertens conjecture, J. Reine Angew. Math. 286/287 (1976), 322-340. JUTILA,M. Zero-density estimates for L-functions, Acta Arith. 32 (1977), 52-62. (2] Zeros of the zeta-function near the critical line, Studies in pure mathe-matics, to the memory of Paul Turdn, 385-394, (Birkhaii.ser, Basel-Stuttgart, 1982). Riemann's zeta-function and the divisor problem, Arkiv {Dr Mat. 21 (1983), 75-96. [ 4] On the value distribution of the zeta-function on the critical line, Bull. London Math. Soc. 15 (1983), 513-518. KARATSUBA, A. A. Estimates of trigonometric sums by Vinogradov's method, and some apPlications, Proc. Steklov. Inst. Math, 119 (1971), 241-255. Principles of analytic number theory, (Russian), (b.dat. 'Nauka ', Moscow, 1975). KoLESNIK, G. The improvement of the error term in the divisor problem, Mat. Zcmetki, 6 (1969), 545-554. [2) On the estimation of certain trigonometric sums, Acta Arith. 25 (1973), 7-30. On the estimation of multiple exponential sums, Recent progress in analytic number theory, Vol I, 231-246, (Academic Press, London, 1981). On the order of ((!+it) and 6(R), Pacific J. Math. 82 (1982), 107-122. On the method of exponent pairs, Acta Arith. 45 (1985), 115-143. KoKOBOV, N. M. Estimates of trigonometric sums and their applications, Uspehi Mat. Nauk, 13 (1958). 185-192. KUBOTA, T., and LEOPOLDT, H. W. (1] Eine p-adische Theorie der Zetawerte. I. Einfii.hrung der p-adischen Dirichletschen L-funktionen, J. Reine Angew. Math. 214/215 (1964), 328-339. LAVR1K, A. F. The functional equation for Dirichlet L-functions and the problem of divisors in arithmetic progressions, Izv. Akad. Nauk SSSR Ser. Mat. 30 (196E), 433-448. ... LAVRIK, A. F., and SomRov, A. S. On the remainder term in the elementary proof of the prime number theorem, Dokl. Akad. Nauk SSSR, 211 (1973), 534-536. LEVINSON, N. !l-theorems for the Riemann zeta-function, Acta Arith. 20 (1972), 319-332. More than one third of the zeros of Riemann's zeta-function are on C1 "" f, Adv. Math. 13 (1974), 383--436. 410 FURTHER REFERENCES Zeros of derivative of Riemann's e-function, Bull. Amer. Math. Soc. 80 (1974), 951-954. A simplification of the proof that N0(T) > J-N(T) for Riemann's zeta-function, Adv. Math. 18 (1975), 239-242. Deduction of semi-optimal mollifier for obtaining lower bounds for N0 (T) for Riemann's zeta-function, Proc. Nat. Acad. Sci. USA, 72 (1975), 294-297. Almost all roots of,(s) =a are arbitrarily close to a=!, Proc. Nat. Acad. &i. USA, 72 (1975), 1322-1324. LEVINSON, N., and MoNTGOMERY, H. L. Zeros of the derivative of the Riemann zeta-function, Acta Math. 133 (1974), 49--65. Lou, 8.-T. A lower bound for the number of zeros of Riemann's zeta-function on a=!. Recent progress in analytic number theory, 319-324, Vol I, (Academic press, London, 1981). MONTGOMERY, H. L. Topics in multiplicative number theory, Lecture Notes in Math. 227 (Springer, Berlin, 1971). The pair correlation of zeros of the zeta-function, Analytic number theory (Proc. Symp. Pure Math. Vol XXIV), 181-193, (Amer. Math. Soc., Providence, R. 1., 1973). Extreme values of the Riemann zeta-function, Comment. Math. Helv. 52 (1977), 511--518. MoNTGOMERY, H. L., and 0DLYZKO, A.M. [I) Gaps between zeros of the zeta-function, Topics in classical number theory, (Coli. Math. Soc. Jimos Bolyai, 34), 1079--1106, (Budapest, 1981). MoNTGOMERY, H. L., and VAUGHAN, R. C. Hilbert's inequality, J. London Math. Soc. (2), 8 (1974), 73-82. MOTOHASHI, Y. An elementary proof of Vinogradov's zero-free region for the Riemann zeta-function, Recent progress in analytic number theory, Vol I, 257-267, (Academic Press, London, 1981). [2) A note on the approximate functional equation for ' 2(s), Proc. Japan Acad. 59A (1983), 392-396. A note on the approximate functional equation for '2(s).ll, Proc. Japan Acad. 59A (1983), 469-472. MUELLER,J. H. On the difference between the consecutive zeros of the Riemann zeta-function, J. Number Theory, 14 (1982), 327-331. On the Riemann Zeta-function ((s)-gaps between sign changes of S(t), Mathematika, 29 (1983), 264-269. OnLYZKO, A. M., and TE RIELE, H. J. J. (1 J Disproof of Mertens conjecture, J. Reine Angew. Math. 357 (1985), 138-160. OSTROWSKI, A. Ober Dirichiletscbe Reiben und algebraische Differentialgleicbungen, Math. Zeit. 8 (1920), 115-143. FURTHER REFERENCES PINTZ,.J. On the remainder term of the prime number formula and the zeros of Riemann's zeta-function, Number theory, Noordwijkerhout 1983, 186--197, Lecture Notes in Math. 1068, (Springer, Berlin, 1984). RAMACHANDRA, K. On the zeros of the Riemann zeta-function and L-series, Acta Arith. 34 (1978), 211-218. (2] Some remarks on a theorem of Montgomery and Vaughan, J. Number Theory, 11 (1980), 465--471. Some remarks on the mean value of the Riemann Zeta-function and other Dirichlet series-ll, Hardy-Ramanujan J. 3 (1980), 1-24. [ 4] Some remarks on the mean value of the Riemann Zeta-function and other Dirichlet series-III, Ann. Acad. &i. Fenn. Ser. AI Math. 5 (1980), 145-158. [5) Mean-value of the Riemann Zeta-function and other remarks- II, Trudy Mat. Inst. Steklov. 163 (1984), 200--204. RANK1N, R. A. Vander Corput's method and the theory of exponent pairs, QU4rt. J. Math. Oxford (2), 6 (1955), 147-153. REICH, A. Zetafunktion und Differenzen-Differentialgleichungen, Arch. Math. (Basel), 38 (1982), 226-235. RICHERT, H.-E. Versch8.rfung der Absch8.tzung beim Dirichiletschen Teilerproblem, Math. Zeit. 58 (1953), 204-218. Einfiihrung in die Theorie der starken Rieszscben Summierbarkeit von Dirichletreihen, Nachr. Akad. Wiss. Gottingen (Math.-Phys.) Kl. II, 1960,17-75. Zur Absch8.tzung der Riemannschen Zetafunktion in der N8.be der Vertikalen u = 1, Math. Ann. 169 (1967), 97-101. SELBERG, A. [1) The zeta-function and the Riemann Hypothesis, Skandinavske Mathe-matikerkongres, 10 (1946), 187-200. An elementary proof of the prime number theorem, Ann. Math. (2), 50 (1949), 305--313. Discontinuous groups and harmonic analysis, Proc. Internal. Congr. Mathematicians (Stockholm, 1962), 177-189 (Jnst. Mittag-Leffler, Djursholm, 1963). SPIRA, R. Zeros of sections of the zeta function. II, Math. Comp. 22 (1968), 163-173. SRINIVASAN, B. R. Lattice point problems of many-dimensional hyperboloids III, Math. Ann. 160 (1965), 280-311. STECKIN, s. B. Mean values of the modulus of a trigonometric sum, Trudy Mat. lTUJt. Steklov. 134 (1975), 283-309, 411. TATE, J. T. Fourier analysis in number fields, and Heeke's zeta-functions, Algebraic Number Theory (Proc. Instructional Con{., Brighton, 1965), 305-347, (Thompson, Washington DC, 1967). FURTHER REFERENCES TITCHMARSH, E. C. The Zeta-function of Riemann, Cambr. Tracts in Math. No 26, (Cambridge University Press, 1930). ToNG, K.-C. On divisor problems III, Acta Math. Sinica, 6 (1956), 515-541. TURGANAUEV, R. T. The asymptotic formula for fractional mean value moments of the zeta-function of Riemann, Trudy Mat. lnst. Steklov. 158 (1981), 203-226. VINOGRADOV, I. M. A new estimate for {(1 +it), lzv. Akad. Nauk SSSR, Ser. Mat. 22 (1958), 161-164. Selected works, (Springer, Berlin, 1985). VORONIN, M. On the distribution of nonzero values of the Riemann C-function, Proc. Steklov lnst. Math. 128 (1972), 153-175. Theorem on the "universality" of the Riemann Zeta-function, Math. USSR Izvestija, 9 (1975), 443-453. On the zeros of zeta-functions of quadratic forms, Trudy Mat. lnst. Steklov. 142 (1976), 135--147. WALFISZ,A. Weylsche Exponentialsummen in der NeUf!ren Zahlentheorie, (VEB Deutscher Verlag, Berlin, 1963). WErr.,A. On the Riemann hypothesis in function-fields, Proc. Nat. Acad. &i. USA, 27 (1941), 345-347.
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Skip to main content 18.5: Criteria for Precipitation and its Completeness Last updated : Jul 12, 2023 Save as PDF 18.4: Limitations of the Kₛₚ Concept 18.6: Fractional Precipitation Page ID : 24309 ( \newcommand{\kernel}{\mathrm{null}\,}) The equation that describes the equilibrium between solid calcium carbonate and its solvated ions is: We can establish this equilibrium either by adding solid calcium carbonate to water or by mixing a solution that contains calcium ions with a solution that contains carbonate ions. If we add calcium carbonate to water, the solid will dissolve until the concentrations are such that the value of the reaction quotient is equal to the solubility product (Ksp = 4.8 × 10–9). If we mix a solution of calcium nitrate, which contains Ca2+ ions, with a solution of sodium carbonate, which contains ions, the slightly soluble ionic solid CaCO3 will precipitate, provided that the concentrations of Ca2+ and ions are such that Q is greater than Ksp for the mixture. The reaction shifts to the left and the concentrations of the ions are reduced by formation of the solid until the value of Q equals Ksp. A saturated solution in equilibrium with the undissolved solid will result. If the concentrations are such that Q is less than Ksp, then the solution is not saturated and no precipitate will form. We can compare numerical values of Q with Ksp to predict whether precipitation will occur, as Example shows. (Note: Since all forms of equilibrium constants are temperature dependent, we will assume a room temperature environment going forward in this chapter unless a different temperature value is explicitly specified.) The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of lime, Ca(OH)2, a readily available inexpensive source of OH– ion: The concentration of Mg2+(aq) in sea water is 0.0537 M. Will Mg(OH)2 precipitate when enough Ca(OH)2 is added to give a [OH–] of 0.0010 M? Solution This problem asks whether the reaction: shifts to the left and forms solid Mg(OH)2 when [Mg2+] = 0.0537 M and [OH–] = 0.0010 M. The reaction shifts to the left if Q is greater than Ksp. Calculation of the reaction quotient under these conditions is shown here: Because Q is greater than Ksp (Q = 5.4 × 10–8 is larger than Ksp = 2.1 × 10–13), we can expect the reaction to shift to the left and form solid magnesium hydroxide. Mg(OH)2(s) forms until the concentrations of magnesium ion and hydroxide ion are reduced sufficiently so that the value of Q is equal to Ksp. Exercise Use the solubility products in Appendix J to determine whether CaHPO4 will precipitate from a solution with [Ca2+] = 0.0001 M and = 0.001 M. Answer No precipitation of CaHPO4; Q = 1 × 10–7, which is less than Ksp Precipitation of AgCl upon Mixing Solutions Does silver chloride precipitate when equal volumes of a 2.0 × 10–4-M solution of AgNO3 and a 2.0 × 10–4-M solution of NaCl are mixed? (Note: The solution also contains Na+ and ions, but when referring to solubility rules, one can see that sodium nitrate is very soluble and cannot form a precipitate.) Solution The equation for the equilibrium between solid silver chloride, silver ion, and chloride ion is: The solubility product is 1.8 × 10–10 (see Appendix J). AgCl will precipitate if the reaction quotient calculated from the concentrations in the mixture of AgNO3 and NaCl is greater than Ksp. The volume doubles when we mix equal volumes of AgNO3 and NaCl solutions, so each concentration is reduced to half its initial value. Consequently, immediately upon mixing, [Ag+] and [Cl–] are both equal to: The reaction quotient, Q, is momentarily greater than Ksp for AgCl, so a supersaturated solution is formed: Since supersaturated solutions are unstable, AgCl will precipitate from the mixture until the solution returns to equilibrium, with Q equal to Ksp. Exercise Will KClO4 precipitate when 20 mL of a 0.050-M solution of K+ is added to 80 mL of a 0.50-M solution of ? (Remember to calculate the new concentration of each ion after mixing the solutions before plugging into the reaction quotient expression.) Answer No, Q = 4.0 × 10–3, which is less than Ksp = 1.07 × 10–2 In the previous two examples, we have seen that Mg(OH)2 or AgCl precipitate when Q is greater than Ksp. In general, when a solution of a soluble salt of the Mm+ ion is mixed with a solution of a soluble salt of the Xn– ion, the solid, MpXq precipitates if the value of Q for the mixture of Mm+ and Xn– is greater than Ksp for MpXq. Thus, if we know the concentration of one of the ions of a slightly soluble ionic solid and the value for the solubility product of the solid, then we can calculate the concentration that the other ion must exceed for precipitation to begin. To simplify the calculation, we will assume that precipitation begins when the reaction quotient becomes equal to the solubility product constant. Precipitation of Calcium Oxalate Blood will not clot if calcium ions are removed from its plasma. Some blood collection tubes contain salts of the oxalate ion, , for this purpose (Figure ). At sufficiently high concentrations, the calcium and oxalate ions form solid, CaC2O4•H2O (which also contains water bound in the solid). The concentration of Ca2+ in a sample of blood serum is 2.2 × 10–3 M. What concentration of ion must be established before CaC2O4•H2O begins to precipitate? Solution The equilibrium expression is: For this reaction: (see Appendix J) CaC2O4 does not appear in this expression because it is a solid. Water does not appear because it is the solvent. Solid CaC2O4 does not begin to form until Q equals Ksp. Because we know Ksp and [Ca2+], we can solve for the concentration of that is necessary to produce the first trace of solid: A concentration of = 1.0 × 10–6 M is necessary to initiate the precipitation of CaC2O4 under these conditions. Exercise If a solution contains 0.0020 mol of per liter, what concentration of Ag+ ion must be reached by adding solid AgNO3 before Ag2CrO4 begins to precipitate? Neglect any increase in volume upon adding the solid silver nitrate. Answer: 7.0 × 10–5 M It is sometimes useful to know the concentration of an ion that remains in solution after precipitation. We can use the solubility product for this calculation too: If we know the value of Ksp and the concentration of one ion in solution, we can calculate the concentration of the second ion remaining in solution. The calculation is of the same type as that in Example—calculation of the concentration of a species in an equilibrium mixture from the concentrations of the other species and the equilibrium constant. However, the concentrations are different; we are calculating concentrations after precipitation is complete, rather than at the start of precipitation. Concentrations Following Precipitation Clothing washed in water that has a manganese [Mn2+(aq)] concentration exceeding 0.1 mg/L (1.8 × 10–6 M) may be stained by the manganese upon oxidation, but the amount of Mn2+ in the water can be reduced by adding a base. If a person doing laundry wishes to add a buffer to keep the pH high enough to precipitate the manganese as the hydroxide, Mn(OH)2, what pH is required to keep [Mn2+] equal to 1.8 × 10–6 M? Solution The dissolution of Mn(OH)2 is described by the equation: We need to calculate the concentration of OH– when the concentration of Mn2+ is 1.8 × 10–6 M. From that, we calculate the pH. At equilibrium: or so Now we calculate the pH from the pOH: If the person doing laundry adds a base, such as the sodium silicate (Na4SiO4) in some detergents, to the wash water until the pH is raised to 10.20, the manganese ion will be reduced to a concentration of 1.8 × 10–6 M; at that concentration or less, the ion will not stain clothing. Exercise The first step in the preparation of magnesium metal is the precipitation of Mg(OH)2 from sea water by the addition of Ca(OH)2. The concentration of Mg2+(aq) in sea water is 5.37 × 10–2 M. Calculate the pH at which [Mg2+] is diminished to 1.0 × 10–5 M by the addition of Ca(OH)2. Answer Due to their light sensitivity, mixtures of silver halides are used in fiber optics for medical lasers, in photochromic eyeglass lenses (glass lenses that automatically darken when exposed to sunlight), and—before the advent of digital photography—in photographic film. Even though AgCl (Ksp = 1.6 × 10–10), AgBr (Ksp = 7.7 × 10–13), and AgI (Ksp = 8.3 × 10–17) are each quite insoluble, we cannot prepare a homogeneous solid mixture of them by adding Ag+ to a solution of Cl–, Br–, and I–; essentially all of the AgI will precipitate before any of the other solid halides form because of its smaller value for Ksp. However, we can prepare a homogeneous mixture of the solids by slowly adding a solution of Cl–, Br–, and I– to a solution of Ag+. When two anions form slightly soluble compounds with the same cation, or when two cations form slightly soluble compounds with the same anion, the less soluble compound (usually, the compound with the smaller Ksp) generally precipitates first when we add a precipitating agent to a solution containing both anions (or both cations). When the Ksp values of the two compounds differ by two orders of magnitude or more (e.g., 10–2 vs. 10–4), almost all of the less soluble compound precipitates before any of the more soluble one does. This is an example of selective precipitation, where a reagent is added to a solution of dissolved ions causing one of the ions to precipitate out before the rest. Determining if a Precipitate forms (The Ion Product): Summary A slightly soluble electrolyte begins to precipitate when the magnitude of the reaction quotient for the dissolution reaction exceeds the magnitude of the solubility product. Precipitation continues until the reaction quotient equals the solubility product. A reagent can be added to a solution of ions to allow one ion to selectively precipitate out of solution. The common ion effect can also play a role in precipitation reactions. In the presence of an ion in common with one of the ions in the solution, Le Chatelier’s principle applies and more precipitate comes out of solution so that the molar solubility is reduced. Glossary common ion effect : effect on equilibrium when a substance with an ion in common with the dissolved species is added to the solution; causes a decrease in the solubility of an ionic species, or a decrease in the ionization of a weak acid or base molar solubility : solubility of a compound expressed in units of moles per liter (mol/L) selective precipitation : process in which ions are separated using differences in their solubility with a given precipitating reagent solubility product (Ksp) : equilibrium constant for the dissolution of a slightly soluble electrolyte Contributors and Attributions Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at 18.4: Limitations of the Kₛₚ Concept 18.6: Fractional Precipitation
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Question Video: Finding the Volume of a Pyramid Using Similarity Mathematics If the two given pyramids are similar and the volume of the larger pyramid is 160 m³, find the volume of the smaller one. Video Transcript If the two given pyramids are similar and the volume of the larger pyramid is 160 meters cubed, find the volume of the smaller one. So we’ve been given two pyramids. And the key piece of information in the question is that they’re similar to each other, which means that all of the corresponding lengths between these two pyramids are in the same ratio. We’ve also been given the volume of the larger pyramid and asked to work out the volume of the smaller pyramid. Let’s think about how to approach this. We’re given a pair of corresponding lengths, the perpendicular heights of each the two pyramids. They’re five meters and 10 meters. We can use this pair of lengths in order to work out the scale factor for the lengths between the two pyramids. By dividing the larger length by the smaller, we have that the length scale factor or LSF between the two pyramids is 10 divided by five, which is two. This means that all of the lengths in the larger pyramid are twice as long as the corresponding lengths in the smaller pyramid. Does it follow then that the volume of the larger pyramid is twice as big as the volume of the smaller pyramid? Well, actually no. Length is a one-dimensional measurement, whereas volume is a three- dimensional measurement. And therefore, the relationship between the lengths and the volumes of similar shapes is not exactly the same. However, it is related. If the length scale factor between two similar shapes is 𝐾, then the volume scale factor or VSF is always 𝐾 cubed. So as we know the length scale factor for these two pyramids, we can work out the scale factor between their volumes. It’s two cubed, which is eight. What this means then is that the volume of the larger pyramid is not twice as big as the volume of the smaller pyramid, but in fact it’s eight times as big. So if we want to work out the volume of the smaller pyramid, we need to divide the larger volume by eight. So it’s 160 divided by eight, which is 20. And so we have our answer to the problem: the volume of the smaller pyramid is 20 meters cubed. Lesson Menu Join Nagwa Classes Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher! Nagwa is an educational technology startup aiming to help teachers teach and students learn. Company Content Copyright © 2025 Nagwa All Rights Reserved Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy
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Chapter 4 (Axial Loading) by Mechanics of Materials R.C Hibbeler (9th Edition) Complete Solution by Engr Adnan Rasheed Mechanical - YouTube Back Skip navigation Search Search with your voice Sign in Home HomeShorts ShortsSubscriptions SubscriptionsYou YouHistory History Play all Chapter 4 (Axial Loading) by Mechanics of Materials R.C Hibbeler (9th Edition) Complete Solution by Engr Adnan Rasheed Mechanical by Engr. Adnan Rasheed Mechanical • Playlist•66 videos•91,058 views CHAPTER 4 AXIAL LOADING BY MECHANICS OF MATERIALS R.C HIBBELER (9TH EDITION)...more CHAPTER 4 AXIAL LOADING BY MECHANICS OF MATERIALS R.C HIBBELER (9TH EDITION)...more...more Play all PLAY ALL Chapter 4 (Axial Loading) by Mechanics of Materials R.C Hibbeler (9th Edition) Complete Solution by Engr Adnan Rasheed Mechanical 66 videos 91,058 views Last updated on Mar 15, 2025 Save playlist Shuffle play Share CHAPTER 4 AXIAL LOADING BY MECHANICS OF MATERIALS R.C HIBBELER (9TH EDITION) Complete Solution Step-by-Step , R.C HIBBELER Mechanics of Materials 09th Edition, SI units Complete Solution by Engr Adnan Rasheed Mechanical Show more Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical Subscribe Play all Chapter 4 (Axial Loading) by Mechanics of Materials R.C Hibbeler (9th Edition) Complete Solution by Engr Adnan Rasheed Mechanical by Engr. Adnan Rasheed Mechanical Playlist•66 videos•91,058 views CHAPTER 4 AXIAL LOADING BY MECHANICS OF MATERIALS R.C HIBBELER (9TH EDITION)...more CHAPTER 4 AXIAL LOADING BY MECHANICS OF MATERIALS R.C HIBBELER (9TH EDITION)...more...more Play all 1 14:29 14:29 Now playing 4-1 Determine displacement of B and A | Axial Loading | Mechanics of Materials by R.C Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 34K views • 3 years ago • 2 8:13 8:13 Now playing 4-2 Determine displacement of end A w.r.t D | Axial Loading | Mechanics of Materials by R.C Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 17K views • 3 years ago • 3 9:42 9:42 Now playing 4-3 Determine displacement of A & normal stress| Axial Loading| Mech of Materials by RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 19K views • 3 years ago • 4 5:05 5:05 Now playing 4-4 Determine displacement of B w.r.t to C | Axial Loading | Mech of Materials by R.C Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 7.8K views • 3 years ago • 5 6:22 6:22 Now playing 4-5| Chapter 4 | Axial Loading | Mechanics of Materials by R.C Hibbeler 9th Edition| Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 8.4K views • 3 years ago • 6 8:52 8:52 Now playing 4-6| Chapter 4 | Axial Loading | Mechanics of Materials by R.C Hibbeler 9th Edition| Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 5.4K views • 3 years ago • 7 10:02 10:02 Now playing 4-7| Chapter 4 | Axial Loading | Mechanics of Materials by R.C Hibbeler 9th Edition| Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 4K views • 3 years ago • 8 10:26 10:26 Now playing 4-8| Chapter 4 | Axial Loading | Mechanics of Materials by R.C Hibbeler 9th Edition| Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 3.3K views • 3 years ago • 9 11:20 11:20 Now playing 4-9| Chapter 4 | Axial Loading | Mechanics of Materials by R.C Hibbeler 9th Edition| Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 13K views • 3 years ago • 10 8:02 8:02 Now playing 4-10| Chapter 4 | Axial Loading | Mechanics of Materials by R.C Hibbeler 9th Edition| Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 9.5K views • 3 years ago • 11 27:56 27:56 Now playing 4-11| Chapter 4 | Axial Loading | Mechanics of Materials by R.C Hibbeler 9th Edition| Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 12K views • 2 years ago • 12 15:24 15:24 Now playing 4-12| Chapter 4 | Axial Loading | Mechanics of Materials by R.C Hibbeler 9th Edition| Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 7.8K views • 2 years ago • 13 12:40 12:40 Now playing 4-13 Determine vertical deflection at D | Axial Loading | Mechanics of Materials by R.C Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 10K views • 2 years ago • 14 13:50 13:50 Now playing 4-15 Determine force F at bottom and displacement of top post A | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 4.8K views • 2 years ago • 15 12:16 12:16 Now playing 4-16 Determine the vertical displacement of F | Mechanics of materials rc hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 2.2K views • 2 years ago • 16 8:34 8:34 Now playing 4-17 Determine the magnitude of the load | Axial Loading | Mechanics of materials rc hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.3K views • 2 years ago • 17 10:40 10:40 Now playing 4-18 Determine the vertical displacement of collar | Axial Loading | Mech of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.4K views • 2 years ago • 18 11:16 11:16 Now playing 4-19 Determine the magnitude of P | Axial Loading | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1K views • 2 years ago • 19 13:54 13:54 Now playing 4-20 Determine maximum average normal stress in each pipe | Mech of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 2.5K views • 2 years ago • 20 12:41 12:41 Now playing Example 4.4| Chapter 4 | Axial Loading | Mechanics of Materials by R.C Hibbeler 9th Edition| Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 21K views • 2 years ago • 21 7:55 7:55 Now playing 4-27 Determine the displacement of end A | Axial Loading | Mechanics of Materials by R.C Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.7K views • 2 years ago • 22 9:22 9:22 Now playing 4-28 Determine compression in length of bone | Axial Loading | Mechanics of Materials by RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1K views • 2 years ago • 23 10:39 10:39 Now playing 4-31 Determine stress in concrete & steel | Axial Loading | Mechanics of Materials by R.C Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 11K views • 2 years ago • 24 11:45 11:45 Now playing 4-32 Determine the diameter of steel rod | Axial Loading | Mechanics of Materials by R.C Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 4.9K views • 2 years ago • 25 8:09 8:09 Now playing 4-33 Determine average normal stress in concrete & steel l Axial Loading | Mechanics of Materials Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 9.1K views • 2 years ago • 26 9:06 9:06 Now playing 4-36 Determine the support reactions at rigid supports A and C | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 6.3K views • 2 years ago • 27 12:10 12:10 Now playing 4-37 Determine the support reactions at A and C | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 3.4K views • 2 years ago • 28 13:12 13:12 Now playing 4-21 Determine the displacement of the pipe when it is attached to | Mech of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 2.6K views • 2 years ago • 29 13:15 13:15 Now playing Determine the relative displacement | Pro 4-24 | axial load | Mechanics of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 2.4K views • 1 year ago • 30 11:14 11:14 Now playing Determine force and displacement | Problem 4-14 | Stress | Force | Mech of materials Rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 2K views • 1 year ago • 31 8:02 8:02 Now playing Determine the elongation | Problem 4-25 | axial load | stress | Mech of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 2.5K views • 1 year ago • 32 13:55 13:55 Now playing Show that the displacement of its end | Problem 4-23 | axial load | stress | Mech of materials rc hi Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.6K views • 1 year ago • 33 8:08 8:08 Now playing Determine elongation of the tapered A992 steel shaft |Problem 4-26| Mech of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.1K views • 1 year ago • 34 43:18 43:18 Now playing Axial load | stress | displacement | Chapter 4 | Mechanics of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.2K views • 1 year ago • 35 9:13 9:13 Now playing Determine force developed in each wire | Problem 4-38 | axial load | Mech of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.1K views • 1 year ago • 36 8:02 8:02 Now playing Determine the cross sectional area of AB | Problem 4-39 | axial load | Mech of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 729 views • 1 year ago • 37 12:53 12:53 Now playing Determine average normal stress | Problem 4-34 | axial load | stress | Mech of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.6K views • 1 year ago • 38 11:50 11:50 Now playing Determine maximum allowable floor loading P |Problem 4-35|axial load| Mech of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.5K views • 1 year ago • 39 53:09 53:09 Now playing Axial load | stress | strain | Mechanics of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.3K views • 1 year ago • 40 8:05 8:05 Now playing 4-87 | Determine the maximum normal stress developed in the bar | Mechanics of materials Rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 2.4K views • 1 year ago • 41 8:19 8:19 Now playing 4-88 | Determine the maximum axial force P that can be applied to bar | Mechanics of materials Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 2.7K views • 1 year ago • 42 8:05 8:05 Now playing 4-90 | Determine maximum axial force P that can be applied to steel plate | Mechanics of materials Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.5K views • 1 year ago • 43 8:03 8:03 Now playing 4-89 | Axial Force P Calculation Made Easy for Engineers | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.7K views • 1 year ago • 44 12:02 12:02 Now playing 4-98 | Calculating Bar Elongation with Mechanics of Materials Rc Hibbeler | Axial load | Elongation Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1K views • 1 year ago • 45 8:02 8:02 Now playing 4-95 | How to Find the Maximum P Value for Bolts | Mechanics of Materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.4K views • 1 year ago • 46 11:56 11:56 Now playing 4-97 | Step-by-Step Guide to Determining Wire Forces in Mechanics of Materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 462 views • 1 year ago • 47 9:50 9:50 Now playing 4-96 | Calculating the Maximum Elastic Load for Your Assembly | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 817 views • 1 year ago • 48 8:06 8:06 Now playing 4-94 | Find the stress concentration factor for geometry and P | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 946 views • 1 year ago • 49 8:02 8:02 Now playing 4-91| Determine the maximum axial force P that can be applied to the bar.| Mechanics of materials Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.8K views • 1 year ago • 50 8:04 8:04 Now playing 4-92 | Determine the maximum normal stress | Mechanics of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.5K views • 1 year ago • 51 8:03 8:03 Now playing 4-93 | Determine the maximum normal stress developed in the bar | Mechanics of materials rc Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.6K views • 1 year ago • 52 49:33 49:33 Now playing 8 Problems in 1 Video | axial load | Mechanics | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 784 views • 1 year ago • 53 12:16 12:16 Now playing 4-100 | Find distributed load w causes one wire to yield | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 655 views • 1 year ago • 54 14:12 14:12 Now playing 4-99 | Determine intensity of the distributed load w | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 985 views • 1 year ago • 55 17:37 17:37 Now playing 4-101 Determine the force developed in both wires & elongation | Mechanics of Materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.3K views • 1 year ago • 56 12:54 12:54 Now playing 4-102 | Determine force P that will cause both wire to yield | Mechanics of Materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.1K views • 1 year ago • 57 53:31 53:31 Now playing Axial Load | Mechanics | Mechanics of Materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 880 views • 1 year ago • 58 1:25:49 1:25:49 Now playing Mechanics of materials | axial load | Energy Methods | Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 853 views • 1 year ago • 59 8:24 8:24 Now playing Determine displacement of the end C of the rod | Example 4.1 | Mechanics of materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.6K views • 8 months ago • 60 15:10 15:10 Now playing Determine the displacement of point F on AB | Example 4.2 | Mechanics of Materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 2.1K views • 8 months ago • 61 10:08 10:08 Now playing Example 4.3 | Determine how far its end is displaced due to gravity | Mechanics of Materials RC Hibb Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 724 views • 8 months ago • 62 10:54 10:54 Now playing Example 4.5 | Determine average normal stress in aluminum and brass | Mechanics of materials RC Hib Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 675 views • 8 months ago • 63 13:57 13:57 Now playing 4-29 | Determine maximum intensity P for equilibrium | Axial Load | Mechanics of materials Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 305 views • 6 months ago • 64 14:18 14:18 Now playing 4-30 | Determine the resisting bearing force F for equilibrium | Mechanics of Materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 322 views • 6 months ago • 65 16:29 16:29 Now playing 4-41 | Determine support reactions when axial force of 400 KN is applied | Mechanics of materials Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.1K views • 6 months ago • 66 14:54 14:54 Now playing 4-42 | Determine the support reactions || Mechanics | Mechanics of Materials RC Hibbeler Engr. Adnan Rasheed Mechanical Engr. Adnan Rasheed Mechanical • 1.2K views • 6 months ago • Search Info Shopping Tap to unmute 2x If playback doesn't begin shortly, try restarting your device. • You're signed out Videos you watch may be added to the TV's watch history and influence TV recommendations. To avoid this, cancel and sign in to YouTube on your computer. Cancel Confirm Share - [x] Include playlist An error occurred while retrieving sharing information. Please try again later. Watch later Share Copy link 0:00 / •Watch full video Live • • NaN / NaN [](
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https://simple.wikipedia.org/wiki/Square_number
Jump to content Search Contents 1 Examples 2 Properties 3 Special cases 4 Odd and even square numbers 5 References 6 Further reading 7 Other websites Square number العربية Azərbaycanca বাংলা Български Català Чӑвашла Čeština Cymraeg Dansk Deutsch Ελληνικά Emiliàn e rumagnòl English Español Esperanto فارسی Français Galego ગુજરાતી 한국어 Hornjoserbsce Hrvatski Bahasa Indonesia Íslenska Italiano עברית Latina Latviešu Magyar मराठी Bahasa Melayu Nederlands 日本語 Nordfriisk Norsk bokmål Norsk nynorsk Oromoo Polski Português Romnă Русский Slovenčina Slovenščina کوردی Srpskohrvatski / српскохрватски Suomi தமிழ் తెలుగు Türkçe Українська Tiếng Việt 文言 ייִדיש 粵語 中文 Change links Page Talk Read Change Change source View history Tools Actions Read Change Change source View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Make a book Download as PDF Page for printing In other projects Wikimedia Commons Wikifunctions Wikidata item Appearance From Simple English Wikipedia, the free encyclopedia A square number, sometimes also called a perfect square, is the result of an integer multiplied by itself. 1, 4, 9, 16 and 25 are the first five square numbers. In a formula, the square of a number n is denoted (exponentiation), usually pronounced as "n squared". The name square number comes from the name of the shape; see below. Square numbers are non-negative. Another way of saying that a (non-negative) number is a square number, is that its square root is again an integer. For example, , so 9 is a square number. Examples [change | change source] The squares (sequence A000290 in the OEIS) smaller than 702 are: : 02 = 0 : 12 = 1 : 22 = 4 : 32 = 9 : 42 = 16 : 52 = 25 : 62 = 36 : 72 = 49 : 82 = 64 : 92 = 81 : 102 = 100 : 112 = 121 : 122 = 144 : 132 = 169 : 142 = 196 : 152 = 225 : 162 = 256 : 172 = 289 : 182 = 324 : 192 = 361 : 202 = 400 : 212 = 441 : 222 = 484 : 232 = 529 : 242 = 576 : 252 = 625 : 262 = 676 : 272 = 729 : 282 = 784 : 292 = 841 : 302 = 900 : 312 = 961 : 322 = 1024 : 332 = 1089 : 342 = 1156 : 352 = 1225 : 362 = 1296 : 372 = 1369 : 382 = 1444 : 392 = 1521 : 402 = 1600 : 412 = 1681 : 422 = 1764 : 432 = 1849 : 442 = 1936 : 452 = 2025 : 462 = 2116 : 472 = 2209 : 482 = 2304 : 492 = 2401 : 502 = 2500 : 512 = 2601 : 522 = 2704 : 532 = 2809 : 542 = 2916 : 552 = 3025 : 562 = 3136 : 572 = 3249 : 582 = 3364 : 592 = 3481 : 602 = 3600 : 612 = 3721 : 622 = 3844 : 632 = 3969 : 642 = 4096 : 652 = 4225 : 662 = 4356 : 672 = 4489 : 682 = 4624 : 692 = 4761 There are infinitely many square numbers, as there are infinitely many natural numbers. Properties [change | change source] The number m is a square number if and only if one can compose a square of m equal (lesser) squares: | | | --- | | m = 12 = 1 | | | m = 22 = 4 | | | m = 32 = 9 | | | m = 42 = 16 | | | m = 52 = 25 | | | Note: White gaps between squares serve only to improve visual perception.There must be no gaps between actual squares. | A square with side length n has area . The expression for the nth square number is n2. This is also equal to the sum of the first n odd numbers as can be seen in the above pictures, where a square results from the previous one by adding an odd number of points (shown in magenta). The formula follows: So for example, . A square number can end only with digits 0, 1, 4, 6, 9, or 25 in base 10, as follows: If the last digit of a number is 0, its square ends in an even number of 0s (so at least 00) and the digits preceding the ending 0s must also form a square. If the last digit of a number is 1 or 9, its square ends in 1 and the number formed by its preceding digits must be divisible by four. If the last digit of a number is 2 or 8, its square ends in 4 and the preceding digit must be even. If the last digit of a number is 3 or 7, its square ends in 9 and the number formed by its preceding digits must be divisible by four. If the last digit of a number is 4 or 6, its square ends in 6 and the preceding digit must be odd. If the last digit of a number is 5, its square ends in 25 and the preceding digits must be 0, 2, 06, or 56. A square number cannot be a perfect number. All fourth powers, sixth powers, eighth powers and so on are perfect squares. Special cases [change | change source] If the number is of the form m5 where m represents the preceding digits, its square is n25 where and represents digits before 25. For example the square of 65 can be calculated by which makes the square equal to 4225. If the number is of the form m0 where m represents the preceding digits, its square is n00 where . For example the square of 70 is 4900. If the number has two digits and is of the form 5m where m represents the units digit, its square is AABB where and . Example: To calculate the square of 57, 25 + 7 = 32 and 72 = 49, which means 572 = 3249. Odd and even square numbers [change | change source] Squares of even numbers are even (and in fact divisible by 4), since . Squares of odd numbers are odd, since . It follows that square roots of even square numbers are even, and square roots of odd square numbers are odd. As all even square numbers are divisible by 4, the even numbers of the form are not square numbers. As all odd square numbers are of the form , the odd numbers of the form are not square numbers. Squares of odd numbers are of the form , since and is an even number. References [change | change source] Eric W. Weisstein, Square Number at MathWorld. Further reading [change | change source] Conway, J. H. and Guy, R. K. The Book of Numbers. New York: Springer-Verlag, pp. 30–32, 1996. ISBN 0-387-97993-X Other websites [change | change source] Learn Square Numbers Archived 2008-02-11 at the Wayback Machine. Practice square numbers up to 144 with this children's multiplication game Dario Alpern, Sum of squares. A Java applet to decompose a natural number into a sum of up to four squares. Fibonacci and Square Numbers Archived 2007-09-30 at the Wayback Machine at Convergence Archived 2006-02-12 at the Wayback Machine The first 1,000,000 perfect squares Includes a program for generating perfect squares up to 1015. | v t e Classes of natural numbers | | | Powers and related numbers | | Achilles Power of 2 Power of 3 Power of 10 Square Cube Fourth power Fifth power Sixth power Seventh power Eighth power Perfect power Powerful Prime power | | | | Of the form a × 2b ± 1 | | Cullen Double Mersenne Fermat Mersenne Proth Thabit Woodall | | | | Other polynomial numbers | | Hilbert Idoneal Leyland Loeschian Lucky numbers of Euler | | | | Recursively defined numbers | | Fibonacci Jacobsthal Leonardo Lucas Padovan Pell Perrin | | | | Possessing a specific set of other numbers | | Congruent Knödel Riesel Sierpiński | | | | Expressible via specific sums | | Nonhypotenuse Polite Practical Primary pseudoperfect Ulam Wolstenholme | | | | Figurate numbers | | | | | | | | | --- --- --- | | 2-dimensional | | | | --- | | centered | Centered triangular Centered square Centered pentagonal Centered hexagonal Centered heptagonal Centered octagonal Centered nonagonal Centered decagonal Star | | non-centered | Triangular Square Square triangular Pentagonal Hexagonal Heptagonal Octagonal Nonagonal Decagonal Dodecagonal | | | 3-dimensional | | | | --- | | centered | Centered tetrahedral Centered cube Centered octahedral Centered dodecahedral Centered icosahedral | | non-centered | Tetrahedral Cubic Octahedral Dodecahedral Icosahedral Stella octangula | | pyramidal | Square pyramidal | | | 4-dimensional | | | | --- | | non-centered | Pentatope Squared triangular Tesseractic | | | | | | Combinatorial numbers | | Bell Cake Catalan Dedekind Delannoy Euler Eulerian Fuss–Catalan Lah Lazy caterer's sequence Lobb Motzkin Narayana Ordered Bell Schröder Schröder–Hipparchus Stirling first Stirling second | | | | Primes | | Wieferich Wall–Sun–Sun Wolstenholme prime Wilson | | | | Pseudoprimes | | Carmichael number Catalan pseudoprime Elliptic pseudoprime Euler pseudoprime Euler–Jacobi pseudoprime Fermat pseudoprime Frobenius pseudoprime Lucas pseudoprime Lucas–Carmichael number Somer–Lucas pseudoprime Strong pseudoprime | | | | Arithmetic functions and dynamics | | | | | --- | | Divisor functions | Abundant Almost perfect Arithmetic Betrothed Colossally abundant Deficient Descartes Hemiperfect Highly abundant Highly composite Hyperperfect Multiply perfect Perfect Practical Primitive abundant Quasiperfect Refactorable Semiperfect Sublime Superabundant Superior highly composite Superperfect | | Prime omega functions | Almost prime Semiprime | | Euler's totient function | Highly cototient Highly totient Noncototient Nontotient Perfect totient Sparsely totient | | Aliquot sequences | Amicable Perfect Sociable Untouchable | | Primorial | Euclid Fortunate | | | | | Other prime factor or divisor related numbers | | Blum Cyclic Erdős–Nicolas Erdős–Woods Friendly Giuga Harmonic divisor Lucas–Carmichael Pronic Regular Rough Smooth Sphenic Størmer Super-Poulet Zeisel | | | | Numeral system-dependent numbers | | | | | | | | | | | | | --- --- --- --- --- | | Arithmetic functions and dynamics | Persistence + Additive + Multiplicative | | | --- | | Digit sum | Digit sum Digital root Self Sum-product | | Digit product | Multiplicative digital root Sum-product | | Coding-related | Meertens | | Other | Dudeney Factorion Kaprekar Kaprekar's constant Keith Lychrel Narcissistic Perfect digit-to-digit invariant Perfect digital invariant + Happy | | | P-adic numbers-related | Automorphic + Trimorphic | | Digit-composition related | Palindromic Pandigital Repdigit Repunit Self-descriptive Smarandache–Wellin Strictly non-palindromic Undulating | | Digit-permutation related | Cyclic Digit-reassembly Parasitic Primeval Transposable | | Divisor-related | Equidigital Extravagant Frugal Harshad Polydivisible Smith Vampire | | Other | | | | | | Binary numbers | | Evil Odious Pernicious | | | | Generated via a sieve | | Lucky Prime | | | | Sorting related | | Pancake number Sorting number | | | | Natural language related | | Aronson's sequence Ban | | | | Graphemics related | | | | Retrieved from " Categories: Square numbers Number theory Integer sequences Hidden category: Webarchive template wayback links Square number Add topic
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AlllOM t,Dftb3D uniiea oiaics Department of Commerce Technology Administration National Institute of Standards and Technology NIST Special Publication 811 1995 Edition Guide for the Use of the International System of Units (SI) Barry N. Taylor #Wy <i^-Vy VV'. fhe National Institute of Standards and Technology was established in 1988 by Congress to "assist industry in the development of technology . . . needed to improve product quality, to modernize manufacturing processes, to ensure product reliability ... and to facilitate rapid commercialization ... of products based on new scientific discoveries." NIST, originally founded as the National Bureau of Standards in 1901, works to strengthen U.S. industry's competitiveness; advance science and engineering; and improve public health, safety, and the environment. One of the agency's basic functions is to develop, maintain, and retain custody of the national standards of measurement, and provide the means and methods for comparing standards used in science, engineering, manufacturing, commerce, industry, and education with the standards adopted or recognized by the Federal Government. As an agency of the U.S. Commerce Department's Technology Administration, NIST conducts basic and applied research in the physical sciences and engineering, and develops measurement techniques, test methods, standards, and related services. The Institute does generic and precompetitive work on new and advanced technologies. NlST's research facilities are located at Gaithersburg, MD 20899, and at Boulder, CO 80303. Major technical operating units and their principal activities are listed below. For more information contact the Public Inquiries Desk, 301-975-3058. 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NIST Special Publication 811 1995 Edition Guide for the Use of the International System of Units (SI) Barry N. Taylor Physics Laboratory National Institute of Standards and Technology Gaithersburg, MD 20899-0001 (Supersedes NIST Special Publication 811, September 1991) April 1995 U.S. Department of Commerce Ronald H. Brown, Secretary Technology Administration Mary L. Good, Under Secretary for Technology National Institute of Standards and Technology Arati Prabhakar, Director National Institute of Standards and Technology Special Publication 81 1995 Edition (Supersedes NIST Special Publication 811, September 1991) Natl. Inst. Stand. Technol. Spec. Publ. 811 1995 Ed. 84 pages (April 1 995) CODEN: NSPUE2 U.S. Government Printing Office Washington: 1995 For sale by the Superintendent of Documents U.S. Government Printing Office Washington, DC 20402 Guide for the Use of the International System of Units (SI) Preface The International System of Units, universally abbreviated SI (from the French Le Systeme International d'Unites), is the modern metric system of measurement. Long the dom-inant measurement system used in science, the SI is becoming the dominant measurement system used in international commerce. The Omnibus Trade and Competitiveness Act of August 1988 [Public Lav^' (PL) 100-418] changed the name of the National Bureau of Standards (NBS) to the National Institute of Standards and Technology (NIST) and gave to NIST the added task of helping United States industry increase its competitiveness in the global marketplace. It also recognized the rapidly expanding use of the SI by amending the Metric Conversion Act of 1975 (PL 94-168). In particular, section 5164 (Metric Usage) of PL 100-418 designates the metric system of measurement as the preferred system of weights and measures for United States trade and commerce . . . and requires that each Federal agency, by a date certain and to the extent economically feasible by the end of fiscal year 1992, use the metric system of measurement in its procurements, grants, and other business-related activities, except to the extent that such use is imprac-tical or is likely to cause significant inefficiencies or loss of markets for United States firms . . . In January 1991, the Department of Commerce issued an addition to the Code of Fed-eral Regulations entitled "Metric Conversion Policy for Federal Agencies," 15 CFR 1170, which removes the voluntary aspect of the conversion to the SI for Federal agencies and gives in detail the policy for that conversion. Executive Order 12770, issued in July 1991, reinforces that policy by providing Presidential authority and direction for the use of the metric system of measurement by Federal agencies and departments. Because of the importance of the SI to both science and technology, NIST has over the years published documents to assist NIST authors and other users of the SI, especially to inform them of changes in the SI and in SI usage. For example, this second edition of the Guide replaces the first edition prepared by Arthur O. McCoubrey and published in 1991. That edition, in turn, replaced NBS Letter Circular LC 1120 (1979), which was widely dis-tributed in the United States and which was incorporated into the NBS Communications Manual for Scientific, Technical, and Public Information , a manual of instructions issued in 1980 for the preparation of technical publications at NBS. It is quite natural for NIST to publish documents on the use of the SI. First, NIST coordinates the Federal Government policy on the conversion to the SI by Federal agencies and on the use of the SI by United States industry and the public. Second, NIST provides official United States representation in the various international bodies es-tablished by the Meter Convention (Convention du Metre, often called the Treaty of the Meter in the United States), which was signed in Paris in 1875 by seventeen countries, including the United States (nearly 50 countries are now members of the Convention). • Executive Order 12770 was published in the Federal Register, Vol. 56, No. 145, p. 35801, July 29, 1991; 15 CFR 1 170 was originally published in the Federal Register, Vol. 56, No. 1, p. 160, January 2, 1991 as 15 CFR Part 19, but was redesignated 15 CFR 1170. Both Executive Order 12770 and 15 CFR 1 170 are reprinted in Ref. . (See Appendix D — Bibliography, which begins on p. 72.) iii Guide for the Use of the International System of Units (SI) One body created by the Meter Convention is the General Conference on Weights and Measures (CGPM, Conference Generale des Poids et Mesures), a formal diplomatic organiza-tion. The International System was in fact established by the 11th CGPM in 1960, and it is the responsibility of the CGPM to ensure that the SI is widely disseminated and that it reflects the latest advances in science and technology. This 1995 edition of the Guide corrects a number of misprints in the 1991 edition, incor-porates a significant amount of additional material intended to answer frequently asked questions concerning the SI and SI usage, and updates the bibliography. The added material includes a check list in Chapter 11, which is reproduced immediately after this Preface for easy reference, for reviewing the consistency of NIST manuscripts with the SI. Some changes in format have also been made in an attempt to improve the ease of use of the Guide. In keeping with United States and NIST practice (see Sec. C.3), this edition of the Guide continues to use the dot as the decimal marker rather than the comma, the spellings "meter," "liter," and "deka" rather than "metre," "litre," and "deca," and the name "metric ton" rather than tonne. I should like to take this opportunity to thank James B. McCracken of the NIST Metric Program for his highly capable assistance in the early stages of the preparation of this Guide . See Ref. or for a brief description of the various bodies established by the Meter Convention: The International Bureau of Weights and Measures (BIPM, Bureau International des Poids and Mesures), the Interna-tional Committee for Weights and Measures (CIPM, Comite International des Poids et Mesures), and the CGPM. The BIPM, vv'hich is located in Sevres, a suburb of Paris, France, and which has the task of ensuring worldwide unification of physical measurements, operates under the exclusive supervision of the CIPM, which itself comes under the authority of the CGPM. In addition to a complete description of the SI, Refs. and also give the various CGPM and CIPM resolutions on which it is based. With the exception of Table 8, Tables 1 to 11 of this Guide and their accompanying text are taken or are adapted from these references. March 1995 Barry N. Taylor iv Guide for the Use of the International System of Units (SI) Check List for Reviewing Manuscripts The following check list, which constitutes Chapter 11 of this Guide and is adapted from Ref. , is intended to help NIST authors review the conformity of their manuscripts with proper SI usage and the basic principles concerning quantities and units. (The chapter or section numbers in parentheses indicate where additional information may be found.) (1) O Only units of the SI and those units recognized for use with the SI are used to express the values of quantities. Equivalent values in other units are given in parentheses following values in acceptable units only when deemed necessary for the intended audience. (See Chapter 2.) (2) O Abbreviations such as sec (for either s or second), cc (for either cm^ or cubic centimeter), or mps (for either m/s or meter per second), are avoided and only standard unit symbols, SI prefix symbols, unit names, and SI prefixes are used. (See Sec. 6.1.8.) (3) O The combinations of letters "ppm," "ppb," and "ppt," and the terms part per mil-lion, part per billion, and part per trillion, and the like, are not used to express the values of quantities. The following forms, for example, are used instead: 2.0 iiL/L or 2.0 X lO"^ V, 4.3 nm/m or 4.3 x 10'' /, 7 ps/s or 7 x 10"'' t, where V, I, and t are, respectively, the quantity symbols for volume, length, and time. (See Sec. 7.10.3.) (4) [H Unit symbols (or names) are not modified by the addition of subscripts or other information. The following forms, for example, are used instead. (See Sees. 7.4 and 7.10.2.) Kmax= 1000V but not: ]/- 1000V„,ax a mass fraction of 10 % but not: 10 % (m/m) or 10 % (by weight) (5) CD Statements such as "the length /i exceeds the length I2 by 0.2 %" are avoided be-cause it is recognized that the symbol % represents simply the number 0.01. In-stead, forms such as "/i = /2(1 + 0.2 %)" or "A = 0.2 %" are used, where A is defined by the relation A -(li - /2)//2. (See Sec. 7.10.2.) (6) Information is not mixed with unit symbols (or names). For example, the form "the water content is 20 mL/kg" is used and not "20 mL HaO/kg" or "20 mL of water/kg." (See Sec. 7.5.) (7) n It is clear to which unit symbol a numerical value belongs and which mathematical operation applies to the value of a quantity because forms such as the following are used. (See Sec. 7.7.) 35 cm X 48 cm but not: 35 x 48cm 1 MHz to 10 MHz or (1 to 10) MHz but not: 1 MHz - 10 MHz or 1 to 10 MHz 20 °C to 30 °C or (20 to 30) °C but not : 20 °C - 30 °C or 20 to 30 °C 123g ± 2g or (123 ± 2)g but not: 123 ±2g 70 % ± 5 % or (70 ± 5) % but not: 70 ± 5 % 240 X (1 ± 10%) V but not: 240 V ± 10% (one cannot add 240 V and 10 %) (8) Cn Unit symbols and unit names are not mixed and mathematical operations are not applied to unit names. For example, only forms such as kg/m^ kg • m or kilogram per cubic meter are used and not forms such as kilogram/m\ kg/cubic meter, kilogram/cubic meter, kg per m^, or kilogram per meter\ (See Sees. 6.1.7, 9.5, and 9.8.) V Guide for the Use of the International System of Units (SI) (9) n Values of quantities are expressed in acceptable units using Arabic numerals and the symbols for the units. (See Sec. 7.6.) m = 5 kg but not : m = five kilograms or m = five kg the current was 15 A but not : the current was 15 amperes. (10) O There is a space between the numerical value and unit symbol, even when the value is used in an adjectival sense, except in the case of superscript units for plane angle. (See Sec. 7.2.) a 25 kg sphere but not: a 25-kg sphere an angle of 2°3'4" but not: an angle of 2 °3 '4 " If the spelled-out name of a unit is used, the normal rules of English are applied: "a roll of 35-millimeter film." (See Sec. 7.6, note 3.) (11) O The digits of numerical values having more than four digits on either side of the decimal marker are separated into groups of three using a thin, fixed space counting from both the left and right of the decimal marker. For example, 15 739.012 53 is highly preferred to 15739.01253. Commas are not used to separate digits into groups of three. (See Sec. 10.5.3.) (12) O Equations between quantities are used in preference to equations between nu-merical values, and symbols representing numerical values are different from sym-bols representing the corresponding quantities. When a numerical-value equation is used, it is properly written and the corresponding quantity equation is given where possible. (See Sec. 7.11.) (13) im Standardized quantity symbols such as those given in Refs. and are used, for example, R for resistance and Ai for relative atomic mass, and not words, acronyms, or ad hoc groups of letters. Similarly, standardized mathematical signs and symbols such as are given in Ref. [6: ISO 31-11] are used, for example, "tanx" and not "tgx." More specifically, the base of "log" in equations is speci-fied when required by writing logo a; (meaning log to the base aofx),lhx (meaning log2J:), \nx (meaning \ogex), or lg;c (meaning logiox). (See Sees. 10.1.1 and 10.1.2.) (14) CH Unit symbols are in roman type, and quantity symbols are in italic type with super-scripts and subscripts in roman or italic type as appropriate. (See Sec. 10.2 and Sees. 10.2.1 to 10.2.4.) (15) n When the word "weight" is used, the intended meaning is clear. (In science and technology, weight is a force, for which the SI unit is the newton; in commerce and everyday use, weight is usually a synonym for mass, for which the SI unit is the kilogram.) (See Sec. 8.3.) (16) A quotient quantity, for example, mass density, is written "mass divided by vol-ume" rather than "mass per unit volume." (See Sec. 7.12.) (17) n An object and any quantity describing the object are distinguished. (Note the dif-ference between "surface" and "area," "body" and "mass," "resistor" and "resis-tance," "coil" and "inductance.") (See Sec. 7.13.) (18) D The obsolete term normality and the symbol N, and the obsolete term molarity and the symbol m, are not used, but the quantity amount-of-substance concentra-tion of B (more commonly called concentration of B), and its symbol Cb and SI unit mol/m^ (or a related acceptable unit), are used instead. Similarly, the obsolete term molal and the symbol m are not used, but the quantity molality of solute B, and its symbol 6b or ma and SI unit mol/kg (or a related unit of the SI), are used instead. (See Sees. 8.6.5 and 8.6.8.) VI Guide for the Use of the International System of Units (SI) Contents Preface iii Check List for Reviewing Manuscripts v 1 Introduction 1 1.1 Purpose of Guide 1 1.2 Outline of Guide 1 2 NIST policy on the Use of the SI 2 2.1 Essential data 2 2.1.1 Tables and graphs 2 2.2 Descriptive information 2 3 Other Sources of Information on the SI 3 3.1 Publications 3 3.2 Fundamental Constants Data Center 3 3.3 Metric Program 3 4 The Three Classes of SI Units and the SI Prefixes 3 4.1 SI base units 4 4.2 SI derived units 4 4.2.1 SI derived units with special names and symbols 4 4.2.1.1 Degree Celsius 5 4.2.2 Use of SI derived units with special names and symbols 6 4.3 SI supplementary units 7 4.4 Decimal multiples and submultiples of SI units: SI prefixes 7 5 Units Outside the SI 8 5.1 Units accepted for use with the SI 8 5.1.1 Hour, degree, liter, and the like 8 5.1.2 Neper, bel, shannon, and the like 9 5.1.3 Electronvolt and unified atomic mass unit 9 5.1.4 Natural and atomic units 9 5.2 Units temporarily accepted for use with the SI 10 5.3 Units not accepted for use with the SI 10 5.3.1 COS units 11 5.3.2 Other unacceptable units 11 5.4 The terms "units of the SI" and "acceptable units" 11 6 Rules and Style Conventions for Printing and Using Units 12 6.1 Rules and style conventions for unit symbols 12 6.1.1 Typeface 12 6.1.2 Capitalization 12 6.1.3 Plurals 12 6.1.4 Punctuation 12 6.1.5 Unit symbols obtained by multiplication 12 6.1.6 Unit symbols obtained by division 13 6.1.7 Unacceptability of unit symbols and unit names together 13 6.1.8 Unacceptability of abbreviations for units 13 vii Guide for the Use of the International System of Units (SI) 6.2 Rules and style conventions for SI prefixes 13 6.2.1 Typeface and spacing 13 6.2.2 Capitalization 14 6.2.3 Inseparability of prefix and unit 14 6.2.4 Unacceptability of compound prefixes 14 6.2.5 Use of multiple prefixes 14 6.2.6 Unacceptability of stand-alone prefixes 14 6.2.7 Prefixes and the kilogram 14 6.2.8 Prefixes with the degree Celsius and units accepted for use with the SI 15 7 Rules and Style Conventions for Expressing Values of Quantities 15 7.1 Value and numerical value of a quantity 15 7.2 Space between numerical value and unit symbol 16 7.3 Number of units per value of a quantity 16 7.4 Unacceptability of attaching information to units 16 7.5 Unacceptability of mixing information with units 17 7.6 Symbols for numbers and units versus spelled-out names of numbers and units 17 7.7 Clarity in writing values of quantities 18 7.8 Unacceptability of stand-alone unit symbols 18 7.9 Choosing SI prefixes 19 7.10 Values of quantities expressed simply as numbers: the unit one, symbol 1 19 7.10.1 Decimal multiples and submultiples of the unit one 20 7.10.2 %, percentage by, fraction 20 7.10.3 ppm, ppb, and ppt 20 7.10.4 Roman numerals 21 7.11 Quantity equations and numerical-value equations 21 7.12 Proper names of quotient quantities 22 7.13 Distinction between an object and its attribute 22 7.14 Dimension of a quantity 22 8 Comments on Some Quantities and Their Units 23 8.1 Time and rotational frequency 23 8.2 Volume 23 8.3 Weight 24 8.4 Relative atomic mass and relative molecular mass 24 8.5 Temperature interval and temperature difference 25 8.6 Amount of substance, concentration, molality, and the like 25 8.6.1 Amount of substance 25 8.6.2 Mole fraction of B; amount-of-substance fraction of B 25 8.6.3 Molar volume 26 8.6.4 Molar mass 27 8.6.5 Concentration of B; amount-of-substance concentration of B 27 8.6.6 Volume fraction of B 27 8.6.7 Mass density; density 28 8.6.8 Molality of solute B 28 8.6.9 Specific volume 28 8.6.10 Mass fraction of B 28 viii Guide for the Use of the International System of Units (SI) 8.7 Lx)garithmic quantities and units: level, neper, bel 28 8.8 Viscosity 30 8.9 Massic, volumic, areic, lineic 30 9 Rules and Style Conventions for Spelling Unit Names 31 9.1 Capitalization 31 9.2 Plurals 31 9.3 Spelling unit names with prefixes 31 9.4 Spelling unit names obtained by multiplication 31 9.5 Spelling unit names obtained by division 31 9.6 Spelling unit names raised to powers 32 9.7 Other spelling conventions 32 9.8 Unacceptability of applying mathematical operations to unit names 32 10 More on Printing and Using Symbols and Numbers in Scientific and Technical Documents 32 10.1 Kinds of symbols 32 10.1.1 Standardized quantity symbols 33 10.1.2 Standardized mathematical signs and symbols 33 10.2 Typefaces for symbols 33 10.2.1 Quantities - italic 34 10.2.2 Units — roman 34 10.2.3 Descriptive terms — roman 35 10.2.4 Sample equations showing correct type 35 10.3 Greek alphabet in roman and italic type 35 10.4 Symbols for the elements 36 10.4.1 Typeface and punctuation for element symbols 36 10.4.2 Subscripts and superscripts on element symbols 36 10.5 Printing numbers 36 10.5.1 Typeface for numbers 36 10.5.2 Decimal sign or marker 36 10.5.3 Grouping digits 37 10.5.4 Multiplying numbers 37 11 Check List for Reviewing Manuscripts 38 Appendbc A. Definitions of the SI Base Units and the Radian and Steradian 40 A.l Introduction 40 A.2 Meter 40 A.3 Kilogram 40 A.4 Second 40 A.5 Ampere 40 A.6 Kelvin 40 A.7 Mole 40 A.8 Candela 40 A.9 Radian 40 A.IO Steradian 40 ix Guide for the Use of the International System of Units (SI) Appendix B. Conversion Factors 41 B.l Introduction 41 B.2 Notation 41 B.3 Use of conversion factors 41 B.4 Organization of entries and style 42 B.5 Factor for converting motor vehicle efficiency 43 B.6 U.S. survey foot and mile 43 B.7 Rules for rounding numbers and converted numerical values of quantities 44 B.7.1 Rounding numbers 44 B.7.2 Rounding converted numerical values of quantities 45 B.8 Factors for units listed alphabetically 46 B.9 Factors for units listed by kind of quantity or field of science 57 Appendix C. Comments on the References of Appendix D — Bibliography 69 C. l Official interpretation of the SI for the United States: 55 FR 52242-52245 69 C.2 Defining document for the SI: BIPM SI Brochure 69 C.3 United States version of defining document for the SI: NIST SP 330 . . . . 69 C.4 ISO 1000 69 C.5 ISO 31 69 C.6 IEC27 70 C.7 ANSI/IEEE Std 268 70 C.8 Federal Register notices 70 C.9 Federal Standard 376B 71 C.IO 1986 CODATA values of the fundamental constants 71 C.ll Uncertainty in measurement 71 Appendix D. Bibliography 72 X Guide for the Use of the International System of Units (SI) 1 Introduction 1.1 Purpose of Guide The International System of Units was established in 1960 by the 11th General Confer-ence on Weights and Measures (CGPM — see Preface). Universally abbreviated SI (from the French Le Syst^me International d'Unites), it is the modern metric system of measure-ment used throughout the world. This Guide has been prepared by the National Institute of Standards and Technology (NIST) to assist members of the NIST staff, as well as others who may have need of such assistance, in the use of the SI in their work, including the reporting of results of measurements. 1.2 Outline of Guide The Preface gives the principal Federal Government actions taken since 1988 regarding the SI and introduces the international body — the CGPM — that is responsible for the SI. A check list immediately follows the Preface to help NIST authors review the confor-mity of their manuscripts with proper SI usage and the basic principles concerning quantities and units. A detailed Contents, the aim of which is to simplify the use of the Guide, follows the check list. This introductory chapter gives the purpose of the Guide and its outline, while Chapter 2 summarizes and clarifies the NIST policy on the use of the SI in NIST publications. Chapter 3 notes the existence of a number of publications on the SI and gives the two organizational units at NIST to which questions concerning the SI may be directed and from which additional information about the SI may be obtained. Chapter 4 discusses the fundamental aspects of the SI, including the three current classes of SI units: base, derived, and supplementary; those derived units that have special names and symbols, including the degree Celsius; and the SI prefbces that are used to form decimal multiples and submultiples of SI units. Chapter 5 discusses units that are outside the SI and indicates those that may be used with it and those that may not. It also gives (see Sec. 5.4) precise definitions of the terms "units of the SI" and "acceptable units" as used in this Guide. Chapter 6 gives the rules and style conventions for printing and using units, especially unit symbols and SI prefix symbols. Chapters 7 and 8, which some readers may view as the most important parts of this Guide, provide, respectively, the rules and style conventions for expressing the values of quantities, and clarifying comments on some often troublesome quantities and their units. Chapter 9 gives the rules and style conventions for spelling unit names. Chapter 10 further elaborates on printing and using symbols and numbers in scientific and technical documents and is intended to assist NIST authors prepare manuscripts that are consistent with accepted typesetting practice. Chapter 11 gives the check list that is reproduced immediately after the Preface. Appendw A gives the definitions of the SI base units and the radian and steradian, while Appendix B gives conversion factors for converting values of quantities expressed in units that are mainly unacceptable for use with the SI to values expressed mainly in units of the SI. Appendix B also includes a simplified discussion of rounding numbers and rounding converted numerical values of quantities. 1 Guide for the Use of the International System of Units (SI) Appendix C discusses in some detail most of the references included in Appendix D — Bibliography, which concludes the Guide. 2 NIST Policy on the Use of the SI In accordance with various Federal Acts, the Code of Federal Regulations, and Exe-cutive Order 12770 (see Preface), it is NIST policy that the SI shall be used in all NIST publications.' When the field of application or the special needs of users of NIST publica-tions require the use of other units, the values of quantities shall first be expressed in accept-able units, where it is to be understood that acceptable units include the units of the SI and those units recognized for use with the SI; the corresponding values expressed in the other units shall then follow in parentheses. (For precise definitions of the terms "units of the SI" and "acceptable units" as used in this Guide, see Sec. 5.4.) Exceptions to this policy require the prior approval of the NIST Director. The following three sections — 2.1 Essential data, 2.1.1 Tables and graphs, and 2.2 Descriptive information — elaborate upon this policy. 2.1 Essential data Essential data express or interpret quantitative results. All such data shall be given in acceptable units. In those cases where — the sole use of acceptable units would compromise good communication, or — units other than acceptable units have been specified as a contractual requirement, values of quantities shall be given in acceptable units followed, in parentheses, by the values of the same quantities given in the other units. Exceptions may sometimes be necessary for commercial devices, technical standards, or quantities having special legal significance; examples include commercial weights and mea-sures devices and the related laws and regulations. However, even in such cases, values of quantities expressed in acceptable units should be used when possible with the same values expressed in other units following in parentheses. 2.1.1 Tables and graphs In tables, values of quantities expressed in acceptable units and the corresponding values expressed in other units may be shown in parallel columns, with the acceptable-unit column preceding the other-unit column. In graphs, axes labeled in other units shall be given secondary status. This may preferably be done by placing scale marks on and labeling the left-hand ordinate and bottom absiccsa in acceptable units, and placing scale marks on and labeling the right-hand ordinate and top abscissa in other units. Alternatively, lighter-weight scale marks and smaller type may be employed to indicate other units using the same ordi-nate and abscissa as is used for the acceptable units. 2.2 Descriptive information Descriptive information characterizes arrangements, environments, the generalized dimensions of objects, apparatus, or materials, and other attributes that do not enter directly into calculations or results. When necessary for effective communication, such information may be expressed using customary terms that are widely used and recognized. Examples include common drill sizes and traditional tools used in the United States, U.S. standard fastener sizes, commercial pipe sizes, and other common terms used in the trades, the pro-fessions, the marketplace, sports, and various social activities. When such descriptive infor-mation is given, values in acceptable units are not required. For example, it is permissible to refer to a "36-inch pipeline" or a "half-inch drill" without first giving the value in an acceptable unit. ' The NIST policy on the use of the SI is set forth in the NIST Administration Manual, Chapter 4, Communica-tions, Subchapter 4.09, NIST Technical Communications Program, Appendix D - Use of Metric Units. 2 Guide for the Use of the International System of Units (SI) 3 Other Sources of Information on the SI 3.1 Publications Appendix C briefly describes a number of publications that deal with the SI and related topics; citations for these publications are given in Appendix D — Bibliography. Additional information about the SI is also available from the two NIST organizational units indicated in Sees. 3.2 and 3.3. 3.2 Fundamental Constants Data Center Questions concerning the more fundamental aspects of the SI and subtle aspects of proper SI usage may be directed to: Fundamental Constants Data Center Physics Laboratory National Institute of Standards and Technology Building 245, Room C229 Gaithersburg, MD 20899-0001 Telephone: (301) 975-4220 Fax: (301) 869-7682 3.3 Metric Program Questions concerning Federal Government use of the SI and Federal Government pol-icy on the use of the SI by U.S. industry and the public may be directed to: Metric Program Technology Services National Institute of Standards and Technology Building 411, Room A146 Gaithersburg, MD 20899-0001 Telephone: (301) 975-3690 Fax: (301) 948-1416 4 The Three Classes of SI Units and the SI Prefixes SI units are currently divided into three classes: — base units, — derived units, — supplementary units, which together form what is called "the coherent system of SI units." ^ The SI also includes prefixes to form decimal multiples and submultiples of SI units. 2 According to Ref. [6: ISO 31-0], a system of units is coherent with respect to a system of quantities and equa-tions if the system of units is chosen in such a way that the equations between numerical values have exactly the same form (including the numerical factors) as the corresponding equations between the quantities (see Sees. 7.11 and 7.14). In such a coherent system, of which the SI is an example, no numerical factor other than the number 1 ever occurs in the expressions for the derived units in terms of the base units. It should also be noted that the class of supplementary units is likely to be abolished as a separate class in the SI — see Sec. 4.3. 3 Guide for the Use of the International System of Units (SI) 4.1 SI base units Table 1 gives the seven base quantities, assumed to be mutually independent, on which the SI is founded; and the names and symbols of their respective units, called "SI base units." Definitions of the SI base units are given in Appendix A. The kelvin and its symbol K are also used to express the value of a temperature interval or a temperature difference (see Sec. 8.5). Table 1. SI base units SI base unit Base quantity Name Symbol length meter m mass kilogram kg time second s electric current ampere A thermodynamic temperature kelvin K amount of substance mole mol luminous intensity candela cd 4.2 SI derived units Derived units are expressed algebraically in terms of base units or other derived units (including the radian and steradian which are the two supplementary units — see Sec. 4.3). The symbols for derived units are obtained by means of the mathematical operations of multiplication and division. For example, the derived unit for the derived quantity molar mass (mass divided by amount of substance) is the kilogram per mole, symbol kg/mol. Addi-tional examples of derived units expressed in terms of SI base units are given in Table 2. (The rules and style conventions for printing and using SI unit symbols are given in Sees. 6.1.1 to 6.1.8.) Table 2. Examples of SI derived units expressed in terms of SI base units SI derived unit Derived quantity Name Symbol area square meter m^ volume cubic meter m^ speed, velocity meter per second m/s acceleration meter per second squared m/s^ wave number reciprocal meter m-' mass density (density) kilogram per cubic meter kg/m' specific volume cubic meter per kilogram mVkg current density ampere per square meter A/m^ magnetic field strength ampere per meter A/m amount-of-substance concentration (concentration) mole per cubic meter mol/m' luminance candela per square meter cd/m^ 4.2.1 SI derived units with special names and symbols Certain SI derived units have special names and symbols; these are given in Tables 3a and 3b. As discussed in Sec. 4.3, the radian and steradian, which are the two supplementary units, are included in Table 3a. 4 Guide for the Use of the International System of Units (SI) Table 3a. SI derived units with special names and symbols, including the radian and steradian SI derived unit Derived quantity Expression Expression Special name Special symbol in terms in terms of other SI units of SI base units plane angle radian rad m • m"' = 1 solid angle steradian sr m^-m-^ = 1 frequency hertz Hz s-i force newton N m • kg • s"^ pressure, stress pascal Pa N/m2 m~ • kg' s~^ energy, work, quantity of heat joule J N-m m^ • kg • s"^ power, radiant flux watt W J/s m^ • kg • s"' electric charge. quantity of electricity coulomb C s-A electric potential. potential difference, electromotive force volt V W/A m^-kg-s"^- A"' m"^ • kg"' • s • A^ capacitance farad F c/v electric resistance ohm n V/A m^ • kg • s"-' • A"^ m-^-kg-'-s'-A^ electric conductance Siemens s A/V magnetic flux weber Wb V-s m^ • kg • s"^ • A"' magnetic flux density tesla T Wb/m^ inductance henry H Wb/A m^ • kg • • A-2 Celsius temperature^") degree Celsius °C K luminous flux lumen Im cd • sr cd • sr^") illuminance lux be Im/m^ m-2-cd-srW (") See Sees. 4.2.1.1, 6.2.8, and 7.2. The steradian (sr) is not an SI base unit. However, in photometry the steradian (sr) is maintained in expressions for units (see Sec. 4.3). Table 3b. SI derived units with special names and symbols admitted for reasons of safeguarding human health SI derived unit Derived quantity Special name Special symbol Expression in terms of other SI units Expression in terms of SI base units activity (of a radionuclide) becquerel Bq S-' absorbed dose, specific energy (imparted), kerma gray Gy J/kg ra2-s-2 dose equivalent, ambient dose equivalent, directional dose equivalent, personal dose equivalent, equivalent dose sievert Sv J/kg m^-s-^ The derived quantities to be expressed in the gray and the sievert have been revised in accordance with the recommendations of the International Commision on Radiation Units and Measurements (ICRU); see Ref. . 4.2.1.1 Degree Celsius In addition to the quantity thermodynamic temperature (symbol T), expressed in the unit Icelvin, use is also made of the quantity Celsius temperature (symbol t) defined by the equation t = T-To , where To = 273.15 K by definition. To express Celsius temperature, the unit degree Celsius, symbol °C, which is equal in magnitude to the unit kelvin, is used; in this case, "degree Cel-sius" is a special name used in place of "kelvin." An interval or difference of Celsius tem-perature can, however, be expressed in the unit kelvin as well as in the unit degree Celsius (see Sec. 8.5). (Note that the thermodynamic temperature To is exactly 0.01 K below the thermodynamic temperature of the triple point of water (see Sec. A.6).) 5 Guide for the Use of the International System of Units (SI) 4.2.2 Use of SI derived units with special names and symbols Examples of SI derived units that can be expressed with the aid of SI derived units having special names and symbols (including the radian and steradian) are given in Table 4. Table 4. Examples of SI derived units expressed with the aid of SI derived units having special names and symbols SI derived unit Expression Derived quantity Name Symbol in terms of SI base units angular velocity radian per second rad/s m • m"' • s"' = s~' angular acceleration radian per second squared rad/s^ m • m~' • s"^ = s"^ dynamic viscosity pascal second Pa-s m~' • kg • s"' moment of force newton meter N-m m^'kg's"^ surface tension newton per meter N/m kg-s-2 heat flux density, irradiance watt per square meter W/m^ kg-s ^ radiant intensity watt per steradian W/sr m^-kg-s~^-sr-' <"> radiance watt per square meter steradian W/(m^ • sr) kg-s~^-sr"'<''> hpat canacitv pntrnnv loiilp ripr Icplvin J/K 111 specific heat capacity, joule per kilogram specific entropy kelvin J/(kg-K) m^-s-^-K-' specific energy joule per kilogram J/kg m^ • s~^ thermal conductivity watt per meter kelvin W/(m-K) J/m' m-kg-s-3-K-' energy density joule per cubic meter m"' • kg • s"^ electric field strength volt per meter V/m m • kg • s"^ • A"' electric charge density coulomb per cubic meter C/m' m"' • s • A electric flux density coulomb per square meter C/m^ m"^'S" A permittivity farad per meter F/m m-'-kg-'-s-'-A^ permeability henry per meter H/m m • kg • s"^ • A~^ m^-kg-s"^'mol"' molar energy joule per mole J/mol molar entropy, molar heat capacity joule per mole kelvin J/(mol • K) m^-kg-s-^-K-'-mol-' exposure (x and y rays) coulomb per kilogram C/kg kg"' • s-A absorbed dose rate gray per second Gy/s m^-s-' ^"^ The steradian (sr) is not an SI base unit. However, in radiometiy the steradian (sr) is maintained in expressions for units (see Sec. 4.3). The advantages of using the special names and symbols of SI derived units are apparent in Table 4. Consider, for example, the quantity molar entropy: the unit J/(mol • K) is obvi-ously more easily understood than its SI base-unit equivalent, m^ • kg • s~^ • K~' • mol~'. Nevertheless, it should always be recognized that the special names and symbols exist for convenience; either the form in which special names or symbols are used for certain combi-nations of units or the form in which they are not used is correct. For example, because of the descriptive value implicit in the compound-unit form, communication is sometimes facil-itated if magnetic flux (see Table 3a) is expressed in terms of the volt second (V • s) instead of the weber (Wb). Tables 3a, 3b, and 4 also show that the values of several different quantities are ex-pressed in the same SI unit. For example, the joule per kelvin (J/K) is the SI unit for heat capacity as well as for entropy. Thus the name of the unit is not sufficient to define the quantity measured. A derived unit can often be expressed in several different ways through the use of base units and derived units with special names. In practice, with certain quantities, preference is given to using certain units with special names, or combinations of units, to facilitate the distinction between quantities whose values have identical expressions in terms of SI base units. For example, the SI unit of frequency is specified as the hertz (Hz) rather than the reciprocal second (s~), and the SI unit of moment of force is specified as the newton meter (N • m) rather than the joule (J). 6 Guide for the Use of the International System of Units (SI) Similarly, in the field of ionizing radiation, the SI unit of activity is designated as the becquerel (Bq) rather than the reciprocal second (s"'), and the SI units of absorbed dose and dose equivalent are designated as the gray (Gy) and the sievert (Sv), respectively, rather than the joule per kilogram (J/kg). 4.3 SI supplementary units As previously stated, there are two units in this class: the radian, symbol rad, the SI unit of the quantity plane angle; and the steradian, symbol sr, the SI unit of the quantity solid angle. Definitions of these units are given in Appendix A. The SI supplementary units are now interpreted as so-called dimensionless derived units (see Sec. 7.14) for which the CGPM allows the freedom of using or not using them in expressions for SI derived units. ^ Thus the radian and steradian are not given in a separate table but have been included in Table 3a together with other derived units with special names and symbols (see Sec. 4.2.1). This interpretation of the supplementary units implies that plane angle and solid angle are considered derived quantities of dimension one (so-called dimensionless quantities — see Sec. 7.14), each of which has the unit one, symbol 1 as its coherent SI unit. However, in practice, when one expresses the values of derived quan-tities involving plane angle or solid angle, it often aids understanding if the special names (or symbols) "radian" (rad) or "steradian" (sr) are used in place of the number 1. For example, although values of the derived quantity angular velocity (plane angle divided by time) may be expressed in the unit s"', such values are usually expressed in the unit rad/s. Because the radian and steradian are now viewed as so-called dimensionless derived units, the Consultative Committee for Units (CCU, Comite Consultatif des Unites) of the CIPM (see footnote, p. iv), as a result of a 1993 request it received from ISO/TC 12 (see Ref. ), recommended to the CIPM that it request the CGPM to abolish the class of supplementary units as a separate class in the SI. The CIPM accepted the CCU recommen-dation, and if the abolishment is approved by the CGPM as is likely (the question will be on the agenda of the 20th CGPM, October 1995), the SI will consist of only two classes of units: base units and derived units, with the radian and steradian subsumed into the class of derived units of the SI. (The option of using or not using them in expressions for SI derived units, as is convenient, would remain unchanged.) 4.4 Decimal multiples and submultiples of SI units: SI prefixes Table 5 gives the SI prefixes that are used to form decimal multiples and submultiples of SI units. They allow very large or very small numerical values (see Sec. 7.1) to be avoided. A prefix attaches directly to the name of a unit, and a prefix symbol attaches directly to the symbol for a unit. For example, one kilometer, symbol 1 km, is equal to one thousand meters, symbol 1000 m or 10^ m. When prefixes are attached to SI units, the units so formed are called "multiples and submultiples of SI units" in order to distinguish them from the coher-ent system of SI units. (See footnote 2 for a brief discussion of coherence. The rules and style conventions for printing and using SI prefixes are given in Sees. 6.2.1 to 6.2.8. The special rule for forming decimal multiples and submultiples of the unit of mass is given in Sec. 6.2.7.) Note: Alternative definitions of the SI prefixes and their symbols are not permitted. For example, it is unacceptable to use kilo (k) to represent 2'° = 1024, mega (M) to represent 2^° = 1 048 576, or giga (G) to represent 2^ = 1 073 741 824. ' See Ref. or . This interpretation was given in 1980 by the CIPM (see footnote, p. iv). It was deemed necessary because Resolution 12 of the 11th CGPM, which established the SI in 1960 [2, 3], did not specify the nature of the supplemental^ units. The interpretation is based on two principal considerations: that plane angle is generally expressed as the ratio of two lengths and solid angle as the ratio of an area and the square of a length, and are thus quantities of dimension one (so-called dimensionless quantities); and that treating the radian and steradian as SI base units — a possibility not disallowed by Resolution 12 — could compromise the internal coher-ence of the SI based on only seven base units. (See Ref. [6: ISO 31-0] and also Sec. 7.14 for a discussion of the concept of dimension, and footnote 2 for a brief discussion of coherence.) 7 Guide for the Use of the International System of Units (SI) Table 5. SI prefixes pactor Prefix Svmbol Factor Prefix in24 _ /in3\8 ia21 _ /in3\7 xw V / 10'" = (10')" 10'5 = (10') 10'2 = (10') 10' = (10')' yotts V I in-' deci a ZCttfl z 10-2 centi C exa E 10-' = (10')-' milli m peta P 10-" = (10')-2 micro M-tera T 10-' = (10')-' 10-'2 = (10')- 10-' = (10')- nano n giga G pico P 10 = (10')2 mega M femto f 10' = (10')' kilo k 10-'" = (10')-" atto a 102 hecto h 10-2' = (10')-' 10-2 = (103) -« zepto z 10' deka da yocto y 5 Units Outside the SI Units that are outside the SI may be divided into three categories: — those units that are accepted for use with the SI; — those units that are temporarily accepted for use with the SI; and — those units that are not accepted for use with the SI and thus in the view of this Guide must strictly be avoided. 5.1 Units accepted for use with the SI The following four sections discuss in detail the units this Guide accepts for use with the SI. 5.1.1 Hour, degree, liter, and the like Certain units that are not part of the SI are essential and used so widely that they are accepted by the CIPM, and thus by this Guide, for use with the SI [2, 3]. These units are given in Table 6. The combination of units of this table with SI units to form derived units should be restricted to special cases in order not to lose the advantages of the coherence of SI units. (The use of SI prefixes with the units of Table 6 is discussed in Sec. 6.2.8.) Additionally, this Guide recognizes that it may be necessary on occasion to use time-re-lated units other than those given in Table 6; in particular, circumstances may require that intervals of time be expressed in weeks, months, or years. In such cases, if a standardized symbol for the unit is not available, the name of the unit should be written out in full. (See Sec. 8.1 for a suggestion regarding the symbol for year and Chapter 9 for the rules and style conventions for spelling unit names.) Table 6. Units accepted for use with the SI Name Symbol Value in SI units minute hour day degree minute second . liter metric ton time plane angle mm h 1, 1 min = 60 s 1 h = 60 min = 3600 s Id = 24 h = 86 400 s 1° = (ir/180)rad 1' = (1/60)° = (ir/lO 800) rad 1" = (1/60)' = (-ir/648 000) rad IL It = 10' kg (") See also Sec. 7.2. The alternative symbol for the liter, L, was adopted by the CGPM in order to avoid the risk of confusion between the letter 1 and the number 1 (see Ref. or ). Thus, although both 1 and L are internationally accepted symbols for the liter, to avoid this risk the symbol to be used in the United States is L (see Refs. and ). The script letter € is not an approved symbol for the liter. This is the name to be used for this unit in the United States (see Refs. and ); it is also used in some other English-speaking countries. However, this unit is called "tonne" in Ref. and is the name used in many coun-tries. 8 Guide for the Use of the International System of Units (SI) 5.1.2 Neper, bel, shannon, and the like There are a few highly specialized units not listed in Table 6 that are given by the International Organization for Standardization (ISO) or the International Electrotechnical Commission (lEC) and which in the view of this Guide are also acceptable for use with the SI. They include the neper (Np), bel (B), octave, phon, and sone, and units used in informa-tion technology, including the baud (Bd), bit (bit), erlang (E), hartley (Hart), and shannon (Sh)." It is the position of this Guide that the only such additional units NIST authors may use with the SI are those given in either the International Standards on quantities and units of ISO (Ref. ) or of lEC (Ref. ). 5.1.3 Electronvolt and unified atomic mass unit The CIPM, and thus this Guide, also finds it necessary to accept for use with the SI the two units given in Table 7 [2, 3]. These units are used in specialized fields; their values in SI units must be obtained from experiment and, therefore, are not known exactly. (The use of SI prefixes with the units of Table 7 is discussed in Sec. 6.2.8.) Note : In some fields the unified atomic mass unit is called the dalton, symbol Da; however, this name and symbol are not accepted by the CGPM, CIPM, ISO, or lEC for use with the SI. Similarly, AMU is not an acceptable unit symbol for the unified atomic mass unit. The only allowed name is "unified atomic mass unit" and the only allowed symbol is u. Table 7. Units accepted for use with the SI whose values in SI units are obtained experimentally Name Symbol Definition electronvolt eV unified atomic mass unit u () The electronvolt is the kinetic energy acquired by an electron in passing through a potential difference of 1 V in vacuum; 1 eV = 1.602 177 33 x 10"'^ J with a combined standard uncertainty of 0.000 000 49 x 10-" J [20, 21]. The unified atomic mass unit is equal to 1/12 of the mass of an atom of the nuclide '"'C; 1 u = 1.660 540 2 X 10-" kg with a combined standard uncertainty of 0.000 001 0 x 10"" kg [20, 21]. 5.1.4 Natural and atomic units In some cases, particularly in basic science, the values of quantities are expressed in terms of fundamental constants of nature or so-called natural units. The use of these units with the SI is, in the view of this Guide , permissible when it is necessary for the most effective communication of information. In such cases, the specific natural units that are used must be identified. This requirement applies even to the system of units customarily called "atomic units" used in theoretical atomic physics and chemistry, inasmuch as there are sev-eral different systems that have the appellation "atomic units." Examples of physical quan-tities used as natural units are given in Table 8. This Guide also takes the position that while theoretical results intended primarily for other theorists may be left in natural units, if they are also intended for experimentalists, they must also be given in acceptable units. NIST measurement results must always be given in such units first. The symbol in parentheses following the name of the unit is its internationally accepted unit symbol, but the octave, phon, and sone have no such unit symbols. For additional information on the neper and bel, see Sec. 0.5 of Ref. [6: ISO 31-2], Ref. [7: IEC27-3], and Sec. 8.7 of this Guide. The question of the byte (B) is under inter-national consideration. 9 Guide for the Use of the International System of Units (SI) Table 8. Examples of physical quantities sometimes used as natural units Kind of quantity Physical quantity used as a unit Symbol action Planck constant divided by Itt electric charge elementary charge e energy Hartree energy £h length Bohr radius ao lengin i^ompion waveiengin ^eieciron) Ac magnetic flux magnetic flux quantum magnetic moment Bohr magneton MB magnetic moment nuclear magneton Mn mass electron rest mass WJe mass proton rest mass nip speed speed of electromagnetic waves in vacuum c 5.2 Units temporarily accepted for use with the SI Because of existing practice in certain fields or countries, in 1978 the CIPM considered that it was permissible for the units given in Table 9 to continue to be used with the SI until the CIPM considers that their use is no longer necessary [2, 3]. However, these units must not be introduced where they are not presently used. Further, this Guide strongly discour-ages the continued use of these units by NIST authors except for the nautical mile, knot, are, and hectare; and except for the curie, roentgen, rad, and rem until the year 2000 (the cessation date suggested by the Committee for Interagency Radiation Research and Policy Coordination or CIRRPC, a United States Government interagency group). Table 9. Units temporarily accepted for use with the SI Name Symbol Value in SI units nautical mile 1 nautical mile = 1852 m knot A 1 nautical mile per hour = (1852/3600) m/s Sngstrom 1 A = O.lnm = 10-'" m are('') a 1 a = 1 dam^ = 10^ m^ hectare^'') ha 1 ha = 1 hm^ = 10" m^ barn b lb = lOOfm^ = 10-^ m^ bar bar 1 bar = 0.1 MPa = 100 kPa = 1000 hPa = 10' Pa gal Gal 1 Gal = 1 cm/s^ = IQ-^ m/s^ curie Ci 1 Ci = 3.7 X 10'" Bq 1 R = 2.58x10-" C/kg 1 rad = 1 cGy = 10-^ Gy roentgen R rad rad^^) rem rem 1 rem = 1 cSv = 10-^ Sv See Sec. 5.2 for the position of this Guide regarding the continued use of these units. This unit and its symbol are used to express agrarian areas. '"^^ When there is risk of confusion with the symbol for the radian, rd may be used as the symbol for rad. 5.3 Units not accepted for use with the SI The following two sections briefly discuss units not accepted for use with the SI. 5 In 1993 the CCU (see Sec. 4.3) was requested by ISO/TC 12 (see Ref. ) to consider asking the CIPM to deprecate the use of the units of Table 9 except for the nautical mile and knot, and possibly the are and hectare. The CCU discussed this request at its February 1995 meeting. 10 Guide for the Use of the International System of Units (SI) 5.3.1 CGS units Table 10 gives examples of centimeter-gram-second (CGS) units having special names. These units are not accepted for use with the SI. Further, no other units of the various CGS systems of units, which includes the CGS Electrostatic (ESU), CGS Electromagnetic (EMU), and CGS Gaussian systems, are accepted for use with the SI except such units as the centimeter, gram, and second that are also defined in the SI. Table 10. Examples of CGS units with special names (not accepted for use with the SI) Name Symbol Value in SI units erg erg 1 erg = 10-' J dyne dyn 1 dyn = 10-' N pwise^") P 1 P = 1 dyn • s/cm^ = 0.1 Pa • s 1 St = IcmVs = lO-'mVs stokes^) St gauss^'^) Gs, G 1 Gs corresponds to IQ- T oersted^^) Oe 1 Oe corresponds to (1000/4Tr) A/m 1 Mx corresponds to IQ-" Wb 1 sb = 1 cd/cm^ = lO" cd/m^ maxwell^^^ Mx stilb sb phot ph 1 ph = 10 be ("^The poise (P) is the CGS unit for viscosity (also called dynamic viscosity). The SI unit is the pascal second (Pa-s). The stokes (St) is the CGS unit for kinematic viscosity. The SI unit is the meter squared per second (m^/s). ('^^ This unit is part of the so-called electromagnetic three-dimensional CGS system and cannot strictly speaking be compared to the corresponding unit of the SI, which has four dimensions when only mechanical and electric quantities are considered. 5.3.2 Other unacceptable units There are many units besides CGS units that are outside the SI and not accepted for use with it, including, of course, all of the U.S. customary (that is, inch-pound) units. In the view of this Guide such units must strictly be avoided and SI units, their multiples or submul-tiples, or those units accepted or temporarily accepted for use with the SI (including their appropriate multiples and submultiples), must be used instead. This restriction also applies to the use of unaccepted special names for SI units or special names for multiples or submul-tiples of SI units, such as mho for Siemens (S) and micron for micrometer (|xm). Table 11 gives a few examples of some of these other unacceptable units. Table II. Examples of other unacceptable units Name Symbol Value in SI units fermi fermi 1 fermi = 1 fm = IQ-" m metric carat metric carat 1 metric carat = 200 mg = 2x 10" kg ton-Torr 1 Torr = (101 325/760) Pa standard atmosphere atm 1 atm = 101 325 Pa kilogram-force kgf 1 kgf = 9.806 65 N micron >-1 tJL = 1 M,m = IQ-'^m calorie (various) cal,h (thermochemical) 1 cal.h = 4.184 J X unit XU 1 XU = 0.1002 pm = 1.002 X 10 -'^m stere St 1 St = 1 m^ gamma 7 1 7 = InT = lO-^T gamma (mass) y 1 7 = 1 >ig = 10-" kg lambda (volume) X 1 \ = 1 M,L = 10-" L = 10-" m' 5.4 The terms "units of the SI" and "acceptable units" Consistent with accepted practice [2, 3], this Guide uses the term "units of the SI" to mean the SI units, that is, the SI base units, SI derived units, and SI supplementary units; and multiples and submultiples of these units formed by using the SI prefixes. The term 11 Guide for the Use of the International System of Units (SI) "acceptable units," which is introduced in this Guide for convenience, is used to mean the units of the SI plus (a) those units accepted for use with the SI (see Tables 6 and 7 and Sees. 5.1.1, 5.1.2, and 5.1.3); (b) those units temporarily accepted for use with the SI (see Table 9 and Sec. 5.2); and (c) appropriate multiples and submultiples of such accepted and temporarily accepted units. Because natural and atomic units are not widely recognized for use with the SI, they are not included in the term. However, such units may be used to the extent discussed in Sec. 5.1.4. 6 Rules and Style Conventions for Printing and Using Units 6.1 Rules and style conventions for unit symbols The following eight sections give rules and style conventions related to the symbols for units. 6.1.1 Typeface Unit symbols are printed in roman (upright) type regardless of the type used in the surrounding text. (See also Sec. 10.2 and Sees. 10.2.1 to 10.2.4.) 6.1.2 Capitalization Unit symbols are printed in lower-case letters except that: (a) the symbol or the first letter of the symbol is an upper-case letter when the name of the unit is derived from the name of a person; and (b) the recommended symbol for the liter in the United States is L [see Table 6, footnote Examples : m (meter) s (second) V (volt) Pa (pascal) Im (lumen) Wb (weber) 6.1.3 Plurals Unit symbols are unaltered in the plural. Example : / = 75 cm but not : / = 75 cms Note : / is the quantity symbol for length. (The rules and style conventions for expressing the values of quantities are discussed in detail in Chapter 7.) 6.1.4 Punctuation Unit symbols are not followed by a period unless at the end of a sentence. Example: "Its length is 75 cm." or "It is 75 cm long." but not: "It is 75 cm. long." 6.1.5 Unit symbols obtained by multiplication Symbols for units formed from other units by multiplication are indicated by means of either a half-high (that is, centered) dot or a space. However, this Guide, as does Ref. , prefers the half-high dot because it is less likely to lead to confusion. Example : N • m or N m Notes : 1 A half-high dot or space is usually imperative. For example, m • s "Ms the symbol for the meter per second while ms"Ms the symbol for the reciprocal millisecond (10^ s"' - see Sec. 6.2.3). 12 Guide for the Use of the International System of Units (SI) 2 Reference [6: ISO 31-0] suggests that if a space is used to indicate units formed by multiplication, the space may be omitted if it does not cause confusion. This possibility is reflected in the common practice of using the symbol kWh rather than kW • h or kW h for the kilowatt hour. Nevertheless, this Guide takes the position that a half-high dot or a space should always be used to avoid possible confusion; and that for this same reason, only one of these two allowed forms should be used in any given manuscript. 6.1.6 Unit symbols obtained by division Symbols for units formed from other units by division are indicated by means of a solidus (oblique stroke, /), a horizontal line, or negative exponents. Example: m/s, or m-s"' However, to avoid ambiguity, the solidus must not be repeated on the same line unless parentheses are used. Examples: m/s^ or m-s"^ but not: m/s/s mkg/(s^'A) or m • kg • s"'' • A"' but not: m-kg/s^/A Negative exponents should be used in complicated cases. 6.1.7 Unacceptability of unit symbols and unit names together Unit symbols and unit names are not used together. (See also Sees. 9.5 and 9.8.) Example: C/kg, C • kg"', or coulomb per but not: coulomb/kg; coulomb per kg; kilogram C/kilogram; coulomb • kg"'; C per kg; coulomb /kilogram 6.1.8 Unacceptability of abbreviations for units Because acceptable units generally have internationally recognized symbols and names, it is not permissible to use abbreviations for their unit symbols or names, such as sec (for either s or second), sq. mm (for either mm^ or square millimeter), cc (for either cm^ or cubic centimeter), mins (for either min or minutes), hrs (for either h or hours), lit (for either L or liter), amps (for either A or amperes), AMU (for either u or unified atomic mass unit), or mps (for either m/s or meter per second). Although the values of quantities are normally expressed using symbols for numbers and symbols for units (see Sec. 7.6), if for some reason the name of a unit is more appropriate than the unit symbol (see Sec. 7.6, note 3), the name of the unit should be spelled out in full. 6.2 Rules and style conventions for SI prefixes The following eight sections give rules and style conventions related to the SI prefixes. 6.2.1 Typeface and spacing Prefix symbols are printed in roman (upright) type regardless of the type used in the surrounding text, and are attached to unit symbols without a space between the prefix symbol and the unit symbol. This last rule also applies to prefixes attached to unit names. Examples: mL (milliliter) pm (picometer) GCl (gigaohm) THz (terahertz) 13 Guide for the Use of the International System of Units (SI) 6.2.2 Capitalization The prefix symbols Y (yotta), Z (zetta), E (exa), P (peta), T (tera), G (giga), and M (mega) are printed in upper-case letters while all other prefix symbols are printed in lower-case letters (see Table 5). Prefixes are normally printed in lower-case letters. 6.2.3 Inseparability of prefix and unit The grouping formed by a prefix symbol attached to a unit symbol constitutes a new inseparable symbol (forming a multiple or submultiple of the unit concerned) which can be raised to a positive or negative power and which can be combined with other unit symbols to form compound unit symbols. Examples : 2.3 cm^ = 2.3(cm)' = 2.3(10 "^m)' = 2.3 x 10 "'^m^ Icm" = l(cm)-' = l(lO-^m)-' = lO^m"' SOOOjjLS-i = 5000(jjLs)-i = 5000(10- s)-i = 5000 x 10 s"' = 5 x 10' s"' 1 V/cm = (1 V)/(10 m) = 10^ V/m Prefixes are also inseparable from the unit names to which they are attached. Thus, for ex-ample, millimeter, micropascal, and meganewton are single words. 6.2.4 Unacceptability of compound prefixes Compound prefix symbols, that is, prefbc symbols formed by the juxtaposition of two or more prefix symbols, are not permitted. This rule also applies to compound prefixes. Example: nm (nanometer) but not: m|xm (millimicrometer) 6.2.5 Use of multiple prefixes In a derived unit formed by division, the use of a prefix symbol (or a prefbc) in both the numerator and the denominator may cause confusion. Thus, for example, lOkV/mm is acceptable, but 10 MV/m is often considered preferable because it contains only one prefix symbol and it is in the numerator. In a derived unit formed by multiplication, the use of more than one prefix symbol (or more than one prefix) may also cause confusion. Thus, for example, 10 MV • ms is accept-able, but 10 kV • s is often considered preferable. Note: Such considerations usually do not apply if the derived unit involves the kilogram. For example, 0.13 mmol/g is not considered preferable to 0.13 mol/kg. 6.2.6 Unacceptability of stand-alone prefixes Prefix symbols cannot stand alone and thus cannot be attached to the number 1, the symbol for the unit one. In a similar vein, prefixes cannot be attached to the name of the unit one, that is, to the word "one." (See Sec. 7.10 for a discussion of the unit one.) Example : the number density of Pb atoms is 5 x 10Vm^ but not : the number density of Pb atoms is 5 M/m^ 6.2.7 Prefixes and the kilogram For historical reasons, the name "kilogram" for the SI base unit of mass contains the name "kilo," the SI prefix for 10^ Thus, because compound prefixes are unacceptable (see Sec. 6.2.4), symbols for decimal multiples and submultiples of the unit of mass are formed by attaching SI prefix symbols to g, the unit symbol for gram, and the names of such multiples and submultiples are formed by attaching SI prefbces to the name "gram." Example: 10" kg = 1 mg (1 milligram) but not: 10" kg = 1 jjikg (1 microkilogram) 14 Guide for the Use of the International System of Units (SI) 6.2.8 Prefixes with the degree Celsius and units accepted for use with the SI Prefix symbols may be used with the unit symbol "C and prefixes may be used with the unit name "degree Celsius." For example, 12 m°C (12 millidegrees Celsius) is acceptable. However, to avoid confusion, prefix symbols (and prefixes) are not used with the time-related unit symbols (names) min (minute), h (hour), d (day); nor with the angle-related symbols (names) ° (degree), ' (minute), and " (second) (see Table 6). Prefix symbols (and prefixes) may be used with the unit symbols (names) L (liter), t (metric ton), eV (electronvolt), and u (unified atomic mass unit) (see Tables 6 and 7). However, although submultiples of the liter such as mL (milliliter) and dL (deciliter) are in common use, multiples of the liter such as kL (kiloliter) and ML (megaliter) are not. Similarly, although multiples of the metric ton such as kt (kilometric ton) are commonly used, submultiples such as mt (millimetric ton), which is equal to the kilogram (kg), are not. Examples of the use of prefix symbols with eV and u are 80 MeV (80 megaelectronvolts) and 15 nu (15 nanounified atomic mass units). 7 Rules and Style Conventions for Expressing Values of Quantities 7.1 Value and numerical value of a quantity The value of a quantity is its magnitude expressed as the product of a number and a unit, and the number multiplying the unit is the numerical value of the quantity expressed in that unit. More formally, the value of quantity^ can be written asA = {A}[A ], where {A } is the numerical value of ^4 when the value ofA is expressed in the unit [A ]. The numerical value can therefore be written as {yl } = A /[A ], which is a convenient form for use in figures and tables. Thus, to eliminate the possibility of misunderstanding, an axis of a graph or the head-ing of a column of a table can be labeled '7/°C" instead of "/ (°C)" or "Temperature (°C)." Similarly, an axis or column heading can be labeled "£/(V/m)" instead of "E (V/m)" or "Electric field strength (V/m)." Examples : 1 In the SI, the value of the velocity of light in vacuum is c = 299 792 458 m/s exactly. The number 299 792 458 is the numerical value of c when c is expressed in the unit m/s, and equals c/(m/s). 2 The ordinate of a graph is labeled 7/(10^ K), where T is thermodynamic temper-ature and K is the unit symbol for kelvin, and has scale marks at 0, 1, 2, 3, 4, and 5. If the ordinate value of a point on a curve in the graph is estimated to be 3.2, the corresponding temperature is 7/(10^ K) = 3.2 or T = 3200 K. Notice the lack of ambiguity in this form of labeling compared with "Temperature (10^ K)." 3 An expression such as ln(/?/MPa), where p is the quantity symbol for pressure and MPa is the unit symbol for megapascal, is perfectly acceptable because p/MPa is the numerical value of p when p is expressed in the unit MPa and is simply a number. Notes : 1 For the conventions concerning the grouping of digits, see Sec. 10.5.3. 2 An alternative way of writing c /(m/s) is {c}m/s, meaning the numerical value of c when c is expressed in the unit m/s. 15 Guide for the Use of the International System of Units (SI) 7.2 Space between numerical value and unit symbol In the expression for the value of a quantity, the unit symbol is placed after the numer-ical value and a space is left between the numerical value and the unit symbol. The only exceptions to this rule are for the unit symbols for degree, minute, and second for plane angle: ', and ", respectively (see Table 6), in which case no space is left between the numerical value and the unit symbol. Example: a = 30°22'8" Note : a is a quantity symbol for plane angle. This rule means that: (a) The symbol °C for the degree Celsius is preceded by a space when one expresses the values of Celsius temperatures. Example: f = 30.2°C but not: f = 30.2°C or r=30.2°C (b) Even when the value of a quantity is used in an adjectival sense, a space is left between the numerical value and the unit symbol. (This rule recognizes that unit symbols are not like ordinary words or abbreviations but are mathematical entities, and that the value of a quan-tity should be expressed in a way that is as independent of language as possible — see Sees. 7.6 and 7.10.3.) Examples: aim end gauge but not: a 1-m end gauge a lOkn resistor but not: a lO-kH resistor However, if there is any ambiguity, the words should be rearranged accordingly. For exam-ple, the statement "the samples were placed in 22 mL vials" should be replaced with the statement "the samples were placed in vials of volume 22 mL." Note : When unit names are spelled out, the normal rules of English apply. Thus, for exam-ple, "a roll of 35-millimeter film" is acceptable (see Sec. 7.6, note 3). 7.3 Number of units per value of a quantity The value of a quantity is expressed using no more than one unit. Example: I = 10.234 m but not: I = 10 m 23 cm 4 mm Note : Expressing the values of time intervals and of plane angles are exceptions to this rule. However, it is preferable to divide the degree decimally. Thus one should write 22.20° rather than 22° 12', except in fields such as cartography and astronomy. 7.4 Unacceptability of attaching information to units When one gives the value of a quantity, it is incorrect to attach letters or other symbols to the unit in order to provide information about the quantity or its conditions of measure-ment. Instead, the letters or other symbols should be attached to the quantity. Example: Fmax= 1000V but not: K = 1000Vmax Note : F is a quantity symbol for potential difference. ! I 16 Guide for the Use of the International System of Units (SI) 7.5 Unacceptability of mixing information with units When one gives the value of a quantity, any information concerning the quantity or its conditions of measurement must be presented in such a way as not to be associated with the unit. This means that quantities must be defined so that they can be expressed solely in acceptable units (including the unit one — see Sec. 7.10). Examples : the Pb content is 5 ng/L but not: 5 ng Pb/L or 5 ng of lead/L the sensitivity for NO3 molecules is 5 x 10'°/cm^ but not: the sensitivity is 5 x 10" NO3 molecules/cm^ the neutron emission rate is 5 x 10 7s but not: the emission rate is 5 X lO'^n/s the number density of O2 atoms is 3 x 10'/cm^ but not: the density is 3 x 10' O2 atoms/cm^ the resistance per square is 100 fl but not: the resistance is 100 fl/ square 7.6 Symbols for numbers and units versus spelled-out names of numbers and units This Guide takes the position that the key elements of a scientific or technical paper, particularly the results of measurements and the values of quantities that influence the mea-surements, should be presented in a way that is as independent of language as possible. This will allow the paper to be understood by as broad an audience as possible, including readers with limited knowledge of English. Thus, to promote the comprehension of quantitative in-formation in general and its broad understandability in particular, values of quantities should be expressed in acceptable units using — the Arabic symbols for numbers, that is, the Arabic numerals, not the spelled-out names of the Arabic numerals; and — the symbols for the units, not the spelled-out names of the units. Examples : the length of the laser is 5 m but not : the length of the laser is five meters the sample was annealed at a but not: the sample was annealed at a temperature temperature of 955 K for 12 h of 955 kelvins for 12 hours Notes : 1 If the intended audience for a publication is unlikely to be familiar with a partic-ular unit symbol, it should be defined when first used. 2 Because the use of the spelled-out name of an Arabic numeral with a unit symbol can cause confusion, such combinations must strictly be avoided. For example, one should never write "the length of the laser is five m." 17 Guide for the Use of the International System of Units (SI) 3 Occasionally, a value is used in a descriptive or literary manner and it is fitting to use the spelled-out name of the unit rather than its symbol. Thus this Guide considers acceptable statements such as "the reading lamp was designed to take two 60-watt light bulbs," or "the rocket journeyed uneventfully across 380 000 kilometers of space," or "they bought a roll of 35-millimeter film for their camera." 4 The United States Government Printing Office Style Manual (Ref. , pp. 165-171) gives the rule that symbols for numbers are always to be used when one expresses (a) the value of a quantity in terms of a unit of measurement, (b) time (including dates), and (c) an amount of money. This publication should be consulted for the rules governing the choice between the use of symbols for numbers and the spelled-out names of numbers when numbers are dealt with in general. 7.7 Clarity in writing values of quantities The value of a quantity is expressed as the product of a number and a unit (see Sec. 7.1). Thus, to avoid possible confusion, this Guide takes the position that values of quantities must be written so that it is completely clear to which unit symbols the numerical values of the quantities belong. Also to avoid possible confusion, this Guide strongly recommends that the word "to" be used to indicate a range of values for a quantity instead of a range dash (that is, a long hyphen) because the dash could be misinterpreted as a minus sign. (The first of these recommendations once again recognizes that unit symbols are not like ordinary words or abbreviations but are mathematical entities — see Sec. 7.2.) Examples : 51mm X 51mm X 25mm but not: 51 x 51 x 25mm 225 nm to 2400 nm or (225 to 2400) nm but not: 225 to 2400 nm 0 °C to 100 °C or (0 to 100) °C but not: 0 °C - 100 °C 0 V to 5 V or (0 to 5) V but not: 0 - 5 V (8.2, 9.0, 9.5, 9.8, 10.0) GHz but not: 8.2, 9.0, 9.5, 9.8, 10.0 GHz 63.2 m ± 0.1 m or (63.2 ± 0.1) m but not: 63.2 ± 0.1 m or 63.2 m ± 0.1 129s - 3s = 126s or (129 - 3) s = 126s but not: 129- 3s = 126s Note : For the conventions concerning the use of the multiplication sign, see Sec. 10.5.4. 7.8 Unacceptability of stand-alone unit symbols Symbols for units are never used without numerical values or quantity symbols (they are not abbreviations). Examples : there are 10 mm in 1 km but not: there are many mm in a km it is sold by the cubic meter but not: it is sold by the m^ t/°C, E/(V/m),p/MPa, and the like are perfectly acceptable (see Sec. 7.1) 18 Guide for the Use of the International System of Units (SI) 7.9 Choosing SI prefixes The selection of the appropriate decimal multiple or submultiple of a unit for expressing the value of a quantity, and thus the choice of SI prefix, is governed by several factors. These include — the need to indicate which digits of a numerical value are significant, — the need to have numerical values that are easily understood, and — the practice in a particular field of science or technology. A digit is significant if it is required to express the numerical value of a quantity. In the expression / = 1200 m, it is not possible to tell whether the last two zeroes are significant or only indicate the magnitude of the numerical value of /. However, in the expression / = 1.200 km, which uses the SI prefix symbol for 10^ (kilo, symbol k), the two zeroes are assumed to be significant because if they were not, the value of / would have been written / = 1.2 km. It is often recommended that, for ease of understanding, prefix symbols should be chosen in such a way that numerical values are between 0.1 and 1000, and that only prefix symbols that represent the number 10 raised to a power that is a multiple of 3 should be used. Examples: 3.3 x lO'^Hz may be written as 33 x 10 Hz = 33 MHz 0.009 52 g may be written as 9.52 x 10"^ g = 9.52 mg 2703 W may be written as 2.703 x 10^ W = 2.703 kW 5.8 X 10"^ m may be written as 58 x 10"' m = 58 nm However, the values of quantities do not always allow this recommendation to be followed, nor is it mandatory to try to do so. In a table of values of the same kind of quantities or in a discussion of such values, it is usually recommended that only one prefix symbol should be used even if some of the numerical values are not between 0.1 and 1000. For example, it is often considered prefer-able to write "the size of the sample is 10 mm x 3 mm x 0.02 mm" rather than "the size of the sample is 1 cm x 3 mm x 20 jxm." In certain kinds of engineering drawings it is customary to express all dimensions in millimeters. This is an example of selecting a prefix based on the practice in a particular field of science or technology. 7.10 Values of quantities expressed simply as numbers: the unit one, symbol 1 Certain quantities, such as refractive index, relative permeability, and mass fraction, are defined as the ratio of two mutually comparable quantities and thus are of dimension one (see Sec. 7.14). The coherent SI unit for such a quantity is the ratio of two identical SI units and may be expressed by the number 1. However, the number 1 generally does not appear in the expression for the value of a quantity of dimension one. For example, the value of the refractive index of a given medium is expressed as n = 1.51 x 1 = 1.51. On the other hand, certain quantities of dimension one have units with special names and symbols which can be used or not depending on the circumstances. Plane angle and solid angle, for which the SI units are the radian (rad) and steradian (sr), respectively, are exam-ples of such quantities (see Sec. 4.3). 19 Guide for the Use of the International System of Units (SI) 7.10.1 Decimal multiples and submultiples of the unit one Because SI prefix symbols cannot be attached to the unit one (see Sec. 6.2.6), powers of 10 are used to express decimal multiples and submultiples of the unit one. Example: fir = 1.2 x 10" but not: /ir = 1.2 |x Note : iJLc is the quantity symbol for relative permeability. 7.10.2 %, percentage by, fraction In keeping with Ref. [6: ISO 31-0], this Guide takes the position that it is acceptable to use the internationally recognized symbol % (percent) for the number 0.01 with the SI and thus to express the values of quantities of dimension one (see Sec. 7.14) with its aid. When it is used, a space is left between the symbol % and the number by which it is multiplied [6: ISO 31-0]. Further, in keeping with Sec. 7.6, the symbol % should be used, not the name "percent." Example xb = 0.0025 = 0.25 % but not: xb = 0.0025 = 0.25% or xb = 0.25 percent Note: Xb is the quantity symbol for amount-of-substance fraction of B (see Sec. 8.6.2). Because the symbol % represents simply a number, it is not meaningful to attach infor-mation to it (see Sec. 7.4). One must therefore avoid using phrases such as "percentage by weight," "percentage by mass," "percentage by volume," or "percentage by amount of sub-stance." Similarly, one must avoid writing, for example, "% (m/m)," "% (by weight)," "% (V/V)," "% (by volume)," or "% (mol/mol)." The preferred forms are "the mass frac-tion is 0.10," or "the mass fraction is 10 %," or "wb = 0.10," or "wb = 10 %" (wb is the quantity symbol for mass fraction of B — see Sec. 8.6.10); "the volume fraction is 0.35," or "the volume fraction is 35 %," or "(ps = 0.35," or "v/(km/h). NIST authors should consider using this preferred form instead of the more traditional form "/ = 3.6'^ vt, where / is in meters, v is in kilometers per hour, and t is in seconds." In fact, this form is still ambiguous because no clear distinction is made between a quantity and its numerical value. The correct statement is, for example, "/ = 3.6"' v r, where / is the numerical value of the distance / travelled by a particle in uniform motion when / is expressed in meters, v is the numerical value of the velocity v of the particle when v is expressed in kilometers per hour, and t is the numerical value of the time of travel t of the particle when t is expressed in seconds." Clearly, as is done here, it is important to use different symbols for quantities and their numerical values to avoid confusion. It is the strong recommendation of this Guide that because of their universality, quantity equations should be used in preference to numerical-value equations. Further, if a numerical value equation is used, it should be written in the preferred form given in the above para-graph and if at all feasible, the quantity equation from which it was obtained should be given. Notes : 1 Two other examples of numerical-value equations written in the preferred form are as follows, where Eg is the gap energy of a compound semiconductor and k is the conductivity of an electrolytic solution: £g/eV = 1.425 -1.337x: + 0.nOx\ O^x ^ 0.15, where ;c is an appropriately defined amount-of-substance fraction (see Sec. 8.6.2). K/(S/cm) = 0.065 135 + 1.7140 x 10-'(r/°C) + 6.4141 x 10-'(r/°C)' -4.5028 X 10"(r/°C)^ 0 °C ^ / ^ 50 °C, where t is Celsius temperature. 2 Writing numerical-value equations for quantities expressed in inch-pound units in the preferred form will simplify their conversion to numerical-value equations for the quantities expressed in units of the SI. 7.12 Proper names of quotient quantities Derived quantities formed from other quantities by division are written using the words "divided by" rather than the words "per unit" in order to avoid the appearance of associat-ing a particular unit with the derived quantity. Example : pressure is force divided by area but not : pressure is force per unit area 7.13 Distinction between an object and its attribute To avoid confusion, when discussing quantities or reporting their values, one should distinguish between a phenomenon, body, or substance, and an attribute ascribed to it. For example, one should recognize the difference between a body and its mass, a surface and its area, a capacitor and its capacitance, and a coil and its inductance. This means that although it is acceptable to say "an object of mass 1 kg was attached to a string to form a pendulum," it is not acceptable to say "a mass of 1 kg was attached to a string to form a pendulum." 7.14 Dimension of a quantity Any SI derived quantity Q can be expressed in terms of the SI base quantities length (/), mass (m), time (t), electric current (/), thermodynamic temperature (7), amount of sub-stance (n ), and luminous intensity (h) by an equation of the form ^ j I K = 1 22 Guide for the Use of the International System of Units (SI) where the exponents a, P, y, . . . are numbers and the factors a are also numbers. The dimension of Q is defined to be dimg = L^M^TM^e^NfJ" , where L, M, T, I, 9, N, and J are the dimensions of the SI base quantities length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity, respectively. The exponents a, fi,y, . . . are called "dimensional exponents." The SI derived unit of is m" • kg^ • s^ • A • • moK cd'', which is obtained by replacing the dimensions of the SI base quantities in the dimension of Q with the symbols for the corresponding base units. Example : Consider a nonrelativistic particle of mass m in uniform motion which travels a distance / in a time f . Its velocity is v =l/t and its kinetic energy is Et = = l^mt ~^/2. The dimension of Ev is dim Ek = L^MT~^ and the dimensional expo-nents are 2, 1, and -2. The SI derived unit of Ek is then m^ • kg • s"^ which is given the special name "joule" and special symbol J. A derived quantity of dimension one, which is sometimes called a "dimensionless quan-tity," is one for which all of the dimensional exponents are zero: dim Q = 1. It therefore follows that the derived unit for such a quantity is also the number one, symbol 1, which is sometimes called a "dimensionless derived unit." Example: The mass fraction wb of a substance B in a mixture is given by wb = m^lm, where /tzb is the mass of B and m is the mass of the mixture (see Sec. 8.6.10). The dimension of is dim we = M'M = 1; all of the dimensional exponents of Wb are zero, and its derived unit is kg' • kg"' = 1 also. 8 Comments on Some Quantities and Their Units 8.1 Time and rotational frequency The SI unit of time (actually time interval) is the second (s) and should be used in all technical calculations. When time relates to calendar cycles, the minute (min), hour (h), and day (d) may be necessary. For example, the kilometer per hour (km/h) is the usual unit for expressing vehicular speeds. Although there is no universaally accepted symbol for the year, Ref. [6: ISO 31-1] suggests the symbol a. The rotational frequency /j of a rotating body is defined to be the number of revolutions it makes in a time interval divided by that time interval [6: ISO 31-5]. The SI unit of this quantity is thus the reciprocal second (s~'). However, as pointed out in Ref. [6: ISO 31-5], the designations "revolutions per second" (r/s) and "revolutions per minute" (r/min) are widely used as units for rotational frequency in specifications on rotating machinery. 8.2 Volume The SI unit of volume is the cubic meter (m^) and may be used to express the volume of any substance, whether solid, liquid, or gas. The liter (L) is a special name for the cubic decimeter (dm^) but the CGPM recommends that the liter not be used to give the results of high accuracy measurements of volumes [2, 3]. Also, it is not common practice to use the liter to express the volumes of solids nor to use multiples of the liter such as the kiloliter (kL) [see Sec. 6.2.8, and also Table 6, footnote (6)]. 23 Guide for the Use of the International System of Units (SI) 8.3 Weight In science and technology, the weight of a body in a particular reference frame is de-fined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame [6: ISO 31-3]. Thus the SI unit of the quantity weight defined in this way is the newton (N). When the reference frame is a celestial object, Earth for ex-ample, the weight of a body is commonly called the local force of gravity on the body. Example : The local force of gravity on a copper sphere of mass 10 kg located on the surface of the Earth, which is its weight at that location, is approximately 98 N. Note : The local force of gravity on a body, that is, its weight, consists of the resultant of all the gravitational forces acting on the body and the local centrifugal force due to the rotation of the celestial object. The effect of atmospheric buoyancy is usually excluded, and thus the weight of a body is generally the local force of gravity on the body in vacuum. In commercial and everyday use, and especially in common parlance, weight is usually used as a synonym for mass. Thus the SI unit of the quantity weight used in this sense is the kilogram (kg) and the verb "to weigh" means "to determine the mass of" or "to have a mass of." Examples : the child's weight is 23 kg the briefcase weighs 6 kg Net wt. 227 g Inasmuch as NIST is a scientific and technical organization, the word "weight" used in the everyday sense (that is, to mean mass) should appear only occasionally in NIST publica-tions; the word "mass" should be used instead. In any case, in order to avoid confusion, whenever the word "weight" is used, it should be made clear which meaning is intended. 8.4 Relative atomic mass and relative molecular mass The terms atomic weight and molecular weight are obsolete and thus should be avoided. They have been replaced by the equivalent but preferred terms relative atomic mass, symbol Ar, and relative molecular mass, symbol Mr, respectively [6: ISO 31-8], which better reflect their definitions. Like atomic weight and molecular weight, relative atomic mass and relative molecular mass are quantities of dimension one and are expressed simply as numbers. The definitions of these quantities are as follows [6: ISO 31-8]: Relative atomic mass (formerly atomic weight): ratio of the average mass per atom of an element to 1/12 of the mass of the atom of the nuclide '^C. Relative molecular mass (formerly molecular weight): ratio of the average mass per molecule or specified entity of a substance to 1/12 of the mass of an atom of the nuclide '^C. Examples: Ar(Si) = 28.0855 A/r(H2) = 2.0159 ArC^C) = 12 exactly Notes : 1 It follows from these definitions that if X denotes a specified atom or nuclide and B a specified molecule or entity (or more generally, a specified substance), then AriX) = m(X)/[m ('^C)/12] and Mr(B) = m(B)/[m(''C)/12], where m(X) is the mass of X, m (B) is the mass of B, and m ('^C) is the mass of an atom of the nuclide '^C. It should also be recognized that m('^C)/12 = u, the unified atomic mass unit, which is approximately equal to 1.66 x 10 ~" kg [see Table 7, footnote 2 It follows from the examples and note 1 that the respective average masses of Si, H2,and2Carem(Si) = ylr(Si) u, m(H2) = Mr(H2) u, andmC^C) =^r(''C)u. 24 A. Guide for the Use of the International System of Units (SI) 3 In publications dealing with mass spectrometry, one may encounter a statement such as "the mass-to-charge ratio is 15." What is usually meant in this case is that the ratio of the nucleon number (that is, mass number — see Sec. 10.4.2) of the ion to its number of charges is 15. Thus mass-to-charge ratio is a quantity of dimension one, even though it is commonly denoted by the symbol miz. For example, the mass-to-charge ratio of the ion '^Ct'H? is 91/2 = 45.5. 8.5 Temperature interval and temperature difference As discussed in Sec. 4.2.1.1, Celsius temperature {t) is defined in terms of thermody-namic temperature (7) by the equation t = T — To, where To = 273.15 K by definition. This implies that the numerical value of a given temperature interval or temperature differ-ence whose value is expressed in the unit degree Celsius (°C) is equal to the numerical value of the same interval or difference when its value is expressed in the unit kelvin (K); or in the notation of Sec. 7.1, note 2, {A?}«c = {ATIk. Thus temperature intervals or temperature differences may be expressed in either the degree Celsius or the kelvin using the same nu-merical value. Example : The difference in temperature between the freezing point of gallium and the triple point of water is At = 29.7546 °C = AT = 29.7546 K. 8.6 Amount of substance, concentration, molality, and the like The following section discusses amount of substance, and the subsequent nine sections, which are based on Ref. [6: ISO 31-8] and which are succinctly summarized in Table 12, discuss quantities that are quotients involving amount of substance, volume, or mass. In the table and its associated sections, symbols for substances are shown as subscripts, for exam-ple, xb, aib, ^B. However, it is generally preferable to place symbols for substances and their states in parentheses immediately after the quantity symbol, for example n(H2S04). (For a detailed discussion of the use of the SI in physical chemistry, see the book cited in Ref. , note 5.) 8.6.1 Amount of substance Quantity symbol', n (also v) . SI unit: mole (mol). Definition: See Sec. A.7. Notes : 1 Amount of substance is one of the seven base quantities upon which the SI is founded (see Sec. 4.1 and Table 1). 2 In general, n{xB) = n{B)/x, where a; is a number. Thus, for example, if the amount of substance of H2SO4 is 5 mol, the amount of substance of (1/3)H2S04 is 15 mol: n[(l/3)H2S04] = 3n(H2S04). Example: The relative atomic mass of a fluorine atom is At(F) = 18.9984. The relative molecular mass of a fluorine molecule may therefore be taken as Mr(F2) = 2/4 r(F) = 37.9968. The molar mass of F2 is then A/(F2) = 37.9968 x 10"^ kg/mol = 37.9968 g/mol (see Sec. 8.6.4). The amount of substance of, for example, 100 g of F2 is then n (F2) = 100 g/(37.9968 g/mol) = 2.63 mol. 8.6.2 Mole fraction of B; amount-of-substance fraction of B Quantity symbol: xb (alsoys). SI unit: one (1) (amount-of-substance fraction is a quantity of dimension one). Definition : ratio of the amount of substance of B to the amount of substance of the mixture: Xb — Ws/n. 25 Guide for the Use of the International System of Units (SI) Table 12. Summary description of nine quantities that are quotients involving amount of substance, volume, or mass^") Amount of substance Symbol: n u 0 w SI unit: mol es 0 s 0 c Volume Symbol: V C tity i SI unit: m' C R Mass c Symbol: m SI unit: kg Quantity in numerator Amount of substance Symbol: n SI unit: mol Volume Symbol: V SI unit: m^ Mass Symbol: m SI unit: kg amount-of-substance fraction n SI unit: mol/mol = 1 molar volume SI unit: m-'/mol molar mass M = ^ n SI unit: kg /mol amount-of-substance concentration r CB- y SI unit: moi/m' volume fraction XBVm,B SI unit: mVm^ = 1 mass density m p = v SI unit: kg/m' molality Ob = — niA SI unit: mol/kg specific volume m SI unit: m^/kg mass fraction wjb Wb = — m SI unit: kg/kg = 1 Adapted from Canadian Metric Practice Guide (see Ref. , note 3; the book cited in Ref. , note 5, may also be consulted). Notes : 1 This quantity is commonly called "mole fraction of B" but this Guide prefers the name "amount-of-substance fraction of B" because it does not contain the name of the unit mole (compare kilogram fraction to mass fraction). 2 For a mixture composed of substances A, B, C, n = ha + + nc + ... = X "a. A 3 A related quantity is amount-of-substance ratio ofB (commonly called "mole ratio of solute B"), symbol rs. It is the ratio of the amount of substance of B to the amount of substance of the solvent substance: tb = ns/ns. For a single solute C in a solvent substance (a one-solute solution), rc = xc/(^ - Xc). This follows from the relations n = nc + ns, xc = nc/n, and rc = nc/ns, where the solvent substance S can itself be a mixture. 8.6.3 Molar volume Quantity symbol: Vm-SI unit: cubic meter per mole (m^/mol). Definition : volume of a substance divided by its amount of substance: Vm = VIn . Notes : 1 The word "molar" means "divided by amount of substance." 26 Guide for the Use of the International System of Units (SI) 2 For a mixture, this term is often called "mean molar volume." 3 The amagat should not be used to express molar volumes or reciprocal molar volumes. (One amagat is the molar volume Vm of a real gas atp = 101 325 Pa and T = 273.15 K and is approximately equal to 22.4 x 10"^ mVmol. The name "amagat" is also given to 1/Fm of a real gas atp = 101 325 Pa and T = 273.15 K and in this case is approximately equal to 44.6 mol/m^.) 8.6.4 Molar mass Quantity symbol: M. SI unit: kilogram per mole (kg/mol). Definition : mass of a substance divided by its amount of substance: M = m/n. Notes : 1 For a mixture, this term is often called "mean molar mass." 2 The molar mass of a substance B of definite chemical composition is given by M(B) = Mr(B) X 10-3 kg/mol ^ Mr(B) kg/kmol = g/mol, where A/r(B) is the relative molecular mass of B (see Sec. 8.4). The molar mass of an atom or nuclide X is M(X) = Ar(X) x IQ-^ kg/mol = ^^(X) kg/kmol = y4r(X) g/mol, where y4r(X) is the relative atomic mass of X (see Sec. 8.4). 8.6.5 Concentration of B; amount-of-substance concentration of B Quantity symbol: cb-SI unit: mole per cubic meter (mol/m^). Definition: amount of substance of B divided by the volume of the mixture: cb = ns/V. Notes : 1 This Guide prefers the name "amount-of-substance concentration of B" for this quantity because it is unambiguous. However, in practice, it is often shortened to amount concentration of B, or even simply to concentration of B. Unfortunately, this last form can cause confusion because there are several different "concentra-tions," for example, mass concentration of B, pb = m^lV; and molecular concen-tration of B, Cb = Nb/V, where A^b is the number of molecules of B. 2 The term normality and the symbol should no longer be used because they are obsolete. One should avoid writing, for example, "a 0.5 A'^ solution of H2SO4" and write instead "a solution having an amount-of-substance concentration of c[(l/2)H2S04]) = 0.5 mol/dm^" (or 0.5 kmol/m^ or 0.5 mol/L since 1 mol/dm^ = 1 kmol/m^ = 1 mol/L). 3 The term molarity and the symbol m should no longer be used because they, too, are obsolete. One should use instead amount-of-substance concentration of B and such units as mol/dm\ kmol/m^ or mol/L. (A solution of, for example, 0.1 mol/dm^ was often called a 0.1 molar solution, denoted 0.1 m solution. The mo-larity of the solution was said to be 0.1 M.) 8.6.6 Volume fraction of B Quantity symbol: (ps-SI unit: one (1) (volume fraction is a quantity of dimension one). Definition : for a mixture of substances A, B, C, . . . , ^B Vm,B \}np = (1/2) ln(F/Po); that is Lp = ln(F/Fo) Np Lp = (l/2)ln(F/Fo) Np . One bel (1 B) is the level of a field quantity when F/Fo = V^. that is, when 2 lg(F/Fo) = 1 (note that \gx - \ogwX — see Sec. 10.1.2). Equivalently, 1 B is the level of a power quantity when F/Fo = 10, that is, when lg(F/Fo) = 1. These definitions imply that the numerical value of Lf when Lf is expressed in the unit bel is {Lf }b = 2 \g(F/Fo) and that the numerical value of Lp when Lp is expressed in the unit bel is {L/>}b = lg(F/Fo); that is Lf = 2 lg(F/Fo) B = 20 lg(F/Fo) dB Lp = lg(F/Fo) B = 10 lg(F/Fo) dB . Since the value of Lf (or Lp ) is independent of the unit used to express that value, one may equate Lf in the above expressions to obtain ln(F/Fo)Np = 21g(F/Fo)B, which implies In 10 1 B = Np exactly « 1.151 293 Np 1 dB » 0.115 129 3 Np . When reporting values of Lf and Lp, one must always give the reference level. Accord-ing to Ref. [7: lEC 27-3], this may be done in one of two ways: L^(rexref) or Lx/x,^„ where X is the quantity symbol for the quantity whose level is being reported, for example, electric field strength E or sound pressure p; and jCref is the value of the reference quantity, for example, 1 piV/m for Eo, and 20 ^.Pa for po. Thus L£(re 1 |xV/m) = - 0.58 Np or Lf/d^v/m) = - 0.58 Np 29 Guide for the Use of the International System of Units (SI) means that the level of a certain electric field strength is 0.58 Np below the reference electric field strength Eo = 1 p-V/m. Similarly Lp (re 20 (xPa) = 25 dB or Lp/2o^Pa = 25 dB means that the level of a certain sound pressure is 25 dB above the reference pressure po = 20 n,Pa. Notes : 1 When such data are presented in a table or in a figure, the following condensed notation may be used instead: - 0.58 Np (1 |xV/m); 25 dB (20 |xPa). 2 When the same reference level applies repeatedly in a given context, it may be omitted if its value is clearly stated initially and if its planned omission is pointed out. 3 The rules of Ref. [7: lEC 27-3] preclude, for example, the use of the symbol dBm to indicate a reference level of power of 1 mW. This restriction is based on the rule of Sec. 7.4, which does not permit attachments to unit symbols. 8.8 Viscosity The proper SI units for expressing values of viscosity 77 (also called dynamic viscosity) and values of kinematic viscosity v are, respectively, the pascal second (Pa • s) and the meter squared per second (mVs) (and their decimal multiples and submultiples as appropriate). The CGS units commonly used to express values of these quantities, the poise (P) and the stokes (St), respectively [and their decimal submultiples the centipoise (cP) and the centistoke (cSt)], are not to be used; see Sec. 5.3.1 and Table 10, which gives the relations IP = 0.1 Pa -s and 1 St = lO'^mVs. 8.9 Massic, volumic, areic, lineic Reference [6: ISO 31-0] has introduced the new adjectives "massic," "volumic," "areic," and "lineic" into the English language based on their French counterparts: "mas-sique," "volumique," "surfacique," and "lineique." They are convenient and NIST authors may wish to use them. They are equivalent, respectively, to "specific," "density," "surface . . . density," and "linear . . . density," as explained below. (a) The adjective massic, or the adjective specific, is used to modify the name of a quantity to indicate the quotient of that quantity and its associated mass. Examples: massic volume or specific volume: v = Vim massic entropy or specific entropy: s = S/m (b) The adjective volumic is used to modify the name of a quantity, or the term density is added to it, to indicate the quotient of that quantity and its associated volume. Examples: volumic mass or (mass) density: p = m/V volumic number or number density: n = N/V Note : Parentheses around a word means that the word is often omitted. (c) The adjective areic is used to modify the name of a quantity, or the terms surface . . . density are added to it, to indicate the quotient of that quantity (a scalar) and its associ-ated surface area. Examples : areic mass or surface (mass) density: pA = m/A areic charge or surface charge density: a = Q/A 30 Guide for the Use of the International System of Units (SI) (d) The adjective lineic is used to modify the name of a quantity, or the terms linear . . . density are added to it, to indicate the quotient of that quantity and its associated length. Examples : Hneic mass or linear (mass) density: pi = mil lineic electric current or linear electric current density: A = I lb 9 Rules and Style Conventions for Spelling Unit Names The following eight sections give rules and style conventions related to spelling the names of units. 9.1 Capitalization When spelled out in full, unit names are treated like ordinary English nouns. Thus the names of all units start with a lower-case letter, except at the beginning of a sentence or in capitalized material such as a title. In keeping with this rule, the correct spelling of the name of the unit °C is "degree Celsius" (the unit "degree" begins with a lower case "d" and the modifier "Celsius" begins with an upper-case "C" because it is the name of a person). 9.2 Plurals Plural unit names are used when they are required by the rules of English grammar. They are normally formed regularly, for example, "henries" is the plural of henry. According to Ref. , the following plurals are irregular: Singular — lux, hertz, Siemens; Plural — lux, hertz, Siemens. (See also Sec. 9.7.) 9.3 Spelling unit names with prefixes When the name of a unit containing a prefix is spelled out, no space or hyphen is used between the prefix and unit name (see Sec. 6.2.3). Examples: milligram but not: milli-gram kilopascal but not: kilo-pascal Reference points out that there are three cases where the final vowel of an SI prefix is commonly omitted: megohm {not megaohm), kilohm {not kiloohm), and hectare {not hectoare). In all other cases where the unit name begins with a vowel, both the final vowel of the prefix and the vowel of the unit name are retained and both are pronounced. 9.4 Spelling unit names obtained by multiplication When the name of a derived unit formed from other units by multiplication is spelled out, a space, which is preferred by Ref. and this Guide , or a hyphen is used to separate the names of the individual units. Example: pascal second or pascal-second 9.5 Spelling unit names obtained by division When the name of a derived unit formed from other units by division is spelled out, the word "per" is used and not a solidus. (See also Sees. 6.1.7 and 9.8.) Example: ampere per meter (A/m) but not: ampere/meter 31 Guide for the Use of the International System of Units (SI) 9.6 Spelling unit names raised to powers When the names of units raised to powers are spelled out, modifiers such as "squared" or "cubed" are used and are placed after the unit name. Example: meter per second squared (m/s^) The modifiers "square" or "cubic" may, however, be placed before the unit name in the case of area or volume. Examples: square centimeter (cm^) cubic millimeter (mm^) ampere per square meter (A/m^) kilogram per cubic meter (kg/m^) 9.7 Other spelling conventions A derived unit is usually singular in English, for example, the value 3 m^ • K/W is usually spelled out as "three square meter kelvin per watt," and the value 3 C • m^/V is usually spelled out as "three coulomb meter squared per volt." However, a "single" unit may be plural; for example, the value 5 kPa is spelled out as "five kilopascals," although "five kilopascal" is acceptable. If in such a single-unit case the number is less than one, the unit is always singular when spelled out; for example, 0.5 kPa is spelled out as "five-tenths kilopascal." Note: These other spelling conventions are given for completeness; as indicated in Sec. 7.6, it is the position of this Guide that symbols for numbers and units should be used to express the values of quantities, not the spelled-out names of numbers and units. Reference also requires that a symbol for a number be used whenever the value of a quantity is expressed in terms of a unit of measurement. 9.8 Unacceptability of applying mathematical operations to unit names Because it could possibly lead to confusion, mathematical operations are not applied to unit names but only to unit symbols. (See also Sees. 6.1.7 and 9.5.) Example: joule per kilogram or J/kg or J -kg"' but not: joule/kilogram or joule • kilogram ~ 10 More on Printing and Using Symbols and Numbers in Scientific and Technical Documents By following the guidance given in this chapter, NIST authors can prepare manuscripts that are consistent with accepted typesetting practice. 10.1 Kinds of symbols Letter symbols are of three principal kinds: (a) symbols for quantities, (b) symbols for units, and (c) symbols for descriptive terms. Quantity symbols, which are always printed in italic (that is, sloping) type, are, with few exceptions, single letters of the Latin or Greek alphabets which may have subscripts or superscripts or other identifying signs. Symbols for units, in particular those for acceptable units, have been discussed in detail in earlier por-tions of this Guide . Symbols for descriptive terms include the symbols for the chemical el-ements, certain mathematical symbols, and modifying superscripts and subscripts on quantity symbols. ^ This chapter is adapted in part from Refs. , [6: ISO 31-0], and [6: ISO 31-11]. 32 Guide for the Use of the International System of Units (SI) 10.1.1 Standardized quantity symbols The use of words, acronyms, or other ad hoc groups of letters as quantity symbols should be avoided by NIST authors. For example, use the quantity symbol Zm for mechanical impedance, not MI. In fact, there are nationally and internationally accepted symbols for literally hundreds of quantities used in the physical sciences and technology. Many of these are given in Refs. and , and it is likely that symbols for the quantities used in most NIST publications can be found in these international standards or can readily be adapted from the symbols and principles given in these standards. Because of their international acceptance, NIST authors are urged to use the symbols of Refs. and to the fullest extent possible.' n (solid angle) Zm (mechanical impedance) Lp (level of a power quantity) 4r (relative mass excess) p (pressure) CTtot (total cross-section) Kt (isothermal compressibility) Eu (Euler number) E (electric field strength) Tn (Neel temperature) 10.1.2 Standardized mathematical signs and symbols As is the case for quantity symbols, most of the mathematical signs and symbols used in the physical sciences and technology are standardized. They may be found in Ref. [6: ISO 31-11] and should be used by NIST authors to the fullest possible extent.' Examples : a (conjunction sign, p /\ q meansp and q ) (a ^ b, a is not equal to b) ^ {a b, a is by definition equal to b) (a ~ b, a is approximately equal to b) ~ (a — b, a is proportional to b) arcsin x (arc sine ofx) logaX (logarithm to the base aofx) Ibx (\hx = \0g2x) \nx {\nx = \ogeX) \gx (\gx = logiox) 10.2 Typefaces for symbols Most word processing systems now in use at NIST are capable of producing lightface (that is, regular) or boldface letters of the Latin or Greek alphabets in both roman (upright) and italic (sloping) types. The understandability of NIST typed and typeset scientific and technical publications is facilitated if symbols are in the correct typeface. The typeface in which a symbol appears helps to define what the symbol represents. For example, irrespective of the typeface used in the surrounding text, "A" would be typed or typeset in — italic type for the scalar quantity area: A ; — roman type for the unit ampere: A; — italic boldface for the vector quantity vector potential: A . ^ In addition to Refs. and , quantity symbols may also be found in ANSI/IEEE Std 280-1985, IEEE Standard Letter Symbols for Quantities Used in Electrical Science and Electrical Engineering. Similarly, in addition to Ref. [6: ISO 31-11], mathematical signs and symbols are also given in ANSI/IEEE Std 260.3-1993, Mathematical Signs and Symbols for Use in Physical Sciences and Technology. (See Ref. , note 1.) 33 Guide for the Use of the International System of Units (SI) More specifically, the three major categories of symbols found in scientific and technical publications should be typed or typeset in either italic or roman type, as follows: — symbols for variables and quantities : italic; — symbols for units: roman; — symbols for descriptive terms : roman. These rules imply that a subscript or superscript on a quantity symbol is in roman type if it is descriptive (for example, if it is a number or represents the name of a person or a particle); but it is in italic type if it represents a quantity, or is a variable such as x in Ex or an index such as / in XiXi that represents a number (see Sees. 10.2.1, 10.2.3, and 10.2.4). An index that represents a number is also called a "running number" [6: ISO 31-0]. Notes 1 The above rules also imply, for example, that |x, the symbol for the SI prefix micro (10"), that H, the symbol for the SI derived unit ohm, and that F, the symbol for the SI derived unit farad, are in roman type; but they are in italic type if they represent quantities (/a, O, and F are the recommended symbols for the quantities magnetic moment of a particle, solid angle, and force, respectively). 2 The typeface for numbers is discussed in Sec. 10.5.1. The following four sections give examples of the proper typefaces for these three major categories. 10.2.1 Quantities — italic Symbols for quantities are italic, as are symbols for functions in general, for example, /(a:): r = 3 s t time, s second T = 22 K T temperature, K kelvin r = llcm r radius, cm centimeter A=633 nm A wavelength, nm nanometer Constants are usually physical quantities and thus their symbols are italic; however, in gen-eral, symbols used as subscripts and superscripts are roman if descriptive (see Sec. 10.2.3): Na Avogadro constant, A Avogadro R molar gas constant 6d Debye temperature, D Debye Z atomic number e elementary charge me m mass, e electron Running numbers and symbols for variables in mathematical equations are italic, as are sym-bols for parameters such as a and b that may be considered constant in a given context: m y = ^ XiZi x^ = ay^ + bz^ 1=1 Symbols for vectors are boldface italic, symbols for tensors are sans-serif bold italic, and symbols for matrices are italic: A'B=C (vectors) T (tensors) A = ^^^) (matrices) Symbols used as subscripts and superscripts are italic if they represent quantities: Cp p pressure qm m mass an O solid angle 10.2.2 Units — roman The symbols for units and SI prefixes are roman: m meter g gram L liter cm centimeter |xg microgram mL milliliter 34 1 I I Guide for the Use of the International System of Units (SI) 10.2.3 Descriptive terms — roman Symbols representing purely descriptive terms (for example, the chemical elements) are roman, as are symbols representing mathematical constants that never change (for example, it) and symbols representing explicitly defined functions or well defined operators (for exam-ple, r(x) or div): Chemical elements: Ar argon B boron C carbon Mathematical constants, functions, and operators: e base of natural logarithms X;c, X sum of expx exp exponential of \ogaX loga logarithm to the base a of dx/dt d 1st derivative of sinj: sin sine of Symbols used as subscripts and superscripts are roman if descriptive: Eo"' ir irrational Ek k kinetic Vm m molar, 1 liquid phase /xb B Bohr 10.2.4 Sample equations showing correct type qi qi F = F = ma <PB = :cbK',b/X;caF:.a = RT' d(ln k)/dT A E = mc^ pB = Ab lini(xB/?/AB) p-0 10.3 Greek alphabet in roman and italic type Table 13 shows the proper form, in both roman and lower-case letters of the Greek alphabet. Table 13. Greek alphabet in roman and italic type alpha A OL A a beta B B gamma r 7 r y delta A 5 A s epsilon E e, € E e, € zeta z Z eta H H V theta e e, ^^"^ 0 9, d"' iota I L I kappa K (a) K, y. K lambda A k A A mu M M A nu N V N V xi ^ omicron O o O o pi n IT, XH n 77, VJ rho p P. q"" p sigma 2 a X a tau T T T T upsilon Y V Y V phi <P chi X X X X psi omega ft n to ISO (see Ref. [6: ISO 31-0]) gives these two letters in the reverse order. pV = nRT c, = A-V[exp(c2/Ar)-l] ^= -grad V italic type, of the upper-case and 35 Guide for the Use of the International System of Units (SI) 10.4 Symbols for the elements The following two sections give the rules and style conventions for the symbols for the elements. 10.4.1 Typeface and punctuation for element symbols Symbols for the elements are normally printed in roman type without regard to the type used in the surrounding text (see Sec. 10.2.3). They are not followed by a period unless at the end of a sentence. 10.4.2 Subscripts and superscripts on element symbols The nucleon number (mass number) of a nuclide is indicated in the left superscript position: ^Si. The number of atoms in a molecule of a particular nuclide is shown in the right subscript position: 'H2. The proton number (atomic number) is indicated in the left subscript position: 29CU. The state of ionization or excitation is indicated in the right superscript position, some examples of which are as follows: State of ionization: Ba ^ Co(N02)6"-" or Co(N02)i-or [Co(N02)6]'" Electronic excited state: Ne, CO Nuclear excited state: '^N or '^N™ 10.5 Printing numbers The following three sections give rules and style conventions related to the printing of numbers. 10.5.1 Typeface for numbers Arabic numerals expressing the numerical values of quantities (see Sec. 7.6) are gener-ally printed in lightface (that is, regular) roman type irrespective of the type used for the surrounding text. Arabic numerals other than numerical values of quantities may be printed in lightface or bold italics, or in bold roman type, but lightface roman type is usually pre-ferred. 10.5.2 Decimal sign or marker The recommended decimal sign or marker for use in the United States is the dot on the line [4, 8]. For numbers less than one, a zero is written before the decimal marker. For ex-ample, 0.25 s is the correct form, not .25 s. 36 Guide for the Use of the International System of Units (SI) 10.5.3 Grouping digits Because the comma is widely used as the decimal marker outside the United States, it should not be used to separate digits into groups of three. Instead, digits should be separated into groups of three, counting from the decimal marker towards the left and right, by the use of a thin, fixed space. However, this practice is not usually followed for numbers having only four digits on either side of the decimal marker except when uniformity in a table is desired. Examples: 76 483 522 but not: 76,483,522 43 279.168 29 but not: 43,279.168 29 8012 or 8 012 but not: 8,012 0.491722 3 is highly preferred to: QA9\llli 0.5947 or 0.594 7 but not: 0.59 47 8012.5947 or 8 012.594 7 but not: 8 012.5947 or 8012.594 7 Note : The practice of using a space to group digits is not usually followed in certain special-ized applications, such as engineering drawings and financial statements. 10.5.4 Multiplying numbers When the dot is used as the decimal marker as in the United States, the preferred sign for the multiplication of numbers or values of quantities is a cross (that is, multiplication sign) ( X ), not a half-high (that is, centered) dot ( • ). Examples: 25 x 60.5 but not: 25 - 60.5 53m/s X 10.2 s but not: 53 m/s- 10.2 s 15 X 72 kg but not: 15 • 72 kg Notes : When the comma is used as the decimal marker, the preferred sign for the multiplication of numbers is the half-high dot. However, even when the comma is so used, this Guide prefers the cross for the multiplication of values of quanti-ties. 2 The multiplication of quantity symbols (or numbers in parentheses or values of quantities in parentheses) may be indicated in one of the following ways: ab, a b, O'b, a X b. 37 Guide for the Use of the International System of Units (SI) 11 Check List for Reviewing Manuscripts The following check list, adapted from Ref. , is intended to help NIST authors re-view the conformity of their manuscripts with proper SI usage and the basic principles con-cerning quantities and units. For easy reference, this check list also appears immediately after the Preface. (1) CH Only units of the SI and those units recognized for use with the SI are used to express the values of quantities. Equivalent values in other units are given in parentheses following values in acceptable units only when deemed necessary for the intended audience. (See Chapter 2.) (2) O Abbreviations such as sec (for either s or second), cc (for either cm^ or cubic centimeter), or mps (for either m/s or meter per second), are avoided and only standard unit symbols, SI prefbc symbols, unit names, and SI prefbces are used. (See Sec. 6.1.8.) (3) The combinations of letters "ppm," "ppb," and "ppt," and the terms part per mil-lion, part per billion, and part per trillion, and the like, are not used to express the values of quantities. The following forms, for example, are used instead: 2.0 |xL/L or 2.0 X 10~ V, 4.3 nm/m or 4.3 x 10"' /, 7 ps/s or 7 x 10"'^ t, where V, I, and t are, respectively, the quantity symbols for volume, length, and time. (See Sec. 7.10.3.) (4) n Unit symbols (or names) are not modified by the addition of subscripts or other information. The following forms, for example, are used instead. (See Sees. 7.4 and 7.10.2.) K„ax = 1000 V but not: V = 1000 V„ax a mass fraction of 10 % but not: 10 % (m/m) or 10 % (by weight) (5) EH Statements such as "the length /i exceeds the length 1 2 by 0.2 %" are avoided be-cause it is recognized that the symbol % represents simply the number 0.01. In-stead, forms such as "/i = hil + 0.2 %)" or "A = 0.2 %" are used, where A is defined by the relation A = (li - Iz)/^- (See Sec. 7.10.2.) (6) Information is not mixed with unit symbols (or names). For example, the form "the water content is 20 mL/kg" is used and not "20 mL H20/kg" or "20 mL of water/kg." (See Sec. 7.5.) (7) n It is clear to which unit symbol a numerical value belongs and which mathematical operation applies to the value of a quantity because forms such as the following are used. (See Sec. 7.7.) 35 cm X 48 cm but not: 35 x 48 cm 1 MHz to 10 MHz or (1 to 10) MHz but not : 1 MHz - 10 MHz or 1 to 10 MHz 20 °C to 30 °C or (20 to 30) °C but not : 20 °C - 30 "C or 20 to 30 °C 123g ± 2g or (123 ± 2)g but not: 123 ± 2g 70 % ± 5 % or (70 ± 5) % but not: 70 ± 5 % 240 X (1 ± 10%)V but not: 240 V ±10% (one cannot add 240 V and 10 %) (8) O Unit symbols and unit names are not mixed and mathematical operations are not applied to unit names. For example, only forms such as kg/m^ kg-m"-', or kilogram per cubic meter are used and not forms such as kilogram/m^ kg/cubic meter, kilogram/cubic meter, kg per m\ or kilogram per meter^ (See Sees. 6.1.7, 9.5, and 9.8.) 38 Guide for the Use of the International System of Units (SI) (9) D Values of quantities are expressed in acceptable units using Arabic numerals and the symbols for the units. (See Sec. 7.6.) m = 5 kg but not : m = five kilograms or m = five kg the current was 15 A but not : the current was 15 amperes. (10) [H There is a space between the numerical value and unit symbol, even when the value is used in an adjectival sense, except in the case of superscript units for plane angle. (See Sec. 7.2.) a 25 kg sphere but not : a 25-kg sphere an angle of 2°3'4" but not: an angle of 2 °3 '4 If the spelled-out name of a unit is used, the normal rules of English are applied: "a roll of 35-millimeter film." (See Sec. 7.6, note 3.) (11) CH The digits of numerical values having more than four digits on either side of the decimal marker are separated into groups of three using a thin, fixed space counting from both the left and right of the decimal marker. For example, 15 739.012 53 is highly preferred to 15739.01253. Commas are not used to separate digits into groups of three. (See Sec. 10.5.3.) (12) mi Equations between quantities are used in preference to equations between nu-merical values, and symbols representing numerical values are different from sym-bols representing the corresponding quantities. When a numerical-value equation is used, it is properly written and the corresponding quantity equation is given where possible. (See Sec. 7.11.) (13) n Standardized quantity symbols such as those given in Refs. and are used, for example, R for resistance and Ar for relative atomic mass, and not words, acronyms, or ad hoc groups of letters. Similarly, standardized mathematical signs and symbols such as are given in Ref. [6: ISO 31-11] are used, for example, "tan;c" and not "tgx." More specifically, the base of "log" in equations is speci-fied when required by writing logo a: (meaning log to the base a ofx), \hx (meaning \0g2x), \nx (meaning loge^c), or \gx (meaning logiooc). (See Sees. 10.1.1 and 10.1.2.) (14) Unit symbols are in roman type, and quantity symbols are in italic type with super-scripts and subscripts in roman or italic type as appropriate. (See Sec. 10.2 and Sees. 10.2.1 to 10.2.4.) (15) CD When the word "weight" is used, the intended meaning is clear. (In science and technology, weight is a force, for which the SI unit is the newton; in commerce and everyday use, weight is usually a synonym for mass, for which the SI unit is the kilogram.) (See Sec. 8.3.) (16) [H A quotient quantity, for example, mass density, is written "mass divided by vol-ume" rather than "mass per unit volume." (See Sec. 7.12.) (17) O An object and any quantity describing the object are distinguished. (Note the dif-ference between "surface" and "area," "body" and "mass," "resistor" and "resis-tance," "coil" and "inductance.") (See Sec. 7.13.) (18) n The obsolete term normality and the symbol A^, and the obsolete term molarity and the symbol m, are not used, but the quantity amount-of-substance concentra-tion of B (more commonly called concentration of B), and its symbol cb and SI unit mol/m' (or a related acceptable unit), are used instead. Similarly, the obsolete term molal and the symbol m are not used, but the quantity molality of solute B, and its symbol foe or nis and SI unit mol/kg (or a related unit of the SI), are used instead. (See Sees. 8.6.5 and 8.6.8.) 39 Guide for the Use of the International System of Units (SI) Appendix A. Definitions of tlie SI Base Units and tlie Radian and Steradian A.1 Introduction The following definitions of the SI base units are taken from Ref. ; the definitions of the SI supplementary units, the radian and steradian, which are now interpreted as SI derived units (see Sec. 4.3), are those generally accepted and are the same as those given in Ref. . It should be noted that SI derived units are uniquely defined only in terms of SI base units; for example, 1 V = 1 m^ • kg • s~^ • A"'. A.2 Meter (17th CGPM, 1983) The meter is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second. A-3 Kilogram (3d CGPM, 1901) The kilogram is the unit of mass; it is equal to the mass of the international prototype of the kilogram. A4 Second (13th CGPM, 1967) The second is the duration o/ 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium- 133 atom. A.5 Ampere (9th CGPM, 1948) The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2 x 10 "^ newton per meter of length. A.6 Kelvin (13th CGPM, 1967) The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16 of the thermody-namic temperature of the triple point of water. A7 Mole (14th CGPM, 1971) 1. The mole is the amount of substance of a system which contains as many elementary en-tities as there are atoms in 0.012 kilogram of carbon 12 2. When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other particles, or specified groups of such particles. In the definition of the mole, it is understood that unbound atoms of carbon 12, at rest and in their ground state, are referred to. Note that this definition specifies at the same time the nature of the quantity whose unit is the mole. A.8 Candela (16th CGPM, 1979) The candela is the luminous intensity, in a given direction, of a source that emits monochro-matic radiation offrequency 540 x 10^^ hertz and that has a radiant intensity in that direction of ( l/683j watt per steradian. A.9 Radian The radian is the plane angle between two radii of a circle that cut off on the circumference an arc equal in length to the radius. A.10 Steradian The steradian is the solid angle that, having its vertex in the center of a sphere, cuts off an area of the surface of the sphere equal to that of a square with sides of length equal to the radius of the sphere. 40 Guide for the Use of the International System of Units (SI) Appendix B. Conversion Factors B.l Introduction Sections B.8 and B.9 give factors for converting values of quantities expressed in various units — predominantly units outside the SI that are unacceptable for use with it — to values expressed either in (a) SI units, (b) units that are accepted for use with the SI (especially units that better reflect the nature of the unconverted units), (c) units formed from such accepted units and SI units, or (d) decimal multiples or submultiples of the units of (a) to (c) that yield numerical values of convenient magnitudes. An example of (d) is the following: the values of quantities expressed in Sngstroms, such as the wavelengths of visible laser radiations, are usually converted to values expressed in nanometers, not meters. More generally, if desired, one can eliminate powers of 10 that appear in converted values as a result of using the conversion factors (or simply factors for brevity) of Sees. B.8 and B.9 by selecting an appropriate SI prefix (see Sec. B.3). B.2 Notation The factors given in Sees. B.8 and B.9 are written as a number equal to or greater than 1 and less than 10, with 6 or fewer decimal places. The number is followed by the letter E, which stands for exponent, a plus ( + ) or minus ( - ) sign, and two digits which indicate the power of 10 by which the number is multiplied. Examples : 3.523 907 E - 02 means 3.523 907 x 10 = 0.035 239 07 3.386 389 E + 03 means 3.386 389 x 10^ = 3386.389 A factor in boldface is exact. All other factors have been rounded to the significant digits given in accordance with accepted practice (see Sees. 7.9, B.7.2, and Refs. [6: ISO 31-0] and ). Where less than six digits after the decimal place are given, the unit does not war-rant a greater number of digits in its conversion. However, for the convenience of the user, this practice is not followed for all such units, including the cord, cup, quad, and teaspoon. B.3 Use of conversion factors Each entry in Sees. B.8 and B.9 is to be interpreted as in these two examples: To convert from to Multiply by atmosphere, standard (atm) pascal (Pa) 1.013 25 E+05 cubic fcx)t per second (ft'/s) cubic meter per second (m'/s) 2.831 685 E-02 means 1 atm = 101 325 Pa (exactly); 1 ftVs = 0.028 316 85 mVs. Thus to express, for example, the pressure p (Pa), write p = 11.8 atm X 101 325Pa/atm = 11.8 standard atmospheres (atm) in pascals and obtain the converted numerical value 11.8 x 101 325 = 1 195 635 and the converted value p = 1.20 MPa. Appendix B is a significantly revised version of Appendix C of the 1991 edition of this NIST Special Publication (see Preface). Appendix C of the 1991 edition was reprinted from ANSI/IEEE Std 268-1982, American National Standard Metric Practice, ®1982 by the Institute of Electrical and Electronics Engineers, Inc., with the permission of the IEEE. The origin of this material is E. A. Mechtly, TTie International System of Units — Physical Constants and Conversion Factors, NASA SP-7012, Second Revision, National Aeronautics and Space Administration (U.S. Government Printing Office, Washington, DC, 1973). 41 Guide for the Use of the International System of Units (SI) Notes : 1 Guidance on rounding converted numerical values of quantities is given in Sec. B.7.2. 2 If the value of a quantity is expressed in a unit of the center column of Sec. B.8 or B.9 and it is necessary to express it in the corresponding unit of the first column, divide by the factor. The factors for derived units not included in Sees. B.8 and B.9 can readily be found from the factors given. Examples : To find the factor for converting values in lb • ft/s to values in kg • m/s, obtain from Sec. B.8 or B.9 lib = 4.535 924 E- 01 kg 1 ft = 3.048 E-01 m and substitute these values into the unit lb • ft/s to obtain 1 lb • ft/s = 0.453 592 4 kg x 0.3048 m/s = 0.138 255 0kg -m/s and the factor is 1.382 550 E-01. To find the factor for converting values in (avoirdupois) oz • in^ to values in kg • m^, obtain from Sec. B.8 or B.9 loz = 2.834 952 E- 02 kg 1 in^ = 6.4516 E-04 m^ and substitute these values into the unit oz • in^ to obtain 1 oz • in^ = 0.028 349 52 kg x 0.000 645 16 m^ = 0.000 018 289 98 kg • m' and the factor is 1.828 998 E- 05. B.4 Organization of entries and style In Sec. B.8 the units for which factors are given are listed alphabetically, while in Sec B.9 the same units are listed alphabetically within the following alphabetized list of kinds of quantities and fields of science: ACCELERATION ANGLE AREA AND SECOND MOMENT OF AREA CAPACITY (see VOLUME) DENSITY (that is, MASS DENSITY -(see MASS DIVIDED BY VOLUME) ELECTRICITY and MAGNETISM ENERGY (includes WORK) ENERGY DIVIDED BY AREA TIME FLOW (see MASS DIVIDED BY TIME or VOLUME DIVIDED BY TIME) FORCE FORCE DIVIDED BY AREA (see PRESSURE) FORCE DIVIDED BY LENGTH HEAT Available Energy Coefficient of Heat Transfer Density of Heat Density of Heat Flow Rate Fuel Consumption Heat Capacity and Entropy Heat Flow Rate Specific Heat Capacity and Specific Entropy Thermal Conductivity Thermal Diffusivity Thermal Insulance Thermal Resistance Thermal Resistivity 42 Guide for the Use of the International System of Units (SI) LENGTH LIGHT MASS and MOMENT OF INERTIA MASS DENSITY (see MASS DIVIDED BY VOLUME) MASS DIVIDED BY AREA MASS DIVIDED BY CAPACITY (see MASS DIVIDED BY VOLUME) MASS DIVIDED BY LENGTH MASS DIVIDED BY TIME (includes FLOW) MASS DIVIDED BY VOLUME (includes MASS DENSITY and MASS CONCENTRATION) MOMENT OF FORCE or TORQUE MOMENT OF FORCE or TORQUE, DIVIDED BY LENGTH PERMEABILITY POWER PRESSURE or STRESS (FORCE DIVIDED BY AREA) RADIOLOGY SPEED (see VELOCITY) STRESS (see PRESSURE) TEMPERATURE TEMPERATURE INTERVAL TIME TORQUE (see MOMENT OF FORCE) VELOCITY (includes SPEED) VISCOSITY, DYNAMIC VISCOSITY, KINEMATIC VOLUME (includes CAPACITY) VOLUME DIVIDED BY TIME (includes FLOW) WORK (see ENERGY) In Sees. B.8 and B.9, the units in the left-hand columns are written as they are often used customarily; the rules and style conventions recommended in this Guide are not neces-sarily observed. Further, many are obsolete and some are not consistent with good technical practice. The corresponding units in the center columns are, however, written in accordance with the rules and style conventions recommended in this Guide. B.5 Factor for converting motor vehicle efficiency The efficiency of motor vehicles in the United States is commonly expressed in miles per U.S. gallon, while in most other countries it is expressed in liters per one hundred kilome-ters. To convert fuel economy stated in miles per U.S. gallon to fuel consumption expressed in L/(100 km), divide 235.215 by the numerical value of the stated fuel economy. Thus 24 miles per gallon corresponds to 9.8 L/(100 km). B.6 U.S. survey foot and mile The U.S. Metric Law of 1866 gave the relationship 1 m = 39.37 in (in is the unit symbol for the inch). From 1893 until 1959, the yard was defined as being exactly equal to (3600/ 3937) m, and thus the foot was defined as being exactly equal to (1200/3937) m. In 1959 the definition of the yard was changed to bring the U.S. yard and the yard used in other countries into agreement. Since then the yard has been defined as exactly equal to 0.9144 m, and thus the foot has been defined as exactly equal to 0.3048 m. At the same time it was decided that any data expressed in feet derived from geodetic surveys within the United States would continue to bear the relationship as defined in 1893, namely, 1 ft = (1200/3937) m (ft is the unit symbol for the foot). The name of this foot is "U.S. survey foot," while the name of the new foot defined in 1959 is "international foot." The two are related to each other through the expression 1 international foot = 0.999 998 U.S. survey foot exactly. 43 Guide for the Use of the International System of Units (SI) In Sees. B.8 and B.9, the factors given are based on the international foot unless other-wise indicated. Users of this Guide may also find the following summary of exact relation-ships helpful, where for convenience the symbols ft and mi, that is, ft and mi in italic type, indicate that it is the U.S. survey foot or U.S. survey mile that is meant rather than the inter-national foot (ft) or international mile (mi), and where rd is the unit symbol for the rod and fur is the unit symbol for the furlong. 1ft = (1200/3937) m 1 ft = 0.3048 m 1 ft = 0.999 998 ft 1 rd, pole, or perch = 16jft 40 rd = 1 fur = 660 ft 8 fur = 1 U.S. survey mile (also called "statute mile") = 1 mi = 5280 ft 1 fathom = 6ft 1 international mile = 1 mi = 5280 ft 2721/4yf' = Ird' 160 rd^ = 1 acre = 43 560 640 acre = Imi^ B.7 Rules for rounding numbers and converted numerical values of quantities Rules for rounding numbers are discussed in Refs. [6: ISO 31-0] and ; the latter ref-erence also gives rules for rounding the converted numerical values of quantities whose val-ues expressed in units that are not accepted for use with the SI (primarily customary or inch-pound units) are converted to values expressed in acceptable units. This Guide gives the principal rules for rounding numbers in Sec. B.7.1, and the basic principle for rounding con-verted numerical values of quantities in Sec. B.7.2. The cited references should be consulted for additional details. B.7.1 Rounding numbers To replace a number having a given number of digits with a number (called the rounded number) having a smaller number of digits, one may follow these rules: (1) If the digits to be discarded begin with a digit less than 5, the digit preceding the 5 is not changed. Example : 6.974 951 5 rounded to 3 digits is 6.97 (2) If the digits to be discarded begin with a 5 and at least one of the following digits is greater than 0, the digit preceding the 5 is increased by 1. Examples : 6.91A 951 5 rounded to 2 digits is 7.0 6.974 951 5 rounded to 5 digits is 6.9750 (3) If the digits to be discarded begin with a 5 and all of the following digits are 0, the digit preceding the 5 is unchanged if it is even and increased by 1 if it is odd. (Note that this means that the final digit is always even.) Examples: 6.91A 951 5 rounded to 7 digits is 6.974 952 6.974 950 5 rounded to 7 digits is 6.974 950 44 Guide for the Use of the International System of Units (SI) B.7.2 Rounding converted numerical values of quantities The use of the factors given in Sees. B.8 and B.9 to convert values of quantities was demonstrated in Sec. B.3. In most cases the product of the unconverted numerical value and the factor will be a numerical value with a number of digits that exceeds the number of significant digits (see Sec. 7.9) of the unconverted numerical value. Proper conversion pro-cedure requires rounding this converted numerical value to the number of significant digits that is consistent with the maximum possible rounding error of the unconverted numerical value. Example : To express the value / = 36 ft in meters, use the factor 3.048 E—01 from Sec. B.8 or Sec. B.9 and write / = 36 ft X 0.3048 m/ft = 10.9728 m = 11.0 m. The final result, / = 11.0 m, is based on the following reasoning: The numerical value "36" has two significant digits, and thus a relative maximum possible rounding error (abbreviated RE in this Guide for simplicity) of ± 0.5/36 = ± 1.4 % because it could have resulted from rounding the number 35.5, 36.5, or any number between 35.5 and 36.5. To be consistent with this RE, the converted numerical value "10.9728" is rounded to 11.0 or three significant digits because the number 11.0 has an RE of ± 0.05/11.0 = ± 0.45 %. Although this ± 0.45 % RE is three times smaller than the ± 1.4 % RE of the unconverted numerical value "36," if the converted numerical value "10.9728" had been rounded to 11 or two sig-nificant digits, information contained in the unconverted numerical value "36" would have been lost. This is because the RE of the numerical value "11" is ± 0.5/11 = ± 4.5 %, which is three times larger than the ± 1.4 % RE of the unconverted numerical value "36." This example therefore shows that when selecting the number of digits to retain in the numerical value of a converted quantity, one must often choose between discarding information or providing unwarranted information. Consideration of the end use of the converted value can often help one decide which choice to make. Note: Consider that one had been told initially that the value / = 36 ft had been rounded to the nearest inch. Then in this case, since / is known to within ± 1 in, the RE of the numerical value "36" is ± 1 in/(36 ft x 12 in/ft) = ± 0.23 %. Although this is less than the ± 0.45 % RE of the number 11.0, it is comparable to it. Therefore, the result / = 11.0 m is still given as the converted value. (Note that the numerical value "10.97" would give excessive unwarranted information because it has an RE that is 5 times smaller than ± 0.23 %.) 45 Guide for the Use of the International System of Units (SI) B.8 Factors for units listed alphabetically Caution: The units listed in column 1 are in general not to be used in NIST publications, with the exception of those few in italic type. Factors in boldface are exact To convert from to Multiply by abampere ampere (A) 1.0 E+01 abcoulomb coulomb (C) 1.0 E+01 abfarad farad (F) 1.0 E+09 abhenry henry (H) 1,0 E-09 abmho Siemens (S) 1.0 E+09 abohm ohm (O) 1.0 E-09 abvolt volt (V) 1.0 E-08 acceleration of free fall, standard (gn) meter per second squared (m/s^) 9.806 65 E+00 acre (based on U.S. survey foot)' square meter (m^) 4.046 873 E + 03 acre foot (based on U.S. survey foot)' cubic meter (m^) 1.233 489 E + 03 ampere hour (A • h) coulomb (C) 3.6 E+03 Sngstrom (A) meter (m) 1.0 E— 10 Angstrom (A) nanometer (nm) 1.0 E—01 are (a) square meter (m^) 1.0 E+02 astronomical unit (AU) meter (m) 1.495 979 E + 11 atmosphere, standard (atm) pascal (Pa) 1.013 25 E+05 atmosphere, standard (atm) kilopascal (kPa) 1.013 25 E+02 atmosphere, technical (at)^° pascal (Pa) 9.806 65 E+04 atmosphere, technical (at)'° kilopascal (kPa) 9.806 65 E+01 bar (bar) pascal (Pa) 1.0 E+05 bar (bar) kilopascal (kPa) 1.0 E+02 barn(b) square meter (m^) 1.0 E-28 barrel for petroleum, 42 gallons (U.S.). . .cubic meter (m^) 1.589 873 E-01 barrel for petroleum, 42 gallons (U.S.). . .liter (L) 1.589 873 E + 02 biot(Bi) ampere (A) 1.0 E+01 British thermal unitir (Bturr)^^ joule (J) 1.055 056 E + 03 British thermal unit.h (Btu.h)^^ joule (J) 1.054 350 E + 03 British thermal unit (mean) (Btu) joule (J) 1.055 87 E + 03 British thermal unit (39 °F) (Btu) joule (J) 1.059 67 E + 03 British thermal unit (59 °F) (Btu) joule (J) 1.054 80 E + 03 British thermal unit (60 °F) (Btu) joule (J) 1.054 68 E + 03 British thermal unitir foot per hour square foot degree Fahrenheit [BtUiT • ft/(h • ft^ • °F)] watt per meter kelvin [W/(m • K)] 1.730 735 E + 00 British thermal unit,h foot per hour square foot degree Fahrenheit [Btu.h • ft/(h • ft^ • °F)] watt per meter kelvin [W/(m • K)] 1.729 577 E + 00 British thermal unitir inch per hour square foot degree Fahrenheit [Btuir • in/(h • ft^ • °F)] watt per meter kelvin [W/(m • K)] 1.442 279 E-01 British thermal unit,h inch per hour square foot degree Fahrenheit [Btu.h • in/(h • ft^ • °F)] watt per meter kelvin [W/(m • K)] 1.441 314 E-01 British thermal unitir inch per second square foot degree Fahrenheit [Btu,T-in/(s-ft2-°F)] watt per meter kelvin [W/(m-K)] 5.192 204 E + 02 ' See Sec. B.6. " One technical atmosphere equals one kilogram-force per square centimeter (1 at = 1 kgf/cm^). " The Fifth International Conference on the Properties of Steam (London, July 1956) defined the International Table calorie as 4.1868 J. Therefore the exact conversion factor for the International Table Btu is 1.055 055 852 62 kJ. Note that the notation for International Table used in this listing is subscript "IT". Similarily, the notation for thermochemical is subscript "th." Further, the thermochemical Btu, Btu,h, is based on the thermochemical calorie, cal,h, where calih = 4.184 J exactly. 46 Guide for the Use of the International System of Units (SI) To convert from to Multiply by British thermal unittb inch per second square foot degree Fahrenheit [Btu.h • in/(s • ft^ • "F)] watt per meter keivin [W/(m • K)] 5.188 732 E + 02 British thermal unitix per cubic foot (Btuix/ft^) joule per cubic meter (J/m^) 3.725 895 E + 04 British thermal unitth per cubic foot (Btu,h/ft^) joule per cubic meter (J/m^) 3.723 403 E + 04 British thermal unitir per degree Fahrenheit (Bturr/°F) joule per keivin (J/k) 1.899 101 E + 03 British thermal unit,h per degree Fahrenheit (Btu,h/°F) joule per keivin (J/k) 1.897 830 E + 03 British thermal unitir per degree Rankine (BtuiT/°R) joule per keivin (J/k) 1.899 101 E + 03 British thermal unitih per degree Rankine (Btu,h/°R) joule per keivin (J/k) 1.897 830 E + 03 British thermal unitix per hour (Btuix/h) watt (W) 2.930 711 E-01 British thermal unit,h per hour (Btu,h/h) watt (W) 2.928 751 E-01 British thermal unitir per hour square foot degree Fahrenheit [BtUiT/(h • ft^ • °F)] watt per square meter keivin [W/(m2 • K)] 5.678 263 E + 00 British thermal unit,h per hour square foot degree Fahrenheit [Btu,h/(h • ft^ • °F)] watt per square meter keivin [W/(m2 • K)] 5.674 466 E + 00 British thermal unit.h per minute (Btu,h/min) ..watt (W) 1.757 250 E + 01 British thermal unitix per pound (Btuix/lb) joule per kilogram (J/kg) 2326 E+03 British thermal unitih per pound (Btu,h/lb) joule per kilogram (J/kg) 2.324 444 E + 03 British thermal unitix per pound degree Fahrenheit [Btuix/(lb • °F)] joule per kilogram keivin (J/(kg • K)] 4.1868 E+03 British thermal unit,h per pound degree Fahrenheit [Btu,h/(lb • °F)] joule per kilogram keivin [J/(kg • K)] 4.184 E+03 British thermal unitix per pound degree Rankine [Btu,x/(lb • "R)] joule per kilogram keivin [J/(kg • K)] 4.1868 E+03 British thermal unit,h per pound degree Rankine [Btu,h/(lb • °R)] joule per kilogram keivin [J/(kg K)] 4.184 E+03 British thermal unitix per second (Btuir/s) watt (W) 1.055 056 E + 03 British thermal unit.h per second (Btu,h/s) watt (W) 1.054 350 E + 03 British thermal unitix per second square foot degree Fahrenheit [Btuix/(s • ft^ • °F)] watt per square meter keivin [W/(m2 • K)] 2.044 175 E + 04 British thermal unit.h per second square foot degree Fahrenheit [Btu,h/(s • ft^ • °F)] watt per square meter keivin [W/(m2 • K)] 2.042 808 E + 04 British thermal unitix per square foot (Btuix/ft^) joule per square meter (J/m^) 1.135 653 E + 04 British thermal unit,h per square foot (Btu,h/ft^) joule per square meter (J/m^) 1.134 893 E + 04 British thermal unitix per square foot hour [(Btuix/(ft^ • h)] watt per square meter (W/m^) 3.154 591 E + 00 British thermal unitth per square foot hour [Btu,h/(ft^ • h)] watt per square meter (W/m^) 3.152 481 E + 00 British thermal unit,h per square foot minute [Btu,h/(ft^ • min)] watt per square meter (W/m^) 1.891 489 E + 02 British thermal unitix per square foot second [(Btuix/(ft^ • s)] watt per square meter (W/m^) 1.135 653 E + 04 British thermal unitth per square foot second [Btuth/(ft^ • s)] watt per square meter (W/m^) 1.134 893 E + 04 British thermal unit,h per square inch second [BtUth/(in2 • s)] watt per square meter (W/m^) 1.634 246 E + 06 47 Guide for the Use of the International System of Units (SI) To convert from to Multiply by bushel (U.S.) (bu) cubic meter (m') 3.523 907 E - 02 bushel (U.S.) (bu) liter (L) 3.523 907 E + 01 calorieiT (calir)^^ joule (J) 4.1868 E+00 calorie,h (cal.h)^^ joule (J) 4.184 E+00 calorie (cal) (mean) joule (J) 4.190 02 E + 00 calorie (15 °C) (calis) joule (J) 4.185 80 E + 00 calorie (20 °C) (cabo) joule (J) 4.181 90 E + 00 calorieiT, kilogram (nutrition) joule (J) 4.1868 E+03 calorieib, kilogram (nutrition) joule (J) 4.184 E+03 calorie (mean), kilogram (nutrition)^^ joule (J) 4.190 02 E + 03 calorie,h per centimeter second degree Celsius [cal,h/(cm • s • °C)] watt per meter kelvin [W/(m • K)] 4.184 E+02 calorieiT per gram (caliy/g) joule per kilogram (J/kg) 4.1868 E+03 calorie,h per gram (caU/g) joule per kilogram (J/kg) 4.184 E+03 calorieiT per gram degree Celsius [caliT/(g • °C)] joule per kilogram kelvin [J/(kg • K)] 4.1868 E+03 caloricih per gram degree Celsius [cal,h/(g • °C)] joule per kilogram kelvin [J/(kg • K)] 4.184 E+03 calorieiT per gram kelvin [caliT/(g • K)] joule per kilogram kelvin [J/(kg K)] 4.1868 E+03 caloricth per gram kelvin [ca!,h/(g • K)] joule per kilogram kelvin [J/(kg • K)] 4.184 E+03 calorieih per minute (cal,h/min) watt (W) 6.973 333 E-02 calorie,h per second (cai,h/s) watt (W) 4.184 E+00 calorieih per square centimeter (calth/cm^) joule per square meter (J/m^) 4.184 E+04 calorieih per square centimeter minute [cal,h/(cm^ • min)] watt per square meter (W/m^) 6.973 333 E + 02 calorieih per square centimeter second [calih/(cm^ • s)] watt per square meter (W/m^) 4.184 E+04 candela per square inch (cd/in^) candela per square meter (cd/m^) 1.550 003 E + 03 carat, metric kilogram (kg) 2.0 E-04 carat, metric gram (g) 2.0 E—01 centimeter of mercury (0 °C) pascal (Pa) 1.333 22 E + 03 centimeter of mercury (0 °C) kilopascal (kPa) 1.333 22 E + 00 centimeter of mercury, conventional (cmHg)^^. . .pascal (Pa) 1.333 224 E + 03 centimeter of mercury, conventional (cmHg)^^. . .kilopascal (kPa) 1.333 224 E + 00 centimeter of water (4 °C) pascal (Pa) 9.806 38 E + 01 centimeter of water, conventional (cmH20)^-' . . .pascal (Pa) 9.806 65 E+01 centipoise (cP) pascal second (Pa • s) 1.0 E— 03 centistokes (cSt) meter squared per second (m^/s) 1.0 E—06 chain (based on U.S. survey foot) (ch)^ meter (m) 2.011 684 E + 01 circular mil square meter (m^) 5.067 075 E-10 circular mil square millimeter (mm^) 5.067 075 E — 04 clo square meter kelvin per watt (m^ • K/W). . . 1.55 E — 01 cord (128 ft^) cubic meter (m') 3.624 556 E + 00 cubic foot (ft') cubic meter (m') 2.831 685 E-02 cubic foot per minute (ft'/min) cubic meter per second (m'/s) 4.719 474 E-04 cubic foot per minute (ft'/min) liter per second (L/s) 4.719 474 E — 01 cubic foot per second (ft'/s) cubic meter per second (m'/s) 2.831 685 E-02 The kilogram calorie or "large calorie" is an obsolete term used for the kilocalorie, which is the calorie used to express the energy content of foods. However, in practice, the prefix "kilo" is usually omitted. Conversion factors for mercury manometer pressure units are calculated using the standard value for the acceleration of gravity and the density of mercury at the stated temperature. Additional digits are not justified because the definitions of the units do not take into account the compressibility of mercury or the change in density caused by the revised practical temperature scale, ITS-90. Similar comments also apply to water manometer pressure units. Conversion factors for conventional mercury and water manometer pressure units are based on Ref. [6: ISO 31-3]. 48 Guide for the Use of the International System of Units (SI) To convert from to Multiply by cubic inch (in') cubic meter (m^) 1.638 706 E-05 cubic inch per minute (in'/min) cubic meter per second (m'/s) 2.731 177 E-07 cubic mile (mi^) cubic meter (m^) 4.168 182 E + 09 cubic yard (yd') cubic meter (m') 7.645 549 E-01 cubic yard per minute (yd'/min) cubic meter per second (m'/s) 1.274 258 E — 02 cup (U.S.) cubic meter (m') 2.365 882 E-04 cup (U.S.) liter (L) 2.365 882 E-01 cup (U.S.) milliliter (mL) 2.365 882 E + 02 curie (Ci) becquerel (Bq) 3.7 E + 10 darcy^^ meter squared (m^) 9.869 233 E - 13 day (d) second (s) 8.64 E + 04 day (sidereal) second (s) 8.616 409 E + 04 debye (D) coulomb meter (C- m) 3.335 641 E-30 degree (angle) (") radian (rad) 1.745 329 E-02 degree Celsius (temperature) ("C) kelvin (K) T/K = t/°C + 273.15 degree Celsius (temperature interval) (°C) kelvin (K) '. 1.0 E+00 degree centigrade (temperature)^^ degree Celsius (°C) t fC ~ //deg. cent. degree centigrade (temperature interval)^'' degree Celsius (°C) 1.0 E + 00 degree Fahrenheit (temperature) (°F) degree Celsius (°C) //°C = (//°F - 32)/1.8 degree Fahrenheit (temperature) (°F) kelvin (K) T/K= {t/°F + 459.67)/ 1.8 degree Fahrenheit (temperature interval) (°F)... degree Celsius (°C) 5.555 556 E-01 degree Fahrenheit (temperature interval) (°F) . . .kelvin (K) 5.555 556 E-01 degree Fahrenheit hour per British thermal unitir (°F- h/Btu,T) kelvin per watt (K/W) 1.895 634 E + 00 degree Fahrenheit hour per British thermal unitih ("F- h/Btu,h) kelvin per watt (K/W) 1.896 903 E + 00 degree Fahrenheit hour square foot per British thermal unitrr ("F • h • ft^/BtUiT) square meter kelvin per watt (m^ • K/W) . . 1.761 102 E-01 degree Fahrenheit hour square foot per British thermal unit,h ("F • h • ft^/Btu,h) square meter kelvin per watt (m^- K/W).. 1.762280 E-01 degree Fahrenheit hour square foot per British thermal unitrr inch ["F - h • ftV(Btu,T • in)] meter kelvin per watt (m • K/W) 6.933 472 E + 00 degree Fahrenheit hour square foot per British thermal unitih inch ["F • h • ft^ / (Btu.h - in)] meter kelvin per watt (m • K/W) 6.938 112 E + 00 degree Fahrenheit second per British thermal unitrr ("F-s/BtUir) kelvin per watt (K/W) 5.265 651 E-04 degree Fahrenheit second per British thermal unit,h degree Rankine (°R) kelvin (K) , denier gram per me dyne (dyn) newton (N). dyne per square centimeter (dyn/cm^) 5.269 175 E-04 , r/K = (r/''R)/1.8 , . 5.555 556 E-01 1.111 111 E-07 1.111 111 E-04 1.0 E-05 1.0 E-07 1.0 E-01 eleclronvolt (eV) joule (J) 1.602 177 E- 19 EMU of capacitance (abfarad) farad (F) 1.0 E+09 EMU of current (abampere) ampere (A) 1.0 E+01 EMU of electric potential (abvolt) volt (V) 1.0 E-08 EMU of inductance (abhenry) henry (H) 1.0 E-09 The exact conversion factor is 1.638 706 4 E-05. The darcy is a unit for expressing the permeability of porous solids, not area. The centigrade temperature scale is obsolete; the degree centigrade is only approximately equal to the degree Celsius. 49 Guide for the Use of the International System of Units (SI) To convert from to Multiply by EMU of resistance (abohm) ohm (CI) 1.0 E-09 erg (erg) joule (J) 1.0 E-07 erg per second (erg/s) watt (W) 1.0 E-07 erg per square centimeter second [erg/(cm^ • s)] watt per square meter (W/m^) 1.0 E—03 ESU of capacitance (statfarad) farad (F) 1.112 650 E - 12 ESU of current (statampere) ampere (A) 3.335 641 E— 10 ESU of electric potential (statvolt) volt (V) 2.997 925 E + 02 ESU of inductance (stathenry) henry (H) 8.987 552 E + 1 1 ESU of resistance (statohm) ohm (fl) 8.987 552 E + 11 faraday (based on carbon 12) coulomb (C) 9.648 531 E + 04 fathom (based on U.S. survey foot)^ meter (m) 1.828 804 E + 00 fermi meter (m) 1.0 E-15 fermi femtometer (fm) 1.0 E+00 fluid ounce (U.S.) (fl oz) cubic meter (m') 2.957 353 E-05 fluid ounce (U.S.) (fl oz) milliliter (mL) 2.957 353 E + 01 foot (ft) meter (m) 3.048 E-01 foot (U.S. survey) (ft)' meter (m) 3.048 006 E-01 footcandle lux (be) 1.076 391 E + 01 footlambert candela per square meter (cd/m^) 3.426 259 E + 00 foot of mercury, conventional (ftHg)^^ pascal (Pa) 4.063 666 E + 04 foot of mercury, conventional (ftHg)'^ kilopascal (kPa) 4.063 666 E + 01 foot of water (39.2 °F) pascal (Pa) 2.988 98 E + 03 foot of water (39.2 °F) kilopascal (kPa) 2.988 98 E + 00 foot of water, conventional (ftH20)^^ pascal (Pa) 2.989 067 E + 03 foot of water, conventional (ftH20)^^ kilopascal (kPa) 2.989 067 E + 00 foot per hour (ft/h) meter per second (m/s) 8.466 667 E-05 foot per minute (ft/min) meter per second (m/s) 5.08 E-03 foot per second (ft/s) meter per second (m/s) 3.048 E— 01 foot per second squared (ft/s^) meter per second squared (m/s^) 3.048 E—01 foot poundal joule (J) 4.214 Oil E-02 foot pound force (ft • Ibf) joule (J) 1.355 818 E + 00 foot pound force per hour (ft • Ibf/h) watt (W) 3.766 161 E-04 foot pound force per minute (ft • Ibf/min) watt (W) 2.259 697 E-02 foot pound force per second (ft • Ibf/s) watt (W) 1.355 818 E + 00 foot to the fourth power (ff)^' meter to the fourth power (m'') 8.630 975 E-03 franklin (Fr) coulomb (C) 3.335 641 E - 10 gal (Gal) meter per second squared (m/s^) 1.0 E-02 gallon [Canadian and U.K (Imperial)] (gal) . . . .cubic meter (m^) 4.546 09 E-03 gallon [Canadian and U.K. (Imperial)] (gal) .... liter (L) 4.546 09 E+00 gallon (U.S.) (gal) cubic meter (m') 3.785 412 E-03 gallon (U.S.) (gal) liter (L) 3.785 412 E + 00 gallon (U.S.) per day (gal/d) cubic meter per second (m'/s) 4.381 264 E-08 gallon (U.S.) per day (gal/d) liter per second (L/s) 4.381 264 E-05 gallon (U.S.) per horsefxjwer hour [gal/(hp • h)] cubic meter per joule (mVJ) 1.410 089 E-09 gallon (U.S.) per horsepower hour [gal/(hp-h)] liter per joule (L/ J) 1.410089 E-06 gallon (U.S.) per minute (gpm)(gal/min) cubic meter per second (m'/s) 6.309 020 E-05 gallon (U.S.) per minute (gpm)(gal/min) liter per second (L/s) 6.309 020 E-02 This is a unit for the quantity second moment of area, which is sometimes called the "moment of section" or "area moment of inertia" of a plane section about a specified axis. 50 Guide for the Use of the International System of Units (SI) To convert from to Multiply by gamma (-y) tesla (T) 1.0 E-09 gauss (Gs, G) tesla (T) 1.0 E-04 gilbert (Gi) ampere (A) 7.957 747 E - 01 gill [Canadian and U.K. (Imperial)] (gi) cubic meter (m^) 1.420 653 E-04 gill [Canadian and U.K. (Imperial)] (gi) liter (L) 1.420 653 E - 01 gill (U.S.) (gi) cubic meter (m^) 1.182 941 E-04 gill (U.S.) (gi) liter (L) 1.182 941 E-01 gon(also called grade) (gon) radian (rad) 1.570 796 E-02 gon (also called grade) (gon) degree (angle) (°) 9.0 E-01 grain (gr) kilogram (kg) 6.479 891 E-05 grain (gr) milligram (mg) 6.479 891 E+01 grain per gallon (U.S.) (gr/gal) kilogram per cubic meter (kg/m^) 1.711 806 E-02 grain per gallon (U.S.) (gr/gal) milligram per liter (mg/L) 1.711 806 E + 01 gram-force per square centimeter (gf/cm^) pascal (Pa) 9.806 65 E+01 gram per cubic centimeter (g/cm^) kilogram per cubic meter (kg/m^) 1.0 E+03 hectare (ha) square meter (m^) 1.0 E+04 horsepower (550 ft • Ibf/s) (hp) watt (W) 7.456 999 E + 02 horsepower (boiler) watt (W) 9.809 50 E + 03 horsepower (electric) watt (W) 7.46 E+02 horsepower (metric) watt (W) 7.354 988 E + 02 horsepower (U.K.) watt (W) 7.4570 E + 02 horsepower (water) watt (W) 7.460 43 E + 02 hour (h) second (s) 3.6 E+03 hour (sidereal) second (s) 3.590 170 E + 03 hundredweight (long, 1121b) kilogram (kg) 5.080 235 E + 01 hundredweight (short, 100 lb) kilogram (kg) 4.535 924 E + 01 nch (in) meter (m) 2.54 E-02 nch (in) centimeter (cm) 2.54 E+00 nch of mercury (32 T) pascal (Pa) 3.386 38 E + 03 nch of mercury (32 °F) kilopascal (kPa) 3.386 38 E + 00 nch of mercury (60 °F) pascal (Pa) 3.376 85 E + 03 nch of mercury (60 "F) 1^ kilopascal (kPa) 3.376 85 E + 00 nch of mercury, conventional (inHg)^^ pascal (Pa) 3.386 389 E + 03 nch of mercury, conventional (inHg) kilopascal (kPa) 3.386 389 E + 00 nch of water (39.2 °F) " pascal (Pa) 2.490 82 E + 02 nch of water (60 °F)" pascal (Pa) 2.4884 E + 02 nch of water, conventional (inH20) pascal (Pa) 2.490 889 E + 02 nch per second (in/s) meter per second (m/s) 2.54 E— 02 nch per second squared (in/s^) meter per second squared (m/s^) 2.54 E— 02 nch to the fourth power (in'')'' meter to the fourth power (m") 4.162 314 E-07 kayser(K) reciprocal meter (m~') 1.0 E+02 kelvin (K) degree Celsius (°C) trC = T/K- 273.15 kilocalorierr (kcalix) joule (J) 4.1868 E+03 kiIocalorie,h (kcaU) joule (J) 4.184 E+03 kilocalorie (mean) (kcal) joule (J) 4.190 02 E + 03 kilocaloricth per minute (kcaU/min) watt (W) 6.973 333 E + 01 kilocalorie,h per second (kcaU/s) watt (W) 4.184 E+03 kilogram-force (kgf) newton (N) 9.806 65 E+00 kilogram-force meter (kgf • m) newton meter (N • m) 9.806 65 E+00 51 Guide for the Use of the International System of Units (SI) To convert from to Multiply by kilogram-force per square centimeter (kgf/cm^) pascal (Pa) 9.806 65 E+04 kilogram-force per square centimeter (kgf/cm^) kilopascal (kPa) 9.806 65 E-t-01 kilogram-force per square meter (kgf/m^) pascal (Pa) 9.806 65 E-HOO kilogram-force per square millimeter (kgf/mm^) pascal (Pa) 9.806 65 E-l-06 kilogram-force per square millimeter (kgf/mm^) megapascal (MPa) 9.806 65 E+00 kilogram-force second squared per meter (kgf • s^/m) kilogram (kg) 9.806 65 E-f-00 kilometer per hour (km/h) meter per second (m/s) 2.777 778 E-01 kilopond (kilogram-force) (kp) newton (N) 9.806 65 E-HOO kilowatt hour (kW • h) joule (J) 3.6 E-l-06 kilowatt hour (kW • h) megajoule (MJ) 3.6 E-t-00 kip (1 kip = 1000 Ibf) newton (N) 4.448 222 E -I- 03 kip (1 kip = 1000 Ibf) kilonewton (kN) 4.448 222 E -I- GO kip per square inch (ksi) (kip/in^) pascal (Pa) 6.894 757 E-l-06 kip per square inch (ksi) (kip/in^) kilopascal (kPa) 6.894 757 E + 03 knot (nautical mile per hour) meter per second (m/s) 5.144 444 E-01 lambert^ candela per square meter (cd/m^) 3.183 099 E-l-03 langley (calih/cm^) joule per square meter (J/m^) 4.184 E-l-04 light year (l.y.)l' meter (m) 9.460 73 E -t-15 liter (L)2° cubic meter (m^) 1.0 E-03 lumen per square foot (Im/ft^) lux (bf) 1.076 391 E-t-Ol maxwell (Mx) weber (Wb) 1.0 E-08 mho Siemens (S) 1.0 E-l-00 microinch meter (m) 2.54 E—08 microinch micrometer (|xm) 2.54 E-02 micron (p.) meter (m) 1.0 E—06 micron (|ji) micrometer (jim) 1.0 E+00 mil (0.001 in) meter (m) 2.54 E-05 mil (0.001 in) millimeter (mm) 2.54 E-02 mil (angle) radian (rad) 9.817 477 E - 04 mil (angle) degree (") 5.625 E-02 mile (mi) meter (m) 1.609 344 E-l-03 mile (mi) kilometer (km) 1.609 344 E-l-00 mile (based on U.S. survey foot) (mi)' meter (m) 1.609 347 E + 03 mile (based on U.S. survey foot) (mi)' kilometer (km) 1.609 347 E-l-00 mile, nautical^^ meter (m) 1.852 E+03 mile per gallon (U.S.) (mpg) (mi/gal) meter per cubic meter (m/m') 4.25 1 437 E + 05 mile per gallon (U.S.) (mpg) (mi/gal) kilometer per liter (km/L) 4.251 437 E-01 mile per gallon (U.S.) (mpg) (mi/gal)^^ liter per 100 kilometer (L/lOO km). . . divide 235.215 by number of miles per gallon mile per hour (mi/h) meter per second (m/s) 4.4704 E-01 mile per hour (mi/h) kilometer per hour (km/h) 1.609 344 E+00 The exact conversion factor is 10 /-n. This conversion factor is based on 1 d = 86 400 s; and 1 Julian century = 36 525 d. (See The Astronomical Almanac for the Year 1995, page K6, U.S. Government Printing Office, Washington, DC, 1994). ^" In 1964 the General Conference on Weights and Measures reestablished the name "liter" as a special name for the cubic deci-meter. Between 1901 and 1964 the liter was slightly larger (1.000 028 dm'); when one uses high-accuracy volume data of that time, this fact must be kept in mind. ^' The value of this unit, 1 nautical mile = 1852 m, was adopted by the First International Extraordinary Hydrographic Conference, Monaco, 1929, under the name "International nautical mile." 22 See Sec. B.5. 52 Guide for the Use of the International System of Units (SI) To convert from to Multiply by mile per minute (mi/min) meter per second (m/s) 2.682 24 E+01 mile per second (mi/s) meter per second (m/s) 1.609 344 E+03 millibar (mbar) pascal (Pa) 1.0 E+02 millibar (mbar) kilopascal (kPa) 1.0 E-01 millimeter of mercuiy, conventional (mmHg)^^. . .pascal (Pa) 1.333 224 E + 02 millimeter of water, conventional (mmH20)'^ . . .pascal (Pa) 9.806 65 E+00 minute (angle) (') radian (rad) 2.908 882 E - 04 minute (min) second (s) 6.0 E+01 minute (sidereal) second (s) 5.983 617 E + 01 oersted (Oe) ampere per meter (A/m) 7.957 747 E + 01 ohm centimeter (fl • cm) ohm meter (ft m) 1.0 E-02 ohm circular-mil per foot ohm meter (ft • m) 1.662 426 E - 09 ohm circular-mil per foot ohm square millimeter per meter (ft-mmVm) 1.662 426 E-03 ounce (avoirdupois) (oz) kilogram (kg) 2.834 952 E-02 ounce (avoirdupois) (oz) gram (g) 2.834 952 E + 01 ounce (troy or apothecary) (oz) kilogram (kg) 3.110 348 E-02 ounce (troy or apothecary) (oz) gram (g) 3.110 348 E + 01 ounce [Canadian and U.K. fluid (Imperial)] (fl oz) cubic meter (m') 2.84 1 306 E - 05 ounce [Canadian and U.K. fluid (Imperial)] (fl oz) milliliter (mL) 2.841 306 E + 01 ounce (U.S. fluid) (fl oz) cubic meter (m^) 2.957 353 E-05 ounce (U.S. fluid) (fl oz) millimeter (mL) 2.957 353 E + 01 ounce (avoirdupois)-force (ozf) newton (N) 2.780 139 E-01 ounce (avoirdupois)-force inch (ozf • in) newton meter (N • m) 7.061 552 E-03 ounce (avoirdupois)-force inch (ozf • in) miiiinewton meter (mN • m) 7.061 552 E + 00 ounce (avoirdupois) per cubic inch (oz/in^) kilogram per cubic meter (kg/m') 1.729 994 E + 03 ounce (avoirdupois) per gallon [Canadian and U.K. (Imperial)] (oz/gal) kilogram per cubic meter (kg/m^) 6.236 023 E + 00 ounce (avoirdupois) per gallon [Canadian and U.K. (Imperial)] (oz/gal) gram per liter (g/L) 6.236 023 E + 00 ounce (avoirdupois) per gallon (U.S.)(oz/gal). . .kilogram per cubic meter (kg/m') 7.489 152 E + 00 ounce (avoirdupois) per gallon(U.S.)(oz/gal) . . .gram per liter (g/L) 7.489 152 E + 00 ounce (avoirdupois) per square foot (oz/ft^) . . .kilogram per square meter (kg/m^) 3.051 517 E-01 ounce (avoirdujjois) per square inch (oz/in^). . .kilogram per square meter (kg/m^) 4.394 185 E + 01 ounce (avoirdupois) per square yard(oz/yd^) . . .kilogram per square meter (kg/m^) 3.390575 E-02 parsec (pc) meter (m) 3.085 678 E + 16 peck (U.S.) (pk) cubic meter (m') 8.809 768 E-03 peck (U.S.) (pk) liter (L) 8.809 768 E + 00 pennyweight (dwt) kilogram (kg) 1.555 174 E-03 pennyweight (dwt) gram (g) 1.555 174 E + 00 perm (0 °C) kilogram per pascal second square meter [kg/(Pa • s • m^)] 5.721 35 E-11 perm (23 °C) kilogram per pascal second square meter [kg/(Pa • s • m^)] 5.745 25 E-11 perm inch (0°C) kilogram per pascal second meter [kg/(Pa-s-m)] 1.453 22 E-12 perm inch (23 °C) kilogram per pascal second meter [kg/(Pa • s • m)] 1.459 29 E-12 53 Guide for the Use of the International System of Units (SI) To convert from to Multiply by phot (ph) lux (Ix) 1.0 E+04 pica (computer) (1/6 in) meter (m) 4.233 333 E-03 pica (computer) (1/6 in) millimeter (mm) 4.233 333 E + 00 pica (printer's) meter (m) 4.217 518 E-03 pica (printer's) millimeter (mm) 4.217 518 E + 00 pint (U.S. dry) (dry pt) cubic meter (m^) 5.506 105 E-04 pint (U.S. dry) (dry pt) liter (L) 5.506 105 E - 01 pint (U.S. liquid) (liq pt) cubic meter (m^) 4.731 765 E-04 pint (U.S. liquid) (liq pt) liter (L) 4.731 765 E - 01 point (computer) (1/72 in) meter (m) 3.527 778 E-04 point (computer) (1/72 in) millimeter (mm) 3.527 778 E-01 point (printer's) meter (m) 3.5 14 598 E-04 point (printer's) millimeter (mm) 3.514 598 E-01 jxjise (P) pascal second (Pa • s) 1.0 E—01 pound (avoirdupois) (Ib)^^ kilogram (kg) 4.535 924 E-01 pound (troy or apothecary) (lb) kilogram (kg) 3.732 417 E-01 poundal newton (N) 1.382 550 E-01 poundal per square foot pascal (Pa) 1.488 164 E + 00 poundal second per square foot pascal second (Pa • s) 1.488 164 E + 00 pound foot squared (lb • ft^) kilogram meter squared (kg-m^) 4.214 011 E-02 pound-force ( Ibf )2' newton (N) 4.448 222 E + 00 pound-force foot (ibf - ft) newton meter (N • m) 1.355 818 E + 00 pound-force foot per inch (Ibf • ft/in) newton meter per meter (N • m/m) 5.337 866 E + 01 pound-force inch (Ibf - in) newton meter (N • m) 1.129 848 E-01 pound-force inch per inch (Ibf- in/in) newton meter per meter (N • m/m) 4.448 222 E + 00 pound-force per foot (Ibf/ft) newton per meter (N/m) 1.459 390 E + 01 pound-force per inch (Ibf/in) newton per meter (N/m) 1.75 1 268 E + 02 pound-force per p)ound (Ibf/lb) (thrust to mass ratio) newton per kilogram (N/kg) 9.806 65 E+00 pound-force per square foot (ibf/ft^) pascal (Pa) 4.788 026 E + 01 pound-force per square inch (psi) (Ibf/in^) pascal (Pa) 6.894 757 E + 03 pound-force per square inch (psi) (Ibf/in^) kilopascal (kPa) 6.894 757 E + 00 pound-force second per square foot (Ibf • s/ft^) pascal second (Pa • s) 4.788 026 E + 01 fxjund-force second per square inch (Ibf • s/in^) pascal second (Pa • s) 6.894 757 E + 03 pound inch squared (lb • in^) kilogram meter squared (kg • m^) 2.926 397 E-04 pound per cubic foot (Ib/ft^) kilogram per cubic meter (kg/m^) 1.601 846 E + 01 pound per cubic inch (lb/in') kilogram per cubic meter (kg/m^) 2.767 990 E + 04 jjound per cubic yard (\b/y<P) kilogram per cubic meter (kg/m') 5.932 764 E-01 pound per foot (lb/ft) kilogram per meter (kg/m) 1.488 164 E + 00 pound per foot hour [lb/ (ft • h)] pascal second (Pa • s) 4.133 789 E-04 pound per foot second [lb/(ft • s)] pascal second (Pa • s) 1.488 164 E + 00 pound per gallon [Canadian and U.K. (Imperial)] (Ib/gal) kilogram per cubic meter (kg/m^) 9.977 637 E + 01 pound per gallon [Canadian and U.K. (Imperial)] (lb/gal) kilogram per liter (kg/L) 9.977 637 E-02 pound per gallon (U.S.) (lb/gal) kilogram per cubic meter (kg/m') 1.198 264 E + 02 pound per gallon (U.S.) (Ib/gal) kilogram per liter (kg/L) 1.198 264 E-01 pound per horsepower hour [lb/(hp h)] kilogram per joule (kg/J) 1.689 659 E — 07 pound per hour (Ib/h) kilogram per second (kg/s) 1.259 979 E-04 The exact conversion factor is 4.535 923 7 E-OI. All units in Sees. B.8 and B.9 that contain the pound refer to the avoirdupois pound. ^ If the local value of the acceleration of free fall is taken as g„ = 9.806 65 m/s^ (the standard value), the exact conversion factor is 4.448 221 615 260 5 E + 00. 54 Guide for the Use of the International System of Units (SI) To convert from to Multiply by pound per inch (lb/in) kilogram per meter (kg/m) 1.785 797 E + 01 pound per minute (Ib/min) kilogram per second (kg/s) 7.559 873 E — 03 pound per second (ib/s) kilogram per second (kg/s) 4.535 924 E — 01 pound per square foot (Ib/ft^) kilogram per square meter (kg/m^) 4.882 428 E + 00 pound per square inch {nol jx)und-force) (Ib/in^) kilogram per square meter (kg/m^) 7.030 696 E + 02 pound per yard (lb/yd) kilogram per meter (kg/m) 4.960 546 E-01 psi (pound-force per square inch) (Ibf/in^) pascal (Pa) 6.894 757 E + 03 psi (pound-force per square inch) (Ibf/in^) kilopascal (kPa) 6.894 757 E-l-00 quad (10'' Bturr)^ joule (J) 1.055 056 E+ 18 quart (U.S. dry) (dry qt) cubic meter (m^) 1.101 221 E-03 quart (U.S. dry) (dry qt) liter (L) 1.101 221 E + 00 quart (U.S. liquid) (liq qt) cubic meter (m') 9.463 529 E-04 quart (U.S. liquid) (liq qt) liter (L) 9.463 529 E-01 rad (absorbed dose) (rad) gray (Gy) 1.0 E-02 rem (rem) sievert (Sv) 1.0 E-02 revolution (r) radian (rad) 6.283 185 E-l-00 revolution per minute (rpm) (r/min) radian per second (rad/s) 1.047 198 E — 01 rhe reciprocal pascal second [(Pa-s)"'] 1.0 E-l-01 rod (based on U.S. survey foot) (rd)^ meter (m) 5.029 210 E + 00 roentgen (R) coulomb per kilogram (C/kg) 2.58 E— 04 rpm (revolution per minute) (r/min) radian per second (rad/s) 1.047 198 E-01 second (angle) (") radian (rad) 4.848 137 E - 06 second (sidereal) second (s) 9.972 696 E-01 shake second (s) 1.0 E-08 shake nanosecond (ns) 1.0 E+01 slug (slug) kilogram (kg) 1.459 390 E + 01 slug per cubic foot (slug/ft') kilogram per cubic meter (kg/m^) 5.153 788 E + 02 slug per foot second [slug/(ft • s)] pascal second (Pa • s) 4.788 026 E + 01 square foot (ft^) square meter (m^) 9.290 304 E-02 square foot per hour (ft^/h) square meter per second (m^/s) 2.580 64 E— 05 square foot per second (ft^/s) square meter per second (m^/s) 9.290 304 E-02 square inch (in^) square meter (m^) 6.4516 E— 04 square inch (in^) square centimeter (cm^) 6.4516 E+00 square mile (mi^) square meter (m^) 2.589 988 E + 06 square mile (mi^) square kilometer (km^) 2.589 988 E + 00 square mile (based on U.S. survey foot) (mi^)^ square meter (m^) 2.589 998 E + 06 square mile (based on U.S. survey foot) (mi^f square kilometer (km^) 2.589 998 E + 00 square yard (yd^) square meter (m^) 8.361 274 E — 01 statampere ampere (A) 3.335 641 E- 10 statcoulomb coulomb (C) 3.335 641 E- 10 statfarad farad (F) 1.112 650 E- 12 stathenry henry (H) 8.987 552 E + 1 1 statmho Siemens (S) 1.112650 E-12 statohm ohm (fl) 8.987 552 E + 1 1 statvolt volt (V) 2.997 925 E + 02 stere (st) cubic meter (m^) 1.0 E+00 stilb (sb) candela per square meter (cd/m^) 1.0 E+04 stokes (St) meter squared per second (mVs) 1.0 E-04 55 Guide for the Use of the International System of Units (SI) To convert from to Multiply by tablespoon cubic meter (m^) 1.478 676 E-05 tablespoon milliliter (mL) 1.478 676 E+01 teaspoon cubic meter (m') 4.928 922 E - 06 teaspoon milliliter (mL) 4.928 922 E + 00 tex kilogram per meter (kg/m) 1.0 E-06 therm (EC)^^ joule (J) 1.055 06 E+08 therm (U.S.)^^ joule (J) 1.054 804 E+08 ton, assay (AT) kilogram (kg) 2.916 667 E - 02 ton, assay (AT) gram (g) 2.916 667 E + 01 ton-force (2000 Ibf ) newton (N) 8.896 443 E + 03 ton-force (2000 Ibf) kilonewton (kN) 8.896 443 E + 00 ton, long (2240 lb) kilogram (kg) 1.016 047 E + 03 ton, long, per cubic yard kilogram per cubic meter (kg/m') 1.328 939 E + 03 ton, metric (t) kilogram (kg) 1.0 E+03 tonne (called "metric ton" in U.S.) (t) kilogram (kg) 1.0 E+03 ton of refrigeration (12 000 Btu,T/h) watt (W) 3.516 853 E + 03 ton of TNT (energy equivalent)^ joule (J) 4.184 E+09 ton, register cubic meter (m^) 2.831 685 E+00 ton, short (2000 lb) kilogram (kg) 9.071 847 E + 02 ton, short, per cubic yard kilogram per cubic meter (kg/m') 1.186 553 E + 03 ton, short, per hour kilogram per second (kg/s) 2.519 958 E-01 ton- (Torr) pascal (Pa) 1.333 224 E + 02 unit pole weber (Wb) 1.256 637 E - 07 watt hour (W • h) joule (J) 3.6 E+03 watt per square centimeter (W/cm^) watt per square meter (W/m^) 1.0 E+04 watt per square inch (W/in^) watt per square meter (W/m^) 1.550 003 E + 03 watt second (W • s) joule (J) 1.0 E+00 yard (yd) meter (m) 9.144 E-01 year (365 days) second (s) 3.1536 E+07 year (sidereal) second (s) 3.155 815 E + 07 year (tropical) second (s) 3.155 693 E + 07 I I I J f " The therm (EC) is legally defined in the Council Directive of 20 December 1979, Council of the European Communities (now [ the European Union, EU). The therm (U.S.) is legally defmed in the Federal Register of July 27, 1968. Although the therm (EC), | which is based on the International Table Btu, is frequently used by engineers in the United States, the therm (U.S.) is the legal unit used by the U.S. natural gas industry. I Defined (not measured) value. 56 Guide for the Use of the International System of Units (SI) B.9 Factors for units listed by kind of quantity or fleld of science Caution: The units listed in column 1 are in general not to be used in NIST publications, with the exception of those few in italic type. Factors in boldface are exact To convert from to Multiply by ACCELERATION (m/s^) .... ... 9.806 65 E+00 (m/s^) .... 3.048 E-01 (m/s^) .... .... 1.0 E-02 (m/s2) .... .... 2.54 E-02 ANGLE degree (°) radian (rad) 1.745 329 E-02 gon (also called grade) (gon) radian (rad) 1.570 796 E-02 gon (also called grade) (gon) degree ( ° ) 9.0 E-01 mil radian (rad) 9.817 477 E-04 mil degree (°) 5.625 E-02 minute (') radian (rad) 2.908 882 E-04 revolution (r) radian (rad) 6.283 185 E + 00 second (") radian (rad) 4.848 137 E-06 AREA AND SECOND MOMENT OF AREA acre (based on U.S. survey foot)^ square meter (m^) 4.046 873 E + 03 are (a) square meter (m^) 1.0 E+02 barn (b) square meter (m^) 1.0 E—28 circular mil square meter (m^) 5.067075 E-10 circular mil square millimeter (mm^) 5.067 075 E — 04 foot to the fourth power (ff)^'' meter to the fourth power (m) 8.630 975 E-03 hectare (ha) square meter (m^) 1.0 E+04 inch to the fourth power (in")' meter to the fourth power (m") 4.162 314 E-07 square foot (ft^) square meter (m^) 9.290 304 E-02 square inch (in^) square meter (m^) 6.4516 E-04 square inch (in") square centimeter (cm^) 6.4516 E+00 square mile (mi^) square meter (m^) 2.589 988 E + 06 square mile (mi^) square kilometer (km^) 2.589 988 E + 00 square mile (based on U.S. survey foot) (mi^)' square meter (m^) 2.589 998 E + 06 square mile (based on U.S. survey foot) (mi^)^ square kilometer (km^) 2.589 998 E + 00 square yard (yd^) square meter (m^) 8.361 274 E-01 CAPACITY (see VOLUME) DENSITY (that is, MASS DENSITY - see MASS DIVIDED BY VOLUME) ELECTRICITY and MAGNETISM abampere ampere (A) 1.0 E+01 abcoulomb coulomb (C) 1.0 E+01 abfarad farad (F) 1.0 E+09 abhenry henry (H) 1.0 E-09 abmho Siemens (S) 1.0 E+09 abohm ohm (fl) 1.0 E-09 abvolt voh (V) 1.0 E-08 ampere hour (A-h) coulomb (C) 3.6 E+03 57 Guide for the Use of the International System of Units (SI) To convert from to Multiply by biot(Bi) ampere (A) 1.0 E+01 EMU of capacitance (abfarad) farad (F) 1.0 E+09 EMU of current (abampere) ampere (A) 1.0 E+01 EMU of electric potential (abvolt) volt (V) 1.0 E-08 EMU of inductance (abhenry) henry (H) 1.0 E-09 EMU of resistance (abohm) ohm (fl) 1.0 E-09 ESU of capacitance (statfarad) farad (F) 1.112 650 E - 12 ESU of current (statampere) ampere (A) 3.335 641 E- 10 ESU of electric potential (statvolt) volt (V) 2.997 925 E + 02 ESU of inductance (stathenry) henry (H) 8.987 552 E+ 11 ESU of resistance (statohm) ohm (CI) 8.987 552 E + 11 faraday (based on carbon 12) coulomb (C) 9.648 531 E + 04 franklin (Fr) coulomb (C) 3.335 641 E - 10 gamma (7) tesla (T) 1.0 E-09 gauss (Gs, G) tesla (T) 1.0 E-04 gilbert (Gi) ampere (A) 7.957 747 E - 01 maxwell (Mx) weber (Wb) 1.0 E-08 mho Siemens (S) 1.0 E+00 oersted (Oe) ampere per meter (A/m) 7.957 747 E + 01 ohm centimeter (n-cm) ohm meter (fl • m) 1.0 E—02 ohm circular-mil per foot ohm meter (ft • m) 1.662 426 E-09 ohm circular-mil per foot ohm square millimeter per meter (fl-mm^/m) 1.662 426 E-03 statampere ampere (A) 3.335 641 E-10 statcoulomb coulomb (C) 3.335 641 E-10 statfarad farad (F) 1.112 650 E- 12 stathenry henry (H) 8.987 552 E + 1 statmho Siemens (S) 1.112650 E-12 statohm ohm (ft) 8.987 552 E + 1 statvolt volt (V) 2.997 925 E + 02 unit pole weber (Wb) 1.256 637 E-07 ENERGY (includes WORK) British thermal unitir (Btuix)^^ . joule (J) 1.055 056 E + 03 British thermal unit.h (Btu.h)^^ joule (J) 1.054 350 E + 03 British thermal unit (mean) (Btu) joule (J) 1.055 87 E + 03 British thermal unit (39 "F) (Btu) joule (J) 1.059 67 E + 03 British thermal unit (59 °F) (Btu) joule (J) 1.054 80 E + 03 British thermal unit (60 °F) (Btu) joule (J) 1.054 68 E + 03 calorierr (calrr)^^ joule (J) 4.1868 E+00 calorie.h (caU)^^ joule (J) 4.184 E+00 calorie (mean) (cal) joule (J) 4.190 02 E + 00 calorie (15 °C) (calis) joule (J) 4.185 80 E + 00 calorie (20 "C) (caho) joule (J) 4.181 90 E + 00 calorieiT, kilogram (nutrition) joule (J) 4.1868 E+03 calorieih, kilogram (nutrition) joule (J) 4.184 E+03 calorie (mean), kilogram (nutrition)^^ joule (J) 4.190 02 E + 03 electronvolt (eV) joule (J) 1.602 177 E- 19 erg (erg) joule (J) 1.0 E-07 foot poundal joule (J) 4.214 Oil E-02 foot pound-force (ft • Ibf) joule (J) 1.355 818 E + 00 kilocalorieiT (kcalir) joule (J) 4.1868 E+03 kilocalorie.h (kcal.h) joule (J) 4.184 E+03 kilocalorie (mean) (kcal) joule (J) 4.190 02 E + 03 58 Guide for the Use of the International System of Units (SI) To convert from to Multiply by kilowalt hour (kW • h) joule (J) 3.6 E+06 kilowatt hour (kW • h) megajoule (MJ) 3.6 E+00 quad (10' Bturr)^^ joule (J) 1.055 056 E+18 therm (EC)^^ joule (J) 1.055 06 E+08 therm (U.S.)^ joule (J) 1.054 804 E+08 ton of TNT (energy equivalent)^^ joule (J) 4.184 E+09 watt hour (W • h) joule (J) 3.6 E+03 watt second (W • s) joule (J) 1.0 E+00 ENERGY DIVIDED BY AREA TIME erg per square centimeter second [erg/(cm^ • s)] watt per square meter (W/m^) 1.0 E-03 watt per square centimeter (W/cm^) watt per square meter (W/m^) 1.0 E+04 watt per square inch (W/in^) watt per square meter (W/m^) 1.550 003 E + 03 FLOW (see MASS DIVIDED BY TIME or VOLUME DIVIDED BY TIME) FORCE dyne(dyn) newton (N) 1.0 E-OS kilogram-force (kgf) newton (N) 9.806 65 E+00 kilopond (kilogram-force) (kp) newton (N) 9.806 65 E+00 kip (1 kip = 1000 Ibf) newton (N) 4.448 222 E + 03 kip (1 kip = 1000 Ibf) kilonewton (kN) 4.448 222 E + 00 ounce (avoirdupois)-force (ozf) newton (N) 2.780 139 E-01 poundal newton (N) 1.382 550 E-01 pound-force (Ibf)24 newton (N) 4.448 222 E + 00 pound-force per pound (Ibf/lb) (thrust to mass ratio) newton per kilogram (N/kg) 9.806 65 E+00 ton-force (2000 Ibf) newton (N) 8.896 443 E + 03 ton-force (2000 Ibf) kilonewton (kN) 8.896 443 E + 00 FORCE DIVIDED BY AREA (see PRESSURE) FORCE DIVIDED BY LENGTH pound-force per foot (Ibf/ft) newton per meter (N/m) 1.459 390 E + 01 pound-force per inch (Ibf/in) newton per meter (N/m) 1.751 268 E + 02 HEAT Available Energy British thermal unitix per cubic foot (Btuix/ft^) joule per cubic meter (J/m^) 3.725 895 E + 04 British thermal unit,h per cubic foot (Btu,h/ft^) joule per cubic meter (J/m^) 3.723 403 E + 04 British thermal unitiT per pound (Btuix/lb) joule per kilogram (J/kg) 2326 E+03 British thermal unit,h per pound (Btu,h/lb) joule per kilogram (J/kg) 2.324 444 E + 03 calorieiT per gram (calix/g) joule per kilogram (J/kg) 4.1868 E+03 caloricih per gram (cal,h/g) joule per kilogram (J/kg) 4.184 E+03 Coefficient of Heat Transfer British thermal unitir per hour square foot degree Fahrenheit [BtUiT/(h • ft^ • °F)] watt per square meter kelvin [W/(m2-K)] 5.678 263 E + 00 British thermal unit,h per hour square foot degree Fahrenheit [BtUth/(h • ft^ • °F)] watt per square meter kelvin [W/ (m^ • K)] 5.674 466 E + 00 British thermal unitir per second square foot degree Fahrenheit [BtUiT/(s • ft^ • °F)] watt per square meter kelvin [W/(m^ • K)] 2.044 175 E + 04 British thermal unitih per second square foot degree Fahrenheit [Btu,h/(s • ft^ • °F)] watt per square meter kelvin [W/(m2 • K)] 2.042 808 E + 04 59 Guide for the Use of the International System of Units (SI) To convert from to Multiply by Density of Heat British thermal unitir per square foot (Btuir/ft^) joule per square meter (J/m^) 1.135 653 E + 04 British thermal unitih per square foot (Btu,h/ft^) joule per square meter (J/m^) 1.134 893 E + 04 calorie,h per square centimeter (cal,h/cm^) joule per square meter (J/m^) 4.184 E+04 langley (cal,h/cm^) joule per square meter (J/m^) 4,184 E+04 Density of Heat Flow Rate British thermal unitir per square foot hour [Btu,T/(ft^ • h)] watt per square meter (W/m^) 3.154 591 E + 00 British thermal unitih per square foot hour [Btu,h/(ft^ • h)] watt per square meter (W/m^) 3.152 481 E + 00 British thermal unitth per square foot minute [Btu,h/(ft^ • min)] watt per square meter (W/m^) 1.891 489 E + 02 British thermal unitix per square foot second [BtuiT/(ft^ • s)] watt per square meter (W/m^) 1.135 653 E + 04 British thermal unit,h per square foot second [Btu,h/(ft^ • s)] watt per square meter (W/m^) 1.134 893 E + 04 British thermal unitth per square inch second [Btu,h/(in^ • s)] watt per square meter (W/m^) 1.634 246 E + 06 calorie,h per square centimeter minute [cal,h/(cm^ • min)] watt per square meter (W/m^) 6.973 333 E + 02 caloricth per square centimeter second [cal,h/(cm^ • s)] watt per square meter (W/m^) 4.184 E+04 Fuel Consumption gallon (U.S.) per horsepower hour [gal/(hp • h)] cubic meter per joule (m'/J) 1.410 089 E-09 gallon (U.S.) per horsepower hour [gal/(hp-h)] liter per joule (L/J) 1.410 089 E-06 mile per gallon (U.S.) (mpg) (mi/gal) meter per cubic meter (m/m^) 4.251 437 E + 05 mile per gallon (U.S.) (mpg) (mi/gal) kilometer per liter (km/L) 4.251 437 E-01 mile per gallon (U.S.) (mpg) (mi/gal)^^ liter per 100 kilometer (L/lOO km). . . divide 235.215 by number of miles per gallon pound per horsepower hour [lb/(hp • h)] kilogram per joule (kg/ J) 1.689 659 E-07 Heat Capacity and Entropy British thermal unitix per degree Fahrenheit (BtuiT/°F) joule per kelvin (J/k) 1.899 101 E + 03 British thermal unitth per degree Fahrenheit (BtUth/°F) joule per kelvin (J/k) 1.897 830 E + 03 British thermal unitir per degree Rankine (Btu,T/°R) joule per kelvin (J/k) 1.899 101 E + 03 British thermal unitth per degree Rankine (Btu,h/''R) joule per kelvin (J/k) 1.897 830 E + 03 Heat Flow Rate British thermal unitir per hour (Bturr/h) watt (W) 2.930 711 E-01 British thermal unitth per hour (BtUth/h) watt (W) 2.928 751 E-01 British thermal unit,h per minute (Btu,h/min) . . watt (W) 1.757 250 E + 01 British thermal unitir per second (Btuir/s) watt (W) 1.055 056 E + 03 British thermal unit,h per second (BtUth/s) watt (W) 1.054 350 E + 03 caIorie,h per minute (cal,h/min) watt (W) 6.973 333 E — 02 caIorie,h per second (cal,h/s) watt (W) 4.184 E+00 kilocaloricih per minute (kcalih/min) watt (W) 6.973 333 E + 01 kiIocalorie,h per second (kcal,h/s) watt (W) 4.184 E+03 ton of refrigeration (12 000 Btuir/h) watt (W) 3.516 853 E + 03 60 ! Guide for the Use of the International System of Units (SI) To convert from to Multiply by Specific Heat Capacity and Specific Entropy British thermal unitir per pound degree Fahrenheit [Btu,T/(lb • °F)] joule per kilogram kelvin [J/(kg • K)] 4.1868 E+03 British thermal unit,h per pound degree Fahrenheit [Btu,h/(lb • "F)] joule per kilogram kelvin [J/(kg • K)] 4,184 E+03 British thermal unitiT per pound degree Rankine [BtuiT/(lb • °R)] joule per kilogram kelvin [J/(kg • K)] 4.1868 E+03 British thermal unit,h per pound degree Rankine [Btu,h/(lb • °R)] joule per kilogram kelvin [J/(kg • K)] 4.184 E+03 calorierr per gram degree Celsius [cal,T/(g • °C)] joule per kilogram kelvin [J/(kg • K)] 4.1868 E+03 calorie,h per gram degree Celsius [cal,h/(g • "C)] joule per kilogram kelvin [J/(kg • K)] 4.184 E+03 calorieiT per gram kelvin [calrr/(g • K)] joule per kilogram kelvin [J/(kg • K)] 4.1868 E+03 calorie,h per gram kelvin [cal,h/(g • K)] joule per kilogram kelvin [J/(kg • K)] 4.184 E+03 Thermal Conductivity Britsh thermal unitiT foot per hour square foot degree Fahrenheit [Btu,T-ft/(h-ft2-°F)] watt per meter kelvin [W/(m-K)] 1.730 735 E + 00 Britsh thermal unit,h foot per hour square foot degree Fahrenheit [Btu.h • ft/(h • ft^ • °F)] watt per meter kelvin [W/(m • K)] 1.729 577 E + 00 Britsh thermal unitij inch per hour square foot degree Fahrenheit [BtUiT • in/(h • ft^ • °F)] watt per meter kelvin [W/(m • K)] 1.442 279 E-01 Britsh thermal unit,h inch per hour square foot degree Fahrenheit [Btu.h • in/(h • ft^ • °F)] watt per meter kelvin [W/(m • K)] 1.441 314 E-01 Britsh thermal unitir inch per second square foot degree Fahrenheit [BtUiT • in/(s • ft^ • °F)] watt per meter kelvin [W/(m-K)] 5.192 204 E + 02 Britsh thermal unit,h inch per second square foot degree Fahrenheit [Btu.h • in/(s • ft^ • °F)] watt per meter kelvin [W/(m • K)] 5.188 732 E + 02 calorieth per centimeter second degree Celsius [cal,h/(cm • s • °C)] watt per meter kelvin [W/(m • K)] 4.184 E+02 Thermal Diffusivity square foot per hour (ft^/h) square meter per second (m^/s) 2.580 64 E-05 Thermal Insulance clo square meter kelvin per watt (m^ • K/W) . . 1.55 E — 01 degree Fahrenheit hour square foot per British thermal unitir (°F- h • ft^/BtUir) square meter kelvin per watt (m^ • K/W) . . 1.761 102 E-01 degree Fahrenheit hour square foot per British thermal unitih ("F-h-ft^/Btu.h) square meter kelvin per watt (m^ • K/W) . . 1.762 280 E-01 Thermal Resistance degree Fahrenheit hour per British thermal unitir ("F- h/BtUir) kelvin per watt (K/W) 1.895 634 E + 00 degree Fahrenheit hour per British thermal unit,h CF-h/Btu.h) kelvin per watt (K/W) 1.896 903 E + 00 degree Fahrenheit second per British thermal unitir ("F-s/BtUir) kelvin per watt (K/W) 5.265 651 E-04 degree Fahrenheit second per British thermal unit,h ("F • s/Btu,h) kelvin per watt (K/W) 5.269 175 E-04 Thermal Resistivity degree Fahrenheit hour square foot per British thermal unitir inch [''F-h-ftV(Btuir-in)] meter kelvin per watt (m-K/W) 6.933 472 E + 00 degree Fahrenheit hour square foot per British thermal unitu, inch ["F • h • ftV(Btu,h • in)] meter kelvin per watt (m • K/W) 6.938 112 E + 04 61 Guide for the Use of the International System of Units (SI) To convert from to Multiply by LENGTH Sngstrom (A) meter (m) 1.0 E-10 Sngstrom (A) nanometer (nm) 1.0 E-01 astronomical unit (AU) meter (m) 1.495 979 E+ 11 chain (based on U.S. survey foot) (ch)^ meter (m) 2.011 684 E + 01 fathom (based on U.S. survey foot)^ meter (m) 1.828 804 E + 00 fermi meter (m) 1.0 E— 15 fermi femtometer (fm) 1.0 E+00 foot (ft) meter (m) 3.048 E-01 foot (U.S. survey) (ft)' meter (m) 3.048 006 E-01 inch (in) meter (m) 2.S4 E—02 inch (in) centimeter (cm) 2.54 E+00 kayser(K) reciprocal meter (m"') 1 E+02 light year (l.y.) meter (m) 9.460 73 E + 15 microinch meter (m) 2.54 E-08 microinch micrometer (jim) 2.54 E— 02 micron (p.) meter (m) 1.0 E— 06 micron (jt) micrometer ((im) 1.0 E+00 mil (0.001 in) meter (m) 2.54 E-05 mil (0.001 in) millimeter (mm) 2.54 E-02 mile (mi) meter (m) 1.609 344 E+03 mile (mi) kilometer (km) 1.609 344 E+00 mile (based on U.S. survey foot) (mi)' meter (m) 1.609 347 E + 03 mile (based on U.S. survey foot) (mi)' kilometer (km) 1.609 347 E + 00 mile, nautical^^ meter (m) 1.852 E+03 parsec (pc) meter (m) 3.085 678 E+ 16 pica (computer) (1/6 in) meter (m) 4.233 333 E-03 pica (computer) (1/6 in) millimeter (mm) 4.233 333 E + 00 pica (printer's) meter (m) 4.217 518 E-03 pica (printer's) millimeter (mm) 4.217 518 E + 00 point (computer) (1/72 in) meter (m) 3.527 778 E - 04 point (computer) (1/72 in) millimeter (mm) 3.527 778 E-01 point (printer's) meter (m) 3.514 598 E - 04 point (printer's) millimeter (mm) 3.514 598 E-01 rod (based on U.S. survey foot) (rd)' meter (m) 5.029 210 E + 00 yard (yd) meter (m) 9.144 E-01 LIGHT candela per square inch (cd/in^) candela per square meter (cd/m^) 1.550 003 E + 03 footcandle lux (Ix) 1.076391 E + 01 footlambert candela per square meter (cd/m^) 3.426 259 E + 00 lambert candela per square meter (cd/m^) 3.183 099 E + 03 lumen per square foot (Im/ft^) lux (Ix) 1.076 391 E + 01 phot (ph) lux (Ix) 1.0 E+04 stilb (sb) candela per square meter (cd/m^) 1.0 E+04 MASS and MOMENT OF INERTIA carat, metric kilogram (kg) 2.0 E—04 carat, metric gram (g) 2.0 E—01 grain (gr) kilogram (kg) 6.479 891 E-05 grain (gr) milligram (mg) 6.479 891 E+01 hundredweight (long, 112 lb) kilogram (kg) 5.080 235 E + 01 hundredweight (short, 100 lb) kilogram (kg) 4.535 924 E + 01 62 Guide for the Use of the International System of Units (SI) To convert from to Multiply by kilogram-force second squared per meter (kgf • sVm) kilogram (kg) 9.806 65 E+00 ounce (avoirdufKjis) (oz) kilogram (kg) 2.834 952 E - 02 ounce (avoirdupois) (oz) gram (g) 2.834 952 E + 01 ounce (troy or apothecary) (oz) kilogram (kg) 3.110 348 E-02 ounce (troy or apothecary) (oz) gram (g) 3.110 348 E + 01 pennyweight (dwt) kilogram (kg) 1.555 174 E-03 pennyweight (dwt) gram (g) 1.555 174 E + 00 pound (avoirdupois) (Ib)^^ kilogram (kg) 4.535 924 E-01 pound (troy or apothecary) (lb) kilogram (kg) 3.732 417 E-01 pound foot squared (lb • ft^) kilogram meter squared (kg • m^) 4.214 Oil E-02 pound inch squared (lb • in^) kilogram meter squared (kg-m^) 2.926 397 E — 04 slug (slug) kilogram (kg) 1.459 390 E + 01 ton, assay (AT) kilogram (kg) 2.916 667 E-02 ton, assay (AT) gram (g) 2.916 667 E + 01 ton, long (2240 lb) kilogram (kg) 1.016 047 E + 03 ton, metric (t) kilogram (kg) 1.0 E+03 tonne (called "metric ton" in U.S.) (t) kilogram (kg) 1.0 E+03 ton, short (2000 lb) kilogram (kg) 9.071 847 E + 02 MASS DENSITY (see MASS DIVIDED BY VOLUME) MASS DIVIDED BY AREA ounce (avoirdupois) per square foot (oz/ft^) kilogram per square meter (kg/m^) 3.051 517 E — 01 ounce (avoirdupois) per square inch (oz/in^) . . .kilogram per square meter (kg/m^) 4.394 185 E + 01 ounce (avoirdupois) per square yard (oz/yd^). . .kilogram per square meter (kg/m^) 3.390 575 E — 02 pound per square foot (Ib/ft^) kilogram per square meter (kg/m^) 4.882 428 E + 00 pound per square inch {not pound force) (Ib/in^) kilogram per square meter (kg/m^) 7.030 696 E + 02 MASS DIVIDED BY CAPACITY (see MASS DIVIDED BY VOLUME) MASS DIVIDED BY LENGTH denier kilogram per meter (kg/m) 1.111111 E-07 denier gram per meter (g/m) 1.111 111 E — 04 pound per foot (lb/ft) kilogram per meter (kg/m) 1.488 164 E + 00 pound per inch (lb/in) kilogram per meter (kg/m) 1.785 797 E + 01 pound per yard (lb/yd) kilogram per meter (kg/m) 4.960 546 E — 01 tex kilogram per meter (kg/m) 1.0 E— 06 MASS DIVIDED BY TIME (includes FLOW) pound per hour (Ib/h) kilogram per second (kg/s) 1.259 979 E-04 pound per minute (Ib/min) kilogram per second (kg/s) 7.559 873 E-03 pound per second (Ib/s) kilogram per second (kg/s) 4.535 924 E-01 ton, short, per hour kilogram per second (kg/s) 2.519 958 E-01 MASS DIVIDED BY VOLUME (includes MASS DENSITY and MASS CONCENTRATION) grain per gallon (U.S.) (gr/gal) kilogram per cubic meter (kg/m') 1.711 806 E-02 grain per gallon (U.S.) (gr/gal) milligram per liter (mg/L) 1.711 806 E + 01 gram per cubic centimeter (g/cm^) kilogram per cubic meter (kg/m') 1.0 E+03 ounce (avoirdupois) per cubic inch (oz/in') kilogram per cubic meter (kg/m') 1.729 994 E + 03 ounce (avoirdupois) per gallon [Canadian and U.K. (Imperial)] (oz/gal) kilogram per cubic meter (kg/m') 6.236 023 E + 00 ounce (avoirdupois) per gallon [Canadian and U.K. (Imperial)] (oz/gal) gram per liter (g/L) 6.236 023 E + 00 ounce (avoirdupois) per gallon (U.S.) (oz/gal). . .kilogram per cubic meter (kg/m') 7.489 152 E + 00 ounce (avoirdupois) per gallon (U.S.) (oz/gal). . .gram per liter (g/L) 7.489 152 E + 00 63 Guide for the Use of the International System of Units (SI) To convert from to Multiply by pound per cubic foot (lb/ft') kilogram per cubic meter (kg/m^) 1.601 846 pound per cubic inch (lb/in') kilogram per cubic meter (kg/m') 2.767 990 pound per cubic yard (Ib/yd^) kilogram per cubic meter (kg/m') 5.932 764 pound per gallon [Canadian and U.K. (Imperial)] (lb/gal) kilogram per cubic meter (kg/m') 9.977 637 pound per gallon [Canadian and U.K. (Imperial)] (lb/gal) kilogram per liter (kg/L) 9.977 637 pound per gallon (U.S.) (lb/gal) kilogram per cubic meter (kg/m^) 1.198 264 pound per gallon (U.S.) (lb/gal) kilogram per liter (kg/L) 1.198 264 slug per cubic foot (slug/ft^) kilogram per cubic meter (kg/m^) 5.153 788 ton, long, per cubic yard kilogram per cubic meter (kg/m^) 1.328 939 ton, short, per cubic yard kilogram per cubic meter (kg/m^) 1.186 553 MOMENT OF FORCE or TORQUE dyne centimeter (dyn • cm) newton meter (N m) 1.0 kilogram-force meter (kgf • m) newton meter (N m) 9.806 65 ounce (avoirdupois)-force inch (ozf • in) newton meter (N • m) 7.061 552 ounce (avoirdupois)-force inch (ozf - in) millinewton meter (mN • m) 7.061 552 pound-force foot (Ibf • ft) newton meter (N • m) 1.355 818 pound-force inch (Ibf • in) newton meter (N • m) 1.129 848 MOMENT OF FORCE or TORQUE, DIVIDED BY LENGTH pound-force foot per inch (Ibf -ft/in) newton meter per meter (N-m/m) 5.337 866 pound-force inch per inch (Ibf in/in) newton meter per meter (N • m/m) 4.448 222 PERMEABILITY darcy^^ meter squared (m^) 9.869233 perm (0 °C) kilogram per pascal second square meter [kg/ (Pa • s • m^)] 5.721 35 perm (23 °C) kilogram per pascal second square meter [kg/(Pa • s • m^)] 5.745 25 perm inch (0 °C) kilogram per pascal second meter [kg/ (Pa • s • m)] 1.453 22 perm inch (23 °C) kilogram per pascal second meter [kg/(Pa • s • m)] 1.459 29 E-l-01 E-l-04 E-01 E-f-01 E-02 E-l-02 E-01 E + 02 E + 03 E + 03 E-07 E-l-00 E-03 E-l-00 E-l-00 E-01 E-l-01 E + 00 E-13 E-11 E-11 E-12 E-12 POWER erg per second (erg/s) watt foot pound-force per hour (ft • Ibf/h) watt foot pound-force per minute (ft • Ibf/min) watt foot pound-force per second (ft • Ibf/s) watt horsepower (550 ft • Ibf/s) watt horsepower (boiler) watt horsepower (electric) watt horsepower (metric) watt horsepower (U.K.) watt horsepower (water) watt W) 1.0 W) 3.766 161 W) 2.259 697 W) 1.355 818 W) 7.456 999 W) 9.809 50 W) 7.46 W) 7.354 988 W) 7.4570 W) 7.460 43 PRESSURE or STRESS (FORCE DIVIDED BY AREA) atmosphere, standard (atm) pascal (Pa) 1.013 25 atmosphere, standard (atm) kilopascal (kPa) 1.013 25 atmosphere, technical (at)^" pascal (Pa) 9.806 65 atmosphere, technical (at)^° kilopascal (kPa) 9.806 65 bar (bar) pascal (Pa) 1.0 bar (bar) kilopascal (kPa) 1.0 E-07 E-04 E-02 E-l-00 E-l-02 E + 03 E+02 E + 02 E + 02 E + 02 E+OS E+02 E+04 E+Ol E+OS E+02 64 Guide for the Use of the International System of Units (SI) To convert from to Multiply by centimeter of mercury (0 °C) " pascal (Pa) 1.333 22 E + 03 centimeter of mercuiy (0 °C) kilopascal (kPa) 1 .333 22 E + 00 centimeter of mercuiy, conventional (cmHg)^^. . .pascal (Pa) 1.333 224 E + 03 centimeter of mercuiy, conventional (cmHg)^^. . .kilopascal (kPa) 1.333 224 E + 00 centimeter of water (4 °C) pascal (Pa) 9.806 38 E + 01 centimeter of water, conventional (cmH20)^^ . .pascal (Pa) 9.806 65 E+01 dyne per square centimeter (dyn/cm^) pascal (Pa) 1,0 E— 01 foot of mercury, conventional (ftHg)^ pascal (Pa) 4.063 666 E + 04 foot of mercury, conventional (ftHg) kilopascal (kPa) 4.063 666 E + 01 foot of water (39.2 T) pascal (Pa) 2.988 98 E + 03 foot of water (39.2 "F)^^ kilopascal (kPa) 2.988 98 E + 00 foot of water, conventional (ftH20)^3 pascal (Pa) 2.989 067 E + 03 foot of water, conventional (ftH20)^^ kilopascal (kPa) 2.989 067 E + 00 gram-force per square centimeter (gf/cm^) pascal (Pa) 9.806 65 E+01 inch of mercuiy (32 °F) pascal (Pa) 3.386 38 E + 03 inch of mercuiy (32 °F) kilopascal (kPa) 3.386 38 E + 00 inch of mercury (60 T) pascal (Pa) 3.376 85 E + 03 inch of mercury (60 °F)l^ kilopascal (kPa) 3.376 85 E + 00 inch of mercury, conventional (inHg)^^ pascal (Pa) 3.386 389 E + 03 inch of mercury, conventional (inHg)^^ kilopascal (kPa) 3.386 389 E + 00 inch of water (39.2 °F) pascal (Pa) 2.490 82 E + 02 inch of water (60 °F) pascal (Pa) 2.4884 E + 02 inch of water, conventional (inH20) pascal (Pa) 2.490 889 E + 02 kilogram-force per square centimeter (kgf/cm^) pascal (Pa) 9.806 65 E+04 kilogram-force per square centimeter (kgf/cm^) kilopascal (kPa) 9.806 65 E+01 kilogram-force per square meter (kgf/m^) pascal (Pa) 9.806 65 E+00 kilogram-force per square millimeter (kgf/mm^) pascal (Pa) 9.806 65 E+06 kilogram-force per square millimeter (kgf/mm^) megapascal (MPa) 9.806 65 E+00 kip per square inch (ksi) (kip/in^) pascal (Pa) 6.894 757 E + 06 kip per square inch (ksi) (kip/in^) kilopascal (kPa) 6.894 757 E + 03 millibar (mbar) pascal (Pa) 1.0 E+02 millibar (mbar) kilopascal (kPa) 1.0 E-01 millimeter of mercuiy, conventional (mmHg) . . . pascal (Pa) 1.333 224 E + 02 millimeter of water, conventional (mmH20)^ pascal (Pa) 9.806 65 E+00 poundal per square foot pascal (Pa) 1.488 164 E + 00 pound-force per square foot (Ibf/ft^) pascal (Pa) 4.788 026 E + 01 pound-force per square inch (psi) (Ibf/in^) pascal (Pa) 6.894 757 E + 03 pound-force per square inch (psi) (Ibf/in^) kilopascal (kPa) 6.894 757 E + 00 psi (pound-force per square inch) (Ibf/in^) pascal (Pa) 6.894 757 E + 03 psi (pound-force per square inch) (Ibf/in^) kilopascal (kPa) 6.894 757 E + 00 torr (Torr) pascal (Pa) 1.333 224 E + 02 RADIOLOGY curie (Ci) becquerel (Bq) 3.7 E+10 rad (absorbed dose) (rad) gray (Gy) 1.0 E—02 rem (rem) sievert (Sv) 1.0 E-02 roentgen (R) coulomb per kilogram (C/kg) 2.58 E-04 SPEED (see VELOCITY) STRESS (see PRESSURE) 65 Guide for the Use of the International System of Units (SI) To convert from to Multiply by TEMPERATURE degree Celsius (°C) kelvin (K) r/K = //°C + 273.15 degree centigrade degree Celsius (°C) //°C » </deg. cent. degree Fahrenheit (°F) degree Celsius ("C) //°C = (//"F - 32)/1.8 degree Fahrenheit (°F) kelvin (K) T/K = (//°F + 4S9.67)/1.8 degree Rankine (°R) kelvin (K) T/K= (r/''R)/1.8 kelvin (K) degree Celsius (°C) trc =T/K- 273.15 TEMPERATURE INTERVAL degree Celsius (°C) kelvin (K) 1.0 E+00 degree centrigrade^^ degree Celsius (°C) 1.0 E + 00 degree Fahrenheit (°F) degree Celsius (°C) 5.555 556 E-01 degree Fahrenheit (°F) kelvin (K) 5.555 556 E-01 degree Rankine (°R) kelvin (K) 5.555 556 E-01 TIME day (d) second (s) 8.64 E+04 day (sidereal) second (s) 8.616 409 E + 04 hour (h) second (s) 3.6 E+03 hour (sidereal) second (s) 3.590 170 E + 03 minute (min) second (s) 6.0 E+01 minute (sidereal) second (s) 5.983 617 E + 01 second (sidereal) second (s) 9.972 696 E-01 shake second (s) 1.0 E-08 shake nanosecond (ns) 1.0 E+01 year (365 days) second (s) 3.1536 E+07 year (sidereal) second (s) 3.155 815 E + 07 year (tropical) second (s) 3. 155 693 E + 07 TORQUE (see MOMENT OF FORCE) VELOCITY (includes SPEED) foot per hour (ft/h) meter per second (m/s) 8.466 667 E-05 foot per minute (ft/min) meter per second (m/s) 5.08 E-03 foot per second (ft/s) meter per second (m/s) 3.048 E-01 inch per second (in/s) meter per second (m/s) 2.54 E-02 kilometer per hour (km/h) meter per second (m/s) 2.777 778 E-01 knot (nautical mile per hour) meter per second (m/s) 5.144 444 E-01 mile per hour (mi/h) meter per second (m/s) 4.4704 E— 01 mile per hour (mi/h) kilometer per hour (km/h) 1.609 344 E+00 mile per minute (mi/min) meter per second (m/s) 2.682 24 E+01 mile per second (mi/s) meter per second (m/s) 1.609 344 E+03 revolution per minute (rpm) (r/min) radian per second (rad/s) 1.047 198 E-01 rpm (revolution per minute) (r/min) radian per second (rad/s) 1.047 198 E-01 VISCOSITY, DYNAMIC centipoise (cP) pascal second (Pa • s) 1.0 E—03 poise (P) pascal second (Pa • s) 1.0 E—01 poundal second per square foot pascal second (Pa • s) 1.488 164 E + 00 pound-force second per square foot (Ibf • s/ft^) pascal second (Pa • s) 4.788 026 E + 01 pound-force second per square inch (Ibf • s/in^) pascal second (Pa • s) 6.894 757 E + 03 pound per foot hour [lb/(ft • h)] pascal second (Pa • s) 4.133 789 E-04 pound per foot second [lb/(ft • s)] pascal second (Pa • s) 1.488 164 E + 00 rhe reciprocal pascal second [(Pa-s)"'] 1.0 E+01 slug per foot second [slug/(ft • s)] pascal second (Pa • s) 4.788 026 E + 01 66 Guide for the Use of the International System of Units (SI) To convert from to Multiply by VISCOSITY, KINEMATIC centistokes (cSt) meter squared per second (m^/s) 1.0 square foot per second (ft^/s) meter squared per second (m^/s) 9.290 304 stokes (St) meter squared per second (m^/s) 1.0 VOLUME (includes CAPACITY) acre-foot (based on U.S. survey foot)^ cubic meter barrel for petroleum, 42 gallons (U.S.). . .cubic meter barrel for petroleum, 42 gallons (U.S.). . .liter (L) . . . bushel (U.S.) (bu) cubic meter bushel (U.S.) (bu) liter (L)... cord (128 ft^) cubic meter cubic foot (ft') cubic meter cubic inch (in' )^' cubic meter cubic mile (mi') cubic meter cubic yard (yd') cubic meter cup (U.S.) cubic meter cup (U.S.) liter (L) cup (U.S.) milliliter (mL) fluid ounce (U.S.) (fl oz) cubic meter fluid ounce (U.S.) (fl oz) milliliter (mL) 2 gallon [Canadian and U.K. (Imperial)] (gal) cubic meter gallon [Canadian and U.K. (Imperial)] (gal) liter (L) gallon (U.S.) (gal) cubic meter gallon (U.S.) (gal) liter (L) . . . gill [Canadian and U.K. (Imperial)] (gi) cubic meter gill [Canadian and U.K. (Imperial)] (gi) liter (L) gill (U.S.) (gi) cubic meter gill (U.S.) (gi) liter (L)... liter (L)^" cubic meter ounce [Canadian and U.K. fluid (Imperial)] (fl oz) .cubic meter (m') 2.841 306 ounce [Canadian and U.K. fluid (Imperial)] (fl oz) milliliter (mL) , 2 ounce (U.S. fluid) (fl oz) cubic meter (m') 2 ounce (U.S. fluid) (fl oz) milliliter (mL) 2 peck (U.S.) (pk) cubic meter peck (U.S.)(pk) liter (L)..., pint (U.S. dry) (dry pt) cubic meter pint (U.S. dry) (dry pt) liter (L) . . . , pint (U.S. liquid) (liq pt) cubic meter pint (U.S. liquid) (liq pt) liter (L) . . . , quart (U.S. dry) (dry qt) cubic meter quart (U.S. dry) (dry qt) liter (L) . . . , quart (U.S. liquid) (liq qt) cubic meter quart (U.S. liquid) (liq qt) liter (L) . . . . stere (st) cubic meter tablespoon cubic meter 8, 5 5, 4, 4 1 1 9, 9. 1, 1, tablespoon milliliter (mL) 1. 4 4 2 teasf)oon cubic meter teaspoon milliliter (mL) ton, register cubic meter m'). m'). m'). m'). m'). m'). m'). m'). m') m') m'). m'). m'). m'). m'). m'). 233 489 589 873 589 873 523 907 523 907 624 556 ,831 685 .638 706 .168 182 .645 549 .365 882 .365 882 .365 882 .957 353 .957 353 ,546 09 ,546 09 .785 412 .785 412 .420 653 .420 653 .182 941 .182 941 ,0 m'). m'). m'). m'). m'). m'). m'). m') m') m'). .841 306 .957 353 .957 353 809 768 809 768 506 105 506 105 731 765 731 765 101 221 101 221 463 529 463 529 0 478 676 478 676 928 922 928 922 831 685 E-06 E-02 E-04 E + 03 E-01 E + 02 E-02 E + Ol E + 00 E-02 E-05 E + 09 E-01 E-04 E-01 E + 02 E-05 E + 01 E-03 E+00 E-03 E + 00 E-04 E-01 E-04 E-01 E-03 E-05 E + 01 E-05 E + 01 E-03 E + 00 E-04 E-01 E-04 E-01 E-03 E + 00 E-04 E-01 E+00 E-05 E + 01 E-06 E + 00 E + 00 67 Guide for the Use of the International System of Units (SI) To convert from to Multiply by VOLUME DIVIDED BY TIME (includes FLOW) cubic foot per minute (ft^/min) cubic meter per second (m'/s) 4.719 474 E — 04 cubic foot per minute (ft^/min) liter per second (L/s) 4.719 474 E-01 cubic foot per second (ft^/s) cubic meter per second (m'/s) 2.831 685 E-02 cubic inch per minute (in'/min) cubic meter per second (m'/s) 2.731 177 E-07 cubic yard per minute (yd'/min) cubic meter per second (m^/s) 1.274 258 E-02 gallon (U.S.) per day (gal/d) cubic meter per second (m^/s) 4.381 264 E-08 gallon (U.S.) per day (gal/d) liter per second (L/s) 4.381 264 E-05 gallon (U.S.) per minute (gpm) (gal/min) cubic meter per second (m'/s) 6.309 020 E-05 gallon (U.S.) per minute (gpm) (gal/min) liter per second (L/s) 6.309 020 E-02 WORK (see ENERGY) 68 Guide for the Use of the International System of Units (SI) Appendix C. Comments on the References of Appendix D — Bibliography C.1 OfTicial interpretation of the SI for the United States: 55 FR 52242-52245 The official interpretation of the International System of Units for the United States, which is the responsibility of the United States Department of Commerce, is stated in the Federal Register, Vol. 55, No. 245, p. 52242, December 20, 1990 . This Federal Register Notice is reprinted in NIST Special Publication 814 , together with the Federal Register Notice that states the metric conversion policy for Federal Agencies and the Executive Order on metric usage in Federal Government programs . C.2 Defining document for the SI: BIPM SI Brochure The defining document for the International System of Units is the Brochure published by the International Bureau of Weights and Measures (BIPM) in French, followed by an English translation . This document is revised from time to time in accordance with the decisions of the General Conference on Weights and Measures (CGPM). C.3 United States version of defining document for the SI: NIST SP 330 The United States edition of the English translation in the BIPM SI Brochure (see Sec. C.2) is published by the National Institute of Standards and Technology as NIST Special Publication 330 ; it differs from the translation in the BIPM publication in the following details: — the dot is used as the decimal marker, in keeping with recommended United States practice (see Sees. C.l and C.7); — the spelling of English-language words — for example, "meter," "liter," and "deka" are used instead of "metre," "litre," and "deca" — is in accordance with the United States Government Printing Office Style Manual , which follows Webster's Third New Interna-tional Dictionary rather than the Oxford Dictionary used in many English-speaking coun-tries. This spelling also reflects recommended United States practice (see Sees. C.l and C.7); — editorial notes regarding the use of the SI in the United States are added; — the index is moderately expanded. Inasmuch as NIST Special Publication 330 is consistent with Ref. (see Sec. C.l), SP 330 is the authoritative source document on the SI for the purposes of this Guide. C.4 ISO 1000 ISO 1000:1992 is an international consensus standard published by the International Organization for Standardization (ISO) to promote international uniformity in the technical interpretation of the actions of the CGPM as they are published by the BIPM in Ref. (see Sec. C.2). C.5 ISO 31-0 ISO 31-0:1992 — ISO 31-13:1992 constitute a series of international consensus stan-dards published by ISO to promote international uniformity in the practical use of the SI in various fields of science and technology, and in particular to standardize the symbols for various quantities and the units in which the values of these quantities are expressed. These standards are compatible with Ref. published by the BIPM (see Sec. C.2). 69 Guide for the Use of the International System of Units (SI) C.6 IEC27 lEC 27-1— lEC 27-4 constitute a series of international consensus standards pub-lished by the International Electrotechnical Commission (lEC) to promote international uniformity in the practical use of the SI in electrical technology, and in particular to standardize the symbols for various quantities used in electrotechnology and the units in which the values of these quantities are expressed. These lEC standards are also compatible with Ref. published by the BIPM (see Sec. C.2), and they are coordinated with the ISO standards cited in Sec. C.5 (Ref. ). The lEC standards should be regarded as more author-itative than the corresponding ISO standards only in connection with electrical technology. C.7 ANSI/IEEE Std 268 ANSI/IEEE Std 268-1992 is an American National Standard for Metric Practice; it is based on the International System of Units as interpreted for use in the United States (see Sees. C.l and C.3). It has been approved by a consensus of providers and consumers that includes interests in industrial organizations, government agencies, and scientific asso-ciations. This standard was developed by the Institute of Electrical and Electronics Engineers (IEEE), and approved as an American National Standard by the American National Standards Institute (ANSI)." ANSI/IEEE Std 268-1992 has been adopted for use by the United States Department of Defense (DoD) and serves as the basis of Ref. (see Sec. C.9); it is recommended as a comprehensive source of authoritative information for the practical use of the SI in the United States. (Similar documents have also been developed by other United States technical organizations; see Ref. , note 2.) C.8 Federal Register notices Important details concerning United States customary units of measurement and the interpretation of the SI for the United States are published from time to time in the Federal Register; these notices have the status of official United States Government policy. A Federal Register notice of July 1, 1959 states the values of conversion factors to be used in technical and scientific fields to obtain the values of the United States yard and pound from the SI base units for length and mass, the meter and the kilogram. These conversion factors were adopted on the basis of an agreement of English-speaking countries to reconcile small differences in the values of the inch-pound units as they were used in different parts of the world. This action did not affect the value of the yard or foot used for geodetic surveys in the United States. Thus, at that time, it became necessary to recognize on a temporary basis a small difference between United States customary units of length for "international measure" and "survey measure." A Federal Register notice of July 19, 1988 announced a tentative decision not to adopt the international foot of 0.3048 meters for surveying and mapping activities in the United States. A final decision to continue the use of the survey foot indefinitely is pending the completion of an analysis of public comments on the tentative decision; this decision will also be announced in the Federal register. Even if a final decision affirms the continued use of the survey foot in surveying and mapping services of the United States, it is significant to note that the Office of Charting and Geodetic Services of the National Ocean Service in the National Oceanic and Atmospheric Administration uses the meter exclusively for the North American Datum . The North American Datum of 1983, the most recent definition and adjustment of this information, was announced in a Federal Register notice of June 14, 1989 . ^^The American National Standards Institute, Inc. (11 West 42nd Street, New York, NY 10036) is a private sector organization that serves as a standards coordinating body, accredits standards developers that follow proce-dures sanctioned by ANSI, designates as American National Standards those standards submitted for and receiving approval, serves as the Untied States Member Body of the International Organization for Standardization (ISO), and functions as the administrator of the United States National Committee for the International Electrotechnical Commission (lEC). 70 Guide for the Use of the International System of Units (SI) The definitions of the international foot and yard and the corresponding survey units are also addressed in a Federal Register Notice published on February 3, 1975 . A Federal Register notice of July 27, 1968 provides a list of the common customary measurement units used in commerce throughout the United States, together with the con-version factors that link them with the meter and the kilogram. A recent Federal Register notice concerning the SI is a restatement of the interpre-tation of the International System for use in the United States, and it updates the corre-sponding information published in earlier notices. A Federal Register notice of January 2, 1991 removes the voluntary aspect of the conversion to the SI for Federal agencies and provides policy direction to assist Federal agencies in their transition to the use of the metric system of measurement. A Federal Register notice of July 29, 1991 provides Presidential authority and direction for the use of the metric system of measurement by Federal departments and agencies in their programs. C.9 Federal Standard 376B Federal Standard 376B was developed by the Standards and Metric Practices Subcommittee of the Metrication Operating Committee, which operates under the Inter-agency Council on Metric Policy. Specified in the Federal Standardization Handbook and issued by, and available from, the General Services Administration, Washington, DC, 20406, it is the basic Federal standard that lists preferred metric units for use throughout the Federal Government. It gives guidance on the selection of metric units required to comply with PL 94-168 (see Preface) as amended by PL 100-418 (see Preface), and with Executive Order 12770 (see Sec. C.8). The basis of Fed. Std. 376B is ANSI/IEEE Std. 268-1992 (see Sec. C.7). C.IO 1986 CODATA values of the fundamental constants The set of self-consistent recommended values of the fundamental physical constants resulting from the 1986 Committee on Data for Science and Technology (CODATA) least-squares adjustment of the constants, the most up-to-date set currently available, is given in Ref. . The next CODATA adjustment of the constants is planned for completion in 1996; some of the considerations relevant to that adjustment may be found in B. N. Taylor and E. R. Cohen, Recommended Values of the Fundamental Physical Constants: A Status Report, J. Res. Natl. Inst. Stand. Technol., Vol. 95, No. 5, p. 497 (September-October 1990). C.ll Uncertainty in measurement Reference cites two publications that describe the evaluation and expression of uncertainty in measurement based on the approach recommended by the CIPM in 1981 and which is currently being adopted worldwide. 71 Guide for the Use of the International System of Units (SI) Appendix D. Bibliography Interpretation of the SIfor the United States and Metric Conversion Policyfor FederalAgen-cies, Ed. by B. N. Taylor, Natl. Inst. Stand. Technol. Spec. Publ. 814 (U.S. Government Printing Office, Washington, DC, October 1991). Le Systeme International d'Unites, The International System of Units, 6th Edition (Bur. Intl. Folds et Mesures, Sevres, France, 1991). Note : This publication, which is commonly called the SI Brochure, consists of the official French text followed by an English translation. The International System of Units (SI) , Ed. by B. N. Taylor, Natl. Inst. Stand. Technol. Spec. Publ. 330, 1991 Edition (U.S. Government Printing Office, Washington, DC, August 1991). Note: This publication is the United States edition of the English translation in Ref. . United States Government Printing Office Style Manual (U.S. Government Printing Office, Washington, DC, 1984). SI units and recommendations for the use of their multiples and of certain other units, ISO 1000:1992 (International Organization for Standardization, Geneva, Switzerland, 1992). Notes : 1 ISO publications are available in the United States from the sales department of the American National Standards Institute (ANSI), 105-111 South State Street, Hackensack, NJ 07601. 2 See the note at the end of Ref. . The following 14 Standards, which are cited in the text in the form [6: ISO 31-. . . ], are published by the International Organization for Standardization (ISO) Geneva, Switzerland: Quantities and units Quantities and units Quantities and units Quantities and units Quantities and units Quantities and units Quantities and units Quantities and units Quantities and units Quantities and units Quantities and units Quantities and units Quantities and units Quantities and units 72 - Part 0 General principles, ISO 31-0:1992. - Part 1 Space and time, ISO 31-1:1992. - Part 2 Periodic and related phenomena, ISO 31-2:1992. - Part 3 Mechanics, ISO 31-3:1992. - Part 4 Heat, ISO 31-4:1992. - Part 5 Electricity and magnetism, ISO 31-5:1992. - Part 6 Light and related electromagnetic radiations , ISO 31-6:1992. - Part 7 Acoustics, ISO 31-7:1992. - Part 8 Physical chemistry and molecular physics , ISO 31-8:1992. - Part 9 Atomic and nuclear physics, ISO 31-9:1992. — Part 10: Nuclear reactions and ionizing radiations , ISO 31-10:1992. — Part II: Mathematical signs and symbols for use in physical sciences and technology, ISO 31-11:1992. - Part 12: Characteristic numbers, ISO 31-12:1992. - Part 13: Solid state physics, ISO 31-13:1992. Guide for the Use of the International System of Units (SI) Note: ISO 31-0:1992 - ISO 31-13:1992 and ISO 1000:1992 are reprinted in the ISO Standards Handbook Quantities and units (International Organization for Standardization, Geneva, Switzerland, 1993). (The availability of ISO publica-tions in the United States is discussed in Ref. , note 1.) The following four standards, which are cited in the text in the form [7: lEC 27-. . . ], are published by the International Electrotechnical Commission (lEC), Geneva, Switzerland. Note: lEC publications are available in the United States from the American National Standards Institute — see Ref. , note 1. Letter symbols to be used in electrical technology, Part 1: General, lEC 27-1 (1991). Letter symbols to be used in electrical technology. Part 2: Telecommunications and elec-tronics, lEC 27-2 (1972) [including lEC 27-2A (1975) and lEC 27-2B (1980), first and second supplements to lEC 27-2 (1972)]. Letter symbols to be used in electrical technology, Part 3: Logarithmic quantities and units, lEC 27-3 (1989). Letter symbols to be used in electrical technology, Part 4: Symbols for quantities to be used for rotating electrical machines, lEC 27-4 (1985). American National Standard for Metric Practice, ANSI/IEEE Std 268-1992 (Institute of Electrical and Electronics Engineers, New York, NY, October 1992). Notes : 1 IEEE publications are available from the Institute of Electrical and Electronics Engineers, Service Center, 445 Hoes Lane, P.O. Box 1331 Piscataway, NJ 08855-1331. 2 A number of similar standards for metric practice are published by United States technical organizations. They include: Standard Practice for Use of the International System of Units (SI) (The Modern-ized Metric System), E 380-93 (American Society for Testing and Materials, Philadelphia, PA, 1993). Note: ASTM publications are available from the Customer Service Department, American Society for Testing and Materials, 1916 Race Street, Philadelphia, PA 19103. Rules for SAE Use of SI (Metric) Units, SAE J916 MAY 91 (Society of Automotive Engineers, Warrendale, PA, May 1991). Note: SAE publications are available from the Society of Automotive Engineers, 400 Commonwealth Drive, Warrendale, PA 15096. 3 The Canadian Standards Association, 178 Rexdale Boulevard, Rexdale (Toronto), Ontario, Canada, M9W 1R3, publishes CAN/CSA-Z234. 1-89, Canadian Metric Practice Guide , a Canadian National Standard. It is similar in scope to ANSI/IEEE Std 268-1992 (Ref. ). 4 A joint ASTM-IEEE effort is currently underway to consolidate ANSI/IEEE Std. 268-1992 and ASTM E 380-93 into a single ANSI standard. 5 The application of the SI to physical chemistry is discussed in Quantities, Units and Symbols in Physical Chemistry, prepared by I. Mills, T. CvitaS, K. Homann, N. Kallay, and K. Kuchitsu, Second Edition (International Union of Pure and Applied Chemistry, Blackwell Scientific Publications, Oxford, 1993). 73 Guide for the Use of the International System of Units (SI) Federal Register, Vol. 24, No. 128, p. 5348, July 1, 1959. Federal Register, Vol. 53, No. 138, p. 27213, July 19, 1988. Federal Register, Vol. 42, No. 57, p. 8847, March 24, 1977. Federal Register, Vol. 54, No. 113, p. 25318, June 14, 1989. Federal Register, Vol. 40, No. 23, p. 5954, February 3, 1975. Federal Register, Vol. 33, No. 146, p. 10755, July 27, 1968. Federal Register, Vol. 55, No. 245, p. 52242, December 20, 1990. Federal Register, Vol. 56, No. 1, p. 160, January 2, 1991. Federal Register, Vol. 56, No. 145, p. 35801, July 29, 1991. Preferred Metric Units for General Use by the Federal Government, Federal Standard 376B (General Services Administration, Washington, DC, 1993). Radiation Quantities and Units, ICRU Report 33, 1980; and Quantities and Units in Radiation Protection and Dosimetry, ICRU Report 51, 1993 (International Commission on Radiation Units and Measurements, 7910 Woodmont Avenue, Bethesda, MD, 20814). E. R. Cohen and B. N. Taylor, The 1986 adjustment of the fundamental physical con-stants. Rev. Mod. Phys., Vol. 59, No. 4, p. 1121 (October, 1987). The term combined standard uncertainty used in the footnotes to Table 7 of this Guide, and the related terms expanded uncertainty and relative expanded uncertainty used in some of the examples of Sec. 7.10.3, are discussed in ISO, Guide to the Expression of Uncertainty in Measurement (International Organization for Standardization, Geneva, Switzerland, 1993); and in B. N. Taylor and C. E. Kuyatt, Guidelines for Evaluating and Expressing the Uncertainty of NIST Measurement Results , Natl. Inst. Stand. Technol. Spec. Publ. 1297, 1994 Edition (U.S. Government Printing Office, Washington, DC, September 1994). A. J. Thor, Secretary, International Organization for Standardization (ISO) Technical Committee (TC) 12, Quantities, units, symbols, conversion factors (private communica-tion, 1993). ISO/TC 12 is responsible for the ISO International Standards cited in Refs. and . 74 f wfw f Technical Publications Periodical Journal of Research of the National Institute of Standards and Technology—Reports NIST research and development in those disciplines of the physical and engineering sciences in which the Institute is active. These include physics, chennistry, engineering, mathematics, and computer sciences. Papers cover a broad range of subjects, with major emphasis on measurement methodology and the basic technology underlying standardization. Also included from time to time are survey articles on topics closely related to the Institute's technical and scientific programs. Issued six times a year. Nonperiodicals Monographs—Major contributions to the technical literature on various subjects related to the Institute's scientific and technical activities. Handbooks—Recommended codes of engineering and industrial practice (including safety codes) developed in cooperation with interested industries, professional organizations, and regulatory bodies. Special Publications—Include proceedings of conferences sponsored by NIST NIST annual reports, and other special publications appropriate to this grouping such as wall charts, pocket cards, and bibliographies. National Standard Reference Data Series—Provides quantitative data on the physical and chemical properties of materials, compiled from the world's literature and critically evaluated. Developed under a worldwide program coordinated by NIST under the authority of the National Standard Data Act (Public Law 90-396). NOTE: The Journal of Physical and Chemical Reference Data (JPCRD) is published bi-monthly for NIST by the American Chemical Society (ACS) and the American Institute of Physics (AlP). Subscriptions, reprints, and supplements are available from ACS, 1 155 Sixteenth St., NW, Washington, DC 20056. Building Science Series—Disseminates technical information developed at the Institute on building materials, components, systems, and whole structures. The series presents research results, test methods, and performance criteria related to the structural and environmental functions and the durability and safety characteristics of building elements and systems. Technical Notes—Studies or reports which are complete in themselves but restrictive in their treatment of a subject. Analogous to monographs but not so comprehensive in scope or definitive in treatment of the subject area. Often serve as a vehicle for final reports of work performed at NIST under the sponsorship of other government agencies. Voluntary Product Standards—Developed under procedures published by the Department of Commerce in Part 1 0, Title 1 5, of the Code of Federal Regulations. The standards establish nationally recognized requirements for products, and provide all concerned interests with a basis for common understanding of the characteristics of the products. NIST administers this program in support of the efforts of private-sector standardizing organizations. Order the following NIST publications—FIPS and NISTIRs—from the National Technical Information Service, Springfield, VA 22161. Federal Information Processing Standards Publications (FIPS PUB)—Publications in this series collectively constitute the Federal Information Processing Standards Register. The Register serves as the official source of information in the Federal Government regarding standards issued by NIST pursuant to the Federal Property and Administrative Services Act of 1949 as amended. Public Law 89-306 (79 Stat. 1127), and as implemented by Executive Order 11717 (38 FR 12315, dated May 11, 1973) and Part 6 of Title 15 CFR (Code of Federal Regulations). NIST Interagency Reports (NISTIR)—A special series of interim or final reports on work performed by NIST for outside sponsors (both government and nongovernment). In general, initial distribution is handled by the sponsor; public distribution is by the National Technical Information Service, Springfield, VA 22161, in paper copy or microfiche form. . o> o o c o • O JC W -s • to CO 3 Z O 8 s CO .E c (A a. CD £ O Q-
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https://pmc.ncbi.nlm.nih.gov/articles/PMC6732798/
Mid-upper arm circumference cut-offs for screening thinness and severe thinness in Indian adolescent girls aged 10–19 years in field settings - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide View on publisher site Download PDF Add to Collections Cite Permalink PERMALINK Copy As a library, NLM provides access to scientific literature. 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Learn more: PMC Disclaimer | PMC Copyright Notice Public Health Nutr . 2019 Aug;22(12):2189–2199. doi: 10.1017/S1368980019000594 Search in PMC Search in PubMed View in NLM Catalog Add to search Mid-upper arm circumference cut-offs for screening thinness and severe thinness in Indian adolescent girls aged 10–19 years in field settings Vani Sethi Vani Sethi 1 UNICEF, India Country Office, UNICEF House, 73 Lodi Estate, New Delhi – 110002, India Find articles by Vani Sethi 1,, Neha Gupta Neha Gupta 2 Department of Mathematical Demography and Statistics, International Institute of Population Sciences, Mumbai, India Find articles by Neha Gupta 2, Sarang Pedgaonkar Sarang Pedgaonkar 2 Department of Mathematical Demography and Statistics, International Institute of Population Sciences, Mumbai, India Find articles by Sarang Pedgaonkar 2, Abhishek Saraswat Abhishek Saraswat 2 Department of Mathematical Demography and Statistics, International Institute of Population Sciences, Mumbai, India Find articles by Abhishek Saraswat 2, Konsam Dinachandra Singh Konsam Dinachandra Singh 2 Department of Mathematical Demography and Statistics, International Institute of Population Sciences, Mumbai, India Find articles by Konsam Dinachandra Singh 2, Hifz Ur Rahman Hifz Ur Rahman 2 Department of Mathematical Demography and Statistics, International Institute of Population Sciences, Mumbai, India Find articles by Hifz Ur Rahman 2, Arjan de Wagt Arjan de Wagt 1 UNICEF, India Country Office, UNICEF House, 73 Lodi Estate, New Delhi – 110002, India Find articles by Arjan de Wagt 1, Sayeed Unisa Sayeed Unisa 2 Department of Mathematical Demography and Statistics, International Institute of Population Sciences, Mumbai, India Find articles by Sayeed Unisa 2 Author information Article notes Copyright and License information 1 UNICEF, India Country Office, UNICEF House, 73 Lodi Estate, New Delhi – 110002, India 2 Department of Mathematical Demography and Statistics, International Institute of Population Sciences, Mumbai, India Corresponding author: Email vsethi@unicef.org Received 2018 Jun 28; Revised 2019 Jan 18; Accepted 2019 Feb 1. © The Authors 2019 This is an Open Access article, distributed under the terms of the Creative Commons Attribution licence ( which permits unrestricted re-use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC6732798 PMID: 31111811 Abstract Objective: (i) To assess diagnostic accuracy of mid-upper arm circumference (MUAC) for screening thinness and severe thinness in Indian adolescent girls aged 10–14 and 15–19 years compared with BMI-for-age Z-score (BAZ) <−2 and <−3 as the gold standard and (ii) to identify appropriate MUAC cut-offs for screening thinness and severe thinness in Indian girls aged 10–14 and 15–19 years. Design: Cross-sectional, conducted October 2016–April 2017. Setting: Four tribal blocks of two eastern India states, Chhattisgarh and Odisha. Participants: Girls (n 4628) aged 10–19 years. Measurements included height, weight and MUAC to calculate BAZ. Standard diagnostic accuracy tests, receiver–operating characteristic curves and Youden index helped arrive at MUAC cut-offs at BAZ < −2 and <−3, as gold standard. Results: Mean MUAC and BMI correlation was positive (0·78, P = 0·001 and r 2 = 0·61). Among 10–14 years, MUAC cut-off corresponding to BAZ < −2 and BAZ < −3 was ≤19·4 and ≤18·9 cm. Among 15–19 years, corresponding values were ≤21·6 and ≤20·7 cm. For both BAZ < −2 and BAZ < −3, specificity was higher in 15–19 v. 10–14 years. State-wise variations existed. MUAC cut-offs ranged from 17·7 cm (10 years) to 22·5 cm (19 years) for BAZ < −2, and from 17·0 cm (10 years) to 21·5 cm (19 years) for BAZ < −3. Single-age area under the curve range was 0·82–0·97. Conclusions: Study provides a case for use of year-wise and sex-wise context-specific MUAC-cut-offs for screening thinness/severe thinness in adolescents, rather than one MUAC cut-off across 10–19 years, depending on purpose and logistic constraints. Keywords: Anthropometry, BMI, Undernutrition, Mid-upper arm circumference, Adolescent The WHO recommends a BMI-for-age Z-score (BAZ) of <−2 and <−3, respectively, for classifying thinness and severe thinness in adolescents aged 10–19 years(1). In field settings which are remote or where availability of resources (skilled manpower and financial resources) is a challenge, mid-upper arm circumference (MUAC) has also been used as a field-friendly alternative for screening thinness in adolescents (Table 1)(2–9). Table 1. Sample country-specific mid-upper arm circumference (MUAC) cut-offs for adolescents for screening severe thinness | Author, year | Country | Age (years) | MUAC cut-off (cm) | Results | Reference | :--- --- --- | | WHO, 2011 | – | – | <16·0 | Admission criteria for therapeutic feeding | 2 | | National CMAM guidelines, Sudan, 2017 | Sudan, South Sudan | 10–18 | <16·0 | Admission criteria for therapeutic feeding | 3 | | National CMAM guidelines, Somalia, 2010 | Somalia | 10–18 | <16·0 | Admission criteria for therapeutic feeding | 4 | | National CMAM guidelines, Ethiopia, 2007 | Ethiopia | 6 months–18 years | <11·0 | Admission criteria for therapeutic feeding | 5 | | Bahwere, 2017 | Syria | 10–14 | <16·0 | Admission criteria for therapeutic feeding | 6 | | 15–17 | <20·0 | | | | ≥18 | <22·0 | | Martin et al., 2009 | Western Australia | 12–17 | <20·0 | For initiation for special nutrition care | 7 | | MoHFW, 2017 | India | 10–18 | <16·0 | For nutrition support | 8 | | FANTA, 2018 | From a sample of countries | 10–14 | < 16·0 | <16·0 cm: SAM | 9 | | ≥16·0 to <18·5 cm: MAM | | ≥18·5 cm: normal | | FANTA, 2018 | DRC | 10–14 | <16·0 | For detecting SAM | 9 | | FANTA, 2018 | Malawi | 10–11 | <16·0 | For detecting SAM | 9 | | 12–14 | <16·0 | | 15–18 | <18·5 | | FANTA, 2018 | Mozambique | 11–14 | <16·0 | For detecting SAM | 9 | | 15–18 | <21·0 | | FANTA, 2018 | Namibia | 10–14 | <16·0 | For detecting SAM | 9 | | FANTA, 2018 | Tanzania | 10–14 | <16·0 | For detecting SAM | 9 | | ≥15 | <18·5 | | FANTA, 2018 | Uganda | 10–14 | <16·0 | For detecting SAM | 9 | | 15–17 | <18·5 | | FANTA, 2018 | Zambia | 10–14 | <16·0 | For detecting SAM | 9 | | 15–17 | <18·5 | Open in a new tab CMAM, community management of acute malnutrition; MoHFW, Ministry of Health and Family Welfare, FANTA, Food and Nutrition Technical Assistance Project; DRC, Democratic Republic of Congo; SAM, severe acute malnutrition; MAM, moderate acute malnutrition. According to the WHO’s IMAI (Integrated Management of Adolescent and Adult Illness) hospital care for adolescents and adults guidelines for the management of illnesses with limited resources, adolescents can be classified as having severe malnutrition (‘severe undernutrition’) if they have MUAC < 160 mm or MUAC = 161–185 mm plus one of the following: pitting oedema up to the knees on both sides, or cannot stand, or sunken eyes(2). Several countries have adopted their country-specific cut-offs (Table 1). India’s National Nutrition Support for Tuberculosis programme uses an MUAC cut-off of <160 mm for classifying severe thinness in adolescents and determining those eligible for inpatient nutrition rehabilitation and support(8). India is home to approximately 120 million adolescent girls, or about 20 % of the world’s population of adolescent girls aged 10–19 years(10). According to an analysis of a 2015–16 nationally representative survey, 10·6 % of Indian unmarried girls aged 15–19 years are thin (BAZ < −2) and 1·8 % are severely thin (BAZ < −3; International Institute for Population Sciences and UNICEF India, unpublished results). India’s national health programmes have provision for routine nutrition screening for adolescents through utilizing schools and outreach adolescent village health days(11). The WHO (2007) BAZ charts(1) have not been adopted in adolescent health programmes. The nutrition assessment is conducted annually by public health workers, using clinical signs, biochemical indicators (Hb) and anthropometry (weight, height). In resource-poor field settings, availability of standardized well-calibrated equipment to measure weight and height and calculation of BMI/BAZ by field workers in the absence of field charts and/or calculators are often challenging. Once identified as severely thin, at present, there is also no policy for provision of nutrition support owing to a lack of dialogue on types and modalities of nutrition support to severely thin adolescents in school/community settings. Use of MUAC for screening severely thin adolescents followed by medical nutrition therapy is restricted to hospital settings in tuberculosis wards only(8). In order to use MUAC as a field-friendly proxy for BAZ, there is a need to assess its diagnostic accuracy compared with BAZ, ascertain whether this diagnostic accuracy differs by age or age bracket (early or late adolescents) and initiate a dialogue on modalities and opportunities to provide nutrition support to those severely thin adolescents, as mere screening will not serve the purpose if an intervention is not in place. We found four India-based studies that compared MUAC measurements in adolescents aged 10–19 years with BMI/BAZ as gold standard (Table 2)(12–15), which showed mean MUAC and mean BMI have a correlation (r) of 0·35–0·822 (P< 0·001). Out of four studies, one study that calculated BAZ reported that MUAC < 18·5 cm and MUAC < 16 cm were in agreement with BAZ of <−2 and <−3, respectively(15). However, none of these studies were in community settings or identified year-wise MUAC cut-offs for thinness and severe thinness for Indian adolescent girls aged 10–19 years. Table 2. Studies on the correlation between mid-upper arm circumference (MUAC) and BMI/BMI-for-age Z-score (BAZ) in India | Author, year | Location | Age (years) | Sample size | Results | Reference | :--- --- --- | | Dasgupta et al., 2010 | Kolkata | 10–19 | 194 | Burden of thinness: 60·30 % (MUAC < 5th percentile) and 47·9 % (BMI < 5th percentile). | 12 | | Strong correlation between measurements of MUAC and BMI (r = 0·822, se = 0·035, 95 % CI 0·8045, 0·8395, P = 0·000000, r 2 = 0·74). MUAC as a marker was 94·6 % sensitive and 71·2 % specific | | De Kankana, 2016 | Paschim Medinipur | 10–19 | 1009 | Burden of thinness: 40 % (MUAC) and 24 % (BMI) | 13 | | BMI and MUAC showed significant correlation (r = 0·350, P = 0·000) | | MUAC < 22·9 cm showed: SN = 53·4 %, SP = 79·9 %, PPV = 80·0 % and NPV = 53·6 % | | Jeyakumar et al., 2013 | Pune, Maharashtra | 16–18 | 565 | Burden of thinness: 5·0 % (MUAC < 5th percentile) and 4·8 % (BMI < 5th percentile) | 14 | | BMI highly correlated with MUAC (r = 0·593) | | MUAC as a screening tool showed SN = 28·57 % and SP = 96·46 % | | Gupta et al., 2016 | Delhi and Haryana | 10–19 | 4183 | Power of association (r) between MUAC (cm) and BAZ was 0·68 (P< 0·001). MUAC < 18·5 cm was in agreement with BAZ < −2 (κ = 0·34; 95 % CI 0·31, 0·38). With BAZ < −2 as gold standard, SN and SP of MUAC < 18·5 cm was 73 and 79 %, respectively | 15 | | | | | | MUAC < 16 cm was compared with BAZ < −3 as gold standard and showed agreement (κ = 0·38; 95 % CI 0·31, 0·38). SN and SP of MUAC < 16 cm was 62·6 and 97·3 %, respectively | | Open in a new tab SN, sensitivity; SP, specificity; PPV, positive predictive value; NPV, negative predictive value. Thus, the present study was conducted in two eastern India states (Chhattisgarh and Odisha) to assess the diagnostic accuracy of MUAC compared with BAZ and to provide MUAC cut-offs for screening thinness and severe thinness in adolescent girls, by year and age group (10–14 and 15–19 years). We consider this as a first step towards opening a dialogue on the need for simplified year-wise MUAC field charts for screening of severe thinness in Indian adolescents and thereafter arriving at protocols for management of severe thinness in Indian adolescents as part of adolescent health programmes. Methodology Data collection We conducted a cross-sectional study on adolescent girls aged 10–19 years in tribal-dominated districts of Odisha (Koraput and Angul districts) and Chhattisgarh (Bastar district) between October 2016 and April 2017. Overall, in these states, ~8 % of adolescent girls aged 15–19 years have BAZ < −2 (7·8 % in Chhattisgarh and 7·7 % in Odisha) as per an analysis of the Fourth Round of the National Family Health Survey (International Institute for Population Sciences and UNICEF India, unpublished results). Our cross-sectional study was a part of a baseline survey for the evaluation of Swabhimaan (meaning ‘self-respect’ or ‘self-pride’), an integrated multisectoral strategy to improve girls’ and women’s nutrition before conception, during pregnancy and after birth. Swabhimaan is a collaboration between UNICEF and the State Rural Livelihood Missions of Chhattisgarh and Odisha. A description of the Swabhimaan strategy and its impact evaluation design is available elsewhere(16). The location and sample size for our cross-sectional study were guided by the design of the Swabhimaan strategy, outcome indicators and the change envisaged. Location was four tribal-dominated blocks: Bastar and Bakawand (Bastar district), Pallara (Angul district) and Koraput Sadar (Koraput district). The eligible participants were non-married and non-pregnant adolescent girls aged 10–19 years residing in the study areas. The sample size of eligible participants to be covered was estimated at 3256 adolescent girls (2196 in Chhattisgarh and 1060 in Odisha). Temporary residents, i.e. those adolescents present in homes at the time of the house-to-house census but who said they would migrate within two months of the census, were excluded. Married and/or pregnant adolescent girls, included in other surveys, were also excluded as all variables considered in the current analysis were not available for them. The target sample was collected using simple random sampling. Although we estimated a sample size of 3256 adolescent girls as eligible participants, we interviewed a total of 4648 eligible participants (2921 in Chhattisgarh and 1727 in Odisha). Of the total of 4648 eligible participants interviewed, fourteen adolescents were not given anthropometric measurements and six adolescents’ BAZ was flagged, therefore 4628 adolescent girls were included in the analysis. Data collection was carried out by thirty investigators, who were supervised by six supervisors. Paper-based method for data collection was used. Written informed consent was obtained from all participants over 18 years of age. For those under 18 years, written consent was taken from their parent(s) or guardian(s) and verbal consent was also taken from the respondents. Utmost confidentiality of information and anonymity of respondents was ensured to prevent linking to any individual. All interviewers participated in a standardization exercise in which they took repeated measurements of ten adolescents in three teams of ten interviewers each. Each interviewer took two height, weight and MUAC measurements for ten participants. We then compared these with supervisors’ (n 6) measurements, as well as within teams. The technical error of measurement(17) for weight was 0·99 and for height was 0·95. Supervisors conducted back-checks for 10 % of interviews. Ethical approval was obtained from the Institutional Ethics Committees of the All India Institute of Medical Sciences in Chhattisgarh and Odisha. The impact evaluation has been registered with the Registry for International Development Impact Evaluations (RIDIE-STUDY-ID-58261b2f46876)(16). A common interview on sociodemographic and household characteristics was administered to all adolescents’ guardians using a pre-tested, structured, bilingual questionnaire (English and Hindi in Chhattisgarh; English and Odia in Odisha). The adolescent girls’ interviews covered sociodemographics and anthropometric measurements (weight, height and MUAC). Anthropometric measurements were conducted using standard techniques(18). Weight to the nearest 0·1 kg was recorded using a SECA electronic weighing scale with minimal clothing. Height was taken barefoot to the nearest 0·1 cm using a stadiometer. MUAC was measured to the nearest 0·1 cm with a non-stretchable measuring tape (procured from UNICEF supply department). The tape was placed firmly but gently on the arm to avoid compression of soft tissue. The weighing scales and stadiometer were calibrated on a weekly basis prior to data collection with standard weights (1, 2 and 5 kg) and a metre rod (100 cm). The mean se of measurements for height, weight and MUAC across all the data collection teams were insignificant and ranged between 0·001 and 0·025 (95 % CI −0·004, 0·042; P< 0·10). The non-response rate was negligible (Chhattisgarh 0·3 %; Odisha 0·4 %). BAZ was calculated using the WHO reference (Stata macro) and classified as <−3 (severe thinness) and <−2 (thinness)(1). Statistical methods Primary data were entered in CS-Pro version 4·1. Descriptive statistics were generated using the statistical software package IBM SPSS Statistics version 20. MUAC cut-offs for screening thinness and severe thinness and year-wise MUAC cut-offs, as well as those for younger (10–14 years) and older (15–19 years) adolescents, were determined using BAZ < −2 and BAZ < −3, respectively, as the gold standard. Diagnostic accuracy of MUAC compared with BAZ was assessed using sensitivity (SN), specificity (SP), negative predictive value (NPV) and positive predictive value (PPV), whose values were calculated using the proportion of true positives (TP), false positives (FP), true negatives (TN) and false negatives (FN) using a 2 × 2 table as shown below: | Thin according to MUAC cut-offs generated in the paper | Thin according to BAZ | :---: | | Yes | No | | Yes | TP | FP | | No | FN | TN | Open in a new tab SN measures the percentage of true positives (thin/severely thin adolescents) calculated as TP/(TP + FN); SP measures the percentage of true negatives (not thin/severely thin adolescents) calculated as TN/(TN + FP); NPV tells us how likely an adolescent is to not be thin if the test is negative, calculated as TN/(TN + FN); and PPV tells us how likely an adolescent is to be thin if the test is positive, calculated as TP/(TP + FP). FP (%) is calculated as FP/(FP + TP) and FN (%) is calculated as FN/(TN + FN). Unlike SN and SP, the NPV and PPV are largely dependent on disease prevalence in an examined population. Values of SN, SP, PPV and NPV were calculated for MUAC cut-offs points against BAZ < −2 and BAZ < −3. Receiver-operating characteristic (ROC) curve analysis was undertaken to determine the area under the curve (AUC), along with its 95 % CI, to establish the optimal cut-off values of MUAC to identify thinness and severe thinness. The shape of the ROC curve and the AUC determine how high is the discriminative power of a test. The AUC can have any value between 0 and 1 and it is a good indicator of the goodness of the test. The categories used to summarize accuracy of AUC in ROC curve analysis are as follows: excellent (0·9–1·0), good (0·8–0·9), fair (0·7–0·8), poor (0·6–0·7) and fail (0·5–0·1). A test with AUC ≥ 0·85 is considered an accurate test(19). Although AUC gives an overall picture of the behaviour of a diagnostic test across all cut-off values, there remains a necessity to ascertain the specific cut-off value that could be used for screening and for this purpose Youden’s index (YI) is used(20). The YI is equivalent to the AUC subtended by a single operating point in the ROC curve(21). We calculated YI by deducting 1 from the sum of the test’s SN and SP expressed not as a percentage but as part of a whole number: (SN + SP) − 1. It is one of the oldest measures for diagnostic accuracy, being used for the evaluation of overall discriminative power of a diagnostic procedure and for comparison of this test with other tests(22). For a test with poor diagnostic accuracy, YI equals 0, and in a perfect test YI equals 1. The YI was calculated using MedCalc software version 17.9.7. Single-age MUAC cut-offs as well as MUAC cut-offs for the groups of younger (10–14 years) and older adolescents (15–19 years) at BAZ < −2 and BAZ < −3 were determined on the basis of the highest corresponding value of YI. Results The analytical sample comprised 4628 adolescents (2910 were from Chhattisgarh and 1718 from Odisha). Their mean age was 14·26 (sd 2·55) years. Sociodemographic characteristics of the participants are described in Table 3. Only 68 % reported being currently enrolled in school. Nearly all (97·2 %) participants belonged to the Hindu religion. Caste-wise, 94 % of participants were from backward castes. Table 3. Sociodemographic characteristics of the sample of adolescent girls aged 10–19 years (n 4628) from two eastern India states (Chhattisgarh and Odisha), October 2016–April 2017 | Characteristic | Pooled (n 4628) | --- | | Chhattisgarh (n 2920) | Odisha (n 1728) | | | MUAC (cm) | BMI (kg/m 2) | BAZ | | n | % | n | % | n | % | Mean | sd | Mean | sd | Mean | sd | | Age (years) | | 10 | 186 | 6·4 | 147 | 8·6 | 333 | 7·2 | 18·7 | 2·42 | 15·3 | 2·98 | −1·0 | 1·28 | | 11 | 307 | 10·6 | 181 | 10·5 | 488 | 10·5 | 19·2 | 2·19 | 15·4 | 2·09 | −1·1 | 1·14 | | 12 | 355 | 12·2 | 179 | 10·4 | 534 | 11·5 | 20·3 | 2·34 | 16·1 | 2·12 | −1·1 | 1·11 | | 13 | 371 | 12·8 | 175 | 10·2 | 546 | 11·8 | 21·5 | 2·67 | 17·1 | 2·16 | −0·9 | 1·06 | | 14 | 394 | 13·5 | 197 | 11·5 | 591 | 12·8 | 22·3 | 2·25 | 17·8 | 2·33 | −0·9 | 1·04 | | 15 | 325 | 11·2 | 190 | 11·1 | 515 | 11·1 | 22·9 | 2·12 | 18·2 | 2·07 | −0·9 | 0·91 | | 16 | 323 | 11·1 | 193 | 11·2 | 516 | 11·2 | 23·5 | 2·09 | 18·8 | 2·12 | −0·8 | 0·85 | | 17 | 308 | 10·6 | 196 | 11·4 | 504 | 10·9 | 23·6 | 2·23 | 18·7 | 2·01 | −1·0 | 0·88 | | 18 | 294 | 10·1 | 178 | 10·4 | 472 | 10·2 | 23·8 | 1·98 | 18·9 | 2·06 | −0·9 | 0·81 | | 19 | 47 | 1·6 | 82 | 4·8 | 129 | 2·8 | 23·9 | 2·36 | 18·6 | 2·15 | −1·1 | 0·90 | | Age group (years) | | 10–14 | 1613 | 55·4 | 879 | 51·2 | 2492 | 53·9 | 20·6 | 2·71 | 16·5 | 2·49 | −1·0 | 1·12 | | 15–19 | 1297 | 44·6 | 839 | 48·8 | 2136 | 46·2 | 23·5 | 2·15 | 18·6 | 2·09 | −0·9 | 0·87 | | Religion | | Hindu | 2859 | 98·3 | 1637 | 95·3 | 4496 | 97·2 | – | – | – | | Non-Hindu | 51 | 1·8 | 81 | 4·7 | 132 | 2·9 | – | – | – | | Caste | | Scheduled caste | 70 | 2·4 | 266 | 15·5 | 336 | 7·3 | – | – | – | | Scheduled tribe | 1895 | 65·1 | 928 | 54·0 | 2823 | 61·0 | – | – | – | | Other backward caste | 811 | 27·9 | 389 | 22·6 | 1200 | 25·9 | – | – | – | | General | 134 | 4·6 | 135 | 7·9 | 269 | 5·8 | – | – | – | | Currently attending school | | Yes | 2179 | 74·9 | 2179 | 74·9 | 3165 | 68·4 | – | – | – | | No | 688 | 23·6 | 688 | 23·6 | 1267 | 27·4 | – | – | – | | Never gone to school | 43 | 1·5 | 43 | 1·5 | 196 | 4·2 | – | – | – | Open in a new tab MUAC, mid-upper arm circumference; BAZ, BMI-for-age Z-score. Mean BMI of the adolescent girls aged 10–14 and 15–19 years was 16·5 (sd 2·5) and 18·6 (sd 2·1) kg/m 2, respectively. Corresponding figures for mean BAZ were −1·0 (sd 1·12) and −0·9 (sd 0·87), and those for MUAC were 20·6 (sd 2·7) cm and 23·5 (sd 2·2) cm. Correlation between BMI and mid-upper arm circumference Figure1 shows the correlation between BMI and MUAC in the pooled data as well as for Chhattisgarh and Odisha, separately. A significant positive correlation was found between measurements of MUAC and BMI (r = 0·78, P = 0·001, r 2 = 0·61), whereas in state-wise correlation, higher correlation was obtained for Chhattisgarh (r = 0·82, P = 0·001) than for Odisha (r = 0·77, P = 0·001). In the age-wise correlation, the lowest correlation was found at 10 years of age (r = 0·41, P = 0·001) and the highest at 15 years (r = 0·81, P = 0·001), with all other correlations lying between 0·63 and 0·79 (P = 0·001). Fig. 1. Open in a new tab Scatter plots showing the correlation between BMI and mid-upper arm circumference (MUAC), overall and by state, in adolescent girls aged 10–19 years (n 4628) from two eastern India states, October 2016–April 2017: (a) pooled (correlation coefficient (r) = 0·78, P = 0·001); (b) Chhattisgarh (r = 0·82, P = 0·001); (c) Odisha (r = 0·77, P = 0·001) Diagnostic accuracy of mid-upper arm circumference cut-offs for thinness and severe thinness Table 4 summarizes the MUAC cut-offs for BAZ < −2. The optimal MUAC cut-off to detect thinness among girls aged 10–14 years was ≤19·4 cm (SN = 84·0 %, SP = 75·4 %) and among older adolescents it was ≤21·6 cm (SN = 81·4 %, SP = 87·1 %). Single-age MUAC cut-offs in adolescent girls ranged between 17·7 cm (10 years) and 22·5 cm (19 years) for identifying thinness (BAZ < −2). Overall, the optimal MUAC cut-off for screening thinness among adolescent girls aged 10–19 years was ≤20·9 cm (YI = 0·56, SN = 83·3–86·1 %, SP = 70·3–72·8 %, AUC = 0·85–0·86, P = 0·001). Comparing state-wise, the optimal MUAC cut-off for screening thinness among adolescent girls aged 10–19 years was ≤20·9 cm in Chhattisgarh and ≤21·3 cm in Odisha (data not shown). The SN and SP of all the single-age MUAC cut-offs ranged from 70 to 90 % depending on the true positives and true negatives that the age-specific cut-offs could identify. The AUC ranged from 0·84 to 0·94 (P = 0·001), signifying good/excellent diagnostic power of the identified single-age cut-offs. At all single-age MUAC cut-offs, the NPV was much higher than the PPV, signifying that the MUAC cut-offs were able to correctly exclude adolescents without thinness. We also did an additional analysis where we obtained the single-age and grouped cut-offs for moderate thinness (BAZ < −2 and ≥−3; data not shown). There was no significant difference obtained in the cut-offs for thinness and moderate thinness. Table 4. Diagnostic test accuracy measures for varying cut-offs of mid-upper arm circumference (MUAC) for predicting thinness (BMI-for-age Z-score < −2 as gold standard) among adolescent girls aged 10–19 years (n 4628) from two eastern India states (Chhattisgarh and Odisha), October 2016–April 2017 | Age (years) | MUAC cut-off (cm) | SN (%) | SP (%) | YI | FN (%) | FP (%) | PPV (%) | NPV (%) | AUC | 95 % CI | --- --- --- --- --- | 10 | ≤17·7 | 87·3 | 71·1 | 0·58 | 4·0 | 58·6 | 41·4 | 96·0 | 0·84 | 0·79, 0·87 | | 11 | ≤18·1 | 81·6 | 81·2 | 0·62 | 5·4 | 48·0 | 52·0 | 94·6 | 0·87 | 0·84, 0·90 | | 12 | ≤19·0 | 87·3 | 86·0 | 0·73 | 4·1 | 35·8 | 64·2 | 95·9 | 0·92 | 0·89, 0·94 | | 13 | ≤20·1 | 90·0 | 82·4 | 0·72 | 1·7 | 57·2 | 42·8 | 98·3 | 0·92 | 0·89, 0·94 | | 14 | ≤20·6 | 77·9 | 85·4 | 0·63 | 3·7 | 55·5 | 44·5 | 96·2 | 0·89 | 0·87, 0·92 | | 15 | ≤20·8 | 71·6 | 93·9 | 0·65 | 3·3 | 42·5 | 57·5 | 96·6 | 0·92 | 0·89, 0·94 | | 16 | ≤21·6 | 79·4 | 88·0 | 0·67 | 1·8 | 64·7 | 35·3 | 98·1 | 0·91 | 0·89, 0·94 | | 17 | ≤21·7 | 91·3 | 88·2 | 0·79 | 0·9 | 56·3 | 43·7 | 99·1 | 0·94 | 0·92, 0·96 | | 18 | ≤22·3 | 83·6 | 81·3 | 0·65 | 2·2 | 65·8 | 34·1 | 97·8 | 0·89 | 0·86, 0·91 | | 19 | ≤22·5 | 72·2 | 82·3 | 0·54 | 5·2 | 60·7 | 39·3 | 94·8 | 0·86 | 0·79, 0·92 | | 10–14 | ≤19·4 | 84·0 | 75·4 | 0·59 | 4·2 | 58·7 | 41·3 | 95·8 | 0·86 | 0·84, 0·87 | | 15–19 | ≤21·6 | 81·4 | 87·1 | 0·68 | 2·2 | 60·0 | 40·0 | 97·8 | 0·91 | 0·89, 0·92 | Open in a new tab SN, sensitivity; SP, specificity; YI, Youden index; FN, false negative; FP, false positive; PPV, positive predictive value; NPV, negative predictive value; AUC, area under the (receiver-operating characteristic) curve. Table 5 summarizes the MUAC cut-offs for BAZ < −3. Single-age MUAC cut-offs in adolescent girls ranged between 17·0 cm (10 years) and 21·5 cm (19 years) for severe thinness (BAZ < −3). The optimal MUAC cut-off to detect severe thinness among girls aged 10–14 years was ≤18·9 cm (SN = 84·6 %, SP = 74·1 %) and among older adolescents it was ≤20·7 cm (SN = 86·1 %, SP = 93·1 %). The SN and SP of the single-age MUAC cut-offs ranged from 70 to 100 %. The AUC were in the range of 0·84–0·97, signifying good/excellent diagnostic power of the identified single-age cut-offs. The PPV of the single-age MUAC cut-offs ranged from 8 to 30 %, while the NPV was >99 % for all ages. Thus, the PPV was lower for single-age MUAC cut-offs with BAZ < −3 as the gold standard compared with MUAC cut-offs with BAZ < −2, while the NPV was higher. The optimal MUAC cut-off for screening severe thinness among adolescent girls aged 10–19 years was ≤19·5 cm (YI = 0·52–0·69, SN = 0·63–87·7 %, SP = 81·3–88·9 %, AUC = 0·83–0·90, P = 0·001). Comparing state-wise, the optimal MUAC cut-off for screening severe thinness among adolescent girls aged 10–19 years was ≤19·4 cm in Chhattisgarh and ≤18·5 cm in Odisha (data not shown). Table 5. Diagnostic test accuracy measures for varying cut-offs of mid-upper arm circumference (MUAC) for predicting severe thinness (BMI-for-age Z-score < −3 as gold standard) among adolescent girls aged 10–19 years (n 4628) from two eastern India states (Chhattisgarh and Odisha), October 2016–April 2017 | Age (years) | MUAC cut-off (cm) | SN (%) | SP (%) | YI | FN (%) | FP (%) | PPV (%) | NPV (%) | AUC | 95 % CI | --- --- --- --- --- | 10 | ≤17·0 | 87·5 | 79·6 | 0·67 | 0·3 | 90·4 | 9·6 | 99·7 | 0·87 | 0·83, 0·91 | | 11 | ≤17·6 | 94·7 | 79·3 | 0·74 | 0·2 | 84·4 | 15·6 | 99·8 | 0·90 | 0·87, 0·92 | | 12 | ≤18·2 | 91·3 | 84·5 | 0·75 | 0·4 | 79·0 | 21·0 | 99·6 | 0·94 | 0·91, 0·96 | | 13 | ≤18·8 | 85·0 | 92·8 | 0·77 | 0·6 | 69·0 | 31·0 | 99·4 | 0·91 | 0·88, 0·93 | | 14 | ≤20·3 | 71·4 | 84·6 | 0·56 | 1·2 | 85·4 | 14·6 | 98·8 | 0·84 | 0·81, 0·87 | | 15 | ≤20·3 | 100·0 | 93·4 | 0·93 | 0·0 | 71·7 | 28·3 | 100·0 | 0·97 | 0·95, 0·98 | | 16 | NA | NA | NA | NA | NA | NA | NA | NA | NA | NA | | 17 | ≤20·7 | 76·9 | 94·1 | 0·71 | 0·6 | 74·3 | 25·7 | 99·4 | 0·91 | 0·89, 0·94 | | 18 | ≤21·1 | 75·0 | 92·5 | 0·67 | 0·2 | 92·1 | 7·9 | 99·8 | 0·97 | 0·95, 0·98 | | 19 | ≤21·5 | 80·0 | 87·3 | 0·67 | 0·9 | 80·0 | 20·0 | 99·1 | 0·84 | 0·76, 0·90 | | 10–14 | ≤18·9 | 84·6 | 74·1 | 0·58 | 0·7 | 89·1 | 10·9 | 99·3 | 0·86 | 0·83, 0·86 | | 15–19 | ≤20·7 | 86·1 | 93·1 | 0·79 | 0·2 | 82·2 | 17·8 | 99·8 | 0·93 | 0·92, 0·94 | Open in a new tab SN, sensitivity; SP, specificity; YI, Youden index; FN, false negative; FP, false positive; PPV, positive predictive value; NPV, negative predictive value; AUC, area under the (receiver-operating characteristic) curve. NA = cases insufficient to estimate a reliable MUAC cut-off for severe thinness. Each ROC curve shows the trade-off between sensitivity and specificity. The AUC for thinness among girls aged 10–14 and 15–19 years was 0·86 and 0·91, respectively (P< 0·0001; Fig.2). Similarly, the AUC for severe thinness among girls aged 10–14 and 15–19 years was 0·86 and 0·93, respectively (P< 0·0001; Fig.3). The AUC values were highly significant and the curves were closer to top left corner, indicating high accuracy of the test to detect thinness and severe thinness. Fig. 2. Open in a new tab Receiver-operating characteristic curves () of mid-upper arm circumference to identify thinness (BMI Z-score < −2), by age group, among adolescent girls aged 10–19 years (n 4628) from two eastern India states (Chhattisgarh and Odisha), October 2016–April 2017: (a) 10–14 years (sensitivity (SN) = 84·5 %; specificity (SP) = 75·1 %; criterion = ≤19·45 cm; area under the curve (AUC) = 0·863; P< 0·001); (b) 15–19 years (SN = 82·0 %; SP = 87·0 %; criterion = ≤21·65 cm; AUC = 0·911; P< 0·001). () represent the 95 % CI and () represents the line of no discrimination Fig. 3. Open in a new tab Receiver-operating characteristic curves () of mid-upper arm circumference to identify severe thinness (BMI Z-score < −3), by age group, among adolescent girls aged 10–19 years (n 4628) in two eastern India states (Chhattisgarh and Odisha), October 2016–April 2017: (a) 10–14 years (sensitivity (SN) = 85·6 %; specificity (SP) = 74·1 %; criterion = ≤18·95 cm; area under the curve (AUC) = 0·860; P< 0·001); (b) 15–19 years (SN = 86·1 %; SP = 93·1 %; criterion = ≤20·70 cm; AUC = 0·934; P< 0·001). () represent the 95 % CI and () represents the line of no discrimination Prevalence of thinness and severe thinness The burden of thinness (BMI < −2) and severe thinness (BMI < −3) among adolescent girls aged 10–14 years was 17·1 and 3·6 %, respectively. Corresponding figures for adolescent girls aged 15–19 years were 9·6 and 1·7 % (Table 6). By BAZ, the prevalence of thinness and severe thinness was highest among adolescents aged 12 years which counted for 22·6 and 4·5 %, respectively; while the lowest prevalence was found in age 16 years where 7·5 % were thin. According to MUAC, the burden of thinness among adolescent girls aged 10–14 years (MUAC ≤ 19·4 cm) and 15–19 years (MUAC ≤ 21·6 cm) was 41·3 and 40·0 %, respectively; whereas the burden of severe thinness among adolescent girls aged 10–14 years (MUAC ≤ 18·9 cm) and 15–19 years (MUAC ≤ 20·7 cm) was 11·0 and 17·7 %, respectively. The burden of thinness and severe thinness was higher when assessed with MUAC compared with BAZ in both age groups (Table 6). Table 6. The burden of thinness and severe thinness based on mid-upper arm circumference (MUAC) and BMI-for-age Z-score (BAZ) among adolescent girls aged 10–19 years (n 4628) from two eastern India states (Chhattisgarh and Odisha), October 2016–April 2017 | | Thinness (BAZ < −2) | Severe thinness (BAZ < −3) | --- | Age (years) | MUAC cut-off (cm) | MUAC-based prevalence (%) | BAZ-based prevalence (%) | MUAC cut-off (cm) | MUAC-based prevalence (%) | BAZ-based prevalence (%) | | 10 | ≤17·7 | 41·3 | 19·2 | ≤17·0 | 9·6 | 2·4 | | 11 | ≤18·1 | 52·8 | 20·3 | ≤17·6 | 16·4 | 4·1 | | 12 | ≤19·0 | 66·2 | 22·6 | ≤18·2 | 22·0 | 4·5 | | 13 | ≤20·1 | 43·0 | 12·9 | ≤18·8 | 29·3 | 3·6 | | 14 | ≤20·6 | 44·4 | 13·3 | ≤20·3 | 15·0 | 3·5 | | 15 | ≤20·8 | 57·5 | 10·3 | ≤20·3 | 30·4 | 2·9 | | 16 | ≤21·6 | 35·1 | 7·5 | NA | NA† | 0·2 | | 17 | ≤21·7 | 47·8 | 9·5 | ≤20·7 | 11·5 | 2·6 | | 18 | ≤22·3 | 34·2 | 10·6 | ≤21·1 | 7·5 | 0·8 | | 19 | ≤22·5 | 65·0 | 13·7 | ≤21·5 | 20·0 | 3·8 | | 10–14 | ≤19·4 | 41·3 | 17·1 | ≤18·9 | 11·0 | 3·6 | | 15–19 | ≤21·6 | 40·0 | 9·6 | ≤20·7 | 17·7 | 1·7 | Open in a new tab NA = cases insufficient to estimate a reliable MUAC cut-off for severe thinness. † NA = cases insufficient to estimate a reliable MUAC cut-off for severe thinness. Hence, prevalence cannot be estimated. Discussion The present study has several important inferences pertaining to prevalence of thinness and severe thinness in adolescents aged 10–19 years using BAZ and MUAC as well as year-wise MUAC cut-offs for screening thinness and severe thinness. First, the burden of thinness (BAZ < −2) among adolescent girls using identified the MUAC cut-off was fourfold higher compared with BAZ (e.g. MUAC-based 40 %, BAZ-based 9·6 %, at 15–19 years). There is limited information from India on use of MUAC for detection of thinness among adolescent girls. Community-based studies reported a burden of thinness among adolescent girls between 24 and 48 % (according to BMI) and between 40 and 60 % (according to MUAC)(12,13). However, one study reported lower burden of thinness as 4·8 % (BMI) and 5·0 % (MUAC), respectively(16). Second, a significant correlation was found between BAZ and MUAC measurements (r = 0·78, P = 0·001, r 2 = 0·61). Our study results are comparable with all three studies which also reported a significant correlation between BMI and MUAC measurements(12,13,16). The study by Dasgupta et al. documented that MUAC is highly sensitive (97 %) and specific (71 %) in the screening of malnourishment among adolescents (10–19 years)(13). De Kankana(13) found that the mean MUAC was 21·7 cm and implied that BMI and MUAC have higher and significant correlation. MUAC can be a useful and efficient index for the screening of thinness, generally assessed from BMI. Gupta et al.(15) showed that the power of association between MUAC and BAZ < −2 was considerably high, and that MUAC can be a gold standard in assessment of thinness. They also found that MUAC < 18·5 cm was in agreement with BAZ < −2 and MUAC cut-off of <16·5 cm with BAZ < −3. A survey conducted among 565 adolescent girls (16–18 years) from Pune, Maharashtra, warranted that MUAC had high specificity but low level of sensitivity(14). Third, our results show that the MUAC cut-off to detect thinness and severe thinness in young adolescent girls (10–14 years) was ≤19·4 and ≤18·9 cm, respectively, and it was ≤21·6 and ≤20·7 cm for late adolescent girls (15–19 years). The specificity of the MUAC cut-off (≤20·7 cm) among 15–19-year-old adolescents for BAZ < −3 (93·1 %) was higher than the specificity of the MUAC cut-off (≤21·6 cm) for BAZ < −2 (87·1 %), signifying that MUAC can be more specific for diagnosis of thinness in a severely malnourished population. For both BAZ < −2 and BAZ < −3, the specificity was higher for older adolescents (15–19 years) compared with younger adolescents (10–14 years; 75·4 and 87·1 % for BAZ < −2; 74·1 and 93·1 % for BAZ < −3), with sensitivity being similar for both. Hence, the MUAC cut-off is more specific for diagnosis of thinness in older adolescents, probably because they are fully grown, so there is less variation in the cut-offs. The results of the present study were corroborated by previous studies. Gupta et al.(15) found that with BAZ < −2 as the gold standard, MUAC cut-off of <18·5 cm had a sensitivity and specificity of 73 and 79 %, respectively, for detecting thinness in 10–19-year-old adolescent girls. In the same study, higher specificity (97·3 %) was obtained with BAZ < −3 as the gold standard and MUAC cut-off of <16 cm(15). Another study reported similar findings on 10–19-year-old adolescent girls, with MUAC cut-off of <22·9 cm showing a sensitivity of 53·4 % and a much higher specificity of 79·9 %(13). A survey conducted with 565 adolescent girls (16–18 years) from Pune, Maharashtra, also warranted that MUAC had high specificity (96·5 %) but low level of sensitivity (28·5 %)(14). Thus, results indicate that MUAC has higher specificity and lower sensitivity, particularly for detecting thinness below the BAZ cut-off of −3. It is, therefore, important to relate MUAC to age and sexual maturity of individual girls for a meaningful identification of thinness(23), which was missed in the present study. From a programmatic perspective, it is not feasible to define a single MUAC cut-off to identify nutritionally at-risk adolescent girls between the ages of 10 and 19 years. Fourth, we found that with BAZ < −2 as the gold standard, the NPV was much higher than the PPV at all single-age MUAC cut-offs (PPV = 35–65 %, NPV > 95 %) signifying that the MUAC cut-offs were able to correctly exclude adolescents without thinness. Similar findings (NPV (>99 %) > PPV (10–30 %) at all cut-offs) were obtained with BAZ < −3 as the gold standard. Thus, MUAC cut-offs were able to correctly exclude adolescents who were not thin according to BAZ. We also found that PPV was lower at MUAC cut-offs with BAZ < −3 as the gold standard compared with MUAC cut-offs with BAZ < −2, while the NPV was higher. This is because, unlike sensitivity and specificity, predictive values are largely dependent on disease prevalence in the examined population. The prevalence of thinness would be much higher with the BAZ < −2 cut-off than with the BAZ < −3 cut-off(17). Hence the PPV was much higher at BAZ < −2 than at BAZ < −3. However, PPV and NPV from one study should not be transferred to some other setting with a different prevalence of the disease in the population. Hence it is better to use SN and SP indicators for comparing results across different populations. There was no difference in the MUAC cut-offs, SN and SP values before and after adjusting for the outliers in the data. Although our study identified both single-age MUAC cut-offs and cut-offs for younger (10–14 years) and older (15–19 years) adolescent girls, it is preferable to use single-age MUAC cut-offs due to wide variations in the cut-offs. For instance, the optimal MUAC cut-off to detect thinness (BAZ < −2) among adolescent girls aged 10–14 years was found to be ≤19·4 cm, while the cut-off at 10 and 14 years was ≤17·7 and ≤20·6 cm, respectively, having a wide variation of 2·9 cm. Thus, the chances of classifying a 10-year-old adolescent as thin is higher using the cut-off of 19·4 cm as compared with the single-age cut-off of 17·7 cm. Hence, it is important that single-age MUAC cut-offs are used in field settings for identifying thinness among adolescents. However, in cases where the age of the adolescent is not known, the grouped cut-offs for younger and older adolescents can be used: MUAC ≤ 19·4 cm (10–14 years) and MUAC ≤ 21·6 cm (15–19 years). There are other considerations about what should be the nutrition support provided to adolescent girls with severe thinness, such as an extra meal or linkage with social protection, for which the assessment measure may be used to determine the response plan. This was outside the scope of the present paper, however it requires deliberation. While MUAC is particularly useful in remote areas, where it is not possible to carry the weighing machine or stadiometer over long distances and so calculation of BMI/BAZ is not feasible (hence MUAC tapes become handy in such places), there are some aspects of MUAC that should be kept in mind. First, MUAC changes substantially with age during adolescence, especially at the younger ages when growth patterns and physical maturity differ largely between individual girls. As a result, different cut-offs must be used for adolescents of different ages. This requires an accurate age for each survey subject in order to judge whether she falls above or below an age-specific cut-off. Second, despite the convenience and ease of measurement of MUAC, it requires careful training and supervision in order to prevent wrapping the measuring tape too tightly or too loosely, which results in an erroneous estimate and some degree of observer variability. Strengths and limitations The present community-based study was conducted on a reasonable sample size with good quality control and monitoring. The sample was drawn systematically from rural deprived areas where programmes and interventions to improve nutrition are intended. However, the following limitations merit consideration. Unlike previous studies, the current study has not defined one single MUAC cut-off to identify nutritionally at-risk adolescent girls between the ages of 10 and 19 years; rather single-age MUAC cut-offs were established for detection of thinness among adolescent girls. The study falls short in evaluating any health consequences and, therefore, cannot compare one method over another. The sample being selected from only two geographical areas is another limiting factor of the study, limiting generalizability of the results. Since we found differences in MUAC between the two states, we need large-scale data to arrive at national year-wise cut-offs for MUAC in Indian adolescents (both girls and boys) for appropriate interventions in emergency situations, field settings and outpatient therapeutic clinics as MUAC field charts. Moreover, since the sample was majorly drawn from poverty pockets in India, the analysis cannot be used to derive at MUAC cut-offs for overweight and obesity. Finally, the present study did not cover adolescent boys aged 10–19 years. Conclusion To conclude, MUAC cut-off points with good predictive ability to detect thinness among adolescent girls aged 10–14 years (young adolescents) and 15–19 years (late adolescents) were ≤19·4 and ≤21·6 cm, respectively. The age-wise, sex-wise and context-specific MUAC cut-offs should be preferred in place of one MUAC cut-off across 10–19 years, based upon several considerations including purpose, burden and logistic resources. Availability of MUAC field charts by year, for adolescent boys and girls, by context/region, will prove useful in settings where BAZ/BMI is not available. However, prior evidence from large-scale representative survey data wherein MUAC and BAZ measurements have been taken for adolescents is needed to prove/disprove MUAC diagnostic accuracy and suitability for the specific region/context compared with BAZ. Acknowledgements Acknowledgements: The authors would like to thank Dr Neha Sareen and Dr H.P.S. Sachdev for review and editing support of preliminary drafts. Financial support: This work was supported by UNICEF, India. V.S. and A.d.W. are employed by UNICEF. However, the views expressed are those of authors and not necessarily those of UNICEF. Conflict of interest: None. Author contribution: V.S. and S.U. conceptualized and supervised the study. N.G. wrote the manuscript and revised it critically. S.P., A.S., K.D.S. and H.U.R. coordinated the data collection and entry and carried out data analysis. A.d.W. and V.S. reviewed and supported additional analyses and revised the manuscript critically. All authors reviewed the manuscript and have approved the final version. Ethics of human subject participation: This study was conducted according to the guidelines laid down in the Declaration of Helsinki and all procedures involving human subjects were approved by the Institutional Ethics Committee of the All India Institute of Medical Sciences (AIIMS), Chhattisgarh and AIIMS, Odisha. Written informed consent was obtained from all subjects. 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https://www.youtube.com/watch?v=ypK-eJNwpvU
Buoyant force on a hot air balloon problem: find the volume of hot air required to lift a balloon. Zak's Lab 15800 subscribers 46 likes Description 6164 views Posted: 22 Dec 2022 00:00 The physics of hot air balloons: In this hot air balloon problem, we use physics to learn how a hot air balloon works. 🧠 Access full flipped physics courses with video lectures and examples at We are given the mass of the balloon material, equipment and passengers, and we want to find the volume of hot air required to lift a balloon. Given the density of the cold outside air and given the density of the hot air inside the balloon, we plan to use Archimedes' principle to compute the buoyant force on a hot air balloon then calculate the minimum volume of hot air required to balance the force of gravity. We work the problem twice: once using the approximation that the buoyant force on the passengers and equipment is negligible, then we relax this assumption to make a small correction to the result for the balloon volume. 00:53 A couple important points before we start the physics: first, the mass of the hot air inside the balloon cannot be ignored. It actually turns out that the vast majority of the mass of the balloon system belongs to the hot air! Second, we get a short reminder of the definition of density rho=mass/volume, which means we can quickly compute masses by writing mass=densityvolume. These calculations happen repeatedly in Archimedes' principle problems. 01:28 Balance forces to start out the solution: we have gravity pulling down on the balloon system and buoyant force pointing up, and to get the balloon to float we use the minimum volume of hot air to balance the vertical forces on the balloon. The buoyant force is approximated by pretending the balloon has all the volume (ignoring the buoyant force on passengers and equipment for this first approximation). So the buoyant force is equal to the weight of displaced cold air, or density of cold airVolume of the balloong. The weight comes from two parts: the mass of the equipment and passengers and the weight of the hot air in the balloon. The weight of the hot air is density of hot airVolumeg. With the forces balanced, we can solve for the volume of the balloon. 05:05 The mass of the hot air is not negligible! Now that we have the volume of the hot air balloon, we can actually compute the mass of the hot air by using density of hot airVolumeg, and it turns out that the hot air accounts for the vast majority of the mass of the balloon system! 05:40 Relax the assumption that the buoyant force on the passengers and equipment is negligible: we improve on our first answer by repeating the calculation, but this time we include the buoyant force on the passengers and equipment. We are given the density of these parts and we know the mass, so we use the density formula solved for volume=mass/density to get the volume of the passengers and equipment. This allows us to use Archimedes' principle to get the buoyant force on the passengers and equipment, and we put this small correction into the force equation. Solving for V once again, we find a small but noticeable correction to the volume of hot air required to lift the balloon. 4 comments Transcript: The physics of hot air balloons: In this hot air balloon problem, we use physics to learn how a hot air balloon works. in this video we're looking at a classic buoyancy problem involving a hot air balloon so we're given the total mass of the balloon material the basket the passengers the equipment we're told that's 500 Kilograms we're given the density of the cold outside air has 1.24 kilograms per cubic meter and the density of the hot air inside the balloon has 0.95 kilograms per cubic meter in part A we're going to compute the volume of hot air required to lift this balloon and we're making an approximation in part A we're ignoring the buoyant force on the passengers and equipment so the passengers and Equipment they're displacing a certain amount of the surrounding air that should generate a small buoyant force but it turns out to be a very small percentage compared to the buoyant force exerted on the balloon itself in Part B we're going to relax that assumption and get a slightly more precise answer to the question one extremely important point about this problem A couple important points before we start the physics: first, the mass of the hot air inside the balloon cannot be ignored. It actually turns out that the vast majority of the mass of the balloon system belongs to the hot air! Second, we get a short reminder of the definition of density rho=mass/volume, which means we can quickly compute masses by writing mass=densityvolume. These calculations happen repeatedly in Archimedes' principle problems. is that we can't ignore the mass of the hot air itself that is part of the balloon system gravity pulls down on that hot air and it actually turns out to be the majority of the weight of the balloon system and one more note before we get started we call it density has the definition mass per volume mass per unit volume if we turn that around to solve for Mass we get mass is density times volume and that's a little calculation we're going to use repeatedly in the solution of this problem it's just nice to be prepared to do that quickly so to start out the physics of Balance forces to start out the solution: we have gravity pulling down on the balloon system and buoyant force pointing up, and to get the balloon to float we use the minimum volume of hot air to balance the vertical forces on the balloon. The buoyant force is approximated by pretending the balloon has all the volume (ignoring the buoyant force on passengers and equipment for this first approximation). So the buoyant force is equal to the weight of displaced cold air, or density of cold airVolume of the balloong. The weight comes from two parts: the mass of the equipment and passengers and the weight of the hot air in the balloon. The weight of the hot air is density of hot airVolumeg. With the forces balanced, we can solve for the volume of the balloon. the solution here we'll put in our Force vectors and that downward Force Vector that's the force of gravity on all the mass contained in the balloon system including all the equipment and passengers and the air inside the balloon and I'll just call that mg for now we'll split that up into a couple smaller parts the upward Force that's the buoyant force on the balloon and remember Archimedes principle says that's going to be equal to the weight of the displaced fluid in other words the weight of the cold outer air that's been pushed out of the way by the volume in this balloon so the basic physics of the problem starts out by saying if we're going to get this thing to float these forces should balance each other and the buoyant force will be equal to the total weight but then we have to get into the details of each of these two forces we'll start with the buoyant force again that's going to be equal to the weight of displaced fluid in other words what's the weight of all the cold air pushed out of the way by the volume of this balloon well that's going to be the mass of the cold air pushed out of the way multiplied by G and the mass of the cold air pushed out of the way again we use this density formula solved for Mass very frequently the mass of that cold air is going to be its density so I'll write it as real cold times the volume of cold air pushed out of the way I'm going to call that V that's the volume of the balloon times G okay so those are the details of the left side of our equation on the right side we had mg where m is the total mass of the entire balloon system including all the equipment passengers and balloon material and so on and the mass of the hot air inside of it so that little m is the mass of the equipment I'll call it meq that was 500 Kilograms that's given to us plus the mass of the hot air and the mass of that hot air again is given by the density of the hot air multiplied by the volume of the hot air which again we're using a capital V for the volume of the balloon okay so then we plug in these details back into our original Force equation and this is our buoyant force it's the density of the cold air times the volume of the balloon times G and then on the right hand side the total mass is going to be m for the equipment plus the density of the hot air times the volume of the balloon all multiplied by G now the G's cancel out of this and I notice there's two places where the unknown volume of the balloon occurs once over here on the left hand side where it was involved in the calculation of buoyant force once over here on the right hand side where it was involved in the calculation of the mass of the hot air so we subtract that term on the right hand side moving it over to the left and we'll factor a v out on the left this leaves me with rho called the density of the cold air minus rho Hots the density of the hot air multiplied by V and on the right hand side we're left with M for the equipment and we'll go ahead and solve symbolically for this volume before we plug numbers in I get mass of the equipment divided by the cold air density minus hot air density and now we can plug numbers in and we get 500 Kilograms the given mass of the equipment divided by 1.24 kilograms per cubic meter the density of the cold air minus 0.95 kilograms per cubic meter the density of the hot air and when we run the numbers on this we get about 1724 cubic meters for the volume of the balloon now before we move on to Part B I wanted to The mass of the hot air is not negligible! Now that we have the volume of the hot air balloon, we can actually compute the mass of the hot air by using density of hot airVolumeg, and it turns out that the hot air accounts for the vast majority of the mass of the balloon system! reiterate how important it was for us to include the mass of the hot air itself let's go ahead and calculate how much that was so the mass of the hot air again just using this density formula turned around is going to be the density of the hot air times the volume of the hot air so that's 0.95 kilograms per cubic meter for the density multiplied by this 1724 cubic meter volume for the balloon and this turns out to be about 1 640 kilograms it's actually the majority of the mass far more than the actual equipment so let's improve on Relax the assumption that the buoyant force on the passengers and equipment is negligible: we improve on our first answer by repeating the calculation, but this time we include the buoyant force on the passengers and equipment. We are given the density of these parts and we know the mass, so we use the density formula solved for volume=mass/density to get the volume of the passengers and equipment. This allows us to use Archimedes' principle to get the buoyant force on the passengers and equipment, and we put this small correction into the force equation. Solving for V once again, we find a small but noticeable correction to the volume of hot air required to lift the balloon. our answer by including the buoyant force on the passengers and Equipment themselves so we were given a total mass there of 500 kilograms now we're given the average density of all the people and Equipment as 1500 kilograms per cubic meter well to find the buoyant force caused by these people and Equipment simply being immersed in cold air we need to get their total volume first and this time we have the mass of all that stuff and the density which allows us to get the volume so I'm going to write down just a reminder of the definition of density mass per unit volume which means if you solve for volume you get Mass over density so we're going to call this V EQ for the volume of the equipment and people and so on and I get 500 Kilograms over a mean density of 1500 kilograms per cubic meter and this gives me exactly one-third or we can write down as a decimal about 0.333 cubic meters of volume so to get the buoyant force generated by this .33 cubic meters of volume we need the mass of the cold air pushed out of the way by the equipment and people so I'm going to call this contribution to buoyant 4 so I'll call it FB EQ and that's the mass of the cold air pushed out of the way so it's going to be the density of the cold air times the volume of the equipment people times G then we get back into the same type of force analysis but this time we realize that our total buoyant force has two different contributions there's the main buoyant force on the balloon itself I'll just call that FB and then there's also FB EQ the small additional buoyant force on the equipment itself and of course pointed down here is the total force of gravity that's mg so to get the balloon to float we balance the forces and then we start putting in the details the buoyant force on the balloon itself again is going to be the density of the cold air times volume of the balloon times G that's the weight of the displaced cold air plus our additional small buoyant force on the equipment and passengers themselves again they're pushing cold air out of the way so density of the cold air times V EQ the volume of equipment times G that's equal to our total weight which again splits into two pieces I have the mass of the equipment plus the mass of the hot air which is going to the density of the hot air times the volume of the balloon all multiplied by G so this is basically what we did before except we're adding in this little helper term that's helping the buoyant force just a little bit the G's cancel out again and again we're going to subtract the v-containing term from the right hand side and Factor it out that gives me a v times row cold minus rho hot and then on the right hand side what I'm doing simultaneously here I have M EQ over there but I wanted to subtract this new term row cold times V EQ so I have minus row cold V EQ now we can solve for V and I get M EQ minus row cold veq volume of the equipment divided by the difference in the two densities again row cold minus rho hot so now we can plug in our numbers and the mass of the equipment was 500 Kilograms the density of the cold air 1.24 kilograms per cubic meter the volume of the equipment in passengers 0.333 cubic meters and we're dividing by 1.24 minus 0.95 for the difference in densities and when I run the numbers on this I end up with 1723 cubic meters in other words improving our approach to the problem to include the buoyant force on the passengers and Equipment themselves made a difference of one single cubic meter in how much we need to inflate this balloon if you enjoyed this video or at least found it useful check out another one by clicking one of the links on the left or click the zaxlab logo on the right to explore dozens of physics and math playlists as always you can leave your questions comments and requests in the comments section below and I'll get back to you within 24 hours thanks for watching Zack's lab and best of luck on your math and physics Journey
8497
https://bioprinciples.biosci.gatech.edu/module-4-genes-and-genomes/4-2-4-mendelian-genetics/
Skip to content Mendelian Genetics Learning Objectives Define and use the terms needed to discuss genetic inheritance including gene, allele, dominant, recessive, gamete, genotype, phenotype, homozygote, heterozygote, carrier Explain how chromosomal separation at meiosis leads to segregation of alleles in gametes Explain how alignment at metaphase results in independent assortment of (unlinked) genes Construct and use a Punnett square for a single trait and for two traits using appropriate terminology Determine possible offspring types and phenotypic ratios using probability rules The terminology of Mendelian inheritance Gregor Mendel is famous for discovering “particulate inheritance” or the idea that hereditary elements are passed on in discrete units rather than “blended” together at each new generation. Today we call those discrete units genes. A gene is a hereditary factor that determines (or influences) a particular trait. A gene is comprised of a specific DNA sequence and is located on a specific region of a specific chromosome. Because of its specific location, a gene can also be called a genetic locus. An allele is a particular variant of a gene, in the same way that chocolate and vanilla are particular variants of ice cream. An organism’s genotype is the particular collection of alleles found in its DNA. An organism with two of the same alleles for a particular gene is homozygous at that locus; an organism with two different alleles for a particular gene is heterozygous at that locus. An organism’s phenotype is its observable traits. An organism can have a heterozygous genotype at a particular locus but have a phenotype that looks like only one of the two alleles. This is because one allele is dominant and masks the effect of the other, recessive allele in a dominant/recessive pattern. A dominant allele produces its phenotype whether the organism is homozygous or heterozygous at that locus. For example, in humans the allele for brown eyes is dominant to the allele for blue eyes, so a person who is heterozygous at the eye color locus will have brown eyes. A recessive allele produces its phenotype only when homozygous at the locus; its phenotype is masked if the locus is heterozygous. For example, a person must have two copies of the blue eye color allele to have blue eyes. Sometimes specific recessive alleles are associated with diseases or conditions. A person who is heterozygous for the gene will be phenotypically normal but carry a copy of the recessive, disease-associated allele. This person is said to be a carrier and can pass on the disease allele to his or her offspring. Crosses with a single trait and the principle of segregation All of the concepts above are illustrated in the types of experiments that Mendel carried out with pea plants. Pea plants do not sound like a particularly exciting organism to study, but they were very useful in figuring out basic patterns of inheritance! The reason they were so useful is that they have a lot of traits that are caused by a single gene with a simple dominant/recessive inheritance pattern (this is actually pretty rare in general – but more on that later). So what does that statement in bold mean? A classic example is pea color. Peas can be either yellow or green, but not any other colors. Whether they are yellow or green is controlled by a single gene with two alleles, and the yellow allele is dominant to the green allele. The inheritance pattern if you cross homozygous yellow and homozygous green pea plants is illustrated here: You can see in the first generation (F1) that all offspring produce yellow seeds, even though they have both the yellow and green alleles. If the F1 generation self-fertilizes (pea plants – like most plants – produce both male and female gametes can can mate with themselves), then you now see some offspring that produce yellow peas and some that produce green peas. The yellow:green pea producers exist in approximately a 3:1 ratio, illustrated by constructing a Punnett square. Punnett squares illustrate the fact that each pea plant gamete contains only one allele for each trait. Even though adult pea plants have two copies of each allele, these two alleles separate into different gametes. Thus when two gametes come together to create a new plant, each gamete carries one allele, resulting in two alleles in the new plant. The idea that each gamete carries only one allele for each trait is the principle of segregation; that is, the two alleles for a particular trait are segregated into different gametes. Crosses with two traits and the principle of independent assortment Pea plants have a lot of other traits beyond seed color, and Mendel studied seven other traits. Things become more complex when you follow more than one trait at at time. Here is a cross looking at both pea color (yellow or green) and pea shape (round or wrinkly). Follow the logic of the cross below to see why offspring demonstrate a 9:3:3:1 ratio of different phenotypes. Punnett squares that show two or more traits illustrate the idea that alleles for different traits (different genes) are segregated independently of each other. Yellow seeds are not always round, and green seeds are not always wrinkly; there can be yellow wrinkly seeds, yellow round seeds, green wrinkly seeds, and green round seeds. The idea that alleles for different traits are segregated independently is the principle of independent assortment. Mendel’s laws and meiosis Mendel’s laws or principles of segregation and independent assortment are both explained by the physical behavior of chromosomes during meiosis. Segregation occurs because each gamete inherits only one copy of each chromosome. Each chromosome has only one copy of each gene; therefore each gamete only contains one or the two possible parental alleles. Segregation occurs when the homologous chromosomes separate during meiotic anaphase I. This principle is illustrated here: Independent assortment occurs because homologous chromosomes are randomly segregated into different gametes; ie, one gamete does not only get all maternal chromosomes while the other gets all paternal chromosomes. Independent assortment occurs when homologous chromosomes align randomly at the metaphase plate during meiotic metaphase I. This principle is illustrated here: Random, independent assortment during metaphase I can be demonstrated by considering a cell with a set of two chromosomes (n = 2). In this case, there are two possible arrangements at the equatorial plane in metaphase I. The total possible number of different gametes is 2n, where n equals the number of chromosomes in a set. In this example, there are four possible genetic combinations for the gametes. With n = 23 in human cells, there are over 8 million possible combinations of paternal and maternal genotypes in a potential offspring. Determine possible offspring types and phenotypic ratios using probability rules Because there can be so many ways for genes to influence the phenotype, in class we will use the information discussed above to determine possible offspring types and phenotypic ratios using simple probability rules. For crosses that involve 2 or more independently assorting traits, using probability rules can be much faster and easier than using 4 x 4 Punnett squares (for 2-factor crosses) or 8 x 8 Punnett squares (for 3-factor crosses). The number of possible gametes is 2N, where N is the number of factors (genes), and the size of the Punnett square needed is 2N x 2N! So instead, we can calculate the outcomes for each factor or gene, then multiply the results. The two simple probability rules are the product rule and the sum rule | Product rule | Sum rule | --- | | For independent events X and Y, the probability (P) of them both occurring (X and Y) is [P(X) x P(Y)]. | For mutually exclusive events X and Y, the probability (P) that one will occur (X or Y) is [P(X) + P(Y)]. | Image credit: Khan Academy. Example: AaBbCcDd x AaBbCcDd is a cross where A, B, C and D are 4 different genes, with the dominant alleles given as A, B, C, and D, and the recessive alleles are a, b, c, and d, respectively. What proportion of the progeny will have the dominant phenotype for A and B, and recessive for c and d? If we look at just Aa x Aa, we know that 3/4 of the progeny will have the dominant A phenotype. Similarly, for just Bb x Bb, 3/4 of the progeny will have the dominant B phenotype. For Cc x Cc, 1/4 of the progeny will have the recessive c phenotype (cc genotype). For Dd x Dd, 1/4 of the progeny will have the recessive d phenotype (dd genotype). The rules of probability say that, if these genes are sorting independently, we can just multiply these proportions: The proportion of ABcd phenotype among the progeny = 3/4 x 3/4 x 1/4 x 1/4 = 9/256 Here’s a quick summary of many of these ideas from Ted Ed: and here is Khan Academy’s take: Sustainable Development Goal UN Sustainable Development Goal (SDG) 3: Good Health and Wellbeing- Understanding genetic inheritance is important for identifying, diagnosing, and treating genetic disorders. 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8498
https://www.litcharts.com/lit/through-the-looking-glass/characters/humpty-dumpty
Humpty Dumpty Character Analysis in Through the Looking-Glass | LitCharts Through the Looking-Glass Introduction + Context Plot Summary Detailed Summary & Analysis Chapter 1: Looking-Glass HouseChapter 2: The Garden of Live FlowersChapter 3: Looking-Glass InsectsChapter 4: Tweedledum and TweedledeeChapter 5: Wool and WaterChapter 6: Humpty DumptyChapter 7: The Lion and the UnicornChapter 8: "It's My Own Invention"Chapters 9-10: Queen Alice; ShakingChapters 11-12: Waking; Which Dreamed It? Themes All ThemesYouth, Identity, and Growing UpAdulthood and the Adult WorldRules and EtiquetteSense, Nonsense, and Language Quotes Characters All CharactersAliceThe White QueenThe Red QueenHumpty DumptyThe White KingThe White KnightThe GnatHaighaTweedledeeTweedledumThe UnicornHattaThe FawnThe Sheep Symbols All SymbolsRushesAlice's CrownThe Poem “Jabberwocky” Literary Devices All Literary DevicesAllusionsAnthropomorphismDramatic IronyFallacyFoilForeshadowingFrame StoryGenreIdiomsImageryIronyMoodMotifsParadoxParodyPersonificationSatireSettingSimilesSituational IronyStyleTone Quizzes All QuizzesChapter 1: Looking-Glass House QuizChapter 2: The Garden of Live Flowers QuizChapter 3: Looking-Glass Insects QuizChapter 4: Tweedledum and Tweedledee QuizChapter 5: Wool and Water QuizChapter 6: Humpty Dumpty QuizChapter 7: The Lion and the Unicorn QuizChapter 8: "It's My Own Invention" QuizChapters 9-10: Queen Alice; Shaking QuizChapters 11-12: Waking; Which Dreamed It? Quiz Download PDF Download Teacher Edition AI Tools Ask LitCharts AIDiscussion Question GeneratorEssay Prompt GeneratorQuiz Question Generator Guides Literature GuidesPoetry GuidesShakespeare TranslationsLiterary Terms Sign InSign up for A+ AI Tools Ask LitCharts AI Discussion Question Generator Essay Prompt Generator Quiz Question Generator Guides Literature Guides Poetry Guides Shakespeare Translations Literary Terms Sign InSign up for A+Sign up Through the Looking-Glass by Lewis Carroll PDF Upgrade to A+ Introduction Intro Plot Summary Plot Summary & Analysis Chapter 1: Looking-Glass House Chapter 2: The Garden of Live Flowers Chapter 3: Looking-Glass Insects Chapter 4: Tweedledum and Tweedledee Chapter 5: Wool and Water Chapter 6: Humpty Dumpty Chapter 7: The Lion and the Unicorn Chapter 8: "It's My Own Invention" Chapters 9-10: Queen Alice; Shaking Chapters 11-12: Waking; Which Dreamed It? Themes All Themes Youth, Identity, and Growing Up Adulthood and the Adult World Rules and Etiquette Sense, Nonsense, and Language Quotes Characters All Characters Alice The White Queen The Red Queen Humpty Dumpty The White King The White Knight The Gnat Haigha Tweedledee Tweedledum The Unicorn Hatta The Fawn The Sheep Symbols All Symbols Rushes Alice's Crown The Poem “Jabberwocky” Lit Devices All Literary Devices Allusions Anthropomorphism Dramatic Irony Fallacy Foil Foreshadowing Frame Story Genre Idioms Imagery Irony Mood Motifs Paradox Parody Personification Satire Setting Similes Situational Irony Style Tone Quizzes All Quizzes Chapter 1: Looking-Glass House Quiz Chapter 2: The Garden of Live Flowers Quiz Chapter 3: Looking-Glass Insects Quiz Chapter 4: Tweedledum and Tweedledee Quiz Chapter 5: Wool and Water Quiz Chapter 6: Humpty Dumpty Quiz Chapter 7: The Lion and the Unicorn Quiz Chapter 8: "It's My Own Invention" Quiz Chapters 9-10: Queen Alice; Shaking Quiz Chapters 11-12: Waking; Which Dreamed It? Quiz Theme Wheel Theme Viz Download this Chart (PDF)Download the Teacher Edition Download this Chart (PDF) Previous The Red QueenHumpty Dumpty Character Analysis ================================Next The White King The egg-shaped individual from the nursery rhyme "Humpty Dumpty." He sits high on a wall when Alice meets him. Humpty Dumpty is rude, imperious, and self-important. He insists that he can make words mean whatever he wants them to (though he pays them more for extra work) and he reprimands Alice for not being properly polite. Helpfully, he does agree to decode the first verse of the poem "Jabberwocky" for Alice, though more than anything, this is an opportunity for Humpty Dumpty to lord his knowledge and expertise over Alice. Humpty Dumpty takes major issue with the fact that Alice didn't stop growing up at age seven and allowed herself to age six months. He offers her riddles when Alice insists that she can't stop growing. As in the nursery rhyme, Humpty Dumpty falls off his wall after Alice leaves him, while the White King sends all his horses and men except two to help him. Humpty Dumpty Quotes in Through the Looking-Glass The Through the Looking-Glass quotes below are all either spoken by Humpty Dumpty or refer to Humpty Dumpty. For each quote, you can also see the other characters and themes related to it (each theme is indicated by its own dot and icon, like this one: ). Chapter 6: Humpty Dumpty Quotes "My name is Alice, but—" "It's a stupid name enough!" Humpty Dumpty interrupted impatiently. "What does it mean?" "Must a name mean something?" Alice asked doubtfully. "Of course it must," Humpty Dumpty said with a short laugh: "my name means the shape I am—a good handsome shape it is, too. With a name like yours, you might be any shape, almost." Related Characters:Alice (speaker), Humpty Dumpty (speaker) Related Themes: Page Number and Citation:182 Cite this Quote Explanation and Analysis: Unlock explanations and citation info for this and every other Through the Looking-Glass quote. Plus so much more... Get LitCharts A+ Already a LitCharts A+ member?Sign in! "Seven years and six months!" Humpty Dumpty repeated thoughtfully. "An uncomfortable sort of age. Now if you'd asked my advice, I'd have said 'Leave off at seven'—but it's too late now." "I never ask advice about growing," Alice said indignantly. "Too proud?" the other enquired. Alice felt even more indignant at this suggestion. "I mean," she said, "that one ca'n't help growing older." "One ca'n't, perhaps," said Humpty Dumpty; "but two can. With proper assistance, you might have left off at seven." Related Characters:Humpty Dumpty (speaker), Alice (speaker) Related Themes: Page Number and Citation:184 Cite this Quote Explanation and Analysis: Unlock with LitCharts A+ "But 'glory' doesn't mean 'a knock-down argument,'" Alice objected. "When I use a word," Humpty Dumpty said, in rather a scornful tone, "it means just what I choose it to mean—neither more nor less." Related Characters:Humpty Dumpty (speaker), Alice (speaker) Related Themes: Page Number and Citation:186 Cite this Quote Explanation and Analysis: Unlock with LitCharts A+ "As to poetry, you know," said Humpty Dumpty, stretching out one of his great hands, "I can repeat poetry as well as other folk, if it comes to that—" "Oh, it needn't come to that!" Alice hastily said, hoping to keep him from beginning. "The piece I'm going to repeat," he went on without noticing her remark, "was written entirely for your amusement." Alice felt that in that case she really ought to listen to it; so she sat down, and said "Thank you" rather sadly. Related Characters:Humpty Dumpty (speaker), Alice (speaker) Related Themes: Page Number and Citation:189 Cite this Quote Explanation and Analysis: Unlock with LitCharts A+ Get the entire Through the Looking-Glass LitChart as a printable PDF. "My students can't get enough of your charts and their results have gone through the roof." -Graham S. Download Humpty Dumpty Quotes in Through the Looking-Glass The Through the Looking-Glass quotes below are all either spoken by Humpty Dumpty or refer to Humpty Dumpty. For each quote, you can also see the other characters and themes related to it (each theme is indicated by its own dot and icon, like this one: ). Chapter 6: Humpty Dumpty Quotes "My name is Alice, but—" "It's a stupid name enough!" Humpty Dumpty interrupted impatiently. "What does it mean?" "Must a name mean something?" Alice asked doubtfully. "Of course it must," Humpty Dumpty said with a short laugh: "my name means the shape I am—a good handsome shape it is, too. With a name like yours, you might be any shape, almost." Related Characters:Alice (speaker), Humpty Dumpty (speaker) Related Themes: Page Number and Citation:182 Cite this Quote Explanation and Analysis: Unlock explanations and citation info for this and every other Through the Looking-Glass quote. Plus so much more... Get LitCharts A+ Already a LitCharts A+ member?Sign in! "Seven years and six months!" Humpty Dumpty repeated thoughtfully. "An uncomfortable sort of age. Now if you'd asked my advice, I'd have said 'Leave off at seven'—but it's too late now." "I never ask advice about growing," Alice said indignantly. "Too proud?" the other enquired. Alice felt even more indignant at this suggestion. "I mean," she said, "that one ca'n't help growing older." "One ca'n't, perhaps," said Humpty Dumpty; "but two can. With proper assistance, you might have left off at seven." Related Characters:Humpty Dumpty (speaker), Alice (speaker) Related Themes: Page Number and Citation:184 Cite this Quote Explanation and Analysis: Unlock with LitCharts A+ "But 'glory' doesn't mean 'a knock-down argument,'" Alice objected. "When I use a word," Humpty Dumpty said, in rather a scornful tone, "it means just what I choose it to mean—neither more nor less." Related Characters:Humpty Dumpty (speaker), Alice (speaker) Related Themes: Page Number and Citation:186 Cite this Quote Explanation and Analysis: Unlock with LitCharts A+ "As to poetry, you know," said Humpty Dumpty, stretching out one of his great hands, "I can repeat poetry as well as other folk, if it comes to that—" "Oh, it needn't come to that!" Alice hastily said, hoping to keep him from beginning. "The piece I'm going to repeat," he went on without noticing her remark, "was written entirely for your amusement." Alice felt that in that case she really ought to listen to it; so she sat down, and said "Thank you" rather sadly. Related Characters:Humpty Dumpty (speaker), Alice (speaker) Related Themes: Page Number and Citation:189 Cite this Quote Explanation and Analysis: Unlock with LitCharts A+ Cite This Page Choose citation style: MLA Brock, Zoë. "Through the Looking-Glass Characters: Humpty Dumpty." LitCharts. LitCharts LLC, 12 Aug 2019. Web. 28 Sep 2025. Copy to Clipboard Close Previous The Red Queen Previous The Red Queen Next The White King Next The White King Cite This Page Close Company About UsOur StoryJobs Support Help CenterContact Us Connect FacebookTwitter Legal Terms of ServicePrivacy PolicyPrivacy Request HomeAboutContactHelp Copyright © 2025 All Rights Reserved TermsPrivacyPrivacy Request Save time. Stress less. Sign up! AI Tools for on-demand study help and teaching prep. Quote explanations, with page numbers, for over 49,143 quotes. PDF downloads of all 2,206 LitCharts guides. 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https://research.uaeu.ac.ae/en/publications/on-alternating-sums-of-binomial-and-q-binomial-coefficients
On alternating sums of binomial and q-binomial coefficients - United Arab Emirates University Skip to main navigation Skip to search Skip to main content United Arab Emirates University Home Home Research units Researchers Projects Research output Datasets 0 More Activities Press/Media Prizes Impacts Search by expertise, name or affiliation On alternating sums of binomial and q-binomial coefficients Mohamed E.L. Bachraoui Department of Mathematical Sciences Research output: Contribution to journal › Article › peer-review Overview Fingerprint Abstract In this paper we shall evaluate two alternating sums of binomial coefficients by a combinatorial argument. Moreover, by combining the same combinatorial idea with partition theoretic techniques, we provide ^-analogues involving the g-binomial coefficients. Our results are a generalisation of some previous work by Guo and Zhang. | Original language | English | | Pages (from-to) | 257-272 | | Number of pages | 16 | | Journal | Ars Combinatoria | | Volume | 151 | | Publication status | Published - Jul 2020 | Keywords And phrases, binomial coefficients Gaussian binomial coefficients Integer partitions Q-binomial coefficients ASJC Scopus subject areas General Mathematics Other files and links Link to publication in Scopus Link to the citations in Scopus Fingerprint Dive into the research topics of 'On alternating sums of binomial and q-binomial coefficients'. Together they form a unique fingerprint. Q-binomial Coefficients Keyphrases 100% Binomial Coefficient Mathematics 100% Sums of Binomial Coefficients Keyphrases 50% View full fingerprint Cite this APA Standard Harvard Vancouver Author BIBTEX RIS Bachraoui, M. E. L. (2020). On alternating sums of binomial and q-binomial coefficients. Ars Combinatoria, 151, 257-272. On alternating sums of binomial and q-binomial coefficients. / Bachraoui, Mohamed E.L. In: Ars Combinatoria, Vol. 151, 07.2020, p. 257-272. Research output: Contribution to journal › Article › peer-review Bachraoui, MEL 2020, 'On alternating sums of binomial and q-binomial coefficients', Ars Combinatoria, vol. 151, pp. 257-272. Bachraoui MEL. On alternating sums of binomial and q-binomial coefficients. Ars Combinatoria. 2020 Jul;151:257-272. Bachraoui, Mohamed E.L. / On alternating sums of binomial and q-binomial coefficients. In: Ars Combinatoria. 2020 ; Vol. 151. pp. 257-272. @article{06e36b2c868e46ff9775cd30e8935096, title = "On alternating sums of binomial and q-binomial coefficients", abstract = "In this paper we shall evaluate two alternating sums of binomial coefficients by a combinatorial argument. Moreover, by combining the same combinatorial idea with partition theoretic techniques, we provide \textasciicircum{}-analogues involving the g-binomial coefficients. Our results are a generalisation of some previous work by Guo and Zhang.", keywords = "And phrases, binomial coefficients, Gaussian binomial coefficients, Integer partitions, Q-binomial coefficients", author = "Bachraoui, {Mohamed E.L.}", note = "Publisher Copyright: {\textcopyright} 2020 Charles Babbage Research Centre. All rights reserved.", year = "2020", month = jul, language = "English", volume = "151", pages = "257--272", journal = "Ars Combinatoria", issn = "0381-7032", publisher = "Charles Babbage Research Centre", } TY - JOUR T1 - On alternating sums of binomial and q-binomial coefficients AU - Bachraoui, Mohamed E.L. N1 - Publisher Copyright: © 2020 Charles Babbage Research Centre. All rights reserved. PY - 2020/7 Y1 - 2020/7 N2 - In this paper we shall evaluate two alternating sums of binomial coefficients by a combinatorial argument. Moreover, by combining the same combinatorial idea with partition theoretic techniques, we provide ^-analogues involving the g-binomial coefficients. Our results are a generalisation of some previous work by Guo and Zhang. AB - In this paper we shall evaluate two alternating sums of binomial coefficients by a combinatorial argument. Moreover, by combining the same combinatorial idea with partition theoretic techniques, we provide ^-analogues involving the g-binomial coefficients. Our results are a generalisation of some previous work by Guo and Zhang. KW - And phrases, binomial coefficients KW - Gaussian binomial coefficients KW - Integer partitions KW - Q-binomial coefficients UR - UR - M3 - Article AN - SCOPUS:85108551930 SN - 0381-7032 VL - 151 SP - 257 EP - 272 JO - Ars Combinatoria JF - Ars Combinatoria ER - Powered by Pure, Scopus&Elsevier Fingerprint Engine™ All content on this site: Copyright © 2025 United Arab Emirates University, its licensors, and contributors. All rights are reserved, including those for text and data mining, AI training, and similar technologies. For all open access content, the relevant licensing terms apply We use cookies to help provide and enhance our service and tailor content. By continuing you agree to the use of cookies Log in to Pure About web accessibility Report vulnerability Contact us