Datasets:
multi-train
/
index

Dataset card Data Studio Files
xet
 Community
1
id
stringlengths
1
6
url
stringlengths
16
1.82k
content
stringlengths
37
9.64M
15600
https://www.engineeringtoolbox.com/density-gravity-t_64.html
Engineering ToolBox - Resources, Tools and Basic Information for Engineering and Design of Technical Applications! Densities Densities of solids, liquids and gases. Definitions and convertion calculators. My Short List Related Documents Air - Density and Specific Volume vs. Altitude Air - Properties at Gas-Liquid Equilibrium Conditions Air Density, Specific Weight, and Thermal Expansion Coefficients at Varying Temperatures and Pressures Air Properties - Density, Viscosity, Heat Capacity, Thermal Conductivity, and more Air Properties: Temperature, Pressure & Density Data Alcohols and Carboxylic Acids - Physical Data Ammonia - Thermophysical Properties Ammonia Gas - Density vs. Temperature and Pressure API Gravity: Definition, Calculation, and Converter Argon - Density and Specific Weight Benzene - Density and Specific Weight vs. Temperature and Pressure Benzene - Thermophysical properties Brick Densities Butane - Density and Specific Weight vs. Temperature and Pressure Calcium Chloride Water Solutions Carbon Dioxide (CO₂) Properties & Characteristics: Density, Thermal Data & More Carbon dioxide - Density and Specific Weight vs. Temperature and Pressure Carbon Monoxide - Density and Specific Weight vs. Temperature and Pressure Coal - Classification Crude Oil - Density vs. Temperature Crude Oil Viscosity vs. Gravity Densities of Aqueous Solutions of Inorganic Chlorides Densities of Aqueous Solutions of Inorganic Potassium Salts Densities of Aqueous Solutions of Inorganic Sodium Salts Densities of Aqueous Solutions of Organic Acids Densities of common Products Density Converter Density of Aqueous Solutions of Organic Substances as Sugars and Alcohols Density of Aqueous Solutions of some Inorganic Substances Density of Gases Density of Wood Species: Data & Material Guide Density, Specific Weight, and Specific Gravity – Definitions & Calculator Dirt and Mud - Densities Ditch Filling Materials - Densities Dry Air and Water Vapor - Density and Specific Volume vs. Temperature - Imperial Units Dry Air Thermodynamic and Physical Properties – Density, Specific Heat, Viscosity Elements of the Periodic System Energy Storage Density Ethane - Density and Specific Weight vs. Temperature and Pressure Ethanol - Density and Specific Weight vs. Temperature and Pressure Ethanol - Thermophysical properties Ethylene - Density and Specific Weight vs. Temperature and Pressure Ethylene - Thermophysical Properties Ethylene Glycol Heat-Transfer Fluid Properties: Density, Data & Charts Food Products - Bulk Densities Fuel Oils - Densities vs. Temperature Fuels - Densities and Specific Volumes Heavy Water - Thermophysical Properties Helium - Density and Specific Weight vs. Temperature and Pressure How to Calculate the Density of Slurries: Formula and Examples Hydrocarbons - Physical Data Hydrocarbons, Linear Alcohols and Acids - Densities Hydrogen - Density and Specific Weight vs. Temperature and Pressure Ice - Thermal Properties Inorganic Compounds in Water - Melting and Boiling Temperature, Density and Solubility ISO Grade Oils - Viscosities and Densities Jet Fuel - Density vs. Temperature Liquid Densities Liquid-Liquid Solution - Shrinkage and Estimation of Density Liquids - Densities vs. Pressure and Temperature Change Liquids - Specific Gravities Liquids - Specific Gravities Lubricating Oil - Densities vs. Temperature Mercury - Thermodynamic Properties Metals and Alloys - Densities Methane - Density and Specific Weight vs. Temperature and Pressure Methanol - Density and Specific Weight vs. Temperature and Pressure Minerals - Densities Moist Air - Density vs. Pressure Moist Air - Density vs. Water Content and Temperature Nitrogen - Thermophysical Properties Organic Nitrogen Compounds - Physical Data Organic Sulfur Compounds - Densities Oxygen - Density and Specific Weight vs. Temperature and Pressure Pentane - Density and Specific Weight vs. Temperature and Pressure Pentane - Thermophysical Properties Propane - Density and Specific Weight vs. Temperature and Pressure SAE Multigrade Oils - Viscosities and Densities Seawater - Properties Solids - Densities Toluene - Density and Specific Weight vs. Teemperature and Pressure Toluene - Thermophysical Properties Volumetric (Cubic) Thermal Expansion Water - Properties at Gas-Liquid Equilibrium Conditions Water - Properties vs. Temperature and Pressure Water - Specific Gravity vs. Temperature Water - Specific Volume vs. Temperature Water Density, Specific Weight and Thermal Expansion Coefficients - Temperature and Pressure Dependence Water Thermal Properties – Density, Heat Capacity, and Conductivity Wood - Density vs. Moisture Content Wood Species - Moisture Content and Weight Search Search is the most efficient way to navigate the Engineering ToolBox. Our Mission The Engineering ToolBox provides a wide range of free tools, calculators, and information resources aimed at engineers and designers. It offers detailed technical data and calculations for various fields such as fluid mechanics, material properties, HVAC systems, electrical engineering, and more. The site includes resources for common engineering tasks, such as calculating physical properties (e.g., density, viscosity, thermal conductivity), converting units, and designing systems like heating and water distribution. With sections on everything from acoustics to hydraulics, it serves as a comprehensive tool for both students and professionals in technical and engineering disciplines. Engineering ToolBox - SketchUp Extension - Online 3D modeling! Add standard and customized parametric components - like flange beams, lumbers, piping, stairs and more - to your Sketchup model with the Engineering ToolBox - SketchUp Extension - enabled for use with older versions of the amazing SketchUp Make and the newer "up to date" SketchUp Pro . Add the Engineering ToolBox extension to your SketchUp Make/Pro from the Extension Warehouse ! Translate this Page Translate this page to Your Own Language . Privacy Policy We don't collect information from our users. More about We use a third-party to provide monetization technology for our site. You can review their privacy and cookie policy here. You can change your privacy settings by clicking the following button: Manage Consent. Advertise in the ToolBox If you want to promote your products or services in the Engineering ToolBox - please use Google Adwords. You can target the Engineering ToolBox by using AdWords Managed Placements. Citation This page can be cited as Modify the access date according your visit. Unit Converter 9.19.10 Make Shortcut to Home Screen?
15601
https://newsnetwork.mayoclinic.org/discussion/mayo-clinic-minute-why-breakfast-may-be-key-to-trimming-your-belly/
Mayo Clinic Minute: Why breakfast may be key to trimming your belly - Mayo Clinic News Network Skip to Content English Español العربية 简体中文 Português Br Journalist Pass Sign In Appointments Search Request appointment Sign in News Releases Arizona Minnesota Florida International Health Topics Overview Cancer Cardiovascular Children’s Center COVID-19 Education Gastroenterology Health & Wellness Infectious Diseases Mayo Clinic Minute Medical Innovation Neurosciences Orthopedics/Sports Research Science Saturday Sharing Mayo Clinic Transplant Medical Research Overview Aging AI and Digital Health Biotherapeutics Cancer Clinical Trials Discovery Science Healthcare Delivery Individualized Medicine Translational Science Media Contacts About Mayo Clinic Minute: Why breakfast may be key to trimming your belly By Ian Roth April 25, 2018 Share this: If you're wanting to lose pounds and your belly, breakfast may be your best friend. "For reasons that we don't quite understand yet, eating breakfast seems to be a marker of, No. 1, less likelihood of having gained weight recently, and, No. 2, ... a smaller belly circumference and less visceral fat," Dr. Virend Somers, a Mayo Clinic cardiologist, says. Watch: The Mayo Clinic Minute Journalists: Broadcast-quality video pkg (1:00) is in the downloads. Read the script. Dr. Somers helped oversee a study that showed people who ate breakfast daily gained less weight than those who didn't eat breakfast. "Those who ate breakfast very frequently put on less than 3 pounds in the past year," Dr. Somers says. "Those who ate breakfast maybe one to four times a week put on about 5 pounds. The ones who didn't eat breakfast at all put on about 8 pounds in the year prior to them seeing us." But Dr. Naima Covassin, who led the study, says the correlation between breakfast and less belly fat is even more important to your health. Dr. Covassin is a researcher in Mayo Clinic's Cardiovascular Physiology Laboratory. "That is the dangerous fat, the fat that is really bad for your health because it's more consistently associated with hypertension, high blood pressure, diabetes, heart disease. And that's the fat that we don't want to have," Dr. Covassin says. Both doctors warn this isn't necessarily about eating more or fewer calories. It's more likely about eating the same number of calories but earlier in the day. So, to lose pounds and belly fat, remember an old saying, "It's eat breakfast like a king, lunch like a prince, and dinner like a beggar," Dr. Somer says. Don’t ignore these 7 serious symptoms Mayo Clinic to sponsor National Prescription Drug Take Back Day event on April 28 Related Articles Health & Wellness Mayo Clinic Q&A: Health benefits of counting steps Cardiovascular Mayo Clinic study reveals hidden causes of heart attacks in younger adults, especially women Cardiovascular Mayo Clinic Q&A: What makes the Mediterranean diet so healthy for your heart? About the News Network Newsbureau@mayo.edu News Releases Cancer Cardiovascular Gastroenterology Neurosciences Transplant Research Mayo Clinic Minute Health and Wellness Orthopedics/Sports Children's Center Topics Terms and Conditions Privacy Policy Manage Cookies © 2025. Mayo Clinic News Network. All Rights Reserved. Loading... Loading... Feedback Mayo Clinic Privacy Policy Mayo Clinic and our partners use technologies such as cookies to collect information from your browser to deliver relevant advertising on our site, in emails and across the Internet, personalize content and perform site analytics. Please visit our Privacy Policy for more information about our use of data and your rights. Click here to update your preferences I Agree x Customize Cookie Settings Trusted These technologies are used in operating the site, including remembering your preferences, ensuring security and providing certain features. They do not track your activity online. [x] Website Analytics These technologies collect information to help us understand how our websites are being used and to make improvements. [x] Advertising These technologies are used to make advertising messages more relevant to you. They perform functions like preventing the same ad from continuously reappearing, ensuring that ads are properly displayed for advertisers and selecting advertisements that are based on your interests. Save Mayo Clinic Privacy Policy Mayo Clinic and our partners use technologies such as cookies to collect information from your browser to deliver relevant advertising on our site, in emails and across the Internet, personalize content and perform site analytics. Please visit our Privacy Policy for more information about our use of data and your rights. Click here to update your preferences I Agree x Customize Cookie Settings Trusted These technologies are used in operating the site, including remembering your preferences, ensuring security and providing certain features. They do not track your activity online. [x] Website Analytics These technologies collect information to help us understand how our websites are being used and to make improvements. [x] Advertising These technologies are used to make advertising messages more relevant to you. They perform functions like preventing the same ad from continuously reappearing, ensuring that ads are properly displayed for advertisers and selecting advertisements that are based on your interests. Save
15602
https://www.uniprot.org/uniprotkb/P10443/entry
P10443 · DPO3A_ECOLI DNA polymerase III subunit alpha UniProtKB reviewed (Swiss-Prot) Escherichia coli (strain K12) Amino acids 1160 (go to sequence) Protein existence Evidence at protein level Annotation score Add a publication Entry feedback Function function DNA polymerase III is a complex, multichain enzyme responsible for most of the replicative synthesis in bacteria (PubMed:2932432). This DNA polymerase also exhibits 3' to 5' exonuclease activity. The alpha chain is the DNA polymerase catalytic subunit (PubMed:2932432). It is tethered to replicating DNA by the beta sliding clamp (dnaN), which confers extremely high processivity to the catalytic subunit, copying a 5.4 kb genome in 11 seconds, a speed of at least 500 nucleotides/second at 30 degrees Celsius (PubMed:2413035). Catalytic activity Rhea 22508 DNA(n) + a 2'-deoxyribonucleoside 5'-triphosphate = DNA(n+1) + diphosphate EC:2.7.7.7 (UniProtKB | ENZYME | Rhea) Gene Ontology GO annotations GO-CAM models Gene Ontology (GO) annotations organized by slimming set. | | Aspect | Term | --- | | Molecular Function | DNA binding | | | Molecular Function | DNA-directed DNA polymerase activity | | | Biological Process | DNA-templated DNA replication | | | Biological Process | lagging strand elongation | | | Biological Process | leading strand elongation | Access the complete set of GO annotations on QuickGO Keywords Molecular function #DNA-binding #DNA-directed DNA polymerase #Nucleotidyltransferase #Transferase Biological process #DNA replication Enzyme and pathway databases BioCyc EcoCyc:EG10238-MONOMER MetaCyc:EG10238-MONOMER ENZYME Search… Names & Taxonomy Protein names Recommended name DNA polymerase III subunit alpha EC number EC:2.7.7.7 (UniProtKB | ENZYME | Rhea) Gene names Name dnaE + Synonyms polC + Ordered locus names b0184, JW0179 Organism names Taxonomic identifier 83333 (NCBI) Organism Escherichia coli (strain K12) Strains K12 / W3110 / ATCC 27325 / DSM 5911 K12 / MG1655 / ATCC 47076 Taxonomic lineage Bacteria > Pseudomonadati > Pseudomonadota > Gammaproteobacteria > Enterobacterales > Enterobacteriaceae > Escherichia Accessions Primary accession P10443 Proteomes Identifier UP000000625 + Component Chromosome Subcellular Location UniProt Annotation GO Annotation Cytoplasm cytoplasm cytosol DNA polymerase III complex DNA polymerase III, core complex replisome Complete GO annotation on QuickGO Keywords Cellular component #Cytoplasm Phenotypes & Variants Features Showing features for mutagenesis. | | Type | ID | Position(s) | Description | | --- --- --- | | | Mutagenesis | | 920-924 | Loss of interaction with beta sliding clamp (dnaN). | | | | | | Mutagenesis | | 921-923 | Increases binding to beta sliding clamp (dnaN), increases stability of enzyme complex. | | | | | | Mutagenesis | | 923-924 | Loss of interaction with beta sliding clamp (dnaN). | | | | Chemistry ChEMBL CHEMBL4621 Interaction Subunit The DNA polymerase III holoenzyme complex contains at least 10 different subunits organized into 3 functionally essential subassemblies: the Pol III core, the beta sliding clamp processivity factor and the clamp-loading complex. The Pol III core (subunits alpha, epsilon and theta) contains the polymerase and the 3'-5' exonuclease proofreading activities (PubMed:2040637). The polymerase is tethered to the template via the dimeric beta sliding clamp processivity factor. The clamp loader (also called gamma complex) assembles the beta sliding clamp onto the primed template and plays a central role in the organization and communication at the replication fork. The clamp-loading complex contains delta, delta', psi and chi, and 3 copies of either or both of two different DnaX proteins, gamma and tau. The DNA replisome complex has a single clamp loader (3 tau and 1 each of delta, delta', psi and chi subunits) which binds 3 Pol III cores (1 core on the leading strand and 2 on the lagging strand) each with a beta sliding clamp dimer. Additional proteins in the replisome are other copies of gamma, psi and chi, Ssb, DNA helicase and RNA primase (PubMed:20413500, PubMed:22157955). Interacts with the beta sliding-clamp subunit via the peptide Gln-Ala-Asp-Met-Phe (residues 920-924) (PubMed:11573000). Binary interactions | | Type | Entry 1 | Entry 2 | Number of experiments | IntAct | --- --- --- | | | BINARY | P10443 | dnaN P0A988 | 19 | EBI-549111, EBI-542385 | | | BINARY | P10443 | dnaQ P03007 | 25 | EBI-549111, EBI-549131 | | | BINARY | P10443 | dnaX P06710 | 12 | EBI-549111, EBI-549140 | | | BINARY | P10443 | holA P28630 | 5 | EBI-549111, EBI-549153 | | | BINARY | P10443 | rplL P0A7K2 | 3 | EBI-549111, EBI-543702 | | | BINARY | P10443 | ybdL P77806 | 3 | EBI-549111, EBI-543661 | Complex viewer View interactors in UniProtKB View CPX-1925 in Complex Portal Protein-protein interaction databases BioGRID 4262228258 interactors 8492775 interactors ComplexPortal CPX-1925DNA polymerase III core complex DIP DIP-9458N FunCoup P10443552 interactors IntAct P1044370 interactors MINT P10443 STRING 511145.b0184 Chemistry BindingDB P10443 Structure 3D structure databases AlphaFoldDB P10443 EMDB EMD-3198 EMD-3201 EMD-3202 EMD-4141 SMR P10443 ModBase Search… PDBe-KB Search… Miscellaneous EvolutionaryTrace P10443 Family & Domains Sequence similarities Belongs to the DNA polymerase type-C family. DnaE subfamily. Phylogenomic databases HOGENOM CLU_001600_0_2_6 InParanoid P10443 OMA DFCMDGR OrthoDB 9803237at2 PhylomeDB P10443 eggNOG COG0587Bacteria Family and domain databases View all family and domain features for this entry's canonical sequence in the UniParc Feature Viewer. CDD cd04485DnaE_OBF 1 hit cd07433PHP_PolIIIA_DnaE1 1 hit FunFam 1.10.10.1600:FF:000001DNA polymerase III subunit alpha 1 hit 1.10.150.870:FF:000001DNA polymerase III subunit alpha 1 hit 2.40.50.140:FF:000106DNA polymerase III subunit alpha 1 hit 3.20.20.140:FF:000028DNA polymerase III subunit alpha 1 hit Gene3D 1.10.150.8701 hit 1.10.10.1600Bacterial DNA polymerase III alpha subunit, thumb domain 1 hit 3.20.20.140Metal-dependent hydrolases 1 hit 2.40.50.140Nucleic acid-binding proteins 1 hit InterPro View protein in InterPro IPR011708DNA_pol3_alpha_NTPase_dom IPR041931DNA_pol3_alpha_thumb_dom IPR040982DNA_pol3_finger IPR048472DNA_pol_IIIA_C NCBIfam TIGR00594polc 1 hit NF004226PRK05673.1 1 hit PANTHER PTHR32294DNA POLYMERASE III SUBUNIT ALPHA 1 hit PTHR32294:SF0DNA POLYMERASE III SUBUNIT ALPHA 1 hit Pfam View protein in Pfam PF07733DNA_pol3_alpha 1 hit PF17657DNA_pol3_finger 1 hit PF20914DNA_pol_IIIA_C 1 hit PF14579HHH_6 1 hit SMART View protein in SMART SM00481POLIIIAc 1 hit SUPFAM SSF89550PHP domain-like 1 hit MobiDB Search… Sequence Sequence status Complete See also sequence in UniParc or sequence clusters in UniRef Length 1,160 Mass (Da) 129,905 Last updated 1989-07-01 v1 MD5 Checksum MSEPRFVHLRVHSDYSMIDGLAKTAPLVKKAAALGMPALAITDFTNLCGLVKFYGAGHGAGIKPIVGADFNVQCDLLGDELTHLTVLAANNTGYQNLTLLISKAYQRGYGAAGPIIDRDWLIELNEGLILLSGGRMGDVGRSLLRGNSALVDECVAFYEEHFPDRYFLELIRTGRPDEESYLHAAVELAEARGLPVVATNDVRFIDSSDFDAHEIRVAIHDGFTLDDPKRPRNYSPQQYMRSEEEMCELFADIPEALANTVEIAKRCNVTVRLGEYFLPQFPTGDMSTEDYLVKRAKEGLEERLAFLFPDEEERLKRRPEYDERLETELQVINQMGFPGYFLIVMEFIQWSKDNGVPVGPGRGSGAGSLVAYALKITDLDPLEFDLLFERFLNPERVSMPDFDVDFCMEKRDQVIEHVADMYGRDAVSQIITFGTMAAKAVIRDVGRVLGHPYGFVDRISKLIPPDPGMTLAKAFEAEPQLPEIYEADEEVKALIDMARKLEGVTRNAGKHAGGVVIAPTKITDFAPLYCDEEGKHPVTQFDKSDVEYAGLVKFDFLGLRTLTIINWALEMINKRRAKNGEPPLDIAAIPLDDKKSFDMLQRSETTAVFQLESRGMKDLIKRLQPDCFEDMIALVALFRPGPLQSGMVDNFIDRKHGREEISYPDVQWQHESLKPVLEPTYGIILYQEQVMQIAQVLSGYTLGGADMLRRAMGKKKPEEMAKQRSVFAEGAEKNGINAELAMKIFDLVEKFAGYGFNKSHSAAYALVSYQTLWLKAHYPAEFMAAVMTADMDNTEKVVGLVDECWRMGLKILPPDINSGLYHFHVNDDGEIVYGIGAIKGVGEGPIEAIIEARNKGGYFRELFDLCARTDTKKLNRRVLEKLIMSGAFDRLGPHRAALMNSLGDALKAADQHAKAEAIGQADMFGVLAEEPEQIEQSYASCQPWPEQVVLDGERETLGLYLTGHPINQYLKEIERYVGGVRLKDMHPTERGKVITAAGLVVAARVMVTKRGNRIGICTLDDRSGRLEVMLFTDALDKYQQLLEKDRILIVSGQVSFDDFSGGLKMTAREVMDIDEAREKYARGLAISLTDRQIDDQLLNRLRQSLEPHRSGTIPVHLYYQRADARARLRFGATWRVSPSDRLLNDLRGLIGSEQVELEFD Sequence caution The sequence AAA70369.1 differs from that shown. Reason: Erroneous initiation Extended N-terminus. Keywords Technical term #3D-structure #Reference proteome Sequence databases C28390DJEC3A NP_414726.1 NC_000913.3 WP_001294757.1 NZ_STEB01000032.1 | | Nucleotide Sequence | Protein Sequence | Molecule Type | Status | --- --- | | M19334 EMBL· GenBank· DDBJ | AAC36920.1 EMBL· GenBank· DDBJ | Genomic DNA | | | | U70214 EMBL· GenBank· DDBJ | AAB08613.1 EMBL· GenBank· DDBJ | Genomic DNA | | | | U00096 EMBL· GenBank· DDBJ | AAC73295.1 EMBL· GenBank· DDBJ | Genomic DNA | | | | AP009048 EMBL· GenBank· DDBJ | BAA77859.1 EMBL· GenBank· DDBJ | Genomic DNA | | | | S52931 EMBL· GenBank· DDBJ | AAB24889.1 EMBL· GenBank· DDBJ | mRNA | | | | M96394 EMBL· GenBank· DDBJ | AAA70369.1 EMBL· GenBank· DDBJ | Genomic DNA | Different initiation | | | D49445 EMBL· GenBank· DDBJ | BAA08424.1 EMBL· GenBank· DDBJ | Genomic DNA | | Genome annotation databases AAC73295 b0184 ecj:JW0179 eco:b0184 ecoc:C3026_00845 UniRef clusters Orthologs & paralogs Disclaimer Any medical or genetic information present in this entry is provided for research, educational and informational purposes only. It is not in any way intended to be used as a substitute for professional medical advice, diagnosis, treatment or care. Our staff consists of biologists and biochemists that are not trained to give medical advice. This website requires cookies, and the limited processing of your personal data in order to function. By using the site you are agreeing to this as outlined in our Privacy Notice.
15603
https://www.quora.com/What-do-boys-wear-daily-but-girls-wear-once-a-year
What do boys wear daily, but girls wear once a year? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Survey Question Fashion and Style Females Young Boys Word Riddles Gender Differences Clothing Item Group Generalization Ques... Young Girls 5 What do boys wear daily, but girls wear once a year? All related (45) Sort Recommended Lamine Sow Feb 26 From my own understanding about this question. Question: Men Wear it everyday while women wear it one in a year. The perfect answer is that “Men Carry Sperm everyday in their life and Women carry it one time a year Which is PREGNANCY ✅ Upvote · 9 8 9 2 Related questions More answers below What is one thing men do every day and woman do one time in our lives? What is the thing that women have once in a year? Can boys wear girls tights, yes or no? Is it okay for a boy to wear girl clothes? Why would a little boy want to wear girl clothes? Shweta Mirani Chromosomally - XX · Author has 73 answers and 645.7K answer views ·9y Unwashed jeans! Did you see what I did there? I'm sorry boys! But you have very limited clothing options. And unlike us girls, you are not concerned about repeating clothes and matching accessories (at least the men I hang out with,don't ) You men are totally okay with wearing the same pair of jeans for over a year while we women, nah nah, we can choose from skirts to shorts to leggings to summer dresses. So, it's safe to assume that your shopping includes three decent pairs of jeans and 12 t-shirts. And, us? Well, let's not even go there! Answer justified? Love! Upvote · 99 36 9 8 Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 210 Hadmat Singh Rajput Doctor of Medicine. AB in Psychology. · Author has 277 answers and 1.2M answer views ·Updated 7y The ‘tag of your age’. We, the boys, never hesitate or be shy in revealing our age. We wear it with pride. Girls, they don't reveal it. Unless, it's her birthday. People will ask or count the number of candles on the cake. They have to wear this so called “age-tag” on one day in whole year. ✌. Upvote · 99 24 9 1 Sumedha Gupta Software Engineer at Tech Mahindra (2016–present) ·9y The answer to this question is not at all a serious one as given by many Quorans. :p This indeed is a puzzle which has no meaning and is asked by kids for fun. ;) The thing that men wear daily and a women wear once in a year is NOTHING. :) Sometimes its better not to use mind. I know my answer is childish but this is the only possible answer for this question. :D Upvote · 99 32 9 6 Related questions More answers below What has every male done at least once in their life? What is something that women do every day that men do once a year? Are boys made to wear girl's panties? Should boys wear girls’ underwear to school? Should boys wear panties? Ishaat Rahman Studied at Bangladesh Army University of Engineering and Technolgy ·7y Hi There! If I'm not mistaken, I think this came from an Arabic riddle that was translated into English. "What is brighter than day darker than night, if eaten, halal if drunk, man uses thrice in a day, woman uses once in life" The answer is Graveyards (Maqabir in Arabic). Note: Due to the many Arabic variations of this riddle, the translation in English seems meaningless. Upvote · 9 5 Sponsored by JetBrains DataGrip, a powerful GUI tool for SQL. Smart code completion, on-the-fly analysis, quick-fixes, refactorings that work in SQL files, and more. Download 999 740 Max Prajapati Lives in Rajkot, Gujarat, India (2016–present) ·1y Actually no answer for this question. But on a adverse, the approximate answer and the most possible one is. South Indian women need to wear only white Lungi half or evenfolded & Vest or Angavastram once in while sometimes or even once a year but their male Correlative wear very regularly and almost daily as it is their traditional clothing. Upvote · 9 1 9 1 Pallavi Software Engineer at Wipro (Indian company) · Author has 392 answers and 586.8K answer views ·Jul 9 I guess it's a tie. Upvote · 9 5 Sponsored by RedHat Know what your AI knows, with open source models. Your AI should keep records, not secrets. Learn More 99 36 Unnati Sajwan Lives in Dehradun, Uttarakhand, India (2022–present) ·Apr 3 A tie 😂 Upvote · 9 8 9 5 Pooja Sharma Lives in New Delhi ·10y Originally Answered: Which is the thing that boys wear every day but girls wear once in the year? · The thing that a boy wear it regular bt a girl wear it once. It is a part of riddle written in the Quran. Quran is the holy book of the Muslims. The answer to that riddle is: Maqabir (Graveyards) Men visit graveyards daily when they bury someone. Upvote · 99 50 9 6 Sponsored by State Bank of India This Festive Season, Make Your Dream Home a Reality! Apply for an SBI Home Loan—easy, quick & hassle-free. Celebrate in a home you can call your own. Learn More 99 68 Rahisha Khatoon Is Live Rahisha khatoon is live · Author has 637 answers and 627.5K answer views ·Jan 13 The answer to this riddle is the letter "Y". Boys wear the letter "Y" daily because it's found in the word "boys," while girls wear it once a year because it's ... Explanation: Technically there's nothing that boys wear every day and girls once a year. Upvote · 9 1 S P Mukherjee Author of Crime novel My Boyfriend’s Fingerprint · Author has 96 answers and 287.1K answer views ·9y Strange but true that while boys wear their 'male ego' everyday, girls get a chance to wear their ego only once and that too under dire circumstances. Every single day of their lives girls face this so called inflated ego and some of us muster up the courage to wear it on too. Upvote · 9 1 Deepika Sai Chitkara When your mind completes COCO with CHANEL.. · Author has 1.2K answers and 4.4M answer views ·Updated 7y Nothing really, though dirty underwear comes pretty close. I really don't get it how can anyone wear the same inner wears for long. Sleeping wear jeans, i get it. Goinf commando, I get it. But dirty underwear.. Kudos to you guys :) Upvote · 9 6 9 6 Molly.Yang Studied at Perking University · Author has 493 answers and 1.5M answer views ·1y The answer to this riddle is the letter "Y". Boys wear the letter "Y" daily because it's found in the word "boys," while girls wear it once a year because it's found in the word "year." Upvote · 99 26 9 4 9 1 Related questions What is one thing men do every day and woman do one time in our lives? What is the thing that women have once in a year? Can boys wear girls tights, yes or no? Is it okay for a boy to wear girl clothes? Why would a little boy want to wear girl clothes? What has every male done at least once in their life? What is something that women do every day that men do once a year? Are boys made to wear girl's panties? Should boys wear girls’ underwear to school? Should boys wear panties? Can 14-year-old boys wear girls’ fluffy pyjamas? Can boys wear girl's clothes and lingerie? Do you think it is okay if a boy wears girl clothes because it is comfortable? Is it a sin for a boy to wear girl clothes? Why does a man do a thing continuously but a woman does the same thing once in her lifetime? Related questions What is one thing men do every day and woman do one time in our lives? What is the thing that women have once in a year? Can boys wear girls tights, yes or no? Is it okay for a boy to wear girl clothes? Why would a little boy want to wear girl clothes? What has every male done at least once in their life? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15604
https://dergipark.org.tr/en/pub/turje/issue/49522/522839
Turkish Journal of Education » Submission » A diagnostic assessment to middle school students’ proportional reasoning Admin PanelUser PanelCoordinator of Scientific Publishing PanelMy JournalsMy ResearchMy FollowersProfile Logout Login Turkish Journal of Education ============================ Research Article TR EN PDF Add To My Library × Create ResearchClose BibTexRISCite Ortaokul öğrencilerinin orantısal akıl yürütmeleri üzerine tanısal bir değerlendirme Year 2019, Volume: 8 Issue: 4, 237 - 257, 31.10.2019 Muhammet Arıcan Cited By: 11 Abstract Bu çalışmada ortaokul öğrencilerinin orantısal akıl yürütmeleri araştırılmış ve oran ve orantı konuları için güçlü ve zayıf yönlerinin bilişsel bir tanısal değerlendirmesi sağlanmıştır. Yirmi iki çoktan seçmeli madde içeren bir orantısal akıl yürütme testi log-linear bilişsel tanı modeli perspektifinden faydalanılarak geliştirilmiştir. Test, ortaokul öğrencilerinin oran ve orantı problemlerini çözmeleri için gerekli olan dört temel bilişsel beceri etrafında tasarlanmıştır. Bu beceriler sırasıyla oran, doğru orantılı ilişki, ters orantılı ilişki ve orantısal olmayan ilişki kavramlarını anlamayı içermektedir. Test 282 yedinci sınıf öğrencisine uygulanmış ve toplanan veriler Mplus yazılımı kullanılarak analiz edilmiştir. Yapılan analizler neticesinde öğrencilerin en çok (yaklaşık 62%) doğru orantılı ilişkileri tanıma becerisine ve en az (yaklaşık 48%) ters orantılı ilişkileri tanıma becerisine sahip oldukları görülmüştür. Ayrıca, öğrencilerin 25%’inin dört temel becerinin hiçbirisine sahip olmadıkları, 39,1%’inin ise bütün becerilere sahip oldukları görülmüştür. Bunlara ek olarak, pek çok öğrencinin orantısal ilişkileri orantısal olmayanlardan ayırt etmede zorlandıkları görülmüştür. Elde edilen bulgular yorumlanarak öğrencilerin güçlü ve zayıf yönleri ile ilgili tanısal geri bildirimler verilmiştir. Keywords Bilişsel tanı modelleri, Ortaokul öğrencileri, Oran ve orantı, Orantısal akıl yürütme, Tanısal değerlendirme References Arican, M. (2018). Preservice middle and high school mathematics teachers’ strategies when solving proportion problems. International Journal of Science and Mathematics Education, 16(2), 315–335. DOI: 10.1007/s10763-016-9775-1 Arican, M. (2019). Preservice mathematics teachers’ understanding of and abilities to differentiate proportional relationships from nonproportional relationships. International Journal of Science and Mathematics Education, 17(7), 1423–1443. DOI: 10.1007/s10763-018-9931-x Arican, M., & Kuzu, O. (2019). Diagnosing preservice teachers’ understanding of statistics and probability: Developing a test for cognitive assessment. International Journal of Science and Mathematics Education, 1–20. DOI: 10.1007/s10763-019-09985-0 Atabas, S., & Oner, D. (2017). An examination of Turkish middle school students’ proportional reasoning. Boğaziçi University Journal of Education, 33(1), 63–85. Ayan, R., & Isiksal-Bostan, M. (2018). Middle school students’ proportional reasoning in real life contexts in the domain of geometry and measurement. International Journal of Mathematical Education in Science and Technology, 1–17. DOI: 10.1080/0020739X.2018.1468042 Beckmann, S. (2011). Mathematics for elementary teachers (3rd. ed.). Boston, MA: Pearson. Bradshaw, L., & Cohen, A. (2010). Accuracy of multidimensional item response model parameters estimated under small sample sizes. In A. Izsák (Chair), Using cognitive attributes to develop mathematics assessments, opportunities, and challenges. Symposium conducted at the annual American Educational Research Association conference in Denver, CO. Bradshaw, L., Izsak, A., Templin, J., & Jacobson, E. (2014). Diagnosing teachers’ understandings of rational numbers: Building a multidimensional test within the diagnostic classification framework. Educational Measurement: Issues and Practice, 33(1), 2–14. DOI: 10.1111/emip.12020 Choi, K. M., Lee, Y. S., & Park, Y. S. (2015). What CDM can tell about what students have learned: An analysis of TIMSS eighth grade mathematics. Eurasia Journal of Mathematics, Science & Technology Education, 11(6), 1563–1577. DOI: 10.12973/eurasia.2015.1421a Common Core State Standards Initiative. (2010). The common core state standards for mathematics. Washington, D.C.: Author. Cramer, K., & Post, T. (1993). Making connections: A case for proportionality. Arithmetic Teacher, 60(6), 342–346. Cramer, K., Post, T., & Currier, S. (1993). Learning and teaching ratio and proportion: Research implications. In D. Owens (Ed.), Research ideas for the classroom: Middle grades mathematics (pp. 159–178). New York, NY: Macmillan. De Bock, D., Verschaffel, L., & Janssens, D. (1998). The predominance of the linear model in secondary school students’ solutions of word problems involving length and area of similar plane figures. Educational Studies in Mathematics, 35(1), 65–83. DOI: 10.1023/A:1003151011999 Degrande, T., Van Hoof, J., Verschaffel, L., & Van Dooren, W. (2017). Open word problems: Taking the additive or the multiplicative road? ZDM, 50(1), 1–12. DOI: 10.1007/s11858-017-0900-6 de la Torre, J. (2008). An empirically based method of Q‐matrix validation for the DINA model: Development and applications. Journal of Educational Measurement, 45(4), 343–362. DOI: 10.1111/j.1745-3984.2008.00069.x de la Torre, J. (2011). The generalized DINA model framework. Psychometrika, 76(2), 179–199. DOI: 10.1007/s11336-011-9207-7 DiBello, L. V., Stout, W. F., & Roussos, L. A. (1995). Unified cognitive/psychometric diagnostic assessment likelihood–based classification techniques. In P. Nichols, S. Chipman, & R. Brennan (Eds.), Cognitively diagnostic assessment (pp. 361–390). Hillsdale, NJ: Lawrence Erlbaum. Dogan, E., & Tatsuoka, K. (2008). An international comparison using a diagnostic testing model: Turkish students’ profile of mathematical skills on TIMSS–R. Educational Studies in Mathematics, 68(3), 263–272. DOI: 10.1007/s10649-007-9099-8 Fisher, L. C. (1988). Strategies used by secondary mathematics teachers to solve proportion problems. Journal for Research in Mathematics Education, 19(2), 157–168. Hartz, S. (2002). A Bayesian framework for the Unified Model for assessing cognitive abilities: Blending theory with practice (Unpublished doctoral dissertation). University of Illinois at Urbana–Champaign. Henson, R., & Douglas, J. (2005). Test construction for cognitive diagnostics. Applied Psychological Measurement, 29(4), 262–277. DOI: 10.1177/0146621604272623 Henson, R., Roussos, L., Douglas, J., & He, X. (2008). Cognitive diagnostic attribute–level discrimination indices. Applied Psychological Measurement, 32(4), 275–288. DOI: 10.1177/0146621607302478 Henson, R., Templin, J., & Willse, J. (2009). Defining a family of cognitive diagnosis models using log–linear models with latent variables. Psychometrika, 74(2), 191–210. DOI: 10.1007/s11336-008-9089-5 Huebner, A. (2010). An overview of recent developments in cognitive diagnostic computer adaptive assessments. Practical Assessment, Research & Evaluation, 15(3), 1–7. Im, S., & Park, H. J. (2010). A comparison of US and Korean students’ mathematics skills using a cognitive diagnostic testing method: Linkage to instruction. Educational Research and Evaluation, 16(3), 287–301. DOI: 10.1080/13803611.2010.523294 Izsák, A., & Jacobson, E. (2017). Preservice teachers’ reasoning about relationships that are and are not proportional: A knowledge-in-pieces account. Journal for Research in Mathematics Education, 48(3), 300–339. DOI: 10.5951/jresematheduc.48.3.0300 Junker, B. W., & Sijtsma, K. (2001). Cognitive assessment models with few assumptions, and connections with nonparametric item response theory. Applied Psychological Measurement, 25(3), 258–272. DOI: 10.1177/01466210122032064 Jurich, D. P., & Bradshaw, L. P. (2014). An illustration of diagnostic classification modeling in student learning outcomes assessment. International Journal of Testing, 14(1), 49–72. DOI: 10.1080/15305058.2013.835728 Kilpatrick, J., Swafford, J., & Findell, B. (2001). Adding it up: Helping children learn mathematics. Washington, DC: National Academy Press. Kuzu, O. (2017). Matematik ve fen bilgisi öğretmen adaylarının integral konusundaki kazanımlarının incelenmesi. Ahi Evran Üniversitesi Kırşehir Eğitim Fakültesi Dergisi, 18(3), 948–970. DOI: 10.29299/kefad.2017.18.3.049 Lamon, S. (2007). Rational numbers and proportional reasoning: Toward a theoretical framework for research. In F. K. Lester, Jr. (Ed.), Second handbook of research on mathematics teaching and learning (Vol 1, pp. 629–667). Charlotte, NC: Information Age Publishing. Lee, Y. S., Park, Y. S., & Taylan, D. (2011). A cognitive diagnostic modeling of attribute mastery in Massachusetts, Minnesota, and the US national sample using the TIMSS 2007. International Journal of Testing, 11(2), 144–177. DOI: 10.1080/15305058.2010.534571 Lei, P. W., & Li, H. (2016). Fit indices’ performance in choosing cognitive diagnostic models and Q-matrices. Paper presented at the annual meeting of the National Council on Measurement in Education (NCME), Philadelphia, PA. Lesh, R., Post, T., & Behr, M. (1988). Proportional reasoning. In J. Hiebert & M. Behr (Eds.), Number concepts and operations in the middle grades (pp. 93–118). Reston, VA: National Council of Teachers of Mathematics. Lim, K. (2009). Burning the candle at just one end: Using nonproportional examples helps students determine when proportional strategies apply. Mathematics Teaching in the Middle School, 14(8), 492–500. Milli Eğitim Bakanlığı (2018). Matematik dersi öğretim programı (1, 2, 3, 4, 5, 6, 7 ve 8. Sınıflar) [Mathematics curriculum (1, 2, 3, 4, 5, 6, 7, and 8. Grades]. Ankara: Talim ve Terbiye Kurulu Başkanlığı. Misailadou, C., & Williams, J. (2003). Measuring children’s proportional reasoning, the “tendency” for an additive strategy and the effect of models. In N. A. Pateman, B. J. Dougherty, & J. T. Zilliox (Eds.), Proceedings of the 27th Conference of the International Group for the Psychology of Mathematics Education (Vol. 3, pp. 293–300). Honolulu, HI: University of Hawaii. Modestou, M., & Gagatsis, A. (2007). Students’ improper proportional reasoning: A result of the epistemological obstacle of “linearity”. Educational Psychology, 27(1), 75–92. DOI: 10.1080/01443410601061462 Muthen, L. K., & Muthen, B. O. (2011). Mplus user’s guide (6th ed.). Los Angeles, CA: Muthen & Muthen. National Council of Teachers of Mathematics (2000). Principles and standards for school mathematics. Reston, VA: Author. Ranjbaran, F., & Alavi, S. M. (2017). Developing a reading comprehension test for cognitive diagnostic assessment: A RUM analysis. Studies in Educational Evaluation, 55, 167–179. DOI: 10.1016/j.stueduc.2017.10.007 Ravand, H., & Robitzsch, A. (2015). Cognitive diagnostic modeling using R. Practical Assessment, Research & Evaluation, 20(11), 1–12. Ravand, H., & Robitzsch, A. (2018). Cognitive diagnostic model of best choice: A study of reading comprehension. Educational Psychology, 38(10), 1255–1277. DOI: 10.1080/01443410.2018.1489524 R Core Team. (2017). R: A language and environment for statistical computing. Vienna, Austria: R Foundation for Statistical Computing. Retrieved from Rupp, A., & Templin, J. (2008). Effects of Q–matrix misspecification on parameter estimates and misclassification rates in the DINA model. Educational and Psychological Measurement, 68(1), 78–98. DOI: 10.1177/0013164407301545 Rupp, A., Templin, J., & Henson, R. A. (2010). Diagnostic measurement: Theory, methods, and applications. Guilford Press. Satorra, A., & Bentler, P. M. (2010). Ensuring positiveness of the scaled difference chi–square test statistic. Psychometrika, 75(2), 243–248. DOI: 10.1007/s11336-009-9135-y Sen, S., & Arican, M. (2015). A diagnostic comparison of Turkish and Korean students’ mathematics performances on the TIMSS 2011 assessment. Journal of Measurement and Evaluation in Education and Psychology, 6(2), 238–253. DOI: 10.21031/epod.65266 Stemn, B. S. (2008). Building middle school students’ understanding of proportional reasoning through mathematical investigation. Education 3–13, 36(4), 383–392. DOI: 10.1080/03004270801959734 Tatsuoka, K. (1985). A probabilistic model for diagnosing misconceptions by the pattern classification approach. Journal of Educational Statistics, 10(1), 55–73. DOI: 10.3102/10769986010001055 Templin, J. (2008). Test construction item discrimination. Lecture presented at the Diagnostic Modelling Seminar at the University of Georgia, Athens. Retrieved from Templin, J., & Bradshaw, L. (2013). Measuring the reliability of diagnostic classification model examinee estimates. Journal of Classification, 30(2), 251–275. DOI: 10.1007/s00357-013-9129-4 Templin, J., & Henson, R. (2006). Measurement of psychological disorders using cognitive diagnosis models. Psychological Methods, 11(3), 287–305. DOI: 10.1037/1082-989X.11.3.287 Terzi, R., & Sen, S. (2019). A nondiagnostic assessment for diagnostic purposes: Q-matrix validation and item-based model fit evaluation for the TIMSS 2011 assessment. SAGE Open, 1–11. DOI: 10.1177/2158244019832684 Toker, T., & Green, K. (2012). An application of cognitive diagnostic assessment on TIMMS–2007 8th grade mathematics items. Paper presented at the annual meeting of the American Educational Research Association, Vancouver, British Columbia, Canada. Van Dooren, W., De Bock, D., Hessels, A., Janssens, D., & Verschaffel, L. (2005). Not everything is proportional: Effects of age and problem type on propensities for overgeneralization. Cognition and Instruction, 23(1), 57–86. DOI: 10.1207/s1532690xci2301_3 Van Dooren, W., De Bock, D., Janssens, D., & Verschaffel, L. (2007). Pupils’ overreliance on linearity: A scholastic effect? British Journal of Educational Psychology, 77(2), 307–321. DOI: 10.1348/000709906X115967 von Davier, M. (2005). A general diagnostic model applied to language testing data. ETS Research Report. Princeton, NJ: Educational Testing Service. Werner, C., & Schermelleh-Engel, K. (2010). Deciding between competing models: Chi–square difference tests. In Introduction to Structural Equation Modeling with LISREL (pp. 1–3). Frankfurt, Germany: Goethe University. A diagnostic assessment to middle school students’ proportional reasoning Year 2019, Volume: 8 Issue: 4, 237 - 257, 31.10.2019 Muhammet Arıcan Cited By: 11 Abstract This study investigated Turkish middle school students’ proportional reasoning and provided a diagnostic assessment of their strengths and weaknesses on the ratio and proportion concepts. A proportional reasoning test with 22 multiple-choice items was developed from the context of the log-linear cognitive diagnosis model. The test was developed around four core cognitive skills (attributes) that required in solving middle school ratio and proportion problems. These skills included understanding ratios, directly, inversely, and nonproportional relationships. The test was applied to 282 seventh grade students, and the collected data were analyzed using the Mplus software. The analysis showed that approximately 62% of the students were able to recognize directly proportional relationships. Whereas, roughly 48% of them were able to recognize inversely proportional relationships. Moreover, while 25% of the students did not master any of the four cognitive skills, 39.1% mastered all four of these skills. In addition, many students had difficulty distinguishing proportional relationships from nonproportional relationships. Diagnostic feedbacks on the students’ strengths and weaknesses were provided based on the findings. Keywords Diagnostic assessment, Diagnostic classification models, Middle school students, Ratios and proportions, Proportional reasoning Thanks I would like to thank Dr. Sedat Şen and Dr. Ragıp Terzi for their valuable feedback. References Arican, M. (2018). Preservice middle and high school mathematics teachers’ strategies when solving proportion problems. International Journal of Science and Mathematics Education, 16(2), 315–335. DOI: 10.1007/s10763-016-9775-1 Arican, M. (2019). Preservice mathematics teachers’ understanding of and abilities to differentiate proportional relationships from nonproportional relationships. International Journal of Science and Mathematics Education, 17(7), 1423–1443. DOI: 10.1007/s10763-018-9931-x Arican, M., & Kuzu, O. (2019). Diagnosing preservice teachers’ understanding of statistics and probability: Developing a test for cognitive assessment. International Journal of Science and Mathematics Education, 1–20. DOI: 10.1007/s10763-019-09985-0 Atabas, S., & Oner, D. (2017). An examination of Turkish middle school students’ proportional reasoning. Boğaziçi University Journal of Education, 33(1), 63–85. Ayan, R., & Isiksal-Bostan, M. (2018). Middle school students’ proportional reasoning in real life contexts in the domain of geometry and measurement. International Journal of Mathematical Education in Science and Technology, 1–17. DOI: 10.1080/0020739X.2018.1468042 Beckmann, S. (2011). Mathematics for elementary teachers (3rd. ed.). Boston, MA: Pearson. Bradshaw, L., & Cohen, A. (2010). Accuracy of multidimensional item response model parameters estimated under small sample sizes. In A. Izsák (Chair), Using cognitive attributes to develop mathematics assessments, opportunities, and challenges. Symposium conducted at the annual American Educational Research Association conference in Denver, CO. Bradshaw, L., Izsak, A., Templin, J., & Jacobson, E. (2014). Diagnosing teachers’ understandings of rational numbers: Building a multidimensional test within the diagnostic classification framework. Educational Measurement: Issues and Practice, 33(1), 2–14. DOI: 10.1111/emip.12020 Choi, K. M., Lee, Y. S., & Park, Y. S. (2015). What CDM can tell about what students have learned: An analysis of TIMSS eighth grade mathematics. Eurasia Journal of Mathematics, Science & Technology Education, 11(6), 1563–1577. DOI: 10.12973/eurasia.2015.1421a Common Core State Standards Initiative. (2010). The common core state standards for mathematics. Washington, D.C.: Author. Cramer, K., & Post, T. (1993). Making connections: A case for proportionality. Arithmetic Teacher, 60(6), 342–346. Cramer, K., Post, T., & Currier, S. (1993). Learning and teaching ratio and proportion: Research implications. In D. Owens (Ed.), Research ideas for the classroom: Middle grades mathematics (pp. 159–178). New York, NY: Macmillan. De Bock, D., Verschaffel, L., & Janssens, D. (1998). The predominance of the linear model in secondary school students’ solutions of word problems involving length and area of similar plane figures. Educational Studies in Mathematics, 35(1), 65–83. DOI: 10.1023/A:1003151011999 Degrande, T., Van Hoof, J., Verschaffel, L., & Van Dooren, W. (2017). Open word problems: Taking the additive or the multiplicative road? ZDM, 50(1), 1–12. DOI: 10.1007/s11858-017-0900-6 de la Torre, J. (2008). An empirically based method of Q‐matrix validation for the DINA model: Development and applications. Journal of Educational Measurement, 45(4), 343–362. DOI: 10.1111/j.1745-3984.2008.00069.x de la Torre, J. (2011). The generalized DINA model framework. Psychometrika, 76(2), 179–199. DOI: 10.1007/s11336-011-9207-7 DiBello, L. V., Stout, W. F., & Roussos, L. A. (1995). Unified cognitive/psychometric diagnostic assessment likelihood–based classification techniques. In P. Nichols, S. Chipman, & R. Brennan (Eds.), Cognitively diagnostic assessment (pp. 361–390). Hillsdale, NJ: Lawrence Erlbaum. Dogan, E., & Tatsuoka, K. (2008). An international comparison using a diagnostic testing model: Turkish students’ profile of mathematical skills on TIMSS–R. Educational Studies in Mathematics, 68(3), 263–272. DOI: 10.1007/s10649-007-9099-8 Fisher, L. C. (1988). Strategies used by secondary mathematics teachers to solve proportion problems. Journal for Research in Mathematics Education, 19(2), 157–168. Hartz, S. (2002). A Bayesian framework for the Unified Model for assessing cognitive abilities: Blending theory with practice (Unpublished doctoral dissertation). University of Illinois at Urbana–Champaign. Henson, R., & Douglas, J. (2005). Test construction for cognitive diagnostics. Applied Psychological Measurement, 29(4), 262–277. DOI: 10.1177/0146621604272623 Henson, R., Roussos, L., Douglas, J., & He, X. (2008). Cognitive diagnostic attribute–level discrimination indices. Applied Psychological Measurement, 32(4), 275–288. DOI: 10.1177/0146621607302478 Henson, R., Templin, J., & Willse, J. (2009). Defining a family of cognitive diagnosis models using log–linear models with latent variables. Psychometrika, 74(2), 191–210. DOI: 10.1007/s11336-008-9089-5 Huebner, A. (2010). An overview of recent developments in cognitive diagnostic computer adaptive assessments. Practical Assessment, Research & Evaluation, 15(3), 1–7. Im, S., & Park, H. J. (2010). A comparison of US and Korean students’ mathematics skills using a cognitive diagnostic testing method: Linkage to instruction. Educational Research and Evaluation, 16(3), 287–301. DOI: 10.1080/13803611.2010.523294 Izsák, A., & Jacobson, E. (2017). Preservice teachers’ reasoning about relationships that are and are not proportional: A knowledge-in-pieces account. Journal for Research in Mathematics Education, 48(3), 300–339. DOI: 10.5951/jresematheduc.48.3.0300 Junker, B. W., & Sijtsma, K. (2001). Cognitive assessment models with few assumptions, and connections with nonparametric item response theory. Applied Psychological Measurement, 25(3), 258–272. DOI: 10.1177/01466210122032064 Jurich, D. P., & Bradshaw, L. P. (2014). An illustration of diagnostic classification modeling in student learning outcomes assessment. International Journal of Testing, 14(1), 49–72. DOI: 10.1080/15305058.2013.835728 Kilpatrick, J., Swafford, J., & Findell, B. (2001). Adding it up: Helping children learn mathematics. Washington, DC: National Academy Press. Kuzu, O. (2017). Matematik ve fen bilgisi öğretmen adaylarının integral konusundaki kazanımlarının incelenmesi. Ahi Evran Üniversitesi Kırşehir Eğitim Fakültesi Dergisi, 18(3), 948–970. DOI: 10.29299/kefad.2017.18.3.049 Lamon, S. (2007). Rational numbers and proportional reasoning: Toward a theoretical framework for research. In F. K. Lester, Jr. (Ed.), Second handbook of research on mathematics teaching and learning (Vol 1, pp. 629–667). Charlotte, NC: Information Age Publishing. Lee, Y. S., Park, Y. S., & Taylan, D. (2011). A cognitive diagnostic modeling of attribute mastery in Massachusetts, Minnesota, and the US national sample using the TIMSS 2007. International Journal of Testing, 11(2), 144–177. DOI: 10.1080/15305058.2010.534571 Lei, P. W., & Li, H. (2016). Fit indices’ performance in choosing cognitive diagnostic models and Q-matrices. Paper presented at the annual meeting of the National Council on Measurement in Education (NCME), Philadelphia, PA. Lesh, R., Post, T., & Behr, M. (1988). Proportional reasoning. In J. Hiebert & M. Behr (Eds.), Number concepts and operations in the middle grades (pp. 93–118). Reston, VA: National Council of Teachers of Mathematics. Lim, K. (2009). Burning the candle at just one end: Using nonproportional examples helps students determine when proportional strategies apply. Mathematics Teaching in the Middle School, 14(8), 492–500. Milli Eğitim Bakanlığı (2018). Matematik dersi öğretim programı (1, 2, 3, 4, 5, 6, 7 ve 8. Sınıflar) [Mathematics curriculum (1, 2, 3, 4, 5, 6, 7, and 8. Grades]. Ankara: Talim ve Terbiye Kurulu Başkanlığı. Misailadou, C., & Williams, J. (2003). Measuring children’s proportional reasoning, the “tendency” for an additive strategy and the effect of models. In N. A. Pateman, B. J. Dougherty, & J. T. Zilliox (Eds.), Proceedings of the 27th Conference of the International Group for the Psychology of Mathematics Education (Vol. 3, pp. 293–300). Honolulu, HI: University of Hawaii. Modestou, M., & Gagatsis, A. (2007). Students’ improper proportional reasoning: A result of the epistemological obstacle of “linearity”. Educational Psychology, 27(1), 75–92. DOI: 10.1080/01443410601061462 Muthen, L. K., & Muthen, B. O. (2011). Mplus user’s guide (6th ed.). Los Angeles, CA: Muthen & Muthen. National Council of Teachers of Mathematics (2000). Principles and standards for school mathematics. Reston, VA: Author. Ranjbaran, F., & Alavi, S. M. (2017). Developing a reading comprehension test for cognitive diagnostic assessment: A RUM analysis. Studies in Educational Evaluation, 55, 167–179. DOI: 10.1016/j.stueduc.2017.10.007 Ravand, H., & Robitzsch, A. (2015). Cognitive diagnostic modeling using R. Practical Assessment, Research & Evaluation, 20(11), 1–12. Ravand, H., & Robitzsch, A. (2018). Cognitive diagnostic model of best choice: A study of reading comprehension. Educational Psychology, 38(10), 1255–1277. DOI: 10.1080/01443410.2018.1489524 R Core Team. (2017). R: A language and environment for statistical computing. Vienna, Austria: R Foundation for Statistical Computing. Retrieved from Rupp, A., & Templin, J. (2008). Effects of Q–matrix misspecification on parameter estimates and misclassification rates in the DINA model. Educational and Psychological Measurement, 68(1), 78–98. DOI: 10.1177/0013164407301545 Rupp, A., Templin, J., & Henson, R. A. (2010). Diagnostic measurement: Theory, methods, and applications. Guilford Press. Satorra, A., & Bentler, P. M. (2010). Ensuring positiveness of the scaled difference chi–square test statistic. Psychometrika, 75(2), 243–248. DOI: 10.1007/s11336-009-9135-y Sen, S., & Arican, M. (2015). A diagnostic comparison of Turkish and Korean students’ mathematics performances on the TIMSS 2011 assessment. Journal of Measurement and Evaluation in Education and Psychology, 6(2), 238–253. DOI: 10.21031/epod.65266 Stemn, B. S. (2008). Building middle school students’ understanding of proportional reasoning through mathematical investigation. Education 3–13, 36(4), 383–392. DOI: 10.1080/03004270801959734 Tatsuoka, K. (1985). A probabilistic model for diagnosing misconceptions by the pattern classification approach. Journal of Educational Statistics, 10(1), 55–73. DOI: 10.3102/10769986010001055 Templin, J. (2008). Test construction item discrimination. Lecture presented at the Diagnostic Modelling Seminar at the University of Georgia, Athens. Retrieved from Templin, J., & Bradshaw, L. (2013). Measuring the reliability of diagnostic classification model examinee estimates. Journal of Classification, 30(2), 251–275. DOI: 10.1007/s00357-013-9129-4 Templin, J., & Henson, R. (2006). Measurement of psychological disorders using cognitive diagnosis models. Psychological Methods, 11(3), 287–305. DOI: 10.1037/1082-989X.11.3.287 Terzi, R., & Sen, S. (2019). A nondiagnostic assessment for diagnostic purposes: Q-matrix validation and item-based model fit evaluation for the TIMSS 2011 assessment. SAGE Open, 1–11. DOI: 10.1177/2158244019832684 Toker, T., & Green, K. (2012). An application of cognitive diagnostic assessment on TIMMS–2007 8th grade mathematics items. Paper presented at the annual meeting of the American Educational Research Association, Vancouver, British Columbia, Canada. Van Dooren, W., De Bock, D., Hessels, A., Janssens, D., & Verschaffel, L. (2005). Not everything is proportional: Effects of age and problem type on propensities for overgeneralization. Cognition and Instruction, 23(1), 57–86. DOI: 10.1207/s1532690xci2301_3 Van Dooren, W., De Bock, D., Janssens, D., & Verschaffel, L. (2007). Pupils’ overreliance on linearity: A scholastic effect? British Journal of Educational Psychology, 77(2), 307–321. DOI: 10.1348/000709906X115967 von Davier, M. (2005). A general diagnostic model applied to language testing data. ETS Research Report. Princeton, NJ: Educational Testing Service. Werner, C., & Schermelleh-Engel, K. (2010). Deciding between competing models: Chi–square difference tests. In Introduction to Structural Equation Modeling with LISREL (pp. 1–3). Frankfurt, Germany: Goethe University. Show All References Show Less References There are 59 citations in total. Details | Primary Language | English | | Subjects | Studies on Education | | Journal Section | Research Articles | | Authors | Muhammet ArıcanKIRŞEHİR AHİ EVRAN ÜNİVERSİTESİ 0000-0002-0496-9148Türkiye | | Publication Date | October 31, 2019 | | Acceptance Date | October 12, 2019 | | Published in Issue | Year 2019 Volume: 8 Issue: 4 | Cite APA Arıcan, M. (2019). A diagnostic assessment to middle school students’ proportional reasoning. Turkish Journal of Education, 8(4), 237-257. Cited By Matematik ve Fen Bilgisi Öğretmeni Adaylarının İntegral Konusundaki Yeterliklerinin Tanısal Değerlendirilmesi Yuzunci Yil Universitesi Egitim Fakultesi Dergisi Okan KUZU Developing a Two-Tier Proportional Reasoning Skill Test: Validity and Reliability Studies International Journal of Assessment Tools in Education Kübra AÇIKGÜL Erken Orantısal Akıl Yürütmeye Yönelik Öğrenme Rotasının Geliştirilmesi: Bir Tasarı Araştırması Mersin Üniversitesi Eğitim Fakültesi Dergisi Developing reflective competence in prospective mathematics teachers by analyzing textbooks lessons Eurasia Journal of Mathematics, Science and Technology Education The Effect of Error Analysis-Based Learning on Proportional Reasoning Ability of Seventh-Grade Students Frontiers in Education Online Cognitive Diagnostic Assessment for Relationship of Time International Journal of Virtual and Personal Learning Environments A diagnostic comparison of Spanish and Turkish middle school students’ proportional reasoning Asian Journal for Mathematics Education 7. Sınıf Öğrencilerine Yönelik Bir Orantısal Akıl Yürütme Beceri Testi (OAYBT) Geliştirilmesi İnönü Üniversitesi Eğitim Fakültesi Dergisi Orantısal Akıl Yürütmeye İlk Adım: Birimleri Bağlama ve Yinelemedeki Temsiller İnönü Üniversitesi Eğitim Fakültesi Dergisi Beyond the Across Algorithm: Informal Strategies for Inverse Proportional Problems International Journal of Science and Mathematics Education ORANTISAL AKIL YÜRÜTME KONULU LİSANSÜSTÜ TEZLERİN SİSTEMATİK LİTERATÜR TARAMASI Trakya Eğitim Dergisi Download Cover Image Article Files Full Text Journal Home Page Archive Volume: 8 Issue: 2 Volume: 8 Issue: 3 Volume: 8 Issue: 4 Volume: 9 Issue: 1 Volume: 9 Issue: 2 Turkish Journal of Education is licensed under CC BY-NC 4.0
15605
https://www.tutor.com/resources/math/algebra/systems-of-equations%2Finequalities
Systems of Equations/Inequalities - Algebra - Math - Homework Resources - Tutor.com Military Families The official provider of online tutoring and homework help to the Department of Defense. Check Eligibility Higher Education Improve persistence and course completion with 24/7 student support online. How it Works Public Libraries Engage your community with learning and career services for patrons of all ages. Learn More Corporate Partners Support your workforce and their families with a unique employee benefit. Get Started Tutor.com is now part of The Princeton Review! Learn more Testimonials Subjects Become a Tutor Sign In 28,763,090 Sessions and Counting! Higher Education Libraries K-12 Schools Employee Benefits U.S. Military Families Individual Subscriptions Higher Education Libraries K-12 Schools Employee Benefits U.S. Military Families Individual Subscriptions Become a Tutor Sign In Get a Tutor < Back to Systems of Equations/Inequalities (18) Math - Algebra Subtopics.show();$('#resource-subtopic-list-zipper').hide();) Elementary Statistics Expressions Factoring Factors Functions Inequalities Integers Linear Equations Polynomials Powers/Roots Probability Proportions Quadratic and Exponential Functions Quadratic Equations Radical Expressions and Triangles Rational Expressions and Equations Rational Numbers Real Numbers Solving Linear Equations Systems of Equations/Inequalities Word Problems Systems of Equations/Inequalities Sort by: Algebra I: Graphing Inequalities This video includes sample problems and step-by-step explanations of graphing inequalities and testing assertions for the California Standards Test. Khan Academy Word problem involving tickets Shows example problems for the common word problem involving cost of tickets, number of tickets sold, and dollar amount collected. Online Math Learning.com Early Edge: Graphing Inequalities In this Early Edge video lesson, you'll learn more about Graphing Inequalities, so you can be successful when you take on high-school Math & Algebra. The Princeton Review Early Edge: Working with Inequalities In this Early Edge video lesson, you'll learn more about Working with Inequalities, so you can be successful when you take on high-school Math & Algebra. The Princeton Review Algebra Terms Algebra Terms StudyStack Systems of equations with Sal Khan In this video, Salman Khan of Khan Academy explains systems of equations. Khan Academy Algebra I: Systems of Inequalities This video includes sample problems and step-by-step explanations of systems of equations and inequalities for the California Standards Test. Khan Academy Online Math Examples Excellent site showing examples of algebra, trig, calculus, differential equations, and linear algebra. Paul's Online Math Notes Online Graphing Calculator This site provides graphing tools. The GCalc 3 menu has several options for graphing. One nice aspect of this site is that multiple functions can be graphed in different colors on the same screen. Another useful feature is that it is easy to copy and paste or print the graphs that are generated. GCalc - Java Online Graphing Calculator Simultaneous Equations by Elimination An instructional walk through of solving Equations using elimination. Teacher's Choice 1.22Next > Learn more FAQ Our Tutors Our Company Press Blog Contact Us Site Map Homework Resources Subjects Math Science English Social Studies AP Support SAT/Test Prep Proudly Serving Military Families Libraries Colleges & Universities For more information call us at: 800-411-1970 ©2025 Tutor.com / The Princeton Review - Not Affiliated with Princeton University. Privacy Policy|CCPA Privacy Notice|Terms of Use
15606
https://en.wikipedia.org/wiki/Covariance_matrix
Jump to content Search Contents 1 Definition 1.1 Conflicting nomenclatures and notations 2 Properties 2.1 Relation to the autocorrelation matrix 2.2 Relation to the correlation matrix 2.3 Inverse of the covariance matrix 2.4 Basic properties 2.5 Block matrices 3 Partial covariance matrix 4 Standard deviation matrix 5 Covariance matrix as a parameter of a distribution 6 Covariance matrix as a linear operator 7 Admissibility 8 Complex random vectors 8.1 Pseudo-covariance matrix 9 Estimation 10 Applications 10.1 Use in optimization 10.2 Covariance mapping 10.3 Two-dimensional infrared spectroscopy 11 See also 12 References 13 Further reading Covariance matrix العربية Català Čeština Deutsch Español فارسی Français 한국어 Italiano Magyar Nederlands 日本語 Polski Português Русский Slovenčina Slovenščina Sunda Türkçe Українська Tiếng Việt 粵語 中文 Edit links Article Talk Read Edit View history Tools Actions Read Edit 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 Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Measure of covariance of components of a random vector Not to be confused with Cross-covariance matrix. | | | Part of a series on Statistics | | Correlation and covariance | | For random vectors Autocorrelation matrix Cross-correlation matrix Auto-covariance matrix Cross-covariance matrix | | For stochastic processes Autocorrelation function Cross-correlation function Autocovariance function Cross-covariance function | | For deterministic signals Autocorrelation function Cross-correlation function Autocovariance function Cross-covariance function | | v t e | In probability theory and statistics, a covariance matrix (also known as auto-covariance matrix, dispersion matrix, variance matrix, or variance–covariance matrix) is a square matrix giving the covariance between each pair of elements of a given random vector. Intuitively, the covariance matrix generalizes the notion of variance to multiple dimensions. As an example, the variation in a collection of random points in two-dimensional space cannot be characterized fully by a single number, nor would the variances in the and directions contain all of the necessary information; a matrix would be necessary to fully characterize the two-dimensional variation. Any covariance matrix is symmetric and positive semi-definite and its main diagonal contains variances (i.e., the covariance of each element with itself). The covariance matrix of a random vector is typically denoted by , or . Definition [edit] Throughout this article, boldfaced unsubscripted and are used to refer to random vectors, and Roman subscripted and are used to refer to scalar random variables. If the entries in the column vector are random variables, each with finite variance and expected value, then the covariance matrix is the matrix whose entry is the covariance: 177 where the operator denotes the expected value (mean) of its argument. Conflicting nomenclatures and notations [edit] Nomenclatures differ. Some statisticians, following the probabilist William Feller in his two-volume book An Introduction to Probability Theory and Its Applications, call the matrix the variance of the random vector , because it is the natural generalization to higher dimensions of the 1-dimensional variance. Others call it the covariance matrix, because it is the matrix of covariances between the scalar components of the vector . Both forms are quite standard, and there is no ambiguity between them. The matrix is also often called the variance-covariance matrix, since the diagonal terms are in fact variances. By comparison, the notation for the cross-covariance matrix between two vectors is Properties [edit] Relation to the autocorrelation matrix [edit] The auto-covariance matrix is related to the autocorrelation matrix by where the autocorrelation matrix is defined as . Relation to the correlation matrix [edit] Further information: Correlation matrix An entity closely related to the covariance matrix is the matrix of Pearson product-moment correlation coefficients between each of the random variables in the random vector , which can be written as where is the matrix of the diagonal elements of (i.e., a diagonal matrix of the variances of for ). Equivalently, the correlation matrix can be seen as the covariance matrix of the standardized random variables for . Each element on the principal diagonal of a correlation matrix is the correlation of a random variable with itself, which always equals 1. Each off-diagonal element is between −1 and +1 inclusive. Inverse of the covariance matrix [edit] The inverse of this matrix, , if it exists, is the inverse covariance matrix (or inverse concentration matrix[dubious – discuss]), also known as the precision matrix (or concentration matrix). Just as the covariance matrix can be written as the rescaling of a correlation matrix by the marginal variances: So, using the idea of partial correlation, and partial variance, the inverse covariance matrix can be expressed analogously: This duality motivates a number of other dualities between marginalizing and conditioning for Gaussian random variables. Basic properties [edit] For and , where is an -dimensional random variable, the following basic properties apply: is positive-semidefinite, i.e. Proof Indeed, from the property 4 it follows that under linear transformation of random variable with covariation matrix by linear operator s.a. , the covariation matrix is tranformed as : . As according to the property 3 matrix is symmetric, it can be diagonalized by a linear orthogonal transformation, i.e. there exists such orthogonal matrix (meanwhile ), that and are eigenvalues of . But this means that this matrix is a covariation matrix for a random variable , and the main diagonal of consists of variances of elements of vector. As variance is always non-negative, we conclude that for any . But this means that matrix is positive-semidefinite. is symmetric, i.e. For any constant (i.e. non-random) matrix and constant vector , one has If is another random vector with the same dimension as , then where is the cross-covariance matrix of and . Block matrices [edit] The joint mean and joint covariance matrix of and can be written in block form where , and . and can be identified as the variance matrices of the marginal distributions for and respectively. If and are jointly normally distributed, then the conditional distribution for given is given by defined by conditional mean and conditional variance The matrix is known as the matrix of regression coefficients, while in linear algebra is the Schur complement of in . The matrix of regression coefficients may often be given in transpose form, , suitable for post-multiplying a row vector of explanatory variables rather than pre-multiplying a column vector . In this form they correspond to the coefficients obtained by inverting the matrix of the normal equations of ordinary least squares (OLS). Partial covariance matrix [edit] A covariance matrix with all non-zero elements tells us that all the individual random variables are interrelated. This means that the variables are not only directly correlated, but also correlated via other variables indirectly. Often such indirect, common-mode correlations are trivial and uninteresting. They can be suppressed by calculating the partial covariance matrix, that is the part of covariance matrix that shows only the interesting part of correlations. If two vectors of random variables and are correlated via another vector , the latter correlations are suppressed in a matrix The partial covariance matrix is effectively the simple covariance matrix as if the uninteresting random variables were held constant. Standard deviation matrix [edit] Main article: Standard deviation § Standard deviation matrix The standard deviation matrix is the extension of the standard deviation to multiple dimensions. It is the symmetric square root of the covariance matrix . Covariance matrix as a parameter of a distribution [edit] If a column vector of possibly correlated random variables is jointly normally distributed, or more generally elliptically distributed, then its probability density function can be expressed in terms of the covariance matrix as follows where and is the determinant of , the so-called generalized variance. Covariance matrix as a linear operator [edit] Main article: Covariance operator Applied to one vector, the covariance matrix maps a linear combination c of the random variables X onto a vector of covariances with those variables: . Treated as a bilinear form, it yields the covariance between the two linear combinations: . The variance of a linear combination is then , its covariance with itself. Similarly, the (pseudo-)inverse covariance matrix provides an inner product , which induces the Mahalanobis distance, a measure of the "unlikelihood" of c.[citation needed] Admissibility [edit] From basic property 4. above, let be a real-valued vector, then which must always be nonnegative, since it is the variance of a real-valued random variable, so a covariance matrix is always a positive-semidefinite matrix. The above argument can be expanded as follows:where the last inequality follows from the observation that is a scalar. Conversely, every symmetric positive semi-definite matrix is a covariance matrix. To see this, suppose is a symmetric positive-semidefinite matrix. From the finite-dimensional case of the spectral theorem, it follows that has a nonnegative symmetric square root, which can be denoted by M1/2. Let be any column vector-valued random variable whose covariance matrix is the identity matrix. Then Complex random vectors [edit] Further information: Complex random vector § Covariance matrix and pseudo-covariance matrix The variance of a complex scalar-valued random variable with expected value is conventionally defined using complex conjugation: where the complex conjugate of a complex number is denoted ; thus the variance of a complex random variable is a real number. If is a column vector of complex-valued random variables, then the conjugate transpose is formed by both transposing and conjugating. In the following expression, the product of a vector with its conjugate transpose results in a square matrix called the covariance matrix, as its expectation:: 293 The matrix so obtained will be Hermitian positive-semidefinite, with real numbers in the main diagonal and complex numbers off-diagonal. Properties The covariance matrix is a Hermitian matrix, i.e. .: 179 The diagonal elements of the covariance matrix are real.: 179 Pseudo-covariance matrix [edit] For complex random vectors, another kind of second central moment, the pseudo-covariance matrix (also called relation matrix) is defined as follows: In contrast to the covariance matrix defined above, Hermitian transposition gets replaced by transposition in the definition. Its diagonal elements may be complex valued; it is a complex symmetric matrix. Estimation [edit] Main article: Estimation of covariance matrices If and are centered data matrices of dimension and respectively, i.e. with n columns of observations of p and q rows of variables, from which the row means have been subtracted, then, if the row means were estimated from the data, sample covariance matrices and can be defined to be or, if the row means were known a priori, These empirical sample covariance matrices are the most straightforward and most often used estimators for the covariance matrices, but other estimators also exist, including regularised or shrinkage estimators, which may have better properties. Applications [edit] The covariance matrix is a useful tool in many different areas. From it a transformation matrix can be derived, called a whitening transformation, that allows one to completely decorrelate the data or, from a different point of view, to find an optimal basis for representing the data in a compact way[citation needed] (see Rayleigh quotient for a formal proof and additional properties of covariance matrices). This is called principal component analysis (PCA) and the Karhunen–Loève transform (KL-transform). The covariance matrix plays a key role in financial economics, especially in portfolio theory and its mutual fund separation theorem and in the capital asset pricing model. The matrix of covariances among various assets' returns is used to determine, under certain assumptions, the relative amounts of different assets that investors should (in a normative analysis) or are predicted to (in a positive analysis) choose to hold in a context of diversification. Use in optimization [edit] The evolution strategy, a particular family of Randomized Search Heuristics, fundamentally relies on a covariance matrix in its mechanism. The characteristic mutation operator draws the update step from a multivariate normal distribution using an evolving covariance matrix. There is a formal proof that the evolution strategy's covariance matrix adapts to the inverse of the Hessian matrix of the search landscape, up to a scalar factor and small random fluctuations (proven for a single-parent strategy and a static model, as the population size increases, relying on the quadratic approximation). Intuitively, this result is supported by the rationale that the optimal covariance distribution can offer mutation steps whose equidensity probability contours match the level sets of the landscape, and so they maximize the progress rate. Covariance mapping [edit] In covariance mapping the values of the or matrix are plotted as a 2-dimensional map. When vectors and are discrete random functions, the map shows statistical relations between different regions of the random functions. Statistically independent regions of the functions show up on the map as zero-level flatland, while positive or negative correlations show up, respectively, as hills or valleys. In practice the column vectors , and are acquired experimentally as rows of samples, e.g. where is the i-th discrete value in sample j of the random function . The expected values needed in the covariance formula are estimated using the sample mean, e.g. and the covariance matrix is estimated by the sample covariance matrix where the angular brackets denote sample averaging as before except that the Bessel's correction should be made to avoid bias. Using this estimation the partial covariance matrix can be calculated as where the backslash denotes the left matrix division operator, which bypasses the requirement to invert a matrix and is available in some computational packages such as Matlab. Fig. 1 illustrates how a partial covariance map is constructed on an example of an experiment performed at the FLASH free-electron laser in Hamburg. The random function is the time-of-flight spectrum of ions from a Coulomb explosion of nitrogen molecules multiply ionised by a laser pulse. Since only a few hundreds of molecules are ionised at each laser pulse, the single-shot spectra are highly fluctuating. However, collecting typically such spectra, , and averaging them over produces a smooth spectrum , which is shown in red at the bottom of Fig. 1. The average spectrum reveals several nitrogen ions in a form of peaks broadened by their kinetic energy, but to find the correlations between the ionisation stages and the ion momenta requires calculating a covariance map. In the example of Fig. 1 spectra and are the same, except that the range of the time-of-flight differs. Panel a shows , panel b shows and panel c shows their difference, which is (note a change in the colour scale). Unfortunately, this map is overwhelmed by uninteresting, common-mode correlations induced by laser intensity fluctuating from shot to shot. To suppress such correlations the laser intensity is recorded at every shot, put into and is calculated as panels d and e show. The suppression of the uninteresting correlations is, however, imperfect because there are other sources of common-mode fluctuations than the laser intensity and in principle all these sources should be monitored in vector . Yet in practice it is often sufficient to overcompensate the partial covariance correction as panel f shows, where interesting correlations of ion momenta are now clearly visible as straight lines centred on ionisation stages of atomic nitrogen. Two-dimensional infrared spectroscopy [edit] Two-dimensional infrared spectroscopy employs correlation analysis to obtain 2D spectra of the condensed phase. There are two versions of this analysis: synchronous and asynchronous. Mathematically, the former is expressed in terms of the sample covariance matrix and the technique is equivalent to covariance mapping. See also [edit] Covariance function Eigenvalue decomposition Gramian matrix Lewandowski-Kurowicka-Joe distribution Multivariate statistics Principal components Quadratic form (statistics) References [edit] ^ a b c Park, Kun Il (2018). Fundamentals of Probability and Stochastic Processes with Applications to Communications. Springer. ISBN 978-3-319-68074-3. ^ William Feller (1971). An introduction to probability theory and its applications. Wiley. ISBN 978-0-471-25709-7. Retrieved 10 August 2012. ^ Wasserman, Larry (2004). All of Statistics: A Concise Course in Statistical Inference. Springer. ISBN 0-387-40272-1. ^ Taboga, Marco (2010). "Lectures on probability theory and mathematical statistics". ^ Eaton, Morris L. (1983). Multivariate Statistics: a Vector Space Approach. John Wiley and Sons. pp. 116–117. ISBN 0-471-02776-6. ^ a b W J Krzanowski "Principles of Multivariate Analysis" (Oxford University Press, New York, 1988), Chap. 14.4; K V Mardia, J T Kent and J M Bibby "Multivariate Analysis (Academic Press, London, 1997), Chap. 6.5.3; T W Anderson "An Introduction to Multivariate Statistical Analysis" (Wiley, New York, 2003), 3rd ed., Chaps. 2.5.1 and 4.3.1. ^ Das, Abhranil; Wilson S Geisler (2020). "Methods to integrate multinormals and compute classification measures". arXiv:2012.14331 [stat.ML]. ^ Lapidoth, Amos (2009). A Foundation in Digital Communication. Cambridge University Press. ISBN 978-0-521-19395-5. ^ Brookes, Mike. "The Matrix Reference Manual". ^ Kessy, Agnan; Strimmer, Korbinian; Lewin, Alex (2018). "Optimal Whitening and Decorrelation". The American Statistician. 72 (4). Taylor & Francis: 309–314. arXiv:1512.00809. doi:10.1080/00031305.2016.1277159. ^ Shir, O.M.; A. Yehudayoff (2020). "On the covariance-Hessian relation in evolution strategies". Theoretical Computer Science. 801. Elsevier: 157–174. arXiv:1806.03674. doi:10.1016/j.tcs.2019.09.002. ^ L J Frasinski "Covariance mapping techniques" J. Phys. B: At. Mol. Opt. Phys. 49 152004 (2016), doi:10.1088/0953-4075/49/15/152004 ^ a b O Kornilov, M Eckstein, M Rosenblatt, C P Schulz, K Motomura, A Rouzée, J Klei, L Foucar, M Siano, A Lübcke, F. Schapper, P Johnsson, D M P Holland, T Schlatholter, T Marchenko, S Düsterer, K Ueda, M J J Vrakking and L J Frasinski "Coulomb explosion of diatomic molecules in intense XUV fields mapped by partial covariance" J. Phys. B: At. Mol. Opt. Phys. 46 164028 (2013), doi:10.1088/0953-4075/46/16/164028 ^ Noda, I. (1993). "Generalized two-dimensional correlation method applicable to infrared, Raman, and other types of spectroscopy". Appl. Spectrosc. 47 (9): 1329–36. Bibcode:1993ApSpe..47.1329N. doi:10.1366/0003702934067694. Further reading [edit] "Covariance matrix", Encyclopedia of Mathematics, EMS Press, 2001 "Covariance Matrix Explained With Pictures", an easy way to visualize covariance matrices! Weisstein, Eric W. "Covariance Matrix". MathWorld. van Kampen, N. G. (1981). Stochastic processes in physics and chemistry. New York: North-Holland. ISBN 0-444-86200-5. | v t e Statistics | | Outline Index | | | Descriptive statistics | | | | | | | | | | | --- --- --- --- | | Continuous data | | | | --- | | Center | Mean + Arithmetic + Arithmetic-Geometric + Contraharmonic + Cubic + Generalized/power + Geometric + Harmonic + Heronian + Heinz + Lehmer Median Mode | | Dispersion | Average absolute deviation Coefficient of variation Interquartile range Percentile Range Standard deviation Variance | | Shape | Central limit theorem Moments + Kurtosis + L-moments + Skewness | | | Count data | Index of dispersion | | Summary tables | Contingency table Frequency distribution Grouped data | | Dependence | Partial correlation Pearson product-moment correlation Rank correlation + Kendall's τ + Spearman's ρ Scatter plot | | Graphics | Bar chart Biplot Box plot Control chart Correlogram Fan chart Forest plot Histogram Pie chart Q–Q plot Radar chart Run chart Scatter plot Stem-and-leaf display Violin plot | | | | | Data collection | | | | | --- | | Study design | Effect size Missing data Optimal design Population Replication Sample size determination Statistic Statistical power | | Survey methodology | Sampling + Cluster + Stratified Opinion poll Questionnaire Standard error | | Controlled experiments | Blocking Factorial experiment Interaction Random assignment Randomized controlled trial Randomized experiment Scientific control | | Adaptive designs | Adaptive clinical trial Stochastic approximation Up-and-down designs | | Observational studies | Cohort study Cross-sectional study Natural experiment Quasi-experiment | | | | | Statistical inference | | | | | --- | | Statistical theory | Population Statistic Probability distribution Sampling distribution + Order statistic Empirical distribution + Density estimation Statistical model + Model specification + Lp space Parameter + location + scale + shape Parametric family + Likelihood (monotone) + Location–scale family + Exponential family Completeness Sufficiency Statistical functional + Bootstrap + U + V Optimal decision + loss function Efficiency Statistical distance + divergence Asymptotics Robustness | | Frequentist inference | | | | --- | | Point estimation | Estimating equations + Maximum likelihood + Method of moments + M-estimator + Minimum distance Unbiased estimators + Mean-unbiased minimum-variance - Rao–Blackwellization - Lehmann–Scheffé theorem + Median unbiased Plug-in | | Interval estimation | Confidence interval Pivot Likelihood interval Prediction interval Tolerance interval Resampling + Bootstrap + Jackknife | | Testing hypotheses | 1- & 2-tails Power + Uniformly most powerful test Permutation test + Randomization test Multiple comparisons | | Parametric tests | Likelihood-ratio Score/Lagrange multiplier Wald | | | Specific tests | | | | Z-test (normal) Student's t-test F-test | | Goodness of fit | Chi-squared G-test Kolmogorov–Smirnov Anderson–Darling Lilliefors Jarque–Bera Normality (Shapiro–Wilk) Likelihood-ratio test Model selection + Cross validation + AIC + BIC | | Rank statistics | Sign + Sample median Signed rank (Wilcoxon) + Hodges–Lehmann estimator Rank sum (Mann–Whitney) Nonparametric anova + 1-way (Kruskal–Wallis) + 2-way (Friedman) + Ordered alternative (Jonckheere–Terpstra) Van der Waerden test | | | Bayesian inference | Bayesian probability + prior + posterior Credible interval Bayes factor Bayesian estimator + Maximum posterior estimator | | | | | Correlation Regression analysis | | | | | --- | | Correlation | Pearson product-moment Partial correlation Confounding variable Coefficient of determination | | Regression analysis (see also Template:Least squares and regression analysis | Errors and residuals Regression validation Mixed effects models Simultaneous equations models Multivariate adaptive regression splines (MARS) | | Linear regression | Simple linear regression Ordinary least squares General linear model Bayesian regression | | Non-standard predictors | Nonlinear regression Nonparametric Semiparametric Isotonic Robust Homoscedasticity and Heteroscedasticity | | Generalized linear model | Exponential families Logistic (Bernoulli) / Binomial / Poisson regressions | | Partition of variance | Analysis of variance (ANOVA, anova) Analysis of covariance Multivariate ANOVA Degrees of freedom | | | | | Categorical / multivariate / time-series / survival analysis | | | | | --- | | Categorical | Cohen's kappa Contingency table Graphical model Log-linear model McNemar's test Cochran–Mantel–Haenszel statistics | | Multivariate | Regression Manova Principal components Canonical correlation Discriminant analysis Cluster analysis Classification Structural equation model + Factor analysis Multivariate distributions + Elliptical distributions - Normal | | Time-series | | | | --- | | General | Decomposition Trend Stationarity Seasonal adjustment Exponential smoothing Cointegration Structural break Granger causality | | Specific tests | Dickey–Fuller Johansen Q-statistic (Ljung–Box) Durbin–Watson Breusch–Godfrey | | Time domain | Autocorrelation (ACF) + partial (PACF) Cross-correlation (XCF) ARMA model ARIMA model (Box–Jenkins) Autoregressive conditional heteroskedasticity (ARCH) Vector autoregression (VAR) (Autoregressive model (AR)) | | Frequency domain | Spectral density estimation Fourier analysis Least-squares spectral analysis Wavelet Whittle likelihood | | | Survival | | | | --- | | Survival function | Kaplan–Meier estimator (product limit) Proportional hazards models Accelerated failure time (AFT) model First hitting time | | Hazard function | Nelson–Aalen estimator | | Test | Log-rank test | | | | | | | | | | | --- | | Biostatistics | Bioinformatics Clinical trials / studies Epidemiology Medical statistics | | Engineering statistics | Chemometrics Methods engineering Probabilistic design Process / quality control Reliability System identification | | Social statistics | Actuarial science Census Crime statistics Demography Econometrics Jurimetrics National accounts Official statistics Population statistics Psychometrics | | Spatial statistics | Cartography Environmental statistics Geographic information system Geostatistics Kriging | | | | Category Mathematics portal Commons WikiProject | | v t e Matrix classes | | Explicitly constrained entries | Alternant Anti-diagonal Anti-Hermitian Anti-symmetric Arrowhead Band Bidiagonal Bisymmetric Block-diagonal Block Block tridiagonal Boolean Cauchy Centrosymmetric Conference Complex Hadamard Copositive Diagonally dominant Diagonal Discrete Fourier Transform Elementary Equivalent Frobenius Generalized permutation Hadamard Hankel Hermitian Hessenberg Hollow Integer Logical Matrix unit Metzler Moore Nonnegative Pentadiagonal Permutation Persymmetric Polynomial Quaternionic Signature Skew-Hermitian Skew-symmetric Skyline Sparse Sylvester Symmetric Toeplitz Triangular Tridiagonal Vandermonde Walsh Z | | Constant | Exchange Hilbert Identity Lehmer Of ones Pascal Pauli Redheffer Shift Zero | | Conditions on eigenvalues or eigenvectors | Companion Convergent Defective Definite Diagonalizable Hurwitz-stable Positive-definite Stieltjes | | Satisfying conditions on products or inverses | Congruent Idempotent or Projection Invertible Involutory Nilpotent Normal Orthogonal Unimodular Unipotent Unitary Totally unimodular Weighing | | With specific applications | Adjugate Alternating sign Augmented Bézout Carleman Cartan Circulant Cofactor Commutation Confusion Coxeter Distance Duplication and elimination Euclidean distance Fundamental (linear differential equation) Generator Gram Hessian Householder Jacobian Moment Payoff Pick Random Rotation Routh-Hurwitz Seifert Shear Similarity Symplectic Totally positive Transformation | | Used in statistics | Centering Correlation Covariance Design Doubly stochastic Fisher information Hat Precision Stochastic Transition | | Used in graph theory | Adjacency Biadjacency Degree Edmonds Incidence Laplacian Seidel adjacency Tutte | | Used in science and engineering | Cabibbo–Kobayashi–Maskawa Density Fundamental (computer vision) Fuzzy associative Gamma Gell-Mann Hamiltonian Irregular Overlap S State transition Substitution Z (chemistry) | | Related terms | Jordan normal form Linear independence Matrix exponential Matrix representation of conic sections Perfect matrix Pseudoinverse Row echelon form Wronskian | | Mathematics portal List of matrices Category:Matrices (mathematics) | Retrieved from " Categories: Covariance and correlation Matrices (mathematics) Summary statistics Hidden categories: Use American English from April 2019 All Wikipedia articles written in American English Articles with short description Short description matches Wikidata All accuracy disputes Articles with disputed statements from September 2024 All articles with unsourced statements Articles with unsourced statements from February 2012 Covariance matrix Add topic
15607
https://greaterlondoncommunityofpractice.wordpress.com/2018/06/29/idd-in-the-icd-11/
Greater London Learning Disability Community of Practice IDD in the ICD-11 This week the WHO (World Health Organisation) published the first release version of the IDC-11. It will replace the ICD-10 and is expected to be implemented in January 2022. The ICD-11 has been eagerly awaited by the learning and developmental disability community due the anticipated changes in the classification of autism spectrum disorders/conditions. The ICD stands for the International Classification of Diseases and defines diseases, disorders, injuries and health related conditions. It is used for both research and clinical practice. The ICD has been published by the World Health Organisation since 1948 (version 6), but it originally began in 1893 when it was called the International List of Cause of Death. The ICD is used throughout Europe and across the globe providing an alternative to the American DSM-V (Diagnostic and Statistical Manual). You can read more about the ICD, including the science and statistics behind it, on the WHO website. The latest updates include Gaming Disorder within addictive behaviours and Gender Incongruence (transgender) is no longer classified as a mental disorder. In addition, the new version of the ICD supports: Learning Disabilities and Autism Spectrum Disorders are in Section 06 of the ICD 11, under Mental, Behavioural and Neurodevelopmental Disorders. Learning Disabilities in the ICD-11 Learning disabilities fall within Disorders of Intellectual Development under code 6A00. The classifications remain the same as in previous versions: Mild ID, Moderate ID, Severe ID, and Profound ID. A Provisional ID diagnostic category exists for when the individual is too young for assessment, and also an Unspecific ID category. Additional codes can be used to specify known aetiologies. Autism Spectrum Disorders in the ICD-11 The DSM-V initiated the removal of Asperger’s syndrome as a separate diagnosis (however it is recommended that those with an existing diagnosis keep it). The ICD-11, was expected to, and has followed suit by removing the classification of Asperger’s syndrome. The argument for this is, in part, that Asperger’s Syndrome is not associated with a language delay in childhood and there is no presence of a co-associated intellectual/learning disability. As a result, it is considered, for some individuals, that the features of autism, do not meet the criteria for a clinical diagnosis i.e. do not impact negatively upon behaviour, health, functioning or wellbeing to a threshold that requires support. In some ways this could be a positive and the beginning of recognising ASD as a difference rather than disability in itself, however there are numerous criticisms as many feel it neglects the variations of difficulties in social and adaptive functioning experienced by individuals with ASC and could render those who may be in need, ineligible for help. The ICD-11 criteria appears to have made an attempt to take this into account and the new codes distinguish between ASC with an intellectual/learning disability and ASC without (the DSM-V only recognises it occurs), and emphasise differences in impairments in, or an absence of, functional verbal language. Difficulties in initiating and sustaining reciprocal social interaction and restricted, repetitive and inflexible patterns of behaviour and interests remain the key characteristics of ASD with classifications falling under Code 6A02 in the ICD-11. What about Aspies?? Removing a diagnosis not only impacts on widespread understanding but also if the diagnosis has become part of the individual’s identity and reconstructed in social norms e.g. referring to one self as an ‘aspie’ and embracing autism as a positive difference rather than disability. This link has quite an interesting take and suggests that despite the diminished used in medical forums, the term Asperger’s is likely to remain – Also in the news this week Events: 18/7/2018 – The Participatory Autism Research Collective (PARC) Conference at the London South Bank University. PARC is designed to bring autistic people, including scholars and activists, together with early career researchers and practitioners who work with autistic people. They aim to build a community network where those who wish to see more significant involvement of autistic people in autism research can share knowledge and expertise. See the schedule and book your place here. Training: Share this: Related Published by Greater London Learning Disability Community of Practice View all posts by Greater London Learning Disability Community of Practice Leave a comment Cancel reply Δ Post navigation Follow me on Twitter Latest blog posts
15608
https://www.quora.com/What-is-the-difference-between-equation-and-inequality
Something went wrong. Wait a moment and try again. What is the difference between equation and inequality? Kelly Kirk Author has 3.2K answers and 1.9M answer views · 9mo To many people demand equality. do you want equality. The resources you need daily for work are the assets that make up the wealth of someone else. For true equality. All the value of a company would be divided by the number of employees and each person would have to supply an equal amount of value for the company to exist. A company worth $1,000,000 with ten people would cost $100,000each and everyone would receive the same pay. And work the same amount of hours. So someone with a doctorate degree would be paid the same as the Gaurd at the door. But that Gaurd would also have to pay the same To many people demand equality. do you want equality. The resources you need daily for work are the assets that make up the wealth of someone else. For true equality. All the value of a company would be divided by the number of employees and each person would have to supply an equal amount of value for the company to exist. A company worth $1,000,000 with ten people would cost $100,000each and everyone would receive the same pay. And work the same amount of hours. So someone with a doctorate degree would be paid the same as the Gaurd at the door. But that Gaurd would also have to pay the same $100,000 to get the job. im happy a rich person supplied the assets I needed for work. The job I do requires more assets than the same job 50 years ago. So yes the rich person owns more assets which I’m thankful for. Too many ignorant people think asset wealth is cash when it’s equipment and buildings not cash. Philip Lloyd Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.3M answer views · 2y I will use a similar expression in each of these so you can compare more easily… Here is an example of an equation… Here is an example of an inequality… Here is an example of an identity… Now this may “look” like a basic equation where we find a single value of x which makes it true as in the first example above. But this is a special type of equation where the left hand side is identical to the right hand side (even if it is not immediately obvious). An identity is true for all values of x. There is no single answer. If I treat an i I will use a similar expression in each of these so you can compare more easily… Here is an example of an equation… Here is an example of an inequality… Here is an example of an identity… Now this may “look” like a basic equation where we find a single value of x which makes it true as in the first example above. But this is a special type of equation where the left hand side is identical to the right hand side (even if it is not immediately obvious). An identity is true for all values of x. There is no single answer. If I treat an identity like a basic equation we end up with silly things like x = x or eventually 0 = 0. Watch this 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. It works with just a name, and you can also try using their email or phone number if you have it. It’s private, fast, and really helpful if you’re trying to get some peace of mind or just want to know what’s out there. 🔍 HERE IS HOW IT WORK 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. It works with just a name, and you can also try using their email or phone number if you have it. It’s private, fast, and really helpful if you’re trying to get some peace of mind or just want to know what’s out there. 🔍 HERE IS HOW IT WORKS: Start by going to this link 👉 UNCOVER DATING PROFILE Enter the person’s name, email address, or phone number. 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. If you’re in a situation where you need to know whether someone is still acting single online, this is one of the most effective and low-stress ways to find out. 👉 Check it out here if you’re ready to start your search. ALSO HERE ARE OTHER HELPFUL TOOLS: Dating Research Tool – Search a large database to learn more about who you’re dating. Who’s Texting Your Partner – Discover who your partner is texting or calling, including their name, age, location, and social profiles. Verify People Tool – Confirm if someone is really who they say they are. Find Social Profiles – Locate someone's social media and dating profiles. People Search Directory – Look up someone's phone number and contact details. Dating Safety Check – Review your date’s background to help keep you safe. Michael Keffer Have taken math classes. · Author has 149 answers and 62.4K answer views · 9mo Originally Answered: Explain, in your own words, what an inequality is and how it differs from an equation? · At first glance it appears that an inequality (in mathematics) is exactly the opposite of an equality (equation). This isn't true. An inequality simply means not equal to. However, when solving problems involving inequality, the reader is being asked to find the range of input values that will make the inequality true. For inequalities involving greater than, the solution ranges go in opposite dir At first glance it appears that an inequality (in mathematics) is exactly the opposite of an equality (equation). This isn't true. An inequality simply means not equal to. However, when solving problems involving inequality, the reader is being asked to find the range of input values that will make the inequality true. For inequalities involving greater than, the solution ranges go in opposite directions. For inequalities involving less than, the solution range is between two different numbers. Now, some inequality problems also involve an equal to option with the inequality. These problems also involve solution ranges, but they add in at least one included end point to the range. So less than or equal to problems will result in solution ranges that include a maximum value. Greater th... Related questions What is the basic difference between an equation and an inequation? What is the difference between an inequality and a difference? What is the difference between solving equations and inequalities? What is the difference between equations and functions? Can you explain the difference between an equation, inequality, expression and formula? David Joyce Professor Emeritus of Mathematics at Clark University · Author has 9.9K answers and 67.8M answer views · 4y An equation includes an equal sign “=”. An inequality includes one of these signs: “≠”, “<”, “>”, “≤”, or “≥”. An equation says two things are equal. An inequality says they’re not equal in some way. It could say they’re just not equal with “≠”. Or with one of the other four signs, the inequality could say more about how they’re not equal. Assistant Bot · 11mo The primary difference between an equation and an inequality lies in the relationship they express between two mathematical expressions: Equation Definition: An equation states that two expressions are equal. It is written in the form A=B, where A and B can be any algebraic expressions. Example: 2x+3=7 Solution: The goal is to find the value(s) of the variable(s) that make the equation true. For the example, solving gives x=2. Inequality Definition: An inequality expresses a relationship of comparison between two expressions. It can show that one expression is less than, greater than, less t The primary difference between an equation and an inequality lies in the relationship they express between two mathematical expressions: Equation Definition: An equation states that two expressions are equal. It is written in the form A=B, where A and B can be any algebraic expressions. Example: 2x+3=7 Solution: The goal is to find the value(s) of the variable(s) that make the equation true. For the example, solving gives x=2. Inequality Definition: An inequality expresses a relationship of comparison between two expressions. It can show that one expression is less than, greater than, less than or equal to, or greater than or equal to another. It is written in forms such as AB, A≤B, or A≥B. Example: 2x+3<7 Solution: The goal is to find the range of values for the variable(s) that satisfy the inequality. For the example, solving gives x<2. Summary Equations assert equality; inequalities assert a relative comparison (less than, greater than, etc.). The solution to an equation is typically a specific value, while the solution to an inequality is usually a range of values. Michel Paul Former Mathematics Teacher (1986–2016) · Author has 839 answers and 925K answer views · 4y One important difference between an equation and an inequality is that if you subtract one side of an equation from the other, you will get 0. In the case of an inequality, if you subtract one side from the other, you will get something other than 0. That’s an important fact about equations and inequalities that is often utilized in finding solutions. When I was teaching I noticed that students frequently interpreted the = sign to mean ‘find the answer’. It does not mean that. It only means that the expression on the left hand side has the same value as the expression on the right hand side. Stude One important difference between an equation and an inequality is that if you subtract one side of an equation from the other, you will get 0. In the case of an inequality, if you subtract one side from the other, you will get something other than 0. That’s an important fact about equations and inequalities that is often utilized in finding solutions. When I was teaching I noticed that students frequently interpreted the = sign to mean ‘find the answer’. It does not mean that. It only means that the expression on the left hand side has the same value as the expression on the right hand side. Students who think that = means ‘find the answer’ become confused when they see symbols like <, >, ≠, or ≈. A further issue these days has to do with computational technology. This issue has to do with variable assignment. In many programming languages, a statement like x=x+1 has the meaning, “Add one to the current value of x and then reassign that value to x.” However, the equation x=x+1 has no solution in traditional algebra. Programming languages that use = for variable assignment tend to use == to mean equality in our traditional math sense. Though something like the meaning of = might seem trivial and ‘obvious’ to many people, there’s actually a lot of subtle variations of understanding possible, and it is important to be aware of these nuances, because that is where confusions arise for students. This is becoming more true today. Promoted by Betterbuck Anthony Madden Writer for Betterbuck · Mar 12 What are the best websites to use for comparing house and car insurance rates? I honestly couldn’t be bothered calling a bunch of different insurers so I just went on this site here (Coverage.com) and this one (Auto-Savings.com), but there are a ton of different comparison sites you can try. I found an offer from one carrier for the exact same coverage, but it was $1,300 less that what I paid for the previous year. Nuts. I say it’s worth giving it a go. You don’t lose anything just from looking. Worst case scenario - you get confirmation you’ve got a good offer. Best case scenario - you actually save some money. … (more) Related questions What's the difference between inequality equations and regular algebraic equations? What is the difference between a formula and an equation? What is an inequality equation? Why is every equation always equal to zero? What are some examples of equations and inequalities? Howard Ludwig Ph.D. in Physics, Northwestern University (Graduated 1982) · Author has 3K answers and 10.1M answer views · 10mo Originally Answered: What is the difference between an equation or inequality in an algebraic expression? · “What is the difference between an equation or inequality in an algebraic expression?” An equation is a mathematical expression that contains an = sign. An inequality is a mathematical expression that contains at least one of the following: <, ≤, >, ≥, ≠. DWB Former Retired Algebra I Teacher at Lakeland Senior High School (1987–2020) · 9mo Originally Answered: Explain, in your own words, what an inequality is and how it differs from an equation? · 2x + y = 10 is an equation for a line. Any ordered pair (x, y) that makes the equation “true” means that point lies on that line and are referred to as solutions. Some solutions are: (0, 10), (2, 6), (3, 4) If you graphed these points you would get a straight line. 2x + y > 10 is an inequality and any ordered pair must now make this statement true or 2x + y must be larger than 10. Some examples are (5, 2), (7, 0) (-1, 14). If you graphed the inequality as if were a line(you would use a dotted line to indicate < or > because points on the line are not solutions. Greater than or equal to and less 2x + y = 10 is an equation for a line. Any ordered pair (x, y) that makes the equation “true” means that point lies on that line and are referred to as solutions. Some solutions are: (0, 10), (2, 6), (3, 4) If you graphed these points you would get a straight line. 2x + y > 10 is an inequality and any ordered pair must now make this statement true or 2x + y must be larger than 10. Some examples are (5, 2), (7, 0) (-1, 14). If you graphed the inequality as if were a line(you would use a dotted line to indicate < or > because points on the line are not solutions. Greater than or equal to and less than or equal to are graphed with solid lines to indicate points on the line DO satisfy the inequality) If you now plot your solutions, you will need to shade the side of the line where those points are located. Any point on that side of the line will make the inequality true. Sponsored by Grommet The Top 12 Coolest Travel Essentials. It's true — Here are some of the smartest "must have" travel items of 2025. Andy Wright Maths teacher and examiner · Author has 1.1K answers and 1.5M answer views · 9y Originally Answered: What is the difference between solving equations and inequalities? · An inequality is similar to an equation, but the two sides instead of being equal are not equal in some well defined way. What you could do is replace the inequality with an equals, solve it as normal, then put the inequality sign back again, so for the example 3x>15, you put 3x=5, then x=5, then put the inequality back, ie x>5. The exception to this is that every time you divide both sides by a negative number, the inequality flips from > to < and vice versa Elizabeth Jean Stapel tutor (1989 to present), instructor (1991+), author (1998+) · Author has 5.8K answers and 3.2M answer views · 10mo Originally Answered: What is the difference between an equation or inequality in an algebraic expression? · What is the difference between an equation or inequality in an algebraic expression? An equation has an “equals” sign somewhere inside of it. An inequality has an inequality symbol (less than, greater than, etc) somewhere inside of it. Promoted by Budget Hustle Ellie Grant Writer for Budget Hustle · Updated Wed What is a brutally honest truth about money that most people learn too late? Here’s the thing: these money moves have helped so many people get out of debt, lower their bills, and finally feel in control again. The best part? Most of them are easier than you'd expect—and even trying just one could give you a little breathing room. Cancel your car insurance—seriously Many people don’t realize they’re wildly overpaying for car insurance. In fact, many companies count on the fact that you’ll auto-renew without shopping around. A free tool like Insurify shows you real-time quotes based on your ZIP code from dozens of top providers. Some people save $500–$1,000+ a year when Here’s the thing: these money moves have helped so many people get out of debt, lower their bills, and finally feel in control again. The best part? Most of them are easier than you'd expect—and even trying just one could give you a little breathing room. Cancel your car insurance—seriously Many people don’t realize they’re wildly overpaying for car insurance. In fact, many companies count on the fact that you’ll auto-renew without shopping around. A free tool like Insurify shows you real-time quotes based on your ZIP code from dozens of top providers. Some people save $500–$1,000+ a year when they check. Only takes a few minutes. Check your ZIP to see what you could save. Ask this company to help pay off your credit card debt If you're stuck in $20,000+ in credit card debt, you're probably watching most of your payments vanish into interest. That's not just frustrating, it's financially crushing. A program called Accredited Debt Relief may be able to lower your total monthly payment—sometimes by over 40%. Instead of managing multiple bills, you could roll everything into one simple payment. Here’s how to check your eligibility: Make sure you have at least $20,000 in credit card debt. Answer a few simple questions to instantly check if you qualify If you’re eligible, you can reserve your spot in this program today. Check if you qualify now Homeowner? Protect yourself from surprise repair costs If you’re a homeowner, one unexpected breakdown – a fridge, water heater, or plumbing issue – could cost you $1,200 or more. That’s why over a million people use Choice Home Warranty for a plan that protects all of your home’s major systems and appliances for one low monthly rate. Instead of getting hit with a $1,000+ bill out of pocket, Choice will send a qualified technician to your home for your unexpected breakdown. You can skip the stress and big costs. You can even get your first month free when you sign up. See if your ZIP code is eligible for a plan. Have huge companies pay you for your opinion Big brands spend millions on market research, and sites like Survey Junkie help you tap into that by paying users to take quick surveys and give feedback. It’s free to join, easy to use, and you can start earning here—even during downtime on your couch. If you’re 50+, this overlooked membership could save you $450/year Everyone’s heard of AARP—but surprisingly, many 50+ Americans still don’t take full advantage of what it offers. Members get discounts on dining, travel, prescriptions, and more. You could save up to $450 a year on restaurants and meal delivery alone—plus get perks on hotels, eyeglasses, and everyday essentials. Right now, you can get a free welcome gift when you join. It also comes with helpful tools for retirement planning, Social Security, caregiving, and fraud protection. It’s a smart move that pays off fast. Register for a membership here. Use your birthday month to score freebies (and a free meal or two) You’d be surprised how many places offer free stuff just for being born. Restaurants, coffee shops, beauty brands, and even retail stores often dish out birthday deals if you’re signed up for their rewards program. Examples: a free drink from Starbucks, a burger from Red Robin, burrito from Moe’s, free pancake meal from IHOP, or a $5 coupon from stores like CVS or Ulta. Create a throwaway email just for rewards programs and sign up a week or two before your birthday—then enjoy claiming your perks all month long. Make money testing games and apps on your phone Freecash pays users to try out mobile games and complete small tasks. Some users make this a daily routine, earning just by playing casually. No weird hoops, no points-only system—you get paid in actual cash. Here’s where to start. Mohammadreza Sanayi Present Director to Eurasian Polymaths (2019–present) · Author has 2.5K answers and 578.3K answer views · 3y Equality is a related term of equation. As nouns the difference between equality and equation is that equality is (uncountable) the fact of being equal while equation is (mathematics) an assertion that two expressions are equal, expressed by writing the two expressions separated by an equal sign; from which one is to determine a particular quantity. Aquiles Fischer Studied at School of Hard Knocks · Author has 805 answers and 129.5K answer views · 2y A variable, say x, is 5 that would be the answer to an equation. A variable, say, another x, is larger than 5, meaning that the inequality can be 6, 7, or more. It would be the answer to an inequality. Equation, x = 5, no 4 nor 6, but 5. On the other hand, [x > 5] can be 6, 7, 8, and so on or the inequality also can be smaller than 5, [ x < 5] meaning it can be 4, 3, 2, or the boundaries that you may have. Joel Weber Research Chemical Engineer (2017–present) · 7y In short, an equation uses “=” meaning what’s on the left side is exactly the same value (when evaluated) as what’s on the right side. 2x = 4, where x = 2, is an equation. An inequality is normally used to denote that what’s on the left side is not necessarily exactly equivalent to what’s on the right side. An easy real world example is the height requirement to ride a roller coaster. In order to ride this ride, the height of the person, H must obey the following: H > 48″ The person must be at least 48 inches tall to ride. The person can of course be more than 48 inches tall, making this an inequali In short, an equation uses “=” meaning what’s on the left side is exactly the same value (when evaluated) as what’s on the right side. 2x = 4, where x = 2, is an equation. An inequality is normally used to denote that what’s on the left side is not necessarily exactly equivalent to what’s on the right side. An easy real world example is the height requirement to ride a roller coaster. In order to ride this ride, the height of the person, H must obey the following: H > 48″ The person must be at least 48 inches tall to ride. The person can of course be more than 48 inches tall, making this an inequality, because H does not have to equal 48. Related questions What is the basic difference between an equation and an inequation? What is the difference between an inequality and a difference? What is the difference between solving equations and inequalities? What is the difference between equations and functions? Can you explain the difference between an equation, inequality, expression and formula? What's the difference between inequality equations and regular algebraic equations? What is the difference between a formula and an equation? What is an inequality equation? Why is every equation always equal to zero? What are some examples of equations and inequalities? What is the difference between function, expression, and equation? What is the difference between equation and equality? What is the difference between equal and unequal equations? What is the difference between equations and expressions? What is the difference between inequalities and linear inequalities? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15609
https://www.khanacademy.org/math/differential-calculus/dc-diff-intro/dc-secant-lines/v/slope-of-a-secant-line
Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Consent Leg.Interest label label label
15610
https://www.merriam-webster.com/dictionary/analogy
Synonyms of analogy a : a comparison of two otherwise unlike things based on resemblance of a particular aspect b : resemblance in some particulars between things otherwise unlike : similarity : inference that if two or more things agree with one another in some respects they will probably agree in others 3 : correspondence between the members of pairs or sets of linguistic forms that serves as a basis for the creation of another form 4 evolutionary biology : correspondence or similarity in form or function between parts (such as the wings of birds and insects) of unrelated or distantly related species that is the result of convergent evolution compare homology, homoplasy Did you know? Digging Into the Most Common Meaning of Analogy In its most common use, analogy has to do with comparison of things based on those things being alike in some way. For example, one can make or draw an analogy between the seasons of the year and the stages of life. People also reason by way of analogy, asserting, for example, that abandoning a project is like leaving a house partway built. Although an analogy can be summarized quickly, as in these examples, an analogy actually encompasses the comparison or inference itself, and is therefore different from figures of speech, like metaphors and similes, which are forms of expression. Some tests ask you to identify analogies, finding the second of a pair that has the same relationship as a completed pair. Analogy tests often look like this: ice : cold :: steel : ____ a. hard b. loud c. fresh d. small Because the relationship between ice and cold is that coldness is a quality of ice, the word that goes with steel is hard, since hardness is a quality of steel. The word analogy (which comes from analogous) traces back by way of Latin to a Greek word meaning "proportionate." That word has a root in the Greek word logos, meaning "reason." Choose the Right Synonym for analogy likeness, similarity, resemblance, similitude, analogy mean agreement or correspondence in details. likeness implies a closer correspondence than similarity which often implies that things are merely somewhat alike. a remarkable likeness to his late father some similarity between the two cases resemblance implies similarity chiefly in appearance or external qualities. statements that bear little resemblance to the truth similitude applies chiefly to correspondence between abstractions. two schools of social thought showing points of similitude analogy implies likeness or parallelism in relations rather than in appearance or qualities. pointed out analogies to past wars Examples of analogy in a Sentence He does, though, suffer from the occupational deformation of international relations specialists: an enthusiasm for ransacking the past in search of precedents, analogies, patterns, and cycles that might explain the present and forecast the future. —Tony Judt, New York Book Review, 10 Apr. 2003 People who do this call themselves "white-hat" hackers—good people who show other people their vulnerabilities. Take the following analogy: I've designed a great new lock pick, and I'm going to give this great new gadget away to show everyone that the typical door lock is ineffective against my new pick. —John C. Dvorak, PC Magazine, 4 Apr. 2000 It has often been said that movie stars are the royalty of America. (The better analogy, really, is that the royals are the movie stars of Britain.) —Neal Gabler, Life: The Movie, 1998 Parts of the far-infrared sky look like colonies of spiders gone mad. The fine structure seen there is called cirrus, by analogy with filamentary clouds on Earth. —Virginia Trimble et al., Sky & Telescope, January 1995 Recent Examples on the Web Examples are automatically compiled from online sources to show current usage. Read More Opinions expressed in the examples do not represent those of Merriam-Webster or its editors. Send us feedback. The political analogy cautions no: entrenched leaders can undermine institutions and tilt the system toward personal gain. —Joel Shulman, Forbes.com, 29 Aug. 2025 Going back to the billiard ball analogy, imagine replacing the billiard balls with bowling balls and the cue ball with a ping-pong ball. —Robert Lea, Space.com, 28 Aug. 2025 Jinny is also genuinely fun to work with — spirited, quick-witted, and always ready with a food analogy for any occasion. —Nellie Andreeva, Deadline, 26 Aug. 2025 Jinny is also genuinely fun to work with—spirited, quick-witted, and always ready with a food analogy for any occasion. —Jennifer Maas, Variety, 26 Aug. 2025 See All Example Sentences for analogy Word History Etymology Middle English analogie, borrowed from Middle French & Latin; Middle French, borrowed from Latin analogia "ratio, proportion, correspondence in the inflection and derivation of words, the use of this correspondence in word derivation," borrowed from Greek analogía "proportion, relation, correspondence in the inflection and derivation of words," from análogos "proportionate, conformable, analogous" + -ia -y entry 2 First Known Use 15th century, in the meaning defined at sense 2 Time Traveler The first known use of analogy was in the 15th century See more words from the same century Phrases Containing analogy draw/make an analogy between Rhymes for analogy tetralogy See All Rhymes for analogy Browse Nearby Words analogue computer analogy analogy test See all Nearby Words Cite this Entry “Analogy.” Merriam-Webster.com Dictionary, Merriam-Webster, Accessed 2 Sep. 2025. Copy Citation Share Kids Definition analogy noun anal·​o·​gy ə-ˈnal-ə-jē plural analogies 1 a : resemblance in some details between things otherwise unlike : similarity b : comparison based on such resemblance 2 : similarity in function between bodily parts of different structure and origin analogical ˌan-ᵊl-läj-i-kəl adjective analogically -k(ə-)lē adverb Medical Definition analogy noun anal·​o·​gy ə-ˈnal-ə-jē plural analogies : functional similarity between anatomical parts without similarity of structure and origin compare homology sense 1 More from Merriam-Webster on analogy Thesaurus: All synonyms and antonyms for analogy Nglish: Translation of analogy for Spanish Speakers Britannica.com: Encyclopedia article about analogy Last Updated: - Updated example sentences Love words? Need even more definitions? Subscribe to America's largest dictionary and get thousands more definitions and advanced search—ad free! Merriam-Webster unabridged More from Merriam-Webster ### Can you solve 4 words at once? Can you solve 4 words at once? Word of the Day carp See Definitions and Examples » Get Word of the Day daily email! Popular in Grammar & Usage See More ### 31 Useful Rhetorical Devices ### Merriam-Webster’s Great Big List of Words You Love to Hate ### How to Use Em Dashes (—), En Dashes (–) , and Hyphens (-) ### The Difference Between 'i.e.' and 'e.g.' ### Democracy or Republic: What's the difference? See More Popular in Wordplay See More ### Our Best Historical Slang Terms ### Even More Bird Names that Sound Like Insults (and Sometimes Are) ### Words That Turned 100 in 2025 ### Better Ways to Say 'This Sucks' ### 'Za' and 9 Other Words to Help You Win at SCRABBLE See More Popular See More ### 31 Useful Rhetorical Devices ### Our Best Historical Slang Terms ### Even More Bird Names that Sound Like Insults (and Sometimes Are) See More
15611
https://www.dictionary.com/browse/craven
Daily Crossword Word Puzzle Word Finder All games Word of the Day Word of the Year New words Language stories All featured Slang Emoji Memes Acronyms Gender and sexuality All culture Writing tips Writing hub Grammar essentials Commonly confused All writing tips Games Featured Culture Writing tips Advertisement View synonyms for craven craven [krey-vuhn] adjective cowardly; contemptibly timid; pusillanimous. Synonyms: timorous, fearful, dastardly noun a coward. verb (used with object) to make cowardly. craven / ˈkreɪvən/ adjective cowardly; mean-spirited noun a coward Discover More Other Word Forms cravenly adverb cravenness noun uncraven adjective Discover More Word History and Origins Origin ofcraven1 1175–1225; Middle English cravant, cravaunde defeated < Old French craventé, past participle of cravanter to crush, overwhelm (< Vulgar Latin crepantāre ), influenced by Middle English creaunt defeated ( recreant ) Discover More Word History and Origins Origin ofcraven1 C13 cravant, probably from Old French crevant bursting, from crever to burst, die, from Latin crepāre to burst, crack Discover More Idioms and Phrases Idioms cry craven, to yield; capitulate; give up. Discover More Example Sentences Examples are provided to illustrate real-world usage of words in context. Any opinions expressed do not reflect the views of Dictionary.com. Not the fact that his presidency is built on his craven desire for personal retribution instead of pursuing what’s best for the American people. FromSalon In response, the X owner called the Liberal Democrat leader a "craven coward". FromBBC More churlish souls might attribute such stubborn determination to craven commercial interests, but something more is in play. FromLos Angeles Times He’s cornered by guards, the players who fought alongside him are dead, and he’s thrown back in with the remaining players, many of whom survived because they’re the most craven of the group. FromLos Angeles Times This year’s retreat from DEI policies is merely a continuation of that craven approach. FromLos Angeles Times Advertisement Discover More Related Words mean-spirited www.thesaurus.com Definitions and idiom definitions from Dictionary.com Unabridged, based on the Random House Unabridged Dictionary, © Random House, Inc. 2023 Idioms from The American Heritage® Idioms Dictionary copyright © 2002, 2001, 1995 by Houghton Mifflin Harcourt Publishing Company. Published by Houghton Mifflin Harcourt Publishing Company. Advertisement Advertisement Advertisement craveablecraving
15612
https://artofproblemsolving.com/wiki/index.php/Cevian?srsltid=AfmBOoq7GStFnu4LIuZ0ZswTdQ3tIo2LwBfkQWZZ91dQAYGKKYee-G60
Art of Problem Solving Cevian - 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 Cevian Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Cevian Contents [hide] 1 Definition 2 Special Cevians 3 Finding Lengths 4 See also Definition A cevian is a line segment or ray that extends from one vertex of a polygon (usually a triangle) to the opposite side (or the extension of that side). In the below diagram, is a cevian. Special Cevians A median is a cevian that divides the opposite side into two congruent lengths. An altitude is a cevian that is perpendicular to the opposite side. An angle bisector is a cevian that divides the angle the cevian came from in half. Finding Lengths In the diagram, note that because . Thus, See also Ceva's Theorem Angle Bisector Theorem Stewart's Theorem This article is a stub. Help us out by expanding it. Retrieved from " Categories: Stubs Definition Geometry 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.
15613
https://www.acc.org/Latest-in-Cardiology/ten-points-to-remember/2023/08/30/02/53/2023-esc-guidelines-for-cardiomyopathies-esc-2023
< Back to Listings 2023 ESC Guidelines for Management of Cardiomyopathies: Key Points Aug 30, 2023 | Debabrata Mukherjee, MD, FACC Facebook X LinkedIn Email Share Font Size : A : A : A Authors: : Arbelo E, Protonotarios A, Gimeno JR, et al., on behalf of the ESC Scientific Document Group. Citation: : 2023 ESC Guidelines for the Management of Cardiomyopathies: Developed by the Task Force on the Management of Cardiomyopathies of the European Society of Cardiology. Eur Heart J 2023;Aug 25:[Epub ahead of print]. The following are key points to remember from the 2023 European Society of Cardiology (ESC) guidelines for the management of cardiomyopathies: The overarching aim of this guideline is to provide a guide to the diagnostic approach to cardiomyopathies, highlight general evaluation and management issues, and make the reader aware of the relevant evidence base for the recommendations. Evidence suggests that cardiomyopathies are more common than previously thought and they typically require nuanced management that may differ from the conventional approach to patients with arrhythmia or heart failure (HF). Discerning etiology of the cardiomyopathy is fundamental to the management of patients with heart muscle disease, and careful and systematic description of the morphological and functional phenotype is a crucial first step in the diagnostic pathway. Patients with cardiomyopathy may seek medical attention due to symptom onset (HF or arrhythmia related), incidental abnormal findings, or as a result of family screening following the diagnosis in a relative. Multimodality imaging to characterize the cardiac phenotype (morphology and function)—including tissue characterization for nonischemic myocardial scar detection—is necessary, in combination with a detailed personal and family history, clinical examination, electrocardiography, and laboratory investigations. However, imaging results should always be interpreted in the overall clinical context, including genetic testing results, rather than in isolation. Tissue characterization by cardiac magnetic resonance (CMR) is of value in diagnosis, monitoring of disease progression, and risk stratification in each of the main cardiomyopathy phenotypes. 3,3-diphosphono-1,2-propanodicarboxylic acid (DPD)/pyrophosphate (PYP)/hydroxymethylene diphosphonate (HMDP) bone-tracer scintigraphy or single-photon emission computed tomography (SPECT) represent the gold standard for the diagnosis of ATTR-related cardiac amyloidosis. The aim of this multiparametric and systemic approach is to generate a phenotype-based etiological diagnosis, interpreting available data with a cardiomyopathy-oriented mindset that combines cardiological assessment with noncardiac parameters. A multidisciplinary approach to patient care and appropriate transition of care from pediatric to adult cardiomyopathy services is needed. Genetic testing should be performed in patients with cardiomyopathy and may influence risk stratification and management. Genetic counseling, including pre- and post-test counseling, and psychological support are an essential aspect of the multidisciplinary care of patients with cardiomyopathy and their relatives. Pediatric cardiomyopathies largely represent part of the same clinical spectrum as those seen in older adolescents and adults, but infant-onset (in the first year of life) cardiomyopathies are often associated with severe phenotypes and a high rate of HF-related morbidity and mortality. Beyond the first year of life, genetic causes of childhood-onset cardiomyopathies are similar to those in adults. Symptom management, identification, and prevention of disease-related complications (including sudden cardiac death [SCD], HF, and stroke) are the cornerstone of management of all cardiomyopathies. Cardiac myosin inhibitors (mavacamten) should be considered in patients with hypertrophic cardiomyopathy (HCM) and left ventricular outflow tract obstruction (LVOTO) who remain symptomatic despite optimal medical therapy. Validated SCD risk-prediction tools (HCM Risk-SCD and HCM Risk-Kids) are the first step in sudden death prevention in patients with HCM. Pharmacological treatment of dilated cardiomyopathy (DCM) patients does not differ from those recommended in chronic HF. SCD risk of DCM and nondilated left ventricular cardiomyopathy (NDLVC) patients varies depending on the underlying cause and genetic subtype. CMR findings play an important role in guiding implantable cardioverter-defibrillator (ICD) implantation for patients with DCM and NDLVC. Of note, in DCM and NDLVC patients, ICD should be considered for certain genetic forms even if LVEF is >35%. It is of importance to define etiology for a tailored management in patients with syndromic and metabolic cardiomyopathies (i.e., enzyme replacement therapy/chaperone in lysosomal storage disease; tafamidis in ATTRwt, etc.). Pregnancy and the post-partum period are associated with increased cardiovascular risk in women with known cardiomyopathy. A multidisciplinary team should evaluate the patient with cardiomyopathy to assess the risk associated with pregnancy. Beta-blocker therapy on arrhythmic indication can safely be continued during pregnancy; safety data should be checked before initiation of new drugs in pregnancy. Healthy adults of all ages and individuals with known cardiac disease should exercise with moderate intensity, totaling ≥150 minutes per week. All patients with cardiomyopathy should have an individualized risk assessment for exercise prescription. Evaluation should be guided by three principles: Preventing life-threatening arrhythmias during exercise, Symptom management to allow sports, and Preventing sports-induced progression of the arrhythmogenic condition. Patients with high-risk genotypes or associated factors for arrhythmic or HF complications or severe LVOTO should be referred for specialized investigations before undergoing elective noncardiac surgery. Identification and management of risk factors and concomitant diseases is recommended as an integral part of the management of cardiomyopathy patients. Facebook X LinkedIn Email Share Clinical Topics: Arrhythmias and Clinical EP, Heart Failure and Cardiomyopathies, Noninvasive Imaging, Prevention, Sports and Exercise Cardiology, Implantable Devices, SCD/Ventricular Arrhythmias, Atrial Fibrillation/Supraventricular Arrhythmias, Acute Heart Failure, Sports and Exercise and Imaging Keywords: Amyloidosis, Arrhythmias, Cardiac, Cardiomyopathies, Cardiomyopathy, Dilated, Cardiomyopathy, Hypertrophic, Death, Sudden, Cardiac, Defibrillators, Implantable, Diagnostic Imaging, ESC Congress, ESC23, Genetic Testing, Heart Failure, Pregnancy, Secondary Prevention, Sports < Back to Listings You must be logged in to save to your library. < Back to Listings YOU ARE HERE: Home > Latest in Cardiology > 2023 ESC Guidelines for Management of Cardiomyopathies: Key Points Guidelines JACC Journals on ACC.org JACC JACC: Advances JACC: Asia JACC: Basic to Translational Science JACC: CardioOncology JACC: Cardiovascular Imaging JACC: Cardiovascular Interventions JACC: Case Reports JACC: Clinical Electrophysiology JACC: Heart Failure Membership Campaign for the Future Become a Member Renew Your Membership Member Benefits and Resources Member Sections Chapters ACC Member Directory About ACC About UX ACC Innovation Program Our Strategic Direction Diversity and Inclusion Our History Our Bylaws and Code of Ethics Leadership and Governance Annual Report Industry Relations Support the ACC Jobs at the ACC Press Releases Social Media Book Our Conference Center Clinical Topics Acute Coronary Syndromes Anticoagulation Management Arrhythmias and Clinical EP Cardiac Surgery Cardio-Oncology Chronic Angina Congenital Heart Disease and Pediatric Cardiology COVID-19 Hub Diabetes and Cardiometabolic Disease Dyslipidemia Geriatric Cardiology Heart Failure and Cardiomyopathies Hypertriglyceridemia Invasive Cardiovascular Angiography and Intervention Noninvasive Imaging Pericardial Disease Prevention Pulmonary Hypertension and Venous Thromboembolism Sports and Exercise Cardiology Stable Ischemic Heart Disease Valvular Heart Disease Vascular Medicine Latest in Cardiology Clinical Updates & Discoveries Advocacy & Policy Perspectives & Analysis Meeting Coverage ACC Member Publications ACC Podcasts Education and Meetings Online Learning Catalog Understanding MOC Products and Resources Image and Slide Gallery Meetings Certificates and Certifications Annual Scientific Session Tools and Practice Support Quality Improvement for Institutions NCDR CardioSmart Accreditation Services Clinical Solutions Clinician Well-Being Diversity and Inclusion Mobile and Web Apps Cardiology as a Career Path Cardiology Careers Practice Solutions JACC Quality Improvement for Institutions NCDR CardioSmart MedAxiom Accreditation Services Contact Us Heart House 2400 N St. NW Washington, DC 20037 Contact Member Care Phone: 1-202-375-6000 Toll Free: 1-800-253-4636 Fax: 1-202-375-6842 Media Center Advertising & Sponsorship Policy Clinical Content Disclaimer ACC.org Editorial Reviewers Privacy Policy Registered User Agreement Terms of Service Cookie Policy Site Map © 2025 American College of Cardiology Foundation. All rights reserved. Last Updated October 2023 ✓
15614
https://www.droracle.ai/articles/129747/mallory-weiss-tear-vs-boorhave
Select Language ​ ▼ What is the difference between a Mallory-Weiss tear and Boerhaave's syndrome? Medical Advisory Board All articles are reviewed for accuracy by our Medical Advisory Board Educational purpose only • Exercise caution as content is pending human review Article Review Status Submitted Under Review Approved Last updated: May 22, 2025 • View editorial policy From the Guidelines Mallory-Weiss tears and Boerhaave syndrome are distinct esophageal injuries with different management approaches, and the primary distinction lies in their severity and the need for urgent surgical intervention in Boerhaave syndrome. Mallory-Weiss tears are partial-thickness mucosal lacerations at the gastroesophageal junction, typically caused by forceful vomiting or retching, and present with hematemesis after episodes of vomiting or coughing 1. In contrast, Boerhaave syndrome is a full-thickness esophageal rupture, often resulting from an abrupt increase in esophageal pressure during vomiting, and is characterized by severe chest or epigastric pain, subcutaneous emphysema, and Mackler's triad (vomiting, chest pain, and subcutaneous emphysema) 1. Key Differences in Management Mallory-Weiss tears usually heal spontaneously within 48-72 hours with conservative management, including acid suppression with proton pump inhibitors, avoiding NSAIDs, and temporary cessation of oral intake if needed 2. Endoscopic intervention with clips or thermal therapy may be required for persistent bleeding in Mallory-Weiss tears 2. Boerhaave syndrome, on the other hand, requires immediate surgical consultation, broad-spectrum antibiotics, NPO status, IV fluids, and often surgical repair or placement of esophageal stents due to its high mortality rate of 20-40% 1, 2. Recent Guidelines and Recommendations The World Journal of Emergency Surgery guidelines suggest that diagnostic endoscopy is useful in patients with suspected esophageal perforation and doubtful CT findings, but caution is warranted due to potential risks of enlarging the perforation size and aggravating contamination 2. Endoscopic treatment, including clip placement and stent placement, has been shown to be effective in managing esophageal perforations, with success rates of up to 88% and mortality rates as low as 7.5% 2. A recent review published in Clinical Gastroenterology and Hepatology highlights the importance of early recognition and management of esophageal perforations, particularly in high-risk procedures such as esophageal dilation and endoscopic mucosal resection 3. Clinical Implications The key distinction between Mallory-Weiss tears and Boerhaave syndrome lies in their severity and the need for urgent surgical intervention in Boerhaave syndrome. Clinicians should be aware of the different management approaches for these two conditions and prioritize timely diagnosis and treatment to prevent morbidity and mortality. The use of endoscopic treatment, including clip placement and stent placement, should be considered in the management of esophageal perforations, particularly in patients with minimal symptoms and signs of sepsis 2, 3. From the Research Mallory-Weiss Tear vs Boerhaave Syndrome Mallory-Weiss syndrome is a cause of upper gastrointestinal hemorrhage, resulting from an abrupt rise in abdominal pressure due to nausea or vomiting, inducing a tear near the esophagogastric mucosal junction 4. Boerhaave syndrome, on the other hand, is a transmural rupture of the esophagus, often caused by a sudden increase in pressure inside the esophagus, and is a more severe and potentially life-threatening condition 5, 6. Key Differences Location and extent of the tear: Mallory-Weiss tear is typically located on the cardia part of the stomach side and spanning across the esophagogastric mucosal junction, whereas Boerhaave syndrome involves a full-thickness rupture of the esophageal wall 4, 5. Severity and mortality: Boerhaave syndrome has a higher mortality rate, ranging from 10 to 50%, compared to Mallory-Weiss syndrome, which often resolves spontaneously or with endoscopic treatment 4, 6. Relationship Between the Two Conditions A Mallory-Weiss syndrome can potentially evolve into a Boerhaave syndrome if left untreated or if further barotraumas occur, as demonstrated in a case report where an untreated Mallory-Weiss syndrome progressed to a fatal Boerhaave syndrome 5. Diagnosis and Management Diagnosis of both conditions requires a combination of clinical presentation, imaging studies, and endoscopic evaluation 7, 6. Management of Mallory-Weiss syndrome typically involves endoscopic hemostasis, fasting, and inhibitors of acid secretion, while Boerhaave syndrome requires more aggressive treatment, including surgical intervention, broad-spectrum antibiotics, and supportive care 4, 7, 6. References 1 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 2 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 3 Guideline Guideline Directed Topic Overview Dr.Oracle Medical Advisory Board & Editors, 2025 4 Research [Mallory-Weiss syndrome]. Nihon rinsho. Japanese journal of clinical medicine, 1998 5 Research Transition of a Mallory-Weiss syndrome to a Boerhaave syndrome confirmed by anamnestic, necroscopic, and autopsy data: A case report. Medicine, 2018 6 Research Walked in with Boerhaave's... Emergency medicine journal : EMJ, 2007 7 Research The Etiology, Diagnosis, and Management of Esophageal Perforation. Journal of gastrointestinal surgery : official journal of the Society for Surgery of the Alimentary Tract, 2022 Related Questions Where is a Mallory-Weiss tear typically located? How to manage a post-surgical patient with signs of infection and possible anastomotic leak? What is the most appropriate next step for a 2-year-old boy who ingested a button battery (lithium battery) from a travel alarm clock, presenting with no distress and an unremarkable physical examination, with a radiograph available? What antibiotics are recommended for esophageal perforation? What is the best course of action for a 22-year-old male patient with paranoid schizophrenia who has ingested a disc-shaped battery, currently lodged in the distal esophagus, and is being managed with Seroquel (quetiapine)? What are the ultrasound (USG) findings of choledocholithiasis? What is the role of Atenolol (beta blocker) in heart failure (congestive heart failure)? Can gastrointestinal side effects like gastroesophageal reflux disease (GERD) occur later with Trintellix (vortioxetine)? What is Familial Hyperlipidemia (Familial Hypercholesterolemia)? What is the comparative efficacy of Hydroxyethyl Starch (HES) 130 and Hydroxyethyl Starch (HES) 200 solutions? What is the treatment for a moderate ankle sprain? Professional Medical Disclaimer This information is intended for healthcare professionals. Any medical decision-making should rely on clinical judgment and independently verified information. The content provided herein does not replace professional discretion and should be considered supplementary to established clinical guidelines. Healthcare providers should verify all information against primary literature and current practice standards before application in patient care. Dr.Oracle assumes no liability for clinical decisions based on this content. Original text Rate this translation Your feedback will be used to help improve Google Translate
15615
https://www.youtube.com/playlist?list=PLSQl0a2vh4HDJO050aGQGrCU4ggj0whlS
Negative numbers | 6th Grade | Khan Academy - YouTube Back Skip navigation Search Search with your voice Sign in Play all Negative numbers | 6th Grade | Khan Academy by Khan Academy • Playlist•27 videos•1,193 views Play all PLAY ALL Negative numbers | 6th Grade | Khan Academy 27 videos 1,193 views Last updated on Dec 20, 2023 Save playlist Shuffle play Share Show more Khan Academy Khan Academy Subscribe Play all Negative numbers | 6th Grade | Khan Academy by Khan Academy Playlist•27 videos•1,193 views Play all 1 2:24 2:24 Now playing Coordinate plane: quadrants | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 629K views • 14 years ago • 2 3:55 3:55 Now playing Comparing absolute values on number line | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 51K views • 10 years ago • 3 3:11 3:11 Now playing Opposite of a number | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 592K views • 10 years ago • Fundraiser 4 5:01 5:01 Now playing Negative symbol as opposite | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 259K views • 10 years ago • Fundraiser 5 2:08 2:08 Now playing Coordinate plane: plot ordered pairs | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 612K views • 14 years ago • 6 9:36 9:36 Now playing Negative numbers introduction | Negative numbers and absolute value | Pre-Algebra | Khan Academy Khan Academy Khan Academy • 2.1M views • 14 years ago • Fundraiser 7 3:35 3:35 Now playing Coordinate plane: reflecting points | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 524K views • 12 years ago • 8 1:56 1:56 Now playing Coordinate plane: graphing points and naming quadrants | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 919K views • 12 years ago • 9 2:16 2:16 Now playing Sorting absolute values on number line | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 165K views • 10 years ago • Fundraiser 10 4:47 4:47 Now playing Coordinate plane: word problem exercise | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 391K views • 12 years ago • 11 2:47 2:47 Now playing Ordering negative numbers | Negative numbers and absolute value | Pre-Algebra | Khan Academy Khan Academy Khan Academy • 803K views • 14 years ago • Fundraiser 12 6:50 6:50 Now playing Coordinate plane examples | Linear equations and functions | 8th grade | Khan Academy Khan Academy Khan Academy • 846K views • 15 years ago • Fundraiser 13 1:02 1:02 Now playing Coordinate plane: have all the points been graphed? | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 374K views • 10 years ago • Fundraiser 14 3:45 3:45 Now playing Interpreting absolute value | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 240K views • 10 years ago • Fundraiser 15 5:54 5:54 Now playing Values to make absolute value inequality true | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 48K views • 10 years ago • 16 4:07 4:07 Now playing Comparing absolute values | Negative numbers and absolute value | Pre-Algebra | Khan Academy Khan Academy Khan Academy • 505K views • 14 years ago • 17 4:13 4:13 Now playing Ordering rational numbers | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 327K views • 10 years ago • Fundraiser 18 4:54 4:54 Now playing Comparing negatives with variables | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 90K views • 10 years ago • 19 3:08 3:08 Now playing Negative numbers and number line examples | Pre-Algebra | Khan Academy Khan Academy Khan Academy • 423K views • 10 years ago • Fundraiser 20 2:22 2:22 Now playing Absolute value of integers | Negative numbers and absolute value | Pre-Algebra | Khan Academy Khan Academy Khan Academy • 934K views • 14 years ago • Fundraiser 21 1:58 1:58 Now playing Decimals and fractions on a number line | Decimals | Pre-Algebra | Khan Academy Khan Academy Khan Academy • 744K views • 14 years ago • Fundraiser 22 3:07 3:07 Now playing Number opposites practice | Negative numbers | 6th grade | Khan Academy Khan Academy Khan Academy • 87K views • 10 years ago • 23 1:15 1:15 Now playing Placing positive and negative decimals on a number line | Decimals | Pre-Algebra | Khan Academy Khan Academy Khan Academy • 303K views • 12 years ago • 24 3:04 3:04 Now playing Introduction to negative numbers Khan Academy Khan Academy • 12K views • 3 years ago • 25 5:39 5:39 Now playing Compare rational numbers using a number line Khan Academy Khan Academy • 22K views • 3 years ago • 26 3:40 3:40 Now playing Meaning of absolute value Khan Academy Khan Academy • 53K views • 3 years ago • 27 3:43 3:43 Now playing Adding integers numbers on the number line | 6th grade | Khan Academy Khan Academy Khan Academy • 900 views • 1 year ago • [](
15616
https://lms.umb.sk/mod/book/tool/print/index.php?id=234757
Kruh, kružnica - cvičenia | VirtualUMB Preskočiť na hlavný obsah Vytlačiť celú knihu Kruh, kružnica - cvičenia Portál:Virtuálna Univerzita Mateja Bela Kurz:Planimetria a stereometria Kniha:Kruh, kružnica - cvičenia Vytlačil(a):Hosťovský používateľ Dátum:pondelok, 29 septembra 2025, 05:01 Opis Kruh Obsah Seminárne zadania Seminárne zadania Cvičenie. ♠ Vypočítajte veľkosti vnútorných uhlov v trojuholníku, ktorý dostanete ak spojíte na ciferníku hodiniek body vyznačujúce čísla 3, 8, 10. Zadanie Tu ♠ V tetivovom štvoruholníku , platí . Vyjadrite veľkosti zostávajúcich vnútorných uhlov tohto štvoruholníka. Zadanie Tu. Daný je trojuholník a kružnica k so stredom , vpísaná do trojuholníka. Body dotyku kružnice k so stranami trojuholníka sú označené . Veľkosť uhla je 32°. Akú veľkosť má uhol ? Otvorte siZadanie Tu. 4. Je daná kružnicaa jej dotyčnica . Zostrojte všetky kružnice, ktoré sa dotýkajú kružnice tak aj priamky a majú pritom polomer . Zadanie Tu 5. ♥ Je daná kružnica a bod , pre ktorý platí . Zostrojte takú sečnicu kružnice prechádzajúcu bodom , aby jej druhý priesečník bol stredom úsečky . 6. ♥ Kružnica je rozdelená na dva oblúky tak, že obvodový uhol príslušný k väčšiemu oblúku je rovný stredovému uhlu príslušnému k menšiemu oblúku. Určite veľkosť obvodových uhlov príslušných k obom oblúkom. 7. je menší oblúk kružnice s obvodovým uhlom . V bodoch sú zostrojené dotyčnice kružnice a bod je ich priesečník. Vypočítajte veľkosť uhla . 8. ♥ Zostrojte kružnicu, ktorá sa dotýka danej priamky a prechádza dvoma rôznymi bodmi , ktoré neležia na priamke a priamky sú rôznobežné. Úlohu riešte pomocou mocnosti bodu ku kružnici.Otvorte si zadanie Tu. 9. Je daná kružnica. Vnútri kružnice leží bod. Narysujte všetky kružnice s polomerom, ktoré sa dotýkajú kružnicea prechádzajú bodom. )
15617
https://nostarch.com/download/MathHacksForScratch_Chapter2.pdf
Exploring Divisibility and Primes 2 You can add, subtract, or multiply any two integers and get an integer back. But when you divide one integer by another, the answer doesn’t have to be an integer. The special case when the result of the division is an integer is worth noticing. Also notable are those rare cases where a number can’t be cleanly divided by any numbers besides 1 and itself. We call those prime numbers. In this chapter, we’ll investigate these two interesting phenomena. These concepts are fundamental to number theory, the study of the properties and mathematics of integers. Number theory is used for everything from random number generation in computer games and simulations to designing error-correcting codes for data transmission and storage. These real-world applications all start with divisibility and primes. The Divisibility Factor We say that the integer d is a divisor of the integer n if the division n / d results in an integer. We can say it with multiplication as well: the number n is divisible by the number d if we can find an integer k so that n = d · k. Another way to say the same thing is that d is a factor of n. Here are some facts, observations, and vocabulary about divisibility: J Every number is divisible by 1, because we can write n = n · 1. J Every number n is a divisor (or factor) of itself. If we don’t want to include n in the list of divisors, we can specify the others as proper divisors. J Integers are even or odd depending on whether they’re divisible by 2. J Every integer divisible by 5 is guaranteed to have a last digit of 0 or 5. J Every integer divisible by 10 ends in a 0. J The set of positive divisors of 6 is {1, 2, 3, 6}. The number 6 is considered perfect because the sum of its proper divisors, 1 + 2 + 3, is 6 itself. Programming Challenge 2.1 Fizz-Buzz is a game that can be played by any number of players seated in a circle. Players take turns counting up from 1, but if the number they’re supposed to say is divisible by 5, the player says “Fizz” instead of the number. If the number is divisible by 7, the player says “Buzz.” If the number is divisible by both 5 and 7, the player says “Fizz Buzz.” If a player says the wrong thing, they’re out, and the last player left wins. Write a program so Scratch Cat can play Fizz-Buzz with you. Modular Arithmetic Even though dividing one integer by another doesn’t necessarily result in another integer, modular arithmetic gives us a way to express any division operation using integers. The answer to a modular division problem is reported as two separate integers: the quotient itself, with any decimal component removed, and an extra part called the remainder. Symbolically, we say the integer b divided by the pos-itive integer a gives a quotient q and a remainder r, where 0 ≤r < a. The relation-ship is given by the equation b = (q · a) + r. Division is the process of determining a quotient and remainder given b and a. The division algorithm identifies the quotient and remainder. Scratch has a built-in operation to capture the remainder r, called mod. To find the quotient q, we do the division using the / operator and indicate that we want to keep only the integer part of the result by using the floor operation. Figure 2-1 gives an example. Figure 2-1: Calculating the quotient and remainder of 45/7 22 Chapter 2 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays Here, floor of 45 / 7 gives us a quotient of 6, and 45 mod 7 gives us a remainder of 3. To check this is right, we can plug the results into our formula: b = (q · a) + r = (6 · 7) + 3 = 42 + 3 = 45 We say two numbers x and y are congruent modulo n if x mod n = y mod n. In this case, when x and y are divided by n, they have the same remainder r. For ex-ample, 7 and 19 are congruent mod 6, because 7 and 19 divided by 6 both yield a remainder of 1. Congruence isn’t as strong as equality, in that equal numbers must be congruent, but congruent numbers need not be equal. Instead of an equal sign (=), we use the triple bar symbol (≡) for congruence, so we write 7 ≡19 mod 6. Here are some facts connecting modular arithmetic to divisibility: J We can test for divisibility of b by a using Scratch by seeing if b mod a is 0. J Odd numbers are all congruent to 1 mod 2, and even numbers are congru-ent to 0 mod 2. J Numbers that end in 0 are congruent to 0 mod 10. They’re also divisible by 10. J Numbers that end in 0 or 5 are congruent to 0 mod 5. They’re also divisi-ble by 5. J When we represent b = (q · a) + r by the division algorithm, the set of all possible remainders is {0, 1, 2, . . . , a – 1}, a set of a elements. Sometimes it’s more useful to use another set of a elements where every integer is congruent to one element of the set. Since Scratch numbers elements in lists starting from 1, the set {1, 2, 3, . . . , a} is often a good choice. We’ll explore a simple hack that uses modular arithmetic to help check the re-sults of a calculation in the next project. Project #5: A Trick for Checking Your Math Casting out nines is a trick for verifying the answer to a large addition or multipli-cation problem. To see how it works, first notice that every power of 10 leaves a remainder of 1 when it’s divided by 9. For example: 10 = 9 + 1 100 = 99 + 1 = (11 · 9) + 1 1,000 = 999 + 1 = (111 · 9) + 1 This points to a broader rule that when you divide a number n by 9, you get the same remainder as when you divide the sum of the digits of n by 9. Take the case of 347 divided by 9. To determine the remainder, we first sum the digits: 3 + 4 + 7 = 14. At this point, we could notice that 14 = (1 · 9) + 5, giving us a remainder Exploring Divisibility and Primes 23 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays of 5. Or we could do the casting out nines trick a second time to get the result in an easier way: 1 + 4 = 5. (In fact, 347 divided by 9 is 38 remainder 5.) Casting out nines is a good way to check your work after a big addition or multiplication operation, because it’s much easier to do arithmetic mod 9 (by sum-ming a number’s digits) than to keep track of multidigit sums and products. Sup-pose, for example, you calculate 347 + 264 and get the answer 601. We’ve already seen that 347 mod 9 is 5. For 264, 2 + 6 + 4 = 12 and 1 + 2 = 3, so 264 mod 9 is 3. That means (347 + 264) mod 9 should be 5 + 3 = 8. But 601 mod 9 is 6 + 0 + 1 = 7, so something is wrong. It looks like somebody forgot to carry the 1 in the original addition! When we fix the sum to be 611, casting out nines works as expected. Even though adding up the digits of a number is pretty easy mental math, let’s have Scratch Cat do the work for us. The program in Figure 2-2 uses the casting out nines technique to find any number mod 9. Figure 2-2: A program for finding x mod 9 by calculating digit sums 24 Chapter 2 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays The trick is to have Scratch see the number x as a string of digits. The length of operator reports how many digits the number has, and the letter of operator lets us pick off one digit at a time so we can add them up. The code is nested inside a repeat until loop that makes it continue until the length of x is 1, meaning the num-ber has only one digit. If that single digit is in the range 0 through 8, we have our answer. The single digit could also be a 9, though, which is congruent to 0 mod 9. In that case, the last if statement picks 0 as the answer to report instead of 9. The Results Figure 2-3 shows a sample run of the program, using 601 as the input. Figure 2-3: Finding 601 mod 9 The last line of the program uses join operations to make the output pretty, reminding us of the input number and the result of the casting out nines process. Hacking the Code We have the same problem here that we had in Chapter 1: Scratch is happy to run the code on input that isn’t a number. The way the program is written, it even gets in trouble with what ought to be perfectly allowable inputs, like negative integers. For example, the number –3 interpreted as a string has a length of 2, and accord-ing to Scratch, the first character, the minus sign, has a numerical value of 0. So Scratch reports that the sum of the digits of –3 is 0 + 3 = 3. The trouble is that –3 mod 9 is equal to 6, not 3. Because we’ll run into problems with negative integers and non-integer inputs, before we put the code out for general use, we should make it safe by screening possible inputs to allow only the ones we want: positive integers. We can create a custom block to screen the input, as shown in Figure 2-4. Exploring Divisibility and Primes 25 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays Figure 2-4: Making sure the input is a positive integer The Boolean statement round test = test is a hack that lets us kill a few birds with one stone. It screens out non-numeric input (such as the word banana), since trying to round a non-number in Scratch produces 0 as a result. It also screens out numbers with nonzero decimal components, which will no longer be equal to themselves after rounding. Combined with text > 0, our if statement is true if the input test is a positive integer and false otherwise, so we can set the value of the variable positive integer? to true if the two conditions are satisfied. N O T E Some programming languages have special Boolean variables that can take on only the values true or false, but Scratch doesn’t. Here, we simply use the words true and false instead. Some programmers prefer to use the numerical values 1 and 0 to keep track of truth values. Once we have a screening block, we can modify the code in Figure 2-2 to ex-ecute for only appropriate values, as shown in Figure 2-5. Paste the original pro-gram from the repeat until block onward into the empty slot after the if. Figure 2-5: Don’t let Scratch Cat make a mistake! 26 Chapter 2 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays Of course, you don’t actually have to go through the casting out nines process to perform this calculation. You can just use Scratch’s mod block! Still, writing the program is good practice for figuring out how to solve a problem and how to an-alyze a number, one digit at a time. The program also generalizes to other cases, such as the one in Challenge 2.2. Programming Challenges 2.2 Casting out nines gives a test for divisibility by 9, since if the sum of the digits is 0 or 9, the number is divisible by 9. A test for divisibility by 11 works similarly, except instead of adding all the digits, you alter-nately subtract and add them. For example, 1,342 is divisible by 11 because 1 – 3 + 4 – 2 = 0. Program Scratch to calculate the –/+ digit sum for a given number to see if it’s divisible by 11. 2.3 Scratch has an operator that lets you pick a random number in a spe-cified range. Write a program to pick 10 random numbers between 1 and 100. Predict how many are likely to be divisible by 9, then use Scratch to check if your prediction was right. 2.4 Sometimes when you have to enter a number into a computer form (like a credit card number or a book’s ISBN code), the number in-cludes a check digit to make sure you haven’t made a mistake. One way to implement this is to add an extra digit at the end that’s derived from the original number. For example, the extra digit could be the original number mod 9, found by casting out nines as in the program in Figure 2-2. Extend this program to give the original number with its check digit added. 2.5 When copying numbers we sometimes make transposition errors, where two digits are switched. For example, we might miswrite 1,467 as 1,647. Could you use the casting out nines trick to help catch this kind of mistake? Prime Numbers Some integers have many divisors, and some have only a few. The integer 1 is a special case, in that it’s divisible only by itself. For any other number, the smallest number of divisors is two: 1 and the number itself. As mentioned at the beginning of this chapter, numbers with only two divisors are called prime numbers. Num-bers with more than two divisors are called composite numbers. The first few prime numbers are 2, 3, 5, 7, 11, 13, and 17. To find more, we’ll turn to Scratch. Exploring Divisibility and Primes 27 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays Project #6: Is It Prime? One way to determine if a number is prime is to try out possible factors one by one, a process called trial division. If there aren’t any other divisors between 1 and the number, then the number is prime. For the number 5, for example, we would try dividing 5 by 2, then 3, then 4. None of those numbers divide evenly into 5, so 5 is prime. Doing trial division manually quickly gets tedious, so we’ll write a program to make Scratch do it for us. Figure 2-6 shows a simple version of the code that doesn’t worry about improper inputs that could cause incorrect answers (for ex-ample, strings or numbers that aren’t positive integers). 1 2 Figure 2-6: Checking for primes by trial division 28 Chapter 2 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays The code prompts for a number to test and decides if the number is prime by working with the Boolean prime? variable. We perform the trial division in a repeat until loop ‚ by calculating answer mod trial ƒ, where the variable trial is the trial divisor. If the result is 0, we know that we have a divisor and that answer isn’t prime, so we exit the loop. Otherwise, we add 1 to trial and try again. At the end, we report an answer based on whether prime? is true or false. The Results Figure 2-7 shows some sample runs of the trial division program. Figure 2-7: Sample runs of the trial division program The program correctly identifies 29 as prime and 30 as not prime. Hacking the Code We should screen the input so Scratch is considering only positive integers. A cus-tom block like the one we made for the casting out nines program (see Figure 2-4) would work, put into an if statement (as in Figure 2-5). There are a few more con-ditions to put into the screening code, though. First, the integer 1 is neither prime nor composite, but 1 would survive the repeat until loop in our trial division pro-gram and be labeled as prime. The custom block in Figure 2-8 includes an initial if test to disallow an input of 1. Exploring Divisibility and Primes 29 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays Figure 2-8: Limiting the input for the trial division program A more subtle problem is that, as we saw in Chapter 1, integer arithmetic is ex-act only up to flintmax. That means the divisibility test works only for numbers up to 9,007,199,254,740,992. After that, Scratch Cat thinks every number is compos-ite! The check code in Figure 2-8 accounts for this as well by verifying that test is less than flintmax. The block also returns a message variable giving more informa-tion for the program to report when the input can’t reliably be tested. Another consideration with this program is that trial division on large numbers potentially takes many steps—so many that even on a fast computer you might have to wait a long time to get an answer. The test in the repeat until loop ‚ in Figure 2-6 is a hack to speed up the process: we really have to consider only trial divisors up to the square root of the input number. This works because if a num-ber n isn’t prime, it must have a factorization n = a · b other than the trivial factor-ization 1 · n. Since n = √ n · √ n, one of a or b must be bigger than √ n and one must be smaller than √ n. We have to do trial division only up to √ n to find the smaller one, if it exists. This hack provides a huge savings! We can test numbers up to 1,000,000 with no more than √ 1,000,000 = 1,000 trial divisions. To speed up the code even fur-ther, once we’ve checked on divisibility by 2, we could test only for divisibility by odd numbers. This is because if a number n is divisible by any even number, it will also be divisible by 2. All these improvements allow for shorter runtimes, but they also make for a longer, more complicated program. Whether the trade-off is worth it will depend 30 Chapter 2 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays on who will be using your work, and for what. Improvements that make the pro-gram easier to use are usually worth it. Improvements that speed up runtime have to be dramatic to be noticeable, but they may be worthwhile if users will be look-ing for quick results. Project #7: The Sieve of Eratosthenes Trial division isn’t the only way to find prime numbers. In this project, we’ll explore a different technique: looking at a list of all numbers up to some limit and throw-ing away the numbers that are composite. This approach sifts, or sieves, out the primes and is called the sieve of Eratosthenes after the Greek mathematician who first used it. Scratch Cat uses sieving in Figure 2-9, where the numbers 1 through 120 have been arranged in a grid. Figure 2-9: Sieving out the primes by throwing away non-primes First, we cross out 1, which is neither prime nor composite, in red. Then, we cross out all multiples of 2 in green, as shown on the left side of Figure 2-10, and see what’s left. We continue by identifying the next few primes after 2 (3, 5, and 7) and crossing out any multiples of them, as shown on the right side of Figure 2-10. Exploring Divisibility and Primes 31 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays Figure 2-10: Eliminating all the even numbers after 2 (left) and all multiples of 3, 5, and 7 (right) Notice that multiples of 2 and 3 are crossed out with vertical lines down the columns of the grid. This works because the grid is set up to be 6 numbers wide, and 6 is divisible by both 2 and 3. Multiples of 5, crossed out in pink, step back-ward on the diagonal. This is because to get from one multiple of 5 to the next multiple we add 5, which is 6 – 1. So to find the next multiple of 5, we go down one row for the 6 and back one column for the –1. Similarly, to find multiples of 7, crossed out in yellow, we go down one row and step one column to the right (because 7 = 6 + 1), giving us lines along the other diagonal. Figure 2-11: All the primes up to 120, after sieving Here’s the payoff of sieving: if a num-ber n is composite and has a factorization n = a · b where 1 < a ≤b < n, then a ≤√ n. In our example, n = 120, so any composite num-ber in the grid must have a prime factor less than √ 120, or approximately 10.95. Once we’ve sieved up to 7, the next number that hasn’t al-ready been crossed out is 11, which is greater than √ 120, so 7 is as far as we need to sieve. Every number that remains, meaning it hasn’t been crossed off as a multiple of 2, 3, 5, or 7, must be a prime number (see Figure 2-11). This is the second time the square root hack has been useful. First, it made the trial division program in Figure 2-6 run faster. Now, it’s telling us when to stop sieving, allowing us (in this example) to find all primes less than 120 just by sieving up to 7. We could use the same technique to sieve up to a much higher bound. All we have to do 32 Chapter 2 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays is get rid of all the multiples of each prime as they’re discovered, up to the square root of the bound. That’s what we do in the Scratch program in Figure 2-12. 1 2 3 Figure 2-12: The sieve program We start by asking how far to go, then seeding the list primes with that many entries ‚. (Since we’re using a list, our upper bound is limited by the maximum list size that Scratch supports, which is 200,000.) Scratch indexes lists starting with 1, so the list entry at index n will keep track of whether n is a prime. Initially we set each entry to true, but we’ll change non-prime entries to false as we sieve. Exploring Divisibility and Primes 33 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays First, we handle the special case of 1, which is neither prime nor composite ƒ. Then, we look for the next number not crossed out by sieving so far. We leave that number as true but set all multiples of that number to false „. We repeat this pro-cess until the next number not crossed out is greater than the square root of the limit. Once we have a complete list, we can access it and answer questions about the prime numbers we’ve found. Figure 2-13 has a little piece of code to count how many primes there are up to the sieve limit. Figure 2-13: Counting primes with the sieve program Here, we step through the list we’ve built and count how many true entries there are, incrementing the variable primecount each time. Figure 2-14 shows an-other extra piece of code that lists the primes we’ve found. Figure 2-14: Listing primes with the sieve program This block finds the true items in the list and stores their corresponding index numbers in a separate list. 34 Chapter 2 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays Hacking the Code Sometimes it’s useful to have the data that Scratch generates as a separate file so you can import it into a text editor or a spreadsheet. Fortunately, Scratch gives us the option to import and export lists by right-clicking the list in the graphics window (see Figure 2-15). This way, you can take your sieved list of primes out of Scratch to play with it further. Figure 2-15: Saving the list to work on later Text editors, word processors, and spreadsheet programs are happy to work with the text output from Scratch. Try importing your data into a spreadsheet pro-gram such as Excel, Numbers, or Open Office. If you want several entries per row, make sure you have Scratch insert commas separating the entries in your text file (using the join block), and then use the CSV format, short for comma separated values, to import it. The default carriage returns in the file that Scratch produces will list the entries in separate rows in the spreadsheet. Programming Challenges 2.6 Use the sieve program to find how many primes there are between 1 and 10, 100, 1,000, 10,000, and 100,000. Keep track of the ratio between the number of primes and the size of the list, and display your results in a table. How does the relative number of primes ap-pear to be changing as the upper bound increases? 2.7 Write a block to scan the list of integers that the sieve program pro-duces, looking for long sequences of consecutive composite numbers. What’s the longest sequence you can find? (continued) Exploring Divisibility and Primes 35 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays 2.8 Twin primes are pairs of primes that differ by exactly 2; for example, 3 and 5 or 11 and 13. Write a block to scan the sieve program’s output and count how many pairs of twin primes there are up to the sieving limit. 2.9 Rewrite the sieve program in Figure 2-12 so it displays the results in a table six entries wide, like the table in Figure 2-9. Use the language of congruences to explain why the only prime numbers that appear after the first row of the table are in columns 1 and 5. Nothing Common About Common Divisors Given two integers a and b, the set of common divisors refers to all the integers that evenly divide both a and b. There will always be at least one common divi-sor, the number 1, since 1 is a factor of all integers. But larger common divisors might exist as well. Of particular interest is the greatest common divisor (GCD), the largest number that evenly divides a and b. If this largest common divisor is d, we write GCD(a, b) = d. As with identifying primes, there are several methods for finding the GCD of two numbers, with varying degrees of efficiency. We’ll explore two such tech-niques in the next two projects. Project #8: Greatest Common Divisors the Slow Way Here’s one way to find the greatest common divisor between two integers a and b. Starting from 1, try dividing a and b by every number. If it divides evenly into both of them, you’ve found a common divisor. Stop once you reach a or b, whichever comes first. The highest common divisor you’ve found is the GCD. The program in Figure 2-16 uses this approach. We use a custom block to identify the minimum of the two input values, a and b. Then we count up from 1 to the minimum, checking if the mod of both a and b is 0. If it is, we store the current divisor in the variable gcd, which holds our answer when the program finishes running. This technique of testing every number as a possible common divisor is known as a brute-force approach. It’s like trying to guess someone’s computer password by testing out every possible sequence of letters and numbers. For our GCD pro-gram, brute force is fast enough for smaller values of a and b, say up to 1 million, but it’s noticeably slower for larger numbers. As the numbers being screened get closer to flintmax, it becomes especially annoying to wait. Luckily, there’s a bet-ter way. 36 Chapter 2 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays Figure 2-16: Finding the GCD the slow way Project #9: Greatest Common Divisors the Fast Way The Greek mathematician Euclid described a more efficient method for calculating the greatest common divisor of two numbers in his textbook The Elements, which was written around 300 BCE. The Elements covers topics in several different areas of mathematics, focusing on geometry and number theory. The book was so influ-ential that Euclid’s organization of the material was used to teach mathematics for centuries, and it continues to be used today. Euclid’s approach to calculating greatest common divisors is based on the ob-servation that for two positive integers a and b where a < b, any common divisor of a and b is also a divisor of b – a. For example, say a = 330 and b = 876. A com-mon divisor of 330 and 876 is 6, and 6 is also a divisor of 876 – 330 = 546. By extension, if we divide the larger of the two numbers, b, by the smaller, a, and keep track of the division with a quotient and remainder, b = q · a + r, then any Exploring Divisibility and Primes 37 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays common divisor of b and a is also a divisor of a and r. Then we can repeat the pro-cess with a and r, and so on until there’s a last remainder of 0. At this point, the next-to-last remainder is the greatest common divisor of a and b. The sequence of divisions looks like this: b = q1 · a + r1 a = q2 · r1 + r2 r1 = q3 · r2 + r3 . . . rk – 2 = qk · rk–1 + rk rk – 1 = qk+1 · rk + 0 The remainders decrease, so a > r1 > r2 > . . . > rk, with rk = GCD(b, a) and rk+1 = 0. Here are the steps to calculate that 6 is the greatest common divisor of b = 876 and a = 330, interpreted both with the division algorithm and with modular arithmetic. Notice how the values shift positions from right to left as we move from one line to the next: 876 = 2 · 330 + 216 876 mod 330 = 216 330 = 1 · 216 + 114 330 mod 216 = 114 216 = 1 · 114 + 102 216 mod 114 = 102 114 = 1 · 102 + 12 114 mod 102 = 12 102 = 8 · 12 + 6 102 mod 12 = 6 12 = 2 · 6 + 0 12 mod 6 = 0 The Scratch program in Figure 2-17 implements Euclid’s algorithm. 1 2 3 Figure 2-17: Finding the GCD with Euclid’s algorithm 38 Chapter 2 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays The program is organized so all the work of the repeated division occurs in the custom gcd block ‚. The block’s definition is surprisingly short compared to our brute-force GCD program (Figure 2-16). Inside a repeat until loop, we keep taking b mod a ƒ and shuffling the values of a and r back into b and a until we finally get down to a remainder of 0. That’s where the loop stops, and the last value of a can be reported as the GCD „. The Results Figure 2-18 shows a sample run of the GCD program with two very large numbers as inputs. Figure 2-18: A calculation with Euclid’s algorithm Unlike our brute-force approach, the code works very quickly, even for num-bers close to flintmax. Hacking the Code So far, the language we’ve used to talk about how efficient an algorithm is has been pretty general. We talk about a program running quickly or slowly, but it would be good to know just how quickly or slowly that turns out to be on your computer. It would also be useful to see how the program’s performance changes as we go from working with numbers in the tens or hundreds to numbers in the thousands or millions. Scratch has a built-in timer that measures elapsed time in seconds from the moment a program starts executing. It’s accessible via the timer block in the Sens-ing section of the block menu. We can take any program and wrap it in a few lines of code to time how long an algorithm takes to run, as shown in Figure 2-19. Figure 2-19: Timing how fast a program runs Exploring Divisibility and Primes 39 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays Here, the initialize block would contain any setup code that you don’t want to time, such as prompting the user for input, while the run code block would contain the code for the algorithm you want to time. We record the value of timer before and after executing run code, then take the difference between the two times to see how long the execution took. Figure 2-20: Testing a big prime number Figure 2-20 shows the result of wrapping the trial division prime testing program from Figure 2-6 in the timer code, including the value of elapsed time when the program fin-ishes. For a prime close to flintmax, it takes my computer a little over a minute to report. For many programs with small test values, the elapsed time will show as 0, since the algo-rithm takes only a fraction of a second to run. The reported time might also vary across runs because your computer is doing other things in the background, which limits the amount of resources Scratch has available to do its job. To get an accurate time, run the program lots of times in a row and keep track of the cumulative runtime, then divide by the number of times you ran the program to find the average time for each run. Programming Challenges 2.10 Use timing loops to compare the runtimes for the two GCD calculating programs (the brute-force version in Figure 2-16 and the Euclidean version in Figure 2-17). 2.11 Program a counter to count how many steps Euclid’s algorithm takes. Experiment to see what numbers make the algorithm take the highest number of steps to run. Conclusion Computations involving divisibility are much easier and faster to do with com-puter assistance. If I had to work out if a number was prime by doing trial division by hand, I would probably give up after a few dozen calculations. Even if I were punching possible divisors into a calculator, I would get bored pretty quickly and probably start making mistakes (“trial and error” is mostly error!). But Scratch Cat is eager to help out for as long as I want. Scratch is a telescope that lets us look deeper into the universe of numbers than we could ever do ourselves. All we have to do is ask. 40 Chapter 2 Math Hacks for Scratch (Sample Chapter) © 2024 by Michael Mays
15618
https://mathoverflow.net/questions/55585/lower-bound-for-sum-of-binomial-coefficients
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 Lower bound for sum of binomial coefficients? Ask Question Asked Modified 1 year, 4 months ago Viewed 14k times $\begingroup$ Hi! I'm new here. It would be awesome if someone knows a good answer. Is there a good lower bound for the tail of sums of binomial coefficients? I'm particularly interested in the simplest case $\sum_{i=0}^k {n \choose i}$. It would be extra good if the bound is general enough to apply to $\sum_{i=0}^k {n \choose i}(1-\epsilon)^{n-i}\epsilon^i$. For the more commonly used upper bound, variants of Chernoff, Hoeffding, or the more general Bernstein inequalities are used. But for the lower bound, what can we do? One could use Stirling to compute $n!$ and then ${n \choose k}$ and then take the sum: ${n \choose k} = \frac{n!}{k!}{(n-k)!}$, and Stirling's formula (a version due to Robbins) gives $$n! = \sqrt{2\pi}n^{-1/2}e^{n-r(n)}$$ with remainder $r(n)$ satisfying $\frac{1}{12n} \leq r(n) \leq \frac{1}{12n+1}$. For the next step, it's easy to apply Stirling thrice. But, even better, I noticed that Stanica 2001 has a slight improvement to the lower bound that also is simpler to state (but more difficult to prove): $${n \choose k} \geq \frac{1}{\sqrt{2\pi}}2^{nH(k/n)}n^{1/2}k^{-1/2}(n-k)^{-1/2}e^{-\frac{1}{8n}}$$ for $H(\delta) = -\delta \log \delta -(1-\delta)\log(1-\delta)$ being the entropy of a coin of probability $\delta$. Now for step 3. If $k$ is small, it's reasonable to approximate the sum by its largest term, which should be the ${n \choose k}$ term unless $\epsilon$ is even smaller than $k/n$. So that's great, we're done! But wait. This bound is off by a factor of at most $\sqrt{n}$. It would be better to be off by at most $1 + O(n^{-1})$, like we could get if we have the appropriate Taylor series. Is there a nice way to do the sum? Should I compute $\int_{0}^{k/n} 2^{nH(x)}\frac{1}{\sqrt{2\pi}}x^{-1/2}(1-x)^{-1/2}n^{1/2}e^{-1/8n} dx$ and compare that to the discrete sum, and try to bound the difference? (This technique has worked for Stirling-type bounds.) (The terms not dependent on $k$ or $x$ can be moved out of the integral.) Another approach would be to start from Chernoff rather than Stirling (i.e. "How tight is Chernoff guaranteed to be, as a function of n and k/n?") Any ideas or references? Thanks! st.statistics binomial-distribution binomial-coefficients co.combinatorics Share Improve this question edited Feb 16, 2011 at 13:10 user13006user13006 asked Feb 16, 2011 at 6:29 user13006user13006 26311 gold badge22 silver badges88 bronze badges $\endgroup$ 3 $\begingroup$ The terms in your sum do not depend on the index $i$. You might want to proofread what you wrote. $\endgroup$ Felipe Voloch – Felipe Voloch 2011-02-16 07:40:51 +00:00 Commented Feb 16, 2011 at 7:40 $\begingroup$ @cscjb: (Typo.) The first appearance of the entropy $H$ should read $H(k/n)$ and not $H(k)$, I think. $\endgroup$ Did – Did 2011-02-16 08:10:40 +00:00 Commented Feb 16, 2011 at 8:10 $\begingroup$ Mert sent me a message and asked for the Stanica paper title. It is "Good lower and upper bounds on binomial coefficients", J. Inequal. in Pure and Appl. Math volume 2 #3, 2001. $\endgroup$ user13006 – user13006 2011-09-23 03:10:39 +00:00 Commented Sep 23, 2011 at 3:10 Add a comment | 8 Answers 8 Reset to default 17 $\begingroup$ First, what the Stirling bound or Stanica's result give is already a $(1+O(n^{-1}))$ approximation of $\binom nk$, hence the only problem can be with the sum. I don't know how to do that with such precision, but it's easy to compute it up to a constant factor by approximating with a geometric series: $$\sum_{i\le k}\binom ni=\begin{cases}\Theta(2^n)&k\ge n/2-\sqrt n,\\\Theta\left(\left(1-\frac{2k}n\right)^{-1}\binom nk\right)&k\le n/2-\sqrt n.\end{cases}$$ More generally, $$\sum_{i\le k}\binom ni(1-\epsilon)^{n-i}\epsilon^i=\begin{cases}\Theta(1)&k\ge\epsilon n-s,\\\Theta\left(\frac{\epsilon(n-k)}{\epsilon n-k}\binom nk(1-\epsilon)^{n-k}\epsilon^k\right)&k\le\epsilon n-s,\end{cases}$$ where $s=\sqrt{n\epsilon(1-\epsilon)}$. Cf. the appendix to my paper . Share Improve this answer answered Feb 16, 2011 at 13:37 Emil JeřábekEmil Jeřábek 50.1k44 gold badges161161 silver badges223223 bronze badges $\endgroup$ 5 $\begingroup$ You're right. I spoke too soon after waking up. The problem is that the $f(x) = x^{1/x}$ term is immediately multiplied by an $f(\frac{x}{x-1})$ term, and while a single $f(x)f(x/(x-1))$ term is maximized for $x=2$, optimizing sums over this product is not as clear. And if you can't precisely estimate the base of an exponential term, you're in trouble. Your approach is better -- and simple too! In retrospect it's the obvious thing to do. So I see it's covered by Proposition A.4 in your paper, with some slightly different notation. $\endgroup$ user13006 – user13006 2011-02-16 16:51:49 +00:00 Commented Feb 16, 2011 at 16:51 $\begingroup$ So, more detail: geometric series comes from $${n \choose k-1}/{n \choose k} = \frac{k}{n-k+1} > \frac{k-i}{n-k+i+1}$$ Let $x = \frac{k}{n-k+1}$. Then $x \leq \frac{2k}{n}$, and $x > \frac{k-i}{n-k+i+1}$ for $k \geq i > 0$, so geometric series gives lower bound. $$\sum_{i=0}^{k} {n \choose i} \leq {n \choose k} \sum_{j = 0}^{k} x^j \leq {n \choose k}/(1-x)$$ matches your statement for $x = \frac{2k}{n}$. But I believe you must have intended to use $x = \frac{k}{n-k}$ or $\frac{k}{n-k+1}$, since otherwise it is not tight. Setting $\epsilon = 1/2$ in your 2nd version gives correct version. $\endgroup$ user13006 – user13006 2011-02-16 16:54:53 +00:00 Commented Feb 16, 2011 at 16:54 $\begingroup$ I just simplified the expression since there are constant factors missed anyway. Taking $x=k/(n-k+1)$ gives $1-x=(n-2k+1)/(n-k+1)$. However, $n-k+1=\Theta(n)$ by the assumptions, hence $(n-2k+1)/(n-k+1)=\Theta((n-2k)/n)=\Theta(1-2k/n)$. Also, note that this argument gives an upper bound on the sum. The matching lower bound is slightly more complicated to prove, you basically need to show that $(k-i)/(n-k+i+1)$ is not too smaller than $k/(n-k+1)$ for sufficiently many $i$'s so that it works out. $\endgroup$ Emil Jeřábek – Emil Jeřábek 2011-02-16 17:28:28 +00:00 Commented Feb 16, 2011 at 17:28 $\begingroup$ I should probably also point out that the extra terms are relevant only when $k$ is close to $n/2$. If $k\le\alpha n$ for some constant $\alpha<1/2$, then it all boils down to $\sum_{i\le k}\binom ni=\Theta\left(\binom nk\right)$, as Anthony Quas wrote in his answer. $\endgroup$ Emil Jeřábek – Emil Jeřábek 2011-02-16 17:33:17 +00:00 Commented Feb 16, 2011 at 17:33 $\begingroup$ Thanks again. You're right of course. This turned out to be much easier than I anticipated. I think I was so excited to see the latex typeset in real-time that I followed up too soon. I do need the case where the terms are relevant, so I'm happy to know how to do that now. $\endgroup$ user13006 – user13006 2011-02-17 19:13:21 +00:00 Commented Feb 17, 2011 at 19:13 Add a comment | 9 $\begingroup$ In my paper "On Littlewood's estimate for the binomial distribution", Adv. Appl. Prob., 21 (1989) 475-478, copy at , I find sharp exact bounds on this sum. Taking $p=1/2$ in Theorem 2 gives: $$B(k; n,1/2) = \sigma \cdot b(k-1,n-1,{1/ 2})\cdot Y({k-n/2\over \sigma})\cdot \exp({E(k; n,1/2)\over \sigma})$$ where: $b(k-1; n-1,1/2) = {1\over 2^{n-1}}{n-1\choose k-1}$ $B(k; n,1/2) = \sum_{j=k}^n b(j; n,1/2) = {1\over 2^n}\sum_{j=k}^n {n\choose j}$ $Y(x) = Q(x)/\phi(x)$, where: $\phi(x) := {1\over \sqrt{2\pi}} e^{-x^2/2}$ $Q(x) := \int_{x}^\infty \phi(u) du $ $E(k; n,1/2)$ is the error term, which lies between 0 and $\min(\sqrt{\pi/8}, {\sigma/(k-n/2)})$. $\sigma = \sqrt{n}/2$. The relative error is at most $O(n^{-1/2})$ for all $k$, better if $k$ is not close to $n/2$. The above requires $\frac n2\le k\le n$. For $0\le k\lt \frac n2$, use $B(k;n,p) = 1 ˆ’ B(n-k+1; n, 1-p)$. Somewhere on the arXiv there is a paper making numerical comparisons of many such approximations. I can't find it just now, maybe someone else can. Share Improve this answer edited Jan 9, 2018 at 11:55 answered Sep 3, 2011 at 7:36 Brendan McKayBrendan McKay 38.3k33 gold badges8484 silver badges153153 bronze badges $\endgroup$ 1 1 $\begingroup$ Brendan, Thanks. You might be referring to this: Central Binomial Tail Bounds, Matus Telgarsky, 2009. arxiv.org/abs/0911.2077v2 At this point I'd rather avoid the normal distribution and erf(), but I understand it's certainly useful, and it seems that it's hard to get a tight lower bound without it. $\endgroup$ user13006 – user13006 2011-09-23 03:07:50 +00:00 Commented Sep 23, 2011 at 3:07 Add a comment | 5 $\begingroup$ Here is a relevant paper: T. Worsch, 1994. Lower and upper bounds for (sums of) binomial coefficients Share Improve this answer edited May 15, 2024 at 15:46 DroneBetter 12155 bronze badges answered Feb 16, 2011 at 11:02 Max AlekseyevMax Alekseyev 38k55 gold badges8383 silver badges164164 bronze badges $\endgroup$ 3 $\begingroup$ Yes! Lemma 2.3 and 2.4 in particular (and, less importantly but still a part, Lemma 2.8) is what is needed to get a tight bound. Interesting that $e^{1/e}$ makes an appearance (via $\min x^{1/x}$ of course). The paper only considers the sums (i.e. probability $1/2$), but with these issues resolved it should be easy to generalize to any $p$. The paper also gets $O(\ldots)$ bounds instead of determining the constants, but the constants can be determined using some of the bounds that Stanica, for example, cites. Looks great! $\endgroup$ user13006 – user13006 2011-02-16 12:53:30 +00:00 Commented Feb 16, 2011 at 12:53 $\begingroup$ Note that the paper only approximates the sums up to a factor of $k$. $\endgroup$ Emil Jeřábek – Emil Jeřábek 2011-02-16 13:39:43 +00:00 Commented Feb 16, 2011 at 13:39 $\begingroup$ indeed, I spoke too soon. Doing math immediately after waking up doesn't work so well. $\endgroup$ user13006 – user13006 2011-02-16 17:07:52 +00:00 Commented Feb 16, 2011 at 17:07 Add a comment | 4 $\begingroup$ Do you know how good you need it? Provided $k < n/3$ say, a reasonable bound (correct to within a multiplicative factor of 2) is obtained by taking the last term $\binom {n}{k}$ (you see this because you can compute the ratio of each term to the prior term and bound it above by 1/2. Now you can estimate the sum as a geometric series.) For $\sum_{j < k} \binom{n}{j}a^j(1-a)^{n-j}$, bounding by the last term also works quite well as long as $k$ is a good bit smaller than $an$. Share Improve this answer edited Feb 16, 2011 at 8:08 answered Feb 16, 2011 at 7:34 Anthony QuasAnthony Quas 23.5k55 gold badges6464 silver badges9999 bronze badges $\endgroup$ 1 $\begingroup$ Honestly, I only need a bound off by a polynomial factor since the exponential ones dominate, so the original that I started with would be fine. But it's become a sort of personal quest to find the "real" answer -- since I use Chernoff so much, and I know Stirling can be made extremely tight, I thought there should be an answer. I figured it's also something that should be answered somewhere on mathoverflow. So, this is what I started with, but I'd like to be able to get $k$ close to $1/2$. $\endgroup$ user13006 – user13006 2011-02-16 12:58:22 +00:00 Commented Feb 16, 2011 at 12:58 Add a comment | 1 $\begingroup$ Summing binomial coefficients $\sum_{i=0}^k\binom{n}{i}$ can be seen as asking "how many binary strings are close to the length-$n$ all-zero string, differing in at most $k$ places?". One can generalize this to larger alphabets, and this almost captures your question on $\sum_{i=0}^k\binom{n}{i} (1-a)^{n-k}a^k$. So perhaps the coding theory community has more to say on this issue? One place to start is this set of lecture notes by Venkat Guruswami: (see page 3). Share Improve this answer answered Feb 16, 2011 at 8:09 miforbesmiforbes 1,08888 silver badges1313 bronze badges $\endgroup$ 1 $\begingroup$ Indeed, that's precisely the origin of the problem. I hadn't realized the connection between smaller epsilon and larger alphabets. Nice! That means the probabilities can be replaced by larger alphabets, so coding theory becomes relevant. (The lecture notes are also interesting to me for other reasons; there's been much progress in searching for more optimal codes in the last few years, so it's great to see Guruswami's notes from 2010. Thanks!) $\endgroup$ user13006 – user13006 2011-02-16 13:07:14 +00:00 Commented Feb 16, 2011 at 13:07 Add a comment | 1 $\begingroup$ Refer to paper: Approximations for the probability in the tails of the binomial distribution by I. Blake, H. Darabian. Might be useful. Share Improve this answer answered Nov 20, 2013 at 6:35 ArtiArti 1111 bronze badge $\endgroup$ 2 $\begingroup$ Do you have a link? $\endgroup$ Thomas Dybdahl Ahle – Thomas Dybdahl Ahle 2015-09-17 23:25:09 +00:00 Commented Sep 17, 2015 at 23:25 1 $\begingroup$ For convenience, here's the link doi.org/10.1109/TIT.1987.1057298 $\endgroup$ j.c. – j.c. 2018-01-06 15:14:35 +00:00 Commented Jan 6, 2018 at 15:14 Add a comment | 1 $\begingroup$ Let $X_1,\ldots,X_n$ be iid from $\mbox{Bernoulli}(\epsilon)$, with standard deviation $\sigma := (\epsilon(1-\epsilon))^{1/2}$. Define $S_n := \sum_{i=1}^n X_i$. Then by Berry-Esseen, we have $$ \sup_{t \in \mathbb R}|P(S_n - np \le t) - \Phi_\sigma(\sqrt{n}t)| = \mathcal O(n^{-1/2}), $$ where $\Phi_\sigma$ is the CDF of a centered normal distribution with variance $\sigma^2$. Therefore, if $Q_\sigma := 1 - \Phi_\sigma$, then $$ \begin{split} P(S_n \ge k) - P(S_n=k) &= P(S_n > k) = 1 - P(S_n \le k)\ &=1 - P(S_n - np \le k-np)\ &= Q_\sigma((k-np)n^{-1/2}) + \mathcal O(n^{-1/2}). % \&= Q_\sigma(kn^{-1/2}-pn^{1/2}) + \mathcal(n^{-1/2}). \end{split} $$ Then you can use standard concentration / anti-concentration results for the normal distribution to get bounds in different regimes for $k$. For example, if $k \ge np - (n(c\log n))^{1/2}\sigma$ with $0 < c \le 1$, then using Hoeffding's inequality gives $$ \begin{split} P(S_n \ge k) - P(S_n = k) &\le Q_1(c\log n) + \mathcal O(n^{-1/2})\ &\le \mathcal O(n^{-c/2}) + \mathcal O(n^{-1/2})\ &= \mathcal O(n^{-c/2}) \to 0. \end{split} $$ Share Improve this answer edited Aug 31, 2020 at 15:44 answered Aug 31, 2020 at 12:24 dohmatobdohmatob 7,02311 gold badge2020 silver badges8383 bronze badges $\endgroup$ Add a comment | 0 $\begingroup$ If you are willing to compute a few binomial coefficients, then (n+1) choose k + (n+1) choose (k-2) + ... + (n+1) choose (k-2l) is a good lower bound even for small l. ( I'm assuing that your summand terms should have i's where they have k's.) Of course, how good depends on how close k is to n/2, in which case one can look at differences from 2^(n/2). Gerhard "Ask Me About System Design" Paseman, 2011.02.15 Share Improve this answer answered Feb 16, 2011 at 7:08 Gerhard PasemanGerhard Paseman 13.1k33 gold badges3434 silver badges6464 bronze badges $\endgroup$ 1 $\begingroup$ Thanks, I'll have to fix that. And that's definitely an obvious place to start. So the Worsch paper mentioned below makes this more explicit ... in particular $k \approx n/e$ seems to be a point of inflection in some sense for a very precise analysis. $\endgroup$ user13006 – user13006 2011-02-16 13:03:04 +00:00 Commented Feb 16, 2011 at 13:03 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 st.statistics binomial-distribution binomial-coefficients co.combinatorics See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked 0 Lower bound on sum of binomial coefficients Expected cardinality of a randomly chosen element of the family of subsets of ${1,\ldots,n}$ with at most $k$-elements 9 Maximum of the weighted binomial sum $2^{-r}\sum_{i=0}^r\binom{m}{i}$ 5 Maximize a weighed combinatorial sum Lower/Upper bounds for $ \sum\limits_{i=0}^k \binom ni x^i $ 4 Partial Sum of the Binomial Theorem 4 Estimating the distribution of minimal hamming distances within a set of strings? Upper bound of sum of binomial coefficients 2 A combinatorial question about encoding the subsets of logarithmic-bounded cardinality 0 Interpolating asymptotic expression for logarithm of middle binomial sums See more linked questions Related 4 Estimate on sum of squares of multinomial coefficients 2 Upper bound of sum of binomial coefficients 2 Lower bound on number of samples for an epsilon delta approximation matching the Chernoff bound 1 A conjecture on 'truncated joint moments' of binomial coefficients under binomial distribution 3 Sum of product of binomial coefficients 2 Lower bound on sum of independent heavy-tailed random variables Question feed
15619
https://www-nds.iaea.org/
| | | | --- | | | | --- | | | IAEA.org | NDS Mission | Mirrors: India | China | Russia | | | | | | | --- --- | | Hot Topics » IAEA-CIELO • TENDL-2023 • JENDL-5 • ENDF/B-VIII.0 | | News » | Pointwise2020//TENDL-2019 | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | ShowDownload;) Download data, codes, packages ShowQuick Links;) Atomic Mass Data Centre Beta-delayed neutrons Charged particle re­fe­ren­ce cross section Decay Data Library for Actinides ENDF Archive ENDF Retrieval ENDF-6 Codes ENDF-6 Format ENSDF ASCII Files ENSDF programs EPICS Electron & pho­ton interaction data Fission Yields Geant4 Libraries LiveChart of Nuclides Medical isotope browser (Isotopia) Minsk Actinides Library Nuclear Charge Radii Nuclear Electro­magnetic Moments Photon Strength Function database Photonuclear RIPL RNAL SIGACE SPECOMP SPECTER STAYSL PNNL Safeguards Data SigmaCalc Spallation models Specialized Evaluated Libraries Standards Stopping Power Data TALYS Tendl2019 Th-U Thermal Scattering Law Library Generator Thermal neutron capture gamma rays Thin Layer Activation URRPACK WIMSD-IAEA Library Wallet cards X and Gamma-rays standards ZVVIEW | | | | --- | | NEW | TALYS-2.0 - Nuclear reaction model code and related packages [page] IDB - An International Database of Reference Gamma Spectra [page] Stopping Power - Electronic Stopping Power of Matter for Ions [page] Web-API for EXFOR/ENDF/IBANDL - software search/download for EXFOR/ENDF/IBANDL [page] | | | | Main All Reaction Data Structure & Decay by Applications Doc & Codes Index Events Links News | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- | | | | | --- | EXFOR Experimental nuclear reaction data | LiveChart of Nuclides Interactive Chart of Nuclides Mobile App: Isotope Browser | CINDA Nuclear reaction bibliography | | ENDF Evaluated nuclear reaction libraries | ENSDF evaluated nuclear structure and decay data (+XUNDL) | NSR Nuclear Science References | | | | | | --- --- | | NuDat-3 selected evaluated nuclear structure data | RIPL reference parameters for nuclear model calculations | IBANDL Ion Beam Analysis Nuclear Data Library | Charged particle re­fe­ren­ce cross section Beam monitor reactions | | PGAA Prompt gamma rays from neutron capture | FENDL Fusion Evaluated Nuclear Data Library | Photonuclear - IAEA Photonuclear Data Library, 2019 - EPICS Electron & Photon Interaction Data, 2017 | IRDFF-II International Reactor Dosimetry and Fusion File | | NAA Neutron Activation Analysis Portal | Safeguards Data Last updated: May 2021 | Medical Portal Medical Portal | Standards - Neutron cross-sections, 2017 - Decay data, 2005 | | | | | Major Databases EXFOR - Experimental nuclear reaction data International Database of Reference Gamma Spectra - International Database of Reference Gamma Spectra ENSDF - evaluated nuclear structure and decay data (+XUNDL) ENDF - Evaluated nuclear reaction libraries CINDA - Nuclear reaction bibliography LiveChart of Nuclides - Interactive Chart of Nuclides Mobile App: Isotope Browser NSR - Nuclear Science References NuDat-3 - selected evaluated nuclear structure data Nuclear Data Libraries and Files General Atomic Mass Data Centre - 2003 atomic mass evaluation, NUBASE, PC-NUCLEUS, etc. RIPL - reference parameters for nuclear model calculations Thermal neutron capture gamma rays - by target and by energy Wallet cards - ground and metastable state properties PSFdatabase - Photon Strength Function database Other evaluated data libraries in ENDF format Photonuclear - Cross sections and spectra up to 140MeV TSL_LibGen - Thermal Scattering Law (TSL) Library Generator INDL/TSL - IAEA Evaluated Nuclear Data Library / Thermal Scattering Law IRDFF-II - International Reactor Dosimetry and Fusion File DXS - neutron and protons induced displacement and gas production cross-sections Minsk Actinides Library - evaluated neutron reaction data (Maslov et al.) NGATLAS - atlas of neutron capture cross sections PADF-2 - Proton Activation Data File Tendl2019 - processing results for a subset of TENDL-2019 for incident neutrons: 630 materials POINT - Pointwise data of ENDF/B-VII.1, processed into temperature dependent form RNAL - Reference Neutron Activation Library Standards - Neutron Cross-section Standards 2017 Th-U - Evaluated nuclear data for the Thorium-Uranium fuel cycle ADS-Lib - Application test library in ACE and MATXS format for ADS neutronics design ENDF Archive - Download evaluated data in original ENDF (4,5,6) format Evaluated libraries in different formats Charged particle re­fe­ren­ce cross section - Beam monitor reactions FENDL - Fusion Evaluated Nuclear Data Library IBANDL - Ion Beam Analysis Nuclear Data Library MIRD - medical internal radiation dose tables Safeguards Data - Last updated: May 2021 PGAA - Prompt gamma rays from neutron capture SigmaCalc - Evaluated (recommended) differential cross sections for Ion Beam Analysis Stopping Power Data - Graphs, data, programs, compiled by H. Paul X and Gamma-rays standards - Decay data standards for detector calibration WIMSD-IAEA Library - multigroup data library for the WIMS-D code Various Specialized Evaluated Data Libraries in ENDF and other formats - Thin Layer Activation - Thin Layer Activation (TLA) Technique for Wear Measurements ENSDF ASCII Files - Evaluated Nuclear Structure Data File in raw (ASCII) format EPICS Electron & pho­ton interaction data - EADL, EEDL, EPDL, EXDL MIB - Medical isotope browser (Isotopia) Beta-delayed neutrons - Reference Database for Beta-Delayed Neutron Emission Electronic Documents Citation Guidelines - online data service manual and citation guidelines ENDF-6 Formats Manual - ENDF-6 Formats Manual (ENDF-102 June 2009 version) ENDF/B-VI Summary Documentation - ENDF-201 December 1996 version ENSDF and NSR Manuals - ENSDF Feb. 2001 version & NSR Aug. '96 version IAEA Technical Documents - IAEA-TECDOCS (Technical Reports Series) IAEA-NDS Reports - Content and formats of nuclear data libraries NDS Staff Publications - publications authored/co-authored by NDS staff ICTP Workshops - workshop material, codes, programme, etc. INDC Reports - INDC reports sorted by area INDC(NDS) Reports - Reports and summaries of meetings Newsletters - NDS Newsletters Fission Yields Report - Doc: Fission Product Yield Data for the Transmutation of Minor Actinide Nuclear Waste Computer codes EMPIRE-3.2 - system of codes for nuclear reaction calculations (Version 3.2) TALYS - nuclear reaction model code; TALYS-related software and databases ENDF Utility Codes - ENDF-6 Format and Codes: Utility, PREPRO, ENDVER ENSDF programs - ENSDF Analysis and Utility programs (ALPHAD, LOGFT, etc.) ENDVER - ENDF File Verification Support Package GANDR - Global Assessment of Nuclear Data Requirements PREPRO - ENDF Preprocessing Codes URRPACK - URR-PACK: Calculating Self-Shielding in the Unresolved Resonance Energy Range GRUCON - Evaluated data preprocessing package Libraries for Geant4 - Processed ENDF-6 libraries to be used by geant4 Monte Carlo transport code SIGACE - package for generating high temperature ACE files DROSG-2000 - Package with programs and data for 60 neutron source reactions LARELKIN - Two-body relativistic kinematics code (by M. Drosg, Faculty of Physics, University of Vienna, 2008) Missing-levels - Missing level corrections using neutron spacings, by G.E.Mitchell & J.F.Shriner PLOTTAB - non-interactive general purpose plotting code to plot continuous curves and discrete physical data SPECTER - code for radiation damage parameter calculations SPECOMP - code for determining displacement cross sections for compound materials STAYSL PNNL - software Suite for Reactor Dosimetry ZVVIEW - interactive plotting of nuclear data (Windows, Linux, Mac) DICEBOX - gamma decay simulation tool General information Datalinks A+M - Atomic and molecular data AMBDAS, ALADDIN, GENIE, etc. Download data, codes, packages - nuclear data, codes, packages for download NDS information Co-ordinated Research Projects (CRPs) - recently completed and ongoing CRPs Nuclear Data Section - introduction, programme description, activities and staff Nuclear data newsletters - from 1994 to now Other useful links Nuclear data networks and other data sources IAEA-NA - Department of Nuclear Sciences and Applications International network of Nuclear Structure and Decay Data evaluators - the NSDD network Nuclear Reaction Data Centre network - the NRDC network, contacts, webpages, etc. Nuclear Knowledge and Databases NIST Standard Reference databases - of National Institute of Standards and Technology (NIST) HyperPhysics - web resources on Physics and application of Georgia State University KADoNiS - Karlsruhe Astrophysical Database of Nucleosynthesis in Stars BrIcc v2.3S - gamma-rays internal atomic shell conversion coefficient calculator Gamma-ray Spectrometry Center - of the Idaho National Laboratory (INL) Computational Nuclear Physics - Computational Nuclear Physics (LLNL) Low Energy Nuclear Knowledge Base - Low Energy Nuclear Knowledge Base (Dubna) Properties of the Elementary Particles - Properties of the Elementary Particles Tcalc - Calculation of reaction threshold energies (JCPRG) Nuclear Reaction Model, Radiation Transport and Nuclear Transmutation Codes TALYS - software for the simulation of nuclear reactions MCNP - family of a general-purpose Monte Carlo Particle codes EASY - European Activation System for simulation of activation and transmutation processes MERCURY - general-purpose Monte-Carlo radiation transport code EGS - general purpose package for the Monte Carlo simulation of the coupled transport of electrons and photons Penelope - general-purpose Monte Carlo code system for the simulation of coupled electron-photon transport Serpent - continuous-energy Monte Carlo Reactor Physics Burnup calculation code FLUKA - general purpose multi-particle Monte Carlo transport code Geant4 - simulation of the passage of particles through matter PHITS - Particle and Heavy Ion Transport code System MARS - simulation of hadronic, electromagnetic, muon, heavy ions and neutron cascades and transport up to 100 TeV | | | | | | | | | --- --- --- | ShowDatabase Retrieval Systems;) | | | --- | | ENDF Evaluated nuclear reaction libraries | EXFOR Experimental nuclear reaction data | | CINDA Nuclear reaction bibliography | IBANDL Ion Beam Analysis Nuclear Data Library | | EE-View Experimental-Evaluated data Viewer | ENDF-DB-Explorer Flexible ENDF Database Explorer | ShowData Libraries for download;) NGATLAS - atlas of neutron capture cross sections IBANDL - Ion Beam Analysis Nuclear Data Library FENDL - Fusion Evaluated Nuclear Data Library Minsk Actinides Library - evaluated neutron reaction data (Maslov et al.) IRDFF-II - International Reactor Dosimetry and Fusion File Charged particle re­fe­ren­ce cross section - Beam monitor reactions PADF-2 - Proton Activation Data File Tendl2019 - processing results for a subset of TENDL-2019 for incident neutrons: 630 materials POINT - Pointwise data of ENDF/B-VII.1, processed into temperature dependent form Standards - Neutron Cross-section Standards 2017 RNAL - Reference Neutron Activation Library Various Specialized Evaluated Data Libraries in ENDF and other formats - ADS-Lib - Application test library in ACE and MATXS format for ADS neutronics design ENDF Archive - Download evaluated data in original ENDF (4,5,6) format Thin Layer Activation - Thin Layer Activation (TLA) Technique for Wear Measurements PIGE - Reference Database for Particle Induced Gamma-ray Emission DXS - neutron and protons induced displacement and gas production cross-sections TSL_LibGen - Thermal Scattering Law (TSL) Library Generator EPICS Electron & pho­ton interaction data - EADL, EEDL, EPDL, EXDL ShowMiscellaneous;) Nuclear Reaction Data Explorer - Nuclear Reaction Data Explorer Nuclear Reaction Data Centre network - the NRDC network, contacts, webpages, etc. Spallation_models - Benchmark of Spallation Models SigmaCalc - Evaluated (recommended) differential cross sections for Ion Beam Analysis EMPIRE-3.2 - system of codes for nuclear reaction calculations (Version 3.2) TALYS - nuclear reaction model code; TALYS-related software and databases Stopping Power Data - Graphs, data, programs, compiled by H. Paul Libraries for Geant4 - Processed ENDF-6 libraries to be used by geant4 Monte Carlo transport code DROSG-2000 - Package with programs and data for 60 neutron source reactions LARELKIN - Two-body relativistic kinematics code (by M. Drosg, Faculty of Physics, University of Vienna, 2008) DICEBOX - gamma decay simulation tool Qcalc - Q-value Calculator | | | | | | | | | | | | | --- --- --- --- --- | ShowStructure and Decay Data;) | | | --- | | NSR Nuclear Science References | ENSDF evaluated nuclear structure and decay data (+XUNDL) | | NuDat-3 selected evaluated nuclear structure data | LiveChart of Nuclides Interactive Chart of Nuclides Mobile App: Isotope Browser | | Decay Data Library for Actinides Evaluated data with detailed comments and decay schemes | Nuclear Electromagnetic Moments Experimental and recommended nuclear moments | | MIB Medical isotope browser (Isotopia) | Beta-delayed neutrons Reference Database for Beta-Delayed Neutron Emission | | Medical Portal Medical Portal | | ShowMiscellaneous;) ENSDF and NSR Manuals - ENSDF Feb. 2001 version & NSR Aug. '96 version ENSDF programs - ENSDF Analysis and Utility programs (ALPHAD, LOGFT, etc.) ICTP Workshops - workshop material, codes, programme, etc. International network of Nuclear Structure and Decay Data evaluators - the NSDD network ENSDF ASCII Files - Evaluated Nuclear Structure Data File in raw (ASCII) format LOGFT - Analysis Program Nuclear Charge Radii - Table of experimental nuclear charge radii for 909 isotopes | | | | HideReactor Physics (particle transport, fuel cycle, transmutation, shielding);) FENDL-2.1 - Fusion Evaluated Nuclear Data Library, Version 2.1 WIMSD-IAEA Library - multigroup data library for the WIMS-D code Minsk Actinides Library - evaluated neutron reaction data (Maslov et al.) NuDat-3 - selected evaluated nuclear structure data ENDF - Evaluated nuclear reaction libraries MENDL-2 - Russian cross-section data library for transmutation and activation of materials irradiated by neutrons with energies up to 100 MeV. Yu.N. Shubin et al. Fission Yields - Fission Product Yield Data for the Transmutation of Minor Actinide Nuclear Waste Fission Yields Report - Doc: Fission Product Yield Data for the Transmutation of Minor Actinide Nuclear Waste ADS-Lib - Application test library in ACE and MATXS format for ADS neutronics design IRDFF-II - International Reactor Dosimetry and Fusion File DXS - neutron and protons induced displacement and gas production cross-sections TSL_LibGen - Thermal Scattering Law (TSL) Library Generator HideAtomic and molecular data for fusion research;) A+M - Atomic and molecular data AMBDAS, ALADDIN, GENIE, etc. ALADDIN - Atomic and Molecular Data for Fusion AMBDAS - Atomic and molecular bibliographical database HideIon Beam and Thin Layer Activation Analysis;) IBANDL - Ion Beam Analysis Nuclear Data Library SigmaCalc - Evaluated (recommended) differential cross sections for Ion Beam Analysis EXFOR - Experimental nuclear reaction data ENDF - Evaluated nuclear reaction libraries HideDosimetry reactions;) IRDF-2002 - International Reactor Dosimetry File IRDF-90 - The International Reactor Dosimetry File (IRDF-90 Version 2) NMF-90 - Neutron Metrology File IRDFF-II - International Reactor Dosimetry and Fusion File HideNeutron Activation Analysis;) PGAA - Prompt gamma rays from neutron capture NGATLAS - atlas of neutron capture cross sections Thermal neutron capture gamma rays - by target and by energy NuDat-3 - selected evaluated nuclear structure data HideNuclear Medicine;) Charged particle re­fe­ren­ce cross section - Beam monitor reactions MIRD - medical internal radiation dose tables RADAR - Radiation Dose Assessment Resource Radiation dose assessment X and Gamma-rays standards - Decay data standards for detector calibration MIB - Medical isotope browser (Isotopia) HideActivation Related Libraries;) PADF-2 - Proton Activation Data File RNAL - Reference Neutron Activation Library Thin Layer Activation - Thin Layer Activation (TLA) Technique for Wear Measurements HideNeutron Source Reactions;) Alpha-n reactions from JAEA - Alpha-n reactions from JAEA DROSG-2000 - Package with programs and data for 60 neutron source reactions Cf-252 neutron spectrum - Evaluation of the californium-252 spontaneous fission neutron spectrum, by W. Mannhart, PTB Braunschweig, FRG, 1986/8 | | | | ShowElectronic Documents;) Citation Guidelines - online data service manual and citation guidelines ENDF-6 Formats Manual - ENDF-6 Formats Manual (ENDF-102 June 2009 version) ENDF/B-VI Summary Documentation - ENDF-201 December 1996 version ENSDF and NSR Manuals - ENSDF Feb. 2001 version & NSR Aug. '96 version IAEA Technical Documents - IAEA-TECDOCS (Technical Reports Series) IAEA-NDS Reports - Content and formats of nuclear data libraries NDS Staff Publications - publications authored/co-authored by NDS staff ICTP Workshops - workshop material, codes, programme, etc. INDC Reports - INDC reports sorted by area INDC(NDS) Reports - Reports and summaries of meetings Newsletters - NDS Newsletters Fission Yields Report - Doc: Fission Product Yield Data for the Transmutation of Minor Actinide Nuclear Waste ShowComputer codes;) EMPIRE-3.2 - system of codes for nuclear reaction calculations (Version 3.2) TALYS - nuclear reaction model code; TALYS-related software and databases ENDF Utility Codes - ENDF-6 Format and Codes: Utility, PREPRO, ENDVER ENSDF programs - ENSDF Analysis and Utility programs (ALPHAD, LOGFT, etc.) ENDVER - ENDF File Verification Support Package GANDR - Global Assessment of Nuclear Data Requirements PREPRO - ENDF Preprocessing Codes URRPACK - URR-PACK: Calculating Self-Shielding in the Unresolved Resonance Energy Range GRUCON - Evaluated data preprocessing package Libraries for Geant4 - Processed ENDF-6 libraries to be used by geant4 Monte Carlo transport code SIGACE - package for generating high temperature ACE files DROSG-2000 - Package with programs and data for 60 neutron source reactions LARELKIN - Two-body relativistic kinematics code (by M. Drosg, Faculty of Physics, University of Vienna, 2008) Missing-levels - Missing level corrections using neutron spacings, by G.E.Mitchell & J.F.Shriner PLOTTAB - non-interactive general purpose plotting code to plot continuous curves and discrete physical data SPECTER - code for radiation damage parameter calculations SPECOMP - code for determining displacement cross sections for compound materials STAYSL PNNL - software Suite for Reactor Dosimetry ZVVIEW - interactive plotting of nuclear data (Windows, Linux, Mac) DICEBOX - gamma decay simulation tool | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | --- | | | Computational Nuclear Physics | | A | A+M ADS-Lib ALADDIN AMBDAS Alpha-n from JAEA Atomic Mass Data Centre | | B | Beta-delayed neutrons BrIcc v2.3S | | C | CD-Catalog CINDA CINDA Backup COG Cf-252 neutron spectrum Charged particle re­fe­ren­ce cross section Citation Guidelines Co-ordinated Research Projects (CRPs) CoNDERC | | D | DICEBOX DROSG-2000 DXS Data Explorer Decay Data Library for Actinides Download data, codes, packages | | E | EASY-II EE-View EGS EMPIRE-3.2 ENDF Archive ENDF Retrieval ENDF-6 Codes ENDF-6 Format ENDF-DB-Explorer ENDF/B-VI Summary Documentation ENDVER ENSDF ENSDF ASCII Files ENSDF and NSR Manuals ENSDF programs EPICS Electron & pho­ton interaction data EXFOR Engineering Statistics | | F | FENDL FENDL-2.1 FLUKA Fission Yields Fission Yields Report | | G | GANDR GRUCON Gamma-ray Spectrometry Center Geant4 Geant4 Libraries | | H | HyperPhysics | | I | IAEA Technical Documents IAEA-NA IAEA-NDS Reports IBANDL ICTP Workshops IDB INDC Reports INDC(NDS) Reports INDL/TSL IRDF-2002 IRDF-90 IRDFF-II International network of Nuclear Structure and Decay Data evaluators Isotopes/LBNL | | K | KADoNiS Kin-Calc | | L | LARELKIN LBNL/LUNDS LOGFT LiveChart of Nuclides LiveChart of Nuclides Advanced Low Energy Nuclear Knowledge Base | | M | MANREAD MARS MCNP MENDL-2 MERCURY MIRD Medical Portal Medical isotope browser (Isotopia) Minsk Actinides Library Missing-levels | | N | NAA NDS Staff Publications NGATLAS NIST Standard Reference databases NMF-90 NSR Newsletters NuDat-3 Nuclear Charge Radii Nuclear Data Guide Nuclear Data Section Nuclear Electro­magnetic Moments Nuclear Reaction Data Centre network Nuclear data newsletters | | P | PADF-2 PGAA PHITS PHSP PIGE PIGE-CRP PLOTTAB POINT POINT2004 POINT2007 POINT2009 PREPRO Penelope Photon Strength Function database Photon and Electron Interaction Data Photonuclear Photonuclear Properties of the Elementary Particles | | Q | Qcalc | | R | RADAR RIPL RNAL Ref db for NAA | | S | SAID SIGACE SPECOMP SPECTER STAYSL PNNL Safeguards Data Serpent SigmaCalc Spallation models Specialized Evaluated Libraries Standards Stopping Power Data Subscribe to our mailing list | | T | TAGS TALYS TALYS TART2016 TLAPrfl TRIPOLI-4 TUNL Tcalc Tendl2019 Th-U Thermal Scattering Law Library Generator Thermal neutron capture gamma rays Thin Layer Activation | | U | URRPACK Updated Actinide decay data | | W | WIMSD-IAEA Library Wallet cards Wolfram MathWorld | | X | X and Gamma-rays standards | | Z | ZVVIEW | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- | ShowEvents;) | | | --- | | | | ShowRecent and Important Events;) | | | | | --- --- | | Compound-Nuclear Reactions and Related Topics (CNR24) July 08-12, 2024 International Atomic Energy Agency, Vienna, Austria | Neutron Resonance Analysis School May 13-17, 2024 EC-JRC Geel, Belgium | LXXIIth International Conference Nucleus-2022: Fundamental problems and applications July 11-16, 2022 Moscow, Russian Federation | LXXIIth International Conference Nucleus-2022: Fundamental problems and applications July 11-16, 2022 Moscow, Russian Federation | | 15thInternational Conference on Nuclear Data for Science and Technology (ND2022) July 24-29, 2022 Sacramento, California, USA | 14th International Conference on Radiation Shielding (ICRS-14) 21st Topical Meeting of the Radiation Protection & Shielding Division of the American Nuclear Society (RPSD-2022) September 25-29, 2022 Seattle, WA, USA | ND2019 International Conference on Nuclear Data for Science and Technology May 19-24, 2019 Beijing, China | 3rd International Conference on Dosimetry and Applications (ICDA-3) May 27-31, 2019 Instituto Superior Técnico, Lisbon, Portugal | | 5th International Workshop On Nuc­lear Data Evaluation for Re­ac­tor Applications (WONDER-2018) October 8-12, 2018 Grand Hôtel Roi René, Aix-en-Provence, France | 6th International Workshop on Compound-Nuclear Reactions and Related Topics (CNR18) September 24-28, 2018 Berkeley, California, USA, | 15th International Symposium on Nuclei in the Cosmos June 24-29, 2018 Laboratori Nazionali del Gran Sasso, Assergi, Italy | 15th International Conference on Nuclear Reaction Mechanisms June 11-15, 2018 Villa Monastero, Varenna, Italy | | 26th International Seminar on Interactions of Neutrons with Nuclei (ISINN-26) May 28-June 1, 2018 Xi'an Jiaotong Universtity, Xi'an, China | New Frontiers in Nuclear Physics and Astrophysics (NNPA 2018) May 28-June 1, 2018 Akdeniz Universtity, Antalya, Turkey | ND2016 International Conference on Nuclear Data for Science and Technology September 11-16, 2016 Bruges, Belgium | ND2013 International Conference on Nuclear Data for Science and Technology March 4-8, 2013 New York, NY USA | | ND2010 International Conference on Nuclear Data for Science and Technology April 26-30, 2010 Jeju Island, South Korea | ND2007 International Conference on Nuclear Data for Science and Technology April 22-27 2007 Nice, France | ND2004 International Conference on Nuclear Data for Science and Technology Sept 26 - Oct 1, 2004 Santa Fe, NM, USA | ND2001 International Conference on Nuclear Data for Science and Technology Oct. 7-12, 2001 Tsukuba, Ibaraki, Japan | | | | | IAEA International Atomic Energy Agency, Vienna, Austria IAEA NA IAEA Department of Nuclear Applications, Vienna, Austria IAEA NDS IAEA-NDS Nuclear Data Services Vienna, Austria BARC IAEA-NDS Mirror Nuclear Data Services Bhabha Atomic Research Centre, Mumbai, India CNDC IAEA-NDS Mirror Nuclear Data Services China Nuclear Data Center, Beijing, China Atom- standart IAEA-NDS Mirror Nuclear Data Services Rosatom, Moscow, Russia NNDC National Nuclear Data Center, Brookhaven, USA NEA Data Bank Nuclear Energy Agency, Nuclear Data Services, France CJD Russian Nuclear Data Center, Institute for Physics and Power Engineering, Obninsk, Russia T-2 Nuclear Information Service, Los Alamos National Laboratory, USA CDFE Centre for Experimental Photonuclear Data, Moscow State University, Russia RSICC Radiation Safety Information Computational Center, Oak Ridge National Laboratory, USA ShowNuclear Knowledge and Databases;) NIST Standard Reference databases - of National Institute of Standards and Technology (NIST) HyperPhysics - web resources on Physics and application of Georgia State University KADoNiS - Karlsruhe Astrophysical Database of Nucleosynthesis in Stars BrIcc v2.3S - gamma-rays internal atomic shell conversion coefficient calculator Gamma-ray Spectrometry Center - of the Idaho National Laboratory (INL) LBNL/LUNDS - Nuclear Data Dissemination Page Isotopes/LBNL - Isotopes Project Page (LBNL) Computational Nuclear Physics - Computational Nuclear Physics (LLNL) Low Energy Nuclear Knowledge Base - Low Energy Nuclear Knowledge Base (Dubna) TUNL - TUNL Nuclear Data Project Kin-Calc - Online two-body kinematics calculator Properties of the Elementary Particles - Properties of the Elementary Particles SAID - Partial-Wave Analyses at GWU Tcalc - Calculation of reaction threshold energies (JCPRG) TALYS - software for the simulation of nuclear reactions MCNP - family of a general-purpose Monte Carlo Particle codes EASY - European Activation System for simulation of activation and transmutation processes MERCURY - general-purpose Monte-Carlo radiation transport code EGS - general purpose package for the Monte Carlo simulation of the coupled transport of electrons and photons Penelope - general-purpose Monte Carlo code system for the simulation of coupled electron-photon transport Serpent - continuous-energy Monte Carlo Reactor Physics Burnup calculation code FLUKA - general purpose multi-particle Monte Carlo transport code Geant4 - simulation of the passage of particles through matter PHITS - Particle and Heavy Ion Transport code System MARS - simulation of hadronic, electromagnetic, muon, heavy ions and neutron cascades and transport up to 100 TeV TART2016 - coupled Neutron-Photon, 3-D combinatorial geometry and time dependent Monte Carlo Transport Code TRIPOLI-4 - Monte Carlo method particle transport computer code COG - High fidelity multi-particle transport code Wolfram MathWorld - Wolfram MathWorld Engineering Statistics - Engineering Statistics | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | | | | | --- | 2024-01-26 | Codes | TALYS: nuclear reaction model code; TALYS-related software and databases | | 2024-01-25 | Database | IDB - An International Database of Reference Gamma Spectra | | 2024-01-25 | Web | Stopping Power - Electronic Stopping Power of Matter for Ions | | 2023-10-19 | Web-API | Application programming interface for EXFOR/ENDF/IBANDL: search/download data as text/CSV/XML/JSON files | | 2023-08-28 | EXFOR | Full EXFOR library translated to computational format C5 | | 2023-06-30 | EXFOR | Comprehensive presentation of full EXFOR library with supplementary data in X5-json format | | 2023-06-21 | Codes | Empire-3.2.3/2023.06 nuclear reaction model code system for data evaluation: for Linux/MacOS, portable version for Windows | | 2023-02-08 | Web | “EE-View” - Experimental-Evaluated data Viewer: for cross sections, angular distributions | | 2022-09-26 | ENDF | TENDL-2021 TALYS-based Evaluated Nuclear Data Library | | 2022-01-18 | ENDF | JENDL-5 Japanese evaluated nuclear data library 2021, Japan | | 2021-12-21 | Codes | GRUCON-2021: ENDF data processing package (source code, documentation, tests, installers for Windows, Linux, MacOS) | | 2021-11-12 | Codes | Empire-3.2.3/2021.11-src for Linux and MacOSX | | 2021-06-08 | Codes | Portable Empire-3.2.3/2021-win for Windows - nuclear reaction model code system for data evaluation | | 2021-05-12 | ENDF | JENDL/DDF-2015 JENDL Decay Data File 2015, Japan | | 2021-05-10 | Database | β-delayed neutrons: reference database for beta-delayed neutron emission | | 2021-04-07 | Codes | TALYS: nuclear reaction model code; TALYS-related software and databases | | 2021-03-11 | ENSDF | AME2020 /Atomic Mass Evaluation/ and NUBASE2020 /Nubase Evaluation/ release | | 2020-12-22 | Codes | GRUCON-2020: ENDF data processing package (source code, documentation, tests, installers for Windows, Linux, MacOS) | | 2020-11-25 | V&V | CoNDERC: Compilation of Nuclear Data Experiments for Radiation Characterisation | | 2020-06-22 | ENDF | CENDL-3.2 Chinese evaluated neutron data library, issued in 2020 | | 2020-06-02 | Point | Pointwise2020: processing results for a subset of TENDL-2019 for incident neutrons (630 materials) | | 2020-03-26 | ENDF | TENDL-2019 TALYS-based Evaluated Nuclear Data Library | | 2019-12-12 | Codes | GRUCON-2019: ENDF data processing package (source code, documentation, tests for Windows, Linux, MacOS) | | 2019-11-08 | Codes | MIB-2019: Medical Isotope Browser | | 2018-12-31 | Codes | GRUCON-2018: ENDF data processing package (new release) | | 2018-12-20 | ENDF | IAEA/PD-1999, IAEA Photonuclear Data Library, issued in 1999, added to ENDF database retrieval system | | 2018-12-04 | ENDF | MINKS-ACT, Minsk Actinides Library (Maslov et al.), 2011 | | 2018-04-30 | ENDF | TENDL-2017 TALYS-based Evaluated Nuclear Data Library | | 2018-04-30 | ENDF | JENDL/AD-2017, JENDL Activation Cross Section File for Nuclear Decommissioning | | 2018-02-12 | ENDF | ENDF/B-VIII.0, U.S. Evaluated Nuclear Data Library, issued in 2018 | | 2018-02-07 | ENDF | JEFF-3.3, Evaluated nuclear data library of the OECD Nuclear Energy Agency, 2017 | | 2018-01-18 | ENDF | JENDL/PD-2016, JENDL Photonuclear Data File 2016 | | 2017-12-20 | Codes | GRUCON-2017: ENDF data processing package (new release includes source code) | | 2017-06-15 | ENDF | BROND-3.1 Russian evaluated neutron data library, issued in 2016 | | 2017-01-26 | ENSDF | ACT-DDL Decay Data Library for Actinides | | 2016-12-07 | Codes | GRUCON: ENDF data processing package (new release) | | 2016-09-14 | Mirror | New NDS Mirror-site in Russia | | 2016-08-14 | Codes | URR-PACK: Calculating Self-Shielding in the Unresolved Resonance Energy Range | | 2016-04-26 | Memorial | Charles Dunford (1940-2016) | | 2016-01-28 | Codes | GRUCON: ENDF data processing package (IPPE-NRCKI, 1980-2016, Russia) | | 2015-10-19 | ENSDF | Nuclear Electromagnetic Moments database | | 2015-08-11 | DXS | Damage and gas-production database extended by data for alloys SS-316 and Eurofer (ENDF6, ACE formats) | | 2015-07-27 | Staff | Mr. Arjan Koning joined NDS as Section Head (replacing R.Forrest) | | 2015-01-31 | Codes | PREPRO-2015: ENDF/B Pre-Processing Codes | | 2015-01-01 | ENDF | IRDFF - International Reactor Dosimetry and Fusion File v1.05 | | 2014-06-02 | News | 50 year anniversary of NDS | | 2014-04-01 | CD-ROM | CD/DVD-ROMs available for on-line downloading | | 2014-03-12 | ENDF | IRDFF - International Reactor Dosimetry and Fusion File v1.03 | | 2014-02-20 | ENDF | JEFF-3.2 - Joint Evaluated Fission and Fusion File, coord. by NEA Data Bank, 2014 | | 2014-01-21 | Codes | Portable Empire-3.2.2 for Windows - nuclear reaction model code system for data evaluation | | 2013-08-27 | Mirror | New NDS Mirror-site in China | | 2013-07-11 | Android | An app for mobile devices to browse Structure and Decay Data deployed on Google Play | | 2013-06-12 | EXFOR | Milestone: 20,000 experimental works are now in EXFOR database | | 2013-03-21 | IBANDL | New Web interface (by V. Zerkin) | | 2013-02-01 | Codes | PREPRO-2012: ENDF/B Pre-Processing Codes | | 2013-01-18 | CINDA | Computer Index of Nuclear Reaction Data upgrade: automatic import from EXFOR and NSR | | 2012-09-19 | ENDF | JENDL-4.0u Update 2012 of Japanese evaluated nuclear data library 2010 | | 2012-06-25 | Staff | Ms. Paraskevi Dimitriou joined NDS as Nuclear Physicist - Codes Development (replacing Mark A. Kellett) | | 2012-06-14 | ENDF | IRDFF - International Reactor Dosimetry and Fusion File v1.02 | | 2012-06-12 | CINDA | Computer Index of Nuclear Reaction Data upgrade: automatic import from EXFOR and NSR | | 2012-05-30 | ENDF | IRDFF - International Reactor Dosimetry file for Fission and Fusion | | 2012-05-12 | ENDF | ENDF/B-5 via ENDF Web interface. U.S. Evaluated Nuclear Data Library, issued in 1978 (Release 2, 1994) | | 2012-05-11 | News | The Nuclear Data Section web site has been unavailable for the previous 13 days. Following an extensive remediation effort it has been shown that the web site is no longer a potential security threat. I believe that all issues have now been addressed and I now anticipate that we can provide an uninterrupted data service in the future. I sincerely apologise to all users for the unavailability of the NDS Web site both in February and in the recent weeks. Robin Forrest (Section Head) | | 2012-04-23 | ENDF | POINT2012 Pointwise data of ENDF/B-VII.1, processed into temperature dependent form | | 2012-03-30 | ENDF | TENDL-2011 - TALYS Evaluated Nuclear Data Library | | 2012-02-29 | Codes | EMPIRE-3.1 (Rivoli) - System of codes for nuclear reaction calculations, February-2012 | | 2012-02-20 | ENDF | JEFF-3.1.2 - Joint Evaluated Fission and Fusion File, coord. by NEA Data Bank, 2012 | | 2012-02-10 | News | The Nuclear Data Section apologises to all its users for the unavailability of its Web site over the last seven days. This was due to circumstances beyond our control and involved potential security threats. These have now been resolved and I anticipate that we can provide an uninterrupted data service in the future. Robin Forrest (Section Head) | | 2012-01-16 | ENDF | ENDF/B-VII.1 - U.S. Evaluated Nuclear Data Library, issued in December 2011 | | 2011-12-08 | ENDF | POINT2011 Pointwise data of ENDF/B-VII.1-beta4, processed into temperature dependent form | | 2011-12-05 | CINDA | Computer Index of Nuclear Reaction Data upgrade (import from EXFOR and NSR) | | 2011-11-11 | DXS | Damage and gas-production database upgrade | | 2011-11-03 | EXFOR | EXFOR database update (E066, 1373, 3152, A073, C108, D079) | | 2011-10-12 | Data | Stopping Power for Light Ions updated | | 2011-10-10 | Web | LiveChart Data updated (Nuclear Magnetic Dipole and Electric Quadrupole Moments) | | 2011-10-03 | Web | New design & functions for Publications (search, browse, etc.) | | 2011-09 | Web | Output to XML; interpretation EXFOR-XML to HTML using XSL | | 2011-09 | EAF-2010 | Library update: European Activation File (816 materials/60MeV) | | 2011-06 | Web | Interactive Web-constructing a covariance matrix from EXFOR uncertainties | | 2011-06 | Web | Output to C5 computational format (C5 = C4 + statistical and systematic uncertainties) | | 2011-05 | Staff | Mr. Marco Verpelli re-joined NDS (previously employed in NDS 2003-2010) | | 2011-01 | EAF-2010 | European Activation File (816 materials/60MeV), UK | | 2010-10 | Staff | Ms. Valentina Semkova joined NDS as EXFOR/CINDA compiler (replacing S.Dunaeva) | | 2010-07 | Staff | Mr. Stanislav Simakov joined NDS as Head of Nuclear Data Services Unit (replacing A.Mengoni) | | 2010 | RIPL-3 | Reference parameters for nuclear model calculations, 2010 | | 2010 | JENDL-4.0 | Japanese evaluated nuclear data library, 2010 | | 2010 | ROSFOND | ROSFOND-2010 Russian Library of Evaluated Neutron Data, 2010 | | 2009-12 | CENDL-3.1 | Chinese Evaluated Neutron Data Library | | 2009-07 | Staff | Mr. Robin Forrest joined NDS as Section Head (replacing A.Nichols) | | 2009-06 | ENDF | POINT2009 Released | | 2008-09 | | Benchmark of Spallation Models | | 2008-08 | Web | New NDS front page | | 2008-06 | ENDF | ENDF-Archive | | 2007-01 | ENDF/B | ENDF/B-VII.0 US Evaluated Nuclear Data Library, released 2006/12 by NNDC | | 2006-08 | Web | Extended EXFOR/ENDF plotting | | 2005-06 | EXFOR | Global EXFOR Master File | | 2005 | Staff | Nobel Peace Prize Awarded to the IAEA and Director General | | | Database at the IAEA, Vienna Database at the US NNDC IAEA Nuclear Data Section | | | | | | | | | | | | --- --- --- --- --- | IAEA-NDS Mission | Atomic and Molecular Data | Meetings Workshops | | Coordinated Research Projects | Nuclear Reaction Data Center Network | Nuclear Structure & Decay Data Network | International Network of Nuclear Data Evaluators | Technical Documents INDC Reports Publications | Computer Codes | IAEA-NA Department of Nuclear Sciences and Applications | | HideMirrors;) Nuclear Data Services International Atomic Energy Agency Vienna, Austria BARC, India Nuclear Data Services Bhabha Atomic Research Centre Mumbai, India CNDC, China Nuclear Data Services China Institute of Atomic Energy Beijing Russia Nuclear Data Services "Atomstandart" //Rosatom// Moscow HidePartners;) NNDC National Nuclear Data Center, Brookhaven, USA NEA Data Bank Nuclear Energy Agency, Nuclear Data Services, France CJD Institute for Physics and Power Engineering, Obninsk, Russia ShowEvents;) | Last Updated: 09-September-2025 Web design: V.Zerkin, IAEA, 2008
15620
https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-variables-expressions/cc-7th-interpreting-lin-exp/v/more-fancy-expressions-from-word-problems
Writing expressions word problems (video) | Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Florida B.E.S.T. Math: Pre-K - 8th grade Math: Get ready courses Math: High school & college Math: Multiple grades Science Math: Illustrative Math-aligned Math: Eureka Math-aligned Test prep Computing Reading & language arts Economics Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever. Select gift frequency One time Recurring Monthly Yearly Select amount $10 $20 $30 $40 Other Give now By donating, you agree to our terms of service and privacy policy. Skip to lesson content 7th grade math Course: 7th grade math>Unit 6 Lesson 3: Interpreting linear expressions Interpreting linear expressions: diamonds Interpreting linear expressions: flowers Interpreting linear expressions Writing expressions word problems Writing expressions word problems Math> 7th grade math> Expressions, equations, & inequalities> Interpreting linear expressions © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Writing expressions word problems FL.BEST.Math: MA.5.AR.2.4, MA.6.AR.1.1 Google Classroom Microsoft Teams About About this video Transcript Learn how to write expressions with variables to describe situations described in word problems. Skip to end of discussions Questions Tips & Thanks Want to join the conversation? Log in Sort by: Top Voted Glitchx123 a year ago Posted a year ago. Direct link to Glitchx123's post “Helena lost her marbles. ...” more Helena lost her marbles. But then she found them and put them in 4 bags with m marbles in each bag. She had 3 marbles left over that didn't fit in the bags. How many marbles does Helena have in all? Write your answer as an expression. Doesn't losing your marbles mean to go crazy or insane figuratively. That's why I found this very funny. Helena lost her marbles. But then she found them 😀 Answer Button navigates to signup page •10 comments Comment on Glitchx123's post “Helena lost her marbles. ...” (11 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Cat with Paperbag on top 2 months ago Posted 2 months ago. Direct link to Cat with Paperbag on top's post “stop this actually send m...” more stop this actually send me rolling to mars 😭😭😭😭😭😭😭 11 comments Comment on Cat with Paperbag on top's post “stop this actually send m...” (6 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Aftab 2 years ago Posted 2 years ago. Direct link to Aftab's post “bruh ts is confusing” more bruh ts is confusing Answer Button navigates to signup page •8 comments Comment on Aftab's post “bruh ts is confusing” (10 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer destiny rowland 10 months ago Posted 10 months ago. Direct link to destiny rowland's post “all of it confusing” more all of it confusing Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Justin a year ago Posted a year ago. Direct link to Justin's post “isn't it supposed to be 3...” more isn't it supposed to be 34/c + 2? Answer Button navigates to signup page •Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer EDISON HADLEY (2) 6 months ago Posted 6 months ago. Direct link to EDISON HADLEY (2)'s post “Since the team as a whole...” more Since the team as a whole scored 36, that's the number we divide by 1 comment Comment on EDISON HADLEY (2)'s post “Since the team as a whole...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Brandon 2 years ago Posted 2 years ago. Direct link to Brandon's post “How he do that” more How he do that Answer Button navigates to signup page •3 comments Comment on Brandon's post “How he do that” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Stella C 8 months ago Posted 8 months ago. Direct link to Stella C's post “this might be THE most in...” more this might be THE most intelligent khan academy comment of all time Comment Button navigates to signup page (5 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more fixenor 2 years ago Posted 2 years ago. Direct link to fixenor's post “Problem with "Hannah" not...” more Problem with "Hannah" not clear. Hana is a student or who? if she is a student then it's 127-300/z at the beginning i was thinking that hannah is library owner or something similar and then this expersion will be 127+300 (total books in her collectio) Answer Button navigates to signup page •1 comment Comment on fixenor's post “Problem with "Hannah" not...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Luna Lovegood 2 years ago Posted 2 years ago. Direct link to Luna Lovegood's post “Yes, she is a student. If...” more Yes, she is a student. If you have enough space in your house, having that many books is not even that impressive. 127 would probably fill one medium sized bookshelf. Also, it wouldn't be 127+300. Her school was going to donate a total of 300. So 300 divided by the number of students would be how many books each student is going to give. Hannah will give that amount, so that is taking away from her collection, subtracting 300/z from her collection. Comment Button navigates to signup page (3 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Show more... Jack McLennan 4 months ago Posted 4 months ago. Direct link to Jack McLennan's post “$50 for a visit to the de...” more $50 for a visit to the dentist and an additional $100 extra per cavity is ridiculous, the total should be no more then $50 or lower Answer Button navigates to signup page •1 comment Comment on Jack McLennan's post “$50 for a visit to the de...” (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer hcdupont 7 months ago Posted 7 months ago. Direct link to hcdupont's post “Beep beep beep beep beep” more Beep beep beep beep beep Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Leia Moua Lane 3 years ago Posted 3 years ago. Direct link to Leia Moua Lane's post “Look at my profile's pict...” more Look at my profile's picture :) Answer Button navigates to signup page •Comment Button navigates to signup page (2 votes) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer ShiraAnnonymousXDD 5 months ago Posted 5 months ago. Direct link to ShiraAnnonymousXDD's post “I can’t access the video ...” more I can’t access the video cause it’s blocked…. Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer szmakumba a year ago Posted a year ago. Direct link to szmakumba's post “who didi the z go to the ...” more who didi the z go to the work Answer Button navigates to signup page •Comment Button navigates to signup page (1 vote) Upvote Button navigates to signup page Downvote Button navigates to signup page Flag Button navigates to signup page more Answer Show preview Show formatting options Post answer Video transcript [Voiceover] The price of a visit to the dentist is $50. If the dentist fills any cavities, an additional charge of $100 per cavity gets added to the bill. If the dentist finds n cavities, what will the cost of the visit be? Write your answer as an expression. Alright, so we're talking about the cost of the visit. So you're gonna spend $50 no matter what. And then you're gonna get an additional charge of $100 per cavity it tells us. An additional charge of $100 per cavity. So lets see, if you have n cavities it's gonna be $100 times n. I can just write this as 100n. Now lets see if this makes sense. If you have no cavities, if n is zero, then you're just gonna pay your $50. But if you have one cavity, you're gonna pay your $50 plus 100 times one. If you have two cavities, you're gonna pay $50 plus 100 times 2, yeah. This seems to make sense. So lets check our answer. We got it right, lets do another one of these. Sunny earns $12 per hour delivering cakes. She worked of x hours this week. She worked for X hours this week. Unfortunately, she was charged $15 for a late delivery on Tuesday. How much money did Sunny earn this week? So if you see that she earns $12 per hour and she worked for x hours. Okay, she got $12 per hour times x hours. So that's how much she would've made except for the fact that she also had, she was charged, I guess her employer, charges her for late delivery. So she had to take $15 out of her paychecks. So this is what she would've gotten paid based on her hourly wage and the number of hours she worked. But then she has that late fee that she has to pay. So it's gonna be 12 x minus 15 is how much she actually earned this week. Lets do a few more of these. There are c players on the Cougars hockey team. The team scored a total of 36 goals this season. One of the players, Matthew, scored two more goals than the average per player. How many goals did Matthew score? Alright, well lets think about the average per player. Cause we know that he scored two more than the average player. So the average is going to be the total number of goals divided by the number of players. And they tell us that they're c players. So this expression right over here, that will tell us the average, that that's the average goals per player. Now we know Matthew scored two more than this thing over here. So we can just add two to that. And that's how many goals Matthew scores. Lets check our answer. Now the key here, this might seem a little confusing, but remember, this says Matthew scored two more goals. Two more goals than the average player. The average player, the average goals per player is 36 divided by c. Lets do one more of these. Hannah has 127 books in her collection. Her school is hosting a book donation. There are z students at her school and they each plan to donate the same amount of books and reach a total donation of 300 books. How many books will Hannah have in her collection after her donation? Alright, so lets think about it. Hannah has 127 books in her collection. That's how much she has in her collection. Her school is hosting a book donation. There's z students at her school and they each plan to donate the same amount of books and reach a total of 300 books. How many books will Hannah have in her collection after her donation? So 127, and I know I just reread it cause that first time I was like, okay wait, is 127 the amount of books that Hannah has or the amount that her school has would be very clear. 127 is the number of book Hannah has. Then she's gonna donate some of those books. So she's gonna donate some of those books to her school. And we have to figure out how much is she going to donate to her school. And we see that the students, the z students plan on together, they each want to donate the same number of books and together they're gonna donate 300 books. So how much is each student going to donate? Well, if z students together are donating 300, each student is going to donate 300 divided by z. 300 divided by z. And so Hannah starts with 127 books. She's going to donate her share of books to the school. And each student is going to donate 300 divided by z books. So Hannah is going to be left with 127 minus 300 over z books after her donation. Creative Commons Attribution/Non-Commercial/Share-AlikeVideo on YouTube Up next: exercise Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Allow All Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies [x] Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies [x] Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies [x] Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Reject All Confirm My Choices
15621
https://www.mytutor.co.uk/answers/4006/GCSE/Maths/How-can-you-find-the-integral-of-x-1/
How can you find the integral of x^-1 | MyTutor Find a tutorHow it worksPricesResourcesFor schoolsBecome a tutor +44 (0) 203 773 6024 Log inSign up Answers>Maths>GCSE>Article How can you find the integral of x^-1 The traditional rule for integration of a variable raised to a power is to raise the power by 1 and divide the answer by that new exponent. However, this rule leads to an answer of x^0 = 1, which can't be right. It's not. The integral of x^-1 = 1/x is the natural logarithm of x, or lnx. Likewise, the derivative of lnx is x^-1, and this is an especially important rule to remember for things like differential equations in the future. AL Answered by Angus L. •Maths tutor 60642 Views See similar Maths GCSE tutors Need help with Maths? One-to-one online tuition can be a great way to brush up on your Maths knowledge. Have a Free Meeting with one of our hand picked tutors from the UK's top universities Find a tutor Download MyTutor's free revision handbook? This handbook will help you plan your study time, beat procrastination, memorise the info and get your notes in order. 8 study hacks, 3 revision templates, 6 revision techniques, 10 exam and self-care tips. Download handbook Free weekly group tutorials Join MyTutor Squads for free (and fun) help with Maths, Coding & Study Skills. Sign up Related Maths GCSE answers All answers ▸ #### Find the inverse function of f(x) = 5 / (x - 4) Answered by Kristen G. #### Simplify the following expression: √48+√(16 3/9) Give your answer in the form: (a√3)/b where a and b are integers. Answered by Rob W. #### Expand and simplify (3 + √ 2)(5 – √ 2) Answered by Emily J. #### What is 25% of 400? Answered by Michelle M. We're here to help Contact us+44 (0) 203 773 6020 Company Information CareersBlogSubject answersBecome a tutorSchoolsSafeguarding policyFAQsUsing the Online Lesson SpaceTestimonials & pressSitemapTerms & ConditionsPrivacy Policy Popular Requests Maths tutorChemistry tutorPhysics tutorBiology tutorEnglish tutorGCSE tutorsA level tutorsIB tutorsPhysics & Maths tutorsChemistry & Maths tutorsGCSE Maths tutors ##### CLICK CEOP Internet Safety Payment Security Cyber Essentials MyTutor is part of the IXL family of brands: ##### IXL Comprehensive K-12 personalized learning ##### Rosetta Stone Immersive learning for 25 languages ##### Wyzant Trusted tutors for 300 subjects ##### Education.com 35,000 worksheets, games, and lesson plans ##### Vocabulary.com Adaptive learning for English vocabulary ##### Emmersion Fast and accurate language certification ##### Thesaurus.com Essential reference for synonyms and antonyms ##### Dictionary.com Comprehensive resource for word definitions and usage ##### SpanishDictionary.com Spanish-English dictionary, translator, and learning resources ##### FrenchDictionary.com French-English dictionary, translator, and learning ##### Ingles.com Diccionario ingles-espanol, traductor y sitio de apremdizaje ##### ABCya Fun educational games for kids © 2025 by IXL Learning
15622
https://cardiologyres.org/index.php/Cardiologyres/article/view/1531/1499
Clinical and Genetic Characteristics of Arrhythmogenic Right Ventricular Cardiomyopathy Patients: A Single-Center Experience | Al-Ghamdi | Cardiology Research This website uses cookies to ensure you get the best experience on our website. Learn more Got it! Journals | Policy | Permission This journal has moved to cr.elmerpub.com Please go to the new site to make submissions Cardiology Research User Username Password - [x] Remember me Journal Content Search Browse By Issue By Author By Title MOST READ (For the last 6-month published articles) INDEXED IN Elsevier Embase Web of Science Scopus Article Tools AbstractHTMLPDF Print this article;) How to cite item;) Finding References;) Review policy Email this article (Login required) ARTICLE STATISTICS Submission to First Decision 26 Days Acceptance Rate 31% Acceptance to Publication 32 Days Average article statistics from the last 12 months data Home About Log In Register Search Archives Editors Fees Home>Vol. 14, No. 5, Oct 2023>Al-Ghamdi Cardiol Res Cardiology Research, ISSN 1923-2829 print, 1923-2837 online, Open Access Article copyright, the authors; Journal compilation copyright, Cardiol Res and Elmer Press Inc Journal website Original Article Volume 14, Number 5, October 2023, pages 379-386 Clinical and Genetic Characteristics of Arrhythmogenic Right Ventricular Cardiomyopathy Patients: A Single-Center Experience Bandar Saeed Al-Ghamdi a, b, g, Faten Alhadeq c, Aisha Alqahtani c, Nadiah Alruwaili a, Monther Rababh d, Sara Alghamdi e, Waleed Almanea a, f, Zuhair Alhassnan b, c a Heart Centre Department, King Faisal Specialist Hospital & Research Center (KFSH&RC), Riyadh, Saudi Arabia b College of Medicine, Alfaisal University, Riyadh, Saudi Arabia c Cardiovascular Genetics Program, Center for Genomic Medicine, King Faisal Specialist Hospital & Research Centre (KFSH&RC), Riyadh, Saudi Arabia d Ministry of Health, Amman, Jordan e King Saud University, College of Pharmacy f Pediatric Cardiology, Security Forces Hospital, Riyadh, Saudi Arabia g Corresponding Author: Bandar Saeed Al-Ghamdi, Heart Centre Department, King Faisal Specialist Hospital & Research Center (KFSH&RC), Riyadh, Saudi Arabia Manuscript submitted June 18, 2023, accepted August 22, 2023, published online October 21, 2023 Short title: ARVC in Saudi Arabia doi: Abstract Introduction Materials and Methods Results Discussion References Abstract▴Top Background:Arrhythmogenic right ventricular cardiomyopathy (ARVC) is an inherited progressive cardiomyopathy. We aimed to define the long-term clinical outcome and genetic characteristics of patients and family members with positive genetic tests for ARVC in a single tertiary care cardiac center in Saudi Arabia. Methods:We enrolled 46 subjects in the study, including 23 index-patients (probands) with ARVC based on the revised 2010 ARVC Task Force Criteria (TFC) and 23 family members who underwent a genetic test for the ARVC between 2016 and 2020. Results:Of the probands, 17 (73.9%) were males with a mean age at presentation of 24.95 ± 13.9 years (7 to 55 years). Predominant symptoms were palpitations in 14 patients (60.9%), and syncope in 10 patients (43.47%). Sustained ventricular tachycardia (VT) was documented in 12 patients (52.2%). The mean left ventricular ejection fraction (LVEF) by echocardiogram was 52.81±6.311% (30-55%), and the mean right ventricular ejection fraction (RVEF) by cardiac MRI was 41.3±11.37% (23-64%). Implantable cardioverter-defibrillator (ICD) implantation was performed in 17 patients (73.9%), and over a mean follow-up of 13.65 ± 6.83 years, appropriate ICD therapy was noted in 12 patients (52.2%). Genetic variants were identified in 33 subjects (71.7%), 16 patients and 17 family members, with the most common variant of plakophilin 2 (PKP2) in 27 subjects (81.8%). Conclusions:ARVC occurs during early adulthood in Saudi patients. It is associated with a significant arrhythmia burden in these patients. The PKP2 gene is the most common gene defect in Saudi patients, consistent with what is observed in other nations. We reported in this study two novel variants in PKP2 and desmocollin 2 (DSC2) genes. Genetic counseling is needed to include all first-degree family members for early diagnosis and management of the disease in our country. Keywords:Arrhythmogenic right ventricular cardiomyopathy; Genetics; PKP2 variant; Heart failure Introduction▴Top Arrhythmogenic right ventricular cardiomyopathy (ARVC), also known as arrhythmogenic right ventricular dysplasia (ARVD), is an inherited cardiomyopathy . ARVC primarily affects the right ventricle (RV) with akinetic or dyskinetic areas involving the free wall of the ventricle, fibrofatty replacement of the myocardium, and accompanying ventricular arrhythmias (VAs), which commonly originate in the RV. However, ARVC has been described as an isolated left ventricular (LV) disease . The typical age of presentation is between the second and the fourth decade of life. ARVC is characterized by VAs ranging from premature ventricular complexes (PVCs) to ventricular tachycardia (VT), typically of RV origin, and may result in RV failure and progress to congestive heart failure at a later stage. ARVC is a recognized cause of sudden cardiac death (SCD) in young individuals . ARVC variants in at least 13 genes are seen in 30-60% of patients . Most of these genes are involved in the function of desmosomes, which are specialized adhesive junction that interacts with the cytoskeleton and participates in crosstalk with gap and adherens junctions. Desmosomes consist of a symmetrical protein complex with each end residing in the cytoplasm of one of a pair of adjacent cells, anchoring intermediate filaments in the cytoskeleton to the cell . A large majority of variants in ARVC patients have been found in genes encoding different components of the cardiac desmosome, i.e., plakophilin 2 (PKP2), desmocollin 2 (DSC2), desmoglein 2 (DSG2), desmoplakin (DSP), and plakoglobin (JUP), suggesting that ARVC/D is primarily a disease of disturbed desmosomal function. However, variants in other genes (non-desmosomal genes) have also been reported in ARVC, including transmembrane protein 43 (TMEM43), desmin (DES), and titin (TTN), indicating genetic heterogeneity [6, 7]. Several ARVC cases were found to be caused by multiple variants in the same gene (compound heterozygosity) or variants in different genes (digenic inheritance), which could result in an earlier onset and increased disease severity . ARVC is inherited predominantly as an autosomal dominant in its classical form and as autosomal recessive in non-classical form with cardiocutaneous disease such as Naxos disease, which is associated with palmoplantar keratoderma and woolly hair and Carvajal syndrome . We previously described the clinical characteristic of ARVC in Saudi Arabia . However, genetic testing was only performed in 60% of patients in the previous study. In the current study, we include only patients with genetic testing and their family members with available genetic testing results. We will concentrate on the genetics of these patients. Furthermore, the duration of follow-up is longer in the current study. Only 10 patients were included in this study from the previous cohort. This study aimed to define the long-term clinical outcome and genetic characteristics in a cohort of Saudi patients and family members for ARVC in a single tertiary care cardiac center in Saudi Arabia. Materials and Methods▴Top Study population The study population comprised 23 index-patients (probands) fulfilling the 2010 ARVC Task Force Criteria (TFC) for definite diagnosis and 23 family members with available data of genetic testing for ARVC enrolled in the ARVC registry at King Faisal Specialist Hospital & Research Centre, Riyadh. The study has been approved by the Research Ethics Committee (REC) at King Faisal Specialist Hospital & Research Center (KFSH&RC), Riyadh. The study was conducted in compliance with the ethical standards of the responsible institution on human subjects, as well as with the Helsinki Declaration. Clinical analysis A retrospective analysis of clinical and genetic data of ARVC patients and family members, including each individual’s demographic data and medical history was conducted. These data were obtained by reviewing medical records, electronic clinical evaluations, and device clinic charts. Genetic counselors with a special interest in ARVC obtained a detailed family history for pedigree analysis through patient interviews. Molecular genetic analysis All index-patients underwent genetic testing of PKP2, DSP, DSG2, DSC2, and JUP with direct sequencing of the entire coding regions or the next generation multi-gene panel. First-degree family members were screened for the variant found in their respective index patients, if any. Statistical analysis Continuous variables were summarized as mean ± standard deviation (SD), and categorical variables were reported as frequency (%). SPSS statistical software (version 20; SPSS Inc., Chicago, IL) was used for the analyses. Results▴Top Presenting clinical characteristics, clinical course, and long-term outcome Table 1 summarizes the major presenting clinical features of the 23 probands. Our patient cohort consists of 23 probands, 17 were males (73.9 %), with a mean age at the diagnosis of 24.95 ± 13.9 years (ranging from 7 to 55 years). The clinical presentation was palpitations in 14 patients (60.9%), dizziness in four patients (17.4 %), shortness of breath in six patients (26.1%), and syncope in 10 patients (43.5%). Sustained VT was seen in 12 patients (52.2%), and three were survivors of sudden cardiac arrest (13%). Family history of ARVC was present in four patients (17.4%), and a family history of SCD was present in six patients (26.1%). A history of consanguinity was present in six patients (26.1%). Click to viewTable 1.Clinical Characteristics of the Probands The electrocardiogram (ECG) depolarization changes meeting major diagnostic criteria were found in nine patients (39%), and ECG repolarization changes meeting major diagnostic criteria were found in 17 patients (73.9%). Echocardiogram changes meeting major diagnostic criteria were found in nine patients (39%), and cardiac magnetic resonance imaging (MRI) was performed in 11 patients (47.8%) and showed changes compatible with major diagnostic criteria in six patients (54.5%). The mean left ventricular ejection fraction (LVEF) by echocardiogram was 52.8±6.3% (30-55%), and the mean right ventricular ejection fraction (RVEF) by cardiac MRI was 41.3±11.37% (23-64%). LV involvement with LVEF of < 40% was present in three patients (13%). Medical management, antiarrhythmic medications, electrophysiologic studies and ablation procedures are summarized in Table 1. Seventeen patients (73.9%) underwent implantable cardioverter-defibrillator (ICD) implantation, of whom 13 patients (56.5%) had ICD implanted for secondary prevention of SCD (three patients were survivors of SCD, and 10 patients had sustained VT). Of note, two survivors of SCD also had VT during hospitalization. One patient (4.3%) died from non-cardiac causes during the follow-up. Over a mean follow-up of 13.65 ± 6.83 years, 12 of the 17 patients (70.6%) had received appropriate therapy for sustained VAs. The 23 family members consisted of 11 males (47.8%) with a mean age of 34.4 ± 13 years (23 - 50 years) with a history of palpitations in five patients (31.25%). There was no definite case of ARVC in these subjects. Genetic testing result Genetic testing was performed on all patients and available first-degree family members. Out of 113 eligible first-degree family members for ARVC screening, only 23 subjects (20%) underwent workup, including genetic testing, despite having genetic counselors and providing screening of family members for free. Variants were identified in 71.7% of all study subjects (33 subjects, including 16 patients and 17 family members). The most common variants detected were in PKP2 in 81.8% of all positive results (27 subjects, including 13 patients and 14 family members), followed by DSP in 9% (three subjects, including two patients and one family member) and DSC2 in 9% (three subjects, including one patient and two family members) (Table 2) [12-26]. Click to viewTable 2.Probands and Family Members With Positive Genetic Testing The discovered genetic variants in our study were deemed pathogenic in seven probands, likely pathogenic in four, and of uncertain significance in three and novel (it has not been reported in the literature before) in two (Table 2). None of our family members with positive genetic testing have had a definitive ARVC diagnosis. Discussion▴Top In this study, we observed ARVC occurrence in a young age with male predominance. ARVC affects males more than females. About two-thirds of our patients were males. Male gender predominance in ARVC with male to female ratio of 3:1 has been reported previously [27, 28]. ARVC generally affects young patients with a mean age of 30 years, whereas it rarely manifests before the age of 12 or after the age of 60 years [6, 28-31]. However, a higher mean age in the late 30s to 40s years has been noted in another study . Most of our patients presented with palpitations, and VT was a common presenting event. Previous studies have shown that individuals with ARVC present with palpitations (67%), syncope (32%), atypical chest pain (27%), or RV failure (6%). However, some patients are asymptomatic (6%) . ARCV is a leading cause of SCD, accounting for 11-22% of cases of SCD in the young athlete patient population [33, 34]. Thirteen percent of our patients were survivors of cardiac arrest. Regarding diagnostic workup, all our patients were diagnosed based on non-invasive tests, including ECG, echocardiogram, cardiac MRI, and family history. ECG is an essential diagnostic test for the ARVC diagnosis. The presence of epsilon waves (described as post-excitation potentials of small amplitude that occur at the end of the QRS complex) in V1-3 is a major ARVC diagnostic criterion noted in about 30% of ARVC cases . ECG repolarization abnormities with T wave inversion in V1-3 or beyond in individuals > 14 years of age in the absence of complete right bundle-branch block is another major ARVC diagnostic criterion seen in 55% to 94% in different ARVC series [35-37]. ECG repolarization was also frequent in our patients. The echocardiogram is a beneficial, non-invasive diagnostic test to assess structural changes in ARVC [38-40]. Echocardiogram changes meeting major diagnostic criteria present in 39% of our patients. Cardiac MRI has the advantage of assessing the RV (and LV) function, size, global or regional wall motion abnormalities, and quantifying of myocardial wall thinning and hypertrophy [41, 42]. Cardiac MRI was performed in about half of our patients, showing changes compatible with major diagnostic criteria in about half of them. Biventricular involvement with LV fibrofatty replacement and involvement have been found in as much as in 70% of the cases of ARVC [1, 43]. It is usually age-dependent and associated with more severe cardiomegaly, arrhythmogenic events, inflammatory infiltrates, and heart failure [1, 43]. LV involvement with LVEF < 40% was present in 13.6% of our patients. A coronary angiogram or cardiac computed tomography (CT) was used to rule out coronary artery disease in patients with impaired LVEF. Patients with typical dilated cardiomyopathy or other cardiomyopathies changes by cardiac MRI were excluded. Patients with ARVC have a high burden of arrhythmia. ICD therapy is the most effective preventive measure for SCD . Of our patients, 73.9% underwent ICD implantation. Over half of our patients received appropriate ICD therapy for sustained VAs during follow-up. About one-third of our patients underwent electrophysiology studies and PVCs/VT ablations. Furthermore, 77.3% of the patients were on antiarrhythmic medications. The Heart Rhythm Society (HRS) 2019 expert consensus statement on arrhythmogenic cardiomyopathy (ACM) provides a class IIb recommendation for amiodarone and sotalol in individuals with ACM to control arrhythmic symptoms or reduce ICD shocks. In the absence of other antiarrhythmic drugs, flecainide with beta-blockers receives a class IIb recommendation in individuals with ARVC and ICD and preserved LV and RV function to control refractory VAs to other therapies . Genetic evaluation ARVC has been documented in families since the early 1980s. In 1985, three out of five siblings in one family were diagnosed with ARVC, and it was postulated that ARVC has an autosomal dominant inheritance with an incomplete penetrance pattern . We have a low rate of first-degree family members screening. Genetic discrimination, i.e., adverse treatment based solely on the genotype or family history of individuals without disease symptoms or the stigma of having a chronic illness in the family, appears to be responsible for the lack of interest in having ARVC screening in the ARVC family members. Another interesting observation is the absence of clinical ARVC disease in positive family members so far. The clinical picture and natural history of familial ARVC have been reported in 37 ARVC families. Of the 365 subjects enrolled in this study, 151 (41%) were affected, 157 (43%) were unaffected, 17 (5%) were healthy carriers, and 40 (11%) were uncertain . The first ARVC diagnostic criterion published in 1994 included familial disease confirmed at necropsy or surgery as a major criterion . A familial history of premature sudden death (< 35 years of age) due to suspected ARVC or a familial history based on clinically diagnosed disease is included as minor criteria . Furthermore, the 2010 ARVC Task Force included the identification of a pathogenic variant categorized as associated or probably associated with ARVC as a major diagnostic criterion . In 1998, linkage analysis in nine families with Naxos disease found a single mutant gene mapped to 17q21 (homozygous genotype) . Two years later, a homozygous deletion variant in JUP was identified in 19 patients with Naxos disease . Subsequently, other desmosome genes were identified as a cause of ARVC. DSP was confirmed as a causative gene in 2002 , PKP2 in 2004 , and DSG2 and DSC2 variants were reported in 2006 [13, 50]. ARVC may occur due to non-desmosome genes, including transforming growth factor-β3 (TGFB3) , the cardiac ryanodine receptor RYR2 , TTN , TMEM43 , and DES . Additional ARVC genes are identified using candidate gene sequence approaches instead of linkage analysis . The PKP2 gene is responsible for about 70% of all ARVC variants in previous studies [7, 12]. We have similar PKP2 gene predominance in Saudi patients. Strengths and limitations of this study To our knowledge, this is the first study describing long-term follow-up with a concentration on the genetics of ARVC in Saudi Arabia. The main limitation of this study is the small number of subjects, but ARVC is a rare disease. The other limitation is the retrospective nature of the study. It is also a single-center study; however, we are a tertiary care hospital with referrals from the kingdom. Conclusions ARVC occurs during early adulthood in Saudi patients. It is associated with a significant arrhythmia burden. ARVC is familial and genetic testing is essential in all cases. The PKP2 gene is the most common gene defect in Saudi patients, consistent with what is observed in other nations. We reported in this study two novel variants in PKP2 and DSC2 genes. Genetic counseling is needed to include all first-degree family members for early diagnosis and management of the disease in our country. Acknowledgments None to declare. Financial Disclosure None to declare. Conflict of Interest The authors declared no conflict of interest for all authors. Informed Consent Consent was waived by REC as the study is retrospective. Author Contributions Bandar Saeed Al-Ghamdi: conception or design of the work, drafting of the manuscript, and final approval of the version to be published; Faten Alhadeq, Aisha Alqahtani, and Monther Rababh: data collection; Nadiah Alruwaili: data collection, data analysis, and interpretation; Sara Alghamdi: data analysis and interpretation; Waleed Almanea: final approval of the version to be published; Zuhair Alhassnan: conception or design of the work, critical revision of the article, and final approval of the version to be published. Data Availability The authors declare that data supporting the findings of this study are available within the article. Abbreviations ARVC: arrhythmogenic right ventricular cardiomyopathy; ARVD: arrhythmogenic right ventricular dysplasia; DES: desmin; DSC2: desmocollin 2; DSG2: desmoglein 2; DSP: desmoplakin; EF: ejection fraction; ICD: implantable cardioverter-defibrillator; JUP: plakoglobin; LV: left ventricle/ventricular; MRI: magnetic resonance imaging; PKP2: plakophilin 2; PVCs: premature ventricular complexes; RV: right ventricle; SCD: sudden cardiac death; TTN: titin; TMEM43: transmembrane protein 43; VAs: ventricular arrhythmias; VT: ventricular tachycardia References▴Top Corrado D, Basso C, Thiene G, McKenna WJ, Davies MJ, Fontaliran F, Nava A, et al. Spectrum of clinicopathologic manifestations of arrhythmogenic right ventricular cardiomyopathy/dysplasia: a multicenter study. J Am Coll Cardiol. 1997;30(6):1512-1520. doipubmed 2. Groeneweg JA, van der Zwaag PA, Jongbloed JD, Cox MG, Vreeker A, de Boer RA, van der Heijden JF, et al. Left-dominant arrhythmogenic cardiomyopathy in a large family: associated desmosomal or nondesmosomal genotype? Heart Rhythm. 2013;10(4):548-559. doipubmed 3. McRae AT, 3rd, Chung MK, Asher CR. Arrhythmogenic right ventricular cardiomyopathy: a cause of sudden death in young people. Cleve Clin J Med. 2001;68(5):459-467. doipubmed 4. Ackerman MJ, Priori SG, Willems S, Berul C, Brugada R, Calkins H, Camm AJ, et al. HRS/EHRA expert consensus statement on the state of genetic testing for the channelopathies and cardiomyopathies: this document was developed as a partnership between the Heart Rhythm Society (HRS) and the European Heart Rhythm Association (EHRA). Europace. 2011;13(8):1077-1109. doipubmed 5. Delmar M, McKenna WJ. The cardiac desmosome and arrhythmogenic cardiomyopathies: from gene to disease. Circ Res. 2010;107(6):700-714. doipubmed 6. Lazzarini E, Jongbloed JD, Pilichou K, Thiene G, Basso C, Bikker H, Charbon B, et al. The ARVD/C genetic variants database: 2014 update. Hum Mutat. 2015;36(4):403-410. doipubmed 7. Groeneweg JA, Bhonsale A, James CA, te Riele AS, Dooijes D, Tichnell C, Murray B, et al. Clinical presentation, long-term follow-up, and outcomes of 1001 arrhythmogenic right ventricular dysplasia/cardiomyopathy patients and family members. Circ Cardiovasc Genet. 2015;8(3):437-446. doipubmed 8. McKoy G, Protonotarios N, Crosby A, Tsatsopoulou A, Anastasakis A, Coonar A, Norman M, et al. Identification of a deletion in plakoglobin in arrhythmogenic right ventricular cardiomyopathy with palmoplantar keratoderma and woolly hair (Naxos disease). Lancet. 2000;355(9221):2119-2124. doipubmed 9. Norgett EE, Hatsell SJ, Carvajal-Huerta L, Cabezas JC, Common J, Purkis PE, Whittock N, et al. Recessive mutation in desmoplakin disrupts desmoplakin-intermediate filament interactions and causes dilated cardiomyopathy, woolly hair and keratoderma. Hum Mol Genet. 2000;9(18):2761-2766. doipubmed 10. Al-Ghamdi B, Shafquat A, Mallawi Y. Arrhythmogenic right ventricular cardiomyopathy/dysplasia in Saudi Arabia: a single-center experience with long-term follow-up. Ann Saudi Med. 2014;34(5):415-426. doipubmedpmc 11. Marcus FI, McKenna WJ, Sherrill D, Basso C, Bauce B, Bluemke DA, Calkins H, et al. Diagnosis of arrhythmogenic right ventricular cardiomyopathy/dysplasia: proposed modification of the Task Force Criteria. Eur Heart J. 2010;31(7):806-814. doipubmedpmc 12. Gerull B, Heuser A, Wichter T, Paul M, Basson CT, McDermott DA, Lerman BB, et al. Mutations in the desmosomal protein plakophilin-2 are common in arrhythmogenic right ventricular cardiomyopathy. Nat Genet. 2004;36(11):1162-1164. doipubmed 13. Syrris P, Ward D, Evans A, Asimaki A, Gandjbakhch E, Sen-Chowdhry S, McKenna WJ. Arrhythmogenic right ventricular dysplasia/cardiomyopathy associated with mutations in the desmosomal gene desmocollin-2. Am J Hum Genet. 2006;79(5):978-984. doipubmedpmc 14. Stenson PD, Mort M, Ball EV, Shaw K, Phillips A, Cooper DN. The Human Gene Mutation Database: building a comprehensive mutation repository for clinical and molecular genetics, diagnostic testing and personalized genomic medicine. Hum Genet. 2014;133(1):1-9. doipubmedpmc 15. Fressart V, Duthoit G, Donal E, Probst V, Deharo JC, Chevalier P, Klug D, et al. Desmosomal gene analysis in arrhythmogenic right ventricular dysplasia/cardiomyopathy: spectrum of mutations and clinical impact in practice. Europace. 2010;12(6):861-868. doipubmed 16. Walsh R, Thomson KL, Ware JS, Funke BH, Woodley J, McGuire KJ, Mazzarotto F, et al. Reassessment of Mendelian gene pathogenicity using 7,855 cardiomyopathy cases and 60,706 reference samples. Genet Med. 2017;19(2):192-203. doipubmedpmc 17. Rigato I, Bauce B, Rampazzo A, Zorzi A, Pilichou K, Mazzotti E, Migliore F, et al. Compound and digenic heterozygosity predicts lifetime arrhythmic outcome and sudden cardiac death in desmosomal gene-related arrhythmogenic right ventricular cardiomyopathy. Circ Cardiovasc Genet. 2013;6(6):533-542. doipubmed 18. van der Zwaag PA, Jongbloed JD, van den Berg MP, van der Smagt JJ, Jongbloed R, Bikker H, Hofstra RM, et al. A genetic variants database for arrhythmogenic right ventricular dysplasia/cardiomyopathy. Hum Mutat. 2009;30(9):1278-1283. doipubmed 19. van Tintelen JP, Entius MM, Bhuiyan ZA, Jongbloed R, Wiesfeld AC, Wilde AA, van der Smagt J, et al. Plakophilin-2 mutations are the major determinant of familial arrhythmogenic right ventricular dysplasia/cardiomyopathy. Circulation. 2006;113(13):1650-1658. doipubmed 20. Dalal D, Molin LH, Piccini J, Tichnell C, James C, Bomma C, Prakasa K, et al. Clinical features of arrhythmogenic right ventricular dysplasia/cardiomyopathy associated with mutations in plakophilin-2. Circulation. 2006;113(13):1641-1649. doipubmed 21. Lek M, Karczewski KJ, Minikel EV, Samocha KE, Banks E, Fennell T, O'Donnell-Luria AH, et al. Analysis of protein-coding genetic variation in 60,706 humans. Nature. 2016;536(7616):285-291. doipubmedpmc 22. Davies G, Armstrong N, Bis JC, Bressler J, Chouraki V, Giddaluru S, Hofer E, et al. Genetic contributions to variation in general cognitive function: a meta-analysis of genome-wide association studies in the CHARGE consortium (N=53949). Mol Psychiatry. 2015;20(2):183-192. doipubmedpmc 23. Pigors M, Schwieger-Briel A, Cosgarea R, Diaconeasa A, Bruckner-Tuderman L, Fleck T, Has C. Desmoplakin mutations with palmoplantar keratoderma, woolly hair and cardiomyopathy. Acta Derm Venereol. 2015;95(3):337-340. doipubmed 24. National Center for Biotechnology Information. ClinVar [VCV000088762.10]. (accessed Sept. 17, 2022). 25. Basso C, Czarnowska E, Della Barbera M, Bauce B, Beffagna G, Wlodarska EK, Pilichou K, et al. Ultrastructural evidence of intercalated disc remodelling in arrhythmogenic right ventricular cardiomyopathy: an electron microscopy investigation on endomyocardial biopsies. Eur Heart J. 2006;27(15):1847-1854. doipubmed 26. Bauce B, Rampazzo A, Basso C, Mazzotti E, Rigato I, Steriotis A, Beffagna G, et al. Clinical phenotype and diagnosis of arrhythmogenic right ventricular cardiomyopathy in pediatric patients carrying desmosomal gene mutations. Heart Rhythm. 2011;8(11):1686-1695. doipubmedpmc 27. Azaouagh A, Churzidse S, Konorza T, Erbel R. Arrhythmogenic right ventricular cardiomyopathy/dysplasia: a review and update. Clin Res Cardiol. 2011;100(5):383-394. doipubmed 28. Bauce B, Frigo G, Marcus FI, Basso C, Rampazzo A, Maddalena F, Corrado D, et al. Comparison of clinical features of arrhythmogenic right ventricular cardiomyopathy in men versus women. Am J Cardiol. 2008;102(9):1252-1257. doipubmed 29. Lin CY, Chung FP, Lin YJ, Chang SL, Lo LW, Hu YF, Tuan TC, et al. Gender differences in patients with arrhythmogenic right ventricular dysplasia/cardiomyopathy: Clinical manifestations, electrophysiological properties, substrate characteristics, and prognosis of radiofrequency catheter ablation. Int J Cardiol. 2017;227:930-937. doipubmed 30. Dalal D, Nasir K, Bomma C, Prakasa K, Tandri H, Piccini J, Roguin A, et al. Arrhythmogenic right ventricular dysplasia: a United States experience. Circulation. 2005;112(25):3823-3832. doipubmed 31. Corrado D, Fontaine G, Marcus FI, McKenna WJ, Nava A, Thiene G, Wichter T. Arrhythmogenic right ventricular dysplasia/cardiomyopathy: need for an international registry. Study Group on Arrhythmogenic Right Ventricular Dysplasia/Cardiomyopathy of the Working Groups on Myocardial and Pericardial Disease and Arrhythmias of the European Society of Cardiology and of the Scientific Council on Cardiomyopathies of the World Heart Federation. Circulation. 2000;101(11):E101-106. doipubmed 32. Ohno S, Nagaoka I, Fukuyama M, Kimura H, Itoh H, Makiyama T, Shimizu A, et al. Age-dependent clinical and genetic characteristics in Japanese patients with arrhythmogenic right ventricular cardiomyopathy/dysplasia. Circ J. 2013;77(6):1534-1542. doipubmed 33. Thiene G, Nava A, Corrado D, Rossi L, Pennelli N. Right ventricular cardiomyopathy and sudden death in young people. N Engl J Med. 1988;318(3):129-133. doipubmed 34. Mazic S, Lazovic B, Delic M. [Arrhythmogenic right ventricular cardiomyopathy as a cause of sudden death in young people—literature review]. Med Pregl. 2012;65(9-10):396-404. doipubmed 35. Nasir K, Bomma C, Tandri H, Roguin A, Dalal D, Prakasa K, Tichnell C, et al. Electrocardiographic features of arrhythmogenic right ventricular dysplasia/cardiomyopathy according to disease severity: a need to broaden diagnostic criteria. Circulation. 2004;110(12):1527-1534. doipubmed 36. Fontaine G, Umemura J, Di Donna P, Tsezana R, Cannat JJ, Frank R. [Duration of QRS complexes in arrhythmogenic right ventricular dysplasia. A new non-invasive diagnostic marker]. Ann Cardiol Angeiol (Paris). 1993;42(8):399-405. pubmed 37. Peters S, Trummel M. Diagnosis of arrhythmogenic right ventricular dysplasia-cardiomyopathy: value of standard ECG revisited. Ann Noninvasive Electrocardiol. 2003;8(3):238-245. doipubmedpmc 38. Lindstrom L, Wilkenshoff UM, Larsson H, Wranne B. Echocardiographic assessment of arrhythmogenic right ventricular cardiomyopathy. Heart. 2001;86(1):31-38. doipubmedpmc 39. Yoerger DM, Marcus F, Sherrill D, Calkins H, Towbin JA, Zareba W, Picard MH, et al. Echocardiographic findings in patients meeting task force criteria for arrhythmogenic right ventricular dysplasia: new insights from the multidisciplinary study of right ventricular dysplasia. J Am Coll Cardiol. 2005;45(6):860-865. doipubmed 40. Boczar KE, Alqarawi W, Green MS, Redpath C, Burwash IG, Dwivedi G. The echocardiographic assessment of the right ventricle in patients with arrhythmogenic right ventricular cardiomyopathy/dysplasia compared with athletes and matched controls. Echocardiography. 2019;36(4):666-670. doipubmed 41. Tandri H, Calkins H, Nasir K, Bomma C, Castillo E, Rutberg J, Tichnell C, et al. Magnetic resonance imaging findings in patients meeting task force criteria for arrhythmogenic right ventricular dysplasia. J Cardiovasc Electrophysiol. 2003;14(5):476-482. doipubmed 42. Jain A, Tandri H, Calkins H, Bluemke DA. Role of cardiovascular magnetic resonance imaging in arrhythmogenic right ventricular dysplasia. J Cardiovasc Magn Reson. 2008;10(1):32. doipubmedpmc 43. Pinamonti B, Sinagra G, Salvi A, Di Lenarda A, Morgera T, Silvestri F, Bussani R, et al. Left ventricular involvement in right ventricular dysplasia. Am Heart J. 1992;123(3):711-724. doipubmed 44. Towbin JA, McKenna WJ, Abrams DJ, Ackerman MJ, Calkins H, Darrieux FCC, Daubert JP, et al. 2019 HRS expert consensus statement on evaluation, risk stratification, and management of arrhythmogenic cardiomyopathy. Heart Rhythm. 2019;16(11):e301-e372. doipubmed 45. Ruder MA, Winston SA, Davis JC, Abbott JA, Eldar M, Scheinman MM. Arrhythmogenic right ventricular dysplasia in a family. Am J Cardiol. 1985;56(12):799-800. doipubmed 46. Nava A, Bauce B, Basso C, Muriago M, Rampazzo A, Villanova C, Daliento L, et al. Clinical profile and long-term follow-up of 37 families with arrhythmogenic right ventricular cardiomyopathy. J Am Coll Cardiol. 2000;36(7):2226-2233. doipubmed 47. McKenna WJ, Thiene G, Nava A, Fontaliran F, Blomstrom-Lundqvist C, Fontaine G, Camerini F. Diagnosis of arrhythmogenic right ventricular dysplasia/cardiomyopathy. Task force of the working group myocardial and pericardial disease of the european society of cardiology and of the scientific council on cardiomyopathies of the international society and federation of cardiology. Br Heart J. 1994;71(3):215-218. doipubmedpmc 48. Coonar AS, Protonotarios N, Tsatsopoulou A, Needham EW, Houlston RS, Cliff S, Otter MI, et al. Gene for arrhythmogenic right ventricular cardiomyopathy with diffuse nonepidermolytic palmoplantar keratoderma and woolly hair (Naxos disease) maps to 17q21. Circulation. 1998;97(20):2049-2058. doipubmed 49. Rampazzo A, Nava A, Malacrida S, Beffagna G, Bauce B, Rossi V, Zimbello R, et al. Mutation in human desmoplakin domain binding to plakoglobin causes a dominant form of arrhythmogenic right ventricular cardiomyopathy. Am J Hum Genet. 2002;71(5):1200-1206. doipubmedpmc 50. Pilichou K, Nava A, Basso C, Beffagna G, Bauce B, Lorenzon A, Frigo G, et al. Mutations in desmoglein-2 gene are associated with arrhythmogenic right ventricular cardiomyopathy. Circulation. 2006;113(9):1171-1179. doipubmed 51. Beffagna G, Occhi G, Nava A, Vitiello L, Ditadi A, Basso C, Bauce B, et al. Regulatory mutations in transforming growth factor-beta3 gene cause arrhythmogenic right ventricular cardiomyopathy type 1. Cardiovasc Res. 2005;65(2):366-373. doipubmed 52. Tiso N, Stephan DA, Nava A, Bagattin A, Devaney JM, Stanchi F, Larderet G, et al. Identification of mutations in the cardiac ryanodine receptor gene in families affected with arrhythmogenic right ventricular cardiomyopathy type 2 (ARVD2). Hum Mol Genet. 2001;10(3):189-194. doipubmed 53. Taylor M, Graw S, Sinagra G, Barnes C, Slavov D, Brun F, Pinamonti B, et al. Genetic variation in titin in arrhythmogenic right ventricular cardiomyopathy-overlap syndromes. Circulation. 2011;124(8):876-885. doipubmedpmc 54. Merner ND, Hodgkinson KA, Haywood AF, Connors S, French VM, Drenckhahn JD, Kupprion C, et al. Arrhythmogenic right ventricular cardiomyopathy type 5 is a fully penetrant, lethal arrhythmic disorder caused by a missense mutation in the TMEM43 gene. Am J Hum Genet. 2008;82(4):809-821. doipubmedpmc 55. Klauke B, Kossmann S, Gaertner A, Brand K, Stork I, Brodehl A, Dieding M, et al. De novo desmin-mutation N116S is associated with arrhythmogenic right ventricular cardiomyopathy. Hum Mol Genet. 2010;19(23):4595-4607. doipubmed 56. Ohno S. The genetic background of arrhythmogenic right ventricular cardiomyopathy. J Arrhythm. 2016;32(5):398-403. doipubmedpmc This article is distributed under the terms of the Creative Commons Attribution Non-Commercial 4.0 International License, which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. Cardiology Research is published by Elmer Press Inc. Browse Journals Journal of Clinical Medicine ResearchJournal of Endocrinology and MetabolismJournal of Clinical Gynecology and Obstetrics World Journal of OncologyGastroenterology ResearchJournal of Hematology Journal of Medical CasesJournal of Current SurgeryClinical Infection and Immunity Cardiology ResearchWorld Journal of Nephrology and UrologyCellular and Molecular Medicine Research Journal of Neurology ResearchInternational Journal of Clinical PediatricsAI in Clinical Medicine Journal of Current Surgery, quarterly, ISSN 1927-1298 (print), 1927-1301 (online), published by Elmer Press Inc. The content of this site is intended for health care professionals. This is an open-access journal distributed under the terms of the Creative Commons Attribution-NonCommercial 4.0 International License, which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited. Creative Commons Attribution license (Attribution-NonCommercial 4.0 International CC-BY-NC 4.0) This journal follows the International Committee of Medical Journal Editors (ICMJE) recommendations for manuscripts submitted to biomedical journals, the Committee on Publication Ethics (COPE) guidelines, and the Principles of Transparency and Best Practice in Scholarly Publishing. website: www.currentsurgery.org editorial contact: editor@currentsurgery.org elmer.editorial2@hotmail.com Address: 9225 Leslie Street, Suite 201, Richmond Hill, Ontario, L4B 3H6, Canada © Elmer Press Inc. All Rights Reserved. Disclaimer: The views and opinions expressed in the published articles are those of the authors and do not necessarily reflect the views or opinions of the editors and Elmer Press Inc. This website is provided for medical research and informational purposes only and does not constitute any medical advice or professional services. The information provided in this journal should not be used for diagnosis and treatment, those seeking medical advice should always consult with a licensed physician.
15623
https://www.mathnasium.com/math-centers/pointloma/news/mastering-bodmas-rule-why-order-operations-important-pl
Math Centers / Point Loma / News / Mastering the BODMAS Rule: Why Order of Operations Is Important Education Mastering the BODMAS Rule: Why Order of Operations Is Important Nov 26, 2024 | Point Loma Have you ever wondered why math problems must be solved in a specific way? Why can't we just go from left to right and add, subtract, multiply, or divide as we see fit? The answer lies in the BODMAS rule, which defines the order in which mathematical operations should be performed. Mastering the BODMAS rule is crucial because it ensures that everyone solves a math problem the same way and gets the same correct answer. Let’s explore why the BODMAS rule is so important and how it can make calculations clearer and more accurate. What Is the BODMAS Rule? The BODMAS rule is an acronym that stands for: Brackets Orders (also known as exponents, like powers and square roots) Division Multiplication Addition Subtraction This rule dictates the order in which you should carry out operations when solving a math problem. The operations are listed in a specific sequence that you must follow to ensure you get the correct result. Why Is the Order Important? Imagine you were given this simple math problem: 8 + 4 × 2 = ? If you went from left to right, adding 8 and 4 first, you would get 12, then multiplying by 2 would give you 24. However, according to the BODMAS rule, multiplication comes before addition. So, you should first calculate 4 × 2, which is 8, and then add it to 8 to get 16. The correct answer is 16, not 24. The order of operations is important because it eliminates ambiguity in mathematical calculations. Without BODMAS, math problems could be interpreted in multiple ways, leading to different answers. This clarity is crucial, especially in complex calculations, financial transactions, science, and engineering. Breaking Down the BODMAS Rule Brackets: Start with any calculations inside brackets first. This can be parentheses ( ), square brackets [ ], or curly braces { }. Anything within brackets takes the highest priority. Example: In (3 + 5) × 2, solve 3 + 5 first to get 8, then multiply by 2 for a final answer of 16. Orders: Next, handle any exponents or powers. This includes squares (²), cubes (³), square roots (√), and other powers. Example: In 3² + 5, calculate 3² (which is 9) first, then add 5 to get 14. Division and Multiplication: These operations are on the same level of priority. They should be performed from left to right as they appear in the problem. Example: In 12 ÷ 3 × 2, start with 12 ÷ 3 to get 4, then multiply by 2 for an answer of 8. Addition and Subtraction: Lastly, perform any addition or subtraction, also working from left to right. Example: In 10 - 2 + 5, subtract 2 from 10 to get 8, then add 5 for a final answer of 13. Common Mistakes with BODMAS A frequent error students make is performing operations out of order, leading to incorrect answers. For example, in the expression 6 + 2 × (3² - 1), a common mistake would be to add 6 and 2 first. However, according to BODMAS: Solve the brackets: 3² - 1 = 9 - 1 = 8 Perform the multiplication: 2 × 8 = 16 Finally, add: 6 + 16 = 22 Following BODMAS step-by-step ensures you avoid these pitfalls. Tips for Mastering BODMAS Practice Regularly: The more problems you solve using BODMAS, the more familiar the rule will become. Use Parentheses to Clarify: If you're unsure, adding parentheses can help break down complex problems. Double-Check Your Work: After solving, review each step to ensure you followed the correct order. Mastering the BODMAS rule is essential for success in math, as it provides a universal standard for solving problems. Whether you’re working on simple equations or tackling advanced algebra, understanding the importance of order of operations will ensure that you get the right answer every time. So next time you’re faced with a math problem, remember BODMAS, and you’ll be one step closer to math mastery! Related Articles The Math of Spring: Patterns in Nature and Growth Rates Mar 18, 2025 | Point Loma The Math of Spring: Nature’s Numbers in Bloom! ✨🌸 From the spirals of a sunflower to the growth rates of budding plants, math is everywhere in spring! 🌻🌿 Discover how patterns like the Fibonacci sequence and geometric growth shape the beauty of nature. The Magic of Pi: Why 3.14159 Never Ends! Mar 11, 2025 | Point Loma Join us as we delve into the enigma of pi and explore its myriad applications in our latest blog post. Space Exploration: How Astronauts Use Numbers Beyond Earth Feb 25, 2025 | Point Loma 🚀 Math in Space: The Astronaut’s Secret Tool! 🌍✨ Ever wondered how astronauts navigate the vastness of space? 🤔 From calculating rocket trajectories to measuring fuel and oxygen levels, math is their ultimate guide! 🧮🌌 Discover how numbers help astronauts explore the universe in our latest blog! Point Loma (619) 224-6284 Download Parent Pack Mathnasium Method Programs International Locations Contact Us Our Results About Mathnasium Own a Franchise Math Tutors Near Me Careers Newsroom FAQs Blog Mathnasium Method Programs International Locations Contact Us Our Instructors Our Results About Mathnasium Own a Franchise Math Tutors Near Me Careers Newsroom FAQs Blog Accessibility Privacy Policy Do Not Sell or Share My Personal Information Terms of Use ©2025 Mathnasium Franchisor LLC. All rights reserved. Contact Us Point Loma (619) 224-6284 2850 Womble Road, #100 San Diego, CA 92106 Opt out preference signal recognized. How We Use Tracking Technologies We use cookies and other tracking technologies to enhance the user experience on our website, for analytical purposes, and for advertising and marketing. Although we do not sell personal information in exchange for money, some of the ways in which we share personal information for targeted advertising or analytics may be considered "sales" or "sharing" under US state privacy laws. If you would like to opt out of targeted advertising, or sales and sharing of personal information through tracking technologies such as cookies, please use the option provided and click "Confirm My Choices" when done. To opt out of offline sales and sharing of personal information, please use ourOffline Opt-Out Form. More information can be found in ourPrivacy Policy.. Manage Consent Preferences To opt out, please click the toggle above, then click "Confirm My Choices."
15624
https://artofproblemsolving.com/wiki/index.php/Inequality?srsltid=AfmBOorAmuS8gRNeZ0j7CeGWSQckJ7GP6vpfvDmwN0Tqefz6flYocXn-
Art of Problem Solving Inequality - 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 Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Inequality The subject of mathematical inequalities is tied closely with optimization methods. While most of the subject of inequalities is often left out of the ordinary educational track, they are common in mathematics Olympiads. Contents [hide] 1 Overview 2 Solving Inequalities 2.1 Linear Inequalities 2.2 Polynomial Inequalities 2.3 Rational Inequalities 3 Complete Inequalities 4 List of Theorems 4.1 Introductory 4.2 Advanced 5 Problems 5.1 Introductory 5.2 Intermediate 5.3 Olympiad 6 Resources 6.1 Books 6.1.1 Intermediate 6.1.2 Olympiad 6.2 Articles 6.2.1 Olympiad 6.3 Classes 6.3.1 Olympiad 7 See also Overview Inequalities are arguably a branch of elementary algebra, and relate slightly to number theory. They deal with relations of variables denoted by four signs: . For two numbers and : if is greater than , that is, is positive. if is smaller than , that is, is negative. if is greater than or equal to , that is, is nonnegative. if is less than or equal to , that is, is nonpositive. Note that if and only if , , and vice versa. The same applies to the latter two signs: if and only if , , and vice versa. Some properties of inequalities are: If , then , where . If , then , where . If , then , where . Solving Inequalities In general, when solving inequalities, same quantities can be added or subtracted without changing the inequality sign, much like equations. However, when multiplying, dividing, or square rooting, we have to watch the sign. In particular, notice that although , we must have . In particular, when multiplying or dividing by negative quantities, we have to flip the sign. Complications can arise when the value multiplied can have varying signs depending on the variable. We also have to be careful about the boundaries of the solutions. In the example , the value does not satisfy the inequality because the inequality is strict. However, in the example , the value satisfies the inequality because the inequality is nonstrict. Solutions can be written in interval notation. Closed bounds use square brackets, while open bounds (and bounds at infinity) use parentheses. For instance, ![Image 49: $x \in 3,6)$ means . Linear Inequalities Linear inequalities can be solved much like linear equations to get implicit restrictions upon a variable. However, when multiplying/dividing both sides by negative numbers, we have to flip the sign. Polynomial Inequalities The first part of solving polynomial inequalities is much like solving polynomial equations -- bringing all the terms to one side and finding the roots. Afterward, we have to consider bounds. We're comparing the sign of the polynomial with different inputs, so we could imagine a rough graph of the polynomial and how it passes through zeroes (since passing through zeroes could change the sign). Then we can find the appropriate bounds of the inequality. Rational Inequalities A more complex example is . Here is a common mistake: The problem here is that we multiplied by as one of the last steps. We also kept the inequality sign in the same direction. However, we don't know if the quantity is negative or not; we can't assume that it is positive for all real . Thus, we may have to reverse the direction of the inequality sign if we are multiplying by a negative number. But, we don't know if the quantity is negative either. A correct solution would be to move everything to the left side of the inequality, and form a common denominator. Then, it will be simple to find the solutions to the inequality by considering the sign (negativeness or positiveness) of the fraction as varies. We will start with an intuitive solution, and then a rule can be built for solving general fractional inequalities. To make things easier, we test positive integers. makes a good starting point, but does not solve the inequality. Nor does . Therefore, these two aren't solutions. Then we begin to test numbers such as , , and so on. All of these work. In fact, it's not difficult to see that the fraction will remain positive as gets larger and larger. But just where does , which causes a negative fraction at and , begin to cause a positive fraction? We can't just assume that is the switching point; this solution is not simply limited to integers. The numerator and denominator are big hints. Specifically, we examine that when (the numerator), then the fraction is , and begins to be positive for all higher values of . Solving the equation reveals that is the turning point. After more of this type of work, we realize that brings about division by , so it certainly isn't a solution. However, it also tells us that any value of that is less than brings about a fraction that has a negative numerator and denominator, resulting in a positive fraction and thus satisfying the inequality. No value between and (except itself) seems to be a solution. Therefore, we conclude that the solutions are the intervals ![Image 78: $(-\infty,-5)\cup\frac{3}{2},+\infty)$. For the sake of better notation, define the "x-intercept" of a fractional inequality to be those values of that cause the numerator and/or the denominator to be .To develop a method for quicker solutions of fractional inequalities, we can simply consider the "x-intercepts" of the numerator and denominator. We graph them on the number line. Then, in every region of the number line, we test one point to see if the whole region is part of the solution. For example, in the example problem above, we see that we only had to test one value such as in the region , as well as one value in the region ![Image 83: $(-\infty,-5]$]( and ![Image 84: $\frac{3}{2},+\infty)$; then we see which regions are part of the solution set. This does indeed give the complete solution set. One must be careful about the boundaries of the solutions. In the example problem, the value was a solution only because the inequality was nonstrict. Also, the value was not a solution because it would bring about division by . Similarly, any "x-intercept" of the numerator is a solution if and only if the inequality is nonstrict, and every "x-intercept" of the denominator is never a solution because we cannot divide by . Complete Inequalities A inequality that is true for all real numbers or for all positive numbers (or even for all complex numbers) is sometimes called a complete inequality. An example for real numbers is the so-called Trivial Inequality, which states that for any real , . Most inequalities of this type are only for positive numbers, and this type of inequality often has extremely clever problems and applications. List of Theorems Here are some of the more useful inequality theorems, as well as general inequality topics. Introductory Arithmetic Mean-Geometric Mean Inequality Cauchy-Schwarz Inequality Titu's Lemma Chebyshev's Inequality Geometric inequalities Jensen's Inequality Nesbitt's Inequality Rearrangement Inequality Power mean inequality Triangle Inequality Trivial inequality Schur's Inequality Advanced Aczel's Inequality Callebaut's Inequality Carleman's Inequality Hölder's inequality Radon's Inequality Homogenization Isoperimetric inequalities Maclaurin's Inequality Muirhead's Inequality Minkowski Inequality Newton's Inequality Ptolemy's Inequality Can someone fix that Ptolemy's is in Advanced? Problems Introductory Practice Problems on Alcumus Inequalities (Prealgebra) Solving Linear Inequalities (Algebra) Quadratic Inequalities (Algebra) Basic Rational Function Equations and Inequalities (Intermediate Algebra) A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly . During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than . What's the largest number of matches she could've won before the weekend began? (1992 AIME Problems/Problem 3) Intermediate Practice Problems on Alcumus Quadratic Inequalities (Algebra) Advanced Rational Function Equations and Inequalities (Intermediate Algebra) General Inequality Skills (Intermediate Algebra) Advanced Inequalities (Intermediate Algebra) Given that , and show that . (weblog_entry.php?t=172070 Source) Olympiad See also Category:Olympiad Inequality Problems Let be positive real numbers. Prove that (2001 IMO Problems/Problem 2) Resources Books Intermediate Introduction to Inequalities Geometric Inequalities Olympiad Advanced Olympiad Inequalities: Algebraic & Geometric Olympiad Inequalities by Alijadallah Belabess. The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele. Problem Solving Strategies by Arthur Engel contains significant material on inequalities. Inequalities by G. H. Hardy, J. E. Littlewood, G. Pólya. Articles Olympiad Inequalities by MIT Professor Kiran Kedlaya. Inequalities by IMO gold medalist Thomas Mildorf. Classes Olympiad The Worldwide Online Olympiad Training Program is designed to help students learn to tackle mathematical Olympiad problems in topics such as inequalities. See also Mathematics competitions Math books Retrieved from " Categories: Algebra Inequalities 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.
15625
https://artofproblemsolving.com/wiki/index.php/2022_AMC_8_Problems/Problem_24?srsltid=AfmBOopu8fvoUK0wsWv0kbZGzRzquYBkI8UvCwO8M6ydu9w5jaii-uyb
Art of Problem Solving 2022 AMC 8 Problems/Problem 24 - 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 2022 AMC 8 Problems/Problem 24 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 2022 AMC 8 Problems/Problem 24 Contents 1 Problem 2 Solution 3 Remark 4 Video Solution (A Clever Explanation You’ll Get Instantly) 5 Video Solution by Math-X (First understand the problem!!!) 6 Video Solution(🚀Under 2 min🚀 With color-coded folding explanation ✨) 7 Video Solution by OmegaLearn 8 Video Solution 9 Video Solution 10 Video Solution 11 Video Solution 12 Video Solution 13 Video Solution 14 Video Solution by Dr. David 15 See Also Problem The figure below shows a polygon , consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that and . What is the volume of the prism? Solution While imagining the folding, goes on goes on and goes on So, and Also, becomes an edge parallel to so that means Since then So, the area of is If we let be the base, then the height is So, the volume is ~aops-g5-gethsemanea2 Remark After folding polygon on the dotted lines, we obtain the following triangular prism: ~MRENTHUSIASM Video Solution (A Clever Explanation You’ll Get Instantly) ~hsnacademy Video Solution by Math-X (First understand the problem!!!) ~Math-X Video Solution(🚀Under 2 min🚀 With color-coded folding explanation ✨) ~Education, the Study of Everything Video Solution by OmegaLearn ~ pi_is_3.14 Video Solution ~Mathematical Dexterity Video Solution ~Interstigation Video Solution ~David Video Solution ~STEMbreezy Video Solution ~savannahsolver Video Solution Please like and subscribe! Video Solution by Dr. David See Also 2022 AMC 8 (Problems • Answer Key • Resources) Preceded by Problem 23Followed by Problem 25 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 All AJHSME/AMC 8 Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Introductory Geometry 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.
15626
https://math.stackexchange.com/questions/3340486/why-does-the-elimination-method-work
linear algebra - Why does the elimination method work? - Mathematics Stack Exchange Join Mathematics 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… 1. 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 2. 3. current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog 5. Log in 6. Sign up 1. 1. Home 2. Questions 3. Unanswered 4. AI Assist Labs 5. Tags 7. Chat 8. Users 2. Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams 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 Why does the elimination method work? Ask Question Asked 5 years, 11 months ago Modified5 years, 9 months ago Viewed 2k times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. Consider solving the following system of equations: {2 x−4 y=0,4 x+4 y=−4. {2 x−4 y 4 x+4 y=0,=−4. We can use the substitution method or the elimination method, of which the latter has always been unclear to me: why does it work? I know that by eliminating one of the two variables you can solve for the variable you're left with, and consequently solve for the other one, but why do you get the point (x,y)(x,y) at which both lines intersect? Can anyone show me this graphically? linear-algebra systems-of-equations gaussian-elimination Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Sep 25, 2019 at 2:26 Xander Henderson♦ 32.7k 25 25 gold badges 73 73 silver badges 122 122 bronze badges asked Aug 31, 2019 at 20:02 mathomatomathomato 39 4 4 bronze badges 0 Add a comment| 3 Answers 3 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. This is a special case of a more general principle: if S S and T T are the solution sets to the equations f(x)=0 f(x)=0 and g(x)=0 g(x)=0, respectively, then the solution set of every linear combination λ f(x)+μ g(x)=0 λ f(x)+μ g(x)=0 of these equations contains the common solutions S∩T S∩T of the original equations. You should verify this for yourself—the proof is fairly obvious. In the case of a pair of linear equations, the solution to the system is the intersection of the two lines represented by the equations. Since any nontrivial linear combination of two linear equations is also linear, when you eliminate a variable by forming a linear combination of two equations, what you’re effectively doing is finding another line that passes through this intersection. Two distinct lines intersect in at most one point, so doing this doesn’t introduce any extraneous solutions. This interpretation generalizes to more linear equations with more variables, but instead of lines, each individual equation represents a hyperplane. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Sep 1, 2019 at 5:57 answered Aug 31, 2019 at 22:15 amdamd 55.1k 3 3 gold badges 39 39 silver badges 97 97 bronze badges Add a comment| This answer is useful 1 Save this answer. Show activity on this post. why do you get the point (x,y)(x,y) at which both lines intersect? You are solving two equations, so two curves are involved (it can be more, of course, but for simplicity let's stick with two curves). You are looking for points which are on both curves (lines in your example). The only way a point can be on both curves is if the curves intersect at those points. Any other points on either line will _not_ be on both lines (will not satisfy both equations at the same time). So each equation represents one line/curve (or in more complex problems it can be curves or more complex structures in multidimensional spaces) and the set of values that satisfy all equations are, by definition, points on each line/curve. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Aug 31, 2019 at 20:28 StephenG - Help UkraineStephenG - Help Ukraine 753 5 5 silver badges 15 15 bronze badges Add a comment| This answer is useful 1 Save this answer. Show activity on this post. Let's use different colors for individual equations in your system: 2 x−4 y=−0 4 x+4 y=−4 2 x−4 y=−0(eq1)4 x+4 y=−4(eq2) If we express y y from the first equation, we will obtain the equation of the same line e q 1 e q 1 (see the animation below), only in an other form: y=2 4 x y=2 4 x So for every real x x you obtain y=2 4 x y=2 4 x such that the point (x,y)(x,y) is the point of the line e q 1 e q 1: Now, when we substitute this y=2 4 x y=2 4 x into the second equation 4 x+4 y=−4 4 x+4 y=−4(eq2) then because y=2 4 x y=2 4 x, the point (x,y)(x,y) will belong to the line e q 1 e q 1, but because at the same time it fulfills the equation (e q 2)(e q 2), it will be a point of the line e q 2 e q 2, too. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Nov 14, 2019 at 4:55 answered Aug 31, 2019 at 21:42 MarianDMarianD 2,963 1 1 gold badge 17 17 silver badges 27 27 bronze badges 2 1 Nice use of an animated PNG file, and not too big for low bandwidth users either, I think. –StephenG - Help Ukraine Commented Sep 3, 2019 at 1:19 OP is asking about the elimination method and you visualized the substitution method. If you were to use the same strategy to explain the elimination method, then I think you would likely earn the answer. –Christian Westbrook Commented Sep 22, 2023 at 7:54 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 linear-algebra systems-of-equations gaussian-elimination See similar questions with these tags. Featured on Meta Community help needed to clean up goo.gl links (by August 25) Report this ad Linked 2Formal proof for why subtracting two equations in a system of linear equations maintains equivalence with the original system of equations Related 69Necessity/Advantage of LU Decomposition over Gaussian Elimination 7What does solving A x=b A x=b mean? 2Explanation of Gauss-Jordan elimination method. 0Can the Elimination Method be used to solve a system of polynomial equations? 0Backward elimination and back-substitution? 1How to solve a system of 32 XOR equations via Gaussian Elimination? 0Using Gaussian elimination for a parametric solution Hot Network Questions Which passport to use at immigration counter Why is muscle cramp called a “charley horse”? Make a square tikzcd diagram How do I pass a parameter to a locally store web page from the context (right click) menu of Notepad++? ssh client automatically reject unknown host key My Canadian employer is sending me to Germany to work on a project. Do I need a visa or a work permit? Can you ask the editor to send a partial review? How can I separate the 2 wheels in this model into individual objects? What happens as you fall into a realistic black hole? Book series where a family living in the country is attacked by fae Is there a way to find someone to be a co-signer if you have no other option? Two Card Betting Game: Asymmetrical Problem Should 1 Peter 1:5 read ‘a’ or ‘the’ before ‘salvation’, or should it just read ‘salvation’? Since the universe is expanding and spacetime is a single fabric, does time also stretch along with the expansion of space? Solve the crossed ladders problem to suggest that they were married Is it mentioned that the 18-day Mahabharata war remained incomplete? postgres-15 - WAL files being archived, but not recycled/removed from pg_wal Why does bible claim Asa was perfect all his days when he didn't rely on God in the latter years Incoming water pipe has no apparent ground How do proponents of the doctrine of Eternal Security explain the evidence of lifelong Christians renouncing their faith? complex 3-clause conditional: 'I realized that if I do..., I will...' vs. 'I realized that if I did..., I would...' Degree in healthcare A Rubik's Cube game in Python Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.8.21.33163 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear - [x] checkbox label label Apply Cancel Consent Leg.Interest - [x] checkbox label label - [x] checkbox label label - [x] checkbox label label Necessary cookies only Confirm my choices
15627
https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electro-Optics/Book%3A_Electromagnetics_I_(Ellingson)/05%3A_Electrostatics/5.06%3A_Electric_Field_Due_to_an_Infinite_Line_Charge_using_Gauss_Law
Skip to main content 5.6: Electric Field Due to an Infinite Line Charge using Gauss’ Law Last updated : Sep 12, 2022 Save as PDF 5.5: Gauss’ Law - Integral Form 5.7: Gauss’ Law - Differential Form Page ID : 3928 Steven W. Ellingson Virginia Polytechnic Institute and State University via Virginia Tech Libraries' Open Education Initiative ( \newcommand{\kernel}{\mathrm{null}\,}) Section 5.5 explains one application of Gauss’ Law, which is to find the electric field due to a charged particle. In this section, we present another application – the electric field due to an infinite line of charge. The result serves as a useful “building block” in a number of other problems, including determination of the capacitance of coaxial cable (Section 5.24). Although this problem can be solved using the “direct” approach described in Section 5.4 (and it is an excellent exercise to do so), the Gauss’ Law approach demonstrated here turns out to be relatively simple. Example : Electric field associated with an infinite line charge, using Gauss’ Law. Use Gauss’ Law to determine the electric field intensity due to an infinite line of charge along the axis, having charge density (units of C/m), as shown in Figure . Solution Gauss’ Law requires integration over a surface that encloses the charge. So, our first problem is to determine a suitable surface. A cylinder of radius that is concentric with the axis, as shown in Figure , is maximally symmetric with the charge distribution and so is likely to yield the simplest possible analysis. At first glance, it seems that we may have a problem since the charge extends to infinity in the and directions, so it’s not clear how to enclose all of the charge. Let’s suppress that concern for a moment and simply choose a cylinder of finite length . In principle, we can solve the problem first for this cylinder of finite size, which contains only a fraction of the charge, and then later let to capture the rest of the charge. (In fact, we’ll find when the time comes it will not be necessary to do that, but we shall prepare for it anyway.) Here’s Gauss’ Law: where is the electric flux density , is a closed surface with outward-facing differential surface normal , and is the enclosed charge. The first order of business is to constrain the form of using a symmetry argument, as follows. Consider the field of a point charge at the origin (Section 5.5): We can “assemble” an infinite line of charge by adding particles in pairs. One pair is added at a time, with one particle on the axis and the other on the axis, with each located an equal distance from the origin. We continue to add particle pairs in this manner until the resulting charge extends continuously to infinity in both directions. The principle of superposition indicates that the resulting field will be the sum of the fields of the particles (Section 5.2). Thus, we see that cannot have any component in the direction because none of the fields of the constituent particles have a component in that direction. Similarly, we see that the magnitude of cannot depend on because none of the fields of the constituent particles depends on and because the charge distribution is identical (“invariant”) with rotation in . Also, note that for any choice of the distribution of charge above and below that plane of constant is identical; therefore, cannot be a function of and cannot have any component in the direction. Therefore, the direction of must be radially outward; i.e., in the direction, as follows: Next, we observe that on the right hand side of Equation is equal to . Thus, we obtain The cylinder consists of a flat top, curved side, and flat bottom. Expanding the above equation to reflect this, we obtain Examination of the dot products indicates that the integrals associated with the top and bottom surfaces must be zero. In other words, the flux through the top and bottom is zero because is perpendicular to these surfaces. We are left with The side surface is an open cylinder of radius , so , a constant over this surface. Thus: The remaining integral is simply the area of the side surface, which is . Solving for we obtain Remarkably, we see is independent of , So the concern raised in the beginning of this solution – that we wouldn’t be able to enclose all of the charge – doesn’t matter. Completing the solution, we note the result must be the same for any value of (not just ), so and since : This completes the solution. We have found that the electric field is directed radially away from the line charge, and decreases in magnitude in inverse proportion to distance from the line charge. Suggestion: Check to ensure that this solution is dimensionally correct. 5.5: Gauss’ Law - Integral Form 5.7: Gauss’ Law - Differential Form
15628
https://www.ldh.la.gov/assets/oph/Center-EH/envepi/Arsenic_for_Health_Providers_Final.pdf
Section of Environmental Epidemiology & Toxicology Office of Public Health, Louisiana Department of Health & Hospitals (888) 293-7020; www.seet.dhh.la.gov INFORMATION FOR HEALTH CARE PROFESSIONALS ARSENIC EXPOSURE & TOXICITY Revised: January 2009 This document is designed for health care providers and summarizes information on the sources, exposure pathways, laboratory testing, recognition, and reporting of arsenic exposure and/or poisoning. Louisiana Law requires the reporting of all cases of arsenic poisoning and all arsenic laboratory test results to the Louisiana Department of Health and Hospital’s Office of Public Health. EXPOSURE TO ARSENIC Arsenic is a naturally occurring metal released into the environment by natural and unnatural (industrial and commercial) processes. Arsenic compounds can be classified into three major forms: inorganic, organic, and arsine gas. Inorganic arsenic may be formed with either trivalent (+3, arsenite) or pentavalent (+5, arsenate) arsenic. The most common inorganic trivalent arsenic compounds are arsenic trioxide, sodium arsenite, and arsenic trichloride. Pentavalent inorganic compounds are arsenic pentoxide, arsenic acid, and arsenates such as lead arsenate and calcium arsenate. Inorganic arsenic compounds are used in metal alloy manufacturing, electronics manufacturing (semiconductors), and as wood preservatives. They are also formed as by-products during the smelting of metals and coal combustion (ATSDR, 2007). Organic arsenic compounds are formed when arsenic combines with carbon. Organic arsenic compounds may be trivalent (+3, arsenite) or pentavalent (+5, arsenate). Most organic arsenic compounds found in the environment are methylated as a result of biomethylation by organisms in soil or water. Organic arsenics, such as disodium methylarsenate (DSMA) and monosodium methylarsenate (MSMA), are used as herbicides. Organic arsenic compounds are registered for use as medicinal animal feed additives. Arsenobetaine or “fish arsenic”, an organic form of arsenic, is found in seafood and is nontoxic (ATSDR, 2007). Arsine gas is formed by the reaction of hydrogen with arsenic and can have a garlic-like or fishy odor at high concentrations. Arsine gas is used commercially in the semiconductor industry, during the synthesis of organic arsenic compounds, and in the manufacture of crystals for computer chips and fiber optics. Arsine gas can be generated accidentally during the smelting and refining of nonferrous metals and in mining processes. Occupational Exposure: Exposure to arsenic may occur in several industries: refining or smelting of metal ores, microelectronics, wood preservation, wood joinery shops, battery manufacturing, and working in power plants that burn arsenic-rich coal. Occupational exposure occurs primarily through inhalation of dust or fumes containing inorganic arsenic. Ingestion and dermal exposure to inorganic and organic arsenic compounds may occur in certain situations. Most occupational exposure to arsine occurs following inhalation of the gas that has been accidentally generated when arsenic-containing crude ores or metals are treated with acid. Acute high-dose exposures to arsine gas can be particularly harmful. It is rare for workers to be exposed to arsenic alone. The exposure usually occurs in combination with other metals (ATSDR, 2007). 33 As 74.9 Take-Home Exposure: Workers can transport arsenic on shoes and work clothes into their homes or automobiles (NIOSH, 1997). “Take-home” arsenic puts children and spouses at risk for arsenic exposure. Steps to prevent take-home arsenic contamination: • Change shoes and clothes at work before getting into the car or going home. Put dirty work clothes and shoes in a plastic bag. • Wash face and hands prior to leaving work. • Shower and wash hair immediately upon arriving home (or before leaving work). • Wash work clothes separately from all other clothes and run the washing machine again to rinse residual arsenic. Exposure Risks to the General Population: The general population may be exposed to small amounts of arsenic through ingestion of arsenic-containing drinking water and in some food, particularly seafood. High levels of naturally occurring arsenic are found in soil and rocks in some areas of the country; this arsenic may lead to unacceptable levels of arsenic in drinking water that may require action. There are arsenic regulations for public water systems and there are no risks for people drinking from a public water system. Arsenic is present in fish and other seafood, however it exists primarily in an organic form, arsenobetaine, which is non-toxic and is rapidly excreted in urine (ATSDR, 2007). Other non-occupational sources of exposure also exist. Sawing or burning arsenic-treated wood (e.g. CCA-treated wood) could result in inhalation of arsenic-containing sawdust or smoke. Some folk medicines, herbal preparations and non-regulated supplements may contain arsenic. People who live near hazardous waste sites, landfills, or industrial arsenic sources (smelting operations, coal-burning power plants) may also be at an increased risk of exposure Environmental Sources of Arsenic Contaminated drinking water Wood working/ building/ remodeling with arsenic-treated wood Burning of arsenic- treated wood or coal in the home Paints/pigments Asian homeopathic medicines, folk remedies-asiatic pill, yellow root Antiparasitic drugs-carbasone Kelp-containing health foods (mainly organic) Fish (organic) Gardening- herbicides, (organic) This is not a complete list of potential exposure sources. Unless otherwise noted, the above are sources of inorganic arsenic exposure. This information was compiled from ATSDR, 2007. through air, soil, dust, or water contamination (ATSDR, 2007). Another consideration in evaluating arsenic exposure is the possibility of intentional poisoning. Arsenic has been a classic poison for suicidal, malicious or homicidal intent. Arsenic Toxicity in Children: Children may be exposed to arsenic from environmental sources. In the past, some playground equipment was made of green wood, or pressed wood treated with chromated copper arsenate (CCA). It is possible for arsenic to leach from CCA-treated wood into the soil. Children who play on or near these wooden structures may ingest arsenic-containing residue, dust, or soil via hand-to-mouth activity. Ingestion of contaminated soil may also be a concern for children who live near hazardous waste sites, landfills, or industrial arsenic sources. Health effects in children are similar to those in adults and include irritation of the GI tract, vascular damage, skin changes, and reduced nerve function. There is evidence that children may be less efficient at converting inorganic arsenic to the less harmful organic forms. For this reason, children may be more susceptible to health effects from inorganic arsenic than adults (ATSDR, 2007). Special Considerations for Adults of Reproductive Age: There is suggestive evidence that arsenic can interfere with normal fetal development, however, these data are inconclusive. Inorganic arsenic crosses the placenta and has been found in fetal tissue. Umbilical cord blood levels of arsenic are similar to maternal blood levels and arsenic is found at low levels in breast milk. Chronic exposure of humans to inorganic arsenic in drinking water has been associated with an increased incidence of miscarriages, stillbirths, preterm births, and infants with low birth weights. Studies in animals demonstrate large doses of inorganic arsenic that cause illness in pregnant females can also cause low birth weight, fetal malformations, and even fetal death (ATSDR, 2007). TOXICITY OF ARSENIC The toxicity of arsenic varies widely based on route of exposure and form. Ingestion and inhalation are the primary routes of both acute and chronic exposures; inorganic arsenic is not readily absorbed by the skin. Arsine gas is one of the most toxic forms and is readily absorbed into the body by inhalation. Inorganic forms of arsenic are more toxic than organic forms. Of inorganic arsenic compounds, the more toxic trivalent form reacts with sulfhydryl groups causing enzyme disruption and reduced oxidative phosphorylation; the pentavalent form is less toxic and, through phosphorus substitution, “uncouples” oxidative phosphorylation. Toxicology quick facts about arsenic: - Arsine gas is extremely toxic. - Inorganic arsenics are more toxic than the organic forms. - Trivalent arsenic compounds tend to be more toxic than pentavalent arsenic compounds. - Metalloid arsenic is nontoxic because it is insoluble in water and body fluids. - Arsenobetaine is the primary form of arsenic found in fish and is nontoxic. Inorganic Arsenic: Often, there will be a garlic smell to the breath and tissue fluids of people exposed to inorganic arsenic. Effects of acute inorganic arsenic poisoning include fever, anorexia, hepatomegaly, melanosis, cardiac arrhythmia and eventual cardiovascular failure, upper respiratory track symptoms, peripheral neuropathies, gastrointestinal and hematopoietic effects. Sensory loss in the peripheral neurons may appear 1-2 weeks after large exposures. Dermal contact with high concentrations of inorganic arsenic compounds may result in skin irritation, redness, and swelling. High acute exposures may cause cholera-like gastrointestinal symptoms of vomiting (often times bloody) and severe diarrhea (which may be rice-watery in character and often bloody); these patients will experience acute distress, dehydration (often), and hypovolemic shock. Ingestion of large doses of inorganic arsenic (70 to 180 mg) may be fatal (ATSDR, 2007). Chronic toxicity is insidious as the symptoms are vague and take time to appear. Dermal or neurological symptoms may occur. Dermal effects may include hyperpigmentation with a classical "raindrops on a dusty road" appearance, a characteristic pattern of darkened skin patches with scattered pale spots. Hyperkeratosis, may follow with small “corns” or “warts”, appearing most often on the palms, soles and torso. Peripheral neuropathy may appear with painful paresthesia that is symmetrical and stocking-glove in distribution. Neurotoxicity begins with sensory changes, paresthesia and muscle tenderness, followed by weakness which progresses from proximal to distal muscle groups. Whitish lines (Mees lines) may be seen on the fingernails. Chronic hepatic and renal damage is common; jaundice may occur with liver injury. Basic mechanisms of inorganic arsenic toxicity: Trivalent arsenic inhibits numerous cellular enzymes through sulfhydryl group binding. Trivalent arsenic inhibits cellular glucose uptake, gluconeogenesis, fatty acid oxidation, and decreases production of cellular ATP and further production of acetyl CoA. Trivalent arsenic also blocks the production of glutathione, which prevents cellular oxidative damage. Effects of pentavalent inorganic arsenic are partially due to its transformation to trivalent arsenic. The most important mechanism of pentavalent arsenics results from its similarity to inorganic phosphate and the potential for it to substitute for phosphate in glycolytic and cellular respiration pathways. Arsenic cannot form the high-energy phosphate bonds that are needed and oxidative phosphorylation is uncoupled. Therefore, ATP with its high-energy phosphate bonds is not formed. Because arsenic affects these fundamental mechanisms, very few organ systems are unaffected by arsenic. The signs and symptoms of arsenic may vary with the route of exposure, the dose, the duration of exposure, and the time elapsed since the exposure. Arsine Gas: Arsine gas is the most acutely toxic form of arsenic. Symptoms are often delayed 2-24 hours after exposure and manifest with an acute hemolytic anemia and striking chills. Initial symptoms include chills, malaise, dizziness, nausea, abdominal pain and dyspnea, bloody urine, and jaundice. Hemoglobinuria causes the urine to appear black. Hemolysis leads to hypoxia and ultimately renal failure. Chronic exposure to arsine gas is rare. Of the few reported cases, symptoms include a progressive drop in the number of red blood cells, shortness of breath on exertion, and a general feeling of weakness (ATSDR, 2007). Organic Arsenic: Most of the available information regarding arsenic toxicity in humans involves exposure to inorganic arsenic compounds or arsine gas. Laboratory studies suggest organic compounds are less toxic than inorganic compounds and that some organic arsenic compounds (such as those found in fish) are virtually nontoxic. However, very high doses of certain organic compounds may be metabolized to inorganic arsenic and result in some of the same effects as exposure to inorganic compounds: nerve injury, stomach irritation, and lung irritation. Carcinogenesis: Arsenic has been classified as a known human carcinogen by multiple agencies based on the increased prevalence of lung and skin cancer observed in human populations exposed to arsenic. Ingestion of arsenic has been primarily associated with an increased risk of skin cancer. Inhaled arsenic has been associated with an increased risk of lung cancer (ATSDR, 2007). EVALUATION OF ARSENIC POISONING Assessment of a patient for arsenic poisoning requires the following • Work history and identification of possible arsenic exposure sources • Medical history • Laboratory testing for arsenic Work and Exposure History: Often, patients do not recognize that they have been exposed to arsenic unless directly asked about their work environment and activities. A detailed occupational and environmental exposure history is a fundamental step toward acquiring information on possible exposure to arsenic. • A full work history is necessary to identify jobs with possible arsenic exposures. Ask patients about their job (where they work and what they do), potential exposure occurring in current and previous jobs, hygiene practices in the workplace, and use of any personal protective equipment. If the individual is employed in a job that may involve exposure to arsenic, further questions on length of time at the job, frequency of tasks handling arsenic materials, and descriptions of how they carry out their work should be asked. • An environmental history can identify other potential exposure sources. Questions about recent activities associated with arsenic provide information on possible risks. Questions may include: sawing or burning of arsenic-treated wood in the home; use of folk remedies (e.g. “Asiatic pill”, kushta, yellow root); spouse’s occupation and the potential for “take-home” arsenic exposure; and proximity of residence to hazardous waste sites, smelting operations, or other industries. The Agency for Toxic Substances & Disease Registry (ATSDR) offers a self-instructional module on “Taking an Exposure History” which provides an exposure history form in Appendix 1 of the document. Medical History: A medical history may help recognize possible symptoms associated with arsenic poisoning. Health effects of arsenic poisoning may include respiratory irritation, nausea, skin effects, and neurological effects. Laboratory testing for arsenic should be considered if the exposure history indicates possible arsenic exposure and/or the patient is experiencing symptoms that suggest possible arsenic poisoning. A laboratory test is the only means to confirm arsenic exposure. Laboratory Tests: Detection of arsenic in blood and urine will confirm recent arsenic exposure. Interpretation of the results of arsenic testing must take into account toxicokinetics as arsenic is rapidly cleared from the body. Arsenic is deposited in hair and nails, however, the results of hair or fingernail analysis must be interpreted carefully. • Urinary Arsenic is the most reliable means of detecting arsenic exposures occurring within the last several days. A urine spot test can be helpful in identifying elevated arsenic levels and a 24-hour urine collection for total arsenic excretion can be diagnostic. Urinary levels of arsenic may drop rapidly in the first 24 to 48 hours after an acute exposure or cessation of exposure. Elevated arsenic levels over time indicate an ongoing source of exposure. If a single urine test is high, a second test should be repeated within 2 weeks to determine if exposure is ongoing. Critical Interference: Urine tests that monitor total arsenic can be misleading. Nutritional sources of arsenic, particularly seafood, are detected in total urine arsenic measurements and can lead to erroneously high levels of arsenic that may be diagnosed as arsenic poisoning. To determine the level of inorganic arsenic, the laboratory must “speciate” the arsenic into organic and inorganic fractions. Inorganic arsenic is responsible for the adverse health effects. Metabolites of inorganic arsenic can be separated from the nontoxic dietary organic forms (arsenobetaine and arsenocholine). It is imperative to ask the laboratory to speciate the arsenic as this is not routinely done in many laboratories. Failure to speciate and using total arsenic levels may lead to the misdiagnosis of arsenic poisoning and result in unnecessary treatment of patients. It is important to inquire about recent diet or advise individuals not to ingest seafood for at least 3 days prior to testing. If they have, it should be noted. • Arsenic in Blood is rapidly cleared, is only useful for very recent exposures, and is not a useful indicator of chronic low level arsenic exposure. The reported half-life of arsenic in blood immediately following ingestion is hours, while whole-body clearance may be days or months and is dose-related. • Complete Blood Count: As with many heavy metals, microcytic hypochromic anemia is common and a CBC with indices and reticulocyte is useful. Acute hemolytic anemia occurs with arsine gas exposure • Arsenic in Hair and Nails: Arsenic may be detected in hair and nails of chronically exposed people and has been used to effectively identify arsenic poisoning post-mortem. Interpretation of levels of arsenic in hair or nails of living people need to be subjected to several factors: 1) the results are qualitative as there is wide variability in the growth of both hair and nails among individuals, seasons, type and color of hair; 2) the pharmacokinetics of deposition of arsenic in hair and nails has not been established; 3) few laboratories are capable of quantitative analysis due to difficulties in extraction and analysis; 4) there are no standards for comparison to determine relative levels; and 5) external contamination occurs from hair products, air pollution, and other sources. HEALTH-BASED REQUIREMENTS, RECOMMENDATIONS & GUIDELINES Occupational Safety and Health Administration: The Occupational Safety and Health Administration (OSHA) promulgates and enforces regulations for toxic substances in the workplace. These regulations are enforceable by law. OSHA regulations are based on both air monitoring and biological monitoring of the worker. Air monitoring provides data on work place conditions, guides industrial hygiene measures and serves as the basis for requiring medical/biological monitoring of workers. Biological monitoring measures the uptake of arsenic into the body, reflects actual exposure and is used to assess health risk to workers (OSHA, 2004). Inorganic Arsenic Compounds • OSHA promulgates and enforces regulations on inorganic arsenic exposure in the workplace. • Workers exposed to levels of airborne arsenic levels > 5 µg/m3 (eight-hour time-weighted average) for more than 30 days per year must be placed in a medical surveillance program. • Biological monitoring of arsenic levels is not explicitly required by this standard. Organic Arsenic Compounds OSHA is currently developing medical surveillance regulations for individuals occupationally exposed to organic arsenic compounds. Until a complete standard is promulgated, OSHA has put-forth recommended guidelines for protection from the risk of illness resulting from exposure to organic arsenic: • Workers cannot be exposed above the permissible exposure limit (PEL) of 0.5 mg/m3 of air expressed as a time-weighted average concentration for an 8-hour period. • Organic arsenic can be measured in the urine. A biological exposure index (BEI) of 50 µg/gram of creatinine should be used for exposure to organic arsenic compounds. American Conference of Governmental Industrial Hygienists: The American Conference of Governmental Industrial Hygienists (ACGIH) is a non-profit, nongovernmental scientific organization which develops peer-reviewed guidelines for workplace exposures. These guidelines are considered best practices for protecting workers health, but are not legally enforceable (ACGIH, 2003). • ACGIH develops Biological Exposure Indices (BEIs®) as guidance values for assessing biological monitoring results. The BEI® generally indicates a concentration below which nearly all workers should not experience adverse health effects. • BEIs® are considered best practices, but are not legally enforceable. • ACGIH has developed a BEI® of 35µg/L for inorganic arsenic and methylated-arsenic metabolites in urine. Centers for Disease Control and Prevention: The US Public Health Service agencies provide guidance and set national goals for improving health and preventing disease and injury. The Centers for Disease Control and Prevention (CDC) has established clinical case definitions for chemical poisoning involving arsenic compounds for the general population. These definitions are based on biological monitoring results. • A case of inorganic arsenic poisoning is confirmed when urinary arsenic levels are >50 µg/L. • A case of arsine poisoning might be indicated when urinary arsenic levels are > 50 ug/L and signs of hemolysis are observed. ACUTE MEDICAL MANAGEMENT GUIDELINES The Agency for Toxic Substances and Disease Registry (ATSDR) has published Medical Management Guidelines for acute exposure to arsine gas and arsenic trioxide. These guidelines were developed to aid emergency department physicians and other emergency healthcare professionals who manage acute exposures resulting from chemical incidents. They provide instruction on effective decontamination of patients, protection from contamination during treatment, efficient transport of patients to a medical facility, and medical evaluation and treatment recommendations. CONTINUING EDUCATION The Agency for Toxic Substances & Disease Registry (ATSDR) offers self-instructional Case Studies in Environmental Medicine designed to increase the health care provider’s knowledge of arsenic (and other hazardous substances) in the environment and to aid in the evaluation of potentially exposed patients. Continuing medical education credits, continuing nursing education units, and continuing education units are offered by ATSDR in support of this series. SOURCES OF INFORMATION The Agency for Toxic Substances and Disease Registry (ATSDR) is a federal public health agency focused on providing trusted health information to prevent harmful exposures and diseases related to toxic substances. Succinct fact sheets, detailed documents, and educational resources regarding arsenic and arsine exposure and toxicity can be found on their website. The Electronic Library of Construction Occupational Safety and Health (eLCOSH), developed and maintained by The Center for Construction Research and Prevention, provides accurate, user-friendly information about the safety and health of construction workers including two articles explaining the dangers of working with wood that has been preserved with arsenic compounds entitled “Hazards of Pressure-Treated Lumber” and “Wood Worries”. The US Environmental Protection Agency provides the general public and professionals with information about arsenic hazards in drinking water and their prevention. The National Institute for Occupational Safety and Health (NIOSH) is a federal agency established to help assure safe and healthful working conditions by providing research, information, education, and training in the field of occupational safety and health. Information on occupational arsenic exposure and take-home arsenic can be found on their website. The US Department of Labor’s Occupational Safety & Health Administration (OSHA) develops and enforces regulations for toxic substances in the workplace. Occupational standards and medical monitoring standards for occupational arsenic exposure can be found on their website. Louisiana Arsenic Poisoning Reporting Requirements The State of Louisiana mandates that all cases of arsenic poisoning and all laboratory arsenic test results (regardless of the arsenic levels) are reported to the Louisiana Office of Public Health. Arsenic poisoning cases are defined as, “any medical condition/visit resulting from exposure as determined from the exposure history or patient statement and/or acute, subacute, or chronic illness or injury resulting from inhalation, ingestion, dermal exposure or ocular contact.” Cases of arsenic poisoning in adults and children must be reported to the Office of Public Health’s Section of Environmental Epidemiology and Toxicology via fax (504)219-4582 telephone 888-293-7020 (business hours) or (504)219-4518 Reporting Form REFERENCES ACGIH. 2003. TLVs® and BEIs®. Threshold Limit Values for Chemical Substances and Physical Agents and Biological Exposure Indices. American Conference of Governmental Industrial Hygienists. ISBN 1-882417-49-6. Cincinnati (OH). ATSDR, 2007. Toxicological Profile for Arsenic, 2007. U.S. Department of Health and Human Services, Centers for Disease Control and Prevention, Agency for Toxic Substances and Disease Registry. Atlanta, (GA). Available at [accessed 3/10/2008] CDC. 1983. National Occupational Exposure Survey 1981-1983. Centers for Disease Control and Prevention, National Institute for Occupational Safety and Health. Available at www.cdc.gov/noes [accessed 1/31/2008] NIOSH. 1997. Protect Your Family. Reduce Contamination at Home. National Institute for Occupational Safety and Health, Centers for Disease Control and Prevention, Department of Health and Human Services. DHHS (NIOSH) Publication No. 97-125. Atlanta (GA). Available at [accessed 1/31/2008] WHO, 2000. Air Quality Guidelines for Europe, 2nd ed. World Health Organization. Regional Publications, European Series No.91. Copenhagen (Denmark). WHO. 2001. Environmental Health Criteria 224: Arsenic and Arsenic Compounds. World Health Organization. International Programme on Chemical Safety. Geneva (Switzerland). Available at [accessed on 3/13/08] OSHA. 2004. Safety and Health Topics: Arsenic. US Department of Labor, Occupational Safety and Health Administration. Available at [accessed 3/10/2008]
15629
https://search.library.dartmouth.edu/discovery/fulldisplay?vid=01DCL_INST%3A01DCL&search_scope=MyInst_and_CI&tab=All&docid=alma991019414729705706&lang=en&context=L&adaptor=Local%20Search%20Engine&query=creator%2Cexact%2C%20Wilson%2C%20R.%20J.%20&offset=0
Full display result Book ; Introduction to graph theory ; West, Douglas Brent.; , Upper Saddle River, N.J. : Prentice Hall, , ©2001, Send to Get It Logging in allows you to request items and access your patron account. 0" flex="none" class="flex-none"> MaintwoRequest FormRequest reply MaintwoRequest FormRequest reply Locations Filters: Baker-Berry Library location Baker-Berry Library volume volume Other.. year year Other.. description (volume/year) description (volume/year) Other.. Baker-Berry Library Available , Cook ; QA166 .W43 2001 LOCATION ITEMS Baker-Berry Library Available , Cook ; QA166 .W43 2001 (1 copy, 1 available, 0 requests) Filters: Baker-Berry Library location Baker-Berry Library volume volume Other.. year year Other.. description (volume/year) description (volume/year) Other.. Item status: Available Terms of use: Login to request item(s). Links Get it through interlibrary loan Send Us Feedback : report a problem with this page Display Source Record Details Title Introduction to graph theory / Douglas B. West. Introduction to graph theory / Douglas B. West. Introduction to graph theory / Douglas B. West. Creator West, Douglas Brent. West, Douglas Brent. West, Douglas Brent. Subject Graph theory Graph theory Graph theory Contents ch. 1. Fundamental concepts: What is a graph? -- Paths, cycles, and trails -- Vertex degrees and counting -- Directed graphs -- ch. 2. Trees and distance: Basic properties -- Spanning trees and enumeration -- Optimization and trees -- ch. 3. Matchings and factors: Matchings and covers -- Algorithms and applications -- Matchings in general graphs -- ch. 4. Connectivity and paths: Cuts and connectivity -- k-connected graphs -- Network flow problems -- ch. 5. Coloring of graphs: Vertex colorings and upper bounds -- Structure of k-chromatic graphs -- Enumerative aspects -- ch. 6. Planar graphs: Embeddings and Euler's formula -- Characterization of Planar graphs -- Parameters of planarity -- ch. 7. Edges and cycles: Line graphs and edge-coloring -- Hamiltonion cycles -- Planarity, coloring, and cycles -- ch. 8. Additional topics (optional): Perfect graphs -- Matroids -- Ramsey theory -- More extremeal problems -- Random graphs -- Eigenvalues of graphs.) ch. 1. Fundamental concepts: What is a graph? -- Paths, cycles, and trails -- Vertex degrees and counting -- Directed graphs -- ch. 2. Trees and distance: Basic properties -- Spanning trees and enumeration -- Optimization and trees -- ch. 3. Matchings and factors: Matchings and covers -- Algorithms and applications -- Matchings in general graphs -- ch. 4. Connectivity and paths: Cuts and connectivity -- k-connected graphs -- Network flow problems -- ch. 5. Coloring of graphs: Vertex colorings and upper bounds -- Structure of k-chromatic graphs -- Enumerative aspects -- ch. 6. Planar graphs: Embeddings and Euler's formula -- Characterization of Planar graphs -- Parameters of planarity -- ch. 7. Edges and cycles: Line graphs and edge-coloring -- Hamiltonion cycles -- Planarity, coloring, and cycles -- ch. 8. Additional topics (optional): Perfect graphs -- Matroids -- Ramsey theory -- More extremeal problems -- Random graphs -- Eigenvalues of graphs. ch. 1. Fundamental concepts: What is a graph? -- Paths, cycles, and trails -- Vertex degrees and counting -- Directed graphs -- ch. 2. Trees and distance: Basic properties -- Spanning trees and enumeration -- Optimization and trees -- ch. 3. Matchings and factors: Matchings and covers -- Algorithms and applications -- Matchings in general graphs -- ch. 4. Connectivity and paths: Cuts and connectivity -- k-connected graphs -- Network flow problems -- ch. 5. Coloring of graphs: Vertex colorings and upper bounds -- Structure of k-chromatic graphs -- Enumerative aspects -- ch. 6. Planar graphs: Embeddings and Euler's formula -- Characterization of Planar graphs -- Parameters of planarity -- ch. 7. Edges and cycles: Line graphs and edge-coloring -- Hamiltonion cycles -- Planarity, coloring, and cycles -- ch. 8. Additional topics (optional): Perfect graphs -- Matroids -- Ramsey theory -- More extremeal problems -- Random graphs -- Eigenvalues of graphs. Related titles Available in other form: Online version: West, Douglas Brent. Introduction to graph theory. 2nd ed. Upper Saddle River, N.J. : Prentice Hall, ©2001 Available in other form: Online version: West, Douglas Brent. Introduction to graph theory. 2nd ed. Upper Saddle River, N.J. : Prentice Hall, ©2001 Available in other form: Online version: West, Douglas Brent. Introduction to graph theory. 2nd ed. Upper Saddle River, N.J. : Prentice Hall, ©2001 Publisher Upper Saddle River, N.J. : Prentice Hall Upper Saddle River, N.J. : Prentice Hall Upper Saddle River, N.J. : Prentice Hall Creation Date ©2001 ©2001 ©2001 Edition 2nd ed. 2nd ed. 2nd ed. Format xix, 588 pages : illustrations ; 25 cm. xix, 588 pages : illustrations ; 25 cm. xix, 588 pages : illustrations ; 25 cm. Identifier LC : 00039993 LC : 00039993 LC : 00039993 ISBN : 0130144002 ISBN : 0130144002 ISBN : 0130144002 ISBN : 9780130144003 ISBN : 9780130144003 ISBN : 9780130144003 OCLC : (OCoLC)ocm43903794 OCLC : (OCoLC)ocm43903794 OCLC : (OCoLC)ocm43903794 Language English English English Record Source Library Catalog Library Catalog Library Catalog MMS ID 991019414729705706 991019414729705706 991019414729705706 Local ID b65855401 b65855401 b65855401 Virtual Browse Ten applications of graph theory ©1984 Über Kreise in Graphen 1974 Six degrees : the science of a connected age ©2003 Small worlds : the dynamics of networks between order and randomness ©1999 Small Worlds : The Dynamics of Networks between Order and Randomness 2018 Introduction to graph theory ©2001 On construction and identification of graphs 1976 Introduction to graph theory. 1972 Graphs : an introductory approach : a first course in discrete mathematics ©1990 Introduction to graph theory 2010 Graph theory in America : the first hundred years 2023 Introduction to graph theory 1996 Syndetics Unbound Syndetics Unbound Summary Video And Music Video Games About The Author Look Inside Series You May Also Like Librarian Lists Professional Reviews Altmetrics Core Titles Also Available As Book Profile Awards and Honors Awards and Honors Summary This book fills a need for a thorough introduction to graph theory that features both the understanding and writing of proofs about graphs. Verification that algorithms work is emphasized more than their complexity. An effective use of examples, and huge number of interesting exercises, demonstrate the topics of trees and distance, matchings and factors, connectivity and paths, graph coloring, edges and cycles, and planar graphs. For those who need to learn to make coherent arguments in the fields of mathematics and computer science. (read less) This book fills a need for a thorough introduction to graph theory that features both the understanding and writing of proofs about graphs. Verification that algorithms work is emphasized more than their complexity. An effective use of examples, and huge number of interesting exercises, demonstrate the topics of trees and distance, matchings and factors, connectivity and paths, graph coloring,(read more) Video And Music Video Games About The Author Look Inside Preface Graph theory is a delightful playground for the exploration of proof techniques in discrete mathematics, and its results have applications in many areas of the computing, social, and natural sciences. The design of this book permits usage in a one-semester introduction at the undergraduate or beginning graduate level, or in a patient two-semester introduction. No previous knowledge of graph theory is assumed. Many algorithms and applications are included, but the focus is on understanding the structure of graphs and the techniques used to analyze problems in graph theory. Many... See Full ExcerptSee Full Table of ContentsCover Series Librarian Recommends Professional Reviews Altmetrics See more details Mentioned in 1 Q&A threads Altmetrics data is provided by Altmetric.com, a research metrics company who track and collect the online conversations around millions of scholarly outputs. Altmetric continually monitors a variety of non-traditional sources to provide real-time updates on new mentions and shares of individual research outputs, which are collated and presented to users via the Altmetric details pages and badge visualizations. See More Core Titles Also Available As Book Profile Awards and Honors Catalog enrichment powered by Syndetics Unbound Related Persons American graph theorist Born 1953-01-01 Field Of Work Mathematics Occupation Mathematician, Graph theorist Source Library of Congress Name Authority and Wikidata Related reading Recommended items that are related to the record Ask Us Now! × Introduction to graph theory / West, Douglas Brent. This pin indicates your item's location on the map Directions: From Berry Level 1, look for Stair 1 or Elevator 1 which are located at the top of the stairs that lead to the Novack Café. Proceed up to Berry 3. ### Go to the row labelled: Berry, Level 3, Stack 118 ### Find item #: QA166 .W43 2001 Mobile-friendly map (share this link via text or email) Powered by StackMap.com Introduction to graph theory / Douglas B. West. / null This pin indicates your item's location on the map Directions: From Berry Level 1, look for Stair 1 or Elevator 1 which are located at the top of the stairs that lead to the Novack Café. Proceed up to Berry 3. ### Go to the row labelled: Berry, Level 3, Stack 118 ### Find item #: QA166 .W43 2001 Mobile-friendly map (share this link via text or email) Powered by StackMap.com Introduction to graph theory / Douglas B. West. / null This pin indicates your item's location on the map Directions: From Berry Level 1, look for Stair 1 or Elevator 1 which are located at the top of the stairs that lead to the Novack Café. Proceed up to Berry 3. ### Go to the row labelled: Berry, Level 3, Stack 118 ### Find item #: QA166 .W43 2001 Mobile-friendly map (share this link via text or email) Powered by StackMap.com
15630
https://users.math.uoc.gr/~pamfilos/eGallery/problems/OrthodiagonalFromCyclic.html
Orthodiagonal from cyclic For every cyclic quadrilateral q = ABCD, there is associated an [orthodiagonal] quadrilateral p = KLMN with the following properties: 1] p is circumscribed about q. 2] The diagonals LN and KM of p pass from points F and E, intersection points of side-pairs (AD,BC) and (AB,CD) respectively. 3] The intersection point O of the diagonals LN and KM is on the perpendicular from H to line EF, H being the intersection point of the diagonals of q. 4] Lines OA, OB, OC and OD are perpendicular to the corresponding sides of p. 5] The other points of intersection of these lines with p lie on the circumcircle c of q and build a rectangle r = ABCD. | | | --- | | | | | | | It is known that H is the pole of EF (look at ) and the circle (d) with diameter EF is orthogonal to (c). Draw the HG perpendicular to EF and define its intersection point O with the circle (d), on same side with the center I of (c). Draw then orthogonals to OA at A, OB at B, ... etc. defining the quadrangle p = KLMN. The quadrilaterals q1 = OAKD and q2 = OBMC are cyclic and F is on their radical axis, since FAD, FBC are chords of the circumcircle (c). It is easy to see that |FO|² = |FP||FQ| = |FA||FD| = |FB||FC|. This implies that FO is orthogonal to the diameters KO, OM of circumcircles of q1 and q2, i.e. the circumcircles of q1 and q2 are tangent to line FO at O. Hence K, O, M are on a line and FO is perpendicular to this line, hence KM passes through E. Analogously one shows that LN passes through F. The last (5th) assertion is a general property of orthodiagonal quadrangles. The construction of the figure here is, in some sense, the inverse of the procedure explained in the file . See this file for a proof of the 5th assertion. Equality |FO|² = |FP||FQ| is a consequence of the orthogonality of two circle bundles handled in . Some additional facts on these kind of quadrilaterals are discussed in the file . Return to Gallery | | | --- | | Produced with EucliDraw© | |
15631
https://www.cliffsnotes.com/cliffs-questions/1940243
[Solved] Quadratic Equations Activity Purpose: Mathematical models based on quadratic equations describe curved paths through... | CliffsNotes Lit NotesStudy GuidesDocumentsQ&AAsk AIChat PDFLog InSign Up Literature NotesStudy GuidesDocumentsHomework QuestionsChat PDFLog InSign Up Quadratic Equations Activity Purpose: Mathematical models based on quadratic equations describe curved paths through... Answered step-by-step Solved by verified expert See Answer Questions & AnswersMath - Other Quadratic Equations Activity Purpose: Mathematical models based on quadratic equations describe curved paths through... Question Answered step-by-step Asked by GrandGooseMaster733 Quadratic Equations Activity Purpose: Mathematical models based on quadratic equations describe curved paths through space. The purpose of this project is to develop an understanding of techniques used to apply quadratic models to describe motion. Assignment Goal: Describe the motion of a rocket using a quadratic model. Instructions: Respond to each of the questions. Your solutions must document all of your work. Scenario: A toy rocket is launched straight upward from a pad 10 ft above ground level with an initial velocity of 80 ft/sec. Use the general formula for the vertical position (height) of an object moving under the influence of gravity 𝑠 = ― 1 2𝑔𝑡2 + 𝑣0𝑡 + 𝑠0 where s0 is initial vertical position, v0 is initial velocity, and s is vertical position of the object. Define a model (equation) to express the vertical position (height) of the rocket s (in meters) above ground level t seconds after launch. Let g = 32 ft/sec2. Graph your equation with time on the horizontal axis and height on the vertical axis. Let the values for time range from 0 to 10 seconds and the values for the height range from 0 to 200 feet. Include your graph in this response. Refer to your graph in Question 2. Explain what the vertex, x-intercept, and y-intercept represent in the context of this problem. Calculate the time(s) at which the rocket is at a height of 40 ft. (Round to 1 decimal place.) Is this the maximum height the rocket will reach? Justify your response. Answer & Explanation Solved by verified expert Answered by toparch6 Get the explanation ------------------- Unlock access to this and over 10,000 step-by-step explanations See Answer Step-by-step explanation 7 attachments PNG PNG PNG PNG PNG PNG PNG Show other answers (1) Other answer Get the explanation ------------------- Unlock access to this and over 10,000 step-by-step explanations See Answer Related Q&A Q A Final Answer Explained clearly in the below explanation. Please go through it. If you have any queries , please feel free to ask. Solved by verified expertQ Answer all of the questions and show work, and explain what is being done. A 181,704,600 Proven true by induction (See Below). 203 3,060 Solved by verified expertQ Please see attachments for details A Final Answers Explained clearly in the below explanation. Please go through it. If you have any queries , please feel free to ask. Solved by verified expertQ Please see attachments for details A Follow the below content for solution Solved by verified expertQ This is the unit: Can you generate me a mock exam, with answers please thank you. A Final Answer Explained clearly in the below explanation. Please go through it. To enhance clarity and understanding, I used a math app to carry out the symbolic calculations in this problem. Solved by verified expertQ Please see attachments for details A Here am attaching proper solution of first two lesson of chapter 3. All necessary details along with necessary hints are provided inside brackets . If still having doubt at any point feel fre Solved by verified expertAll Questions CliffsNotes study guides are written by real teachers and professors, so no matter what you're studying, CliffsNotes can ease your homework headaches and help you score high on exams. Quick Links Literature NotesStudy GuidesDocumentsHomework Questions Company About CliffsNotesContact us Do Not Sell My Personal Information Legal Service TermsPrivacy policyCopyright, Community Guidelines & other legal resourcesHonor CodeDisclaimer CliffsNotes, a Learneo, Inc. business © Learneo, Inc. 2025 AI homework help Explanations instantly Do Not Sell or Share 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. However, you can choose not to allow certain types of cookies, which may impact your experience of the site and the services we are able to offer. Click on the different category headings to find out more and change our default settings according to your preference. You cannot opt-out of our First Party Strictly Necessary Cookies as they are deployed in order to ensure the proper functioning of our website (such as prompting the cookie banner and remembering your settings, to log into your account, to redirect you when you log out, etc.). For more information about the First and Third Party Cookies used please follow this link. More information Allow All Manage Consent Preferences Essential Cookies Always Active Essential Cookies are required for providing you with features or services that you have requested. For example, certain Cookies enable you to log into secure areas of our Services. Share or Sale of Personal Data Always Active Under the California Consumer Privacy Act, you have the right to opt-out of the sale of your personal information to third parties. These cookies collect information for analytics and to personalize your experience with targeted ads. You may exercise your right to opt out of the sale of personal information by using this toggle switch. If you opt out we will not be able to offer you personalised ads and will not hand over your personal information to any third parties. Additionally, you may contact our legal department for further clarification about your rights as a California consumer by using this Exercise My Rights link. If you have enabled privacy controls on your browser (such as a plugin), we have to take that as a valid request to opt-out. Therefore we would not be able to track your activity through the web. This may affect our ability to personalize ads according to your preferences. Advertising Cookies Always Active Advertising Cookies collect data about your online activity and identify your interests so that we can provide advertising that we believe is relevant to you. Advertising Cookies may include Retargeting Cookies. Analytics Cookies Always Active Analytics Cookies allow us to understand how visitors use our Services. They do this by collecting information about the number of visitors to the Services, what pages visitors view on our Services and how long visitors are viewing pages on the Services. Analytics Cookies also help us measure the performance of our advertising campaigns in order to help us improve our campaigns and the Services’ content for those who engage with our advertising. Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Confirm My Choices
15632
https://www.youtube.com/watch?v=Jsvj51-urdM
Solving Inequalities: Flip Inequality Signs when Multiplying or Dividing by Negative Numbers Dr. Pierce's Physics & Math 3010 subscribers 9 likes Description 1583 views Posted: 18 Sep 2022 Just a short tutorial to remind pre-algabra, algebra 1, algebra 2, and precalculus students to flip the direction of inequality signs when multiplying or dividing both sides of an inequality by a negative number. Failure to do so will result in false mathematical statements. Note that adding or subtracting numbers from both sides of an inequality does not affect the direction of the inequality sign. 4 comments Transcript: so we have an inequality here negative three x is less than 27. and what we're going to do is we're going to look at a wrong way to approach this first so what somebody might think on seeing this is well maybe I will just divide both sides by negative 3. and what that would do is that would remove the negative 3 from the left hand side and leave us with X over there and 27 divided by negative 3 would give us a negative 9. and so we get this statement that uh well to solve it then X needs to be less than negative 9. well let's try this if we try a value that's less than negative 9 like x equals negative 10 let's say and put that back into this original equation what you'd find out is well negative 3 times negative 10 would actually give you positive 30 and you get this statement that 30 is less than 27 which we know isn't the case so that that approach is wrong um So to avoid this sort of thing something that you're going to need to remember to do it the right way is uh I would begin this uh by dividing both sides by negative 3 as we did um but the change you want to make is you want to flip that inequality sign so it points the other way and so the overarching little rule you want to remember is to flip it if you divide by a negative number let's just check if this worked so uh the statement here is that X should be greater than negative 9. so if we try negative 8 which is greater than negative 9. we would get a statement that positive 24 is less than positive 27 and that is true so it does work so the overarching principle here is that you want to flip that inequality sign if you multiply or divide both sides by a negative number let's look at one more case here um what we want to do here is uh see what happens if we have an inequality it looks like this negative 3x plus 6 is less than or equal to 27. well the first little algebra move that most of us would make would be to well add negative 6 to both sides otherwise known as subtracting 6 from both sides and that will leave you with negative 3x on the left because you'll have gotten rid of the six and then 27 minus 6 is 21. so you get Negative 3x less than or equal to 21. you wouldn't flip the inequality sign here because we haven't multiplied or divided by a negative number um and so then what we wanted what we would do here next is divide by that negative three um to get rid of it from the left uh so divide by negative 3 to isolate the X um but notice that what we did here is not only did we divide both sides by negative 3 but we also flipped the inequality sign and so this is the way that you want to approach it the the key is if you multiply or divide both sides of an inequality by a negative number you're going to want to remember to flip that inequality sign
15633
https://byjus.com/maths/lines-and-angles/
Lines And Angles are the basic shapes in geometry. Lines are figures that are made up of infinite points extending indefinitely in both directions. Lines are straight and have negligible depth or width. There are a variety of lines you will learn about, such as perpendicular lines, intersecting lines, transversal lines, etc. An angle is a figure in which two rays emerge from a common point. You may also come across alternate and corresponding angles in this field. Geometry shapes and their properties are the most practical branch of mathematics. Mostly this concept has been taught in Class 7 and Class 9. Definition of Lines and Angles As we have discussed, both lines and angles form the base for any shape in geometry. We cannot draw a two-dimensional to three-dimensional shape without using lines and angles. Thus, it is very necessary to learn the definitions of both terms. Here, the basic definitions and properties of lines and also for angles are given. It will give the students a basic knowledge of these geometrical terms. What are Lines? A line is a straight one-dimensional figure, that extends in the opposite directions infinitely. A line can be horizontal or vertical. It can be drawn from left to right or top to bottom. What are Angles? Angles are the shape that is formed when the endpoints of two rays meet at a single point. They are measured in degrees (°) or radians. A complete rotation is equal to an angle of 360 degrees. It is represented by the symbol ‘∠’. Types of Lines and Angles There are various types of lines and angles in geometry based on the measurements and different scenarios. Let us learn here all those lines and angles along with their definitions. Types of Lines Lines are basically categorized as: Line segment Ray Based on concepts or operations performed on lines, they are; Parallel Lines Perpendicular Lines Transversal Line Segment A line segment is a part of a line with two end-points. It is the shortest distance between two points and has a fixed length. Ray A ray is a part of a line, which has a starting point and extends infinitely in one direction. Perpendicular Lines When two lines form a right angle with each other, by meeting at a single point, are called perpendicular lines. In the figure, you can see, lines AB and CD are perpendicular to each other. Parallel Lines Two lines are said to be parallel when they do not meet at any point in a plane or which do not intersects each other. In the figure, lines PQ and RS are parallel to each other. Transversal Line When a line intersects two lines at distinct points, it is called a transversal. In the figure, a transversal l is intersecting two lines at point P and Q. Types of Angles Angles are basically classified as: Acute Angle(<90°) Right Angle(=90°) Obtuse Angle(>90°) Straight Angle(=180°) And based on the relation between two angles, conceptual wise, they are; Supplementary Angles Complementary Angles Adjacent Angles Vertically Opposite Angles Acute Angle If the inclination between the arms is less than a right angle, it is called an acute angle. Obtuse Angle If the inclination between the arms is more than a right angle, it is called an obtuse angle. Right Angle If the arms form an angle of 90 degrees between them, it is called a right angle. Straight Angle If the arms form an angle of 180 degrees between them, it is called a straight angle. Complementary Angles Two angles which sum up to 90 degrees are called complementary angles. Supplementary Angles Two angles which sum up to 180 degrees are called supplementary angles. Adjacent Angles Two angles which have a common side and a common vertex are called adjacent angles. In the following figure, ∠α and ∠β are adjacent angles. Vertically Opposite Angles Two angles which are formed, opposite to each other, when two lines intersect at a common point or vertex, are called vertically opposite angles. In the figure, given below; ∠POR = ∠SOQ and ∠POS = ∠ROQ Video Lesson on Lines For More Information On Lines, Watch The Below Video: 10,432 Video Lessons on Types and Parts of Angles Types of angles 1,36,490 Parts of Angles 14,976 Properties of Lines and Angles Similar to other shapes and sizes in geometry, lines and angles also have their own properties. Let us see what are they. Properties of Lines Collinear points are a set of three or more points which lie on the same line. The points which do not lie on the same line are called non-collinear points. Note: Three points can be either collinear or non-collinear, but not both together at the same time. Properties of Angles An angle is a figure in which two rays emerge from a common point. This point is called the vertex of the angle and the two rays forming the angle are called its arms or sides. An angle which is greater than 180 degrees but less than 360 degrees is called a reflex angle. If two adjacent angles add up to 180 degrees, they form a linear pair of angles. In the following figure, ∠a and ∠b form a linear pair. When two lines intersect each other, the two opposite pairs of angles formed are called vertically opposite angles. In the following figure, ∠A and ∠B are vertically opposite angles. Another pair is ∠C and ∠D. Related Articles Lines Angles Angles In A Pentagon Angles of a Parallelogram Intersecting Lines And Non-intersecting Lines Parallel Lines and Angle Relationships Lines And Angles Class 7 Lines and Angles Class 9 To solve more problems on Lines and Angles visit our page. Study more by downloading BYJU’S-The Learning App and have a better understanding of the topic. Frequently Asked Questions on Lines And Angles Q1 What are the five types of Angles? The five types of angles are: Acute Angle Right Angle Obtuse Angle Straight Angle Reflex Angle Q2 What are the properties of Lines and Angles? If two parallel lines are intersected by a transversal then: Its vertically opposite angles are always equal Its corresponding angles are equal Its alternate exterior and interior angles are equal Q3 What are Lines and its types? Lines are figures that are made up of infinite points extending indefinitely in both directions. The types of lines are: Horizontal lines Vertical lines Parallel lines Perpendicular lines Test your knowledge on Lines And Angles Q5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Congrats! Visit BYJU’S for all Maths related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted Login To View Results Did not receive OTP? Request OTP on Login To View Results Comments Leave a Comment Cancel reply Soumil Basu August 17, 2019 at 10:30 am Very nice tutors and very useful Reply Drishya October 31, 2019 at 3:27 pm This article is very helpful. Its really very informative. Reply Ayushi Maitray March 27, 2020 at 4:04 pm This is very useful to me. Reply Michael Noel April 28, 2020 at 4:26 am Thank you very much for the information on your site. I’m a Construction Manager student and do not have a strong math background, only basic tape measure stuff. Good luck. Mike Reply Deepika September 30, 2020 at 7:35 pm It is very useful for me . Thank u Reply ALEENA JOJI September 30, 2020 at 10:05 pm VERY USEFUL TO ME Reply Vikas September 30, 2020 at 11:50 pm Good helpers for us. Good teachers. Fast interactive. No bad report regarding byjus . Reply
15634
http://graphic-technologies.co.nz/doc/Linearprogrammingwithverticallines.pdf
Linear programming with vertical lines: converting x = c to y = mx + c This resource was written by Derek Smith with the support of CASIO New Zealand. It may be freely distributed but remains the intellectual property of the author and CASIO. Select GRAPH mode from the main menu by using the arrow keys to highlight the GRAPH icon or pressing 5. Introduction Open the GRAPH-window, and make sure that the V-Window is appropriate to see the graphs that you will be drawing. SHIFT F3 Select F1, in this case for INITial conditions. Now, the EXIT or EXE key to go back to the Graph Func window. Vertical lines x = c, found by menu trail: F3 for TYPE then F4 for x = c formatting. Example: Draw the graphs of x = 2 Answer: Enter into the GRAPH icon Press Type in ‘2’ and then EXE F6 to draw. In linear programming usually the inequalities are required to find the ‘feasible region’ that satisfies the constraints given. You are not able to draw inequalities with x ≥ or < or > or ≤ on the graphic calculator. The menu trail for inequalities is: F3 F6 So, you need to ‘trick’ the calculator into thinking that it is drawing a vertical line but in the y = mx + c format. Type into the ‘Y2’ space ‘1x – 2’ and then F6 to draw. What do you notice? Now try or They both produce a ‘vertical line’ very similar to x = 2 over this Viewing window. But, you have essentially tricked the calculator into drawing a ‘vertical line’ You can check that they are NOT the same by doing a Trace – SHIFT F1 and using the up / down arrows to trace either x = 2 and y = 1000x – 2000 (or similar). Problem: Find the feasible region that satisfies the following constraints over the domain 0 ≤ x ≤ 25 and range 0 ≤ y ≤ 25: x + y ≤ 15 y ≥ 6 4x + y ≤ 24 x ≥ 2 y ≥ 2x Rearranging to make y the ‘subject’ gives: y ≤ 15 – x y ≥ 6 y ≤ 24 - 4x x ≥ 2 y ≥ 2x becomes y ≤ ≤ ≤ ≤ 1000x - 2000 The V-window SHIFT F3 becomes: Enter in the constraints and F6 to draw them. The find the intersection points (vertices) of the lines that intersect, use G-Solve SHIFT F5 and then F5 for ISCT (intersection), selecting two lines at a time and generating the 5 intersection points. Y1 and Y4 gives ( 2 , 13) Y1 and Y3 gives ( 3 , 12 ) Y4 and Y2 gives ( 2 , 6 ) Y2 and Y5 gives ( 3 , 6 ) Y3 and Y5 gives ( 4 , 8 ) [N.B. Some interpretation is required if the substitution line for x = c is not ‘extremely’ vertical on the region the constraints are drawn.] For further tips, more helpful information and software support visit our website www.monacocorp.co.nz/casio
15635
https://byjus.com/maths/simultaneous-equations/
Before starting with simultaneous equations, let’s recall what are equations in maths and the types of equations. In Mathematics, an equation is a mathematical statement in which two things should be equal to each other. An equation consists of two expressions on each side of an equal sign (=). It consists of two or more variables. In short, the L.H.S value should be equal to the R.H.S value. While substituting the values of the variables in an equation, it should prove its equality. There are different types of equations in Maths, such as: Linear Equation Quadratic Equation Polynomial Equations and so on. In this article, we are going to discuss the simultaneous equations which involve two variables along with different methods to solve. What are Simultaneous Equations? The simultaneous equation is an equation that involves two or more quantities that are related using two or more equations. It includes a set of few independent equations. The simultaneous equations are also known as the system of equations, in which it consists of a finite set of equations for which the common solution is sought.To solve the equations, we need to find the values of the variables included in these equations. The system of equations or simultaneous equations can be classified as: Simultaneous linear equations (Or) System of linear equations Simultaneous non-linear equations System of bilinear equations Simultaneous polynomial equations System of differential equations Here, you will learn the methods of solving simultaneous linear equations along with examples. The general form of simultaneous linear equations is given as: ax +by = c dx + ey = f Methods for Solving Simultaneous Equations The simultaneous linear equations can be solved using various methods. There are three different approaches to solve the simultaneous equations such as substitution, elimination, and augmented matrix method. Among these three methods, the two simplest methods will effectively solve the simultaneous equations to get accurate solutions. Here we are going to discuss these two important methods, namely, Elimination Method Substitution Method Apart from those methods, we can also the system of linear equations using Cramer’s rule. If the simultaneous linear equations contain only two variables, we may also use the cross-multiplication method to find their solution. Read more: Algebra Equation Solving system of linear equations in two and three variables Simultaneous Equation Example Let us now understand how to solve simultaneous equations through the above-mentioned methods. We will get the value of a and b to find the solution for the same. x and y are the two variables in these equations. Go through the following problems which use substitution and elimination methods to solve the simultaneous equations. Try Out:Simultaneous Equation Solver Solving Simultaneous Linear Equations Using Elimination Method Go through the solved example given below to understand the method of solving simultaneous equations by the elimination method along with steps. Example:Solve the following simultaneous equations using the elimination method. 4a + 5b = 12, 3a – 5b = 9 Solution: The two given equations are 4a + 5b = 12 …….(1) 3a – 5b = 9……….(2) Step 1: The coefficient of variable ’b’ is equal and has the opposite sign to the other equation. Add equations 1 and 2 to eliminate the variable ‘b’. Step 2:The like terms will be added. (4a+3a) +(5b – 5b) = 12 + 9 7a = 21 Step 3:Bring the coefficient of a to the R.H.S of the equation a = 21/ 7 Step 4:Dividing the R.H. S of the equation, we get a = 3 Step 5: Now, substitute the value a=3 in the equation (1), it becomes 4(3) + 5b = 12, 12 + 5b = 12 5b = 12-12 5b =0 b = 0/5 = 0 Step 6:Hence, the solution for the given simultaneous equations is a = 3 and b = 0. Solving Simultaneous Linear Equations Using Substitution Method Below is the solved example with steps to understand the solution of simultaneous linear equations using the substitution method in a better way. Example:Solve the following simultaneous equations using the substitution method. b= a + 2 a + b = 4. Solution: The two given equations are b = a + 2 ————–(1) a + b = 4 ————–(2) We will solve it step-wise: Step 1:Substitute the value of b into the second equation. We will get, a + (a + 2) = 4 Step 2: Solve for a a +a + 2 = 4 2a + 2 = 4 2a = 4 – 2 a = 2/2 = 1 Step 3: Substitute this value of a in equation 1 b = a + 2 b = 1 + 2 b = 3 step 4:Hence, the solution for the given simultaneous equations is: a = 1 and b = 3 Practice Problems Solve: 5x + 3y = 7 and -3x + 5y = 23 Solve the following simultaneous equations: p x + qy – r² = 0 p²x + q²y – r² = 0 Solve for a and b: 10a – 8b = 6 10a – 9b = -2 To practice more problems and to learn new Mathematical concepts, please visit BYJU’S – The Learning App and download the app to learn with ease. Test your knowledge on Simultaneous Equations Q 5 Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Start Quiz Congrats! Visit BYJU’S for all Maths related queries and study materials Your result is as below 0 out of 0 arewrong 0 out of 0 are correct 0 out of 0 are Unattempted View Quiz Answers and Analysis X Login To View Results Mobile Number Send OTP Did not receive OTP? Request OTP on Voice Call Login To View Results Name Email ID Grade City View Result Register with BYJU'S & Download Free PDFs Send OTP Download Now Register with BYJU'S & Watch Live Videos Send OTP Watch Now
15636
https://math.stackexchange.com/questions/3081930/what-is-the-number-of-elements-in-the-solution-set-of-x2-42-cdotx2-6x-7
algebra precalculus - What is the number of elements in the solution set of $(x^2-4)^2\cdot(x^2-6x-7)=0$? - Mathematics Stack Exchange Join Mathematics 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 Mathematics helpchat Mathematics 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 Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams 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 What is the number of elements in the solution set of (x 2−4)2⋅(x 2−6 x−7)=0(x 2−4)2⋅(x 2−6 x−7)=0? Ask Question Asked 6 years, 8 months ago Modified6 years, 8 months ago Viewed 441 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. (x 2−4)2⋅(x 2−6 x−7)=0(x 2−4)2⋅(x 2−6 x−7)=0 S.S.={x 1,x 2,...,x n}S.S.={x 1,x 2,...,x n} ⇒n=?⇒n=? Answer is given as 4 4. I think it should be 6 6 because of multiplicity of the roots. I debated this problem with my classmates but we can't reach a verdict. What is the number of elements in the Solution Set of this problem? algebra-precalculus quadratics Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jan 21, 2019 at 14:45 Eldar RahimliEldar Rahimli asked Jan 21, 2019 at 14:27 Eldar RahimliEldar Rahimli 848 4 4 silver badges 11 11 bronze badges 6 2 You have 2 double roots, leading you to believe it was 6, but it has 4 unique roots. From stackoverflow.com/questions/10011475/… we have that "a set cannot have duplicate elements by its mere definition."For the love of maths –For the love of maths 2019-01-21 14:29:40 +00:00 Commented Jan 21, 2019 at 14:29 5 Well, I think this is a question of semantics. There are 4 4 distinct real roots, and there are 6 6 real roots if you count with multiplicity.lulu –lulu 2019-01-21 14:29:57 +00:00 Commented Jan 21, 2019 at 14:29 4 If you are asked to write the set of solutions S S, there are really only 4 4 elements.Viera Čerňanová –Viera Čerňanová 2019-01-21 14:33:21 +00:00 Commented Jan 21, 2019 at 14:33 2 In a set each element appears just once.Viera Čerňanová –Viera Čerňanová 2019-01-21 14:34:17 +00:00 Commented Jan 21, 2019 at 14:34 2 This is a really good question and I applaud you asking it. The distinction between "the set of zeros" and "zeros with multiplicity" is important -- the set of zeros tells you what the zeros are, but including multiplicity lets you figure out not just where the polynomial is zero, but how it behaves near that zero.Neal –Neal 2019-01-21 14:46:13 +00:00 Commented Jan 21, 2019 at 14:46 |Show 1 more comment 4 Answers 4 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. This seems to be a matter of semantics. When counted with multiplicity, the equation has 6 6 roots. But the set of roots has only four elements, because there are only four distinct roots. That is to say, {−2,−2,1,2,2,7}={−2,1,2,7},{−2,−2,1,2,2,7}={−2,1,2,7}, because by definition a set does not contain duplicate elements. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 21, 2019 at 14:38 ServaesServaes 68k 8 8 gold badges 83 83 silver badges 174 174 bronze badges Add a comment| This answer is useful 1 Save this answer. Show activity on this post. It comes down to how you're supposed to interpret the "S.S." notation. I think the idea of writing the roots as members of a set was that you should represent the set using the minimal possible value of n,n, hence n=4,n=4, because the solution set can be written {−2,−1,2,7}.{−2,−1,2,7}. If you don't assume that n n is to be minimized, n n could be any integer greater than 3.3. For example, set x 1=−2,x 1=−2,x 2=−1,x 2=−1,x 3=2,x 3=2, and x i=7 x i=7 for i=4,5,…,8.i=4,5,…,8. Then n=8 n=8 and you can write the solution set as {−2,−1,2,7,7,7,7,7},{−2,−1,2,7,7,7,7,7}, which is considered the same set as {−2,−1,2,7}{−2,−1,2,7} (your classmates' answer) or as {−2,−2,−1,2,2,7}{−2,−2,−1,2,2,7} (your answer). This is one of the reasons I think some people ought to use words more often and mathematical symbols less. It's very simple to ask in words, "How many unique roots does this equation have?" The same question is much harder to write in symbolic form. The question as asked does not quite succeed at that. We have to make our best guess as to what meaning was intended, and it seems most people would guess n=4 n=4 (as I also would guess). Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 21, 2019 at 14:48 David KDavid K 110k 8 8 gold badges 91 91 silver badges 243 243 bronze badges Add a comment| This answer is useful 0 Save this answer. Show activity on this post. This polynomial can be simplified to: (x−2)2(x+2)2(x+1)(x−7)(x−2)2(x+2)2(x+1)(x−7) Hence it has repeated roots at x=±2 x=±2 and non-repeated roots at x=1 x=1 and x=7 x=7. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 21, 2019 at 14:39 Rhys HughesRhys Hughes 13.3k 2 2 gold badges 14 14 silver badges 36 36 bronze badges Add a comment| This answer is useful 0 Save this answer. Show activity on this post. It should be 4, because you can expand the whole thing out, and get a sextic equation, and there are 2 answers that each is seen 2 times. The four answers are 2, -2, and the two different roots of x 2−6 x−7=0 x 2−6 x−7=0. Or, just 6-2=4. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jan 22, 2019 at 0:51 answered Jan 21, 2019 at 14:34 Math LoverMath Lover 424 1 1 gold badge 3 3 silver badges 15 15 bronze badges 2 2 The polynomial is not quartic, it is sextic.Servaes –Servaes 2019-01-21 14:36:48 +00:00 Commented Jan 21, 2019 at 14:36 Oh, I didn’t see the squared. But still, you will get 2,-2,and the two different roots of x 2−6 x−7=0 x 2−6 x−7=0.Math Lover –Math Lover 2019-01-22 00:48:48 +00:00 Commented Jan 22, 2019 at 0:48 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 algebra-precalculus quadratics 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 Report this ad Related 11Sum of squares of roots of a polynomial P(x)P(x) 0Verification of solutions to some polynomial prob/ 0Unique solution of nolinear equation set 2How to expand x 3 1+x 3 2 x 1 3+x 2 3 with the parameters of quadratic equation 0Finding parameter for quadratic equation 2Find x n 1+x n 2 x 1 n+x 2 n on any quadratic equation, general case. 2The values of x 1⋅x 2 x 1⋅x 2 from x 2−2 m x+2 m 2−2 m=0 x 2−2 m x+2 m 2−2 m=0 1Interval that the sum of the roots of a quadratic equation can be in. 4Do we count only distinct roots in Descartes' rule of signs? Hot Network Questions Is it safe to route top layer traces under header pins, SMD IC? Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator Repetition is the mother of learning Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward" Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations? Is encrypting the login keyring necessary if you have full disk encryption? What NBA rule caused officials to reset the game clock to 0.3 seconds when a spectator caught the ball with 0.1 seconds left? What’s the usual way to apply for a Saudi business visa from the UAE? How different is Roman Latin? Cannot build the font table of Miama via nfssfont.tex Is existence always locational? "Unexpected"-type comic story. Aboard a space ark/colony ship. Everyone's a vampire/werewolf Overfilled my oil Change default Firefox open file directory The rule of necessitation seems utterly unreasonable Does the curvature engine's wake really last forever? Exchange a file in a zip file quickly Why do universities push for high impact journal publications? In Dwarf Fortress, why can't I farm any crops? Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? Checking model assumptions at cluster level vs global level? alignment in a table with custom separator Does the mind blank spell prevent someone from creating a simulacrum of a creature using wish? Do we need the author's permission for reference Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
15637
https://www.splashlearn.com/math-vocabulary/x-and-y-axis
Log inSign up Parents Parents Explore by Grade Preschool (Age 2-5)KindergartenGrade 1Grade 2Grade 3Grade 4Grade 5 Explore by Subject Math ProgramEnglish Program More Programs Homeschool ProgramSummer ProgramMonthly Mash-up Helpful Links Parenting BlogSuccess StoriesSupportGifting Educators Educators Teach with Us For Teachers For Schools and Districts Data Protection Addendum Impact Success Stories Resources Lesson Plans Classroom Tools Teacher Blog Help & Support More Programs SpringBoard Summer Learning Our Library Our Library All All By Grade PreschoolKindergartenGrade 1Grade 2Grade 3Grade 4Grade 5 By Subject MathEnglish By Topic CountingAdditionSubtractionMultiplicationPhonicsAlphabetVowels Games Games By Grade Preschool GamesKindergarten GamesGrade 1 GamesGrade 2 GamesGrade 3 GamesGrade 4 GamesGrade 5 Games By Subject Math GamesReading GamesArt and Creativity GamesGeneral Knowledge GamesLogic & Thinking GamesMultiplayer GamesMotor Skills Games By Topic Counting GamesAddition GamesSubtraction GamesMultiplication GamesPhonics GamesSight Words GamesAlphabet Games Worksheets Worksheets By Grade Preschool WorksheetsKindergarten WorksheetsGrade 1 WorksheetsGrade 2 WorksheetsGrade 3 WorksheetsGrade 4 WorksheetsGrade 5 Worksheets By Subject Math WorksheetsReading Worksheets By Topic Addition WorksheetsMultiplication WorksheetsFraction WorksheetsPhonics WorksheetsAlphabet WorksheetsLetter Tracing WorksheetsCursive Writing Worksheets Lesson Plans Lesson Plans By Grade Kindergarten Lesson PlansGrade 1 Lesson PlansGrade 2 Lesson PlansGrade 3 Lesson PlansGrade 4 Lesson PlansGrade 5 Lesson Plans By Subject Math Lesson PlansReading Lesson Plans By Topic Addition Lesson PlansMultiplication Lesson PlansFraction Lesson PlansGeometry Lesson PlansPhonics Lesson PlansGrammar Lesson PlansVocabulary Lesson Plans Teaching Tools Teaching Tools By Topic Math FactsMultiplication ToolTelling Time ToolFractions ToolNumber Line ToolCoordinate Graph ToolVirtual Manipulatives Articles Articles By Topic Prime NumberPlace ValueNumber LineLong DivisionFractionsFactorsShapes Log inSign up Skip to content X and Y Axis in Graph – Definition, Graph, Facts, Examples Home » Math Vocabulary » X and Y Axis in Graph – Definition, Graph, Facts, Examples Plotting Points on X- and Y-Axis Representing a Linear Equation on X- and Y-Axis Solved Questions Practice Problems Frequently Asked Questions An axis in mathematics is defined as a line that is used to make or mark measurements. The x and y-axis are two important lines of the coordinate plane. The x-axis is a horizontal number line and the y-axis is a vertical number line. These two axes intersect perpendicularly to form the coordinate plane. The x-axis is also called the abscissa and the y-axis is called the ordinate. Any point on the coordinate plane can be located or represented using these two axes in the form of an ordered pair of the form (x,y). Here, x represents the location of the point with respect to the x-axis and y represents the location of the point with respect to the y-axis. The origin is where the two axes intersect and is written as (0,0). Plotting Points on X and Y Axis Let us learn how to plot a point on the graph by using the X- and Y-axis. For example: Let’s try to plot the point B(3,4) on the graph. Here, the x-coordinate of B is 3. So we will start from the origin and move 3 units to the right on x-axis. Now, the y-coordinate of B(3,4) is 4, so we will go 4 spaces up from this point. And thus we have plotted our point B(3,4) on the graph using the axes. Recommended Games Add 2-Digit Numbers By Regrouping Game Play Add 3 Numbers in Any Order Game Play Add and Find the Amount of Money in Given Word Problems Game Play Add by Making a 10 Game Play Add by Regrouping Ones and Tens Game Play Add by Regrouping Ones Game Play Add by Regrouping Tens Game Play Add by Regrouping the Ones Game Play Add by Regrouping the Tens Game Play Add Coins of Different Types and Compare Game Play More Games Representing a Linear Equation on X- and Y-Axis To understand how to represent a linear equation on the graph using the X- and Y-axis, let us consider a linear equation, y = x + 1. Now, let’s build a table to represent the corresponding values of y for different values of x and create their ordered pairs: | | | | --- | x | y | Ordered pair | | 0 | 1 | (0,1) | | 1 | 2 | (1,2) | | 2 | 3 | (2,3) | | 3 | 4 | (3,4) | The next step is to plot these ordered pairs on the coordinate plane graph. As a final step, we will join these points to form a straight line and that will be the representation of the equation y = x + 1. Recommended Worksheets More Worksheets Solved Questions On X and Y Axis Question 1: Which of the following points lie on the x-axis? (0, 1) (4, 0) (7, 7) (−5, 0) (−4, 4) (0, −5) (8, 0) (6, 0) Answer: Sincethe coordinates lying on x-axis have their y coordinate zero (0), the following points will lie on x-axis: (4, 0) (−5, 0) (8, 0) (6, 0) Question 2: Two different points are to be plotted on a graph. If the given points are (3,2) and (2,3), then plot these two points on the X- and Y-axis. Also, find out the point where the straight line going through these points meets the x-axis. Answer: For (3,2), as we can see, the x-coordinate point is 3, and the y-coordinate point is 2. Similarly we can plot the point (2,3). Now, we can join both points with a straight line when we have plotted both points. After extending the straight line, we see that this line intersects the x-axis at point (5,0). Question 3: For a linear equation y = 2x + 6, find the point where the straight line meets y-axis on the graph. Answer: On y-axis, the x-coordinate of the point is 0. Therefore, we can find the intersection point of y-axis and y = 2x + 6 by simply putting the value of x as 0 and finding the value of y. y = 2(0)+6 = 0 + 6 = 6. So the straight line of the equation y = 2x + 6 meets the y-axis at (0,6). Practice Problems On X and Y Axis X And Y Axis Attend this Quiz & Test your knowledge. 1 What is the x-axis called? Ordinate Abscissa Applicate None of the above CorrectIncorrect Correct answer is: Abscissa x - axis is also called abscissa. 2 What is the correct way of representing a point on a graph? (X-coordinate, X-coordinate) (Y-coordinate, X-coordinate) (Y-coordinate, Y-coordinate) (X-coordinate, Y-coordinate) CorrectIncorrect Correct answer is: (X-coordinate, Y-coordinate) (X-coordinate, Y-coordinate) is the correct way to represent a point. 3 How is the origin point represented on a graph? (0,0) (0,x) (y,0) (x,y) CorrectIncorrect Correct answer is: (0,0) (0, 0) is the coordinates of origin on the graph. 4 A point (0,5) will be lie on the None of the above CorrectIncorrect Correct answer is: Y-axis As abscissa (x- coordinate) is 0. So, given point lies on the y-axis. Frequently Asked Questions On X and Y Axis Why are the X- and Y-axis important? How is the coordinate plane formed? What are quadrants in a graph? How are the X- and Y-axis different?
15638
https://math.stackexchange.com/questions/278923/is-f-surjective-from-mathbbr-rightarrow-mathbbr?lq=1
real analysis - is $f$ surjective from $\mathbb{R}\rightarrow\mathbb{R}$ - Mathematics Stack Exchange Join Mathematics 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 Mathematics helpchat Mathematics 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 Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams 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 is f f surjective from R→R R→R [duplicate] Ask Question Asked 12 years, 8 months ago Modified12 years, 8 months ago Viewed 439 times This question shows research effort; it is useful and clear 2 Save this question. Show activity on this post. This question already has answers here: Closed 12 years ago. Possible Duplicate: how to prove, f f is onto if f f is continuous and satisfying |f(x)−f(y)|≥|x−y||f(x)−f(y)|≥|x−y| for all x,y x,y in R R we are given that f f is continous on reals, and |f(x)−f(y)|≥|x−y|∀x,y|f(x)−f(y)|≥|x−y|∀x,y, we need to find out: is f f surjective from R→R R→R? I think the statement is false, is it not violating definition of continuity itself?I have picked it from a question paper. thank you. real-analysis Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Apr 13, 2017 at 12:20 CommunityBot 1 asked Jan 14, 2013 at 22:51 MyshkinMyshkin 37k 28 28 gold badges 172 172 silver badges 349 349 bronze badges 2 4 It's surely not violating the definition of continuity since f(x)=x f(x)=x satisfies exactly that condition.JSchlather –JSchlather 2013-01-14 22:55:08 +00:00 Commented Jan 14, 2013 at 22:55 If it violated the definition of continuity, then the statement would be neither false nor true: it would be ill-defined.user14972 –user14972 2013-01-15 01:35:00 +00:00 Commented Jan 15, 2013 at 1:35 Add a comment| 3 Answers 3 Sorted by: Reset to default This answer is useful 10 Save this answer. Show activity on this post. Assume f f is bounded from above, M:=sup f<∞M:=sup f<∞. Select x 0 x 0 with f(x 0)>M−1 f(x 0)>M−1. Then |f(x 0±1)−f(x 0)|≥1|f(x 0±1)−f(x 0)|≥1 implies f(x 0±1)≤f(x 0)−1 f(x 0±1)≤f(x 0)−1 as we cannot have values ≥f(x 0)+1≥f(x 0)+1. By the intermediate value theorem, there is x 1∈(x 0−1,x 0)x 1∈(x 0−1,x 0) with f(x 1)=f(x 0)−1 2 f(x 1)=f(x 0)−1 2. Similarly, there is x 2∈(x 0,x 0+1)x 2∈(x 0,x 0+1) with f(x 2)=f(x 0)−1 2 f(x 2)=f(x 0)−1 2. But then 0=|f(x 1)−f(x 2)|≥|x 1−x 2|>0 0=|f(x 1)−f(x 2)|≥|x 1−x 2|>0, contradiction. We conclude that f f is not bounded from above. Similarly, f f is not bounded from below. Now given y∈R y∈R, we therefore find x 3 x 3 with f(x 3)y f(x 4)>y and by the intermediate value theorem again, some x 5 x 5 with f(x 5)=y f(x 5)=y. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 14, 2013 at 23:20 Hagen von EitzenHagen von Eitzen 384k 33 33 gold badges 379 379 silver badges 686 686 bronze badges Add a comment| This answer is useful 5 Save this answer. Show activity on this post. f f is 1-1 as x≠y x≠y implies |f(x)−f(y)|≥|x−y|>0.|f(x)−f(y)|≥|x−y|>0. Any 1-1 continuous function from R to R is strictly monotonic (). If f f is strictly increasing then for any x,y x,y with x>y x>y, f(x)≥f(y)+(x−y)f(x)≥f(y)+(x−y), keeping y y fixed and letting x x tend into ∞∞, we have lim x→∞f(x)=∞lim x→∞f(x)=∞, and keeping x x fixed and letting y y tend to −∞−∞ we get lim y→−∞f(y)=−∞.lim y→−∞f(y)=−∞. Hence f f is surjective. For strictly decreasing f f redo the argument with −f−f instead of f f. () For completeness: Let f f mapping reals to reals be continuous and 1-1, then g(x,y)=(x−y)(f(x)−f(y))g(x,y)=(x−y)(f(x)−f(y)) maps the connected set {(x,y):x>y}{(x,y):x>y} to a connected subset of R−{0}R−{0}, hence the image of g lies in (0,∞)(0,∞) in which case f is strictly increasing, or it lies (−∞,0)(−∞,0) in which case f f is strictly decreasing. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Jan 15, 2013 at 1:20 answered Jan 15, 2013 at 1:08 Arin ChaudhuriArin Chaudhuri 6,569 2 2 gold badges 29 29 silver badges 46 46 bronze badges 1 I love your proof that 1-1 and continuous implies monotonic! :)Bruno Joyal –Bruno Joyal 2013-02-19 07:19:22 +00:00 Commented Feb 19, 2013 at 7:19 Add a comment| This answer is useful 1 Save this answer. Show activity on this post. I claim that either f(y)−f(x)≤−(y−x)f(y)−f(x)≤−(y−x) for all y>x y>x or f(y)−f(x)≥(y−x)f(y)−f(x)≥(y−x) for all y>x y>x. For suppose not. Then you could find x 1<y 1 x 1<y 1 with f(y 1)−f(x 1)≤−(y 1−x 1)<0 f(y 1)−f(x 1)≤−(y 1−x 1)<0 and x 20 f(y 2)−f(x 2)≥(y 2−x 2)>0. So by the intermediate value theorem on t∈[0,1]t∈[0,1] there would be some t∈(0,1)t∈(0,1) such that f(t y 1+(1−t)y 2)−f(t x 1+(1−t)x 2)=0 f(t y 1+(1−t)y 2)−f(t x 1+(1−t)x 2)=0. Since t y 1+(1−t)y 2>t x 1+(1−t)x 2 t y 1+(1−t)y 2>t x 1+(1−t)x 2 this would contradict the condition you're given. So replacing f f by −f−f if necessary assume that f(y)−f(x)≥(y−x)f(y)−f(x)≥(y−x) for all x0 y>0, so lim y→∞f(y)=∞lim y→∞f(y)=∞. Similarly f(y)≤−y+f(0)f(y)≤−y+f(0) for all y<0 y<0, so lim y→−∞f(y)=−∞lim y→−∞f(y)=−∞. Since f f is continuous the intermediate value theorem tells you the range of f f is all of R R Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Jan 14, 2013 at 23:34 ZarraxZarrax 46.7k 2 2 gold badges 71 71 silver badges 128 128 bronze badges Add a comment| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions real-analysis 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 Report this ad Linked 11how to prove, f f is onto if f f is continuous and satisfying |f(x)−f(y)|≥|x−y||f(x)−f(y)|≥|x−y| for all x,y x,y in R R Related 7Lie group homomorphism from R→S 1 R→S 1 1Prove that f:[a,b]→R f:[a,b]→R is a bijection from [a,b][a,b] to [f(a),f(b)][f(a),f(b)] 0show that f n→f f n→f uniformly on R R 1Prove f is continuous if f n→f f n→f is unif continuous? 4[If f:[0,1)→R f:0,1)→R is uniformly continous, f f is bounded 4How to demonstrate that a differentiable function f:Q→Q f:Q→Q is not uniformly continuous 2Prove that the level sets of a continous function f:S∈R m→R f:S∈R m→R are closed in S 0Existence of Continuous Surjective map from [0,π]∪[2 π,3 π][0,π]∪[2 π,3 π] to [0,1][0,1] 6No continuous injective functions from R 2 R 2 to R R 1Surjective function from open interval to closed interval Hot Network Questions Cannot build the font table of Miama via nfssfont.tex Childhood book with a girl obsessed with homonyms who adopts a stray dog but gives it back to its owners Vanishing ext groups of sheaves with disjoint support The geologic realities of a massive well out at Sea How do you emphasize the verb "to be" with do/does? What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? ConTeXt: Unnecessary space in \setupheadertext Xubuntu 24.04 - Libreoffice Passengers on a flight vote on the destination, "It's democracy!" Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations? Do we need the author's permission for reference How many color maps are there in PBR texturing besides Color Map, Roughness Map, Displacement Map, and Ambient Occlusion Map in Blender? What were "milk bars" in 1920s Japan? Origin of Australian slang exclamation "struth" meaning greatly surprised Is direct sum of finite spectra cancellative? Where is the first repetition in the cumulative hierarchy up to elementary equivalence? Bypassing C64's PETSCII to screen code mapping Who is the target audience of Netanyahu's speech at the United Nations? Gluteus medius inactivity while riding How to fix my object in animation The rule of necessitation seems utterly unreasonable An odd question Discussing strategy reduces winning chances of everyone! Fundamentally Speaking, is Western Mindfulness a Zazen or Insight Meditation Based Practice? Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
15639
https://math.stackexchange.com/questions/1996680/finding-the-values-of-m-using-roots
finding the values of m, using roots - Mathematics Stack Exchange Join Mathematics 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 Mathematics helpchat Mathematics 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 Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams 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 finding the values of m, using roots Ask Question Asked 8 years, 11 months ago Modified8 years, 11 months ago Viewed 1k times This question shows research effort; it is useful and clear 0 Save this question. Show activity on this post. Consider the equation 4 x 2−m x−m=0 4 x 2−m x−m=0 (with unknown x). Find the values of m such that the equation has (a) one (double) root; (b) two distinct (real) roots; (c) no real roots. My working out 4 x 2−m x−m=0 4 x 2−m x−m=0 Apply the quadratic formula gives m+/−√m 2+4 m−16/2(4)m+/−m 2+4 m−16−−−−−−−−−−−−√/2(4) a) m 2+4 m−16=0 m 2+4 m−16=0 Through completing the square I get: (m+2)2−20=0(m+2)2−20=0 m=−2 m=−2 b)m 2+4 m−16>0 m 2+4 m−16>0 (m+2)2−20>0(m+2)2−20>0 m+2>0 m+2>0 I can't seem to find more solutions c) m 2+4 m−16<0 m 2+4 m−16<0 roots Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Nov 2, 2016 at 20:56 MaggieMaggie 51 3 3 silver badges 10 10 bronze badges Add a comment| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. HINT: Consider the discriminant D=b 2−4 a c D=b 2−4 a c. Then a x 2+b x+c=0 a x 2+b x+c=0 has two different real solution iff D>0 D>0, double root iff D=0 D=0 and complex roots iff D<0 D<0 The solutions of inequalities are wrong. Try to solve m 2+4 m−16=0 m 2+4 m−16=0 and write m 2+4 m−16=(m−m 1)(m−m 2)m 2+4 m−16=(m−m 1)(m−m 2), where m 1,m 2 m 1,m 2 are the solutions of the equation. Then we have that: m 2+4 m−16>0⟺(m−m 1)>0 and(m−m 2)>0 or(m−m 1)<0 and(m−m 2)<0 m 2+4 m−16>0⟺(m−m 1)>0 and(m−m 2)>0 or(m−m 1)<0 and(m−m 2)<0 m 2+4 m−16=0⟺(m−m 1)=0 or(m−m 2)=0 m 2+4 m−16=0⟺(m−m 1)=0 or(m−m 2)=0 m 2+4 m−16<0⟺(m−m 1)<0 and(m−m 2)>0 or(m−m 1)>0 and(m−m 2)<0 m 2+4 m−16<0⟺(m−m 1)<0 and(m−m 2)>0 or(m−m 1)>0 and(m−m 2)<0 Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Nov 2, 2016 at 21:26 Bill Dubuque 284k 42 42 gold badges 339 339 silver badges 1k 1k bronze badges answered Nov 2, 2016 at 20:57 Stefan4024Stefan4024 36.5k 9 9 gold badges 54 54 silver badges 103 103 bronze badges 7 That makes sense however the equation cannot be factorise easily and requires using completing the square for m 2+4 m−16=(m−m 1)(m−m 2)m 2+4 m−16=(m−m 1)(m−m 2) how would you find the solutions with completing the square Maggie –Maggie 2016-11-02 21:06:15 +00:00 Commented Nov 2, 2016 at 21:06 @Maggie Use the quadratic formula to get that m 1=−2+√20 m 1=−2+20−−√ and m 2=−2−√20 m 2=−2−20−−√Stefan4024 –Stefan4024 2016-11-02 21:07:40 +00:00 Commented Nov 2, 2016 at 21:07 @Maggie By completing the square we have: 0=m 2+4 m−16=(m+2)2−20⟺(m+2)2=20 0=m 2+4 m−16=(m+2)2−20⟺(m+2)2=20 ⟺m+2=±√20⟺m=−2±√20⟺m+2=±20−−√⟺m=−2±20−−√ Stefan4024 –Stefan4024 2016-11-02 21:07:56 +00:00 Commented Nov 2, 2016 at 21:07 That's the discriminant, not determinant.Bill Dubuque –Bill Dubuque 2016-11-02 21:18:12 +00:00 Commented Nov 2, 2016 at 21:18 I am slightly confused so if it is one(double) root b 2−4 a c=0 b 2−4 a c=0, two distinct (real) roots b 2−4 a c>0 b 2−4 a c>0 and no real roots (complex) b 2−4 a c<0 b 2−4 a c<0 Maggie –Maggie 2016-11-02 21:18:45 +00:00 Commented Nov 2, 2016 at 21:18 |Show 2 more comments 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 roots 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 Report this ad Related 4Finding roots of quadratic polynomials, negative square root 9Finding the discriminant and roots of a polynomial 2Roots of quadratic equation by completing the square or other method? 2The number of distinct real roots of a polynomial of degree 4 17Is there anything like “cubic formula”? 3Roots of complex quadratic polynomial 1Find a formula that generates the number of possible combinations of roots for a given polynomial Hot Network Questions Sign mismatch in overlap integral matrix elements of contracted GTFs between my code and Gaussian16 results Bypassing C64's PETSCII to screen code mapping Implications of using a stream cipher as KDF Drawing the structure of a matrix My dissertation is wrong, but I already defended. How to remedy? Is existence always locational? How to home-make rubber feet stoppers for table legs? How to start explorer with C: drive selected and shown in folder list? What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? I have a lot of PTO to take, which will make the deadline impossible Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? Origin of Australian slang exclamation "struth" meaning greatly surprised What happens if you miss cruise ship deadline at private island? Do we need the author's permission for reference Should I let a player go because of their inability to handle setbacks? Can you formalize the definition of infinitely divisible in FOL? The rule of necessitation seems utterly unreasonable If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church? Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward" How to locate a leak in an irrigation system? Can I go in the edit mode and by pressing A select all, then press U for Smart UV Project for that table, After PBR texturing is done? Checking model assumptions at cluster level vs global level? Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? Does the curvature engine's wake really last forever? Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
15640
https://11848055.s21i.faiusr.com/61/ABUIABA9GAAggeOZ8wUohJSrmQU.pdf
1 例题1 7 个人排成一排． （1）甲、乙、丙排在一起，共有多少种排法？ （2）甲、乙相邻，且丙、丁相邻，有多少种排法？ （3）甲、乙、丙排在一起，且都不在两端，有多少种排法？ （4）甲、乙、丙排在一起，且甲在两端，有多少种排法？ （5）甲、乙之间恰有2 人的排法有多少种？ （6）甲、乙之间是丙的排法有多少种？ 例题2 7 个人排成一排． （1）甲、乙、丙互不相邻，共有多少种排法？ （2）甲、乙相邻，丙、丁不相邻有多少种排法 （3）甲不与乙相邻，丙不与乙相邻，有多少种排法？ 例题3 用1,2,3,4,5,6 组成的没有重复数字的6 位偶数中，1 与3 都不与5 相邻的有（ ）个． A．72 B．96 C．108 D．144 常见排列问题 2 【练1】 一排8 个车位，停5 辆不同车，每车位至多停一车． （1）停车的5 个车位相邻有多少种停法？ （2）不停车的3 个空位相邻有多少种停法？ （3）一共有多少种停法？ 【练2】 某会议室第一排共有8 个座位，现有3 人就座，若要求每人左右均有空位，那么不同的坐法种数 为（ ） A．12 B．16 C．24 D．32 【练3】 由0,1,2,3,4,5 这6 个数字可组成 个无重复数字且2,3 不相邻的六位数（用数字作答） ． 例题4 4 男3 女坐成一排，且4 男不等高，4 男自左向右按从高到矮的顺序排列，有多少种不同的排法？ 例题5 8 人排两排，前后两排各4 人，组成 的方阵． （1）甲乙不同排有多少种排法？ （2）甲乙同排有多少种排法？ （3）甲乙同排相邻或前后相邻有多少种排法？ （4）甲乙不在两端有多少种排法？ （5）任意排有多少种排法？ 例题6 3 个“1” ，2 个“2” ，1 个“3” ，排成一行，共有 种不同排法（用数字作答） ． 2 4 ´
15641
https://artofproblemsolving.com/wiki/index.php/Root-Mean_Square-Arithmetic_Mean-Geometric_Mean-Harmonic_mean_Inequality?srsltid=AfmBOopmOyasMG-kHITKtrsbN44COaWETcEiLN0lchGaSKZTXY6DvQ1x
Art of Problem Solving Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality - 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 Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality The Root-Mean Power-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (RMP-AM-GM-HM) or Exponential Mean-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (EM-AM-GM-HM) or Quadratic Mean-Arithmetic Mean-Geometric Mean-Harmonic Mean Inequality (QM-AM-GM-HM), is an inequality of the root-mean power, arithmetic mean, geometric mean, and harmonic mean of a set of positivereal numbers that says: , where , and is the . The geometric mean is the theoretical existence if the root mean power equals 0, which we couldn't calculate using radicals because the 0th root of any number is undefined when the number's absolute value is greater than or equal to 1. This creates the indeterminate form of . Then, we can say that the limit as x goes to 0 is the geometric mean of the numbers. The quadratic mean's root mean power is 2 and the arithmetic mean's root mean power is 1, as and the harmonic mean's root mean power is -1 as . Similarly, there is a root mean cube (or cubic mean), whose root mean power equals 3. When the root mean power approaches , the mean approaches the highest number. When the root mean power reaches , the mean approaches the lowest number. with equality if and only if . This inequality can be expanded to the power mean inequality, and is also known as the Mean Inequality Chain. As a consequence, we can have the following inequality: If are positive reals, then with equality if and only if ; which follows directly by cross multiplication from the AM-HM inequality. This is extremely useful in problem-solving. The Root Mean Power of 2 is also known as the quadratic mean, and the inequality is therefore sometimes known as the QM-AM-GM-HM Inequality. Proof The inequality is a direct consequence of the Cauchy-Schwarz Inequality; Alternatively, the RMS-AM can be proved using Jensen's inequality: Suppose we let (We know that is convex because and therefore ). We have: Factoring out the yields: Taking the square root to both sides (remember that both are positive): The inequality is called the AM-GM inequality, and proofs can be found here. The inequality is a direct consequence of AM-GM; , so , so . Therefore, the original inequality is true. Geometric Proofs The inequality is clearly shown in this diagram for Desmos SlidersDesmos Equation NOTE: The Desmos equation will not show the line when the numbers are negative. (Note how the RMS is "sandwiched" between the minimum and the maximum) Retrieved from " Categories: Algebra Inequalities 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.
15642
https://www.chegg.com/homework-help/questions-and-answers/q2-9-seats-round-table-mr-sit-opposite-stage-assistant-ms-b-must-sit-left-investors-mr-mrs-q100265420
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: Q2. There are 9 seats in a round table. Mr. A has to sit opposite the stage. His assistant Ms. B must sit on his left. The investors Mr. and Mrs. C have to sit together. How many ways are there to arrange the sitting ? Suggest at least 3 ways in linear form (suppose other guests are IU, NSW, UWE, IE, SCM) Since Mr. A has to sit opposite the stage, his position is fixed. That means the position of Ms. B is al… 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
15643
https://stackoverflow.com/questions/871448/how-many-ways-are-there-to-see-if-a-number-is-even-and-which-one-is-the-fastest
Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Asked Modified 3 years, 5 months ago Viewed 2k times This question shows research effort; it is useful and clear Save this question. Show activity on this post. 1 7 Well... you could make a list of every single even number, and then go through the list, checking if your number is in it! :p – DeadHead Commented May 16, 2009 at 1:53 Add a comment | 21 Answers 21 Reset to default This answer is useful 34 Save this answer. Show activity on this post. ``` bool isEven = ((number & 0x01) == 0) ``` The question said "any number", so one could either discard floats or handle them in another manner, perhaps by first scaling them up to an integral value first - watching out for overflow - i.e. change 2.1 to 21 (multiply by 10 and convert to int) and then test. It may be reasonable to assume, however, that by mentioning "any number" the person who posed the question is actually referring to integral values. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications edited May 26, 2009 at 4:54 Chris Lutz 75.8k1616 gold badges132132 silver badges184184 bronze badges answered May 16, 2009 at 1:45 DemiDemi 6,23777 gold badges3838 silver badges3838 bronze badges 6 In the selection between speed and clarity, especially in this unbounded case, I prefer speed. – Demi Commented May 16, 2009 at 1:59 6 @Shane: In all seriousness, I have seen (and used) the & operator far more often than I have the % operator (assuming C/C++/Java - like languages here :-) For myself, (x & 1 == 0) is actually clearer than (x % 2 == 0). This is one reason why I said it's an "odd" question; there's no 'clearest' for something as simple as this. – Dan Breslau Commented May 16, 2009 at 2:00 This is perfectly clear to me also - it should be clear from context what the code is doing - "bool isEven = " is pretty clear. And of course, you can always add a comment. – Blorgbeard Commented May 16, 2009 at 2:14 3 template inline bool is_even( T number ) { return (number & 0x01) == 0; } a little clarity never hurt anyone – deft_code Commented May 16, 2009 at 2:25 1 Agreed, this is very clear. When I saw this question with 8 answers I was anticipating this answer. As programmers, we should be able to interpret this sort of thing. We do want to make money. – Mr. Shickadance Commented May 16, 2009 at 2:35 | Show 1 more comment This answer is useful 28 Save this answer. Show activity on this post. ``` bool isEven = number % 2 == 0; ``` Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered May 16, 2009 at 1:42 sipsorcerysipsorcery 30.8k2525 gold badges108108 silver badges158158 bronze badges 9 7 modulo is slow. Use bit and. If the low bit is 1 then it is odd. If the low bit is 0 it is even. – Demi Commented May 16, 2009 at 1:48 3 A good compiler will convert % 2 to & 1 automatically. gcc will do this even if optimizations are turned off. – bdonlan Commented May 16, 2009 at 2:10 1 The compiler will optimize number % 2 into number & 2, so performance-wise it doesn't matter which you pick. I think mod is a lot more appropriate than bitwise-and for even/odd. – Craig Gidney Commented May 16, 2009 at 2:10 4 Some compilers do this optimization, some don't. Don't trust your compiler when you can ensure this optimization yourself. If the number was the maximum integer value, the difference is speed can be quite significant. – Demi Commented May 16, 2009 at 2:16 7 Relying on a compiler optimization is bad practice. Modulo implies a division, which is mathematically more expensive. You should actually understand what your code is doing and write it in such a way that it clearly explains to any reader what you intended. – Jay Commented May 16, 2009 at 4:26 | Show 4 more comments This answer is useful 10 Save this answer. Show activity on this post. ``` isEven(n) = ((-1) ^ n) == 1 ``` where ^ is the exponentiation/pow function of your language. I didn't say it was fast or clear, but it has novelty value. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered May 16, 2009 at 2:07 OddthinkingOddthinking 25.5k2020 gold badges8585 silver badges125125 bronze badges 1 +1 for a unique solution, although I don't know how practical it is. – Sasha Chedygov Commented May 16, 2009 at 4:48 Add a comment | This answer is useful 10 Save this answer. Show activity on this post. The answer depends on the position being applied for. If you're applying for an Enterprise Architect position, then the following may be suitable: First, you should create a proper Service-Oriented Architecture, as certainly the even-odd service won't be the only reusable component in your enterprise. An SOA consists of a service, interface, and service consumers. The service is function which can be invoked over the network. It exposes an interface contract and is typically registered with a Directory Service. You can then create a Simple Object Access Protocol (SOAP) HTTP Web Service to expose your service. Next, you should prevent clients from directly calling your Web Service. If you allow this, then you will end up with a mess of point-to-point communication, which is very hard to maintain. Clients should access the Web Service through an Enterprise Service Bus (ESB). In addition to providing a standard plug-able architecture, additional components like service orchestration can occur on the bus. Generally, writing a bespoke even/odd service should be avoided. You should write a Request for proposal (RFP), and get several vendors to show you their even/odd service. The vendor's product should be able to plug into your ESB, and also provide you with an Service level agreement (SLA). Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Jun 17, 2009 at 6:30 brianeggebrianegge 30k1313 gold badges7676 silver badges103103 bronze badges 1 sorry, you've gone too far :) (I can hear the architects now... "you had me at SOA" ... ) – Jeffrey Kemp Commented Jun 17, 2009 at 12:06 Add a comment | This answer is useful 5 Save this answer. Show activity on this post. This is even easier in ruby: ``` isEven = number.even? ``` Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered May 16, 2009 at 1:52 BlorgbeardBlorgbeard 104k5050 gold badges237237 silver badges276276 bronze badges 5 only Numeric-based (Numeric, Fixnum, Bignum) types. Not Decimal types – Demi Commented May 16, 2009 at 2:10 Wonder if a function lookup and call is faster on ruby than a logical and? – zaratustra Commented May 26, 2009 at 5:11 I doubt it, but I can't test this, because I get "-:1: undefined method `even?' for 1:Fixnum (NoMethodError)" when I try to run anything resembling this code. – Chris Lutz Commented May 26, 2009 at 5:20 For what it's worth (this is just guesswork), regardless of function lookup and call times, the function is probably implemented using the logical and trick (or with the modulus trick), so I seriously doubt it will be faster than either one of those. – Chris Lutz Commented May 26, 2009 at 5:24 I didn't say it was faster :P It works here: tryruby.hobix.com - try 1.even? – Blorgbeard Commented May 26, 2009 at 9:53 Add a comment | This answer is useful 4 Save this answer. Show activity on this post. Yes.. The fastest way is to check the 1 bit, because it is set for all odd numbers and unset for all even numbers.. Bitwise ANDs are pretty fast. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered May 16, 2009 at 1:48 John ChristmanJohn Christman 60355 silver badges1111 bronze badges Add a comment | This answer is useful 4 Save this answer. Show activity on this post. If your type 'a' is an integral type, then we can define, `` even :: Integral a => a -> Bool even n = nrem` 2 == 0 ``` according to the Haskell Prelude. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered May 16, 2009 at 1:59 Don StewartDon Stewart 138k3636 gold badges372372 silver badges471471 bronze badges 1 this is equivalent to n % 2 in other languages isn't it? – David Johnstone Commented Jun 17, 2009 at 5:59 Add a comment | This answer is useful 4 Save this answer. Show activity on this post. For floating points, of course within a reasonable bound. ``` modf(n/2.0, &intpart, &fracpart) return fracpart == 0.0 ``` With some other random math functions: ``` return gcd(n,2) == 2 return lcm(n,2) == n return cos(npi) == 1.0 ``` Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications edited May 16, 2009 at 3:07 answered May 16, 2009 at 2:39 UnknownUnknown 46.9k2929 gold badges142142 silver badges183183 bronze badges Add a comment | This answer is useful 4 Save this answer. Show activity on this post. If int is 32 bits then you could do this: ``` bool is_even = ((number << 31) >> 31) == 0; ``` With using bit shifts you'll shift the right-most bit to the left-most position and then back again, thus making all other bits 0's. Then the number you're left with is either 0 or 1. This method is somewhat similar to "number & 1" method where you again make all bits 0's except the first one. Another approach, similar to this one is this: ``` bool is_even = (number << 31) == 0; ``` or ``` bool is_odd = (number << 31) < 0; ``` If the number is even (the right-most bit is 0), then shifting it 31 positions will make the whole number 0. If the bit is 1, i.e. the number is odd, then the resulting number would be negative (every integer with left-most bit 1 is negative except if the number is of type unsigned, where it won't work). To fix signed/unsigned bug, you can just test: ``` bool is_odd = (number << 31) != 0; ``` Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Nov 16, 2012 at 18:21 KatieK 13.9k1919 gold badges7979 silver badges9191 bronze badges answered Nov 16, 2012 at 18:02 evgenidbevgenidb 4111 bronze badge Add a comment | This answer is useful 3 Save this answer. Show activity on this post. Actually I think (n % 2 == 0) is enough, which is easy to understand and most compilers will convert it to bit operations as well. I compiled this program with gcc -O2 flag: ``` include int main() { volatile int x = 310; printf("%d\n", x % 2); return 0; } ``` and the generated assembly code is ``` main: pushl %ebp movl %esp, %ebp andl $-16, %esp subl $32, %esp movl $310, 28(%esp) movl 28(%esp), %eax movl $.LC0, (%esp) movl %eax, %edx shrl $31, %edx addl %edx, %eax andl $1, %eax subl %edx, %eax movl %eax, 4(%esp) call printf xorl %eax, %eax leave ret ``` which we can see that % 2 operation is already converted to the andl instruction. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered May 26, 2009 at 4:44 ZelluXZelluX 73.3k2020 gold badges7676 silver badges106106 bronze badges 1 I hate to beat a dead horse, but whether or not the optimization is done depends on what compiler you use. If yours does it, that's awesome. But what if mine doesn't? – Chris Lutz Commented May 26, 2009 at 4:58 Add a comment | This answer is useful 3 Save this answer. Show activity on this post. Similar to DeadHead's comment, but more efficient: ``` include bool isEven(int num) { bool arr[UINT_MAX] = { 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, // ...and so on }; return arr[num]; } ``` As fast as an array index, which may or may not be faster than bitwise computations (it's difficult to test because I don't want to write the full version of this function). For what it's worth, that function above only has enough filled in to find even numbers up to 442, but would have to go to 4294967295 to work on my system. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered May 26, 2009 at 5:09 Chris LutzChris Lutz 75.8k1616 gold badges132132 silver badges184184 bronze badges 1 the memory requirements for a version that works on INTEGER_MAX is going to cause most systems to grind to a halt... – Chii Commented Sep 21, 2009 at 13:10 Add a comment | This answer is useful 2 Save this answer. Show activity on this post. With reservations for limited stack space. ;) (Is this perhaps a candidate for tail calls?) ``` public static bool IsEven(int num) { if (num < 0) return !IsEven(-num - 1); if (num == 0) return true; return IsEven(-num); } ``` Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications edited May 17, 2009 at 7:48 answered May 16, 2009 at 5:25 sisvesisve 19.8k33 gold badges5555 silver badges9696 bronze badges 3 you are clearly missing a case. How about including: if (num == 1) return false; – Demi Commented May 16, 2009 at 7:03 1 @Demi, IsEven(1) would return !IsEven(1-1) = !IsEven(0) = !true = false. – sisve Commented May 16, 2009 at 8:26 I don't think tail calls will work b/c the return values of IsEven need to be inverted (by the ! on the 2nd line). But I likes the solution. – Trey Jackson Commented Jun 2, 2009 at 16:54 Add a comment | This answer is useful 2 Save this answer. Show activity on this post. a % 2. It's clear It's fast on every decent compiler. Everyone who cries "But! But! What if compiler doesn't optimize it" should find normal compiler, shut up and read about premature optimization, read again, read again. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications edited Jun 17, 2009 at 6:02 Dominic Rodger 100k3737 gold badges203203 silver badges218218 bronze badges answered Jun 17, 2009 at 5:46 anonymousanonymous Add a comment | This answer is useful 1 Save this answer. Show activity on this post. If it's low level check if the last (LSB) bit is 0 or 1 :) 0 = Even 1 = Odd Otherwise, +1 @sipwiz: "bool isEven = number % 2 == 0;" Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered May 16, 2009 at 3:57 WhyNotHugoWhyNotHugo 9,94366 gold badges6767 silver badges7373 bronze badges Add a comment | This answer is useful 1 Save this answer. Show activity on this post. Assumming that you are dealing with an integer, the following will work: if ((testnumber & -2)==testnumber) then testnumber is even. basically, -2 in hex will be FFFE (for 16 bits) if the number is even, then anding with with -2 will leave it unchanged. Tom Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered May 16, 2009 at 4:36 TomTom 78611 gold badge55 silver badges1212 bronze badges 2 representation of negative integers can't be trusted inherently. One should refer to the language they are using. Your answer assumes "two's complement" representation. In "one's complement", -2 in hex is 0xFFFD, which would not give the correct answer. Another representation is "sign and magnitude", which, for 16 bits, would give 0x8002 - also an incorrect answer. Negative values are not as they would seem, oftentimes. – Demi Commented May 16, 2009 at 4:56 Demi, your points are well taken, but the most common storage system is still 2's complement. The programmer in question is always responsible to know how data is stored in their environment. – Tom Commented May 17, 2009 at 17:03 Add a comment | This answer is useful 0 Save this answer. Show activity on this post. You can either using integer division and divide it by two and inspect the remainder or use a modulus operator and mod it by two and inspect the remainder. The "fastest" way depends on the language, compiler, and other factors but I doubt there are many platforms for which there is a significant difference. Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered May 16, 2009 at 1:43 BobbyShaftoeBobbyShaftoe 28.5k77 gold badges5555 silver badges7575 bronze badges 2 Many languages don't give you the remainder when doing integer division; I doubt this will ever be faster than straight modulo, as the compiler can always 'optimize' modulo to division-with-remainder – bdonlan Commented May 16, 2009 at 2:11 Right, that's why I said it depends on the language. – BobbyShaftoe Commented May 16, 2009 at 4:41 Add a comment | This answer is useful 0 Save this answer. Show activity on this post. Recursion! ``` function is_even (n number) returns boolean is if n = 0 then return true elsif n = 1 then return false elsif n < 0 then return is_even(n -1) else return is_even(n - 2) end if end ``` Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Jun 17, 2009 at 5:54 Jeffrey KempJeffrey Kemp 60.4k1515 gold badges111111 silver badges161161 bronze badges 2 @Jeffrey - I guess you'd have to first define what made a non-integral number "even". – Dominic Rodger Commented Jun 17, 2009 at 6:04 1 I'd insert an additional "elsif" just before the final "else": "elsif n < 1 then return FILE_NOT_FOUND" :) – Jeffrey Kemp Commented Jun 17, 2009 at 6:20 Add a comment | This answer is useful 0 Save this answer. Show activity on this post. Continuing the spirit of "how many ways are there...": ``` function is_even (n positive_integer) returns boolean is i := 0 j := 0 loop if n = i then return (j = 0) end if; i := i + 1 j := 1 - j end loop end ``` Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Jun 17, 2009 at 6:27 Jeffrey KempJeffrey Kemp 60.4k1515 gold badges111111 silver badges161161 bronze badges Add a comment | This answer is useful 0 Save this answer. Show activity on this post. In response to Chris Lutz, an array lookup is significantly slower than a BITWISE_AND operation. In an array lookup you're doing a memory lookup which will always be slower than a bitwise operation because of memory latency. This of course doesn't even factor in the problem of putting all possible int values into your array which has a memory complexity of O(2^n) where n is your bus size (8,16,32,64). The odd/even property is only defined in integers. So any answer dealing with floating point is invalid. The abstract representation of this problem is Int -> bool (to use Haskell notation). Share CC BY-SA 2.5 Improve this answer Follow this answer to receive notifications answered Sep 21, 2009 at 13:05 dhargadharga 2,21733 gold badges2626 silver badges3434 bronze badges Add a comment | This answer is useful 0 Save this answer. Show activity on this post. Another useless novelty solution: ``` if (2 (n/2) == n) return true; else return false; ``` Only with integers, and it depends on how the langugage handles integer division. n/2 == n/2 if it's even or n/2-.5 if it's odd. So 2(n/2) == n if it's even or n - 1 if it's odd. Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications edited Sep 20, 2011 at 16:55 answered Sep 6, 2011 at 13:59 AlexQueueAlexQueue 6,58155 gold badges3939 silver badges4747 bronze badges Add a comment | This answer is useful 0 Save this answer. Show activity on this post. Here's a recursive way to do it in python: ``` def is_even(n: int) -> bool: if n == 0: return True else: return is_odd(n-1) def is_odd(n: int) -> bool: if n == 0: return False else: return is_even(n-1) ``` Of course, you can add in logic to check if n is negative as well. Share CC BY-SA 4.0 Improve this answer Follow this answer to receive notifications answered Mar 9, 2022 at 23:27 Jac FrallJac Frall 42388 silver badges1515 bronze badges Add a comment | Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions algorithm math See similar questions with these tags. By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy.
15644
https://www.chegg.com/homework-help/questions-and-answers/question-3-refrigerant-r134a-throttled-flowing-insulated-duct-pressure-temperature-upstrea-q74267506
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: Question # 3 Refrigerant R134a is throttled while flowing through an insulated duct. The pressure and temperature upstream of the throttle valve are 200 kPa and 60°C. The pressure downstream of the valve is atmospheric and may be assumed to be 100 kPa. Calculate the temperature drop across the valve. (15 points) If you wish to attain a higher cooling effect This AI-generated tip is based on Chegg's full solution. Sign up to see more! Consult the refrigerant R134a property tables for the temperature corresponding to the downstream pressure. Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & ServicesChegg Products & Services Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
15645
https://www.zippia.com/adjunct-faculty-member-jobs/demographics/
Adjunct Faculty Member Demographics and Statistics : Number Of Adjunct Faculty Members In The US Automatically apply for jobs. Get interviews, not rejections. Your free AI job assistant. Learn more Jobs Career Companies Sign in Sign In Jobs Career Companies Resume Builder Career quiz Career advice Explore majors Questions and answers Interview questions Post job Sign In Explore jobs Jobs near me Remote jobs Full time jobs Part time jobs Entry level jobs Work from home jobs Find specific jobs $15 per hour jobs $20 per hour jobs Hiring immediately jobs High school jobs H1b visa jobs LGBTQ jobs Explore careers Sales Healthcare Business And Financial Architecture And Engineering Computer And Mathematical Explore professions Salaries What they do Certifications Skills Education Demographics Best companies Health care Media Start-up Non-profit Fortune 500 Explore companies Reviews Salaries History Locations CEO and executies Resume Builder Career advice Explore majors Questions and answers Interview questions OverviewJobsSalaryResumeSkillsWhat They DoEducationCertifications Demographics Best CompaniesBest StatesBest CitiesJob OutlookComparePart Time jobs Adjunct faculty member demographics and statistics in the US Adjunct faculty member demographics research summary. Zippia estimates adjunct faculty member demographics and statistics in the United States by using a database of 30 million profiles. Our adjunct faculty member estimates are verified against BLS, Census, and current job openings data for accuracy. Zippia's data science team found the following key facts about adjunct faculty members after extensive research and analysis: There are over 251,096 adjunct faculty members currently employed in the United States. 55.1% of all adjunct faculty members are women, while 44.9% are men. The average adjunct faculty member age is 46 years old. The most common ethnicity of adjunct faculty members is White (66.2%), followed by Asian (11.3%), Hispanic or Latino (10.2%) and Black or African American (7.1%). Adjunct faculty members are most in-demand in Fairmont, WV. Hoboken, NJ pays an annual average wage of $76,844, the highest in the US. In 2022, women adjunct faculty members earned 96% of what men earned. Oregon is the best state for adjunct faculty members to live. Adjunct faculty members are 95% more likely to work at education companies in comparison to private companies. On this page Adjunct faculty member demographics by: Gender Race Age Education Statistics See yourself in this field? Zippi will automatically apply for you. Completely free Adjunct faculty member gender ratio Male - 45% Female - 55% See all gender info Adjunct faculty member racial demographics White - 66.2% Asian - 11.3% See all race info Average adjunct faculty member age 46.1 See all age info Average adjunct faculty member salary $60,817 See all heatmaps Adjunct faculty member gender statistics 55.1% of adjunct faculty members are women and 44.9% of adjunct faculty members are men. Job title Female, 55.1% Male, 44.9% Adjunct faculty member gender ratio | Gender | Percentages | --- | | Female | 55.1% | | Male | 44.9% | Adjunct faculty member gender pay gap Women earn 96¢ for every $1 earned by men Male income $55,981 Female income $53,494 Adjunct faculty member gender ratio over time This data breaks down the percentage of men and women in adjunct faculty member positions over time. Currently, 55.1% of adjunct faculty members are female. Male Female 100% 80% 60% 40% 20% 0% 2010 2011 2012 2013 2014 2015 2016 2017 2018 2019 2020 2021 Adjunct faculty member gender ratio by year | Year | Male | Female | --- | 2010 | 45.56% | 54.44% | | 2011 | 46.09% | 53.91% | | 2012 | 45.77% | 54.23% | | 2013 | 45.33% | 54.67% | | 2014 | 44.22% | 55.78% | | 2015 | 45.05% | 54.95% | | 2016 | 44.71% | 55.29% | | 2017 | 45.30% | 54.70% | | 2018 | 44.72% | 55.28% | | 2019 | 44.30% | 55.70% | | 2020 | 43.66% | 56.34% | | 2021 | 44.94% | 55.06% | Adjunct faculty member related jobs gender ratio We compared adjunct faculty members with similar job titles to see how gender ratios vary. As you can see, hvac/r instructor and assistant professor of air conditioning and refrigeration have the biggest gender ratio gaps. Compare jobs | Job title | Male | Female | --- | Aesthetics Instructor | 8% | 92% | | Cosmetology Instructor | 9% | 91% | | Adjunct Nursing Faculty | 9% | 91% | | Adjunct Faculty Member | 45% | 55% | | Diesel Technology Instructor | 97% | 3% | | Assistant Professor Of Air Conditioning And Refrigeration | 97% | 3% | | Hvac/R Instructor | 97% | 3% | Male Female 100% 75% 50% 25% 0% Hvac/R Instructor Hvac/R Instructor: 97% Hvac/R Instructor: 3% Assistant Professor Of Air Conditioning And Refrigeration Assistant Professor Of Air Conditioning And Refrigeration: 97% Assistant Professor Of Air Conditioning And Refrigeration: 3% Diesel Technology Instructor Diesel Technology Instructor: 97% Diesel Technology Instructor: 3% Adjunct Faculty Member Adjunct Faculty Member: 45% Adjunct Faculty Member: 55% Adjunct Nursing Faculty Adjunct Nursing Faculty: 9% Adjunct Nursing Faculty: 91% Cosmetology Instructor Cosmetology Instructor: 9% Cosmetology Instructor: 91% Aesthetics Instructor Aesthetics Instructor: 8% Aesthetics Instructor: 92% 0% 25% 50% 75% 100% Adjunct faculty member demographics by race The most common ethnicity among adjunct faculty members is White, which makes up 66.2% of all adjunct faculty members. Comparatively, 11.3% of adjunct faculty members are Asian and 10.2% of adjunct faculty members are Hispanic or Latino. Job title White, 66.2% Asian, 11.3% Hispanic or Latino, 10.2% Black or African American, 7.1% Unknown, 4.9% American Indian and Alaska Native, 0.3% Adjunct faculty member race | Adjunct faculty member race | Percentages | --- | | White | 66.2% | | Asian | 11.3% | | Hispanic or Latino | 10.2% | | Black or African American | 7.1% | | Unknown | 4.9% | | American Indian and Alaska Native | 0.3% | Adjunct faculty member race and ethnicity over time See how adjunct faculty member racial and ethnic diversity trended since 2010 according to the United States Census Bureau data. White Asian Hispanic or Latino Black or African American Adjunct faculty member race and ethnicity by year | Year | White | Black or African American | Asian | Hispanic or Latino | --- --- | 2010 | 72.62% | 7.28% | 9.19% | 8.30% | | 2011 | 72.32% | 7.13% | 9.92% | 8.06% | | 2012 | 71.91% | 7.42% | 9.79% | 8.30% | | 2013 | 71.05% | 7.78% | 10.08% | 8.55% | | 2014 | 71.14% | 7.40% | 9.74% | 8.94% | | 2015 | 69.59% | 7.63% | 10.69% | 9.23% | | 2016 | 70.08% | 7.56% | 10.41% | 9.12% | | 2017 | 69.16% | 7.30% | 11.58% | 9.17% | | 2018 | 70.94% | 7.04% | 9.90% | 9.37% | | 2019 | 69.27% | 7.31% | 10.85% | 9.89% | | 2020 | 67.35% | 7.02% | 10.49% | 10.05% | | 2021 | 66.20% | 7.14% | 11.28% | 10.17% | Adjunct faculty member wage gap by race According to our data, asian adjunct faculty members have the highest average salary compared to other ethnicities. Black or african american adjunct faculty members have the lowest average salary at $54,968. $58,000 $57,000 $56,000 $55,000 $54,000 Unknown White Black or African American Asian Hispanic or Latino | Ethnicity | Salary | --- | | Unknown | $55,665 | | White | $55,743 | | Black or African American | $54,968 | | Asian | $57,553 | | Hispanic or Latino | $56,009 | How much do adjunct faculty member make? Adjunct faculty member jobs Work From Home Distance Salary Job type Job level Education Date posted Filters AMN Healthcare Remote or Maywood, IL Radiology - Body Imaging Faculty - Fully Remote Skills: Leadership, Image Scanning, Radiology, Biomedical, Colleges, MRI, Ultrasound, Translational 1d ago new $50k-121k yearly est. Average adjunct faculty member age See how the average age of adjunct faculty members varies based on ethnicity and gender. Male Female Black or African American Unknown Asian Hispanic or Latino White American Indian and Alaska Native 40 44 48 52 Adjunct faculty member age by race and gender | Race | Male age | Female age | --- | Black or African American | 49 | 45 | | Unknown | 44.5 | 44 | | Asian | 41 | 41 | | Hispanic or Latino | 43 | 41 | | White | 48 | 46 | | American Indian and Alaska Native | 54 | 48 | Adjunct faculty member age breakdown The average age of adjunct faculty members is 40+ years years old, representing 67% of the adjunct faculty member population. Compare jobs 40+ years 30-40 years 20-30 years 0% 20% 40% 60% 80% 100% Adjunct Faculty Member: 67% Adjunct Faculty Member: 21% Adjunct Faculty Member: 11% Adjunct faculty member age breakdown | Adjunct Faculty Member Years | Percentages | --- | | 40+ years | 67% | | 30-40 years | 21% | | 20-30 years | 11% | Adjunct faculty member educational attainment The most common degree for adjunct faculty members is bachelor's degree, with 49% of adjunct faculty members earning that degree. The second and third most common degree levels are master's degree degree at 33% and master's degree degree at 9%. Job title Bachelor's, 49% Master's, 33% Doctorate, 9% Associate, 7% Other Degrees, 2% Adjunct faculty members by degree level | Adjunct faculty member degree | Percentages | --- | | Bachelor's | 49% | | Master's | 33% | | Doctorate | 9% | | Associate | 7% | | Other Degrees | 2% | Adjunct faculty member jobs you might like Adjunct faculty member wage gap by degree level $73,000 $70,000 $67,000 $64,000 $61,000 $58,000 $55,000 Master's Degree Doctorate Degree | Education | Salary | --- | | Master's Degree | $57,358 | | Doctorate Degree | $71,332 | Learn more about adjunct faculty member required education Adjunct faculty member employment statistics Most adjunct faculty members work for a education in the education industry. Company size where adjunct faculty members work Below, you can see the size of companies where adjunct faculty members work. Compare jobs < 50 employees 50 - 100 employees 100 - 500 employees 500 - 1,000 employees 1,000 - 10,000 employees 10,000 0% 20% 40% 60% 80% 100% Adjunct Faculty Member: 2% Adjunct Faculty Member: 10% Adjunct Faculty Member: 22% Adjunct Faculty Member: 19% Adjunct Faculty Member: 41% Adjunct Faculty Member: 5% Adjunct faculty member jobs by employer size | Company size | Percentages | --- | | < 50 employees | 2% | | 50 - 100 employees | 10% | | 100 - 500 employees | 22% | | 500 - 1,000 employees | 19% | | 1,000 - 10,000 employees | 41% | | > 10,000 employees | 5% | Adjunct faculty member jobs by company type Employees with the adjunct faculty member job title have their preferences when it comes to working for a company. For instance, most adjunct faculty members prefer to work at education companies over private companies. Compare jobs Education Private Government 0% 20% 40% 60% 80% 100% Adjunct Faculty Member: 95% Adjunct Faculty Member: 3% Adjunct Faculty Member: 1% Adjunct faculty member jobs by sector | Company type | Percentages | --- | | Education | 95% | | Public | 0% | | Private | 3% | | Government | 1% | Adjunct faculty member jobs by industry The most common industries for adjunct faculty members are education, health care and government. Compare jobs Education Health Care Manufacturing Government Show More 0% 20% 40% 60% 80% 100% Adjunct Faculty Member: 93% Adjunct Faculty Member: 2% Adjunct Faculty Member: 1% Adjunct Faculty Member: 1% Adjunct faculty member industry statistics | Industry | Percentages | --- | | Education | 93% | | Health Care | 2% | | Manufacturing | 1% | | Government | 1% | Automatically apply for jobs with Zippi. Upload your resume to get started. Upload your resume Adjunct faculty member turnover and employment statistics Adjunct faculty member unemployment rate over time The unemployment rate for adjunct faculty members between 2008 and the most recent data has varied, according to the Bureau of Labor Statistics. Unemployment rate Adjunct faculty member unemployment rate by year | Year | Adjunct faculty member unemployment rate | --- | | 2010 | 3.13% | | 2011 | 3.11% | | 2012 | 3.14% | | 2013 | 3.31% | | 2014 | 2.86% | | 2015 | 2.61% | | 2016 | 2.42% | | 2017 | 2.51% | | 2018 | 2.40% | | 2019 | 2.23% | | 2020 | 2.58% | | 2021 | 2.98% | Average adjunct faculty member tenure The average adjunct faculty member stays at their job for 1-2 years, based on the 3,107 adjunct faculty members resumes in Zippia's database. Compare jobs Less than one year 1-2 years 3-4 years 5-7 years 8-10 years 11+ years 0% 20% 40% 60% 80% 100% Adjunct Faculty Member: 15% Adjunct Faculty Member: 27% Adjunct Faculty Member: 12% Adjunct Faculty Member: 16% Adjunct Faculty Member: 8% Adjunct Faculty Member: 22% Adjunct faculty member tenure statistics | Number of years | Percentages | --- | | Less than 1 year | 15% | | 1-2 years | 27% | | 3-4 years | 12% | | 5-7 years | 16% | | 8-10 years | 8% | | 11+ years | 22% | Adjunct faculty member LGBT ratio compared to similar jobs Review the percent of adjunct faculty members that are members of the LGBT community. You can see how this compares to similar jobs in the chart below. Compare jobs | Job title | LGBT | Job openings | --- | Graduate Student Assistant | 6.04% | 20,764 | | Music Teacher | 9.52% | 54,429 | | Reference Librarian | 10.75% | 1,058 | | Special Education Teacher For Grades | 11.12% | 64,392 | | Library Technician | 13.69% | 24,941 | | Adjunct Faculty Member | 15.52% | Home Advisor | 16.25% | 77,547 | Home Advisor: 16.25% Home Advisor Adjunct Faculty Member: 15.52% Adjunct Faculty Member Library Technician: 13.69% Library Technician Special Education Teacher For Grades: 11.12% Special Education Teacher For Grades Reference Librarian: 10.75% Reference Librarian Music Teacher: 9.52% Music Teacher Graduate Student Assistant: 6.04% Graduate Student Assistant 0% 4% 8% 12% 16% 20% Similar adjunct faculty member jobs LGBT demographics | Profession | Percentages of LGBT | Job openings | --- | Graduate Student Assistant | 6.04% | 20,764 | | Music Teacher | 9.52% | 54,429 | | Reference Librarian | 10.75% | 1,058 | | Special Education Teacher For Grades | 11.12% | 64,392 | | Library Technician | 13.69% | 24,941 | | Home Advisor | 16.25% | 77,547 | Foreign languages spoken by adjunct faculty members The most common foreign language among adjunct faculty members is Spanish at 40.2%. The second-most popular foreign language spoken is French at 14.5% and Italian is the third-most popular at 7.7%. Job title Spanish, 40.2% French, 14.5% Italian, 7.7% Arabic, 6.0% German, 5.1% Other, 26.5% Adjunct faculty member languages spoken | Foreign language | Percentages | --- | | Spanish | 40.2% | | French | 14.5% | | Italian | 7.7% | | Arabic | 6.0% | | German | 5.1% | | Other | 26.5% | Adjunct faculty member jobs Work From Home Distance Salary Job type Job level Education Date posted Filters Auto-Apply Angelo State University Remote or San Angelo, TX Adjunct (Cybersecurity) Skills: Student Learning, Operation System, Artificial Intelligence, Non-Majors, National Security, Computer Science, Graduate Courses, Software Safety 51d ago $42k-74k yearly est. Looking for a job? Let Zippi find it for you. Auto-apply Auto-Apply Angelo State University Remote or San Angelo, TX Adjunct (Cybersecurity) Skills: Student Learning, Operation System, Artificial Intelligence, Non-Majors, National Security, Computer Science, Graduate Courses, Software Safety 51d ago $42k-74k yearly est. Auto-Apply Goucher College Remote or Baltimore, MD Adjunct Lecturer-Digital Audience Research Skills: Study, Graduate Courses, Analytics, Social Media Sites, Data Visualization, Digital Medium, Syllabus, Data Analysis 16d ago $3.8k-4.3k monthly Auto-Apply P Penn State University Remote or Parkesburg, PA Adjunct Clinical Lecturer, Nursing - Fall 2025 Skills: Statistical Physics, Achievements, Class Content, Technical Service, Social Science, Child Abuse, Business, Business Business 18h ago new $56k-119k yearly est. Auto-Apply University of Michigan-Flint Remote or Flint, MI Adjunct Clinical Lecturer (NUR 910) DNP Scholarly Project (Fall 2025) - School of Nursing - Hybrid Skills: Sustainability, Research Projects, Semester, DNP, Urgent Care, Research, Public Health, Disseminates 25d ago $46k-61k yearly est. Auto-Apply Columbia University In The City of New York Remote or New York, NY Adjunct Lecturer, Technology Management Skills: Problem Solving, Software Engineering, Field, Professional Studies, Science, Student Workshops, Technology Management, Classroom Discussions 19h ago new $10k-15k monthly Auto-Apply The Pennsylvania State University Remote or University Park, FL Adjunct Lecturer, Part-Time Skills: Non-Tenure, Laboratory Classes, Background Checks, Sexual Assault, Institutional Policies, Technical Service, Thermodynamics, Engineering Design 60d+ ago $35k-64k yearly est. Auto-Apply State of South Carolina Remote or Charleston, SC Health Information Management Adjunct Instructor Skills: Course Objectives, Technology, Management System, Learning Management System, Human Resources, D2L, Academic Support, Field 2d ago new $26k-42k yearly est. Auto-Apply T Trident Technical College Foundation, Incorporated Remote or Charleston, SC Health Information Management Adjunct Instructor Skills: Course Objectives, Management System, Learning Management System, Human Resources, D2L, Academic Support, Field, Professional Knowledge 2d ago new $37k-45k yearly est. Auto-Apply West Virginia University Remote or Morgantown, WV Remote Part-time Adjunct Instructor - Department of Industrial and Management Systems Engineering Skills: Research, Health NET, IV, Industrial Training, Management, Laboratory, Review Process, Active Research 18h ago new $36k-64k yearly est. Auto-Apply West Virginia University Research Corporation Human Resources Remote or Morgantown, WV (Remote) Part-time Adjunct Instructor - Department of Industrial and Management Systems Engineering Skills: Health NET, Integrations, Active Research, IV, Industrial Training, Technology, Investigation, Ms 18h ago new $36k-64k yearly est. Auto-Apply Seattle Colleges Remote or Seattle, WA Part-Time Faculty- Applied Environmental Science Skills: Student Learning, Zoom, Video Conferences, Study, Collective Bargaining, Canvas, Colleges, Field 52d ago $32k-53k yearly est. Auto-Apply South Puget Sound Community College Remote or Olympia, WA Adjunct Instructor - College and Career Success Skills: General Education, Career Exploration, Learning Experiences, Integrated Marketing, CCS, Professional Development, Child Care, Learning Environment 18h ago new $58k-72k yearly est. Auto-Apply Renton Technical College Remote or Renton, WA Adjunct Phlebotomy Instructor Skills: Infection Control, Program Training, Organizational -Ability, Legal Issues, Learning Objectives, Career Education, Stipend, Statistical Physics 18h ago new $61k-74k yearly est. Auto-Apply Unitek Learning Remote or Phoenix, AZ Remote BSN Adjunct Instructor, WFD Skills: Ob, Mental Health, Workforce Development, LMS, DNP, Zoom, Geriatrics, Student Body 19h ago new $34k-64k yearly est. Auto-Apply Anderson University, Sc Remote or Anderson, SC Part-Time Faculty - Computer Science (Graduate Level - Online) Skills: Data Structures, College-Level Teaching, Computer Engineering, Computer Science, Architecture, Scholarship, Algorithms, Philosophy 60d+ ago $23k-33k yearly est. Auto-Apply Troy University Remote or Alabama Adjunct Instructor - MSN Program Skills: Semester, Courses Taught 2d ago new $41k-58k yearly est. Auto-Apply California Institute of Applied Technology Remote or San Diego, CA Workflow Deployment- ServiceNow Adjunct Instructor Skills: Instructional Design, Role Model, Hr, Management, Business Administration, Curriculum Development, Professional +, Instructional Delivery 4d ago new $38-46 hourly Auto-Apply Liberty University Remote or Lynchburg, VA Online Adjunct, Gerontology - School of Health Sciences Skills: Excellent Interpersonal, Learning Outcomes, Collaborative Learning, Organizational -Ability, Geriatrics, Management, Educational Process, Student Satisfaction 3d ago new $35k-45k yearly est. Auto-Apply Charter Oak State College Remote or New Britain, CT SLPA-S Adjunct Instructor Skills: Language, Attendance Reports, Courses Taught, Validates, Academic Achievement, Education Programs, Discussion Boards, Student Development 11d ago $49k-72k yearly est. See yourself in this field? Zippi will automatically apply for you. Completely free Adjunct faculty member gender ratio Male - 45% Female - 55% See all gender info Adjunct faculty member racial demographics White - 66.2% Asian - 11.3% See all race info Average adjunct faculty member age 46.1 See all age info Average adjunct faculty member salary $60,817 See all heatmaps Recently added adjunct faculty member jobs Work From Home Distance Salary Job type Job level Education Date posted Filters Auto-Apply P Penn State University Remote or Parkesburg, PA Adjunct Clinical Lecturer, Nursing - Fall 2025 Skills: Statistical Physics, Achievements, Class Content, Technical Service, Social Science, Child Abuse, Business, Business Business 18h ago new $56k-119k yearly est. Looking for a job? Let Zippi find it for you. Auto-apply Auto-Apply Columbia University In The City of New York Remote or New York, NY Adjunct Lecturer, Technology Management Skills: Problem Solving, Software Engineering, Field, Professional Studies, Science, Student Workshops, Technology Management, Classroom Discussions 19h ago new $10k-15k monthly Auto-Apply West Virginia University Remote or Morgantown, WV Remote Part-time Adjunct Instructor - Department of Industrial and Management Systems Engineering Skills: Research, Health NET, IV, Industrial Training, Management, Laboratory, Review Process, Active Research 18h ago new $36k-64k yearly est. Auto-Apply West Virginia University Research Corporation Human Resources Remote or Morgantown, WV (Remote) Part-time Adjunct Instructor - Department of Industrial and Management Systems Engineering Skills: Health NET, Integrations, Active Research, IV, Industrial Training, Technology, Investigation, Ms 18h ago new $36k-64k yearly est. Auto-Apply South Puget Sound Community College Remote or Olympia, WA Adjunct Instructor - College and Career Success Skills: General Education, Career Exploration, Learning Experiences, Integrated Marketing, CCS, Professional Development, Child Care, Learning Environment 18h ago new $58k-72k yearly est. Auto-Apply Renton Technical College Remote or Renton, WA Adjunct Phlebotomy Instructor Skills: Infection Control, Program Training, Organizational -Ability, Legal Issues, Learning Objectives, Career Education, Stipend, Statistical Physics 18h ago new $61k-74k yearly est. Auto-Apply Unitek Learning Remote or Phoenix, AZ Remote BSN Adjunct Instructor, WFD Skills: Ob, Mental Health, Workforce Development, LMS, DNP, Zoom, Geriatrics, Student Body 19h ago new $34k-64k yearly est. Auto-Apply Southern Oregon University Remote or Oregon Adjunct Instructor/Music Skills: Student Support, Music Courses, Instrumentation, Professional Development, Field, Curriculum Planning, Philosophy, Seminar 18h ago new $40k-57k yearly est. Auto-Apply Fox Valley Technical College Remote or Wautoma, WI Adjunct Instructor - Horticulture Community Class (Non-Credit) Skills: Classroom Management, Instructional Technology, Writing Intern, Professional Development, Student Learning, Student Engagement, Written Communication, Instructional Strategies 1d ago new $25 hourly Auto-Apply Association for Institutional Research In The Upper Midwest Remote or Appleton, WI Adjunct Instructor - Horticulture Community Class (Non-Credit) Skills: Student Engagement, Instructional Technology, Certifications, Written Communication, Professional Development, Classroom Management, Instructional Strategies, Writing Intern 18h ago new $25 hourly AMN Healthcare Remote or Maywood, IL Radiology - Body Imaging Faculty - Fully Remote Skills: Leadership, Image Scanning, Radiology, Biomedical, Colleges, MRI, Ultrasound, Translational 1d ago new $50k-121k yearly est. Auto-Apply University of Charleston Remote or Charleston, WV Adjunct Faculty - School of Business Skills: Field, Student Learning, Technology, Assessment, Classroom Lectures, Economics, Student Inquiries, Business Administration 1d ago new $62k-102k yearly est. Auto-Apply Barry University Remote or Miami Shores, FL Adjunct Faculty, Department of Social Work Skills: Curriculum Development, Telephone Calls, Student Engagement, Classroom Management, Written Communication, Course Syllabus, Academic Performance, Social Work 18h ago new $73k-110k yearly est. Auto-Apply Pima Community College Remote or Tucson, AZ Adjunct Faculty Pool - Fitness and Wellness Skills: Performance Contracts, Management, Course Objectives, Classroom Discipline, Various Training, Field, Student Performance, Assessment 18h ago new $98k-199k yearly est. Auto-Apply P Penn State University Remote or Parkesburg, PA Part Time - Intensive English Communication Program - Adjunct Faculty Skills: Institutional Policies, Graduate Studies, English Communication, Professional Development, Application Processes, Technical Service, Sexual Assault, English Language 18h ago new $58k-112k yearly est. Auto-Apply Baker College Remote or Newport Beach, CA Adjunct Faculty - St Francis School of Law Skills: Canvas, Disciplinary Actions, American Association, Zoom, Mastery, Study, Professional +, ST 1d ago new $500 weekly Auto-Apply M McPhs University-Boston Remote Adjunct Faculty - Professional Communication for Life Sciences II Skills: Writing Intern, Science Program, Critical Thinking, Learning Management System, Timelines, Discussion Boards, Professional Studies, Student Management 1d ago new $82k-130k yearly est. Auto-Apply Colorado Christian University Remote Adjunct/Affiliate Faculty - Special Education - College of Adult and Graduate Studies Skills: Undergraduate Classes, Class Preparation, Alumni, Graduate Studies, Assignment, Course Maintenance, Course Objectives, Quality Standards 6h ago new $58k-67k yearly est. Auto-Apply Monmouth University Remote or West Long Branch, NJ Adjunct, Nursing Lab Skills: Technology, Semester, Medication Administration, Field, Care Plan, Student Learning, Clinical Rotations, Vital Signs 1d ago new $30-35 hourly Auto-Apply U Uagc Remote Online Associate Faculty (Adjunct) Faculty- BS in Health Information Management (REMOTE) Skills: Faculty Recruitment, Statistical Physics, Technology, Classroom Discussions, Healthcare, Professional Development, Course Development, Student Inquiries 18h ago new $19.3-30 hourly Adjunct faculty member location demographics Where do adjunct faculty members earn the most? Adjunct faculty members earn the most in Ohio, where the average adjunct faculty member salary is $81,909. The map here shows where adjunct faculty members earn the highest salaries in the U.S. The darker areas across the 50 states highlight the highest salaries. Average salary Embed on your website $35k Lowest Highest $91k+ Which city has the most Adjunct Faculty Members? By looking through more than 3,107 resumes, we found that the most popular places for adjunct faculty members are Phoenix, AZ and Chicago, IL. Number of Adjunct Faculty Members Embed on your website 0 Lowest Highest 9500+ Adjunct faculty member jobs by state | Rank | State | Number of jobs | Average salary | --- --- | | 1 | New York | 1,392 | $69,703 | | 2 | Texas | 1,295 | $56,718 | | 3 | California | 1,226 | $73,937 | | 4 | Virginia | 1,037 | $62,882 | | 5 | Washington | 987 | $71,040 | | 6 | Oregon | 886 | $76,534 | | 7 | Pennsylvania | 697 | $57,260 | | 8 | Colorado | 587 | $54,418 | | 9 | New Jersey | 579 | $76,798 | | 10 | North Carolina | 542 | $64,082 | | 11 | Florida | 505 | $64,223 | | 12 | Indiana | 463 | $58,894 | | 13 | Georgia | 447 | $57,051 | | 14 | Ohio | 423 | $81,909 | | 15 | Illinois | 413 | $48,458 | | 16 | Michigan | 390 | $53,300 | | 17 | South Carolina | 384 | $35,761 | | 18 | Maryland | 363 | $70,052 | | 19 | Kentucky | 355 | $44,020 | | 20 | Massachusetts | 334 | $69,644 | | 21 | District of Columbia | 251 | $82,695 | | 22 | Missouri | 220 | $68,679 | | 23 | Alabama | 218 | $50,234 | | 24 | Wisconsin | 215 | $52,592 | | 25 | Tennessee | 209 | $55,337 | | 26 | Iowa | 206 | $68,884 | | 27 | New Hampshire | 205 | $67,652 | | 28 | Minnesota | 195 | $58,147 | | 29 | Louisiana | 190 | $55,120 | | 30 | Arizona | 169 | $53,008 | | 31 | Alaska | 167 | $75,420 | | 32 | Utah | 151 | $62,160 | | 33 | Delaware | 149 | $68,771 | | 34 | Mississippi | 137 | $55,039 | | 35 | Maine | 127 | $51,172 | | 36 | North Dakota | 125 | $56,264 | | 37 | Montana | 115 | $44,613 | | 38 | Vermont | 108 | $64,113 | | 39 | West Virginia | 104 | $48,892 | | 40 | South Dakota | 101 | $43,773 | | 41 | Nebraska | 97 | $50,758 | | 42 | Hawaii | 79 | $60,248 | | 43 | Connecticut | 64 | $71,358 | | 44 | Oklahoma | 64 | $46,631 | | 45 | New Mexico | 62 | $60,557 | | 46 | Kansas | 47 | $47,052 | | 47 | Rhode Island | 46 | $71,553 | | 48 | Arkansas | 34 | $46,447 | | 49 | Nevada | 31 | $65,891 | | 50 | Idaho | 17 | $45,758 | | 51 | Wyoming | 15 | $46,486 | Where are adjunct faculty members in high demand? Adjunct faculty members are in high demand in Fairmont, WV. There are currently 2 job openings for adjunct faculty members here. We looked over job openings to determine where in the country adjunct faculty members are in the most demand. Jobs openings Embed on your website 0 Lowest Highest 1000+ Adjunct faculty member demographics FAQs How many adjunct faculty member are there in the US? What percentage of Adjunct Faculty Members are black? What race are most adjunct faculty members? Are adjunct faculty member jobs male - dominated? How old is the average Adjunct Faculty Member? Have more questions? See all answers to common education, training, and library questions. Search for adjunct faculty member jobs Find jobs Previous:Certifications Next: Best Companies Zippia Careers Education, Training, and Library Industry Adjunct Faculty Member Adjunct Faculty Member Demographics Updated January 8, 2025 Browse education, training, and library jobs A B C D E F G H I J K L M N O P Q R S T U V W X Y Z Email: support@zippia.com Copyright © 2025 Zippia, Inc. Zippia Zippi AI Job Assistant About us Terms of use Privacy policy Contact us Careers & jobs Resume Builder Job search Q&A Full time jobs Part time jobs Work from home jobs Zippi your career sidekick
15646
https://artofproblemsolving.com/wiki/index.php/Trigonometry?srsltid=AfmBOoormBE2YJaQL7dm-tEcsMsVn5RD4j2vkAV-9XO1iafzzBhTWVwB
Art of Problem Solving Trigonometry - 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 Trigonometry Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Trigonometry Trigonometry is the study of relations between the side lengths and angles of triangles through the trigonometric functions. It is a fundamental branch of mathematics, and its discovery paved the way towards countless famous results. In contest math, trigonometry is an integral subfield of both geometry and algebra. Many essential results in geometry are written in terms of the trigonometric functions, such as the Law of Sines and the Law of Cosines; many more, such as Stewart's Theorem, are most easily proven using trigonometry. In algebra, expressions involving the trigonometric functions appear frequently on contests. These are solved by clever usage of the trigonometric functions' countless identities, which can simplify otherwise unwieldy equations. Outside of competition math, trigonometry is the backbone of much of analysis. In particular, Fourier Analysis is written almost entirely in the language of the trigonometric functions. Contents [hide] 1 Definitions 1.1 Right triangle definition 1.2 Unit circle definition 1.3 Taylor series definition 2 Applications in Geometry 2.1 Law of Sines 2.2 Law of cosines 3 Trigonometric identities 4 See also Definitions The trigonometric functions can be defined in several equivalent ways. The definition usually taught first is the right triangle definition, for its ease of access. An intermediate to olympiad geometry course usually uses the unit circle definition of trigonometry. Beyond the scope of contest math, the Taylor series definition of trigonometry is preferred in order to extend trigonometry to a complex domain. Right triangle definition The right triangle definition of trigonometry involves the ratios between edges of a right triangle, with respect to a given angle. The definitions below will be referring to angle , with side lengths specified in the diagram. Because angle must be less than for the triangle to stay right, these definitions only work for acute angles. Sine: The sine of angle , denoted , is defined as the ratio of the side opposite to the hypotenuse. Cosine: The cosine of angle , denoted , is defined as the ratio of the side adjacent to the hypotenuse. Tangent: The tangent of angle , denoted , is defined as the ratio of the side opposite to the side adjacent to . A common mnemonic to remember this is SOH-CAH-TOA, where Sine = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent More uncommon are the reciprocals of the trigonometric functions, listed below. Cosecant: The cosecant of angle , denoted , is defined as the reciprocal of the sine of . Secant: The secant of angle , denoted , is defined as the reciprocal of the cosine of . Cotangent: The cotangent of angle , denoted , is defined as the reciprocal of the tangent of . The right triangle definition is most commonly taught in introductory geometry classes for its simplicity. However, it has its limitations. It only works if is right, which means that the trigonometric functions are only defined when angle is acute. Even though it is defined using right triangles, trigonometry is just as useful when used on acute and obtuse triangles. The Law of Sines and Law of Cosines mentioned below generalize the right triangle definition to include all triangles. Unit circle definition Consider the unit circle, the circle with radius one centered at the origin. Starting at , walk a distance counterclockwise around the unit circle, as shown in the diagram. The coordinates of this point are defined to be . As for the other trigonometric functions, is defined to be the ratio of to , and cosecant, secant, and cotangent are defined to be the reciprocals of sine, cosine, and tangent, respectively. The benefit of this definition is that it matches the right triangle definition for acute angles, but extends their domain from acute angles to all real-valued angles. As such, this definition is usually preferred in intermediate to olympiad geometry settings. Taylor series definition The Taylor series for sine and cosine are used as their definitions in all higher mathematics. This meets the rigorous standards of real analysis, and gives a concrete way to extend the definitions of trigonometric functions from the real numbers to the full complex plane. The Taylor series for sine and cosine are shown below: These formulas are not used in high school math competitions. However, they do appear on the Putnam and other undergraduate competitions. Applications in Geometry While trigonometry is useful at any level, intermediate competitions are particularly fond of geometry problems demanding trigonometry. In addition to those mentioned, here are some highlights of the applications of trigonometry to geometry: Law of Sines The Law of Sines states that in any , where is the side opposite to , opposite to , opposite to , and is the circumradius of . The law of sines is particularly handy in problems involving the circumradius, seeing extremely wide usage in intermediate geometry. Law of cosines The Law of Cosines states that in any , where is the side opposite to , opposite to , and opposite to . It is a generalization of the Pythagorean Theorem and is used to prove several famous results, such as Heron's Formula and Stewart's Theorem. However, it sees limited applicability compared to the Law of Sines, as usage of the Law of Cosines can get algebra-heavy. It is helpful to memorize common, "nicer" values of sine and cosine as it can come in handy in contests, especially if you wish to apply either this or the Law of Sines to problems. Trigonometric identities Trigonometric identities are expressions true for all inputs involving the trigonometric functions. Due to the natural relationship between their definitions, these identities run numerous. In contest math, the most useful of these are: Pythagorean identities Angle addition identities Double angle identities Half angle identities Sum-to-product identities Product-to-sum identities See also Trigonometric identities Law of Sines Law of Cosines Stewart's Theorem Retrieved from " Categories: Trigonometry Definition 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.
15647
https://theoremoftheweek.wordpress.com/2012/02/04/theorem-39-eulers-criterion/
Theorem of the week Expositions of interesting mathematical results « Number Theory: Lecture 24 Theorem 40: Sperner’s lemma about coloured triangles » Theorem 39: Euler’s criterion I’ve previously written about Fermat’s Little Theorem, which tells us that if is a prime and is not divisible by , then . That’s not terribly exciting when , so let’s concentrate on odd primes in this post. Then is a whole number, so it makes sense (in the modular arithmetic world) to let . Then Fermat‘s Little Theorem tells us that . How many possibilities for are there (modulo , of course), if we know that ? Well, that congruence tells us that . But, in the jargon, “there are no zero-divisors modulo a prime”. That is, if , then or . (Note that we really do need to be prime here — try some examples!) So if is such that , then . Which is pretty nice. (Again, we really needed to be prime here. Try working modulo 8 instead, and see what happens there!) Back to where we started. We said that if is not divisible by the odd prime , then , so we know know that . That narrows down the possibilities a lot. It’s no good trying to tell me that — I simply shan’t believe you (even without doing any calculations). But tell me that , or that , and I have a problem — how do I know which is correct? This is where Euler‘s criterion comes in. I think that I’ll just state it now, and then say a little about how one might prove it afterwards. Theorem (Euler’s Criterion) Let be an odd prime, and let be coprime to . Then . Here the symbol on the right-hand side of the congruence is the Legendre symbol. So not only is certainly congruent to 1 or -1 modulo , but we can tell which — it just depends whether or not is a quadratic residue modulo . Looking at this another way, it gives us a way to compute the Legendre symbol (which appears slightly mysterious) — it’s just a suitable power. That can be very useful indeed. How might we try to prove this? I’ll outline two approaches here, each using a bit of background theory. One approach would be to make the most of the last Theorem of the Week, which tells us that there is a primitive root modulo . That is, there is some such that each non-zero number modulo is a power of . How does that help? Well, our number is coprime to , so we must have for some . Now . So we have if and only if divides (using the fact that has order ). And we can check that this happens if and only if is even. But that’s exactly the criterion for to be a quadratic residue, and so we’re done. Alternatively, we could think about the polynomial modulo . A theorem of Lagrange (an analogue of the fundamental theorem of algebra) tells us that this has at most roots modulo . The same applies to the polynomial modulo . But we have values that satisfy one or other of these polynomials, so in fact they must each have roots. But if is a quadratic residue modulo , then for some and so , so the quadratic residues modulo must be the roots of the polynomial modulo . Why is Euler’s criterion a good thing? Partly, of course, because it’s intrinsically interesting in its own right. But also it can be applied to a bunch of number theory problems. One quick consequence, for example, is the fact that if is an odd prime then is a quadratic residue modulo if and only if . It also has relevance in the field of primality testing. We say that is an Euler pseudoprime to the base if it `behaves like a prime’ as far as Euler’s criterion is concerned: if . It turns out that if is an odd composite number then is an Euler pseudoprime to at most half of all possible bases, and this gives rise to the Solovay-Strassen primality test. This is very different from the situation for Fermat pseudoprimes, where there are Carmichael numbers (numbers that cannot be distinguished from primes just using this notion of a pseudoprime). Further reading This is another result that occurs in introductory number theory books such as The Higher Arithmetic by Davenport, and A concise introduction to the theory of numbers by Baker (to name but two examples). Share this: Click to share on Facebook (Opens in new window) Facebook Click to share on X (Opens in new window) X Like Loading... Related This entry was posted on February 4, 2012 at 9:32 pm and is filed under Cambridge Maths Tripos, II Number Theory, Theorem. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site. 6 Responses to “Theorem 39: Euler’s criterion” Number Theory: Lecture 6 « Theorem of the week Says: February 4, 2012 at 9:33 pm […] 20 (Euler‘s criterion): Let be an odd prime, and let be an integer. Then . This is interesting only when is […] 2. Gareth Says: February 6, 2012 at 9:43 am I always liked the following proof of Euler’s Criterion, as it is very much like the standard proof of Wilson’s Theorem, so it’s easy to remember and recreate. Say that and are “associates” if . Clearly each has a unique associate, and there are two self-associate elements if is a quadratic residue (assuming ), and none otherwise. We multiply out , which is congruent to by Wilson’s Theorem. If is not a quadratic residue then in the product we get all numbers pairing up with their associates, that is copies of , and so . On the other hand, if is a quadratic residue, say with , then in the product we get numbers pairing up with their associates, along with not pairing up. So the product is , and so . 3. Luis Guzman Says: February 26, 2012 at 3:53 pm Reblogged this on Guzman's Mathematics Weblog and commented: 4. Theorem 41: Gauss’s lemma « Theorem of the week Says: July 31, 2012 at 10:41 am […] going to write this post on the assumption that you know about Fermat’s little theorem and Euler’s criterion. Fortunately, those links take you to places on this blog where you can read everything you need […] 5. Number Theory: Lecture 6 | Theorem of the week Says: October 23, 2013 at 10:51 am […] Theorem 20 (Euler’s criterion): Let be an odd prime, and let be an integer. Then . We proved this by first showing that and then using Lagrange’s theorem to show that the even powers of a primitive root ‘use up’ all the possible solutions to . […] 6. Number Theory: Lecture 7 | Theorem of the week Says: October 25, 2013 at 10:50 am […] Proposition 23 (Gauss‘s Lemma): Let be an odd prime, and let be coprime to . Then , where . We proved this by showing that , , …, correspond to , , …, in some order, where each of these has a definite sign, and then multiplying them all together to show that . Then we finished off using Euler’s criterion. […] Leave a comment Blog at WordPress.com. Entries (RSS) and Comments (RSS). Comment Reblog Subscribe Subscribed Theorem of the week Already have a WordPress.com account? Log in now. Theorem of the week Subscribe Subscribed Sign up Log in Copy shortlink Report this content View post in Reader Manage subscriptions Collapse this bar Design a site like this with WordPress.com Get started
15648
https://www.youtube.com/watch?v=YTbGMGcyLPo
Tag Questions – English Grammar Lessons Ellii (formerly ESL Library) 800000 subscribers 8103 likes Description 430132 views Posted: 22 Jul 2022 Discover how to form tag questions with both positive and negative verbs. You will also learn how to respond to tag questions correctly! This video includes a practice exercise at the end. Happy studying! LearnEnglish #ESL #English #Ellii #Grammar #Questions #TagQuestions Chapters: 0:00 Introduction 0:32 Function 1:11 Form 4:19 Intonation 5:17 Responding to Tag Questions 6:22 Practice 436 comments Transcript: Introduction Ellii Tag Questions What are tag questions? Tag questions include a short phrase, called the “tag,” at the end of a positive or negative sentence stem. For example, in the question “That was a good movie, wasn’t it?” “that was a good movie” is the sentence stem, and “wasn’t it?” is the tag. Function Why do we use tag questions in English? We use tag questions for emphasis or confirmation. Tag questions are very common in spoken English. When we are sure that someone will agree with what we are saying, we can use a tag question. For example, if my friend and I are looking at a small kitten and I think it is cute, I would assume that my friend also thinks it is cute. I can ask, "That kitten is cute, isn’t it?" and my friend will probably answer, “Yes, it is!” Form How do we form tag questions in English? Look at the position of the subject and verb in the sentence stem. Then form the tag by reversing the order. In other words, put the verb before the subject in the tag. For example, in the question “She is tired, isn’t she?" the subject “she” comes before the verb “is” in the stem. In the tag, the verb “isn’t” comes before the subject “she.” The tag always takes a subject pronoun. If the subject in the sentence stem of the question is a pronoun, it remains the same in the tag. If the subject in the stem is a noun, it becomes a pronoun in the tag. For example, in the question “They are hungry, aren’t they?” the subject of the stem is “they,” so the subject of the tag doesn’t change. But in the question “The children are hungry, aren’t they?” the subject of the stem is “children,” so the subject of the tag changes to “they.” The tag verb is always opposite of the verb in the sentence stem. If the verb in the stem is positive, the tag is negative. If the verb in the stem is negative, the tag is positive. For example, in the question “He speaks French fluently, doesn’t he?” the verb in the stem is positive, so the verb in the tag is negative. But in the question “He doesn’t speak Spanish fluently, does he?” the verb in the stem is negative, so the verb in the tag is positive. If the sentence stem uses the Be verb, the tag uses the Be verb too. For example: This pasta is delicious, isn't it? The baseball game wasn’t canceled, was it? If the sentence stem uses a simple present verb other than Be, the tag uses “don’t,” “do,” “doesn’t,” or “does.” For example: You like ice cream, don’t you? They don’t like broccoli, do they? He likes pasta, doesn’t he? If the sentence stem uses a simple past verb other than Be, the tag uses “didn’t” or “did.” For example: We ran for hours in gym class, didn’t we? You didn’t finish your homework, did you? For all other tenses and modals, the tag uses the auxiliary verb or modal that was used in the sentence stem. For example: Your neighbors have sold their house, haven’t they? She will call me back, won’t she? You can’t play the guitar, can you? Intonation What is intonation? “Intonation” refers to the rising or falling sound of someone’s voice. For tag questions, we use rising intonation in the tag when we are about 80–90% sure that someone will agree with us. We use falling intonation in the tag when we are about 95% sure that someone will agree with us. For example, if my friend and I watched a really good action movie, but I know my friend prefers comedies, then I’m only about 80–90% sure that my friend will agree with me. In this case, I would use rising intonation. I would say, "That was a great movie, wasn’t it?" But if I know my friend loves action movies as much as I do, then I’m about 95% sure that my friend will agree with me, so I would use falling intonation. I would say, "That was a great movie, wasn’t it?" How do we respond to tag questions in English? Responding to Tag Questions Make sure you respond to the sentence stem, not the tag. If you agree with the sentence stem, use the same type of verb in your response as the stem. Respond with “yes” for a positive sentence and “no” for a negative sentence. For example: You like hot weather, don’t you? Yes, I do. I like hot weather. You don’t like cold weather, do you? No, I don’t. I don’t like cold weather. If you disagree with the sentence stem, use the opposite type of verb in your response as the stem. Respond with “no” for a positive sentence and “yes” for a negative sentence. For example: You like hot weather, don’t you? No, I don’t. I don’t like hot weather. You don’t like cold weather, do you? Yes, I do. I like cold weather. Practice Look at the picture and the sentence stem. Say or write the correct tag. Try saying each question twice. First, imagine you are 80–90% sure that someone will agree with you and use rising intonation. Then imagine you are 95% sure that someone will agree with you and use falling intonation. For example: You don’t feel well today, do you? You don’t feel well today, do you? Now you try! They speak Spanish fluently, don’t they? They speak Spanish fluently, don’t they? Your brother didn’t like that spicy dish, did he? Your brother didn’t like that spicy dish, did he? The dogs haven’t gone outside today, have they? The dogs haven’t gone outside today, have they? She is playing baseball right now, isn’t she? She is playing baseball right now, isn’t she? This quiz is going to be easy, isn't it? This quiz is going to be easy, isn't it? Thanks for watching! Ellii
15649
https://pmc.ncbi.nlm.nih.gov/articles/PMC4985807/
Comprehensive evidence-based review on European antitussives - 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 BMJ Open Respir Res . 2016 Aug 5;3(1):e000137. doi: 10.1136/bmjresp-2016-000137 Search in PMC Search in PubMed View in NLM Catalog Add to search Comprehensive evidence-based review on European antitussives Alyn Morice Alyn Morice 1 Head of the Centre for Cardiovascular and Metabolic Research, Hull York Medical School, University of Hull, Castle Hill Hospital, Cottingham, UK Find articles by Alyn Morice 1, Peter Kardos Peter Kardos 2 Group Practice and Centre for Allergy, Respiratory and Sleep Medicine at Red Cross Maingau Hospital, Frankfurt, Germany Find articles by Peter Kardos 2 Author information Article notes Copyright and License information 1 Head of the Centre for Cardiovascular and Metabolic Research, Hull York Medical School, University of Hull, Castle Hill Hospital, Cottingham, UK 2 Group Practice and Centre for Allergy, Respiratory and Sleep Medicine at Red Cross Maingau Hospital, Frankfurt, Germany ✉ Correspondence to Dr Alyn Morice; A.H.Morice@hull.ac.uk Received 2016 Mar 21; Accepted 2016 Jun 28; Collection date 2016. Published by the BMJ Publishing Group Limited. For permission to use (where not already granted under a licence) please go to This is an Open Access article distributed in accordance with the terms of the Creative Commons Attribution (CC BY 4.0) license, which permits others to distribute, remix, adapt and build upon this work, for commercial use, provided the original work is properly cited. See: PMC Copyright notice PMCID: PMC4985807 PMID: 27547407 Abstract Acute cough caused by viral respiratory tract infections is probably the most common illness to afflict mankind. Despite the widespread but ineffective prescribing of antibiotics, there is no specific therapy. Home remedies and over-the-counter medicines are the mainstay for treatment of this short-lived but debilitating condition where cough is a major troublesome symptom. Across Europe, there are large variations in the recommendations made by healthcare professionals for the treatment of acute cough. This has arisen through custom and practice based on the evidence of historical studies performed to standards well short of what would be considered legitimate today. Acute cough is particularly difficult to study in a controlled setting because of the high rate of spontaneous remission and a large placebo effect. Here we detail the validated modern methodology used to assess the efficacy of antitussives and review the drugs commonly used in Europe against these standards. Keywords: Cough/Mechanisms/Pharmacology, Viral infection Introduction Acute cough is the most common symptom for which medical advice is sought. It is responsible for over 50% of new patient attendance in primary care and is the major source of consultation in pharmacy practice. Indeed, since symptomatic therapy is the mainstay of management of this generally benign and self-limiting illness, the pharmacist is the key player in the treatment of this condition. Unfortunately, much of the over-the-counter (OTC) therapy currently recommended throughout Europe is based on custom and practice and is not supported by clinical studies of sufficient quality to meet the standards of modern evidence-based medicine. Here we review the diagnosis and therapeutic options available for the treatment of what is perhaps the most common ailment to afflict mankind. Acute cough in common cold and acute bronchitis A number of overlapping terms are used throughout the world to describe the clinical syndrome of acute viral upper respiratory tract infection (URTI). We suggest that the terminology below really describes different aspects of the same common syndrome. The common cold is defined as an acute viral URTI, with symptoms of sore throat, sneezing, chilliness, nasal discharge, nasal obstruction, cough and malaise.1 Acute cough, that is a cough arbitrarily defined as being of <2 weeks duration, is one of the most common reasons for patient visits to ambulatory care.2 Acute bronchitis is a clinical term implying a self-limited inflammation of the large airways of the lung that is characterised by cough without pneumonia, the latter being diagnosed by focal consolidation on examination or on chest X-ray.3 It is now recognised that distinguishing between acute cough due to acute bronchitis and/or common cold is not practicable.45 Only slight pathological differences, if any, exist due to the principal localisation of viruses infecting the respiratory tract. Epidemiological surveys have shown that acute cough in otherwise healthy adults is a self-limiting disease with an average duration of the main symptom, cough of 14 days.6 In children, however, acute cough can last an average of 25 days.7 Acute bronchitis is caused by viruses (∼50% rhinovirus infection) in at least 90% of cases.8 For these infections, no curative (antiviral) treatment exists and antibiotic therapy has been repeatedly shown to be ineffective in patients without pre-existing lung disease.9 Despite being a self-limiting disease, acute bronchitis poses both a high symptom burden to individuals and a high financial burden to society, mainly due to work and school absenteeism. Over 50% of new patient consultations to primary care are due to acute cough and up to 85% of cases are erroneously treated with antibiotics—with no impact on recovery.10 Apparent success is due to rapid spontaneous recovery and a huge placebo effect.11 Unnecessary and uncontrolled use of antibiotics in acute bronchitis contributes to an impending doom of antibiotic resistance.12 Acute cough due to viral respiratory tract infections In viral respiratory tract infections, sore throat, headache, sneezing, runny nose and nasal congestion appear early in the course of the disease; cough emerges on day 2 or 3 only, but subsequently, from day 4 cough becomes the most bothersome and by far the longest lasting symptom until day 14.13–15 Viral infections of the respiratory epithelium cause early release of many inflammatory mediators disrupting the respiratory epithelium, sensitising chemosensitive cough receptors and the neuronal pathway of the cough reflex.1617 Thus, hypersensitivity of the afferent sensory nerves is thought to be the major mechanism causing cough in acute bronchitis, not the production of excessive mucus. Where mild-to-moderate mucus hypersecretion occurs, it is through the superficial goblet cells and submucosal glands.18 The incidence of mucus production, if any, seems to be present in common colds in just the first 48–72 hours. An evaluation of the placebo arms (n=774) of several studies in common cold after day 1 show no increase in sputum production.19 Thus, in viral respiratory tract infections, sputum expectoration, if any, lasts for a short time and the amount of secretion is small.20 From the therapeutic aspect, the treatment of wet and dry cough remains the same and recently a call for the removal of this classification has been made.21 Therefore, antitussives with proven efficacy might be the most appropriate treatment to relieve debilitating cough, of whatever character, in acute respiratory tract infections. Worsening bronchial obstruction may only be a risk in patients with pre-existing chronic airway obstruction.21 Much of the evidence supporting drug therapy in acute cough is old and of poor quality. There is little randomised controlled trial-based evidence which is of a modern standard. There are also well-known geographical differences in prescribing. For example, in Germany, OTC secretolytics and mucolytics such as ambroxol and N-acetylcysteine (NAS) are by far the most popular treatment with a market share as high as 47.4% of the entire common cold OTC market (source: IMS OTC Report). In contrast, in North America, OTC oral decongestant/first-generation (sedating) H1 antihistamines are used most frequently. Both strategies have little supporting evidence. Degrading mucus polymers and lowering mucus viscosity by mucolytic drugs has not been proven effective in treatment of cough in acute bronchitis.22 While first-generation antihistamines such as diphenhydramine might be effective in the treatment of cough,23 the second-generation ones are not. How can we assess the efficacy of antitussive medications? Since acute bronchitis and acute cough are by definition self-limiting illnesses lasting a few days, it is extremely difficult to distinguish between spontaneous remissions because of the patient getting better naturally from the effect of any medicine which has been administered. Basically, three tools have been used over the years to examine the antitussive activity of the currently marketed drugs. Subjective measures such as the visual analogue scale (VAS) or simply asking the patient whether they think their cough has improved were originally the favoured efficacy measure and many long-established preparations obtained their licence on this basis. Unfortunately, many of the studies are poorly designed with an inadequate number of patients and frequently using a mixed bag of diseases such as chronic bronchitis, tuberculosis and even lung cancer! Clearly, such studies would not be permissible in the modern era. Thus, the evidence base for many traditional antitussive preparations is extremely poor and, in our opinion, would be insufficient to make any claims of antitussive activity in terms of modern ‘evidence-based medicine’. Two objective methods of assessing cough have been developed. First, in the 1950s, cough challenge was introduced and has been perfected as a highly accurate tool for assessing the cough reflex. The participant inhales an increasing concentration of a protussive substance such as citric acid or capsaicin—the pungent extract of red peppers. The effect of drug on their cough reflex sensitivity is then compared with that of placebo. This methodology is excellent at assessing the characteristics of the study drug, such as its time course, and is frequently used in the development of novel therapies; indeed, it is recommended by the Food and Drug Administration (FDA) as part of the submission portfolio. However, it does not always correlate with subjective measures. For example, morphine has been demonstrated to have a highly effective activity in suppressing cough in some patients, but does not seem to alter cough reflex sensitivity. The third is a recently developed modality of assessing cough using cough counting.24 It has required a number of strides in technical development, particularly in computing power, to establish a reliable methodology using cough counters. Cough counting is now recognised as the ‘gold standard’ for assessing antitussive efficacy by the FDA. Unfortunately, since it is a recently developed technique, very few of the current OTC antitussive medications have been studied using cough counting. Indeed, only a single agent, dextromethorphan, has been demonstrated to be efficacious in this arena.25 It is best to consider the various methodologies for assessing cough as the three overlapping circles of Venn diagram (figure 1). Of the three, subjective measures have proven to be the least reliable and with a few notable exceptions have not been rigorously evaluated. We consider therefore that claims made of antitussive activity solely using subjective criteria provide insufficient evidence of efficacy, a view currently supported by the FDA. Figure 1. Open in a new tab The three methods for studying cough and the relationship between them. LCQ, Leicester cough questionnaire. Therefore, in an attempt to promote rational prescribing, we have reviewed the evidence for frequently used treatments in acute cough, particularly from a European perspective. We examined three aspects of drug efficacy on acute cough: the effect on the cough reflex using cough challenge, and both objective (cough recording) and subjective (ie, symptom scores, specific quality of life tools) effects on clinical outcomes. The efficacy of antitussive drugs Codeine Codeine is often considered the archetypal antitussive, yet there is little evidence that it has any intrinsic activity of its own. In man, codeine acts as a prodrug, being converted to morphine in the liver by the enzyme cytochrome P450 2D6.26 Morphine has been used for centuries in the treatment of cough and indeed has been demonstrated to have efficacy in randomised controlled trials (RCTs). Experience in chronic cough suggests that morphine is only efficacious in about a third to half of patients, others having no effective relief of the symptom. Whether this is also true in the acute bronchitis and cough in common cold is unknown. Despite its widespread use, there is very little clinical evidence supporting significant antitussive activity for orally administered codeine. In some studies, it has been reported to have no effect on cough challenge or on the sensation of urge to cough, whereas others have reported a small but significant effect.27 In two well-designed studies investigating cough due to URTIs, codeine 30 mg, followed by 4 days of dosing four times a day, had no effect greater than placebo syrup, either on an objective initial cough recording or on a subsequent self-reported cough.28 In the second study, oral codeine (50 mg) was compared with placebo syrup in 82 participants in a parallel group design using all three measures of cough assessment; again, no effect greater than that of placebo was observed.29 The cytochrome system which converts the prodrug codeine to morphine is highly polymorphic.26 Some patients are fast metabolisers converting the majority of codeine to morphine at first pass through the liver.26 In others, the slow metabolisers, very little codeine is converted. Thus, when prescribing codeine to an individual patient who has not previously used the drug, it is impossible to predict the degree of opiate effects or indeed side effects. Both overdosing or underdosing occurs in an unpredictable fashion. The European Medicines Agency has restricted the use of codeine in children for precisely this reason while the FDA are currently reviewing the use of codeine cough-and-cold medicines in children.3031 Children who are fast metabolisers were observed to have dangerous levels of sedation and suppression of respiration.26 We believe that this is not just a problem in the young and that the dangers of codeine far outweigh the limited evidence of efficacy in clinical studies. Dextromethorphan In the domain of cough counting, which is widely regarded as being the gold standard for assessing cough by regulatory authorities such as the FDA, only dextromethorphan has been demonstrated to significantly suppress acute cough using objective measures. In the three studies reported by Parvez et al,27 451 patients were observed using acoustic cough monitors. Compared with placebo, there was a highly significant reduction in cough counts with the dose of 30 mg dextromethorphan. To demonstrate true drug effect, dextromethorphan was administered within a capsule form, thus removing the demulcent effect of syrup. This also probably explains the relatively slow onset of action seen in this study. Subsequent studies with dextromethorphan have been performed with the syrup formulation, thus combining the rapid onset of action of demulcent activity with assessed efficacy of the drug. These positive results have been confirmed in a subsequent meta-analysis.32 In the second aspect of the assessment of efficacy of antitussive medications, dextromethorphan again passes. There have been multiple studies of the pharmacodynamics and pharmacokinetics of dextromethorphan performed under a variety of cough challenge conditions.27 Citric acid challenge is the most common modality, but recently dextromethorphan was found to be superior in the capsaicin challenge model.33 Dextromethorphan is revealed to be a drug with a relatively slow onset of action peaking in efficacy after ∼2 hours. Owing to its relatively slow penetration through the blood–brain barrier and consequent retention within the central nervous system, dextromethorphan may have a prolonged antitussive activity, being significantly better than placebo after as long as 24 hours.34 Some challenge studies also show that a higher than recommended daily dose of 30 mg may be even more efficacious against cough. It has proven to be more difficult to obtain subjective evidence for the effect of dextromethorphan in acute cough resulting from common cold infection. As with so much of the literature assessing subjective antitussive effects with a wide range of products, many of the studies are far from rigorous in their execution, using small numbers of participants, often with diverse disease and measuring symptoms without a validated methodology. Perhaps the biggest problem in any subjective measures in cough with common cold is the rapid rate of spontaneous remission in this acute illness, the large placebo effect and the demulcent effect of the syrups. In current preparations containing dextromethorphan, all of these ancillary options are used to enhance activity. Combining the three strands of evidence, it has been estimated that there is an excess antitussive activity due to dextromethorphan at the dose of 30 mg of ∼17%.32 Pentoxyverine Pentoxyverine citrate is in use as a non-opioid central acting antitussive with very little evidence of clinical efficacy, seen in poorly designed >50 years old clinical studies. Animal studies, however, show efficacy on evoked cough by electrical stimulation or citric acid challenge.3536 In our experience, animal studies are extremely poor at predicting clinical effectiveness of antitussives. Butamirate Butamirate preparations are widely used in Europe as OTC antitussives. Butamirate is thought to have a central mechanism which is neither chemically nor pharmacologically related to that of the opioid alkaloids. Butamirate also possesses non-specific anticholinergic and thus bronchodilator effects. Butamirate is claimed to be effective by the manufacturer in a number of double-blind, randomised, parallel group trials with codeine and other comparators, none of which were placebo controlled.37–39 The single placebo-controlled study remains unpublished and is held on file. The effects of butamirate on cough reflex sensitivity, as demonstrated by capsaicin inhalational cough challenge in normal participants, were recently studied in a placebo-controlled six-way randomised cross-over study with dextromethorphan as the positive control. All four doses of butamirate failed to demonstrate greater cough reflex suppression than placebo, whereas dextromethorphan was significantly effective.33 Levodropropizine Levodropropizine is suggested to be a peripherally acting antitussive which is widely used in southern Europe, particularly Italy. The clinical trials supporting its use in children and adults are summarised in a recent open access meta-analysis.40 There were four studies in children and three in adults. Only two studies were placebo comparisons. The paediatric study contained 12 children41 with asthma and the adult study (n=40) is not reported in full but is contained in another meta-analysis and appears to have been performed in hospitalised patients, the majority of whom were suffering from chronic bronchitis.42 There are thus no placebo-controlled studies demonstrating the efficacy of levodropropizine in acute cough. Of the other comparator studies, only two were in acute cough.4344 By far the largest, and thus contributing most to the results of the meta-analysis, was a non-randomised open observation in children.43 All treatments were claimed to be equally effective in reducing subjective measures, but since the comparators have also not been shown to be effective against placebo, little can be made of this claim. Ambroxol Ambroxol is the active metabolite of bromhexine and the most popular drug on the German OTC market (in 2015, 24% of the expectorant market share with an additional 1.7% for bromhexine, source: IMS OTC Report). Most references arise from the 1970s to 1980s and are related to long-term use in obstructive lung disease to prevent exacerbations or are in chronic bronchitis45 to ease expectoration. A recent review of ambroxol clinical data46 stated that, based on acceptability of study design (ie, randomised, double-blind, controlled) for short-term use in adults, only 3 out of 24 studies were acceptable.154748 Only the Matthys et al15 study investigated acute respiratory tract infection in a large, four parallel arm (some 170 patients in each arm) double-blind quadruple dummy randomised design the effect of 3×30 mg ambroxol days 1–3, 2×30 mg days 4–14, 4×300 mg myrtol (a standardised phytotherapeutic distillate containing 1,18 cineol) 1–14 days and 2×250 mg cefuroxime 1–6 days versus placebo over 2 weeks. Among secondary outcomes were diary data on nightly cough and coughing bouts during the day assessed. All three treatments were similarly effective and significantly better than placebo. The remaining two studies assessed short-term treatment of chronic conditions.4748 Studies in children were conducted only without a control group or versus an active comparator in an open design. Based on these data, symptomatic efficacy of ambroxol versus placebo on cough is proven in a single RCT. N-acetylcysteine NAC is the second most popular drug for acute cough in Germany with 23.5% of the OTC expectorant market share in 2015 (source: IMS OTC Report), a fact sharply contrasting with the available evidence for this indication. A Cochrane Library meta-analysis of three RCTs with cough at day 7 as the main outcome is available for acute upper and lower respiratory tract infections in a paediatric population.49 Statistically significant benefit was seen but the authors felt they were ‘of little clinical relevance’. Another Cochrane review for OTC medications for acute cough in 2014 did not find any references for NAS50 nor did a MEDLINE search by the authors of this paper (search terms of N-Acetylcystein AND Cough; N-Acetylcystein AND bronchitis; Acetylcystein AND cough; Acetylcystein AND Bronchitis). Oxomemazine There are no published placebo-controlled, double-blind studies supporting the use of oxomemazine in cough. In a study by Pujet et al,51 oxomemazine with guaifenesin was compared in a single-blind study with clobutinol in 130 patients with ‘infectious cough’. Cough intensity as assessed by VAS was rapidly reduced in the oxomemazine group, although there was no difference in the overall rate of resolution of cough. In an uncontrolled study in 46 infants under the age of 2,52 progress was described as ‘bonne’ in half. Chapuis et al53 report uncontrolled observations of the ‘novel antihistamine’ on cough with other allergic conditions without any patient details supplied in the manuscript. Helicidine Helicidine is a mucoglycoprotein extracted from the snail Helix pomatia. Helicidine has been used for >50 years in France as a cough medicine. In vivo animal studies showed antitussive efficacy in the cats; however, this study was not published.54 A placebo-controlled study in adult hospitalised patients with various diagnoses and an observational study in children were also not published.54 Studies from the 1950s claim antibacterial effects against Haemophilus (now called Bordetella) pertussis.55 A placebo syrup-controlled study54 with objective overnight cough counting in the sleep laboratory in n=30 patients with chronic obstructive pulmonary disease and objectively documented night-time cough was also performed. For the co-primary outcomes cough frequency and cough duration, an almost 50% higher reduction was demonstrated, while for secondary subjective outcomes no significant statistical difference was reported. Thus, there is no published clinical evidence to support helicidine's action in acute cough or acute bronchitis. Menthol Menthol is monoterpene produced by the peppermint plant Mentha arvensis from which most of the naturally occurring peppermint oil is extracted. Menthol's cooling activity is through the specific ‘cold’ receptor TRPM8, a member of the transient receptor potential family of nociceptors.56 It is primarily located on afferent sensory neurons and is anti-irritant by blockade of voltage-gated sodium channels. Menthol has an ancient history and has become a stock ingredient of many OTC preparations. Antitussive activity was commercialised by the development of a topical rub by Lunsford Richardson in 1890,57 and recent evidence indicates that the antitussive activity of menthol may reside in the activation of nasal as opposed to lung sensory afferents.58 Clinical evidence of menthol's activity is sparse with few clinical studies performed to modern standards. Challenge studies in normal participants produce a short-lived reduction in evoked cough. In a small and poorly controlled study, menthol vapour produced a decrease in capsaicin-induced cough.59 Cough induced by inhalation of citric acid was reduced in adults by inhalation of menthol vapour compared with air and pine oil control60 and in children compared with baseline challenge, but failed to reach significance when compared with placebo.61 Surprisingly, there appears to be no published clinical studies on the effect of menthol or of the many products containing it in acute cough or bronchitis. Diphenhydramine Diphenhydramine is a first-generation H1 antihistamine approved in some countries as an OTC antitussive, including the USA and the UK. In Germany, diphenhydramine 50 mg is approved as a hypnotic or antiemetic only. First-generation antitussives in combination with oral decongestants are recommended by the American College of Chest Physicians Evidence Based Guidelines for the treatment of cough in common cold and in the so-called upper airway cough syndrome.62 However, despite the title of those guidelines, this recommendation is based on expert opinion.63 In cough challenge studies in healthy participants64 and patients with acute viral respiratory infection (diphenhydramine combination syrup with decongestant) in adults efficacy could have been established.65 However, no symptom or objective cough monitoring-based studies are available for acute cough. There is a clear-cut discrepancy between evidence of efficacy and broad clinical use of diphenhydramine/decongestant combinations for acute cough—despite an important sedative effect (dizziness)—especially in the USA. Table 1 evaluates how European antitussives match up to the modern metrics in cough research. Table 1. RCT-proven efficacy of antitussives by three different cough measurement methods in acute bronchitis | RCT-evidence for drug efficacy | Subjective clinical symptoms | Objective cough recording | Cough challenge | Remarks | :--- :--- | Codeine | − | − | − | No convincing evidence of efficacy | | Dextromethorphan | + | + | ++ | Well characterised in objective studies | | Pentoxyverine | − | − | − | Only animal studies via 3 clinical studies >50 years old | | Butamirate | − | − | − | No placebo-controlled study published | | Levodropropizine | + | − | + | 6 short-term placebo or active comparator controlled studies n=174 | | Ambroxol | + | − | − | Many additional non-interventional studies | | N-acetylcysteine | Children + Adults − | − | − | Many studies for COPD, chronic cough, antioxidant properties | | Oxomemazine | − | − | − | Only observational studies | | Menthol | − | − | + | Widely used. Vapour antitussive via TRPM8 | | Helicidine | − | − | − | No clinical evidence of efficacy in acute cough | | Diphenhydramine | − | − | − | Broad clinical use | Open in a new tab COPD, chronic obstructive pulmonary disease; RCT, randomised controlled trial. Antitussives in combination cold therapy A popular strategy to combat the multiple symptoms in acute viral URTI has been to combine active ingredients. Such a strategy is entirely logical when symptoms require different therapeutic approaches. Thus, the addition of paracetamol to an antitussive to deal both with the cough and headache or myalgia makes therapeutic sense. Similarly, the use of a sedating antihistamine for a nocturnal preparation in combination with the antitussive may well give additional benefit and symptom relief. Some combinations on the market are, however, illogical and based on a poor understanding of the pathophysiology. It is becoming increasingly recognised that there is little evidence to support expectorant activity and, indeed, some agents classified as expectorants have been reported to have anti-inflammatory,66 antioxidant67 or antitussive activity in challenge studies.68 Similarly, expectorants, although widely prescribed in combination treatments, may actually work by decreasing the cough reflex hypersensitivity and thus relieving the sensation of mucus hypersecretion. Perhaps the most interesting studies to provide insight into the mode of action of ambroxol are recent investigations into its ability to block voltage-gated sodium channels located on sensory nerves.69 Such an activity is likely to underlie the clinically important local anaesthetic properties which support the use of ambroxol in its indication for sore throat. Sodium channel blockade may also explain some of the other activities of this class of agents through the blockade of neurogenic inflammation. Perhaps the most important consideration when using combination products is the possibility of a drug interaction, and therefore well-conducted safety and efficacy studies are required. Only studies of adequate power should be considered. As an example, Mizoguchi et al70 studied 432 participants in a placebo-controlled study of a syrup containing 15 mg dextromethorphan hydrobromide, 7.5 mg doxylamine succinate, 600 mg paracetamol and 8 mg ephedrine sulfate. The primary end point (composite of nasal congestion/runny nose/cough/pain relief scores 3 hours postdosing) showed a highly significant beneficial effect in the group given active treatment (p=0.0002). Each individual symptom score also showed statistically significant improvement 3 hours postdosing (p≤0.017). The next morning active treatment continued to show clinically and statistically significant benefits (p≤0.003). Evidence of benefit with the test syrup was also seen in the higher score for overall night-time relief (p<0.0001) and greater satisfaction on sleep (p=0.002). Adverse events were reported at half the frequency in the active treatment group compared with the placebo and there were no reported events >1% in the population. We suggest that only by the use of large well-controlled and well-designed studies such as this can combination products be recommended with any surety. Recommended treatment strategy URTIs are benign and self-limiting, and therefore patients with milder symptoms can be safely reassured. The demulcent effect of a simple linctus/syrup such as a home remedy of honey and lemon may bring a significant reduction in the cough, albeit of a relatively short duration. This strategy should be the first choice, particularly in children. In isolated dry or minimally productive cough, dextromethorphan 30–60 mg per day has the best evidence base. When other symptoms are also present, a combination product containing adequate amounts of dextromethorphan should be considered. A cough persisting longer than 2 weeks requires additional diagnostic evaluation. Conclusions Acute cough is perhaps the most common symptom to afflict mankind. While it is usually a benign and self-limiting illness, the amount of morbidity endured as a consequence of acute viral respiratory tract infection has enormous consequences on humanity. By the use of evidence-based treatment, significant improvement in patient outcomes can be achieved. While there are many gaps in the knowledge of our therapy in acute cough, our improved understanding of the mechanism of cough hypersensitivity brings rational treatment choices a step closer. A greater understanding of how drugs may normalise this aberrant reflex, thus bringing relief and shortening the duration of illness, may be of enormous benefit to the whole of society. Footnotes Funding: Procter & Gamble. Competing interests: At the time of drafting this manuscript, AM and PK were paid consultants of PGT Healthcare but were responsible for full manuscript drafting. PGT Healthcare also funded the article publishing charge. Provenance and peer review: Not commissioned; externally peer reviewed. Data sharing statement: No additional data are available. References 1.Eccles R. Understanding the symptoms of the common cold and influenza. Lancet Infect Dis 2005;5:718–25. doi:10.1016/S1473-3099(05)70270-X [DOI] [PMC free article] [PubMed] [Google Scholar] 2.Burt CW, Schappert SM. Ambulatory care visits to physician offices, hospital outpatient departments, and emergency departments: United States, 1999–2000. Vital Health Stat 2004;157:1–70. [PubMed] [Google Scholar] 3.Wenzel RP, Fowler AA III. Clinical practice. Acute bronchitis. N Engl J Med 2006;355:2125–30. doi:10.1056/NEJMcp061493 [DOI] [PubMed] [Google Scholar] 4.Braman SS. Chronic cough due to acute bronchitis: ACCP evidence-based clinical practice guidelines. Chest 2006;129:95S–103S. [DOI] [PMC free article] [PubMed] [Google Scholar] 5.Hueston WJ, Mainous AG III, Dacus EN et al. Does acute bronchitis really exist? A reconceptualization of acute viral respiratory infections. J Fam Pract 2000;49:401–6. [PubMed] [Google Scholar] 6.Albert RH. Diagnosis and treatment of acute bronchitis. Am Fam Physician 2010;82:1345–50. [PubMed] [Google Scholar] 7.Thompson M, Vodicka TA, Blair PS et al. , TARGET Programme Team. Duration of symptoms of respiratory tract infections in children: systematic review. BMJ 2013;347:f7027 doi:10.1136/bmj.f7027 [DOI] [PMC free article] [PubMed] [Google Scholar] 8.Acute Cough Illness (Acute Bronchitis) (accessed Jan 2016). 9.Fahey T, Stocks N, Thomas T. Quantitative systematic review of randomised controlled trials comparing antibiotic with placebo for acute cough in adults. BMJ 1998;316:906–10. [DOI] [PMC free article] [PubMed] [Google Scholar] 10.Butler CC, Hood K, Verheij T et al. Variation in antibiotic prescribing and its impact on recovery in patients with acute cough in primary care: prospective study in 13 countries. BMJ 2009;338:b2242. [DOI] [PMC free article] [PubMed] [Google Scholar] 11.Lee PC, Jawad MS, Hull JD et al. The antitussive effect of placebo treatment on cough associated with acute upper respiratory infection. Psychosom Med 2005;67:314–17. doi:10.1097/01.psy.0000155667.59662.92 [DOI] [PubMed] [Google Scholar] 12.Antibiotic resistance: a final warning. Lancet 2013;382:1072 doi:10.1016/S0140-6736(13)62008-5 [DOI] [PubMed] [Google Scholar] 13.Witek TJ, Ramsey DL, Carr AN et al. The natural history of community-acquired common colds symptoms assessed over 4-years. Rhinology 2015;53:81–8. [DOI] [PubMed] [Google Scholar] 14.Gwaltney JM Jr, Hendley JO, Simon G et al. Rhinovirus infections in an industrial population. II. Characteristics of illness and antibody response. JAMA 1967;202:494–500. [PubMed] [Google Scholar] 15.Matthys H, de MC, Carls C et al. Efficacy and tolerability of myrtol standardized in acute bronchitis. A multi-centre, randomised, double-blind, placebo-controlled parallel group clinical trial vs. cefuroxime and ambroxol. Arzneimittelforschung 2000;50: 700–11. [DOI] [PubMed] [Google Scholar] 16.Abdullah H, Heaney LG, Cosby SL et al. Rhinovirus upregulates transient receptor potential channels in a human neuronal cell line: implications for respiratory virus-induced cough reflex sensitivity. Thorax 2014;69:46–54. doi:10.1136/thoraxjnl-2013-203894 [DOI] [PubMed] [Google Scholar] 17.Atkinson SK, Sadofsky LR, Morice AH. How does rhinovirus cause the common cold cough? BMJ Open Resp Res 2016;3:e000118 doi:10.1136/bmjresp-2015-000118 [DOI] [PMC free article] [PubMed] [Google Scholar] 18.Rubin BK. Mucolytics, expectorants, and mucokinetic medications. Respir Care 2007;52:859–65. [PubMed] [Google Scholar] 19.Lehrl S, Matthys H, Kamin W et al. The BSS—a valid clinical instrument to measure the severity of acute bronchitis. Pulm Respir Res 2014;1:00016. [Google Scholar] 20.Morice AH, Widdicombe J, Dicpinigaitis P et al. Understanding cough. Eur Respir J 2002;19:6–7. doi:10.1183/09031936.02.00281002 [DOI] [PubMed] [Google Scholar] 21.Morice AH, Kantar A, Dicpinigaitis PV et al. Treating acute cough: wet versus dry—have we got the paradigm wrong? ERJ Open Res 2015;1:00055–2015. doi:10.1183/23120541.00055-2015 [DOI] [PMC free article] [PubMed] [Google Scholar] 22.Nadel JA. Mucous hypersecretion and relationship to cough. Pulm Pharmacol Ther 2013;26:510–13. doi:10.1016/j.pupt.2013.02.003 [DOI] [PubMed] [Google Scholar] 23.Gwaltney JM, Druce HM. Efficacy of brompheniramine maleate for the treatment of rhinovirus colds. Clin Infect Dis 1997;25: 1188–94. doi:10.1086/516105 [DOI] [PubMed] [Google Scholar] 24.Morice AH, Fontana GA, Belvisi MG et al. ERS guidelines on the assessment of cough. Eur Respir J 2007;29:1256–76. doi:10.1183/09031936.00101006 [DOI] [PubMed] [Google Scholar] 25.Parvez L, Vaidya M, Sakhardande A et al. Evaluation of antitussive agents in man. Pulm Pharmacol 1996;9:299–308. doi:10.1006/pulp.1996.0039 [DOI] [PubMed] [Google Scholar] 26.Gasche Y, Daali Y, Fathi M et al. Codeine intoxication associated with ultrarapid CYP2D6 metabolism. N Engl J Med 2004;351:2827–31. [DOI] [PubMed] [Google Scholar] 27.Dicpinigaitis PV, Morice AH, Birring SS et al. Antitussive drugs—past, present, and future. Pharmacol Rev 2014;66:468–512. doi:10.1124/pr.111.005116 [DOI] [PMC free article] [PubMed] [Google Scholar] 28.Eccles R, Morris S, Jawad M. Lack of effect of codeine in the treatment of cough associated with acute upper respiratory tract infection. J Clin Pharm Ther 1992;17:175–80. doi:10.1111/j.1365-2710.1992.tb01289.x [DOI] [PubMed] [Google Scholar] 29.Freestone C, Eccles R, Morris S et al. Assessment of the antitussive efficacy of codeine using cough sound pressure levels as a means of measuring cough. Pulm Pharmacol 1996;9:365 doi:10.1006/pulp.1996.0048 [DOI] [PubMed] [Google Scholar] 30.European Medicines Agency. Codeine not to be used in children below 12 years for cough and cold. 24 April 2015, EMA/249413/2015. 31.U.S. Food and Drug Administration Drug Safety Communications. FDA evaluating the potential risks of using codeine cough-and-cold medicines in children. 1 July 2015. 32.Pavesi L, Subburaj S, Porter-Shaw K. Application and validation of a computerized cough acquisition system for objective monitoring of acute cough: a meta-analysis. Chest 2001;120:1121–8. doi:10.1378/chest.120.4.1121 [DOI] [PubMed] [Google Scholar] 33.Faruqi S, Wright C, Thompson R et al. A randomized placebo controlled trial to evaluate the effects of butamirate and dextromethorphan on capsaicin induced cough in healthy volunteers. Br J Clin Pharmacol 2014;78:1272–80. doi:10.1111/bcp.12458 [DOI] [PMC free article] [PubMed] [Google Scholar] 34.Abdul-Manap R, Wright CE, Rostami-Hodjegan A et al. Cytochrome P450 2D6 (CYP2D6) activity in relation to the antitussive effect of dextromethorphan. Br J Clin Pharmacol 1999;47:586–7. [DOI] [PMC free article] [PubMed] [Google Scholar] 35.Brown C, Fezoui M, Selig WM et al. Antitussive activity of sigma-1 receptor agonists in the guinea-pig. Br J Pharmacol 2004;141:233–40. doi:10.1038/sj.bjp.0705605 [DOI] [PMC free article] [PubMed] [Google Scholar] 36.Talbott MW, Barch GK, Gabel LP. A new method for evaluating antitussives in cats using an electrode-cannula. Eur J Pharmacol 1975;34:59–63. doi:10.1016/0014-2999(75)90225-3 [DOI] [PubMed] [Google Scholar] 37.Lejeune J, Weibel MA. Comparison of 2 antitussive agents in pediatrics (butamirate citrate in drinkable solution and zipeprol syrup). Rev Med Suisse Romande 1990;110:181–5. [PubMed] [Google Scholar] 38.Germouty J, Weibel MA. Clinical comparison of butamirate citrate with a codeine-based antitussive agent. Rev Med Suisse Romande 1990;110:983–6. [PubMed] [Google Scholar] 39.Charpin J, Weibel MA. Comparative evaluation of the antitussive activity of butamirate citrate linctus versus clobutinol syrup. Respiration 1990;57:275–9. doi:10.1159/000195855 [DOI] [PubMed] [Google Scholar] 40.Zanasi A, Lanata L, Fontana G et al. Levodropropizine for treating cough in adult and children: a meta-analysis of published studies. Multidiscip Respir Med 2015;10:19 doi:10.1186/s40248-015-0014-3 [DOI] [PMC free article] [PubMed] [Google Scholar] 41.Fiocchi A, Arancio R, Murgo P et al. Levodropropizine effectiveness on nocturnal cough in asthmatic children. Eur Res J 1991;4:594. [Google Scholar] 42.Allegra L, Bossi R. Clinical trials with the new antitussive levodropropizine in adult bronchitic patients. Arzneimittelforschung 1988;38:1163–6. [PubMed] [Google Scholar] 43.Banderali G, Riva E, Fiocchi A et al. Efficacy and tolerability of levodropropizine and dropropizine in children with non-productive cough. J Int Med Res 1995;23:175–83. [DOI] [PubMed] [Google Scholar] 44.Catena E, Daffonchio L. Efficacy and tolerability of levodropropizine in adult patients with non-productive cough, comparison with dextromethorphan. Pulm Pharmacol Ther 1997;10:89–96. doi:10.1006/pupt.1997.0083 [DOI] [PubMed] [Google Scholar] 45.Terminology, definitions, and classification of chronic pulmonary emphysema and related conditions: a report of the conclusions of a CIBA guest symposium. Thorax 1959;14:286–99. [Google Scholar] 46.Malerba M, Ragnoli B. Ambroxol in the 21st century: pharmacological and clinical update. Expert Opin Drug Metab Toxicol 2008;4:1119–29. doi:10.1517/17425255.4.8.1119 [DOI] [PubMed] [Google Scholar] 47.Germouty J, Jirou-Najou JL. Clinical efficacy of ambroxol in the treatment of bronchial stasis. Respiration 1987;51(Suppl 1):37–41. [DOI] [PubMed] [Google Scholar] 48.Ericsson CH, Juhasz J, Jonsson E et al. Ambroxol therapy in simple chronic bronchitis: effects on subjective symptoms and ventilatory function. Eur J Respir Dis 1986;69:248–55. [PubMed] [Google Scholar] 49.Chalumeau M, Duijvestijn YC. Acetylcysteine and carbocysteine for acute upper and lower respiratory tract infections in paediatric patients without chronic broncho-pulmonary disease. Cochrane Database Syst Rev 2013;5:CD003124 doi:10.1002/14651858.CD003124.pub4 [DOI] [PMC free article] [PubMed] [Google Scholar] 50.Smith SM, Schroeder K, Fahey T. Over-the-counter (OTC) medications for acute cough in children and adults in community settings. Cochrane Database Syst Rev 2014;(11):CD001831 doi:10.1002/14651858.CD001831.pub5 [DOI] [PMC free article] [PubMed] [Google Scholar] 51.Pujet JC, Keddad K, Sévenier F et al. Comparative study of two antitussive drugs in the treatment of acute dry cough of infectious origin (prospective, randomized, single blind study). Therapie 2002;57:457–63. [PubMed] [Google Scholar] 52.Alison M. Therapeutic trials in children and particularly in infants of a new expectorant cough syrup with oxomemazine base. J Med Lyon 1965;46:1683–6. [PubMed] [Google Scholar] 53.Chapuis, Dalmais, Belpois, et al. [Oxomemazine (new antihistaminic). Its value in general medicine]. Lyon Med 1965;214:413–16. [PubMed] [Google Scholar] 54.Sergysels R, Art G. A double-masked, placebo-controlled polysomnographic study of the antitussive effects of helicidine. Curr Ther Res Clin Exp 2001;62:35–47. doi:10.1016/S0011-393X(01)80040-0 [Google Scholar] 55.Mainil J. Inhibition of Hemophilus pertussis in vitro and lysis of mucoid substances which it secretes, with an extract of secretions of Gastropoda. Ann Pharm Fr 1950;8:734–6. [PubMed] [Google Scholar] 56.Bautista DM, Siemens J, Glazer JM et al. The menthol receptor TRPM8 is the principal detector of environmental cold. Nature 2007;448:204–8. doi:10.1038/nature05910 [DOI] [PubMed] [Google Scholar] 57.Al Aboud K. The founder of Vicks: Lunsford Richardson (1854–1919). Skinmed 2010;8:100–1. [PubMed] [Google Scholar] 58.Plevkova J, Biringerova Z, Gavliakova S et al. The role of nasal trigeminal nerves expressing TRP channels in modulation of cough threshold and urge to cough—possible clinical application. Clin Transl Allergy 2013;3(Suppl 2):O17. [Google Scholar] 59.Wise PM, Breslin PA, Dalton P. Sweet taste and menthol increase cough reflex thresholds. Pulm Pharmacol Ther 2012;25:236–41. doi:10.1016/j.pupt.2012.03.005 [DOI] [PMC free article] [PubMed] [Google Scholar] 60.Morice AH, Marshall AE, Higgins KS et al. Effect of inhaled menthol on citric acid induced cough in normal subjects. Thorax 1994;49:1024–6. doi:10.1136/thx.49.10.1024 [DOI] [PMC free article] [PubMed] [Google Scholar] 61.Kenia P, Houghton T, Beardsmore C. Does inhaling menthol affect nasal patency or cough? Pediatr Pulmonol 2008;43:532–7. doi:10.1002/ppul.20797 [DOI] [PubMed] [Google Scholar] 62.Irwin RS, Baumann MH, Bolser DC et al. (ACCP) ACoCP. Diagnosis and management of cough executive summary: ACCP evidence-based clinical practice guidelines. Chest 2006;129:1S–23S. [DOI] [PMC free article] [PubMed] [Google Scholar] 63.Chronic cough. Lancet 1981;2:907–8. [PubMed] [Google Scholar] 64.Packman EW, Ciccone PE, Wilson J et al. Antitussive effects of diphenhydramine on the citric acid aerosol-induced cough response in humans. Int J Clin Pharmacol Ther Toxicol 1991;29:218–22. [PubMed] [Google Scholar] 65.Dicpinigaitis PV, Dhar S, Johnson A et al. Inhibition of cough reflex sensitivity by diphenhydramine during acute viral respiratory tract infection. Int J Clin Pharm 2015;37:471–4. doi:10.1007/s11096-015-0081-8 [DOI] [PMC free article] [PubMed] [Google Scholar] 66.Beeh KM, Beier J, Esperester A et al. Antiinflammatory properties of ambroxol. Eur J Med Res 2008;13:557–62. [PubMed] [Google Scholar] 67.Rahman I, Skwarska E, MacNee W. Attenuation of oxidant/antioxidant imbalance during treatment of exacerbations of chronic obstructive pulmonary disease. Thorax 1997;52:565–8. doi:10.1136/thx.52.6.565 [DOI] [PMC free article] [PubMed] [Google Scholar] 68.Dicpinigaitis PV, Gayle YE. Effect of guaifenesin on cough reflex sensitivity. Chest 2003;124:2178–81. doi:10.1378/chest.124.6.2178 [DOI] [PubMed] [Google Scholar] 69.Kamei J, Nakanishi Y, Ishikawa Y et al. Possible involvement of tetrodotoxin-resistant sodium channels in cough reflex. Eur J Pharmacol 2011;652:117–20. doi:10.1016/j.ejphar.2010.11.019 [DOI] [PubMed] [Google Scholar] 70.Mizoguchi H, Wilson A, Jerdack GR et al. Efficacy of a single evening dose of syrup containing paracetamol, dextromethorphan hydrobromide, doxylamine succinate and ephedrine sulfate in subjects with multiple common cold symptoms. Int J Clin Pharmacol Ther 2007;45:230–6. doi:10.5414/CPP45230 [DOI] [PubMed] [Google Scholar] Articles from BMJ Open Respiratory Research are provided here courtesy of BMJ Publishing Group ACTIONS View on publisher site PDF (925.1 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract Introduction The efficacy of antitussive drugs Conclusions Footnotes References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
15650
https://www.masterorganicchemistry.com/2018/05/07/nucleophilicity-of-amines/
Skip to content Comments Comment section 24 thoughts on “Nucleophilicity of Amines” Hello, I am trying to get Si-Cl and -NH-Si- to react together in order to form Si-N-Si + HCl. I would like to know if the NH is nucleophilic enough to have the reaction between the electrophilic Si-Cl and the amine. this reaction works with Si-OH and Si-Cl so I would like to transpose it to amine counterpart. Reply Yes, it will definitely work, for example look up the synthesis of hexamethyldisilazane. Reply 2. Hi James, thank you for a really interesting topic ! I’m currently working with nitriles, and I’ve been unable to find a good quantification method of their nuclophilicity, weither looking into Mayr’s of Hammet’s quantifying methods. Do you know if there’s any ? Thank you in advance ! Guillaume Reply Hi Guillaume – nitriles or cyanide ion? Nitriles themselves are really bad nucleophiles and will only work in Ritter-type reactions where they are reacting with an extremely good electrophile. Think about hybridization. Acetylene C-H is more acidic than alkanes because sp-hybridization stabilizes negative charge. Likewise the nitrogen of nitriles, being sp-hybridized makes that nitrogen lone pair very stable (and a very poor base / nucleophile) relative to the sp3 hybridized nitrogen in amines. Reply 3. I am stuck with carrying out a nucleophilic substitution reaction of triethylamine with chloride atoms in Polyvinyl chloride. Could you gove some ideas please? Reply How do you know it’s going to do a substitution and not elimination? Reply 4. Hi James, I routinely conjugate proteins via their primary amines to substrates and tags containing NHS esters at pH 7.2. It just occurred to me that at pH 7.2 the primary amine should be protonated to R-NH3(+), since the pKa of primary amines in proteins is approximately 9 and 10.5, for alpha and epsilon amines, respectively. How can the positively charged, protonated amine act as a nucleophile?? Thanks in advance, and I love reading your page! Rich Reply The pKa values you cite are for the conjugate acids of the amines, not the amines themselves. pH 7.2 is a neutral solution. The neutral amines will not be protonated to any appreciable extent until pH4 or below, and even then they will be in equilibrium. Reply 5. Very good explanation but could you clarify the influence of H2O in section about quantifying nucleophilicity? Water is polar protic solvent so why should the nucleophilicity of the amines (NH3 vs secondary vs tertiary) increase in water? Shouldn’t it be the other way around due to H-bonding? Reply Right. The nucleophilicity will certainly be solvent dependent. If you look at the tables from Mayr they all specify solvent and only should be compared directly when the numbers were obtained in the same solvent. You will also note that tertiary amines don’t have good nucleophilicity numbers in water, because they’re just not very soluble in polar solvents. I don’t have a good clear answer for you regarding trends. Hydrogen bonding is one factor, but so is the increased electron density that comes with additional alkyl groups. It’s not a one-dimensional quantity, it’s multivariable. Making a measurement is the best way to sort it all out. Reply 6. What can be done to increase the nucleophilicity of amines? what kind of base can be used for this purpose? Reply One way would be to deprotonate a primary or secondary amine to yield its conjugate base, but you’d likely run into problems with competing acid-base reactions. Reply 7. Very helpful posts！ How to compare the nucleophilic ability of phenol，thiophenol and aniline？ Reply 8. Hello I have the task where tertiary anime reacts with 3-clorpentane. So if I am correct it is an SN2 reaction ant the product will be amonium salt. But I am not sure why the reaction rate would increase if we replace solvents in this order: Tertiary-buthanol, ethanol, methanol and water? If you could help me I would be very grateful :) Reply Think about the influence of solubility on reaction rate and how each of those solvents will affect the solubility of a tertiary amine. Reply 9. I have a question about the using N-methyl hydrazine as a nucleophile. Both nitrogens experience the alpha effect from each other, right? The primary “amine” nitrogen is less sterically hindered but the secondary “amine” nitrogen experiences the sigma donating effect of the methyl group. Which of the two nitrogens is more nucleophilic then? Thanks! Reply I haven’t looked at the nucleophilicity table for mono-methyl hydrazine but I would expect that the methylated nitrogen will be more nucleophilic. Mayr has looked a bit at “alpha effects” and it seems to be complicated. Reply 10. Why does -NH2 shows more mesomeric effect than that of -NHR and -NR2 (where R denotes alkyl group)? Reply This is actually a great question. In order for the lone pair to donate into the aromatic ring, it has to adopt a conformation where the substituents on nitrogen are in the same plane as the aromatic ring. As steric bulk increases on the nitrogen, this conformation will become less favorable due to allylic strain. So it’s not so much that NH2 is more nucleophilic, but that the conformations of NHR and NR2 are less favorable. Reply 11. Hi James, First, I think the blog is fantastic. I’ve been reading it for years and your passion comes off the page. Sometimes there is a turn of phrase or a description that helps when thinking about something differently, making it all the clearer. Second, here are a couple of papers from Mayr relating to, specifically, the alpha effect and HSAB for interest. Reply Thank you very much for this – I love Mayr’s work. Reply 12. how strong is the nitrogen in acetonitrile? I’ve always had that question. why is it so much weaker compared to the oxygen in carbonyl groups? Thanks! Reply Very poor nucleophile. Hard to compare with sp2 hybridized oxygen, but likely similar. Clear basicity trend for nitrogen is: sp < sp2 < sp3 nitrogen i.e. nitrogen of nitrile < nitrogen of imine < nitrogen of amine Why? Less s-character, lone pair held more closely to nucleus, more stable. More stable = less basic (recall how terminal alkynes are easier to deprotonate than alkanes. Same phenomenon going on here!) Reply 2. The triple bond also causes a reasonable amount of electrostatic repulsion from the neighboring to electrophile groups. For the same reason acetylenides will only do SN2 with primary alkyl halides but won’t substitute the secondary (will react as a base instead). Reply Leave a Reply
15651
https://www.zhihu.com/question/4593694157?write
信息熵文氏图里面的阴影和圆圈分别代表什么意思？ - 知乎 关注推荐热榜专栏圈子 New付费咨询知学堂 ​ 直答 切换模式 登录/注册 信息熵文氏图里面的阴影和圆圈分别代表什么意思？ 关注问题​写回答 登录/注册 信息论 信息论基础（书籍） 信息论与编码（书籍） 信息论要义 信息熵文氏图里面的阴影和圆圈分别代表什么意思？ [图片]显示全部 ​ 关注者 8 被浏览 207 关注问题​写回答 ​邀请回答 ​好问题 ​3 条评论 ​分享 ​ 4 个回答 默认排序 ludabizi ​​ 国元证券股份有限公司 内刊执行主编 ​ 关注 谢邀 @小熊猫 1 人赞同了该回答 在信息熵的文氏图中： 圆圈通常代表不同的事件集合或者随机变量。例如有两个圆圈，可能分别代表事件A和事件B，每个圆圈内包含了该事件发生的所有可能情况。 阴影部分的意义取决于具体的场景和所研究的内容。如果是两个事件A和B的文氏图，阴影部分可能代表A和B的交集（A∩B），意味着两个事件同时发生的情况；也可能代表A和B的差集，比如A - B（仅A发生而B不发生的情况）或者B - A；还可能代表A和B的对称差（A∪B）-（A∩B），即要么A发生要么B发生，但A和B不同时发生的情况。这些不同的集合对应的信息熵计算方式也不同。 发布于 2024-11-20 01:44 ​赞同 1​​1 条评论 ​分享 ​收藏​喜欢 智一失 ​ 信息论与编码理论 ​ 关注 可以看作随机变量的信息熵到集合的映射 编辑于 2024-11-20 09:28 ​赞同​​添加评论 ​分享 ​收藏​喜欢 中国老年经济研究 ​​ 广东万发房地产开发有限公司 高级技术顾问 ​ 关注 可能包含 AI 创作内容 在信息熵的文氏图中，阴影和圆圈被用来直观地表示信息论中的几个关键概念：熵、条件熵、互信息和联合熵。这些概念在信息论中起着至关重要的作用，它们描述了信息的不确定性、相关性以及信息之间的依赖关系。下面，我将详细解释文氏图中阴影和圆圈的具体含义。 一、文氏图的基本构成 文氏图（Venn diagram）是一种用来表示集合之间关系的图示方法。在信息熵的文氏图中，通常包含两个或更多的圆圈，每个圆圈代表一个随机变量的熵（即不确定性）。这些圆圈可以重叠，重叠部分表示随机变量之间的共同不确定性或相关性。 二、圆圈的含义 在信息熵的文氏图中，每个圆圈通常代表一个随机变量的熵。例如，在图中有两个随机变量X和Y，那么可以分别用两个圆圈来表示它们的熵H(X)和H(Y)。这两个圆圈的大小反映了各自随机变量的不确定性大小，圆圈越大表示不确定性越高。 三、阴影的含义 条件熵的阴影： • 条件熵H(X|Y)表示在已知随机变量Y的条件下，随机变量X的不确定性。在文氏图中，这可以通过在表示H(X)的圆圈中去掉与H(Y)重叠的部分来表示。具体来说，可以在H(X)的圆圈内画一个阴影区域，这个阴影区域就是H(X)中去掉与H(Y)重叠部分后的剩余部分，即H(X|Y)。 互信息的阴影： • 互信息I(X;Y)表示两个随机变量X和Y之间的相关性。它可以通过计算H(X)减去H(X|Y)来得到，即I(X;Y) = H(X) - H(X|Y)。在文氏图中，互信息可以用两个圆圈重叠部分的阴影来表示。这个阴影区域既属于H(X)也属于H(Y)，它反映了两个随机变量之间的共同不确定性或相关性。 联合熵的阴影（虽未直接提问，但为完整解释文氏图，可提及）： • 联合熵H(X,Y)表示两个随机变量X和Y同时发生的不确定性。在文氏图中，联合熵可以用两个圆圈及其重叠部分的总和来表示。虽然联合熵本身没有直接的阴影表示，但可以通过观察两个圆圈及其重叠部分来理解其含义。 四、文氏图在信息论中的应用 信息熵的文氏图不仅有助于直观地理解熵、条件熵、互信息和联合熵等概念，还可以用于推导这些概念之间的数学关系。例如，通过观察文氏图，我们可以很容易地理解为什么互信息可以表示为H(X) - H(X|Y)，以及为什么联合熵可以表示为H(X) + H(Y) - I(X;Y)（尽管后者在文氏图中没有直接的阴影表示，但可以通过逻辑推理得出）。 此外，文氏图还可以用于解释信息论中的一些基本原理和定理，如数据处理不等式、熵的链式法则等。这些原理和定理在通信、信号处理、机器学习等领域具有广泛的应用价值。 综上所述，信息熵的文氏图通过阴影和圆圈直观地表示了信息论中的关键概念及其之间的关系。这种表示方法不仅有助于我 丨们深入理解这些概念的本质含义，还可以用于推导和解释信息论中的基本原理和定理。 展开阅读全文​ ​赞同​​添加评论 ​分享 ​收藏​喜欢 查看剩余 1 条回答 ​写回答 下载知乎客户端 与世界分享知识、经验和见解 相关问题 信息熵公式怎么运用?例如，50万字的书，常用汉字7000个，如字的概率相同，则每个字需约13比特。？ 2 个回答 如何初步理解信息熵这一概念？ 2 个回答 为什么信息熵要借用“熵”这个概念？ 10 个回答 如何理解信息熵的概念？ 4 个回答 广告 帮助中心 知乎隐私保护指引申请开通机构号联系我们 举报中心 涉未成年举报网络谣言举报涉企侵权举报更多 关于知乎 下载知乎知乎招聘知乎指南知乎协议更多 京 ICP 证 110745 号 · 京 ICP 备 13052560 号 - 1 · 京公网安备 11010802020088 号 · 互联网新闻信息服务许可证：11220250001 · 京网文2674-081 号 · 药品医疗器械网络信息服务备案（京）网药械信息备字（2022）第00334号 · 广播电视节目制作经营许可证:（京）字第06591号 · 互联网宗教信息服务许可证：京（2022）0000078 · 服务热线：400-919-0001 · Investor Relations · © 2025 知乎 北京智者天下科技有限公司版权所有 · 违法和不良信息举报：010-82716601 · 举报邮箱：jubao@zhihu.com 想来知乎工作？请发送邮件到 jobs@zhihu.com 登录知乎，问答干货一键收藏 打开知乎App 在「我的页」右上角打开扫一扫 其他扫码方式：微信 下载知乎App 无障碍模式 验证码登录 密码登录 开通机构号 中国 +86 获取短信验证码 获取语音验证码 登录/注册 其他方式登录 未注册手机验证后自动登录，注册即代表同意《知乎协议》《隐私保护指引》 扫码下载知乎 App 关闭二维码
15652
https://www.quora.com/What-is-the-easy-way-to-find-3-common-multiples-of-1000-1200-1500
What is the easy way to find 3 common multiples of 1000,1200,1500? - 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 Common Multiple Factors and Multiples Arithmetic Function Mathematics Number Theory Arithmetic Multiple Numbers Math Theory 5 What is the easy way to find 3 common multiples of 1000,1200,1500? All related (33) Sort Recommended Assistant Bot · 1y To find common multiples of the numbers 1000, 1200, and 1500, you can follow these steps: Find the Least Common Multiple (LCM) of the three numbers. The LCM is the smallest number that is a multiple of all three. Calculate the LCM: First, factor each number into its prime factors: 1000=2 3×5 3 1000=2 3×5 3 1200=2 4×3 1×5 2 1200=2 4×3 1×5 2 1500=2 2×3 1×5 3 1500=2 2×3 1×5 3 Next, take the highest power of each prime that appears in these factorizations: For 2 2: max is 2 4 2 4 (from 1200) For 3 3: max is 3 1 3 1 (from both 1200 and 1500) For 5 5: max is 5 3 5 3 (from both 1000 and 1500) Therefore, the LCM is: LCM=2^LCM=2^ Continue Reading To find common multiples of the numbers 1000, 1200, and 1500, you can follow these steps: Find the Least Common Multiple (LCM) of the three numbers. The LCM is the smallest number that is a multiple of all three. Calculate the LCM: First, factor each number into its prime factors: 1000=2 3×5 3 1000=2 3×5 3 1200=2 4×3 1×5 2 1200=2 4×3 1×5 2 1500=2 2×3 1×5 3 1500=2 2×3 1×5 3 Next, take the highest power of each prime that appears in these factorizations: For 2 2: max is 2 4 2 4 (from 1200) For 3 3: max is 3 1 3 1 (from both 1200 and 1500) For 5 5: max is 5 3 5 3 (from both 1000 and 1500) Therefore, the LCM is: L C M=2 4×3 1×5 3=16×3×125=6000 L C M=2 4×3 1×5 3=16×3×125=6000 Find the first three common multiples: The common multiples are simply the LCM multiplied by integers. The first three common multiples of 1000, 1200, and 1500 are: 1×6000=6000 1×6000=6000 2×6000=12000 2×6000=12000 3×6000=18000 3×6000=18000 Thus, the first three common multiples of 1000, 1200, and 1500 are 6000, 12000, and 18000. Upvote · Related questions More answers below The product of two successive multiples of 10 is 1200. What are their multiples? How do you find the common multiples of 4 and 6? Is it possible to find the highest common multiple of two or more numbers? Why? What is the relation between the lowest common multiple, and it's numbers? How do you find common multiples quickly? Ishan Agnihotri suffering from Lethologica except for this precise moment ·12y Originally Answered: What is the easy way to find 3 common multiples of 1000,1200,1500 ? · For this question specifically it can be done very easily. Forget about the two zeroes at the end of each number. You are left with 10, 12, 15. Find the lcm for them which is 60. This should also not require any calculation as the lcm will have zero at the end (has to be a multiple of 10). Find such a number for 12 and 15. After having found 60 as lcm, using its table find the next two multiples as 120 and 180. Now just add back the two zeroes we removed in the beginning. You get 6000, 12000 and 18000 as the common multiples. Upvote · 9 1 Promoted by Grammarly Grammarly Great Writing, Simplified ·Aug 18 Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do Continue Reading There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Docs – Your all-in-one writing surface Think of docs as your smart notebook meets your favorite editor. It’s a writing surface where you can brainstorm, draft, organize your thoughts, and edit—all in one place. It comes with a panel of smart tools to help you refine your work at every step of the writing process and even includes AI Chat to help you get started or unstuck. Expert Review – Your built-in subject expert Need to make sure your ideas land with credibility? Expert Review gives you tailored, discipline-aware feedback grounded in your field—whether you're writing about a specific topic, looking for historical context, or looking for some extra back-up on a point. It’s like having the leading expert on the topic read your paper before you submit it. AI Grader – Your predictive professor preview Curious what your instructor might think? Now, you can get a better idea before you hit send. AI Grader simulates feedback based on your rubric and course context, so you can get a realistic sense of how your paper measures up. It helps you catch weak points and revise with confidence before the official grade rolls in. Citation Finder – Your research sidekick Not sure if you’ve backed up your claims properly? Citation Finder scans your paper and identifies where you need sources—then suggests credible ones to help you tighten your argument. Think fact-checker and librarian rolled into one, working alongside your draft. Reader Reactions – Your clarity compass Writing well is one thing. Writing that resonates with the person reading it is another. Reader Reactions helps you predict how your audience (whether that’s your professor, a TA, recruiter, or classmate) will respond to your writing. With this tool, easily identify what’s clear, what might confuse your reader, and what’s most likely to be remembered. All five tools work together inside Grammarly’s document editor to help you grow your skills and get your writing across the finish line—whether you’re just starting out or fine-tuning your final draft. The best part? It’s built for school, and it’s ready when you are. Try these features and more for free at Grammarly.com and get started today! Upvote · 999 200 99 34 9 3 Gangadher Matta math graduate ·8y Originally Answered: What is the easy way to find 3 common multiples of 1000,1200,1500 ? · Take two ZEROES out from 1000, 1200, 1500. Now Common multiple of 10, 12, 15 is LCM, which is 60. Hence common multiples of 1000, 1200, 1500 are : 6000, 60000, 600000 Upvote · Bijay Sigdel 10y Related What is the least common multiple of 4 and 6? multiples of 4 = 4,8,12,16,20,24,.... multiples of 6 = 6,12,18,24,30,.... So, least common multiple of 4 and 6 = 12 Upvote · 9 9 Related questions More answers below What are the common multiples of 3, 4, and 6? Which number is the LCM PF 1200, 1500, and 2000? What is the beat way to find greatest common multiple? How do you find the least common multiple by listing common multiples? The highest common factor of two numbers is 3. The lowest common multiple of two number is 45. Why is each number a multiple of 3? Gokul Author has 169 answers and 746.6K answer views ·8y Related How do I find the sum of all the multiples of 7 that are between 1 and 1,000? Multiples of 7 that are between 1 and 1000 are 7+14+21+28…..+994 The sequence is in Arithematic Progression. Hope it helps Continue Reading Multiples of 7 that are between 1 and 1000 are 7+14+21+28…..+994 The sequence is in Arithematic Progression. Hope it helps Upvote · 99 21 9 1 Laura Kay Posey Education, K-6th from SWT (Graduated 1966) · Author has 6.5K answers and 2M answer views ·2y Related How do you find the least common multiple for two, three, or four numbers? ANSWER: RULE: Find LCM of all denominators. Line up all common denominators of fractions & whole numbers, etc 1 EXAMPLE: A. 1/ 2, 3/4, 6 / 8, 7 / 10 LCM = 40 1/ 2 = 0, 2, 4, 6, 8, 10, 12, 14……………..40 3 / 4 = 0, 4, 8, 12, 16, 20, 24,28,32,36,40 7 / 8 = 0, 8, 16, 24, 32, 40 7 / 10 = 0, 10, 20, 30, 40 B. 1 / 2 = 20 / 40 3 / 4 = 30 / 40 7 / 8 = 35 / 40 7 /10 = 28 / 40 C. 1 / 2 < 3 / 4 7 / 8 > 7 / 10 3 / 4 < 7 / 8 II. EXAMPLE LCM =.60 4, 6, 12, 5 , 3 4 = 0, 4, 8, 12, 16, 20, 24, 28, 32, 36…..60 6 = 0, 6, 12, 18, 24, 30, 36, 42, 48……….60 12 = 0, 12, 24, 36, 48, 60 5 = 0,B5, 10, 15, 20, 25, 30, 35, 40, ..…60 3 = 0, 3, 6, 9, 1 Continue Reading ANSWER: RULE: Find LCM of all denominators. Line up all common denominators of fractions & whole numbers, etc 1 EXAMPLE: A. 1/ 2, 3/4, 6 / 8, 7 / 10 LCM = 40 1/ 2 = 0, 2, 4, 6, 8, 10, 12, 14……………..40 3 / 4 = 0, 4, 8, 12, 16, 20, 24,28,32,36,40 7 / 8 = 0, 8, 16, 24, 32, 40 7 / 10 = 0, 10, 20, 30, 40 B. 1 / 2 = 20 / 40 3 / 4 = 30 / 40 7 / 8 = 35 / 40 7 /10 = 28 / 40 C. 1 / 2 < 3 / 4 7 / 8 > 7 / 10 3 / 4 < 7 / 8 II. EXAMPLE LCM =.60 4, 6, 12, 5 , 3 4 = 0, 4, 8, 12, 16, 20, 24, 28, 32, 36…..60 6 = 0, 6, 12, 18, 24, 30, 36, 42, 48……….60 12 = 0, 12, 24, 36, 48, 60 5 = 0,B5, 10, 15, 20, 25, 30, 35, 40, ..…60 3 = 0, 3, 6, 9, 12, 15, 18, 21, 24, 27…….60 Upvote · 9 1 Sponsored by Stake.com Stake | Exclusive Sports Betting Odds! Your ultimate sports betting hub is here! Get access to live odds, expert insights, and thrilling action. Play Now 99 36 Anil Bapat Lives in Mumbai, Maharashtra, India · Author has 2.8K answers and 3.8M answer views ·Apr 6 Related What is the method for finding the least common multiple of three or more unlike fractions? What is the method for finding the least common multiple of three or more unlike fractions? There is a well-defined method for fractions, when it comes to finding Least Common Multiple (LCM) of fractions. We start with finding the LCM of any two fractions, to be explained later. Taking the LCM from the previous calculation, we can go with the next fraction to come up with new LCM. And this process can continue until all the fractions are consumed, we started with. The last outcome is our required LCM of the given 3 or more fractions. Let me demonstrate it for you. Let the 4 4 fractions be \dfrac\dfrac Continue Reading What is the method for finding the least common multiple of three or more unlike fractions? There is a well-defined method for fractions, when it comes to finding Least Common Multiple (LCM) of fractions. We start with finding the LCM of any two fractions, to be explained later. Taking the LCM from the previous calculation, we can go with the next fraction to come up with new LCM. And this process can continue until all the fractions are consumed, we started with. The last outcome is our required LCM of the given 3 or more fractions. Let me demonstrate it for you. Let the 4 4 fractions be 1 2,3 4,5 8 and 7 12 1 2,3 4,5 8 and 7 12. All the fractions have been randomly chosen and there’s no loss of generality. If you want find the LCM of 2 2 fractions, the formula says, Divide the LCM of the Numerators by the HCF of the Denominators. The Numerators are 1 1 and 3 3 and LCM here is 3 3 and the Denominators are 2 2 and 4 4 and the HCF here is 2 2. Therefore, the LCM is 3 2 3 2. Now, the LCM of 3 2 3 2 and 5 8 5 8 is 15 2 15 2 and finally, the LCM of 15 2 15 2 and 7 12 7 12 is 105 2 105 2. As such, the formula could be applied over all the fractions simultaneously. The LCM of the Numerators (1,3,5 and 7)(1,3,5 and 7) is: 105 105 The HCF of the Denominators (2,4,8 and 12)(2,4,8 and 12) is: 2 2 The required LCM of the fractions is: 105 2 105 2. So, why did I use the approach that worked with only 2 2 fractions at a time. That’s because that’s an approach, that’s best suited for computers and even while working manually, we are dealing with 2 numbers at a time and it’s more manageable. Anyway, one can use the strategy that works best for him / her. Verification: Upon separately dividing the LCM by individual fractions, we get 105,70,84 105,70,84 and 90 90. All are integers and there’s no common factor other than 1 1, amongst 105,70,84 105,70,84 and 90 90. Thus, 105 2 105 2 is indeed the least expression that satisfies all the properties of LCM and therefore that’s the required LCM. Upvote · Peter Groot B.S. in Mathematics, Massachusetts Institute of Technology (Graduated 1971) · Author has 9.3K answers and 3.5M answer views ·3y Related How do you find common multiples quickly? Factor each number into primes, such as 90 = 2 3 3 5 and 75 = 3 5 5 Then the common multiple of a list of numbers will contain each prime factor to the highest power found for one of the numbers. So the common multiple is 2 3 3 5 5 = 450 = 90 5 = 75 6 Upvote · 9 1 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 David Townsend Staff Software Engineer (1982–present) · Author has 29.5K answers and 11M answer views ·3y Related How do you find common multiples quickly? For small numbers, you can factor each number into it’s prime factors and then identify the common factors in each factorization. This works well for small numbers where finding the factorization ( or determining if the number is actually prime ) but is not efficient for very large numbers where finding the prime factors is an expensive operation. The “GCD algorithm” based upon Euclid’s division algorithm is more efficient since it does not require identifying the prime factors, even very large numbers can be handled in a small number of steps. At the end of the algorithm you have the GCD of t Continue Reading For small numbers, you can factor each number into it’s prime factors and then identify the common factors in each factorization. This works well for small numbers where finding the factorization ( or determining if the number is actually prime ) but is not efficient for very large numbers where finding the prime factors is an expensive operation. The “GCD algorithm” based upon Euclid’s division algorithm is more efficient since it does not require identifying the prime factors, even very large numbers can be handled in a small number of steps. At the end of the algorithm you have the GCD of the two numbers, usually it is a simple matter to extract the prime factors of the GCD and hence the common divisors of the two numbers. Upvote · Sol Warda Author has 5.8K answers and 2.5M answer views ·Updated 2y Related What's the smallest common multiple of all numbers from 1 to 1,000? You mean the “Least Common Multiple” or LCM. What do you need it for? It is a huge number with 433 digits long. Here it is in scientific notation: LCM(listfor(n, 1, 1000, n))==7.12886527466509 E+432 Upvote · 9 3 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 999 624 Brunilda Ortiz BA from University of Puerto Rico at Río Piedras (Graduated 1975) · Author has 111 answers and 90K answer views ·1y Related What are the first three common multiples of 2 and 4? What are the first three common multiples of 2 and 4? To find the multiples of 2 and 4, ask your self if 4 is divisible by 2. 4 is an even number and 4 is the square of 2, so 4 is a multiple of 2. All ecmcen numbers are multiples of 2, but not all of them are multiples of 4.That's the first common multiple of 2 and 4.What are the next multiples of 2? Multiples of 2 = {2, 4, _, , _, , _, _ , …} Now make a list of the next 2 multiples of 4. Multiples of 4{ 4, _, , _, , _, _ …} The intersection of the 2 sets is the set of common multiples of 2 and 4. Common multi Continue Reading What are the first three common multiples of 2 and 4? To find the multiples of 2 and 4, ask your self if 4 is divisible by 2. 4 is an even number and 4 is the square of 2, so 4 is a multiple of 2. All ecmcen numbers are multiples of 2, but not all of them are multiples of 4.That's the first common multiple of 2 and 4.What are the next multiples of 2? Multiples of 2 = {2, 4, _, , _, , _, _ , …} Now make a list of the next 2 multiples of 4. Multiples of 4{ 4, _, , _, , _, _ …} The intersection of the 2 sets is the set of common multiples of 2 and 4. Common multiples of 2 & 4: { 4, _, , _, , _, _ …} Which ones are the first e common multiples of 3& 4? That is the answer to your question. I suggest you write the prime factorization of the numbers in the last lists and compare the powers of the prime numbers. Remember prime numbers are integers greater than 1 that have only 2 different factors. I is not prime because it only has one factor, itself. Prime numbers {2, 3, _, , _, , _, , _, , ____, _, _, ____ ...} If you are a student, look up the definition of prime numbers in your book or s Mathematics dictionary. and compare it to what you might have found using Google. If it says “A number that only has 2 factors, 1 and the number itself”, change it when you copy it. 1 is not a prime number because 1 has only 1 factor, “1”. Upvote · Janet Heberling Lives in San Francisco, CA (2022–present) · Author has 21.5K answers and 9.4M answer views ·1y Related What are the first three common multiples of 9 and 21? 9=3 2 9=3 2 21=3⋅7 21=3⋅7 LCM=3 2⋅7=63 LCM=3 2⋅7=63 63⋅2=126 63⋅2=126 63⋅3=189 63⋅3=189 Upvote · 9 1 Related questions The product of two successive multiples of 10 is 1200. What are their multiples? How do you find the common multiples of 4 and 6? Is it possible to find the highest common multiple of two or more numbers? Why? What is the relation between the lowest common multiple, and it's numbers? How do you find common multiples quickly? What are the common multiples of 3, 4, and 6? Which number is the LCM PF 1200, 1500, and 2000? What is the beat way to find greatest common multiple? How do you find the least common multiple by listing common multiples? The highest common factor of two numbers is 3. The lowest common multiple of two number is 45. Why is each number a multiple of 3? Can there be two different lowest common multiples for the same set of numbers? If so, how would this be possible? The third common multiple of 2 whole numbers is 177. What is the 6th common multiple? What's an example of a number that has no least common multiples, or LCM's? What is the sum of all the multiples of 3 or 5 below 1000? What is the best way to find the lowest common multiple? Related questions The product of two successive multiples of 10 is 1200. What are their multiples? How do you find the common multiples of 4 and 6? Is it possible to find the highest common multiple of two or more numbers? Why? What is the relation between the lowest common multiple, and it's numbers? How do you find common multiples quickly? What are the common multiples of 3, 4, and 6? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
15653
https://bioresources.cnr.ncsu.edu/resources/determination-of-thermal-conductivity-properties-in-some-wood-species-obtained-from-turkey/
NC State Home Campus Directory MyPack Portal Campus Map Libraries Campus Resources : University Housing NC State Dining Bookstores Wolfline Buses Calendar Diversity Academic Resources : Undergraduate Admissions Academic Departments WolfWare/Moodle Global Engagement Distance Education Academic Calendar Other Resources : Student Services Center Cashier's Office Registration and Records Scholarships and Financial Aid Career Development Center NC State Help Desk : Coronavirus Updates Giving NC State News Graduate School Gmail Access NC State Social Media ###### BioResources Determination of thermal conductivity properties in some wood species obtained from Turkey Çavuş, V., Şahin, S., Esteves, B., and Ayata, U. (2019). "Determination of thermal conductivity properties in some wood species obtained from Turkey," BioRes. 14(3), 6709-6715. Abstract With the increased awareness of thermal insulation of buildings, the knowledge of thermal conductivity of non-structural materials applied for roughing, cladding or flooring has become more important. The objective of this study was to investigate the thermal conductivity of 31 different wood species originated from the region of Izmir in Turkey. Thermal conductivity of air dried boards was determined in accordance to ASTM 5334 standard which measures this property on the interior of wood rather than on the surface. Thermal conductivity varied from 0.090 to 0.197 W/mK. The highest thermal conductivity was obtained for oak and the lowest for Canadian poplar. A linear relation was obtained between wood density and thermal conductivity. Download PDF Full Article Determination of Thermal Conductivity Properties in Some Wood Species Obtained from Turkey Vedat Çavuş,a, Sırrı Şahin,b Bruno Esteves,c and Ümit Ayata d With the increased awareness of thermal insulation of buildings, the knowledge of thermal conductivity of non-structural materials applied for roughing, cladding or flooring has become more important. The objective of this study was to investigate the thermal conductivity of 31 different wood species originated from the region of Izmir in Turkey. Thermal conductivity of air dried boards was determined in accordance to ASTM 5334 standard which measures this property on the interior of wood rather than on the surface. Thermal conductivity varied from 0.090 to 0.197 W/mK. The highest thermal conductivity was obtained for oak and the lowest for Canadian poplar. A linear relation was obtained between wood density and thermal conductivity. Keywords: Thermal Conductivity; Wood Contact information: a: Izmir Katip Celebi University, Faculty of Forestry, Forest Industry Engineering Department, Cigli/Izmir, Turkey; b: Department of Agricultural Buildings and Irrigation, Agriculture Faculty, Ataturk University, Erzurum, Turkey; c: Superior School of Technology, Polytechnic Institute of Viseu, Viseu, Portugal; d: Bayburt University, Faculty of Art and Design, Interior Architecture and Environmental Design, Bayburt, Turkey; Corresponding author: vedatcavus@hotmail.com INTRODUCTION The variations of exterior temperature between night and day and between summer and winter seasons make thermal conductivity (TC) of wood an important property when applied as cladding, roughing, or flooring in building construction. Moreover, with the increase in energy costs, consumers are becoming more aware of the importance of a good thermal insulation of the materials used in construction. It follows that the knowledge of the thermal conductivity of the most commonly used wood species is vital. With respect to wood, the TC is highly dependent on wood density and moisture but also on the direction of the measurements, on the kind and amount of extractives or other chemical substances, on the relative density and proportion of earlywood/latewood, and also on wood defects (MacLean 1941). Generally, higher density leads to higher TC, and good linear correlations have been reported before (Narayanamurti and Ranganathan 1941; Gu and Zink-Sharp 2007; Yu et al.2011; Vay et al.2015). For instance Pelit et al. (2014) densified fir wood and concluded that after densification the TC had increased by about 50%. Moreover, Yapici et al. (2011), who determined the thermal conductivity of several species, obtained higher thermal conductivities for more dense woods, with the highest TC achieved for oak (0.8 g/cm3), followed by fir (0.45 g/cm3), beech (0.6 g/cm3), chestnut (0.52 g/cm3), and Scots pine (0.47 g/cm3). Water is a good heat conductor, and therefore higher amounts of water in wood increase the thermal conductivity. According to some authors (MacLean 1941; Vay et al.2015) below the fiber saturation point there is a linear correlation between moisture in wood and thermal conductivity. The direction of the measurements is also important for thermal conductivity, which is generally higher in the axial direction (Samuel et al. 2012). This is due to the orientation of the molecular chains within the cell wall (Suleiman et al. 1999). According to Kotlarewski et al. (2014) the rate of heat flow in the axial direction is two and a half times greater than the rate through the other directions. Although cellulose microfibrils have different orientations, the majority are aligned with the longitudinal axis. Vay et al. (2015), supported by different studies (Griffiths and Kaye 1923; Rowley 1933; Bučar and Straže 2008), stated that the thermal conductivity is about 2 to 3 times higher in the longitudinal direction than in the radial or tangential directions. Although smaller, there is also a difference between radial and tangential directions. Thermal conductivity in the radial direction is about 5% to 10% higher than in tangential direction (Griffiths and Kaye 1923; Faouel et al.2012). Some studies show that hardwoods that have a high amount of rays usually have higher thermal conductivity, since rays serve as paths for the heat transport, making radial thermal conductivity higher than tangential (Rowley 1933; Vay et al. 2015). Wood porosity is also an important factor because air is a poor thermal conductor compared to wood material. Therefore porous woods have lower thermal conductivity. For example, Vasubsbu et al. (2015) tested the thermal conductivity of several Indian trees and observed that the lowest TC were obtained for the most porous woods. The curry tree presented almost 73% porosity and had the lowest TC, around 1.47 x 10-4 cal/(s·cm ºC). EXPERIMENTAL Materials Boards of 31 different species commonly used in Turkey were used in this study. The species were: walnut (Juglans regia), maun (Swietenia mahagoni), black locust (Robinia pseudoacacia L.), chestnut (Castanea sativa Mill.), oak (Quercus petraea Liebl.), apple (Malus domestica), eucalyptus (E. camaldulensis Dehnh.), avocado (Persea americana), fig (Ficuscarica), European larch (Larix decidua), Monterey cypress (Cupressus macrocarpa), black pine (Pinus nigra), fir (Abies bornmuelleriana), beey (Morus Sp.), cedar (Cedrus libani), Scots pine (Pinus sylvestris L.), red pine (Pinus brutia Ten.), ash (Fraxinus excelsior), Mediterranean cypress (Cupressus sempervirens), lime (Tilia cordata), juniper (Juniperus communis L.), plum (Prunus domestica), olive (Olea europaea), iroko (Chlorophora excelsa), hornbeam (Carpinus betulus L.), peach (Prunus persica), Canadian poplar (Populus canadensis), black poplar (Populus nigra), Russian olive (Elaeagnus angustifolia), plane (Platanus orientalis L.), and white oak (Quercus alba). The wood samples came from various lumber sales sites, in Izmir City, Turkey. The samples were air dried until an initial moisture content of around 12% (ISO 554, 1976). After the drying period 5 samples with dimensions 5 cm x 5 cm x 15 cm (radial x tangential x longitudinal) were cut from each board. The density of all the samples was determined at 12% moisture content by weighing and measuring the dimensions of the samples with a calliper. Thermal Conductivity Measurement Thermal conductivity measurements were made with a THERM 2227–2, ALHBORN thermal conductivity meter (Fig. 1) in accordance with ASTM 5334-08. Although this method is more suitable for isotropic materials, it has already been used by Kotlarewski et al. (2014) to determine the TC of balsa wood. In order to make the measurements, a 14 cm long hole was drilled in each sample along longitudinal direction. After introducing the still pin in the hole, three measurements were made for each sample. The device is done measuring when a balance of 30 to 36 °C degrees is obtained, which takes 10 min. Fig. 1. Thermal conductivity measurement (Model THERM 2227–2, ALHBORN) Statistical Analysis A statistical analysis was made by using SPSS 17 Software (Sun Microsystems Inc., Santa Clara, CA, USA). For thermal conductivity (W/mK) the average value of fifteen replicates was recorded. RESULTS AND DISCUSSION Table 1 presents the results of the variance analysis of thermal conductivity made on the 31 different wood species. Results show that the wood species had a significant effect on thermal conductivity, which makes the selection of wood species important when wood is applied to building construction. Table 1. Thermal Conductivity Variance Analysis Table 2 presents the thermal conductivities of the 31 species measured in this work. The lowest thermal conductivity was obtained for Canadian poplar (0.090 W/mK), followed by Monterey cypress (0.093 W/mK), black poplar (0.109 W/mK), and fir (0.11 W/mK). The highest was for oak (0.197 W/mK) followed by olive (0.195 W/mK), Mediterranean cypress (0.195 W/mK), and plum (0.179 W/mK). The lowest density was obtained for Canadian poplar (0.340 g/cm3), Monterey cypress (0.405 g/cm3), and fir (0.410 g/cm3), and the highest density was olive (0.894 g/cm3), followed by oak (0.841 g/cm3) and plum (0.799 g/cm3). There wasn’t much information available about thermal conductivity of the species studied; however some authors reported comparable thermal conductivities for some of them. For example Kol and Sefil (2011) reported a thermal conductivity of 0.1297 W/mK and 0.1362 W/mK in tangential and radial directions for fir (Abies bornmüllerianaMattf.), which is a little higher than the value obtained here (0.110 W/mK); nevertheless the samples in the cited study had 0.457 g/cm3density, which was also higher than the samples of the present study (0.410 g/cm3). However, Dündar et al. (2012) presented a thermal conductivity of 0.111 W/mK, which is almost the same as the value obtained here, for samples with 0.388 g/cm3 density. Surprisingly, Yapici et al.(2011) reported a much higher thermal conductivity perpendicular to the grain for this fir (0.195 W/mK) with 0.450 g/cm3 density. These authors also reported a thermal conductivity of 0.182 W/mK for Scots pine and 0.196 W/mK for chestnut, which were a little higher than the values obtained here of 0.132 W/mK and 0.114 W/mK. Table 2. SPSS Analysis Results for Thermal Conductivity of the Studied Species and density Figure 2 presents the relation between thermal conductivity and density of the tested woods. Results show that there was a clear linear relation between density and thermal conductivity, as stated before (MacLean 1941). A similar relation was reported by Mason et al.(2016) for several kinds of woods reported in literature. Nevertheless, there are some woods that present a higher thermal conductivity than expected in accordance to density. This is the case for Mediterranean cypress. This is probably due to the differences in anatomical features such as porosity or amount and kind of extractives, which are known to influence the thermal conductivity of wood (MacLean 1941; Vasubsbu et al. 2015). Fig. 2. Relation between thermal conductivity and density Based on looking at thermal conductivity alone, species like fir or Canadian and black poplar would be the ideal choices for interior cladding, roughing, or flooring since they would have the best insulation performance. Although it is known that low density species are not suitable for flooring due to their low hardness, this property is not that important when used for example in roughing. On the other hand, for example oak, one of the most used species for flooring, has the highest thermal conductivity of this list, showing that it is not the best choice in terms of energy consumption. CONCLUSIONS Results show that wood species have a significant effect on thermal conductivity, establishing once more that it is important to select the right wood species for application to building construction. The highest thermal conductivity was obtained for oak (0.197 W/mK) and the lowest for Canadian poplar (0.090 W/mK). A linear relationship was achieved between thermal conductivity and density of wood. Results show that a more careful selection of wood species for non-structural applications can be made in order to decrease energetic consumption. REFERENCES CITED ASTM D5334-08. (2008). “Standard test method for determination of thermal conductivity of soil and soft rock by thermal needle probe procedure,” ASTM International, West Conshohocken, PA. Bučar, B., and Straže, A. (2008). “Determination of the thermal conductivity of wood by the hot plate method: The influence of morphological properties of fir wood (Abies albaMill.) to the contact thermal resistance,” Holzforschung 62(3), 362-367. Dündar, T., Kurt, Ş., As, N., and Uysal, B. (2012). “Nondestructive evaluation of wood strength using thermal conductivity,” BioResources 7(3), 3306-3316. Faouel, J., Mzali, F., Jemni, A., and Nasrallah, S. B. (2012). “Thermal conductivity and thermal diffusivity measurements of wood in the three anatomic directions using the transient hot-bridge method,” Special Topics & Reviews in Porous Media: An International Journal 3(3). Griffiths, E., and Kaye, G. W. C. (1923). “The measurement of thermal conductivity,” Proc. R. Soc. Lond. A, 104(724), 71-98. Gu, H., and Zink-Sharp, A. (2007). “Geometric model for softwood transverse thermal conductivity. Part I,” Wood and Fiber Science 37(4), 699-711. ISO 554. (1976). “Standard atmospheres for conditioning and/or testing – Specifications.” Kol, H. Ş., and Sefil, Y. (2011). “The thermal conductivity of fir and beech wood heat treated at 170, 180, 190, 200, and 212°C,” Journal of Applied Polymer Science 121(4), 2473-2480. DOI: 10.1002/app.33885 Kotlarewski, N. J., Ozarska, B., and Gusamo, B. K. (2014). “Thermal conductivity of Papua New Guinea balsa wood measured using the needle probe procedure,” BioResources 9(4), 5784-5793. MacLean, J. D. (1941). “Thermal conductivity of wood,” Heating, Piping & Air Conditioning 13(6), 380-391. Mason, P. E., Darvell, L. I., Jones, J. M., and Williams, A. (2016). “Comparative study of the thermal conductivity of solid biomass fuels,” Energy & Fuels 30(3), 2158-2163. Narayanamurti, D., and Ranganathan, V. (1941). “The thermal conductivity of Indian timbers,” in: Proceedings of the Indian Academy of Sciences-Section A, Springer, 300-315. Pelit, H., Sönmez, A., and Budakçı, M. (2014). “Effects of ThermoWood® process combined with thermo-mechanical densification on some physical properties of scots pine (Pinus sylvestris L.),” BioResources 9(3), 4552-4567. Rowley, F. B. (1933). “The heat conductivity of wood at climatic temperature differences,” Heating, Piping, and Air Conditioning 5, 313-323. Samuel, O. S., Ramon, B. O., and Johnson, Y. O. (2012). “Thermal conductivity of three different wood products of Combretaceae family; Terminalia superba, Terminalia ivorensis and Quisqualis indica,” Journal of Natural Sciences Research 2(4). Suleiman, B. M., Larfeldt, J., Leckner, B., and Gustavsson, M. (1999). “Thermal conductivity and diffusivity of wood,” Wood Science and Technology 33(6), 465-473. Vasubsbu, M., Nagaraju, B., Kumar, J. V., and Kumar, R. J. (2015). “Experimental measurement of thermal conductivity of wood species in india: effect of density and porosity,” International Journal of Science, Environment and Technology 4(5), 1360-1364. Vay, O., De Borst, K., Hansmann, C., Teischinger, A., and Müller, U. (2015). “Thermal conductivity of wood at angles to the principal anatomical directions,” Wood Science and Technology 49(3), 577-589. DOI: 10.1007/s00226-015-0716-x Yapici, F. I., Ozcifci, A., Esen, R., and Kurt, S. (2011). “The effect of grain angle and species on thermal conductivity of some selected wood species,” BioResources 6(3), 2757-2762. DOI: 10.15376/biores.6.3.2757-2762 Yu, Z.-T., Xu, X., Fan, L.-W., Hu, Y.-C., and Cen, K.-F. (2011). “Experimental measurements of thermal conductivity of wood species in China: Effects of density, temperature, and moisture content,” Forest Products Journal 61(2), 130-135. DOI: 10.13073/0015-7473-61.2.130 Article submitted: October 17, 2018; Peer review completed: December 15, 2018; Revisions accepted: June 11, 2018; Published: July 3, 2019. DOI: 10.15376/biores.14.3.6709-6715
15654
https://saylordotorg.github.io/text_introductory-chemistry/s19-04-uses-of-radioactive-isotopes.html
Uses of Radioactive Isotopes Previous Section Table of Contents Next Section 15.4 Uses of Radioactive Isotopes Learning Objective Learn some applications of radioactivity. Radioactive isotopes have a variety of applications. Generally, however, they are useful because either we can detect their radioactivity or we can use the energy they release. Radioactive isotopes are effective tracers because their radioactivity is easy to detect. A tracerA substance that can be used to follow the pathway of that substance through a structure. is a substance that can be used to follow the pathway of that substance through some structure. For instance, leaks in underground water pipes can be discovered by running some tritium-containing water through the pipes and then using a Geiger counter to locate any radioactive tritium subsequently present in the ground around the pipes. (Recall that tritium is a radioactive isotope of hydrogen.) Tracers can also be used to follow the steps of a complex chemical reaction. After incorporating radioactive atoms into reactant molecules, scientists can track where the atoms go by following their radioactivity. One excellent example of this is the use of carbon-14 to determine the steps involved in photosynthesis in plants. We know these steps because researchers followed the progress of carbon-14 throughout the process. Radioactive Dating Radioactive isotopes are useful for establishing the ages of various objects. The half-life of radioactive isotopes is unaffected by any environmental factors, so the isotope acts like an internal clock. For example, if a rock is analyzed and is found to contain a certain amount of uranium-235 and a certain amount of its daughter isotope, we can conclude that a certain fraction of the original uranium-235 has radioactively decayed. If half of the uranium has decayed, then the rock has an age of one half-life of uranium-235, or about 4.5 × 10 9 y. Many analyses like this, using a wide variety of isotopes, have indicated that age of the earth itself is over 4 × 10 9 y. In another interesting example of radioactive dating, hydrogen-3 dating has been used to verify the stated vintages of some old fine wines. One isotope, carbon-14, is particularly useful in determining the age of once-living artifacts. A tiny amount of carbon-14 is produced naturally in the upper reaches of the atmosphere, and living things incorporate some of it into their tissues, building up to a constant, albeit very low, level. Once a living thing dies, it no longer acquires carbon-14; as time passes the carbon-14 that was in the tissues decays. (The half-life of carbon-14 is 5,370 y.) If a once-living artifact is discovered and analyzed many years after its death and the remaining carbon-14 is compared to the known constant level, an approximate age of the artifact can be determined. Using such methods, scientists determined that the age of the Shroud of Turin (Figure 15.3 "Shroud of Turin"; purported by some to be the burial cloth of Jesus Christ and composed of flax fibers, a type of plant) is about 600–700 y, not 2,000 y as claimed by some. Scientists were also able to use radiocarbon dating to show that the age of a mummified body found in the ice of the Alps was 5,300 y. Figure 15.3 Shroud of Turin In 1989, several groups of scientists used carbon-14 dating to demonstrate that the Shroud of Turin was only 600–700 y. Many people still cling to a different notion, despite the scientific evidence. Irradiation of Food The radiation emitted by some radioactive substances can be used to kill microorganisms on a variety of foodstuffs, extending the shelf life of these products. Produce such as tomatoes, mushrooms, sprouts, and berries are irradiated with the emissions from cobalt-60 or cesium-137. This exposure kills a lot of the bacteria that cause spoilage, so the produce stays fresh longer. Eggs and some meat, such as beef, pork, and poultry, can also be irradiated. Contrary to the belief of some people, irradiation of food does not make the food itself radioactive. Medical Applications Radioactive isotopes have numerous medical applications—diagnosing and treating illness and diseases. One example of a diagnostic application is using radioactive iodine-131 to test for thyroid activity (Figure 15.4 "Medical Diagnostics"). The thyroid gland in the neck is one of the few places in the body with a significant concentration of iodine. To evaluate thyroid activity, a measured dose of 131 I is administered to a patient, and the next day a scanner is used to measure the amount of radioactivity in the thyroid gland. The amount of radioactive iodine that collects there is directly related to the activity of the thyroid, allowing trained physicians to diagnose both hyperthyroidism and hypothyroidism. Iodine-131 has a half-life of only 8 d, so the potential for damage due to exposure is minimal. Technetium-99 can also be used to test thyroid function. Bones, the heart, the brain, the liver, the lungs, and many other organs can be imaged in similar ways by using the appropriate radioactive isotope. Figure 15.4 Medical Diagnostics Radioactive iodine can be used to image the thyroid gland for diagnostic purposes. Source: Scan courtesy of Myo Han, Very little radioactive material is needed in these diagnostic techniques because the radiation emitted is so easy to detect. However, therapeutic applications usually require much larger doses because their purpose is to preferentially kill diseased tissues. For example, if a thyroid tumor were detected, a much larger infusion (thousands of rem, as opposed to a diagnostic dose of less than 40 rem) of iodine-131 could help destroy the tumor cells. Similarly, radioactive strontium is used to not only detect but also ease the pain of bone cancers. Table 15.5 "Some Radioactive Isotopes with Medical Applications" lists several radioactive isotopes and their medical uses. Table 15.5 Some Radioactive Isotopes with Medical Applications | Isotope | Use | --- | | 32 P | cancer detection and treatment, especially in eyes and skin | | 59 Fe | anemia diagnosis | | 60 Co | gamma ray irradiation of tumors | | 99m Tc | brain, thyroid, liver, bone marrow, lung, heart, and intestinal scanning; blood volume determination | | 131 I | diagnosis and treatment of thyroid function | | 133 Xe | lung imaging | | 198 Au | liver disease diagnosis | | The “m” means that it is a metastable form of this isotope of technetium. | In addition to the direct application of radioactive isotopes to diseased tissue, the gamma ray emissions of some isotopes can be directed toward the tissue to be destroyed. Cobalt-60 is a useful isotope for this kind of procedure. Food and Drink App: Radioactivity in Wines Wine lovers put some stock in vintages, or the years in which the wine grapes were grown before they were turned into wine. Wine can differ in quality depending on the vintage. Some wine lovers willingly pay much more for a bottle of wine with a certain vintage. But how does one verify that a bottle of wine was in fact part of a certain vintage? Is the label a fake? Is that stash of wine found in the basement of a French chateau really from the 1940s, or was it made in 2009? This wine label from a bottle of wine claims a vintage of 1991. Is the wine really from this vintage, or is it a fake? Radioactivity can help determine the answer. Source: Used by permission of Ralph E. Wermuth. Cesium-137 is a radioactive isotope that has a half-life of 30.1 y. It was introduced into the atmosphere in the 1940s and 1950s by the atmospheric testing of nuclear weapons by several countries after World War II. A significant amount of cesium-137 was released during the Chernobyl nuclear disaster in 1986. As a result of this atmospheric contamination, scientists have precise measurements of the amount of cesium-137 available in the environment since 1950. Some of the isotope of cesium is taken up by living plants, including grape vines. Using known vintages, oenologists (wine scientists) can construct a detailed analysis of the cesium-137 of various wines through the years. The verification of a wine’s vintage requires the measurement of the activity of cesium-137 in the wine. By measuring the current activity of cesium-137 in a sample of wine (the gamma rays from the radioactive decay pass through glass wine bottles easily, so there’s no need to open the bottle), comparing it to the known amount of cesium-137 from the vintage, and taking into account the passage of time, researchers can collect evidence for or against a claimed wine vintage. Before about 1950, the amount of cesium-137 in the environment was negligible, so if a wine dated before 1950 shows any measurable cesium-137 activity, it is almost surely a fake, so don’t shell out lots of money for it! It may be a good wine, but it is almost definitely not over 60 years old. Key Takeaway Radioactivity has several practical applications, including tracers, medical applications, dating once-living objects, and preservation of food. Exercises Define tracer and give an example of how tracers work. Name two isotopes that have been used as tracers. Explain how radioactive dating works. Name two isotopes that have been used in radioactive dating. The current disintegration rate for carbon-14 is 14.0 Bq. A sample of burnt wood discovered in an archeological excavation is found to have a carbon-14 disintegration rate of 3.5 Bq. If the half-life of carbon-14 is 5,730 y, approximately how old is the wood sample? A small asteroid crashes to Earth. After chemical analysis, it is found to contain 1 g of technetium-99 to every 3 g of ruthenium-99, its daughter isotope. If the half-life of technetium-99 is 210,000 y, approximately how old is the asteroid? What is a positive aspect of the irradiation of food? What is a negative aspect of the irradiation of food? Describe how iodine-131 is used to both diagnose and treat thyroid problems. List at least five organs that can be imaged using radioactive isotopes. Which radioactive emissions can be used therapeutically? Which isotope is used in therapeutics primarily for its gamma ray emissions? Answers A tracer is a radioactive isotope that can be detected far from its original source to trace the path of certain chemicals. Hydrogen-3 can be used to trace the path of water underground. If the initial amount of a radioactive isotope is known, then by measuring the amount of the isotope remaining, a person can calculate how old that object is since it took up the isotope. 11,500 y increased shelf life (answers will vary) The thyroid gland absorbs most of the iodine, allowing it to be imaged for diagnostic purposes or preferentially irradiated for treatment purposes. gamma rays Previous Section Table of Contents Next Section A substance that can be used to follow the pathway of that substance through a structure.
15655
https://invent.ilmkidunya.com/images/Section/mathematical-analysis-css-book.pdf
PREFACE A glance at the table of contents will reveal that this textbook treats topics in analysis at the "Advanced Calculus" level. The aim has been to provide a develop-ment of the subject which is honest, rigorous, up to date, and, at the same time, not too pedantic. The book provides a transition from elementary calculus to advanced courses in real and complex function theory, and it introduces the reader to some of the abstract thinking that pervades modern analysis. The second edition differs from the first in many respects. Point set topology is developed in the setting of general metric spaces as well as in Euclidean n-space, and two new chapters have been added on Lebesgue integration. The material on line integrals, vector analysis, and surface integrals has been deleted. The order of some chapters has been rearranged, many sections have been completely rewritten, and several new exercises have been added. The development of Lebesgue integration follows the Riesz-Nagy approach which focuses directly on functions and their integrals and does not depend on measure theory. The treatment here is simplified, spread out, and somewhat rearranged for presentation at the undergraduate level. The first edition has been used in mathematics courses at a variety of levels, from first-year undergraduate to first-year graduate, both as a text and as supple-mentary reference. The second edition preserves this flexibility. For example, Chapters 1 through 5, 12, and 13 provide a course in differential calculus of func-tions of one or more variables. Chapters 6 through 11, 14, and 15 provide a course in integration theory. Many other combinations are possible; individual instructors can choose topics to suit their needs by consulting the diagram on the next page, which displays the logical interdependence of the chapters. I would like to express my gratitude to the many people who have taken the trouble to write me about the first edition. Their comments and suggestions influenced the preparation of the second edition. Special thanks are due Dr. Charalambos Aliprantis who carefully read the entire manuscript and made numerous helpful suggestions. He also provided some of the new exercises. Finally, I would like to acknowledge my debt to the undergraduate students of Caltech whose enthusiasm for mathematics provided the original incentive for this work. Pasadena T.M.A. September 1973 LOGICAL INTERDEPENDENCE OF THE CHAPTERS 1 THE REAL AND COM-PLEX NUMBER SYSTEMS 2 SOME BASIC NOTIONS OF SET THEORY 3 ELEMENTS OF POINT SET TOPOLOGY I 4 LIMITS AND CONTINUITY I 5 DERIVATIVES I 6 FUNCTIONS OF BOUNDED VARIATION AND REC-TIFIABLE CURVES 8 INFINITE SERIES AND INFINITE PRODUCTS 7 THE RIEMANN-STIELTJES INTEGRAL 12 MULTIVARIABLE DIF-FERENTIAL CALCULUS 9 SEQUENCES OF FUNCTIONS 10 THE LEBESGUE INTEGRAL I 13 IMPLICIT FUNCTIONS AND EXTREMUM PROBLEMS 14 MULTIPLE RIEMANN INTEGRALS 11 FOURIER SERIES AND FOURIER INTEGRALS 16 15 CAUCHY'S THEOREM AND MULTIPLE LEBESGUE THE RESIDUE CALCULUS INTEGRALS CONTENTS Chapter 1 The Real and Complex Number Systems 1.1 Introduction . . . . . . . . . . . . . . . . . 1 1.2 The field axioms . . . . . . . . . . . . . . . . 1 1.3 The order axioms . . . . . . . . . . . . . . . 2 1.4 Geometric representation of real numbers . . . . . . . . 3 1.5 Intervals . . . . . . . . . . . . . . . . . . 3 1.6 Integers . . . . . . . . . . . . . . . . . . 4 1.7 The unique factorization theorem for integers . . . . . . . 4 1.8 Rational numbers . . . . . . . . . . . . . 6 1.9 Irrational numbers . . . . . . . . . . . . . . . 7 1.10 Upper bounds, maximum element, least upper bound (supremum) . . . . . . . . . . . . . . . . . 8 1.11 The completeness axiom . . . . . . . . . . . . . 9 1.12 Some properties of the supremum . . . . . . . . . . 9 1.13 Properties of the integers deduced from the completeness axiom . . 10 1.14 The Archimedean property of the real-number system . . . . . 10 1.15 Rational numbers with finite decimal representation . . . . . 11 1.16 Finite decimal approximations to real numbers . . . . . . . 11 1.17 Infinite decimal representation of real numbers . . . . . . . 12 1.18 Absolute values and the triangle inequality . . . . . . . . 12 1.19 The Cauchy-Schwarz inequality . . . . . . . . . . . 13 1.20 Plus and minus infinity and the extended real number system R 14 1.21 Complex numbers . . . . . . . . . . . . . . . 15 1.22 Geometric representation of complex numbers . . . . . . . 17 1.23 The imaginary unit . . . . . . . . . . . . . . . 18 1.24 Absolute value of a complex number . . . . . . . . . . 18 1.25 Impossibility of ordering the complex numbers . . . . . . . 19 1.26 Complex exponentials . . . . . . . . . . . . . . 19 1.27 Further properties. of complex exponentials . . . . . . . . 20 1.28 The argument of a complex number . . . . . . . . . . 20 1.29 Integral powers and roots of complex numbers . . . . . . . 21 1.30 Complex logarithms . . . . . . . . . . . . . . 22 1.31 Complex powers . . . . . . . . . . . . . . . 23 1.32 Complex sines and cosines . . . . . . . . ... . . . 24 1.33 Infinity and the extended complex plane C . . . . . . . . 24 Exercises . . . . . . . . . . . . . . . . . . 25 vi Contents Chapter 2 Some Basic Notions of Set Theory 2.1 Introduction . . . . . . . . . . . . . . . . . 32 2.2 Notations . . . . . . . . . . . . . . . . . 32 2.3 Ordered pairs . . . . . . . . . . . . . . . . 33 2.4 Cartesian product of two sets . . . . . . . . . . . . 33 2.5 Relations and functions . . . . . . . . . . . . . 34 2.6 Further terminology concerning functions . . . . . . . . 35 2.7 One-to-one functions and inverses . . . . . . . . . . 36 2.8 Composite functions . . . . . . . . . . . . . . 37 2.9 Sequences . . . . . . . . . . . . . . . . . . 37 2.10 Similar (equinumerous) sets . . . . . . . . . . . . 38 2.11 Finite and infinite sets . . . . . . . . . . . . . . 38 2.12 Countable and uncountable sets . . . . . . . . . 39 2.13 Uncountability of the real-number system . . . . . . . . 39 2.14 Set algebra . . . 40 2.15 Countable collections of countable sets . . . . . . . . . 42 Exercises . . . . . . . . . . . . . . . . . . 43 Chapter 3 Elements of Point Set Topology 3.1 Introduction . . . . . . . . . . . . . . . . . 47 3.2 Euclidean space R" . . . . . . . . . . . . . . . 47 3.3 Open balls and open sets in R" . . . . . . . . . . . 49 3.4 The structure of open sets in R1 . . . . . . . . . . . 50 3.5 Closed sets . . . . . . . . . . . . . . . . . 52 3.6 Adherent points. Accumulation points . . . . . . . . . 52 3.7 Closed sets and adherent points . . . . . . . . . . . 53 3.8 The Bolzano-Weierstrass theorem . . . . . . . . . . 54 3.9 The Cantor intersection theorem . . . . . . . . . . . 56 3.10 The LindelSf covering theorem . . . . . . . . . . . 56 3.11 The Heine-Borel covering theorem . . . . . . . . . . 58 3.12 Compactness in R" . . . . . . . . . . . . . . . 59 3.13 Metric spaces . . . . . . . . . . . . . . . . 60 3.14 Point set topology in metric spaces . . . . . . . . . . 61 3.15 Compact subsets of a metric space . . . . . . . . . . 63 3.16 Boundary of a set . . . . . . . . . . . . . . . 64 Exercises . . . . . . . . . . . . . . . . . . 65 Chapter 4 Limits and Continuity 4.1 Introduction . . . . . . . . . . . . . . . . . 70 4.2 Convergent sequences in a metric space . . . . . . . . . 70 4.3 Cauchy sequences . . . . . . . . . . . . . . . 72 4.4 Complete metric spaces . . . . . . . . . . . . . . 74 4.5 Limit of a function . . . . . . . . . . . . . . . 74 4.6 Limits of complex-valued functions . . . . . . . . . . 76 Contents vii 4.7 Limits of vector-valued functions . . . . . . . . . . . 77 4.8 Continuous functions . . . . . . . . . . . . . . 78 4.9 Continuity of composite functions. . . . . . . . . . . 79 4.10 Continuous complex-valued and vector-valued functions . . . . 80 4.11 Examples of continuous functions . . . . . . . . . . 80 4.12 Continuity and inverse images of open or closed sets . . . . . 81 4.13 Functions continuous on compact sets . . . . . . . . . 82 4.14 Topological mappings (homeomorphisms) . . . . . . . . 84 4.15 Bolzano's theorem . . . . . . . . . . . . . 84 4.16 Connectedness . . . . . . . . . . . . . . . . . 86 4.17 Components of a metric space . . . . . . . . . . . . 87 4.18 Arcwise connectedness . . . . . . . . . . . . . . 88 4.19 Uniform continuity . . . . . . . . . . . . . . 90 4.20 Uniform continuity and compact sets . . . . . . . . . 91 4.21 Fixed-point theorem for contractions . . . . . . . . . 92 4.22 Discontinuities of real-valued functions . . . . . . . . . 92 4.23 Monotonic functions . . . . . . . . . . . . . . 94 Exercises . . . . . . . . . . . . . . . . . . 95 Chapter 5 Derivatives 5.1 Introduction . . . 104 5.2 Definition of derivative . . . . . . . . . . . . . . 104 5.3 Derivatives and continuity . . . . . . . . . . . . . 105 5.4 Algebra of derivatives . . . . . . . . . . . . . . 106 5.5 The chain rule . . . . . . . . . . . . . . . . 106 5.6 One-sided derivatives and infinite derivatives . . . . . . . 107 5.7 Functions with nonzero derivative . . . . . . . . . . 108 5.8 Zero derivatives and local extrema . . . . . . . . . . 109 5.9 Rolle's theorem . . . . . . . . . . . . . . . . 110 5.10 The Mean-Value Theorem for derivatives . . . . . . . . 110 5.11 Intermediate-value theorem for derivatives . . . . . . . . 111 5.12 Taylor's formula with remainder . . . . . . . . . . . 113 5.13 Derivatives of vector-valued functions . . . . . . . . . 114 5.14 Partial derivatives . . . . . . . . . . . . . . . 115 5.15 Differentiation of functions of a complex variable . . . . . . 116 5.16 The Cauchy-Riemann equations . . . . . . . . . . . 118 Exercises . . . . . . . . . . . . . . . . . . 121 Chapter 6 Functions of Bounded Variation and Rectifiable Curves 6.1 Introduction . . . . . . . . . . . . . . . . . 127 6.2 Properties of monotonic functions . . . . . . . . . . 127 6.3 Functions of bounded variation . . . . . . . . . . . 128 6.4 Total variation . . . . . . . . . . . . . . . . 129 6.5 Additive property of total variation . . . . . . . . . . 130 viii Contents 6.6 Total variation on [a, x] as a function of x . . . . . . . . 131 6.7 Functions of bounded variation expressed as the difference of increasing functions . . . . . . . . . . . . . . . 132 6.8 Continuous functions of bounded variation . . . . . . . . 132 6.9 Curves and paths . . . . . . . . . . . . . . . 133 6.10 Rectifiable paths and arc length . . . . . . . . . . . 134 6.11 Additive and continuity properties of arc length . . . . . . . 135 6.12 Equivalence of paths. Change of parameter . . . . . . . 136 Exercises . . . . . . . . . . . . . . . . . . 137 Chapter 7 The Riemann-Stieltjes Integral 7.1 Introduction . . . . . . . . . . . . . . . . . 140 7.2 Notation . . . . . . . . . . . . . . . . . . 141 7.3 The definition of the Riemann-Stieltjes integral . . . . . . . 141 7.4 Linear properties . . . . . . . . . . . . . . . 142 7.5 Integration by parts . . . . . . .. . . . . . . . . 144 7.6 Change of variable in a Riemann-Stieltjes integral . . . . . . 144 7.7 Reduction to a Riemann integral . . . . . . . . . . . 145 7.8 Step functions as integrators . . . . . . . . . . . . 147 7.9 Reduction of a Riemann-Stieltjes integral to a finite sum . . . . 148 7.10 Euler's summation formula . . . . . . . . . . . . 149 7.11 Monotonically increasing integrators. Upper and lower integrals . 150 7.12 Additive and linearity properties of upper and lower integrals 153 7.13 Riemann's condition . . . . . . . . . . . . . . 153 7.14 Comparison theorems . . . . . . . . . . . . . . 155 7.15 Integrators of bounded variation . . . . . . . . . . . 156 7.16 Sufficient conditions for existence of Riemann-Stieltjes integrals 159 7.17 Necessary conditions for existence of Riemann-Stieltjes integrals . 160 7.18 Mean Value Theorems for Riemann-Stieltjes integrals . . . . . 160 7.19 The integral as a function of the interval . . . . . . . . . 161 7.20 Second fundamental theorem of integral calculus . . . . . . 162 7.21 Change of variable in a Riemann integral . . . . . . . . 163 7.22 Second Mean-Value Theorem for Riemann integrals . . . . . 165 7.23 Riemann-Stieltjes integrals depending on a parameter . . . . . 166 7.24 Differentiation under the integral sign . . . . . . . . . 167 7.25 Interchanging the order of integration . . . . . . . . . 167 7.26 Lebesgue's criterion for existence of Riemann integrals . . . . 169 7.27 Complex-valued Riemann-Stieltjes integrals . . . . . . . . 173 Exercises . . . . . . . . . . . . . . . . . . 174 Chapter 8 Infinite Series and Infinite Products 8.1 Introduction . . . . . . . . . . . . . . . . . 183 8.2 Convergent and divergent sequences of complex numbers . . . . 183 8.3 Limit superior and limit inferior of a real-valued sequence . . . 184 8.4 Monotonic sequences of real numbers . . . . . . . . . 185 8.5 Infinite series . . . . . . . . . . . . . . . . . 185 8.6 Contents Inserting and removing parentheses . . . . . . . . . . ix 187 8.7 Alternating series . . . . . . . . . . . . . . . 188 8.8 Absolute and conditional convergence . . . . . . . . . 189 8.9 Real and imaginary parts of a complex series . . . . . . . 189 8.10 Tests for convergence of series with positive terms . . . . . . 190 8.11 The geometric series . . . . . . . . . . . . . . 190 8.12 The integral test . . . . . . . . . . . . . . . . 191 8.13 The big oh and little oh notation . . . . . . . . . . . 192 8.14 The ratio test and the root test . . . . . . . . . . . 193 8.15 Dirichlet's test and Abel's test . . . . . . . . . . . . 193 8.16 Partial sums of the geometric series Y. z" on the unit circle Iz1 = 1 . . 195 8.17 Rearrangements of series . . . . . . . . . . . . . 196 8.18 Riemann's theorem on conditionally convergent series . . . . . 197 8.19 Subseries . . . . . . . . . . . . . . . . . . 197 8.20 Double sequences . . . . . . . . . . . . . . . 199 8.21 Double series . . . . . . . . . . . . . . . . 200 8.22 Rearrangement theorem for double series . . . . . . . . 201 8.23 A sufficient condition for equality of iterated series . . . . . . 202 8.24 Multiplication of series . . . . . . . . . . . . . . 203 8.25 Cesaro summability . . . . . . . . . . . . . . . 205 8.26 Infinite products . . . . . . . . . . . . . . . . 206 8.27 Euler's product for the Riemann zeta function . . . . . . . 209 Exercises . . . . . . . . . . . . . . . . . 210 Chapter 9 9.1 Sequences of Functions Pointwise convergence of sequences of functions . . . . . . 218 9.2 Examples of sequences of real-valued functions . . . . . . . 219 9.3 Definition of uniform convergence . . . . . . . . . . 220 9.4 Uniform convergence and continuity . . . . . . . . . . 221 9.5 The Cauchy condition for uniform convergence . . . . . . 222 9.6 Uniform convergence of infinite series of functions . . . . . . 223 9.7 A space-filling curve . . . . . . . . . . . . . . 224 9.8 Uniform convergence and Riemann-Stieltjes integration . . . . 225 9.9 Nonuniformly convergent sequences that can be integrated term by term . . . . . . . . . . . . . . . . . . . 226 9.10 Uniform convergence and differentiation . . . . . . . . 228 9.11 Sufficient conditions for uniform convergence of a series . . . . 230 9.12 Uniform convergence and double sequences . . . . . . . . 231 9.13 Mean convergence . . . . . . . . . . . . . . . 232 9.14 Power series . . . . . . . . . . . . . . . . . 234 9.15 Multiplication of power series . . . . . . . . . . . . 237 9.16 The substitution theorem . . . . . . . . . . . . . 238 9.17 Reciprocal of a power series . . . . . . . . . . . . 239 9.18 Real power series . . . . . . . . . . . . . . . 240 9.19 The Taylor's series generated by a function . . . . . . . . 241 9.20 Bernstein's theorem . . . . . . . . . . . . . . 242 9.21 The binomial series . . . . . . . . . . . . . . . 244 x Contents 9.22 Abel's limit theorem . . . . . . . . . . . . . . 244 9.23 Tauber's theorem . . . . . . . . . . . . . . . 246 Exercises . . . . . . . . . . . . . . . . . . 247 Chapter 10 10.1 The Lebesgue Integral Introduction . . . . . . . . . . . . . . . . . 252 10.2 The integral of a step function . . . . . . . . . . . 253 10.3 Monotonic sequences of step functions . . . . . . . . . 254 10.4 Upper functions and their integrals . . . . . . . . . . 256 10.5 Riemann-integrable functions as examples of upper functions 259 10.6 The class of Lebesgue-integrable functions on a general interval . 260 10.7 Basic properties of the Lebesgue integral . . . . . . . . . 261 10.8 Lebesgue integration and sets of measure zero . . . . . . . 264 10.9 The Levi monotone convergence theorems . . . . . . . . 265 10.10 The Lebesgue dominated convergence theorem . . . . . . . 270 10.11 Applications of Lebesgue's dominated convergence theorem . . . 272 10.12 Lebesgue integrals on unbounded intervals as limits of integrals on bounded intervals . . . . . . . . . . . . . . . 274 10.13 Improper Riemann integrals . . . . . . . . . . . . 276 10.14 Measurable functions . . . . . . . . . . . . . . 279 10.15 Continuity of functions defined by Lebesgue integrals . . . . . 281 10.16 Differentiation under the integral sign . . . . . . . . . 283 10.17 Interchanging the order of integration . . . . . . . . . 287 10.18 Measurable sets on the real line . . . . . . . . . . . 289 10.19 The Lebesgue integral over arbitrary subsets of R . . . . . . 291 10.20 Lebesgue integrals of complex-valued functions . . . . . . . 292 10.21 Inner products and norms . . . . . . . . . . . . . 293 10.22 The set L2(I) of square-integrable functions . . . . . . . . 294 10.23 The set L2(I) as a semimetric space . . . . . . . . . . 295 10.24 A convergence theorem for series of functions in L2(I) . . . . 295 10.25 The Riesz-Fischer theorem . . . . . . . . . . . . 297 Exercises . . . . . . . . . . . . . . . . . . 298 Chapter 11 Fourier Series and Fourier Integrals 11.1 Introduction . . . . . . . . . . . . . . . . . 306 11.2 Orthogonal systems of functions . . . . . . . . . . . 306 11.3 The theorem on best approximation . . . . . . . . . . 307 11.4 The Fourier series of a function relative to an orthonormal system . . 309 11.5 Properties of the Fourier coefficients . . . . . . . . . . . 309 11.6 The Riesz-Fischer theorem . . . . . . . . . . . . 311 11.7 The convergence and representation problems for trigonometric series 312 11.8 The Riemann-Lebesgue lemma . . . . . . . . . . . 313 11.9 The Dirichlet integrals . . . . . . . . . . . . . . 314 11.10 An integral representation for the partial sums of a Fourier series . 317 11.11 Riemann's localization theorem . . . . . . . . . . . 318 Contents xi 11.12 Sufficient conditions for convergence of a Fourier series at a particular point . . . . . . . . . . . . . . . . . . . 319 11.13 Ceshro summability of Fourier series . . . . . . . . . . 319 11.14 Consequences of Fej6r's theorem . . . . . . . . . . . 321 11.15 The Weierstrass approximation theorem . . . . . . . . . 322 11.16 Other forms of Fourier series . . . . . . . . . . . . 322 11.17 The Fourier integral theorem . . . . . . . . . . . . 323 11.18 The exponential form of the Fourier integral theorem . . . . . 325 11.19 Integral transforms . . . . . . . . . . . . . . . 326 11.20 Convolutions . . . . . . . . . . . . . . . . 327 11.21 The convolution theorem for Fourier transforms . . . . . . 329 11.22 The Poisson summation formula . . . . . . . . . . . 332 Exercises . . . . . . . . . . . . . . . . . . 335 Chapter 12 Multivariable Differential Calculus 12.1 Introduction . . . . . . . . . . . . . . . . . 344 12.2 The directional derivative . . . . . . . . . . . . . 344 12.3 Directional derivatives and continuity . . . . . . . . . 345 12.4 The total derivative . . . . . . . . . . . . . . . 346 12.5 The total derivative expressed in terms of partial derivatives . . . 347 12.6 An application to complex-valued functions . . . . . . . . 348 12.7 The matrix of a linear function . . . . . . . . . . . 349 12.8 The Jacobian matrix . . . . . . . . . . . . . . 351 12.9 The chain rule . . . . . . . . . . . . . . . . 352 12.10 Matrix form of the chain rule . . . . . . . . . . . . 353 12.11 The Mean-Value Theorem for differentiable functions . . . . . 355 12.12 A sufficient condition for differentiability . . . . . . . . 357 12.13 A sufficient condition for equality of mixed partial derivatives . . 358 12.14 Taylor's formula for functions from R" to RI . . . . . . . 361 Exercises . . . . . . . . . . . . . . . . . . 362 Chapter 13 Implicit Functions and Extremum Problems 13.1 Introduction . . . . . . . . . . . . . . . . . 367 13.2 Functions with nonzero Jacobian determinant . . . . . . . 368 13.3 The inverse function theorem . . . . . . . . . . . . 372 13.4 The implicit function theorem . . . . . . . . . . . . 373 13.5 Extrema of real-valued functions of one variable . . . . . . 375 13.6 Extrema of real-valued functions of several variables . . . . . 376 13.7 Extremum problems with side conditions . . . . . . . . 380 Exercises . . . . . . . . . . . . . . . . . . 384 Chapter 14 Multiple Riemann Integrals 14.1 Introduction . . . . . . . . . . . . . . . . . 388 14.2 The measure of a bounded interval in R" . . . . . . . . 388 xii Contents 14.3 The Riemann integral of a bounded function defined on a compact interval in R" . . . . . . . . . . . . . . . . . 389 14.4 Sets of measure zero and Lebesgue's criterion for existence of a multiple Riemann integral . . . . . . . . . . . . . . . 391 14.5 Evaluation of a multiple integral by iterated integration . . . . 391 14.6 Jordan-measurable sets in R" . . . . . . . . . . . . 396 14.7 Multiple integration over Jordan-measurable sets . . . . . . 397 14.8 Jordan content expressed as a Riemann integral . . . . . . 398 14.9 Additive property of the Riemann integral . . . . . . . . 399 14.10 Mean-Value Theorem for multiple integrals . . . . . . . . 400 Exercises . . . . . . . . . . . . . . . . . . 402 Chapter 15 Multiple Lebesgue Integrals 15.1 Introduction . . . . . . . . . . . . . . . 405 15.2 Step functions and their integrals . . . . . . . . . . . 406 15.3 Upper functions and Lebesgue-integrable functions . . . . . 406 15.4 Measurable functions and measurable sets in R'. . . . . . . 407 15.5 Fubini's reduction theorem for the double integral of a step function . 409 15.6 Some properties of sets of measure zero . . . . . . . . . 411 15.7 Fubini's reduction theorem for double integrals . . . . . . 413 15.8 The Tonelli-Hobson test for integrability . . . . . . . . 415 15.9 Coordinate transformations . . . . . . . . . . . . 416 15.10 The transformation formula for multiple integrals . . . . . . 421 15.11 Proof of the transformation formula for linear coordinate transforma-tions . . . . . . . . . . . . . . . . . . . 421 15.12 Proof of the transformation formula for the characteristic function of a compact cube . . . . . . . . . . . . . . . . 423 15.13 Completion of the proof of the transformation formula . . . . 429 Exercises . . . . . . . . . . . . . . . . . . 430 Chapter 16 Cauchy's Theorem and the Residue Calculus 16.1 Analytic functions . . . . . . . . . . . . . . . 434 16.2 Paths and curves in the complex plane . . . . . . . . . 435 16.3 Contour integrals . . 436 16.4 The integral along a circular path as a function of the radius . . . 438 16.5 Cauchy's integral theorem for a circle . . . . . . . . . 439 16.6 Homotopic curves 439 16.7 Invariance of contour integrals under homotopy . . . . . . 442 16.8 General form of Cauchy's integral theorem . . . . . . . . 443 16.9 Cauchy's integral formula . . . . . . . . . . . . . 443 16.10 The winding number of a circuit with respect to a point . . . . 444 16.11 The unboundedness of the set of points with winding number zero . 446 16.12 Analytic functions defined by contour integrals . . . . . . . 447 16.13 Power-series expansions for analytic functions . . . . . . . 449 16.14 Cauchy's inequalities. Liouville's theorem . . . . . . . . 450 16.15 Isolation of the zeros of an analytic function . . . . . . . 451 Contents xiii 16.16 The identity theorem for analytic functions . . . . . . . 452 16.17 The maximum and minimum modulus of an analytic function . . 453 16.18 The open mapping theorem . . . . . . . . . . . . 454 16.19 Laurent expansions for functions analytic in an annulus . . . . 455 16.20 Isolated singularities . . . . . . . . . . . . . . 457 16.21 The residue of a function at an isolated singular point . . . . . 459 16.22 The Cauchy residue theorem . . . . . . . . . . . . 460 16.23 Counting zeros and poles in a region . . . . . . . . . . 461 16.24 Evaluation of real-valued integrals by means of residues . . . . 462 16.25 Evaluation of Gauss's sum by residue calculus . . . . . . . 464 16.26 Application of the residue theorem to the inversion formula for Laplace transforms . . . . . . . . . . . . . . . . . 468 16.27 Conformal mappings . . . . . . . . . . . . . . 470 Exercises . . . . . . . . . . . . . . . . . . 472 Index of Special Symbols . . . . . . . . . . . . . 481 Index . . . . . . . . . . . . . . . . . . . 485 CHAPTER 1 THE REAL AND COMPLEX NUMBER SYSTEMS 1.1 INTRODUCTION Mathematical analysis studies concepts related in some way to real numbers, so we begin our study of analysis with a discussion of the real-number system. Several methods are used to introduce real numbers. One method starts with the positive integers 1, 2, 3, ... as undefined concepts and uses them to build a larger system, the positive rational numbers (quotients of positive integers), their negatives, and zero. The rational numbers, in turn, are then used to construct the irrational numbers, real numbers like -,/2 and iv which are not rational. The rational and irrational numbers together constitute the real-number system. Although these matters are an important part of the foundations of math-ematics, they will not be described in detail here. As a matter of fact, in most phases of analysis it is only the properties of real numbers that concern us, rather than the methods used to construct them. Therefore, we shall take the real numbers themselves as undefined objects satisfying certain axioms from which further properties will be derived. Since the reader is probably familiar with most of the properties of real numbers discussed in the next few pages, the presentation will be rather brief. Its purpose is to review the important features and persuade the reader that, if it were necessary to do so, all the properties could be traced back to the axioms. More detailed treatments can be found in the references at the end of this chapter. For convenience we use some elementary set notation and terminology. Let S denote a set (a collection of objects). The notation x e S means that the object x is in the set S, and we write x 0 S to indicate that x is not in S. A set S is said to be a subset of T, and we write S s T, if every object in S is also in T. A set is called nonempty if it contains at least one object. We assume there exists a nonempty set R of objects, called real numbers, which satisfy the ten axioms listed below. The axioms fall in a natural way into three groups which we refer to as the field axioms, the order axioms, and the completeness axiom (also called the least-upper-bound axiom or the axiom of continuity). 1.2 THE FIELD AXIOMS Along with the-set R of real numbers we assume the existence of two operations, called addition and multiplication, such that for every pair of real numbers x and y 1 2 Real and Complex Numbcr Systems Ax.1 the sum x + y and the product xy are real numbers uniquely determined by x and y satisfying the following axioms. (In the axioms that appear below, x, y, z represent arbitrary real numbers unless something is said to the contrary.) Axiom 1. x + y = y + x, xy = yx (commutative laws). Axiom 2. x + (y + z) = (x + y) + z, x(yz) _ (xy)z (associative laws). Axiom 3. x(y + z) = xy + xz (distributive law). Axiom 4. Given any two real numbers x and y, there exists a real number z such that x + z = y. This z is denoted by y - x; the number x - x is denoted by 0. (It can be proved that 0 is independent of x.) We write - x for 0 - x and call - x the negative of x. Axiom S. There exists at least one real number x 96 0. If x and y are two real numbers with x 0, then there exists a real number z such that xz = y. This z is denoted by y/x; the number x/x is denoted by 1 and can be shown to be independent of x. We write x-1 for 1/x if x 0 and call x-1 the reciprocal of x. From these axioms all the usual laws of arithmetic can be derived; for example, -(-x)=x,(x-1)-1=x, -(x-y)=y-x,x-y=x+(-y),etc. (For a more detailed explanation, see Reference 1.1.) 1.3 THE ORDER AXIOMS We also assume the existence of a relation < which establishes an ordering among the real numbers and which satisfies the following axioms: Axiom 6. Exactly one of the relations x = y, x < y, x > y holds. NOTE. x > y means the same as y < x. Axiom 7. If x < y, then for every z we have x + z < y + z. Axiom 8. If x > 0 and y > 0, then xy > 0. Axiom 9. If x > y and y > z, then x > z. NOTE. A real number x is called positive if x > 0, and negative if x < 0. We denote by R+ the set of all positive real numbers, and by R- the set of all negative real numbers. From these axioms we can derive the usual rules for operating with inequalities. For example, if we have x < y, then xz < yz if z is positive, whereas xz > yz if z is negative. Also, if x > y and z > w where both y and w are positive, then xz > yw. (For a complete discussion of these rules see Reference 1.1.) NOTE. The symbolism x < y is used as an abbreviation for the statement: 46x < y or Y . Th. 1.1 Intervals 3 Thus we have 2 < 3 since 2 < 3; and 2 < 2 since 2 = 2. The symbol >- is similarly used. A real number x is called nonnegative if x >- 0. A pair of simul-taneous inequalities such as x < y, y < z is usually written more briefly as x<y<z. The following theorem, which is a simple consequence of the foregoing axioms, is often used in proofs in analysis. Theorem 1.1. Given real numbers a and b such that a < b + e for every e > 0. (1) Then a S b. Proof. If b < a, then inequality (1) is violated for e = (a - b)/2 because b+e=b+a-b=a+b<a+a=a 2 2 2 Therefore, by Axiom 6 we must have a < b. Axiom 10, the completeness axiom, will be described in Section 1.11. 1.4 GEOMETRIC REPRESENTATION OF REAL NUMBERS The real numbers are often represented geometrically as points on a line (called the real line or the real axis). A point is selected to represent 0 and another to represent 1, as shown in Fig. I.I. This choice determines the scale. Under an appropriate set of axioms for Euclidean geometry, each point on the real line corresponds to one and only one real number and, conversely, each real number is represented by one and only one point on the line. It is customary to refer to the point x rather than the point representing the real number x. 0 1 x y Figure 1.1 The order relation has a simple geometric interpretation. If x < y, the point x lies to the left of the point y, as shown in Fig. 1.1. Positive numbers lie to the right of 0, and negative numbers to the left of 0. If a < b, a point x satisfies the inequalities a < x < b if and only if x is between a and b. 1.5 INTERVALS The set of all points between a and b is called an interval. Sometimes it is important to distinguish between intervals which include their endpoints and intervals which do not. NOTATION. The notation {x: x satisfies P} will be used to designate the set of all real numbers x which satisfy property P. 4 Real and Complex Number Systems Def. 1.2 Definition 1.2. Assume a < b. The open interval (a, b) is defined to be the set (a, b) = {x : a < x < b}. The closed interval [a, b] is the set {x: a < x < b}. The half-open intervals (a, b] and [a, b) are similarly defined, using the inequalities a < x :!5; b and a < x < b, respectively. Infinite intervals are defined as follows: (a, + oo) = {x : x > a), [a, + oo) = {x : x a}, (- oo, a) = {x: x < a}, (- oo, a] = {x : x < a}. The real line R is sometimes referred to as the open interval (- co, + co). A single point is considered as a "degenerate" closed interval. NOTE. The symbols + oo and - oo are used here purely for convenience in notation and are not to be considered as being real numbers. Later we shall extend the real-number system to include these two symbols, but until this is done, the reader should understand that all real numbers are "finite." 1.6 INTEGERS This section describes the integers, a special subset of R. Before we define the integers it is convenient to introduce first the notion of an inductive set. Definition 1.3. A set of real numbers is called an inductive set if it has the following two properties: a) The number 1 is in the set. b) For every x in the set, the number x + 1 is also in the set. For example, R is an inductive set. So is the set V. Now we shall define the positive integers to be those real numbers which belong to every inductive set. Definition 1.4. A real number is called a positive integer if it belongs to every inductive set. The set of positive integers is denoted by V. The set Z+ is itself an inductive set. It contains the number 1, the number 1 + 1 (denoted by 2), the number 2 + 1 (denoted by 3), and so on. Since Z' is a subset of every inductive set, we refer to Z+ as the smallest inductive set. This property of Z+ is sometimes called the principle of induction. We assume the reader is familiar with proofs by induction which are based on this principle. (See Reference 1.1.) Examples of such proofs are given in the next section. The negatiyes of the positive integers are called the negative integers. The positive integers, together with the negative integers and 0 (zero), form a set Z which we call simply the set of integers. 1.7 THE UNIQUE FACTORIZATION THEOREM FOR INTEGERS If n and d are integers and if n = cd for some integer c, we say d is a divisor of n, or n is a multiple of d, and we write d In (read : d divides n). An integer n is called Th. 1.7 Unique Factorization Theorem 5 a prime if n > 1 and if the only positive divisors of n are 1 and n. If n > 1 and n is not prime, then n is called composite. The integer 1 is neither prime nor composite. This section derives some elementary results on factorization of integers, culminating in the unique factorization theorem, also called the fundamental theorem of arithmetic. The fundamental theorem states that (1) every integer n > 1 can be represented as a product of prime factors, and (2) this factorization can be done in only one way, apart from the order of the factors. It is easy to prove part (1). Theorem 1.5. Every integer n > 1 is either a prime or a product of primes. Proof. We use induction on n. The theorem holds trivially for n = 2. Assume it is true for every integer k with 1 < k < n. If n is not prime it has a positive divisor d with 1 < d < n. Hence n = cd, where 1 < c < n. Since both c and d are 0, b > 0 and use induction on n = a + b. If n = 0 then a = b = 0, and we can take d = 0 with x = y = 0. Assume, then, that the theorem has been proved for 0, 1, 2, ... , n - 1. By symmetry, we can assume a >- b. If b = 0 take d = a, x = 1, y = 0. If b >- 1 we can apply the induction hypothesis to a - b and b, since their sum is a = n - b < n - 1. Hence there is a common divisor d of a - b and b of the form d = (a - b)x + by. This d also divides (a - b) + b = a, so d is a common divisor of a and b and we have d = ax + (y - x)b, a linear combination of a and b. To complete the proof we need to show that every common divisor divides d. Since a common divisor divides a and b, it also divides the linear combination ax + (y - x)b = d. This completes the proof if a >- 0 and b Z 0. If one or both ofa and b is negative; apply the result just proved to lal and obi. NOTE. If d is a common divisor of a and b of the form d = ax + by, then - d is also a divisor of the same form, -d = a(- x) + b(-y). Of these two common divisors, the nonnegative one is called the greatest common divisor of a and b, and is denoted by gcd(a, b) or, simply by (a, b). If (a, b) = I then a and b are said to be relatively prime. Theorem 1.7(Euclid's Lemma). If albc and (a, b) = 1, then alc. 6 Real and Complex Number Systems Th. 1.8 Proof. Since (a, b) = 1 we can write 1 = ax + by. Therefore c = acx + bcy. But aiacx and ajbcy, so aic. Theorem 1.8. If a prime p divides ab, then pia or pib. More generally, if a prime p, divides a product a1 . ak, then p divides at least one of the factors. Proof. Assume plab and that p does not divide a. If we prove that (p, a) = 1, then Euclid's Lemma implies pib. Let d = (p, a). Then dip so d = I or d = p. We cannot have d = p because dia but p does not divide a. Hence d = 1. To prove the more general statement we use induction on k, the number of factors. Details are left to the reader. Theorem 1.9 (Unique factorization theorem). Every integer n > I can be repre-sented as a product of prime factors in only one way, apart from the order of the factors. . Proof. We use induction on n. The theorem is true for n = 2. Assume, then, that it is true for all integers greater than 1 and less than n. If n is prime there is nothing more to prove. Therefore assume that n is composite and that n has two factorizations into prime factors, say n = PiP2 ' Ps = g1g2 "' qt (2) We wish to show that s = t and that each p equals some q. Since pt divides the product q 1 q 2 q, it divides at least one factor. Relabel the q's if necessary so that p1l q 1. Then pt = q 1 since both pt and q 1 are primes. In (2) we cancel pt on both sides to obtain n -=P2...PS=g2"'gr P1 Since n .is composite, 1 < n/p1 < n; so by the induction hypothesis the two factorizations of n/p1 are identical, apart from the order of the factors. Therefore the same is true in (2) and the proof is complete. 1.8 RATIONAL NUMBERS Quotients of integers alb (where b : 0) are called rational numbers. For example, 1/2, - 7/5, and 6 are rational numbers. The set of rational numbers, which we denote by Q, contains Z as a subset. The reader should note that all the field axioms and the order axioms are satisfied by Q. We assume that the reader is familiar with certain elementary properties of rational numbers. For example, if a and b are rational, their average (a + b)/2 is also rational and lies between a and b. Therefore between any two rational numbers there are infinitely many rational numbers, which implies that if we are given a certain rational number we cannot speak of the "next largest" rational number. 7b. 1.11 Irrational Numbers 7 1.9 IRRATIONAL NUMBERS Real numbers that are not rational are called irrational. For example, the numbers -,/2, e, ir and e" are irrational. Ordinarily it is not too easy to prove that some particular number is irrational. There is no simple proof, for example, of the irrationality of e. However, the irrationality of certain numbers such as and 3 is not too difficult to establish and, in fact, we easily prove the following : Theorem 1.10. If n is a positive integer which is not a perfect square, then i is irrational. Proof Suppose first that n contains no square factor > 1. We assume that is rational and obtain a contradiction. Let n = a/b, where a and b are integers having no factor in common. Then nb2 = a2 and, since the left side of this equation is a multiple of n, so too is a2. However, if a2 is a multiple of n, a itself must be a multiple of n, since n has no square factors > 1. (This is easily seen by examining the factorization of a into its prime factors.) This means that a = cn, where c is some integer. Then the equation nb2 = a2 becomes nb2 = c2n2, or b2 = nc2. The same argument shows that b must also be a multiple of n. Thus a and b are both multiples of n, which contradicts the fact that they have no factor in common. This completes the proof if n has no square factor > 1. If n has a square factor, we can write n = m2k, where k > 1 and k has no square factor > 1. Then = m,/k; and if '/n were rational, the number fk-would also be rational, contradicting that which was just proved. A different type of argument is needed to prove that the number e is irrational. (We assume familiarity with the exponential e" from elementary calculus and its representation as an infinite series.) Theorem 1.11. If ex = 1 + x + x2/2! + x3/3! + + x"/n! + , then the number e is irrational. Proof. We shall prove that a-1 is irrational. The series for a-1 is an alternating series with terms which decrease steadily in absolute value. In such an alternating series the error made by stopping at the nth term has the algebraic sign of the first neglected term and is less in absolute value than the first neglected term. Hence, if s" = Ek=o (- 1)k/k!, we have the inequality 1 -1 0 < e - s2k_1 < (2k) ! ' from which we obtain 0 < (2k - 1)! (e-1 - s2k-1) < 2k 2' (3) for any integer -k z 1. Now (2k - D! s2k _ 1 is always an integer. If e-' were rational, then we could choose k so large that (2k - 1)! a-1 would also be an 8 Real and Complex Number Systems Def. 1.12 integer. Because of (3) the difference of these two integers would be a number between 0 and 1, which is impossible. Thus a-1 cannot be rational, and hence e cannot be rational. NOTE. For a proof that iv is irrational, see Exercise 7.33. The ancient Greeks were aware of the existence of irrational numbers as early as 500 B.c. However, a satisfactory theory of such numbers was not developed until late in the nineteenth century, at which time three different theories were. introduced by Cantor, Dedekind, and Weierstrass. For an account of the theories of Dedekind and Cantor and their equivalence, see Reference 1.6. 1.10 UPPER BOUNDS, MAXIMUM ELEMENT, LEAST UPPER BOUND (SUPREMUM) Irrational numbers arise in algebra when we try to solve certain quadratic equa-tions. For example, it is desirable to have a real number x such that x2 = 2. From the nine axioms listed above we cannot prove that such an x exists in R because these nine axioms are also satisfied by Q and we have shown that there is no rational number whose square is 2. The completeness axiom allows us to introduce irrational numbers in the real-number system, and it gives the real-number system a property of continuity that is fundamental to many theorems in analysis. Before we describe the completeness axiom, it is convenient to introduce additional terminology and notation. Definition 1.12. Let S be a set of real numbers. If there is a real number b such that x < b for every x in S, then b is called an upper bound for S and we say that S is bounded above by b. We say an upper bound because every number greater than b will also be an upper bound. If an upper bound b is also a member of S, then b is called the largest member or the maximum element of S. There can be at most one such b. If it exists, we write b = max S. A set with no upper bound is said to be unbounded above. Definitions of the terms lower bound, bounded below, smallest member (or minimum element) can be similarly formulated. If S has a minimum element we denote it by min S. Examples 1. The set R+ = (0, + oo) is unbounded above. It has no upper bounds and no max-imum element. It is bounded below by 0 but has no minimum element. 2. The closed interval S = [0, 1 ] is bounded above by 1 and is bounded below by 0. In fact, max S = 1 and min S = 0. 3. The half-open interval S = [0, 1) is bounded above by 1 but it has no maximum element. Its minimum element is 0. Th. 1.14 Some Properties of the Supremum 9 For sets like the one in Example 3, which are bounded above but have no maximum element, there is a concept which takes the place of the maximum ele-ment. It is called the least upper bound or supremum of the set and is defined as follows : Definition 1.13. Let S be a set of real numbers bounded above. A real number b is called a least upper bound for S if it has the following two properties: a) b is an upper bound for S. b) No number less than b is an upper bound for S. Examples. If S = [0, 1 ] the maximum element 1 is also a least upper bound for S. If S = [0, 1) the number 1 is a least upper bound for S, even though S has no maximum element. It is an easy exercise to prove that a set cannot have two different least upper bounds. Therefore, if there is a least upper bound for S, there is only one and we can speak of the least upper bound. It is common practice to refer to the least upper bound of a set by the more concise term supremum, abbreviated sup. We shall adopt this convention and write b = sup S to indicate that b is the supremum of S. If S has a maximum element, then max S = sup S. The greatest lower bound, or infimum of S, denoted by inf S, is defined in an analogous fashion. 1.11 THE COMPLETENESS AXIOM Our final axiom for the real number system involves the notion of supremum. Axiom 10. Every nonempty set S of real 'numbers which is bounded above has a supremum; that is, there is a real number b such that b = sup S. As a consequence of this axiom it follows that every nonempty set of real numbers which is bounded below has an infimum. 1.12 SOME PROPERTIES OF THE SUPREMUM This section discusses some fundamental properties of the supremum that will be useful in this text. There is a corresponding set of properties of the infimum that the reader should formulate for himself. The first property shows that a set with a supremum contains numbers arbi-trarily close to its supremum. Theorem 1.14 (Approximation property). Let S be a nonempty set of real numbers with a supremum, say b = sup S. Then for every a < b there is some x in S such that a<x<b. 10 Real and Complex Number Systems Th. 1.15 Proof. First of all, x < b for all x in S. If we had x < a for every x in S, then a would be an upper bound for S smaller than the least upper bound. Therefore x > a for at least one x in S. Theorem 1.15 (Additive property). Given nonempty subsets A and B of R, let C denote the set C= {x + y:xc-A, yEB}. If each of A and B has a supremum, then C has a supremum and sup C = sup A + sup B. Proof. Let a = sup A, b = sup B. If z e C then z = x + y, where x c- A, y e B, so z = x + y <- a + b. Hence a + b is an upper bound for C, so C has a supremum, say c = sup C, and c < a + b. We show next that a + b < c. Choose any e > 0. By Theorem 1.14 there is an x in A and a y in B such that a - E<x and b - E<y. Adding these inequalities we find a + b - 2E<x+y<c. Thus, a + b < c + 2E for every e > 0 so, by Theorem 1.1, a + b < c. The proof of the next theorem is left as an exercise for the reader. Theorem 1.16 (Comparison property). Given nonempty subsets S and T of R such that s <- t for every s in S and tin T. If T has a supremum then S has a supremum and sup S < sup T. 1.13 PROPERTIES OF THE INTEGERS DEDUCED FROM THE COMPLETENESS AXIOM Theorem 1.17. The set Z+ of positive integers 1, 2, 3, ... is unbounded above. Proof. If Z+ were bounded above then Z+ would have a supremum, say a = sup V. By Theorem 1.14 we would have a - 1 < n for some n in Z+. Then n + 1 > a for this n. Since n + 1 e Z+ this contradicts the fact that a = sup Z+. Theorem 1.18. For every real x there is a positive integer n such that n > x. Proof. If this were not true, some x would be an upper bound for Z+, contra-dicting Theorem 1.17. -1.14 THE ARCHIMEDEAN PROPERTY OF THE REAL NUMBER SYSTEM The next theorem describes the Archimedean property of the real number system. Geometrically, it tells us that any line segment, no matter how long, can be Th. 1.20 Finite Decimal Approximations 11 covered by a finite number of line segments of a given positive length, no matter how small. Theorem 1.19. If x > 0 and if y is an arbitrary real number, there is a positive integer n such that nx > y. Proof. Apply Theorem 1.18 with x replaced by y/x. 1.15 RATIONAL NUMBERS WITH FINITE DECIMAL REPRESENTATION A real number of the form r=ao+++---+ 10 102 10" where ao is a nonnegative integer and al, ... , a are integers satisfying 0 < at < 9, is usually written more briefly as follows: r = afinite decimal representation of r. For example, 1 __ 5 =0.5, 1 __ 2 =0.02, 29=7+ 2+ 5 =7.25. 2 10 50 102 4 10 102 Real numbers like these are necessarily rational and, in fact, they all have the form r = a/10", where a is an integer. However, not all rational numbers can be ex-pressed with finite decimal representations. For example, if } could be so expressed, then we would have I = a/10" or 3a = 10" for some integer a. But this is im-possible since 3 does not divide any power of 10. 1.16 FINITE DECIMAL APPROXIMATIONS TO REAL NUMBERS This section uses the completeness axiom to show that real numbers can be approximated to any desired degree of accuracy by rational numbers with finite decimal representations. Theorem 1.20. Assume x > 0. Then for every integer n > 1 there is a finite decimal r" = ao. ala2 - - a" such that 1 r"<x<r"+-. Ion Proof. Let S be the set of all nonnegative integers <x. Then S is nonempty, since 0 e S, and S is bounded above by x. Therefore S has a supremum, say ao = sup S. It is easily verified that ao e S, so ao is a nonnegative integer. We call ao the greatest integer in x, and we write ao = [x]. Clearly, we have ao<x<ao+1. 12 Real and Complex Number Systems Th. 1.20 Now let a1 = [lox - 10ao], the greatest integer in lOx - 10ao. Since 0 < lOx - 10ao = 10(x - ao) < 10, we have 0 < a1 5 9 and a1<lox -l0ao<a1+l. In other words, a1 is the largest integer satisfying the inequalities ao+io<<x<ao+a11+01 More generally, having chosen a1, ... , an_1 with 0 <- ai 5 9, let a" be the largest integer satisfying the inequalities am a" < x < ao + a1 + + a" ao+0+...+10"o... 10" (4) Then 0 < a" < 9 and we have , r"<x<r"+ion where r" = ao. a1a2 a". This completes the proof. It is easy to verify that x is actually the supremum of the set of rational numbers r1, r2, ... . 1.17 INFINITE DECIMAL REPRESENTATIONS OF REAL NUMBERS The integers ao, a1, a2, ... obtained in the proof of Theorem 1.20 can be used to define an infinite decimal representation of x. We write x = ao. ala2 .. . to mean that a" is the largest integer satisfying (4). For example, if x = } we find ao = 0, a1 = 1, a2 = 2, a3 = 5, and a" = 0 for all n Z 4. Therefore we can write = 0.125000 If we interchange the inequality signs S and < in (4), we obtain a slightly different definition of decimal expansions. The finite decimals r,, satisfy r" < x < r" + 10-" although the digits ao, a1, a2i ... need not be the same as those in (4). For example, if we apply this second definition to x = } we find the infinite decimal representation = 0.124999 The fact that a real number might have two different decimal representations is merely a reflection of the fact that two different sets of real numbers can have the same supremum. 1.18 ABSOLUTE VALUES AND THE TRIANGLE INEQUALITY Calculations with inequalities arise quite frequently in analysis. They are of particular importance in dealing with the notion of absolute value. If x is any real Th. 1.22 Cauchy-Schwarz Inequality number, the absolute value of x, denoted by lxj, is defined as follows: Ixl x, ifx 0, -x, ifx50. 13 A fundamental inequality concerning absolute values is given in the following: Theorem 1.21. If a >- 0, then we have the inequality Ixl 5 a if, and only if, -a<xSa. Proof. From the definition of Ixl, we have the inequality - Ixl 5 x < Ixl, since x = Ixl or x = - lxl. If we assume that lxl 5 a, then we can write -a 5 - Ixl 5 x 5 lxl 5 a and thus half of the theorem is proved. Conversely, let us assume -a 5 x 5 a. Then ifx >_ 0, we have lxl = x 5 a, whereas ifx < 0, we have lxl = -x 5 a. In either case we have lxi 5 a and the theorem is proved. We can use this theorem to prove the triangle inequality. Theorem 1.22. For arbitrary real x and y we have Ix + yl <- Ixl + IYI (the triangle inequality). Proof. We have - lxl -< x 5 lxl and -I yj <- y <- l yl. Addition gives us -(Ixl + IYI) <- x + y 5 lxl + lyl, and from Theorem 1.21 we conclude that Ix + yl 5 lxl + l yl. This proves the theorem. The triangle inequality is often used in other forms. For example, if we take x = a - c and y = c - b in Theorem 1.22 we find la-bl5 la - cl+lc - bl. Also, from Theorem 1.22 we have Ixl >- Ix + yl - IYI Taking x = a + b, y = - b, we obtain Ia + bl z lal - lbl. Interchanging a and b we also find J a + bl > I bI - j al = - (lal - Ibl), and hence la + bl z hlal - IbIl. By induction we can also prove the generalizations Ix1 + x2 + ... + x.l 5 Ix1l + Ixzl + ... + and Ixl + x2 + ... + x.l z Ix1l - Ix21 14-1.19 THE CAUCHY-SCHWARZ INEQUALITY We shall now derive another inequality which is often used in analysis. 14 Real and Complex Number Systems Tb.1.23 Theorem 1.23 (Cauchy-Schwarz inequality). If a1,.. . , an and b1i ... , bn are arbitrary real numbers, we have Ck = 1 akbkl2 - E a)(kr b k) . Moreover, if some a. 0 equality holds if and only if there is a real x such that akx + bk = 0 for each k = 1, 2,,, .. , n. Proof. A sum of squares can never be negative. Hence we have (akx + bk)2 > 0 k=1 for every real x, with equality if and only if each term is zero. This inequality can be written in the form where Ax2+2Bx+C>-0, n n 2 A ak, B akbk, k=1 k=11 n C=Ebk. k=1 If A > 0, put x = -B/A to obtain B2 - AC < 0, which is the desired inequality. If A = 0, the proof is trivial. NOTE. In vector notation the Cauchy-Schwarz inequality takes the form (a b)2 < Ila11211b112, where a = (a1, ... , an), b = (b1i ... , bn) are two n-dimensional vectors, a b = akbk, k=1 is their dot product, and Ilall = (a a)''2 is the length of a. 1.20 PLUS AND MINUS INFINITY AND THE EXTENDED REAL NUMBER SYSTEM R Next we extend the real number system by adjoining two "ideal points" denoted by the symbols + oo and - oo ("plus infinity" and "minus infinity"). Definition 1.24. By the extended real number system R we shall mean the set of real numbers R together with two symbols + co and - oo which satisfy the following properties: a) If x e R, then we have x+(+oo)= +oo, x+(-co)= -oo, x - (+00} = -oo, x - (-00) = +00, x/(+oo)=x/(-co)=0. Def. 1.26 Complex Numbers b) If x > 0, then we have x(+ oo) = + oo, c) If x < 0, then we have x(+oo) = -oo, X(- 00) = -oo. X(- 00) = +oo. d)-,k+ oo) + (+oo) = (+oo)(+oo) = (-oo)(-oo) = +oo, (-co) + (-00) = (+c0)(-o0) = -oo. e) If X E R, then we have - oo < x < + co. 15 NOTATION. We denote R by (- oo, + co) and R by [ - oo, + oo]. The points in R are called "finite" to distinguish them from the "infinite" points + oo and - oo. The principal reason for introducing the symbols + oo and - oo is one of convenience. For example, if we define + oo to be the sup of a set of real numbers which is not bounded above, then every nonempty subset of R has a supremum in R. The sup is finite if the set is bounded above and infinite if it is not bounded above. Similarly, we define - oo to be the inf of any set of real numbers which is not bounded below. Then every nonempty subset of R has an inf in R. For some of the later work concerned with limits, it is also convenient to introduce the following terminology. Definition 1.25. Every open interval (a, + oo) is called a neighborhood of + 00 or a ball with center + co. Every open interval (- oo, a) is called a neighborhood of - o0 or a ball with center - oo. 1.21 COMPLEX NUMBERS It follows from the axioms governing the relation < that the square of a real number is never negative. Thus, for example, the elementary quadratic equation 2 x = - I has no solution among the real numbers. New types of numbers, called complex numbers, have been introduced to provide solutions to such equations. It turns out that the introduction of complex numbers provides, at the same time, solutions to general algebraic equations of the form where the' coefficients ao, a,, ... , a" are arbitrary real numbers. (This fact is known as the Fundamental Theorem of Algebra.) We shall now define complex numbers and discuss them in further detail. Definition 1.26. By a complex number we shall mean an ordered pair of real numbers which we denote by (x,, x2). The first member, x,, is called the real part of the complex number; the second member, x2, is called the imaginary part. Two complex numbers x = (x,, x2) and y = (y,, y2) are called equal, and we write x = y, if, 16 Real and Complex Number Systems Th. 1.27 and only if, x1 = y1 and x2 = Y2. We define the sum x + y and the product xy by the equations x + y = (x1 + Y1, x2 + Y2), xY = (x1 Y1 - x2Y2, x1 Y2 + x2Y1) NOTE-, The set of all complex numbers will be denoted by C. Theorem 1.27. The operations of addition and multiplication just defined satisfy the commutative, associative, and distributive laws. Proof. We prove only the distributive law; proofs of the others are simpler. If x = (x1, x2), y = (Y1, y2), and z = (z1, z2), then we have x(Y + z) = (x1, x2)(Y1 + z1, Y2 + z2) = (x1Y1 + x121 - x2Y2 - x222, x1Y2 + x122 + x2Y1 + x221) = (x1Y1 - x2Y2, x1Y2 + x2Y1) + (x121 - x222, x122 + x221) = xy + xz. Theorem 1.28. (x1, x2) + (0, 0) = (x1, x2), (x1, x2)(0, 0) _ (0, 0), (x1, x2)(1, 0) = (x1, x2), (x1, x2) + (-x1, -x2) = (0, 0) Proof. The proofs here are immediate from the definition, as are the proofs of Theorems 1.29, 1.30, 1.32, and 1.33. ' Theorem 1.29. Given two complex numbers x = (x1, x2) and y = (y1, Y2), there exists a complex number z such that x + z = y. In fact, z = (y1 - x1, Y2 - x2). This z is denoted by y - x. The complex number (-x1, -x2) is denoted by -x. Theorem 1.30. For any two complex numbers x and y, we have (-x)Y = x(- Y) = -(xY) = (-1, 0)(xy) Definition 1.31. If x = (x1i x2) # (0, 0) and y are complex numbers, we define x' = [x1/(xi + xi), -x2/(xi + xz)], and y/x = yx-1. Theorem 1.32. If x and y are complex numbers with x (0, 0), there exists a complex number z such that xz = y, namely, z = yx-1. Of special interest are operations with complex numbers whose imaginary part is 0. Theorem 1.33. (x 1, 0) + (y1, 0) = (x 1 + y 1p 0), (x1, 0)(Y1, 0) = (x1 Y1, 0), (x1, 0)/(Y1, 0) = (x1/Y1, 0), if Y1 # 0. NOTE. It is evident from Theorem 1.33 that we can perform arithmetic operations on complex numbers with zero imaginary part by performing the usual real-num-ber operations on the real parts alone. Hence the complex numbers of the form (x, 0) have the same arithmetic properties as the real numbers. For this reason it is Fig. 1.3 Geometric Representation 17 convenient to think of the real number system as being a special case of the complex number system, and we agree to identify the complex number (x, 0) and the real number x. Therefore, we write x = (x, 0). In particular, 0 = (0, 0) and I = (1, 0). 1.22 GEOMETRIC REPRESENTATION OF COMPLEX NUMBERS Just as real numbers are represented geometrically by points on a line, so complex numbers are represented by points in a plane. The complex number x = (x1i x2) can be thought of as the "point" with coordinates (x1, x2). When this is done, the 'definition of addition amounts to addition by the parallelogram law. (See Fig. 1.2.) x+y=(x1+,1,x2+y2) 0 = (0, 0) X1 = (x1, 0) Figure 1.2 The idea of expressing complex numbers geometrically as points on a plane was formulated by Gauss in his dissertation in 1799 and, independently, by Argand in 1806. Gauss later coined the somewhat unfortunate phrase "complex number." Other geometric interpretations of complex numbers are possible. Instead of using points on a plane, we can use points on other surfaces. Riemann found the sphere particularly convenient for this purpose. Points of the sphere are projected from the North Pole onto the tangent plane at the South Pole and thus there corresponds to each point of the plane a definite point of the sphere. With the exception of the North Pole itself, each point of the sphere corresponds to exactly one point of the plane. This correspondence is called a stereographic projection. (See Fig. 1.3.) 18 Real and Complex Number Systems 1.23 THE IMAGINARY UNIT Def. 1.34 It is often convenient to think of the complex number (x1, x2) as a two-dimensional vector with components x1 and x2. Adding two complex numbers by means of Definition 1.26 is then the same as adding two vectors component by component. The complex number l = (1, 0) plays the same role as a unit vector in the hori-zontal direction. The analog of a unit vector in the vertical direction will now be introduced. Definition 1.34. The complex number (0, 1) is denoted by i and is called the imag-inary unit. Theorem 1.35. Every complex number x = (x1, x2) can be represented in the form x = x1 + 'X2-Proof. x1 = (x1, 0), ix2 = (0, 1)(x2i 0) = (0, x2), x1 + ix2 = (x1, 0) + (0, x2) = (x1, x2). The next theorem tells us that the complex number i provides us with a solution to the equation x2 = -1. Theorem 1.36. i 2 = -1. Proof. i2 = (0, 1)(0, 1) = (-1, 0) _ -1. 1.24 ABSOLUTE VALUE OF A COMPLEX NUMBER We now extend the concept of absolute value to the complex number system. Definition 1.37. If x = (x1, x2), we define the modulus, or absolute value, of x to be the nonnegative real number Ixl given by IxI =v'x2+x2. Theorem 1.38. i) 1(0, 0)1 = Q, and IxI > 0 if x ; 0. ii) Ixyl = IxI IYI. iii) 1x/y1 = Ixl/IYI, if y # 0. iv) I(xl, 0)1 = lxii. Proof Statements (i) and (iv) are immediate. To prove (ii), we write x = x1 + ix2, y = y1 + iy2, so that xy = x1 y1 - x2y2 + i(x1 y2 + x2 y1). Statement (ii) follows from the relation Ixy12 = xiyi + xiy2 + xiy2 + x2yi = (x1 + x2)(y1 + y2) = IxI21Y12 Equation (iii) can be derived from (ii) by writing it in the form IxI = IYI Ix/yl Geometrically, IxI represents the length of the segment joining the origin to the point x. More generally, lx - yl is the distance between the points x and y. Using this geometric interpretation, the following theorem states that one side of a triangle is less than the sum of the other two sides. Def. 1.40 Complex Exponentials Theorem 1.39. If x and y are complex numbers, then we have Ix + y1 < Ixj + IyI (triangle inequality). The proof is left as an exercise for the reader. 19 .1.25 IMPOSSIBILITY OF ORDERING THE COMPLEX NUMBERS As yet we have not defined a relation of the form x < y if x and y are arbitrary complex numbers, for the reason that it is impossible to give a definition of < for complex numbers which will have all the properties in Axioms 6 through 8. To illustrate, suppose we were able to define an order relation < satisfying Axioms 6, 7, and 8. Then, since i # 0, we must have either i > 0 or i < 0, by Axiom 6. Let us assume i > 0. Then taking, x = y = i in Axiom 8, we get i2 > 0, or -1 > 0. Adding 1 to both sides (Axiom 7), we get 0 > 1. On the other hand, applying Axiom 8 to -1 > 0 we find 1 > 0. Thus we have both 0 > 1 and 1 > 0, which, by Axiom 6, is impossible. Hence the assumption i > 0 leads us to a contradiction. [Why was the inequality -1 > 0 not already a contradiction?] A similar argument shows that we cannot have i < 0. Hence the complex numbers cannot be ordered in such a way that Axioms 6, 7, and 8 will be satisfied. 1.26 COMPLEX EXPONENTIALS The exponential ex (x real) was mentioned earlier. We now wish to define eZ when z is a complex number in such a way that the principal properties of the real exponential function will be preserved. The main properties of ex for x real are the law of exponents, ex,ex2 = exl+X2, and the equation e° = 1. We shall give a definition of eZ for complex z which preserves these properties and reduces to the ordinary exponential when z is real. If we write z = x + iy (x, y real), then for the law of exponents to hold we want ex+'y = exe'y. It remains, therefore, to define what we shall mean by e'y. Definition 1.40. If z = x + iy, we define e= = ex+'y to be the complex number e= = ex (cos y + i sin y). This definition agrees with the real exponential function when z is real (that is, y = 0). We prove next that the law of exponents still holds. Several arguments can be given to motivate the equation e'y = cos y + i sin y. For example, let us write e'y = f (y) + ig(y) and try to determine the real-valued functions f and g so that the usual rules of operating with real exponentials will also apply to complex exponentials. Formal differentiation yields e'' = g'(y) - if'(y), if we assume that (e'y)' = ie'y. Comparing the two expressions for e'y, we see that f and g must satisfy the equations f (y) = g'(y), f'(y) = - g(y). Elimination of g yields fly) = - f"(y). Since we want e° = 1, we must have f (O) = 1 and f'(0) = 0. It follows that fly) = cos y and g(y) = -f'(y) = sin y. Of course, this argument proves nothing, but it strongly suggests that the definition e'y = cos y + i sin y is reasonable. 20 Real and Complex Number Systems Th. 1.41 Theorem 1.41. If z1 = x1 + iy1 and z2 = x2 + 1Y2 are two complex numbers, then we have ez'ez2 = ez'+22. Proof ez' = ex'(cos y1 + i sin y1), ez2 = ex2(cos Y2 + i sin Y2), ez'ez2 = ex'ex2[cos YI COS Y2 - sin y1 sin Y2 + i(cos y1 sin Y2 + sin y1 cos Y2)]. Now ex'ex2 = ex'+x2, since xi1 and x2 are both real. Also, cos y1 cos Y2 - sin y1 sin Y2 = cos (Y, + Y2) and cos y1 sin Y2 + sin y1 cos Y2 = sin (y1 + Y2), and hence ezieZ2 = ex'+x2[cos (YI + Y2) + i sin (YI + Y2)] = ez'+Z2. 1.27 FURTHER PROPERTIES OF COMPLEX EXPONENTIALS In the following theorems, z, z1, z2 denote complex numbers. Theorem 1.42. ez is never zero. Proof. eze-z = e° = 1. Hence ez cannot be zero. Theorem 1.43. If x is real, then Ie'xI = 1. Proof. Ie'ii2 = cost x + sin 2 x = 1, and I e'xi > 0. Theorem 1.44. ez = 1 if, and only if, z is an integral multiple of 27ri. Proof. If z = 2irin, where n is an integer, then ez = cos (2irn) + i sin (21rn) = 1. Conversely, suppose that ez = 1. This means that ex cos y = 1 and ex sin y = 0. Since ex 0, we must have sin y = 0, y = k7r, where k is an integer. But cos (k7r) = (- I)k. Hence ex = (-1)k, since ex cos (kir) = 1. Since ex > 0, k must be even. Therefore ex = 1 and hence x = 0. This proves the theorem. Theorem 1.45. ez' = eze if, and only if, z1 - z2 = 2irin (where n is an integer). Proof. ez' = eze if, and only if, ez' 2 = 1. 1.28 THE ARGUMENT OF A COMPLEX NUMBER If the point z -_ (x, y) = x + iy is represented by polar coordinates r and 0, we can write x = r cos 0 and y = r sin 0, so that z = r cos 0 + it sin 0 = re'B Def. 1.49 Integral Powers and Roots 21 The two numbers r and 0 uniquely determine z. Conversely, the positive number r is uniquely determined by z; in fact, r = Iz I. However, z determines the angle 0 only up to multiples of 2n. There are infinitely many values of 0 which satisfy the equations x = Iz I cos 0, y = Iz sin 0 but, of course, any two of them differ by some multiple of 2n. Each such 0 is called an argument of z but one of these values is singled out and is called the principal argument of z. Definition 1.46. Let z = x + iy be a nonzero complex number. The unique real number 0 which satisfies the conditions x=1zIcos0, y=1zIsin0, -7r<0<-+n is called the principal argument of z, denoted by 0 = arg (z). The above discussion immediately yields the following theorem: Theorem 1.47. Every complex number z 0 can be represented in the form z = reie, where r = IzI and 0 = arg (z) + 2nn, n being any integer. NOTE. This method of representing complex numbers is particularly useful in connection with multiplication and division, since we have t iO2 i(81+92) and e = rl ei(91-82) (rleiB)(rie ) = r1r2e and r2efe2 r2 Theorem 1.48. If z1z2 0, then arg (z1z2) = arg (z1) + arg (z2) + 2irn(z1, z2), where 0, f -n < arg (z1) + arg (z2) < +ir, n(z1, z2) = + 1, if - 2n < arg (z1) + arg (z2) 5 - 7r, -1, f it < arg (z1) + arg (z2) < 27r. Proof. Write z1 = Iz11eB', Z2 = I z21 e02, where 01 = arg (z1) and 02 = arg (z2). Then z1z2 = Iziz21ei(01+02) Since -n < 01 < +7t and -7r < 02 < +n, we have -27t < 01 + 02 < 2n. Hence there is an integer n such that -7r < 01 + 02 + 2nn < it. This n is the same as the integer n(z1, z2) given in the theorem, and for this n we have arg (z1z2) = 01 + 02 + 27rn. This proves the theorem. 1.29 INTEGRAL POWERS AND ROOTS OF COMPLEX NUMBERS Definition 1.49. Given a complex number z and an integer n, we define the nth power of z as follows: z0 = 1, zn + 1 = ZnZ, if n 0, z-"=(z-1)", ifz#Oand n>0. Theorem 1.50, which states that the usual laws of exponents hold, can be proved by mathematical induction. The proof is left as an exercise. 22 Real and Complex Number Systems Th. 1.50 Theorem 1.50. Given two integers m and n, we have, for z 0 0, ZnZm = Zn+m and (Z1Z2)" = ZIZZ. Theorem 1.51. If z 0, and if n is a positive integer, then there are exactly n distinct complex numbers zo, z1, ... , zn_1 (called the nth roots of z), such that zk = Z, for each k = 0, 1, 2, ... , n - 1. Furthermore, these roots are given by the formulas zk = Re'mk, where R = 1Z I 1 In, and arg (z) 2nk k= + (k=0, 1,2,...,n-1). n n NOTE. The n nth roots of z are equally spaced on the circle of radius R = IzI',", center at the origin. Proof. The n complex numbers Re'4k, 0 < k < n - 1, are distinct and each is an nth root of z, since (Re'mk)n = Rnei"4k = IzIe'[arg(z)+2,rk] = Z. We must now show that there are no other nth roots of z. Suppose w = Ae" is a complex number such that w" = z. Then I wI" = Iz I, and hence A" = Izi, A = Iz I' I". Therefore, w" = z can be written e'"" = e'larg (0], which implies na - arg (z) = 2nk for some integer k. Hence a = [arg (z) + 27rk]/n. But when k runs through all integral values, w takes only the distinct values z 0 , . . . , zn _ 1. (See Fig. 1.4.) Figure 1.4 1.30 COMPLEX LOGARITHMS By Theorem 1.42, ez is never zero. It is natural to ask if there are other values that ez cannot assume. The next theorem shows that zero is the only exceptional value. Def. 135 Complex Powers 23 Theorem 1.52. If z is a complex number # 0, then there exist complex numbers w such that ew = z. One such w is the complex number log IzI + i arg (z), and any other such w must have the form log IzI + i arg (z) + 2mri, where n is an integer. Proof. Since elog Izl +i arg (z) = elog IzI ei arg (z) = Iz Iei arg (z) = Z, we see that w = log IzI + i arg (z) is a solution of the equation ew = z. But if wl is any other solution, then ew = ewi and hence w - wl = 2niri. Definition 1.53. Let z 0 be a given complex number. If w is a complex number such that ew = z, then w is called a logarithm of z. The particular value of w given by w = log IzI + i arg (z) is called the principal logarithm of z, and for this w we write w = Log z. Examples 1. Since i l = 1 and arg (i) it/2, Log (f) = iir/2. 2. Since I - i l = 1 and arg (- i) _ - x/2, Log (- i) in/2. 3. Since -1 = 1 and arg (-1) = n, Log (-1) = in. 4. If x > 0, Log (x) = log x, since IxI = x and arg (x) = 0. 5. Since 11 + i I = I2 and arg (1 + i) = n/4, Log (1 + i) = log ,l2 + in/4. Theorem 1.54. If zxz2 # 0, then Log (zlz2) = Log zl + Log z2 + 27rin(zl, z2), where n(zl, z2) is the integer defined in Theorem 1.48. Proof. Log (zlz2) = log IZ1z2I + i arg (ztz2) = log IZ1I + log IZ2I + i [arg (z1) + arg (Z2) + 2irn(zl, Z2)] 1.31 COMPLEX POWERS Using complex logarithms, we can now give a definition of complex powers of complex numbers. Definition 1.55. If z # 0 and if w is any complex number, we define Zw = ewLogz 24 Real and Complex Number Systems Th. 1.56 Examples 1. i' = eiLogi = eUin/2) = e-rz/2 2. (-1)i = eiLog(-1) = ei(in) = e-n. 3. If n is an integer, then zi+1 = e(n+l)Logz = enLogzeLogz = znz, so Definition 1.55 does not conflict with Definition 1.49. The next two theorems give rules for calculating with complex powers: Theorem 1.56. zwt ZW2 = Zwt+w2 if z 0. Proof. Zwt+w2 = e(wt+w2)Logz = ew1Logzew2Logz = ZWIZW2. Theorem 1.57. If z1z2 0, then (zlz2)w = z1 2 w we 2aiwn(z1,z2) where n(zl, z2) is the integer defined in Theorem 1.48. Proof. (z1Z2)w = ewLog(ztz2) = ew[Logzt+Logz2+2ain(zt,z2)] 1.32 COMPLEX SINES AND COSINES Definition 1.58. Given a complex number z, we define eiz + e-'z e'z - e- 1Z cos z = - --- , 2 sin z = 2i NOTE. When z is real, these equations agree with Definition 1.40. Theorem 1.59. If z = x + iy, then we have cos z = cos x cosh y - i sin x sinh y, sin z = sin x cosh y + i cos x sinh y. Proof. 2 cos z = e'z + e-1z = e-'(cos x + i sin x) + ey(cos x - i sin x) = cos x(ey + e-') - i sin x(ey - e-y) = 2 cos x cosh y - 2i sin x sinh y. The proof for sin z is similar. Further properties of sines and cosines are given in the exercises. 1.33 INFINITY AND THE EXTENDED COMPLEX PLANE C Next we extend the complex number system by adjoining an ideal point denoted by the symbol oo. Definition 1.60. By the extended complex number system C we shall mean the complex plane C along with a symbol oo which satisfies the following properties: a) If z e C, then we have z + oo = z - oo = oo, z/oo = 0. Def. 1.61 Exercises 25 b) If z e C, but z 0, then z(oo) = oo and z/0 = oo. c) co + oo = (co)(oo) = oo. Definition 1.61. Every set in C of the form {z : Iz I > r > 0) is called a neighbor-hood of oo, or a ball with center at oo. The reader may wonder why two symbols, + oo and - oo, are adjoined to R but only one symbol, oo, is adjoined to C. The answer lies in the fact that there is an ordering relation < among the real numbers, but no such relation occurs among the complex numbers. In order that certain properties of real numbers involving the relation < hold without exception, we need two symbols, + oo and - oo, as defined above. We have already mentioned that in R every nonempty set has a sup, for example. In C it turns out to be more convenient to have just one ideal point. By way of illustration, let us recall the stereographic projection which establishes a one-to-one correspondence between the points of the complex plane and those points on the surface of the sphere distinct from the North Pole. The apparent exception at the North Pole can be removed by regarding it as the geometric representative of the ideal point cc. We then get a one-to-one correspondence between the extended complex plane C and the total surface of the sphere. It is geometrically evident that if the South Pole is placed on the origin of the complex plane, the exterior of a "large" circle in the plane will correspond, by stereographic projection, to a "small" spherical cap about the North Pole. This illustrates vividly why we have defined a.neighborhood of cc by an inequality of the form Iz I > r. EXERCISES Integers 1.1 Prove that there is no largest prime. (A proof was known to Euclid.) 1.2 If n is a positive integer, prove the algebraic identity n-1 an - b" = (a - b) E akbn-l-k. k=0 1.3 If 2" - I is prime, prove that n is prime. A prime of the form 2° - 1, where p is prime, is called a Mersenne prime. 1.4 If 2" + 1 is prime, prove that n is a power of 2. A prime of the form 22"' + 1 is called a Fermat prime. Hint. Use Exercise 1.2. 1.5 The Fibonacci numbers 1, 1, 2, 3, 5, 8, 13.... are defined by the recursion formula xn+1 = x" + x"_1, with x1 = x2 = 1. Prove that (x", xn+1) = I and that x" _ (a" - bn)/(a - b), where a and bare the roots of the quadratic equation x2 - x - I = 0. 1.6 Prove that-every nonempty set of positive integers contains a smallest member. This is called the well-ordering principle. 26 Real and Complex Number Systems Rational and irrational numbers 1.7 Find the rational number whose decimal expansion is 0.3344444.. . 1.8 Prove that the decimal expansion of x will end in zeros (or in nines) if, and only if, x is a rational number whose denominator is of the form 2"5'", where m and n are non-negative integers. 1.9 Prove that s2 + J3 is irrational. 1.10 If a, b, c, d are rational and if x is irrational, prove that (ax + b)l(cx + d) is usually irrational. When do exceptions occur? 1.11 Given any real x > 0, prove that there is an irrational number between 0 and x. 1.12 If alb < c/d with b > 0, d > 0, prove that (a + c)l(b + d) lies between alb and c/d. 1.13 Let a and b be positive integers. Prove that V2 always lies between the two fractions alb and (a + 2b)l(a + b). Which fraction is closer to 2? 1.14 Prove that I n - 1 + jn + I is irrational for every integer n >-1. 1.15 Given a real x and an integer N > 1, prove that there exist integers h and k with 0 < k < N such that jkx - hi < 1/N. Hint. Consider the N + 1 numbers tx - [tx] for t = 0, 1, 2, ... , N and show that some pair differs by at most 1/N. 1.16 If x is irrational prove that there are infinitely many rational numbers h/k with k > 0 such that Ix - h/kI < 1/k2. Hint. Assume there are only a finite number h1/k1,..., hr/kr and obtain a contradiction by applying Exercise 1.15 with N > 1/6, where 6 is the smallest of the numbers Ix - hi/kid. 1.17 Let x be a positive rational number of the form 0 X = E ak a nonnegative integer with ak 5 k - I for k >- 2 and a" > 0. Let [x] denote the greatest integer in x. Prove that al = [x ], that ak = [k! x ] - k [(k - 1) ! x ] for k = 2, ... , n, and that n is the smallest integer such that n! x is an integer. Con-versely, show that every positive rational number x can be expressed in this form in one and only one way. Upper bounds 1.18 Show that the sup and inf of a set are uniquely determined whenever they exist. 1.19 Find the sup and inf of each of the following sets of real numbers: a) All numbers of the form 2-P + 3-q + 5-', where p, q, and r take on all positive integer values. b) S = [x: 3x2 - lOx + 3 < 0}. c) S = {x: (x - a)(x - b)(x - c)(x - d) < 0), where a < b < c < d. 1.20 Prove the comparison property for suprema (Theorem 1.16). 1.21 Let A and B be two sets. of positive numbers bounded above, and let a = sup A, b = sup B. Let C be the set of all products of the form xy, where x e A and y e B. Prove that ab = sup C. Exercises 27 1.22 Given x > 0 and an integer k >- 2. Let ao denote the largest integer <- x and, assuming that ao, a1, ... , an_ 1 have been defined, let an denote the largest integer such that ao+A1+ a2+ ... +Q" <x. k k2 k" a) Prove that 0 5 al <- k - 1 for each i = 1, 2, .. . b) Let r" = ao + a1k-' + a2k-2 + + a"k-" and show that x is the sup of the set of rational numbers r1, r2, .. . NOTE. When k = 10 the integers ao, a1, a2,... are the digits in a decimal representation of x. For general k they provide a representation in the scale of k. Inequalities 1.23 Prove Lagrange's identity for real numbers: k=1 =1 akbk) 2 = k ak) (k1 bk) l sk sn (Rkb; - a;bk)2. Note that this identity implies the Cauchy-Schwarz inequality. 1.24 Prove that for arbitrary real ak, bk, ck we have n 4 ( akbkCk) < k=1 1.25 Prove Minkowski's inequality: k=1 Uo (ak + bk)2)1/2 < R ak)1/2 + \k bk)1/2 This is the triangle inequality Ira + bll <- IIall + IIbIj for n-dimensional vectors, where a = (a1,..., an), b = (b1, ... , bn) and IIaHH = ( ak n 1/2 ) k=1 1.26 If a1 >- a2 >->- an and b1 >- b2 >_ >_ bn, prove that k=1 ak)(k=1 bk) Hint. X15Jsk5n (ak - aj)(bk - bj) >- 0. 5 n F akbk. k=1 Complex numbers 1.27 Express the following complex numbers in the form a + bi. a) (1 + i)3, b) (2 + 3i)/(3 - 4i), c) is + i16, d) 4(1 + i)(1 + i-8). bk)2(E Ck) 1.28 In each case, determine all real x and y which satisfy the given relation. 100 a) x + iy = Ix - iyJ, b) x + iy = (x - iy)2, c) E ik = x + jy. k=0 28 Real and Complex Number Systems 1.29 If z = x + iy, x and y real, the complex conjugate of z is the complex number z = x - iy. Prove that: a) z1 + z2 = Z1 + z2, b) F, -z2 = f, Z2, C) ZZ = 1z12, d) z + z = twice the real part of z, e) (z - 27)/i = twice the imaginary part of z. 1.30 Describe geometrically the set of complex numbers z which satisfies each of the following conditions: a)IzI=1, b)IzI<1, C)IzI_<1, d)z+z= 1, e)z - z=i, f) z+z= IzI2. 1.31 Given three complex numbers z1, z2, z3 such that Iz, I = IZ21 = Iz31 = 1 and z1 + z2 + z3 = 0. Show that these numbers are vertices of an equilateral triangle inscribed in the unit circle with center at the origin. 1.32 If a and b are complex numbers, prove that: a) Ia - b12 < (1 + Ia12)(1 + Ib12). b) If a # 0, then Ia + bI = IaI + IbI if, and only if, b/a is real and nonnegative. 1.33 If a and b are complex numbers, prove that Ia-6I=11-abI if, and only if, Ial = I or IbI = 1. For which a and b is the inequality Ia - bI < I 1 - abI valid? 1.34 If a and c are real constants, b complex, show that the equation azz+bz+bz+c=0 (a:0,z=x+iy) represents a circle in the xy-plane. 1.35 Recall the definition of the inverse tangent: given a real number t, tan-1 (t) is the unique real number 0 which satisfies the two conditions tan0= t. - 2 < 0 < + 2 If z = x + iy, show that a) arg (z) = tan-1 (!) , if x > 0, b) arg (z) = tan-1 (!) + n, if x < 0, y >- 0, c) arg (z) = tan-1 (y) - 7r, x if x < 0, y < 0, d)arg(z)= 2ifx=0,y>0;arg(z)= -2ifx=0,y<0. Exercises 29 1.36 Define the following "pseudo-ordering" of the complex numbers: we say z1 < z2 if we have either i) IZl I < Iz2I or ii) IZl I = Iz21 and arg (Z1) < arg (z2). Which of Axioms 6, 7, 8, 9 are satisfied by this relation? 1.37 Which of Axioms 6, 7, 8, 9 are satisfied if the pseudo-ordering is defined as follows? We say (x1, yl) < (x2, y2) if we have either i) x1 < x2 or ii) x1 = x2 and y1 < y2. 1.38 State and prove a theorem analogous to Theorem 1.48, expressing arg (zl/z2) in terms of arg (z1) and arg (z2). 1.39 State and prove a theorem analogous to Theorem 1.54, expressing Log (z,42) in terms of Log (z1) and Log (Z2)-1.40 Prove that the nth roots of 1 (also called the nth roots of unity) are given by a, a2_., a", where a = e2'ri", and show that the roots :1 satisfy the equation 1 + x + x2 + + x"-1 = 0. 1.41 a) Prove that Izrl < e' for all complex z : 0. b) Prove that there is no constant M > 0 such that Icos z I < M for all complex z. 1.42 If w = u + iv (u, v real), show that zw = euloglzl-varg(z)eitvlogjzI+uarg(z)1 1.43 a) Prove that Log (zw) = w Log z + 2irin, where n is an integer. b) Prove that (zw)° = zwae2nt"a, where n is an integer. 1.44 i) If 0 and a are real numbers, - 7r < 0 < + ir, prove that (cos 0 + i sin O)" = cos (a9) + i sin (aO). ii) Show that, in general, the restriction - n < 0 < + it is necessary in (i) by taking 0 = - ir, a = 1. iii) If a is an integer, show that the formula in (i) holds without any restriction on 0. In this case it is known as DeMoivre's theorem. 1.45 Use DeMoivre's theorem (Exercise 1.44) to derive the trigonometric identities sin 30 = 3 cos' 0 sin 0 - sin3 0, cos 30 = cos3 0 - 3 cos 0 sine 0, valid for real 0. Are these valid when 0 is complex? 1.46 Define tan z = (sin z)/(cos z) and show that for z = x + iy, we have tan z = sin 2x + i sinh 2y cos 2x + cosh 2y 1.47 Let w be a given complex number. If w 96 ± 1, show that there exist two values of z = x + iy satisfying the conditions cos z = w and - it < x < + it. Find these values when w = i and- when w = 2. 30 Real and Complex Number Systems 1.48 Prove Lagrange's identity for complex numbers: k=1 2 n IakI2 Ibkl2 Iak5J - aj5kl2. k=1 k=1 15k<Jsn Use this to deduce a Cauchy-Schwarz inequality for complex numbers. 1.49 a) By equating imaginary parts in DeMoivre's formula prove that sin nO = sin" 0 {(n cotn-1 0 -(1?) coin-3 0 + (n) cot"-5 0 -I) 3 b) If 0 < 0 < n/2, prove that sin (2m + 1)0 = sin2,n+10Pm(cot2 0) where Pm is the polynomial of degree m given by Pm(X) (2m+ 11 (2m+ 1) x"-1 + (2m+ 11 X"_2 - + = 1 /fx"' -3 5 J... Use this to show that P. has zeros at them distinct points Xk = cot2 {nk/(2m + 1)} fork = 1, 2, ... , m. c) Show that the sum of the zeros of P. is given by "2 irk m(2m - 1) E cot = 2m + 1 3 and that the sum of their squares is given by , m cot4 irk = m(2m - 1)(4m2 + lOm - 9) k=1 2m + 1 45 NOTE. These identities can be used to prove that E 1 n-2 = n2/6 and 1 n-4 = n4/90. (See Exercises 8.46 and 8.47.) 1.50 Prove that z" - I = Iik=1 (z - e2nikln) for all complex z. Use this to derive the formula sin kn = n for n >- 2. n 2"-1 k=1 SUGGESTED REFERENCES FOR FURTHER STUDY 1.1 Apostol, T. M., Calculus, Vol. 1, 2nd ed. Xerox, Waltham, 1967. 1.2 Birkhoff, G., and MacLane, S., A Survey of Modern Algebra, 3rd ed. Macmillan, New York, 1965. 1.3 Cohen, L., and Ehrlich, G., The Structure of the Real-Number System. Van Nos-trand, Princeton, 1963. 1.4 Gleason, A., Fundamentals of Abstract Analysis. Addison-Wesley, Reading, 1966. 1.5 Hardy, G.- H., A Course of Pure Mathematics, 10th ed. Cambridge University Press, 1952. References 31 1.6 Hobson, E. W., The Theory of Functions of a Real Variable and the Theory of Fourier's Series, Vol. 1, 3rd ed. Cambridge University Press, 1927. 1.7 Landau, E., Foundations of Analysis, 2nd ed. Chelsea, New York, 1960. 1.8 Robinson, A., Non-standard Analysis. North-Holland, Amsterdam, 1966. 1.9 Thurston, H. A., The Number System. Blackie, London, 1956. 1.10 Wilder, R. L., Introduction to the Foundations of Mathematics, 2nd ed. Wiley, New York, 1965. CHAPTER 2 SOME BASIC NOTIONS OF SET THEORY 2.1 INTRODUCTION In discussing any branch of mathematics it is helpful to use the notation and terminology of set theory. This subject, which was developed by Boole and Cantor in the latter part of the 19th century, has had a profound influence on the develop-ment of mathematics in the 20th century. It has unified many seemingly discon-nected ideas and has helped reduce many mathematical concepts to their logical foundations in an elegant and systematic way. We shall not attempt a systematic treatment of the theory of sets but shall confine ourselves to a discussion of some of the more basic concepts. The reader who wishes to explore the subject further can consult the references at the end of this chapter. A collection of objects viewed as a single entity will be referred to as a set. The objects in the collection will be called elements or members of the set, and they will be said to belong to or to be contained in the set. The set, in turn, will be said to contain or to be composed of its elements. For the most part we shall be inter-ested in sets of mathematical objects; that is, sets of numbers, points, functions, curves, etc. However, since much of the theory of sets does not depend on the nature of the individual objects in the collection, we gain a great economy of thought by discussing sets whose elements may be objects of any kind. It is because of this quality of generality that the theory of sets has had such a strong effect in furthering the development of mathematics. 2.2 NOTATIONS Sets will usually be denoted by capital letters : A, B, C, ... , X, Y, Z, and elements by lower-case letters: a, b, c, ... , x, y, z. We write x e S to mean "x is an element of S," or "x belongs to S." If x does not belong to S, we write x 0 S. We sometimes designate sets by displaying the elements in braces; for example, the set of positive even integers less than 10 is denoted by {2, 4, 6, 8}. If S is the collection of all x which satisfy a property P, we indicate this briefly by writing S = {x: x satisfies P}. From a given set we can form new sets, called subsets of the given set. For example, the set consisting of all positive integers less than 10 which are divisible 32 Def. 2.3 Cartesian Product of Two Sets 33 by 4, namely, {4, 8}, is a subset of the set of even integers less than 10. In general, we say that a set A is a subset of B, and we write A B whenever every element of A also belongs to B. The statement A c B does not rule out the possibility that B c A. In fact, we have both A s B and B c A if, and only if, A and B have the same elements. In this case we shall call the sets A and B equal and we write A = B. If A and B are not equal, we write A : B. If A c B but A B, then we say that A is a proper subset of B. It is convenient to consider the possibility of a set which contains no elements whatever; this set is called the empty set and we agree to call it a subset of every set. The reader may find it helpful to picture a set as a box containing certain objects, its elements. The empty set is then an empty box. We denote the empty set by the symbol 0. 2.3 ORDERED PAIRS Suppose we have a set consisting of two elements a and b; that is, the set {a, b}. By our definition of equality this set is the same as the set {b, a), since no question of order is involved. However, it is also necessary to consider sets of two elements in which order is important. For example, in analytic geometry of the plane, the coordinates (x, y) of a point represent an ordered pair of numbers. The point (3, 4) is different from the point (4, 3), whereas the set {3, 4} is the same as the set {4, 31. When we wish to consider a set of two elements a and b as being ordered, we shall enclose the elements in parentheses: (a, b). Then a is called the first element and b the second. It is possible to give a purely set-theoretic definition of the concept of an ordered pair of objects (a, b). One such definition is the following: Definition 2.1. (a, b) = {{a}, {a, b}}. This definition states that (a, b) is a set containing two elements, {a} and {a, b}. Using this definition, we can prove the following theorem: Theorem 2.2. (a, b) = (c, d) if, and only if, a = c and b = d. This theorem shows that Definition 2.1 is a "reasonable" definition of an ordered pair, in the sense that the object a has been distinguished from the object b. The proof of Theorem 2.2 will be an instructive exercise for the reader. (See Exercise 2.1.) 2.4 CARTESIAN PRODUCT OF TWO SETS Definition 2.3. Given two sets A and B, the set of all ordered pairs (a, b) such that a e A and b e B is called the cartesian product of A and B, and is denoted by A x B. Example. If R -denotes the set of all real numbers, then R x R is the set of all complex numbers. 34 Some Basic Notions of Set Theory 2.5 RELATIONS AND FUNCTIONS Def. 2.4 Let x and y denote real numbers, so that the ordered pair (x, y) can be thought of as representing the rectangular coordinates of a point in the xy-plane (or a com-plex number). We frequently encounter such expressions as xy=1, x2+y2= 1, x2+y2<1, x<y. (a) Each of these expressions defines a certain set of ordered pairs (x, y) of real numbers, namely, the set of all pairs (x, y) for which the expression is satisfied. Such a set of ordered pairs is called a plane relation. The corresponding set of points plotted in the xy-plane is called the graph of the relation. The graphs of the relations described in (a) are shown in Fig. 2.1. xy=1 x2+y2=1 x2+y2<_1 x<y Figure 2.1 The concept of relation can be formulated quite generally so that the objects x and y in the pairs (x, y) need not be numbers but may be objects of any kind. Definition 2.4. Any set of ordered pairs is called a relation. If S is a relation, the set of all elements x that occur as first members of pairs (x, y) in S is called the domain of S, denoted by s(S). The set of second members y is called the range of S, denoted by A(S). The first example shown in Fig. 2.1 is a special kind of relation known as a function. Definition 2.5. A function F is a set of ordered pairs (x, y), no two of which have the same first member. That is, if (x, y) e F and (x, z) a F, then y = z. The definition of function requires that for every x in the domain of F there is exactly one y such that (x, y) e F. It is customary to call y the value of F at x and to write y = F(x) instead of (x, y) e F to indicate that the pair (x, y) is in the set F. As an alternative to describing a function F by specifying the pairs it contains, it is usually preferable to describe the domain of F, and then, for each x in the domain, to describe how the function value F(x) is obtained. In this connection, we have the following theorem whose proof is left as an exercise for the reader. Th. 2.6 Further Termmobgy Theorem 2.6. Two functions F and G are equal if and only if a) 21(F) _ 21(G) (F and G have the same domain), and b) F(x) = G(x) for every x in 21(F). 35 2.6 FURTHER TERMINOLOGY CONCERNING FUNCTIONS When the domain 2(F) is a subset of R, then F is called a function of one real variable. If 2(F) is a subset of C, the complex number system, then F is called a function of a complex variable. If 2(F) is a subset of a cartesian product A x B, then F is called a function of two variables. In this case we denote the function values by F(a, b) instead of F((a, b)). A function of two real variables is one whose domain is a subset of RxR. If S is a subset of 2(F), we say that F is defined on S. In this case, the set of F(x) such that x E S is called the image of S under F and is denoted by F(S). If T is any set which contains F(S), then F is also called a mapping from S to T. This is often denoted by writing F : S -+ T. If F(S) = T, the mapping is said to be onto T. A mapping of S into itself is some-times called a transformation. Consider, for example, the function of a complex variable defined by the equa-tion F(z) = z2. This function maps every sector S of the form 0 < arg (z) < a 5 ir/2 of the complex z-plane onto a sector F(S) described by the inequalities 0 < arg [F(z)] < 2a. (See Fig. 2.2.) Figure 2.2 If two functions F and G satisfy the inclusion relation G c F, we say that G is a restriction of F or that F is an extension of G. In particular, if S is a subset of 21(F) and if G is defined by the equation G(x) = F(x) for all x in S, then we call G_the restriction of F to S. The function G consists of those pairs (x, F(x)) such that x e S. Its domain is S and its range is F(S). 36 Some Basic Notions of Set Theory Def. 2.7 2.7 ONE-TO-ONE FUNCTIONS AND INVERSES Definition 2.7. Let F be a function defined on S. We say F is one-to-one on S if, and only if, for every x and y in S, F(x) = F(y) implies x = y. This is the same as saying that a function which is one-to-one on S assigns distinct function values to distinct members of S. Such functions are also called injective. They are important because, as we shall presently see, they possess inverses. However, before stating the definition of the inverse of a function, it is convenient to introduce a more general notion, that of the converse of a relation. Definition 2.8. Given a relation S, the new relation S defined by S= {(a,b):(b,a)ES} is called the converse of S. Thus an ordered pair (a, b) belongs to S if, and only if, the pair (b, a), with elements interchanged, belongs to S. When S is a plane relation, this simply means that the graph of S is the reflection of the graph of S with respect to the line y = x. In the relation defined by x < y, the converse relation is defined by y < x. Definition 2.9. Suppose that the relation F is a function. Consider the converse relation t, which may or may not be a function. If t is also a function, then P is called the inverse of F and is denoted by F-1. Figure 2.3(a) illustrates an example of a function F for which Fis not a function. In Fig. 2.3(b) both F and its converse are functions. The next theorem tells us that a function which is one-to-one on its domain always has an inverse. (a) (b) Figure 2.3 Def. 2.13 Sequences 37 Theorem 2.10. If the function F is one-to-oneon its domain, then P is also a function. Proof. To show that Fis a function, we must show that if (x, y) E Fand (x, z) e F, then y = z. But (x, y) E F means that (y, x) E F; that is, x = F(y). Similarly, (x, z) e Fmeans that x = F(z). Thus F(y) = F(z) and, since we are assuming that F is one-to-one, this implies y = z. Hence, P is a function. NOTE. The same argument shows that if F is one-to-one on a subset S of 2(F), then the restriction of F to S has art inverse. 2.8 COMPOSITE FUNCTIONS Definition 2.11. Given two functions F and G such that 3P(F) c 21(G), we can form a new function, the composite G o F of G and F, defined as follows: for every x in the domain of F, (G o F)(x) = G[F(x)]. Since M (F) 9 21(G), the element F(x) is in the domain of G, and therefore it makes sense to consider G[F(x)]. In general, it is not true that G o F = F o G. In fact, F o G may be meaningless unless the range of G is contained in the domain of F. However, the associative law, Ho(GoF) = (HoG)oF, always holds whenever each side of the equation has a meaning. (Verification will be an interesting exercise for the reader. See Exercise 2.4.) 2.9 SEQUENCES Among the important examples of functions are those defined on subsets of the integers. Definition 2.12. By a finite sequence of n terms we shall understand a function F whose domain is the set of numbers 11, 2, ... , n}. The range of F is the set {F(l), F(2), F(3),... , F(n)}, customarily written {F1, F2, F3, ... , The elements of the range are called terms of the sequence and, of course, they may be arbitrary objects of any kind. Definition 2.13. By an infinite sequence we shall mean a function F whose domain is the set {1, 2, 3.... } of all positive integers. The range of F, that is, the set {F(l), F(2), F(3), ... }, is also written IF,, F2, F3, ... }, and the function value F. is called the nth term of the sequence. For brevity, we shall occasionally use the notation to denote the infinite sequence whose nth term is F. Let s = be an infinite sequence, and let k be a function whose domain is the set of positive integers and whose range is a subset of the positive integers. 38 Some Basic Notions of Set Theory Def. 2.14 Assume that k is "order-preserving," that is, assume that k(m) < k(n), if m < n. Then the composite function s o k is defined for all integers n >_ 1, and for every such n we have (s o k)(n) = sk(n) Such a composite function is said to be a subsequence of s. Again, for brevity, we often use the notation {sk( )} or {skn} to denote the subsequence of whose nth term is sk( ). Example. Let s = {11n) and let k be defined by k(n) = 2". Then s o k = {1/2}. 2.10 SIMILAR (EQUINUMEROUS) SETS Definition 2.14. Two sets A and B are called similar, or equinumerous, and we write A - B, if and only if there exists a one-to-one function F whose domain is the set A and whose range is the set B. We also say that F establishes a one-to-one correspondence between the sets A and B. Clearly, every set A is similar to itself (take Fto be the "identity" function for which F(x) = x for all x in A). Furthermore, if A - B then B - A, because if F is a one-to-one function which makes A similar to B, then F-1 will make B similar to A. Also, if A - B and if B - C, then A - C. (The proof is left to the reader.) 2.11 FINITE AND INFINITE SETS A set S is called finite and is said to contain n elements if S- {1,2,...,n}. The integer n is called the cardinal number of S. It is an easy exercise to prove that if (1, 2, ... , n} - (1, 2, ... , m} then m = n. Therefore, the cardinal number of a finite set is well defined. The empty set is also considered finite. Its cardinal number is defined to be 0. Sets which are not finite are called infinite sets. The chief difference between the two is that an infinite set must be similar to some proper subset of itself, whereas a finite set cannot be similar to any proper subset of itself. (See Exercise 2.13.) For example, the set Z+ of all positive integers is similar to the proper subset {2, 4, 8, 16, ... } consisting of powers of 2. The one-to-one function F which makes them similar is defined by F(x) = 2" for each x in V. Th. 2.17 Uncamtability of the Real Number System 2.12 COUNTABLE AND UNCOUNTABLE SETS 39 A set S is said to be countably infinite if it is equinumerous with the set of all positive integers; that is, if SN {1,2,3,...}. In this case there is a function f which establishes a one-to-one correspondence between the positive integers and the elements of S; hence the set S can be dis-played as follows: S = {.f(l),.f(2),.f(3), ... }. Often we use subscripts and denote f(k) by ak (or by a similar notation) and we write S = {al, a2, a3, .. . }. The important thing here is that the correspondence enables us to use the positive integers as "labels" for the elements of S. A count-ably infinite set is said to have cardinal number No (read: aleph nought). Definition 2.15. A set S is called countable if it is either finite or countably infinite. A set which is not countable is called uncountable. The words denumerable and nondenumerable are sometimes used in place of countable and uncountable. Theorem 2.16. Every subset of a countable set is countable. Proof. Let S be the given countable set and assume A S. If A is finite, there is nothing to prove, so we can assume that A is infinite (which means S is also in-finite). Let s = be an infinite sequence of distinct terms such that S = {s,, S2, }. Define a function on the positive integers as follows : Let k(l) be the smallest positive integer m such that s, a A. Assuming that k(1), k(2), ... , k(n - 1) have been defined, let k(n) be the smallest positive integer m > k(n - 1) such that s, e A. Then k is order-preserving: m > n implies k(m) > k(n). Form the composite function s o k. The domain of s o k is the set of positive integers and the range of s o k is A. Furthermore, s ° k is one-to-one, since implies s[k(n)] = s[k(m)], Sk(i) = Sk(m), which implies k(n) = k(m), and this implies n = m. This proves the theorem. 2.13 UNCOUNTABILITY OF THE REAL NUMBER SYSTEM The next theorem shows that there are infinite sets which are not countable. Theorem 2.17. - The set of all real numbers is uncountable. 40 Some Basic Notions of Set Theory Th. 2.18 Proof. It suffices to show that the set of x satisfying 0 < x < 1 is uncountable. If the real numbers in this interval were countable, there would be a sequence s = {sn} whose terms would constitute the whole interval. We shall show that this is impossible by constructing, in the interval, a real number which is not a term of this sequence. Write each s as an infinite decimal: S. = O.un,1un 2un,3 ... , where each un,; is 0, 1, ... , or 9. Consider the real number y which has the decimal expansion where v = y = O.v1v2v3..., (1, if 1, 12, if u = 1. Then no term of the sequence can be equal to y, since y differs from s, in the first decimal place, differs from s2 in the second decimal place, . . . , from s in the nth decimal place. (A situation like sn = 0.1999... and y = 0.2000.. . cannot occur here because of the way the v are chosen.) Since 0 < y < 1, the theorem is proved. Theorem 2.18. Let Z+ denote the set of all positive integers. Then the cartesian product Z + x Z + is countable. Proof. Define a function f on Z+ x Z+ as follows: f(m, n) = 2'3 if (m, n) e Z+ x Z+. Then f is one-to-one on Z + x Z + and the range of f is a subset of Z+. 2.14 SET ALGEBRA Given two sets A, and A2, we define a new set, called the union of A, and A2, denoted by A, u A2, as follows: Definition 2.19. The union A, u A2 is the set of those elements which belong either to A, or to A2 or to both. This is the same as saying that A, u A2 consists of those elements which belong to at least one of the sets A,, A2. Since there is no question of order involved in this definition, the union A 1 u A2 is the same as A2 u A, ; that is, set addition is commutative. The definition is also phrased in such a way that set addition is associative: A, U (A2 u A3) = (A, u A2) u A3. Def. 2.22 Set Algebra 41 The definition of union can be extended to any finite or infinite collection of sets: Definition 2.20. If F is an arbitrary collection of sets, then the union of all the sets in F is defined to be the set of those elements which belong to at least one of the sets in F, and is denoted by U A. AeF If F is a finite collection of sets, F = {A1, ... , we write n U A= U Ak=A,uA2u...uAn. AeF k=1 If F is a countable collection, F = {A1, A2, ... }, we write OD U A= UAk=A,UA2v... AeF k=1 Definition 2.21. If F is an arbitrary collection of sets, the intersection of all sets in F is defined to be the set of those elements which belong to every one of the sets in F, and is denoted by n A. AeF The intersection of two sets A 1 and A 2 is denoted by A 1 n A 2 and consists of those elements com,.ion to both sets. If A 1 and A2 have no elements in common, then A 1 n A 2 is the empty set and A 1 and A2 are said to be disjoint. If F is a finite collection (as above), we write n n A = n Ak=A1nA2n...nA., AeF k=1 and if F is a countable collection, we write 00 n A= n Ak=A1nA2n... AeF k=1 If the sets in the collection have no elements in common, their intersection is the empty set. Our definitions of union and intersection apply, of course, even when F is not countable. Because of the way we have defined unions and intersections, the commutative and associative laws are automatically satisfied. Definition 2.22. The complement of A relative to B, denoted by B - A, is defined to be the set B- A = {x:xeB,butx0A}. Note that B - (B - A) = A whenever A c B. Also note that B - A = B if B n A is empty. The notions of union, intersection, and complement are illustrated in Fig. 2.4. 42 Some Basic Notions of Set Theory AuB AnB B-A Th. 2.23 Figure 2.4 Theorem 2.23. Let F be a collection of sets. Then for any set B, we have B-UA=n(B-A), AeF AeF and B- nA=U(B-A). AeF AeF Proof Let S = UAEF A, T = I IAEF (B - A). If x e B - S, then x e B, but x f S. Hence, it is not true that x belongs to at least one A in F; therefore x belongs to no A in F. Hence, for every A in F, x e B - A. But this implies x e T, so that B - S s T. Reversing the steps, we obtain T s B - S, and this proves that B - S = T. To prove the second statement, use a similar argument. 2.15 COUNTABLE COLLECTIONS OF COUNTABLE SETS Definition 2.24. If F is a collection of sets such that every two distinct sets in F are disjoint, then F is said to be a collection of disjoint sets. Theorem 2.25. If F is a countable collection of disjoint sets, say F = {A1, A2, ... such that each set An is countable, then the union Uk 1 Ak is also countable. Proof. Let A. = {a1,,,, a2,e, a3,n .. }, n = 1, 2, ... , and let S = Uk 1 Ak Then every element x of S is in at least one of the sets in F and hence x = for some pair of integers (m, n). The pair (m, n) is uniquely determined by x, since F is a collection of disjoint sets. Hence the function f defined by f(x) = (m, n) if x = a.,,,, x e S, has domain S. The rangef(S) is a subset of V x V (where Z+ is the set of positive integers) and hence is countable. But f is one-to-one and there-fore S - f(S), which means that S is also countable. Theorem 2.26. If F = JA1, A2, ... } is a countable collection of sets, let G = {B1, B2, ... }, where B1 = Al and, for n > U Ak. R-1 k=1 Then G is a collection of disjoint sets, and we have 00 00 U Ak = U Bk. k=1 k=1 Th. 2.27 Exercises 43 Proof. Each set B. is constructed so that it has no elements in common with the earlier sets B1, B2, ... , Bn_1. Hence G is a collection of disjoint sets. Let A = Uk 1 Ak and B = Uk 1 Bk. We shall show that A = B. First of all, if x e A, then x e Ak for some k. If n is the smallest such k, then x e A but x 0 Uk=i Ak, which means that x e B,,, and therefore x e B. Hence A c B. Conversely, if x e B, then x E B for some n, and therefore x E A for this same n. Thus x e A and this proves that B s A. Using Theorems 2.25 and 2.26, we immediately obtain Theorem 2.27. If F is a countable collection of countable sets, then the union of all sets in F is also a countable set. Example 1. The set Q of all rational numbers is a countable set. Proof. Let A denote the set of all positive rational numbers having denominator n. The set of all positive rational numbers is equal to Uk 1 Ak. From this it follows that Q is countable, since each A is countable. Example 2. The set S of intervals with rational endpoints is a countable set. Proof. Let {x1, X2.... } denote the set of rational numbers and let A. be the set of all intervals whose left endpoint is x and whose right endpoint is rational. Then A. is countable and S = Uk 1 Ak. EXERCISES 2.1 Prove Theorem 2.2. Hint. (a, b) = (c, d) means {{a}, {a, b}} = {{c}, {c, d}}. Now appeal to the definition of set equality. 2.2 Let S be a relation and let -Q(S) be its domain. The relation S is said to be i) reflexive if a e -9(S) implies (a, a) e S, ii) symmetric if (a, b) e S implies (b, a) e S, iii) transitive if (a, b) e S and (b, c) e S implies (a, c) a S. A relation which is symmetric, reflexive, and transitive is called an equivalence relation. Determine which of these properties is possessed by S, if S is the set of all pairs of real numbers (x, y) such that a)xsy, b)x<y, c)x<lyl, d) x2 + y2 = 1, e) x2 + y2 < 0, f) x2 + x = y2 + y. 2.3 The following functions F and G are defined for all real x by the equations given. In each case where the composite function G o F can be formed, give the domain of G o F and a formula (or formulas) for (G o F)(x). a) F(x) = I - x, G(x) = x2 + 2x. b) F(x) = x + 5, G(x) = Ixl/x, if x j4 0, G(0) = 1. c) F(x) =- (2x, if 0 < x < 1, G(x) = (x2, if 0 S x <_ (1, otherwise, 0, otherwise. 44 Some Basic Notions of Set Theory Find F(x) if G(x) and G[F(x)] are given as follows: d) G(x) = x3, G[F(x)] = x3 - 3x2 + 3x - 1. e) G(x) = 3 + x + x2, G[F(x)] = x2 - 3x + 5. 2.4 Given three functions F, G, H, what restrictions must be placed on their domains so that the following four composite functions can be defined? GoF, Ho G, H°(GoF), (HoG)oF. Assuming that H o (G o F) and (H o G) o Fcan be defined, prove the associative law: Ho(GoF) = (HoG)oF. 2.5 Prove the following set-theoretic identities for union and intersection: a) Au(BuC) _ (AUB)UC, An(BnC) = (AnB)nC. b) An(BuC) _ (AnB)u(AnC). c) (AuB)n(AuC) = Au(BnC). d) (AvB)n(BvC)n(CvA) _ (AnB)u(AnC)v(BnC). e) A n (B - C) = (A n B) - (A n C). f) (A - C) n (B - C) = (A n B) - C. g) (A - B) u B = A if, and only if, B c A. 2.6 Let f : S -+ T be a function. If A and B are arbitrary subsets of S, prove that f(A u B) = f(A) v f(B) and f(A n B) c f(A) n f(B). Generalize to arbitrary unions and intersections. 2.7 Let f : S -. T be a function. If Y s T, we denote by f -'(Y) the largest subset of S which f maps into Y. That is, f-'(Y) = {x:xeSand f(x)e Y}. The set f -'(Y) is called the inverse image of Y under f. Prove the following for arbitrary subsets X of S and Y of T. a) X S .f-' [f(X)], b) f'[f-'(Y)] Y, c) f'' [Yl U Y2] = f-'(Y1) u.f-'(Y2), d) f-'(Y, n Y2) = f-'(Y1) nf-'(Y2), e)f-'(T- Y) = S-f-'(Y). f) Generalize (c) and (d) to arbitrary unions and intersections. 2.8 Refer to Exercise 2.7. Prove that f [f -' (Y) ] = Y for every subset Y of T if, and only if, T = f(S). 2.9 Let f : S -+ T be a function. Prove that the following statements are equivalent. a) f is one-to-one on S. b) f(A n B) = f(A) n f(B) for all subsets A, B of S. c) f -' [f(A) ] = A for every subset A of S. d) For all disjoint subsets A and B of S, the images f (A) and f'(B) are disjoint. Exercises 45 e) For all subsets A and B of S with B - A, we have f(A - B) = f(A) - f(B). 2.10 Prove that if A B and B - C, then A - C. 2.11 If {l, 2, ... , n} {1, 2, ... , m}, prove that m = n. 2.12 If S is an infinite set, prove that S contains a countably infinite subset. Hint. Choose an element al in S and consider S - {al }. 2.13 Prove that every infinite set S contains a proper subset similar to S. 2.14 If A is a countable set and B an uncountable set, prove that B - A is similar to B. 2.15 A real number is called algebraic if it is a root of an algebraic equation f(x) = 0, where f(x) = ao + a1x + -+ a"x" is a polynomial with integer coefficients. Prove that the set of all polynomials with integer coefficients is countable and deduce that the set of algebraic numbers is also countable. 2.16 Let S be a finite set consisting of n elements and let The the collection of all subsets of S. Show that T is a finite set and find the number of elements in T. 2.17 Let R denote the set of real numbers and let S denote the set of all real-valued func-tions whose domain is R. Show that S and R are not equinumerous. Hint. Assume S - R and let f be a one-to-one function such that f(R) = S. If a e R, let ga = f(a) be the real-valued function in S which corresponds to the real number a. Now define h by the equation h(x) = 1 + gx(x) if x e R, and show that h f S. 2.18 Let S be the collection of all sequences whose terms are the integers 0 and 1. Show that S is uncountable. 2.19 Show that the following sets are countable: a) the set of circles in the complex plane having rational radii and centers with rational coordinates, b) any collection of disjoint intervals of positive length. 2.20 Let f be a real-valued function defined for every x in the interval 0 < x < 1. Suppose there is a positive number M having the following property: for every choice of a finite number of points x1, x2, ... , x" in the interval 0 < x < 1, the sum If(XI) + ... + f(x")l < M. Let S be the set of those x in 0 < x < 1 for which f(x) j4 0. Prove that S is countable. 2.21 Find the fallacy in the following "proof" that the set of all intervals of positive length is countable. Let {x1, x2, ... } denote the countable set of rational numbers and let 1 be any interval of positive length. Then I contains infinitely many rational points x", but among these there will be one with smallest index n. Define a function F by means of the equation F(1) = n, if x" is the rational number with smallest index in the interval 1. This function establishes a one-to-one correspondence between the set of all intervals and a subset of the positive integers. Hence the set of all intervals is countable. 2.22 Let S denote the collection of all subsets of a given set T. Let f : S - R be a real-valued function defined on S. The function f is called additive iff(A U B) = f(A) + f(B) whenever A and B are disjoint subsets of T. If f is additive, prove that for any two subsets 46 Some Basic Notions of Set Theory A and B we have f(A U B) = f(A) + f(B - A) and f(A u B) = f(A) + f(B) - f(A n B). 2.23 Refer to Exercise 2.22. Assume f is additive and assume also that the following relations hold for two particular subsets A and B of T: f(A u B) = f(A') + f(B') - f(A')f(B') f(A n B) = f(A)f(B), f(A) + f(B) 0 f(T), where A' = T - A, B' = T - B. Prove that these relations determine f (T), and com-pute the value of f(T). SUGGESTED REFERENCES FOR FURTHER STUDY 2.1 Boas, R. P., A Primer of Real Functions. Carus Monograph No. 13. Wiley, New York, 1960. 2.2 Fraenkel, A., Abstract Set Theory, 3rd ed. North-Holland, Amsterdam, 1965. 2.3 Gleason, A., Fundamentals of Abstract Analysis. Addison-Wesley, Reading, 1966. 2.4 Halmos, P. R., Naive Set Theory. Van Nostrand, New York, 1960. 2.5 Kamke, E., Theory of Sets. F. Bagemihl, translator. Dover, New York, 1950. 2.6 Kaplansky, I., Set Theory and Metric Spaces. Allyn and Bacon, Boston, 1972. 2.7 Rotman, B., and Kneebone, G. T., The Theory of Sets and Transfinite Numbers. Elsevier, New York, 1968. CHAPTER 3 ELEMENTS OF POINT SET TOPOLOGY 3.1 INTRODUCTION A large part of the previous chapter dealt with "abstract" sets, that is, sets of arbitrary objects. In this chapter we specialize our sets to be sets of real numbers, sets of complex numbers, and more generally, sets in higher-dimensional spaces. In this area of study it is convenient and helpful to use geometric terminology. Thus, we speak about sets of points on the real line, sets of points in the plane, or sets of points in some higher-dimensional space. Later in this book we will study functions defined on point sets, and it is desirable to become acquainted with certain fundamental types of point sets, such as open sets, closed sets, and compact sets, before beginning the study of functions. The study of these sets is called point set topology. 3.2 EUCLIDEAN SPACE Rn A point in two-dimensional space is an ordered pair of real numbers (x1, x2). Similarly, a point in three-dimensional space is an ordered triple of real numbers (x1, x2i x3). It is just as easy to consider an ordered n-tuple of real numbers (x1i x2,... , xn) and to refer to this as a point in n-dimensional space. Definition 3.1. Let n > 0 be an integer. An ordered set of n real numbers (x1, x2i ... , xn) is called an n-dimensional point or a vector with n components. Points or vectors will usually be denoted by single bold face letters; for example, x = (x1, x2, ... , xn) or Y = (Y1, Y2, ... , Yn) The number xk is called the kth coordinate of the point x or the kth component of the vector x. The set of all n-dimensional points is called n-dimensional Euclidean space or simply n-space, and is denoted by R. The reader may wonder whether there is any advantage in discussing spaces of dimension greater than three. Actually, the language of n-space makes many complicated situations much easier to comprehend. The reader is probably familiar enough with three-dimensional vector analysis to realize the advantage of writing the equations of motion of a system having three degrees of freedom as a single vector equation rather than as three scalar equations. There is a similar advantage if the system has n degrees of freedom. 47 48 Elements of Point Set Topology Def. 3.2 Another advantage in studying n-space for a general n is that we are able to deal in one stroke with many properties common to 1-space, 2-space, 3-space, etc., that is, properties independent of the dimensionality of the space. Higher-dimensional spaces arise quite naturally in such fields as relativity, and statistical and quantum mechanics. In fact, even infinite-dimensional spaces are quite common in quantum mechanics. Algebraic operations on n-dimensional points are defined as follows: Definition 3.2. Let x = (x1, ... , x") and y = (yi, ... , y") be in R". We define: a) Equality: b) Sum : x=yif,and only if,x1 X + y = (x1 + Y1, ... , xn + Yn) c) Multiplication by real numbers (scalars): ax = (axi, ... , ax") (a real). d) Difference: x-y=x+(-1)y. e) Zero vector or origin: 0 = (01 ...10). f) Inner product or dot product: x'Y = xkYk g) Norm or length: k=1 IIXiI = (x.x)112 = xk)1/2 k-1 The norm Ilx - YII is called the distance between x and y. NOTE. In the terminology of linear algebra, R" is an example of a linear space. Theorem 3.3. Let x and y denote points in R". Then we have: a) ll x ll >- O, and II x 11 = 0 if, and only if, x = 0. b) Ilaxll = Ial IIx11 for every real a. c) llx - YII = IIY - xll d) Ix'YI < Ilxll IIYII (Cauchy-Schwarz inequality). e) Ilx + YII < Ilxll + IIYII (triangle inequality). Proof. Statements (a), (b) and (c) are immediate from the definition, and the Cauchy-Schwarz inequality was proved in Theorem 1.23. Statement (e) follows Def. 3.6 Open Balls and Open Sets in R" from (d) because n n llx + y112 = E (xk + Yk)2 = E (x 'k + 2Xk Yk + Y20 k=1 k=1 49 = IIx112 + 2x-y + Ily112 <-IIXI12 + 211x11 Ilyll + 11Y112 = (11x11 + Ilyll)2 NOTE. Sometimes the triangle inequality is written in the form llx - ZII <- llx - Y11 + Ily - Z11. This follows from (e) by replacing x by x - y and y by y - z. We also have 111X11 - IIY11l s 11X - yll Definition 3.4. The unit coordinate vector Uk in R" is the vector whose kth com-ponent is I and whose remaining components are zero. Thus, U1 = (1,0,...,0), U2 = (0, 1,0,...,0), ...,U" = (0,0,...,0, 1). If x= (x1, ... , xn) then x = x1u1 + + x"u" and x1 = X'u1, x2 = X u2, ... , x, = x - u". The vectors u1, ... , u,, are also called basis vectors. 3.3 OPEN BALLS AND OPEN SETS IN R" Let a be a given point in R" and let r be a given positive number. The set of all points x in R" such that llx - all < r, is called an open n-ball of radius r and center a. We denote this set by B(a) or by B(a; r). The ball B(a; r) consists of all points whose distance from a is less than r. In R1 this is simply an open interval with center at a. In R2 it is a circular disk, and in R3 it is a spherical solid with center at a and radius r. 3.5 Definition of an interior point. Let S be a subset of R", and assume that a e S. Then a is called an interior point of S if there is an open n-ball with center at a, all of whose points belong to S. In other words, every interior point a of S can be surrounded by an n-ball B(a)s S. The set of all interior points of S is called the interior of S and is denoted by int S. Any set containing a ball with center a is sometimes called a neighborhood of a. 3.6 Definition of an open set. A set S in R" is called open if all its points are interior points. NOTE. A set S_is open.if and only if S = int S. (See Exercise 3.9.) 50 Elements of Point Set Topology Th. 3.7 Examples. In R1 the simplest type of nonempty open set is an open interval. The union of two or more open intervals is also open. A closed interval [a, b] is not an open set because the endpoints a and b are not interior points of the interval. Examples of open sets in the plane are: the interior of a disk; the Cartesian product of two one-dimensional open intervals. The reader should be cautioned that an open interval in R1 is no longer an open set when it is considered as a subset of the plane. In fact, no subset of R1 (except the empty set) can be open in R2, because such a set cannot contain a 2-ball. In R" the empty set is open (Why?) as is the whole space R". Every open n-ball is an open set in R". The cartesian product (a1, b1) x ... x (a", b") of n one-dimensional open intervals (a1, b1), ... , (a", b") is an open set in R" called an n-dimensional open interval. We denote it by (a, b), where a = (a1, ... , a") and b = (bl,..., b"). The next two theorems show how additional open sets in R" can be constructed from given open sets. Theorem 3.7. The union of any collection of open sets is an open set. Proof. Let Fbe a collection of open sets and let S denote their union, S = UAEF A. Assume x e S. Then x must belong to at least one of the sets in F, say x e A. Since A is open, there exists an open n-ball B(x) c A. But A c S, so B(x) c S and hence x is an interior point of S. Since every point of S is an interior point, S is open. Theorem 3.8. The intersection of a finite collection of open sets is open. Proof. Let S = nk=1 Ak where each Ak is open. Assume x E S. (If S is empty, there is nothing to prove.) Then x e Ak for every k = 1, 2, ... , m, and hence there is an open n-ball B(x; rk) c Ak. Let r be the smallest of the positive numbers r1, r2, ... , rm. Then x e B(x; r) c S. That is, x is an interior point, so S is open. Thus we see that from given open sets, new open sets can be formed by taking arbitrary unions or finite intersections. Arbitrary intersections, on the other hand, will not always lead to open sets. For example, the intersection of all open intervals of the form (-1 In, 1 /n), where n = 1, 2, 3, . . . , is the set consisting of 0 alone. 3.4 THE STRUCTURE OF OPEN SETS IN R' In R1 the union of a countable collection of disjoint open intervals is an open set and, remarkably enough, every nonempty open set in R1 can be obtained in this way. This section is devoted to a proof of this statement. First we introduce the concept of a component interval. Th. 3.11 The Structure of Open Sets in R' 51 3.9 Definition of component interval. Let S be an open subset of R'. An open interval I (which may be finite or infinite) is called a component interval of S if I c S and if there is no open interval J # 1 such that I s J s S. In other words, a component interval of S is not a proper subset of any other open interval contained in S. Theorem 3.10. Every point of a nonempty open set S belongs to one and only one component interval of S. Proof Assume x e S. Then x is contained in some open interval I with I S S. There are many such intervals but the "largest" of these will be the desired com-ponent interval. We leave it to the reader to verify that this largest interval is I. = (a(x), b(x)), where a(x) = inf {a: (a, x) 9 S}, b(x) = sup (b: (x, b) s S}. Here a(x) might be - oo and b(x) might be + oo. Clearly, there is no open interval J such that Ix c J T- S, so Ix is a component interval of S containing x. If Jx is another component interval of S containing x, then the union Ix U Jx is an open interval contained in S and containing both Ix and J. Hence, by the defi-nition of component interval, it follows that Ix u Jx = Ix and Ix u Jx = Jx, so Ix=Jx. Theorem 3.11 (Representation theorem for open sets on the real line). Every non-empty open set S in R1 is the union of a countable collection of disjoint component intervals of S. Proof. If x e S, let Ix denote the component interval of S containing x. The union of all such intervals Ix is clearly S. If two of them, Ix and Iy, have a point in common, then their union Ix u Iy is an open interval contained in S and containing both Ix and Iy. Hence I. u Iy = Ix and I. u Iy = Iy so Ix = Iy. Therefore the intervals Ix form a disjoint collection. It remains to show that they form a countable collection. For this purpose, let {x1, x2, x3i .. . } denote the countable set of rational numbers. In each com-ponent interval Ix there will be infinitely many x,,, but among these there will be exactly one with smallest index n. We then define a function F by means of the equation F(II) = n, if x is the rational number in Ix with smallest index n. This function F is one-to-one since F(Ix) = F(ly) = n implies that Ix and IY have x in common and this implies Ix = Iy. Therefore F establishes a one-to-one corre-spondence between the intervals Ix and a subset of the positive integers. This completes the proof. NOTE. This representation of S is unique. In fact, if S is a union of disjoint open intervals, then these intervals must be the component intervals of S. This is an immediate consequence of Theorem 3.10. If S is an open interval, then the representation contains only one component interval, namely S itself. Therefore an open interval in R1 cannot be expressed as 52 Elements of Point Set Topology Def. 3.12 the union of two nonempty disjoint open sets. This property is also described by saying that an open interval is connected. The concept of connectedness for sets in R" will be discussed further in Section 4.16. 3.5 CLOSED SETS 3.12 Definition of a closed set. A set S in R" is called closed if its complement R" - S is open. Examples. A closed interval [a, b] in R' is a closed set. The cartesian product [a1, b1 ] x ... x [an, bn] of n one-dimensional closed intervals is a closed set in R" called an n-dimensional closed interval [a, b]. The next theorem, a consequence of Theorems 3.7 and 3.8, shows how to construct further closed sets from given ones. Theorem 3.13. The union of a finite collection of closed sets is closed, and the intersection of an arbitrary collection of closed sets is closed. A further relation between open and closed sets is described by the following theorem. Theorem 3.14. If A is open and B is closed, then A - B is open and B - A is closed. Proof. We simply note that A - B = A r (R" - B), the intersection of two open sets, and that B - A = B r (R" - A), the intersection of two closed sets. 3.6 ADHERENT POINTS. ACCUMULATION POINTS Closed sets can also be described in terms of adherent points and accumulation points. 3.15 Definition of an adherent point. Let S be a subset of R", and x a point in R", x not necessarily in S. Then x is said to be adherent to S if every n-ball B(x) contains at least one point of S. Examples 1. If x e S, then x adheres to S for the trivial reason that every n-ball B(x) contains x. 2. If S is a subset of R which is bounded above, then sup S is adherent to S. Some points adhere to S because every ball B(x) contains points of S distinct from x. These are called accumulation points. 3.16 Definition of an accumulation point. If S c R" and x e R", then x is called an accumulation point of S if every n-ball B(x) contains at least one point of S distinct from x. Def. 3.19 Closed Sets and Adherent Points 53 In other words, x is an accumulation point of S if, and only if, x adheres to S - {x}. If x e S but x is not an accumulation point of S, then x is called an isolated point of S. Examples 1. The set of numbers of the form 1/n, n = 1, 2, 3, ... , has 0 as an accumulation point. 2. The set of rational numbers has every real number as an accumulation point. 3. Every point of the closed interval [a, b] is an accumulation point of the set of num-bers in the open interval (a, b). Theorem 3.17. If x is an accumulation point of S, then every n-ball B(x) contains infinitely many points of S. Proof. Assume the contrary; that is, suppose an n-ball B(x) exists which contains only a finite number of points of S distinct from x, say a,, a2, ... , am. If r denotes the smallest of the positive numbers Ilx-a,11, Ilx-a211, ..., IIx - amll, then B(x; r/2) will be an n-ball about x which contains no points of S distinct from x. This is a contradiction. This theorem implies, in particular, that a set cannot have an accumulation point unless it contains infinitely many points to begin with. The converse, how-ever, is not true in general. For example, the set of integers {1, 2, 3, . .. } is an infinite set with no accumulation points. In a later section we will show that infinite sets contained in some n-ball always have an accumulation point. This is an important result known as the Bolzano-Weierstrass theorem. 3.7 CLOSED SETS AND ADHERENT POINTS A closed set was defined to be the complement of an open set. The next theorem describes closed sets in another way. Theorem 3.18. A set S in R" is closed if, and only if, it contains all its adherent points. Proof. Assume S is closed and let x be adherent to S. We wish to prove that x e S. We assume x 0 S and obtain a contradiction. If x 0 S then x e R" - S and, since R" - S is open, some n-ball B(x) lies in R" - S. Thus B(x) contains no points of S, contradicting the fact that x adheres to S. To prove the converse, we assume S contains all its adherent points and show that S is closed. Assume x e R" - S. Then x 0 S, so x does not adhere to S. Hence some ball B(x) does not intersect S, so B(x) c R" - S. Therefore R" - S is open, and hence S is closed. 3.19 Definition of closure. The set of all adherent points of a set S is called the closure of S and is denoted by S. 54 Elements of Point Set Topology Th. 3.20 For any set we have S E- S since every point of S adheres to S. Theorem 3.18 shows that the opposite inclusion S c S holds if and only if S is closed. Therefore we have: Theorem 3.20. A set S is closed if and only if S = S. 3.21 Definition of derived set. The set of all accumulation points of a set S is called the derived set of S and is denoted by S'. Clearly, we have S = S u S' for any set S. Hence Theorem 3.20 implies that S is closed if and only if S' S. In other words, we have: Theorem 3.22. A set S in R" is closed if, and only if, it contains all its accumulation points. 3.8 THE BOLZANO-WEIERSTRASS THEOREM 3.23 Definition of'a bounded set. A set Sin R" is said to be bounded if it lies entirely within an n-ball B(a; r) for some r > 0 and some a in R". Theorem 3.24 (Bolzano-Weierstrass). If a bounded set S in R" contains infinitely many points, then there is at least one point in R" which is an accumulation point of S. Proof To help fix the ideas we give the proof first for R1. Since S is bounded, it lies in some interval [ -a, a]. At least one of the subintervals [ - a, 0] or [0, a] contains an infinite subset of S. Call one such subinterval [a1, b1]. Bisect [a1, b1] and obtain a subinterval [a2, b2] containing an infinite subset of S, and continue this process. In this way a countable collection of intervals is obtained, the nth interval [an, bn] being of length b" - an = a/2n-1. Clearly, the sup of the left endpoints an and the inf of the right endpoints bn must be equal, say to x. [Why are they equal?] The point x will be an accumulation point of S because, if r is any positive number, the interval [a", b"] will be contained in B(x; r) as soon as n is large enough so that bn - an < r/2. The interval B(x; r) contains a point of S distinct from x and hence x is an accumulation point of S. This proves the theorem for R1. (Observe that the accumulation point x may or may not belong to S.) Next we give a proof for R", n > 1, by an extension of the ideas used in treating R1. (The reader may find it helpful to visualize the proof in R2 by referring to Fig. 3.1.) Since S is bounded, S lies in some n-ball B(0; a), a > 0, and therefore within the n-dimensional interval J1 defined by the inequalities - a < xk 5 a (k = 1, 2, ... , n). Here J1 denotes the cartesian product J1 = 1(1) x IZ1) x ... X IM. that is, the set of points (x1, . . . , xn), where xk e I,") and where each Ikl) is a one-dimensional interval -a < xk < a. Each interval IV) can be bisected to Th. 3.24 Bolzano-Weierstrass Theorem 55 Figure 3.1 form two subintervals Ikll and I"), defined by the inequalities lill : -a < xk < 0; Ik2 : 0 < xk < a. Next, we consider all possible cartesian products of the form lik, X IZ 2 X X (a) where each k; = 1 or 2. There are exactly 2" such products and, of course, each such product is an n-dimensional interval. The union of these 2" intervals is the original interval J1, which contains S; and hence at least one of the 2" intervals in (a) must contain infinitely many points of S. One of these we denote by J2, which can then be expressed as JZ = Ii 2) x 122) X ... X j,((2), where each Ik2) is one of the subintervals of Ik1) of length a. We now proceed with J2 as we did with J1, bisecting each interval Ik2) and arriving at an n-dimen-sional interval J3 containing an infinite subset of S. If we continue the process, we obtain a countable collection of n-dimensional intervals J1, J2, J3, ... , where the mth interval J," has the property that it contains an infinite subset of S and can be expressed in the form J. = I(-) X I2(M) x . X Iwhere Ikm) Ik1) Writing we have 1(m) = [a(m) b(m)7 k k , k ' bim) - aim) = 2^a _2 (k = 1, 2, . , n). For each fixed k, the sup of all left endpoints aim), (m = 1, 2, ... ), must therefore be equal to the-inf of all right endpoints b(m), (m = 1, 2,... ), and their common value we denote by tk. We now assert that the point t = (t1, t2, ... , t") is an 56 Elements of Point Set Topology 11. 3.25 accumulation point of S. To see this, take any n-ball B(t; r). The point t, of course, belongs to each of the intervals J1, J2, ... constructed above, and when m is such that a/2' -2 < r/2, this neighborhood will include J contains infinitely many points of S, so will B(t; r), which proves that t is indeed an accumulation point of S. 3.9 THE CANTOR INTERSECTION THEOREM As an application of the Bolzano-Weierstrass theorem we prove the Cantor intersection theorem. Theorem 3.25. Let {Q1, Q2, ... } be a countable collection of nonempty sets in R" such that: i) Qk+1 C Qk (k = 1, 2, 3, ... ). ii) Each set Qk is closed and Q1 is bounded. Then the intersection nk 1 Qk is closed and nonempty. Proof. Let S = nk 1 Qk. Then S is closed because of Theorem 3.13. To show that S is nonempty, we exhibit a point x in S. We can assume that each Qk con-tains infinitely many points; otherwise the proof is trivial. Now form a collection of distinct points A = {x1, x2, ... }, where Xk a Qk. Since A is an infinite set contained in the bounded set Q1, it has an accumulation point, say x. We shall show that x e S by verifying that x e Qk for each k. It will suffice to show that x is an accumulation point of each Qk, since they are all closed sets. But every neighborhood of x contains infinitely many points of A, and since all except (possibly) a finite number of the points of A belong to Qk, this neighborhood also contains infinitely many points of Qk. Therefore x is an accumulation point of Qk and the theorem is proved. 3.10 THE LINDELOF COVERING THEOREM In this section we introduce the concept of a covering of a set and prove the Lindelof covering theorem. The usefulness of this concept will become apparent in some of the later work. 3.26 Definition of a covering. A collection F of sets is said to be a covering of a given set S if S c A. The collection F is also said to cover S. If F is a collection of open sets, then F is called an open covering of S. Examples 1. The collection of all intervals of the form 1/n < x < 2/n, (n = 2, 3, 4.... ), is an open covering of the interval 0 < x < 1. This is an example of a countable covering. 2. The real line R1 is covered by the collection of all open intervals (a, b). This covering is not countable. However, it contains a countable covering of R1, namely, all inter-vals of the form (n, n + 2), where n runs through the integers. Th. 3.28 Lindelof Covering Theorem 57 3. Let S = {(x, y) : x > 0, y > 0}. The collection F of all circular disks with centers at (x, x) and with radius x, where x > 0, is a covering of S. This covering is not countable. However, it contains a countable covering of S, namely, all those disks in which x is rational. (See Exercise 3.18.) The Lindelof covering theorem states that every open covering of a set S in W contains a countable subcollection which also covers S. The proof makes use of the following preliminary result : Theorem 3.27 Let G = {A1, A2, ... } denote the countable collection of all n-balls having rational radii and centers at points with rational coordinates. Assume x e R" and let S be an open set in R" which contains x. Then at least one of the n-balls in G contains x and is contained in S. That is, we have x e Ak 9 S for some Ak in G. Proof. The collection G is countable because of Theorem 2.27. If x e R" and if S is an open set containing x, then there is an n-ball B(x; r) S S. We shall find a point y in S with rational coordinates that is "near" x and, using this point as center, will then find a neighborhood in G which lies within B(x; r) and which contains x. Write x = (x1,x2,...,x"), and let yk be a rational number such that IYk - xkl < rl(4n) for each k = 1, 2, ... , n. Then IIY-xII<IY1-x1I+...+IY"-xnI<r 4 Next, let q be a rational number such that r/4 < q < r/2. Then x e B(y; q) and B(y; q) c B(x; r) c S. But B(y; q) e G and hence the theorem is proved. (See Fig. 3.2 for the situation in R2.) B(y; q) Figure 3.2 Theorem 3.28 (Lindelof covering theorem). Assume A c R" and let F be an open covering of A. Then there is a countable subcollection of F which also covers A. Proof. Let G = {A1, A2, ... } denote the countable collection of all n-balls having rational centers and rational radii. This set G will be used to help us extract a countable subcollection of F which covers A. 58 Elements of Point Set Topology Th. 3.29 Assume x e A. Then there is an open set S in F such that x e S. By Theorem 3.27 there is an n-ball Ak in G such that x e Ak S S. There are, of course, infinitely many such At corresponding to each S, but we choose only one of these, for ex-ample, the one of smallest index, say m = m(x). Then we have x e Am(X) c S. The set of all n-balls Am(x) obtained as x varies over all elements of A is a countable collection of open sets which covers A. To get a countable subcollection of F which covers A, we simply correlate to each set Ak(X) one of the sets S of -F which contained Ak(X). This completes the proof. 3.11 THE HEINE-BOREL COVERING THEOREM The Lindelof covering theorem states that from any open covering of an arbitrary set A in R" we can extract a countable covering. The Heine-Borel theorem tells us that if, in addition, we know that A is closed and bounded, we can reduce the covering to a finite covering. The proof makes use of the Cantor intersection theorem. Theorem 3.29 (Heine-Borel). Let F be an open covering of a closed and bounded set A in R". Then a finite subcollection of F also covers A. Proof. A countable subcollection of F, say {I1, I2, ... }, covers A, by Theorem 3.28. Consider, for m > 1, the finite union m Sm U Ik k=1 This is open, since it is the union of open sets. We shall show that for some value of m the union S. covers A. For this purpose we consider the complement R" - Sm, which is closed. Define a countable collection of sets {Q1i Q2,... } as follows: Q1 = A, and for m > 1, Qm=An(R"-Sm). That is, Q. consists of those points of A which lie outside of Sm. Ifwe can show that for some value of m the set Q. is empty, then we will have shown that for this m no point of A lies outside Sm; in other words, we will have shown that some S. covers A. Observe the following properties of the sets Qm : Each set Q. is closed, since it is the intersection of the closed set A and the closed set R" - Sm. The sets Qm are decreasing, since the S. are increasing; that is, Qm+ 1 Ez Qm. The sets Qm, being subsets of A, are all bounded. Therefore, if no set Q. is empty, we can apply the Cantor intersection theorem to conclude that the intersection nk 1 Qk is also not empty. This means that there is some point in A which is in all the sets Qm, or, what is the same thing, outside all the sets Sm. But this is impossible, since A a Uk 1 Sk. _ Therefore some Qm must be empty, and this completes the proof. Th. 3.31 Compactness in R" 59 3.12 COMPACTNESS IN R" We have just seen that if a set S in R" is closed and bounded, then any open covering of S can be reduced to a finite covering. It is natural to inquire whether there might be sets other than closed and bounded sets which also have this property. Such sets will be called compact. 3.30 Definition of a compact set. A set S in R" is said to be compact if, and only if, every open covering of S contains a finite subcover, that is, a finite subcollection which also covers S. The Heine-Borel theorem states that every closed and bounded set in R" is compact. Now we prove the converse result. Theorem 3.31. Let S be a subset of R". Then the following three statements are equivalent: a) S is compact. b) S is closed and bounded. c) Every infinite subset of S has an accumulation point in S. Proof. As noted above, (b) implies (a). If we prove that (a) implies (b), that (b) implies (c) and that (c) implies (b), this will establish the equivalence of all three statements. Assume (a) holds. We shall prove first that S is bounded. Choose a point p in S. The collection of n-balls B(p; k), k = 1, 2, ... , is an open covering of S. By compactness a finite subcollection also covers S and hence S is bounded. Next we prove that S is closed. Suppose S is not closed. Then there is an accumulation point y of S such that y f S. If x e S, let rx = IIx - y11/2. Each rX is positive since y 0 S and the collection {B(x; rx) : x e S} is an open covering of S. By compactness, a finite number of these neighborhoods cover S, say P S U B(xk; rk). k=1 Let r denote the smallest of the radii r1, r2, ... , r,,. Then it is easy to prove that the ball B(y; r) has no points in common with any of the balls B(xk; rk). In fact, if x e B(y; r), then llx - yll < r 5 rk, and by the triangle inequality we have IIY - xkll < IIY - xil + IIx - xkll, So IIx - xkll > - IIY - xkll - IN -Y II = 2rk - IIx - yll > rk Hence x 0 B(xk; rk). Therefore B(y; r) n S is empty, contradicting the fact that y is an accumulation point of S. This contradiction shows that S is closed and hence (a) implies (b). Assume (b) holds. In this case the proof of (c) is immediate, because if T is an infinite subset of S then T is bounded (since S is bounded), and hence by the Bolzano-Weierstrass theorem T has an accumulation point x, say. Now x is also 60 Elements of Point Set Topology Def. 3.32 an accumulation point of S and hence x e S, since S is closed. Therefore (b) implies (c). Assume (c) holds. We shall prove (b). If S is unbounded, then for every m > 0 there exists a point xm in S with I I xm I I > m. The collection T = {x1, x2, ... } is an infinite subset of S and hence, by (c), T has an accumulation point y in S. But form > 1 + IIYII we have Ilxm - YII >- Ilxmll - IIYII > m - IIYII > 1, contradicting the fact that y is an accumulation point of T. This proves that S is bounded. To complete the proof we must show that S is closed. Let x be an accumulation point of S. Since every neighborhood of x contains infinitely many points of S, we can consider the neighborhoods B(x; 1/k), where k = 1 , 2, ... , and obtain a countable set of distinct points, say T = {x1, x2i ... }, contained in S, such that xk a B(x; 1/k). The point x is also an accumulation point of T. Since T is an infinite subset of S, part (c) of the theorem tells us that T must have an accumula-tion point in S. The theorem will then be proved if we show that x is the only accumulation point of T. To do this, suppose that y 0 x. Then by the triangle inequality we have IIY - xll < 11Y -xkil + Ilxk - xII < IIY - xkll + 1/k, if xk a T. If ko is taken so large that 1/k < illy - xII whenever k >- ko, the last inequality leads to II y - x11 < 11 Y - xk II . This shows that xk 0 B(y; r) when k >- ko, if r = illy - xII Hence y cannot be an accumulation point of T. This completes the proof that (c) implies (b). 3.13 METRIC SPACES The proofs of some of the theorems of this chapter depend only on a few properties of the distance between points and not on the fact that the points are in R". When these properties of distance are studied abstractly they lead to the concept of a metric space. 3.32 Definition of a metric space. A metric space is a nonempty set M of objects (called points) together with a function d from M x M to R (called the metric of the space) satisfying the following four properties for all points x, y, z in M: 1. d(x, x) = 0. 2. d(x, y) > O if x # y. 3. d(x, y) = d(y, x). 4. d(x, y) 5 d(x, z) + d(z, y). The nonnegative number d(x, y) is to be thought of as the distance from x to y. In these terms the intuitive meaning ofproperties 1 through 4 is clear. Property 4 is called the triangle inequality. Point Set Topology in Metric Spaces 61 We sometimes denote a metric space by (M, d) to emphasize that both the set M and the metric d play a role in the definition of a metric space. Examples 1. M = R"; d(x, y) = IIx - y1I. This is called the Euclidean metric. Whenever we refer to Euclidean space R", it will be understood that the metric is the Euclidean metric unless another metric is specifically mentioned. 2. M = C, the complex plane; d(zl, z2) = IZ1 - Z2i. As a metric space, C is indistin-guishable from Euclidean space R2 because it has the same points and the same metric. 3. Many nonempty set; d(x, y) = 0 if x = y, d(x, y) = I if x :;6 y. This is called the discrete metric, and (M, d) is called a discrete metric space. 4. If (M, d) is a metric space and if S is any nonempty subset of M, then (S, d) is also a metric space with the same metric or, more precisely, with the restriction of d to S x S as metric. This is sometimes called the relative metric induced by don S, and S is called a metric subspace of M. For example, the rational numbers Q with the metric d(x, y) = Ix - yI form a metric subspace of R. 5. M=R 2 ; d(x, y) = (x1 - yl)2 + 4(x2 - y2)2, where x = (x1, x2) and y = (yl, y2). The metric space (M, d) is not a metric subspace of Euclidean space R2 because the metric is different. 6. M = {(x1, x2) : xi + x2 = 1 }, the unit circle in R2 ; d(x, y) = the length of the smaller arc joining the two points x and y on the unit circle. 7. M = 01, x2, x3) : x1 + x2 + x3 = 1), the unit sphere in R3; d(x, y) = the length of the smaller arc along the great circle joining the two points x and y. 8. M = R";d(x,y) = Ixl - .l +...+ Ixn - yni. 9. M = R"; d(x, y) = max {Ixl - yl i+ , Ixn - ynl } 3.14 POINT SET TOPOLOGY IN METRIC SPACES The basic notions of point set topology can be extended to an arbitrary metric space (M, d). If a e M, the ball B(a; r) with center a and radius r > 0 is defined to be the set of all x in M such that d (x, a) < r. Sometimes we denote this ball by BM(a; r) to emphasize the fact that its points come from M. If S is a metric subspace of M, the ball Bs(a; r) is the intersection of S with the ball BM(a; r). Examples. In Euclidean space R' the ball B(0; 1) is the open interval (-1, 1). In the metric subspace S = [0, 1 ] the ball Bs(0; 1) is the half-open interval [0, 1). NOTE. The geometric appearance of a ball in R" need not be "spherical" if the metric is not the Euclidean metric. (See Exercise 3.27.) If S c M, a point a in S is called an interior point of S if some ball BM(a; r) lies entirely in S. The interior, int S, is the set of interior points of S. A set S is 62 Elements of Point Set Topology Th. 3.33 called open in M if all its points are interior points; it is called closed in M if M - S is open in M. Examples. 1. Every ball BM(a; r) in a metric space M is open in M. 2. In a discrete metric space M every subset S is open. In fact, if x e S, the ball B(x; f) consists entirely of points of S (since it contains only x), so S is open. Therefore every subset of M is also closed! 3. In the metric subspace S = [0, 1 ] of Euclidean space R', every interval of the form [0, x) or (x, 1 ], where 0 < x < 1, is an open set in S. These sets are not open in R'. Example 3 shows that if S is a metric subspace of M the open sets in S need not be open in M. The next theorem describes the relation between open sets in M and those in S. Theorem 3.33. Let (S, d) be a metric subspace of (M, d), and let X be a subset of S. Then X is open in S if, and only if, X=AnS for some set A which is open in M. Proof. Assume A is open in M and let X = A n S. If x e X, then x e A so BM(x; r) A for some r > 0. Hence Bs(x; r) = BM(x; r) n S s A n S = X so X is open in S. Conversely, assume X is open in S. We will show that X = A n S for some open set A in M. For every x in X there is a ball Bs(x; rx) contained in X. Now Bs(x; rx) = BM(x; rx) n S, so if we let A = U BM(x ; rx), xeX then A is open in M and it is easy to verify that A n S = X. Theorem 3.34. Let (S, d) be a metric subspace of (M, d) and let Y be a subset of S. Then Y is closed in S if, and only if, Y = B n S for some set B which is closed in M. Proof If Y = B n S, where B is closed in MM then B = M - A where A is open in M so Y = S n B = S n (M - A) = S - A ; hence Y is closed in S. Conversely, if Y is closed in S, let X = S - Y. Then X is open in S so X = A n S, where A is open in M and Y=S-X=S-(AnS)=S-A=Sn(M-A)=SnB, where B = M - A is closed in M. This completes the proof. If S M, a point x in M is called an adherent point of S if every ball BM(x; r) contains at least one point of S. If x adheres to S - {x} then x is called an accumulation point of S. The closure S of S is the set of all adherent points of S, and the derived set S' is the set of all accumulation points of S. Thus, S = S v S'. Th. 3.38 Compact Subsets of a Metric Space 63 The following theorems are valid in every metric space (M, d) and are proved exactly as they were for Euclidean space W. In the proofs, the Euclidean distance llx - yll need only be replaced by the metric d(x, y). Theorem 3.35. a) The union of any collection of open sets is open, and the inter-section of a finite collection of open sets is open. b) The union of a finite collection of closed sets is closed, and the intersection of any collection of closed sets is closed. Theorem 3.36. If A is open and B is closed, then A - B is open and B - A is closed. Theorem 3.37. For any subset S of M the following statements are equivalent: a) S is, closed in M. b) S contains all its adherent points. c) S contains all its accumulation points. d) S = S. Example. Let M = Q, the set of rational numbers, with the Euclidean metric of R'. Let S consist of all rational numbers in the open interval (a, b), where both a and b are irrational. Then S is a closed subset of Q. Our proofs of the Bolzano-Weierstrass theorem, the Cantor intersection theorem, and the covering theorems of Lindelof and Heine-Borel used not only the metric properties of Euclidean space W but also special properties of R" not gen-erally valid in an arbitrary metric space (M, d). Further restrictions on M are required to extend these theorems to metric spaces. One of these extensions is outlined in Exercise 3.34. The next section describes compactness in an arbitrary metric space. 3.15 COMPACT SUBSETS OF A METRIC SPACE Let (M, d) be a metric space and let S be a subset of M. A collection F of open subsets of M is said to be an open covering of S if S c UAeF A. A subset S of M is called compact if every open covering of S contains a finite subcover. S is called bounded if S c B(a; r) for some r > 0 and some a in M. Theorem 3.38. Let S be a compact subset of a metric space M. Then: i) S is closed and bounded. ii) Every infinite subset of S has an accumulation point in S. Proof. To prove (i) we refer to the proof of Theorem 3.31 and use that part of the argument which showed that (a) implies (b). The only change is that the Euclidean distance llx - yll is to be replaced throughout by the metric d(x, y). 64 Elements of Point Set Topology Th. 3.39 To prove (ii) we argue by contradiction. Let T be an infinite subset of S and assume that no point of S is an accumulation point of T. Then for each point x in S there is a ball B(x) which contains no point of T (if x 0 T) or exactly one point of T (x itself, if x e T). As x runs through S, the union of these balls B(x) is an open covering of S. Since S is compact, a finite subcollection covers S and hence also covers T. But this is a contradiction because T is an infinite set and each ball contains at most one point of T. NOTE. In Euclidean space R", each of properties (i) and (ii) is equivalent to com-pactness (Theorem 3.31). In a general metric space, property (ii) is equivalent to compactness (for a proof see Reference 3.4), but property (i) is not. Exercise 3.42 gives an example of a metric space M in which certain closed and bounded subsets are not compact. Theorem 3.39. Let X be a closed subset of a compact metric space M. Then X is compact. Proof. Let F be an open covering of X, say X c UAEF A. We will show that a finite number of the sets A cover X. Since X is closed its complement M - X is open, so F u {(M - X)} is an open covering of M. But M is compact, so this covering contains a finite subcover which we can assume includes M - X. There-fore This subcover also covers X and, since M - X contains no points of X, we can delete the set M - X from the subcover and still cover X. Thus X c A 1 u u AP so X is compact. 3.16 BOUNDARY OF A SET Definition 3.40. Let S be a subset of a metric space M. A point x in M is called a boundary point of S if every ball BM(x; r) contains at least one point of S and at least one point of M - S. The set of all boundary points of S is called the boundary of S and is denoted by as. The reader can easily verify that BS= Sr M - S. This formula shows that as is closed in M. Example In R", the boundary of a ball B(a; r) is the set of points x such that l!x - all = r. In R', the boundary of the set of rational numbers is all of R'. Further properties of metric spaces are developed in the Exercises and also in Chapter 4. Exercises EXERCISES 65 Open and closed sets in R' and R2 3.1 Prove that an open interval in R' is an open set and that a closed interval is a closed set. 3.2 Determine all the accumulation points of the following sets in R' and decide whether the sets are open or closed (or neither). a) All integers. b) The interval (a, b]. c) All numbers of the form 1/n, (n = 1, 2, 3, ... ). d) All rational numbers. e) All numbers of the form 2-" + 5-1, (m, n = 1, 2, ... ). f) All numbers of the form (- 1)" + (1/m), (m, n = 1, 2, ... g) All numbers of the form (1/n) + (1/m), (m, n = 1, 2, ... ). h) All numbers of the form (-1)"/[1 + (1/n)], (n = 1, 2, ... ). 3.3 The same as Exercise 3.2 for the following sets in R2: a) All complex z such that Iz I > 1. b) All complex z such that Iz I >- 1. c) All complex numbers of the form (1/n) + (i/m), (m, n = 1 , 2, ... ). d) All points (x, y) such that x2 - y2 < 1. e) All points (x, y) such that x > 0. f) All points (x, y) such that x >- 0. 3.4 Prove that every nonempty open set S in R' contains both rational and irrational numbers. 3.5 Prove that the only sets in R' which are both open and closed are the empty set and R' itself. Is a similar statement true for R2? 3.6 Prove that every closed set in R' is the intersection of a countable collection of open sets. 3.7 Prove that a nonempty, bounded closed set S in R' is either a closed interval, or that S can be obtained from a closed interval by removing a countable disjoint collection of open intervals whose endpoints belong to S. Open and closed sets in R" 3.8 Prove that open n-balls and n-dimensional open intervals are open sets in R. 3.9 Prove that the interior of a set in R" is open in R". 3.10 If S 9 R", prove that int S is the union of all open subsets of R" which are contained in S. This is described by saying that int S is the largest open subset of S. 3.11 If S and Tare subsets of R", prove that (int S) n (int T) = int (S n T), and (int S) v (int T) s int (S U T). 66 Elements of Point Set Topology 3.12 Let S' denote the derived set and S the closure of a set S in W. Prove that: a) S' is closed in R"; that is, (S')' si S'. b) If S s T, then S' s T'. c) (S v T)' = S' v T'. d) (S)' = S'. e) S is closed in R". f) S is the intersection of all closed subsets of R" containing S. That is, S is the smallest closed set containing S. 3.13 Let S and T be subsets of R. Prove that S n T s S n T and that S n T S S-r )T if S is open. NOTE. The statements in Exercises 3.9 through 3.13 are true in any metric space. 3.14 A set S in R" is called convex if, for every pair of points x and y in S and every real 6 satisfying 0 < 0 < 1, we have Ox + (1 - 0)y c S. Interpret this statement geometric-ally (in R2 and R3) and prove that: a) Every n-ball in R" is convex. b) Every n-dimensional open interval is convex. c) The interior of a convex set is convex. d) The closure of a convex set is convex. 3.15 Let F be a collection of sets in R", and let S = UA E F A and T = nA E F A. For each of the following statements, either give a proof or exhibit a counterexample. a) If x is an accumulation point of T, then x is an accumulation point of each set A in F. b) If x is an accumulation point of S, then x is an accumulation point of at least one set A in F. 3.16 Prove that the set S of rational numbers in the interval (0, 1) cannot be expressed as the intersection of a countable collection of open sets. Hint. Write S = {x1, X2.... }, assume S = nk 1 Sk, where each Sk is open, and construct a sequence {Q"} of closed intervals such that Q"+ 1 s Q. s S. and such that x" 0 Q. Then use the Cantor inter-section theorem to obtain a contradiction. Covering theorems in R" 3.17 If S s R", prove that the collection of isolated points of S is countable. 3.18 Prove that the set of open disks in the xy-plane with center at (x, x) and radius x > 0, x rational, is a countable covering of the set {(x, y) : x > 0, y > 0}. 3.19 The collection F of open intervals of the form (1/n, 2/n), where n = 2, 3, ... , is an open covering of the open interval (0, 1). Prove (without using Theorem 3.31) that no finite subcollection of Fcovers (0, 1). 3.20 Give an example of a set S which is closed but not bounded and exhibit a countable open covering F such that no finite subset of F covers S. 3.21 Given a set. S in R" with the property that for every x in S there is an n-ball B(x) such that B(x) n S is countable. Prove that S is countable. 3.22 Prove that a collection of disjoint open sets in R" is necessarily countable. Give an example of a collection of disjoint closed sets which is not countable. Exercises 67 3.23 Assume that S Si R". A point x in R" is said to be a condensation point of S if every n-ball B(x) has the property that B(x) n S is not countable. Prove that if S is not count-able, then there exists a point x in S such that x is a condensation point of S. 3.24 Assume that S c R" and assume that S is not countable. Let T denote the set of condensation points of S. Prove that: a) S - T is countable, b) S n T is not countable, c) T is a closed set, d) T contains no isolated points. Note that Exercise 3.23 is a special case of (b). 3.25 A set in R" is called perfect if S = S', that is, if S is a closed set which contains no isolated points. Prove that every uncountable closed set F in R" can be expressed in the form F = A U B, where A is perfect and B is countable (Cantor-Bendixon theorem). Hint. Use Exercise 3.24. Metric spaces 3.26 In any metric space (M, d), prove that the empty set 0 and the whole space M are both open and closed. 3.27 Consider the following two metrics in R": n d1(x, y) = max lxi - ytI, d2(x, y) _ lxi - yil 15isn i=1 In each of the following metric spaces prove that the ball has the geometric appearance indicated : a) In (R2, d1), a square with sides parallel to the coordinate axes. b) In (R2, d2), a square with diagonals parallel to the axes. c) A cube in (R3, d1). d) An octahedron in (R3, d2). 3.28 Let d1 and d2 be the metrics of Exercise 3.27 and let lix - yll denote the usual Euclidean metric. Prove the following inequalities for all x and y in R": d1(x, y) 5 lix - YIl <- d2(x, y) and d2(x, y) -Jnllx - Yll <_ nd1(x, y). 3.29 If (M, d) is a metric space, define d '(x, y) = d (x, y) 1 + d(x, y) Prove that d' is also a metric for M. Note that 0 < d'(x, y) < 1 for all x, y in M. 3.30 Prove that every finite subset of a metric space is closed. 3.31 In a metric space (M, d) the closed ball of radius r > 0 about a point a in M is the set B(a; r) = {x: d(x, a) < r }. a) Prove that f(a; r) is a closed set. b) Give an example of a metric space in which .9(a; r) is not the closure of the open ball B(a; r). 68 Elements of Point Set Topology 3.32 In a metric space M, if subsets satisfy A S S s A, where A is the closure of A, th A is said to be dense in S. For example, the set Q of rational numbers is dense in R. A is dense in S and if S is dense in T, prove that A is dense in T. 3.33 Refer to Exercise 3.32. A metric space M is said to be separable if there is a counta, subset A which is dense in M. For example, R is separable because the set Q of ratios numbers is a countable dense subset. Prove that every Euclidean space R' is separable. 3.34 Refer to Exercise 3.33. Prove that the Lindelof covering theorem (Theorem 3.: is valid in any separable metric space. 3.35 Refer to Exercise 3.32. If A is dense in S and if B is open in S, prove that B c A n Hint. Exercise 3.13. 3.36 Refer to Exercise 3.32. If each of A and B is dense in S and if B is open in S, prc that A n B is dense in S. 3.37 Given two metric spaces (Si, dl) and (S2, d2), a metric p for the Cartesian prods S, x S2 can be constructed from d, and d2 in many ways. For example, if x = (x,, and y = (y,, y2) are in S, X S2, let p(x, y) = d&,, y,) + d2(x2, y2). Prove that u a metric for S, x S2 and construct further examples. Compact subsets of a metric space Prove each of the following statements concerning an arbitrary metric space (M, d) a subsets S, T of M. 3.38 Assume S c T S M. Then S is compact in (M, d) if, and only if, S is compact the metric subspace (T, d). 3.39 If S is closed and T is compact, then S r T is compact. 3.40 The intersection of an arbitrary collection of compact subsets of M is compact 3.41 The union of a finite number of compact subsets of M is compact. 3.42 Consider the metric space Q of rational numbers with the Euclidean metric of Let S consist of all rational numbers in the open interval (a, b), where a and b are it tional. Then S is a closed and bounded subset of Q which is not compact. Miscellaneous properties of the interior and the boundary If A and B denote arbitrary subsets of a metric space M, prove that : 3.43 int A = M - M - A. 3.44 int (M - A) = M - A. 3.45 int (int A) = int A. 3.46 a) int (ni=, A,) = ni=, (int A,), where each At M. b) int (nAEF A) S nAEF (int A), if F is an infinite collection of subsets of M. c) Give an example where equality does not hold in (b). 3.47 a) UAEF (int A) int (UAEF A). b) Give an example of a finite collection F in which equality does not hold in (a) 3.48 a) int (&A) = 0 if A is open or if A is closed in M. b) Give an example in which int (BA) = M. References 3.49 If int A = int B = 0 and if A is closed in M, then int (A U B) = 0. 3.50 Give an example in which int A = int B = 0 but int (A u B) = M. 3.51 OA = A n M - A and aA = a(M - A). 3.52 If A n B = 0, then a(A V B) = aA v aB. SUGGESTED REFERENCES FOR FURTHER STUDY 69 3.1 Boas, R. P., A Primer of Real Functions. Carus Monograph No. 13. Wiley, New York, 1960. 3.2 Gleason, A., Fundamentals of Abstract Analysis. Addison-Wesley, Reading, 1966. 3.3 Kaplansky, I., Set Theory and Metric Spaces. Allyn and Bacon, Boston, 1972. 3.4 Simmons, G. F., Introduction to Topology and Modern Analysis. McGraw-Hill, New York, 1963. CHAPTER 4 LIMITS AND CONTINUITY 4.1 INTRODUCTION The reader is already familiar with the limit concept as introduced in elementary calculus where, in fact, several kinds of limits are usually presented. For example, the limit of a sequence of real numbers {x"}, denoted symbolically by writing lim x = A, means that for every number e > 0 there is an integer N such that Ix - Al < s whenever n >- N. This limit process conveys the intuitive idea that x" can be made arbitrarily close to A provided that n is sufficiently large. There is also the limit of a function, indicated by notation such as lim f(x) = A, x-,p which means that for every c > 0 there is another number S > 0 such that l f(x) - AI < e whenever 0 < Ix - pi < S. This conveys the idea that f(x) can be made arbitrarily close to A by taking x sufficiently close to p. Applications of calculus to geometrical and physical problems in 3-space and to functions of several variables make it necessary to extend these concepts to R". It is just as easy to go one step further and introduce limits in the more general setting of metric spaces. This achieves a simplification in the theory by stripping it of unnecessary restrictions and at the same time covers nearly all the important aspects needed in analysis. First we discuss limits of sequences of points in a metric space, then we discuss limits of functions and the concept of continuity. 4.2 CONVERGENT SEQUENCES IN A METRIC SPACE Definition 4.1. A sequence {x"} of points in a metric space (S, d) is said to converge if there is a point p in S with the following property: For every e > 0 there is an integer N such that d(xn, p) < e whenever n Z N. 70 Th. 4.3 Convergent Sequences in a Metric Space 71 We also say that converges to p and we write x -+ p as n -+ co, or simply x -+ p. If there is no such p in S, the sequence {xx} is said to diverge. NOTE. The definition of convergence implies that x -+ p if and only if d(x,,, p) -+ 0. The convergence of the sequence {d(x,,, p)} to 0 takes place in the Euclidean metric space R'. Examples 1. In Euclidean space R', a sequence is called increasing if x -< for all n. If an increasing sequence is bounded above (that is, if x 5 M for some M > 0 and all n), then converges to the supremum of its range, sup {x1i x2, ... }. Similarly, is called decreasing if xit1 5 x for all n. Every decreasing sequence which is bounded below converges to the infimum of its range. For example, {1/n} converges to 0. 2. If and are real sequences converging to 0, then (a + also converges to 0. If 0 5 c 5 a for all n and if converges to 0, then also converges to 0. These elementary properties of sequences in R1 can be used to simplify some of the proofs concerning limits in a general metric space. 3. In the complex plane C, let z = 1 + n-2 + (2 - 1/n)i. Then converges to 1 + 2i because sod(z,,,I+2i)-+0. Theorem 4.2. A sequence in a metric space (S, d) can converge to at most one point in S. Proof. Assume that x -i p and x -+ q. We will prove that p = q. By the triangle inequality we have 0 5 d(p, q) 5 d(p, d(x,,, q). Since d(p, 0 and d(x1, q) -+ 0 this implies that d(p, q) = 0, so p = q. If a sequence {x.} converges, the unique point to which it converges is called the limit of the sequence and is denoted by lim x or by lim, x,,. Example. In Euclidean space RI we have 1/n = 0. The same sequence in the metric subspace T = (0, 1 ] does not converge because the only candidate for the limit is 0 and 0 0 T. This example shows that the convergence or divergence of a sequence depends on the underlying space as well as on the metric. Theorem 4.3. In a metric space (S, d), assume x - p and let T = {x1, x2i ... } be the range of Then: a) T is bounded. b) p is an adherent point of T. 72 Limits and Continuity Th. 4.4 Proof. a) Let N be the integer corresponding to e = 1 in the definition of con-vergence. Then every x with n z N lies in the ball B(p; 1), so every point in T lies in the ball B(p; r), where r = 1 + max {d(p, x1), ... , d(p, xN_1)}. Therefore T is bounded. b) Since every ball B(p; e) contains a point of T, p is an adherent point of T. NOTE. If T is infinite, every ball B(p; e) contains infinitely many points of T, so p is an accumulation point of T. The next theorem provides a converse to part (b). Theorem 4.4. Given a metric space (S, d) and a subset T S; S. If a point p in S is an adherent point of T, then there is a sequence of points in T which converges to p. Proof. For every integer n Z 1 there is a point x in T with d(p, 5 1/n. Hence d(p, x -+ p. Theorem 4.5. In a metric space (S, d) a sequence converges to p if, and only if, every subsequence converges to p. Proof. Assume x -+ p and consider any subsequence (xk(.)). For every e > 0 there is an N such that n z N implies d(x,,, p) < e. Since is a subsequence, there is an integer M such that k(n) Z N for n z M. Hence n >_ M implies d(xk(fl), p) < e, which proves that xk(n) - p. The converse statement holds trivially since is itself a subsequence. 43 CAUCHY SEQUENCES If a sequence converges to a limit p, its terms must ultimately become close to p and hence close to each other. This property is stated more formally in the next theorem. Theorem 4.6. Assume that converges in a metric space (S, d). Then for every e > 0 there is an integer N such that d(x,,, xm) < e whenever n >_ N and m >_ N. Proof. Let p = lim x,,. Given e > 0, let N be such that d(x., p) < e/2 whenever n >_ N. Then d(xm, p) < e/2 if m >_ N. If both n >_ N and m Z N the triangle inequality gives us = C. d(xn, xm) :!!-. d(x., p) + d(p, xm) < 2 + 2 Th. 4.8 Cauchy Sequences 73 4.7 Definition of a Cauchy Sequence. A sequence {xR} in a metric space (S, d) is called a Cauchy sequence if it satisfies the following condition (called the Cauchy condition) : For every s > 0 there is an integer N such that d(xx, xm) < e whenever n >- N and m >_ N. Theorem 4.6 states that every convergent sequence is a Cauchy sequence. The converse is not true in a general metric space. For example, the sequence (1/n) is a Cauchy sequence in the Euclidean subspace T = (0, 1] of R', but this sequence does not converge in T. However, the converse of Theorem 4.6 is true in every Euclidean space R. Theorem 4.8. In Euclidean space Rk every Cauchy sequence is convergent. Proof. Let {xR} be a Cauchy sequence in Rk and let T = {x1, x2, ... } be the range of the sequence. If T is finite, then all except a finite number of the terms {x.} are equal and hence {xR} converges to this common value. Now suppose T is infinite. We use the Bolzano-Weierstrass theorem to show that T has an accumulation point p, and then we show that {xR} converges to p. First we need to know that T is bounded. This follows from the Cauchy condition. In fact, when s = 1 there is an N such that n >- N implies IIxR - xNll < 1. This means that all points xR with n >- N lie inside a ball of radius 1 about XN as center, so T lies inside a ball of radius 1 + M about 0, where M is the largest of the numbers I)xlII, , IIXNII. Therefore, since T is a bounded infinite set it has an accumulation point p in Rk (by the Bolzano-Weierstrass theorem). We show next that {xR} converges to p. Given e > 0 there is an N such that IIxR - XmII < e/2 whenever n >- N and m >- N. The ball B(p; a/2) contains a point xm with m >- N. Hence if n >- N we have IIXR - PII <- IIxR - Xmll+ Ilxm - PII < -+2=e, 2 so lim xR = p. This completes the proof. Examples 1. Theorem 4.8 is often used for proving the convergence of a sequence when the limit is not known in advance. For example, consider the sequence in RI defined by R 1 xR = 1 - 2 + 3 - 4 + ... + (-n . If m > n >- N, we find (by taking successive terms in pairs) that 1 1 1 -1 +...+ 1 Ixm - XRI= n N n+1 n+2 m 74 Limits and Continuity Def. 4.9 SO I xm - XnI < e as soon as N > 1/c. Therefore {xn} is a Cauchy sequence and hence it converges to some limit. It can be shown (see Exercise 8.18) that this limit is log 2, a fact which is not immediately obvious. 2. Given a real sequence {a"} such that Ian+2 - an+1I < 3Ian+1 - aaI for all n >- 1. We can prove that {an} converges without knowing its limit. Let b" = Ian+1 - al. Then 0 < bn+ 1 < bn/2 so, by induction, bn+ 1 <- b1/2". Hence b" -> 0. Also, if m > n we have m-1 am - an = E (ak+1 - ak) hence k=n m-1 Iam - anI <Ebk<-bn 1+1+...+ 1 <2bn. k=n 2 2m-1-n) This implies that {an } is a Cauchy sequence, so {an } converges. 4.4 COMPLETE METRIC SPACES Definition 4.9. A metric space (S, d) is called complete if every Cauchy sequence in S converges in S. A subset T of S is called complete if the metric subspace (T, d) is complete. Example 1. Every Euclidean space Rk is complete (Theorem 4.8). In particular, R1 is complete, but the subspace T = (0, 1 ] is not complete. Example 2. The space R" with the metric d(x, y) = maxl 5 i sn Ixt - Y,I is complete. The next theorem relates completeness with compactness. Theorem 4.10. In any metric space (S, d) every compact subset T is complete. Proof. Let {xn} be a Cauchy sequence in T and let A = {x1, x2, ... } denote the range of {xn}. If A is finite, then {xn} converges to one of the elements of A, hence {xn} converges in T. If A is infinite, Theorem 3.38 tells us that A has an accumulation point p in T since T is compact. We show next that xn -+ p. Given c > 0, choose N so that n Z N and m Z N implies d(xn, x,,) < c/2. The ball B(p; a/2) contains a point xm with m z N. Therefore if n >_ N the triangle inequality gives us d(xn, p) <_ d(x, Xm) + d(xm, p) < 2 + 2 = E, so xn - p. Therefore every Cauchy sequence in T has a limit in T, so T is complete. 4.5 LIMIT OF A FUNCTION In this section we consider two metric spaces (S, ds) and (T, dT), where ds and dT denote the respective metrics. Let A be a subset of S and let f : A -+ T be a function from A to T. Th. 4.12 Limit of a Function Definition 4.11. If p is an accumulation point ofA and if b e T, the notation lim f(x) = b, X- P 75 (1) is defined to mean the following: For every e > 0 there is a b > 0 such that dT(f (x), b) < e whenever x e A, x p, and ds(x, p) < 6. The symbol in (1) is read "the limit of f(x), as x tends to p, is b," or f(x) approaches b as x approaches p." We sometimes indicate this by writingf(x) - b asx -p. The definition conveys the intuitive idea that f(x) can be made arbitrarily close to b by taking x sufficiently close to p. (See Fig. 4.1.) We require that p be an accumulation point of A to make certain that there will be points x in A sufficiently close to p, with x # p. However, p need not be in the domain of f, and b need not be in the range off. f Figure 4.1 NOTE. The definition can also be formulated in terms of balls. Thus, (1) holds if, and only if, for every ball BT(b), there is a ball Bs(p) such that Bs(p) r A is not empty and such that f (x) a BT(b) whenever x e Bs(p) n A, x# p. When formulated this way, the definition is meaningful whenp or b (or both) are in the extended real number system R or in the extended complex number system C. However, in what follows, it is to be understood that p and b are finite unless it is explicitly stated that they can be infinite. The next theorem relates limits of functions to limits of convergent sequences. Theorem 4.12. Assume p is an accumulation point of A and assume b E T. Then lim f(x) = b, (2) X- p 76 Limits and Continuity T6. 4.13 if, and only if, lim f(xn) = b, (3) n-+co for every sequence {xn} of points in A - { p} which converges to p. Proof. If (2) holds, then for every E > 0 there is a S > 0 such that dT(f (x), b) < E whenever x E A and 0 < ds(x, p) < 6. 4) Now take any sequence {xn} in A - {p} which converges to p. For the S in (4), there is an integer N such that n >- N implies ds(xn, p) < 6. Therefore (4) implies dT(f(xn), b) < E for n >- N, and hence { f(xn)} converges to b. Therefore (2) implies (3). To prove the converse we assume that (3) holds and that (2) is false and arrive at a contradiction. If (2) is false, then for some c > 0 and every S > 0 there is a point x in A (where x may depend on S) such that 0 < ds(x, p) < 6 but dT(f(x), b) >- e (5) Taking 6 = 1/n, n = 1, 2, ... , this means there is a corresponding sequence of points {xn} in A - f p} such that 0 < ds(xn, p) < 1 In but dT(f (xn), b) >- E. Clearly, this sequence {xn} converges to p but the sequence { f(xn)} does not con-verge to b, contradicting (3). NOTE. Theorems 4.12 and 4.2 together show that a function cannot have two different limits as x -+ p. 4.6 LIMITS OF COMPLEX-VALUED FUNCTIONS Let (S, d) be a metric space, let A be a subset of S, and consider two complex-valued functions f and g defined on A, f:A-C, g is defined to be the function whose value at each point x of A is the complex number f(x) + g(x). The difference f - g, the product f - g, and the quotient f/g are similarly defined. It is understood that the quotient is defined only at those points x for which g(x) # 0. The usual rules for calculating with limits are given in the next theorem. Theorem 4.13. Let f and g be complex-valued functions defined on a subset A of a metric space (S, d). Let p be an accumulation point of A, and assume that lim f(x) = a, lim g(x) = b. x-p x-p Th. 4.14 Limits of Vector-Valued Functions Then we also have: 77 a) limx-P [f(x) ± g(x)] = a ± b, b) limx_P f(x)g(x) = ab, c) limx.P f(x)/g(x) = alb if b : 0. Proof. We prove (b), leaving the other parts as exercises. Given a with 0 < e < 1, let e' be a second number satisfying 0 < e' < 1, which will be made to depend on e in a way to be described later. There is a S > 0 such that if x e A and d(x, p) < S, then Then 11(x) - al < e' and I g(x) - bI < e'. If(x)l=la+(f(x)-a)I<lal+e'<lal+1. Writing f(x)g(x) - ab = f(x)g(x) - bf(x) + bf(x) - ab, we have If(x)g(x) - abl <- If(x)l lg(x) - bl + I bI 1f(x) - al < (lal + 1)e' + Ible' = e'(IaI + IbI + 1). If we choose c' = e/(Ial + IbI + 1), we see that If(x)g(x) - abl < e whenever x e A and d(x, p) < 6, and this proves (b). 4.7 LIMITS OF VECTOR-VALUED FUNCTIONS Again, let (S, d) be a metric space and let A be a subset of S. Consider two vector-valued functions f and g defined on A, each with values in R', f : A --> Rk, g : A - R". Quotients of vector-valued functions are not defined (if k > 2), but we can define the sum f + g, the product Af (if A is real) and the inner product f g by the respec-tive formulas (f + g)(x) = f(x) + g(x), (Af)(x) = Af(x), (f-g)(x) = for each x in A. We then have the following rules for calculating with limits of vector-valued functions. Theorem 4.14. Let p be an accumulation point of A and assume that lim f(x) = a, lim g(x) = b. x-'P Then we also have: a) limx.P [f(x) + g(x)] = a + b, b) limx, P ).f(x) = Aa for every scalar A, c) lima-P f(x) . g(x) = a - b, d) lima-P Ilf(x)II = Ilall. 78 Limits and Continuity Def. 4.15 Proof We prove only parts (c) and (d). To prove (c) we write f(x) g(x) - a b = [f(x) - a] [g(x) - b] + a [g(x) - b] + b [f(x) - a]. The triangle inequality and the Cauchy-Schwarz inequality give us Ilf(x) - all llg(x) - bll + Ilall llg(x) - bll + Ilbll Ilf(x) - all. Each term on the right tends to 0 as x - p, so f(x) g(x) - a b. This proves (c). To prove (d) note that IIlf(x)ll - Ilail I < Ilf(x) - all. NOTE. Let f1, ... , f,, be n real-valued functions defined on A, and let f : A - R" be the vector-valued function defined by the equation f(x) = (f, (x), fz(x), ... , f (x)) if x e A. Then f1, ... , f are called the components of f, and we also write f = (f1, ... , to denote this relationship. If a = (a1, ... , then for each r = 1, 2, ... , n we have Ifr(x) - arl < llf(x) - all < E Ifr(x) - arl r=1 These inequalities show that limx+P f(x) = a if, and only if, limx. p fr(x) = ar for each r. 4.8 CONTINUOUS FUNCTIONS The definition of continuity presented in elementary calculus can be extended to functions from one metric space to another. Definition 4.15. Let (S, ds) and (T, dT) be metric spaces and let f : S - T be a function from S to T. The function f is said to be continuous at a point p in S if for every s > 0 there is a 8 > 0 such that dT(f(x), f(p)) < E whenever ds(x, p) < S. If f is continuous at every point of a subset A of S, we say f is continuous on A. This definition reflects the intuitive idea that points close to p are mapped by f into points close to f(p). It can also be stated in terms of balls: A function f is continuous at p if and only if, for every s > 0, there is a S > 0 such that f(Bs(p; S)) e BT(f(p); e). Here Bs(p; 8) is a ball in S; its image under f must be contained in the ball BT(f (p) ; s) in T. (See Fig. 4.2.) If p is an accumulation point of S, the definition of continuity implies that lim f(x) = f(p). x-+p Th. 4.17 Continuity of Composite Functions f Image of Bs(p; S) Figure 4.2 79 If p is an isolated point of S (a point of S which is not an accumulation point of S), then every f defined at p will be continuous at p because for sufficiently small S there is only one x satisfying ds(x, p) < 6, namely x = p, and dT(f(p), f(p)) = 0. Theorem 4.16. Let f : S - T be a function from one metric space (S, ds) to another (T, dT), and assume p E S. Then f is continuous at p if, and only if, for every sequence {xn} in S convergent to p, the sequence {f(xn)} in T converges to f(p); in symbols, fl+ n-00 00 Jim Ax.) = f(lim x,, . J The proof of this theorem is similar to that of Theorem 4.12 and is left as an exercise for the reader. (The result can also be deduced from 4.12 but there is a minor complication in the argument due to the fact that some terms of the sequence {xn} could be equal to p.) The theorem is often described by saying that for continuous functions the limit symbol can be interchanged with the function symbol. Some care is needed in interchanging these symbols because sometimes { f(x )} converges when {xn} diverges. Example If xn - x and y - y in a metric space (S, d), then d(xn, yn) -i d(x, y) (Exercise 4.7). The reader can verify that d is continuous on the metric space (S x S, p), where p is the metric of Exercise 3.37 with Sl = S2 = S. NOTE. Continuity of a function fat a point p is called a local property off because it depends on the behavior off only in the immediate vicinity ofp. A property of f which concerns the whole domain off is called a global property. Thus, continuity off on its domain is a global property. 4.9 CONTINUITY OF COMPOSITE FUNCTIONS Theorem 4.17. Let (S, ds), (T, dT), and (U, du) be metric spaces. Let f : S -+ T and g : f(S) -+ U be functions, and let h be the composite function defined on S by the equation h(x) = g(f(x)) for x in S. Bs(p; S) , BT(f(p); e) If f is continuous at p and if g is continuous at f(p), then h is continuous at p. 80 Limits and Continuity Th. 4.18 Proof. Let b = f(p). Givens > 0, there is a S > 0 such that du(g(y), g(b)) < s whenever dT(y, b) < S. For this S there is a b' such that dT(f(x), f(p)) < 8 whenever ds(x, p) < 8'. Combining these two statements and taking y = f(x), we find that dv(h(x), h(p)) < s whenever ds(x, p) < b', so h is continuous at p. 4.10 CONTINUOUS COMPLEX-VALUED AND VECTOR-VALUED FUNCTIONS Theorem 4.18. Let f and g be complex-valued functions continuous at a point p in a metric space (S, d). Then f + g, f - g, and f -g are each continuous at p. The quotient fig is also continuous at p if g(p) 0 0. Proof. The result is trivial if p is an isolated point of S. If p is an accumulation point of S, we obtain the result from Theorem 4.13. There is, of course, a corresponding theorem for vector-valued functions, which is proved in the same way, using Theorem 4.14. Theorem 4.19. Let f and g be functions continuous at a point p in a metric space (S, d), and assume that f and g have values in R". Then each of the following is continuous at p: the sum f + g, the product )f for every real A,, the inner product f g, and the norm II f ll . Theorem 4.20. Let fl, ... , f" be n real-valued functions defined on a subset A of a metric space (S, ds), and let f = (fl, . . . , Then f is continuous at a point p of A if and only if each of the functions fl, ... , f" is continuous at p. Proof. If p is an isolated point of A there is nothing to prove. If p is an accumula-tion point, we note that f(x) -- f(p) as x - p if and only if fk(x) -+ fk(p) for each k = 1, 2, ...,n. 4.11 EXAMPLES OF CONTINUOUS FUNCTIONS Let S = C, the complex plane. It is a trivial exercise to show that the following complex-valued functions are continuous on C : a) constant functions, defined by f(z) = c for every z in C; b) the identity function defined by f(z) = z for every z in C. Repeated application of Theorem 4.18 establishes the continuity of every poly-nomial : f(z)=ao+a1z+a2Z2+ +az", the a1 being complex numbers. Th. 4.22 Continuity and inverse Images of Open or Closed Sets 81 If S is a subset on C on which the polynomial f does not vanish, then I/f is continuous on S. Therefore a rational function g/f, where g and f are polynomials, is continuous at those points of C at which the denominator does not vanish. The familiar real-valued functions of elementary calculus, such as the ex-ponential, trigonometric, and logarithmic functions, are all continuous wherever they are defined. The continuity of these elementary functions justifies the common practice of evaluating certain limits by substituting the limiting value of the "independent variable"; for example, lim ex = e° = 1. X-0 The continuity of the complex exponential and trigonometric functions is a consequence of the continuity of the corresponding real-valued functions and Theorem 4.20. 4.12 CONTINUITY AND INVERSE IMAGES OF OPEN OR CLOSED SETS The concept of inverse image can be used to give two important global descriptions of continuous functions. 4.21 Definition of inverse image. Let f : S - T be a function from a set S to a set T. If Y is a subset of T, the inverse image of Y under f, denoted by f (Y), is defined to be the largest subset of S which f maps into Y; that is, f-1(Y) _ {x: x e S and f(x) e Y). NOTE. If f has an inverse function f -1, the inverse image of Y under f is the same as the image of Y under f -1, and in this case there is no ambiguity in the notation f -1(Y). Note also that f -1(A) c f -1(B) if A c B c T. Theorem 4.22. Let f : S - T be a function from S to T. If X c S and Y c T, then we have: a) X = f -1(Y) impliesf(X) g Y. b) Y = f(X) implies X e f -1(Y) The proof of Theorem 4.22 is a direct translation of the definition of the sym-bols f-1(Y) and f(X), and is left to the reader. It should be observed that; in general, we cannot conclude that Y = f(X) implies X = f -1(Y). (See the example in Fig. 4.3.) Figure 4.3 82 Limits and Continuity Th. 4.23 Note that the statements in Theorem 4.22 can also be expressed as follows: f[.f-1(Y)] c Y, X c f-1[f(X)] Note also that f -1(A u B) = f -1(A) u f -1(B) for all subsets A and B of T. Theorem 4.23. Let f : S -+ T be a function from one metric space (S, ds) to another (T, dT). Then f is continuous on S if, and only if, for every open set Y in T, the inverse image f -1(Y) is open in S. Proof. Let f be continuous on S, let Y be open in T, and let p be any point of f -1(Y). We will prove that p is an interior point off -1(Y). Let y = f(p). Since Y is open we have BT(y; a) c Y for some e > 0. Since f is continuous at p, there is a S > 0 such that f(Bs(p; S)) I BT(y; B). Hence, Bs(p; S) - f -1 [.f(Bs(p; S))] - f -1 [BT(y; a)] 9 f -' (Y), so p is an interior point off -1(Y). Conversely, assume that f-1(Y) is open in S for every open subset Y in T. Choose p in S and let y = f(p). We will prove that f is continuous at p. For every a > 0, the ball BT(y; a) is open in T, so f -1(BT(y; a)) is open in S. Now, p e f -1(BT(y; a)) so there is a S > 0 such that Bs(p; S) c f -1(BT(y; a)). There-fore, f(Bs(p; 8)) 9 BT(y; a) so f is continuous at p. Theorem 4.24. Let f : S - T be a function from one metric space (S, ds) to another (T, dT). Then f is continuous on S if and only if, for every closed set Y in T, the inverse image f -1(Y) is closed in S. Proof If Y is closed in T, then T - Y is open in T and f-1(T - Y) = S - f-1(Y). Now apply Theorem 4.23. Examples. The image of an open set under a continuous mapping is not necessarily open. A simple counterexample is a constant function which maps all of S onto a single point in R'. Similarly, the image of a closed set under a continuous mapping need not be closed. For example, the real-valued function f(x) = arctan x maps R1 onto the open interval (-ir/2, it/2). 4.13 FUNCTIONS CONTINUOUS ON COMPACT SETS The next theorem shows that the continuous image of a compact set is compact. This is another global property of continuous functions. Theorem 4.25. Let f : S --> T be a function from one metric space (S, ds) to another (T, dT). If f is continuous on a compact subset X of S, then the image f(X) is a compact subset of T; in particular, f(X) is closed and bounded in T. Th. 4.29 Functions Continuous on Compact Sets 83 Proof. Let F be an open covering off(X), so thatf(X) c UAEF A. We will show that a finite number of the sets A cover f(X). Since f is continuous on the metric subspace (X, ds) we can apply Theorem 4.23 to conclude that each set f -1(A) is open in (X, ds). The sets f -1(A) form an open covering of X and, since X is compact, a finite number of them cover X, say X c f -1(A,) v u f -1(Ap). Hence f(X) 9 fU '(A,) u ... u f-'(Ap)] = f[f '(A,)] u ... u f[f-1(Ap)] 9A1u...uAp, so f(X) is compact. As a corollary of Theorem 3.38, we see thatf(X) is closed and bounded. Definition 4.26. A function f : S -> R"is called bounded on S if there is a positive number M such that 11 f(x) II < M for all x in S. Since f is bounded on S if and only if f(S) is a bounded subset of R, we have the following corollary of Theorem 4.25. Theorem 4.27. Let f : S -+ R" be a function from a metric space S to Euclidean space R. If f is continuous on a compact subset X of S, then f is bounded on X. This theorem has important implications for real-valued functions. If f is real-valued and bounded on X, then f(X) is a bounded subset of R, so it has a supremum, sup f(X), and an infimum, inff(X). Moreover, inff(X) < f(x) < sup f(X) for every x in X. The next theorem shows that a continuous f actually takes on the values sup f(X) and inff(X) if X is compact. Theorem 4.28. Let f : S -+ R be a real-valued function from a metric space S to Euclidean space R. Assume that f is continuous on a compact subset X of S. Then there exist points p and q in X such that f(p) = inf f(X) and f(q) = sup f(X). NOTE. Since f(p) < f(x) < f(q) for all x in X, the numbers f(p) and f(q) are called, respectively, the absolute or global minimum and maximum values of f on X. Proof. Theorem 4.25 shows that f(X) is a closed and bounded subset of R. Let m = inff(X). Then m is adherent to f(X) and, since f(X) is closed, m e f(X). Therefore m = f(p) for some p in X. Similarly, f(q) = sup f(X) for some q in X. Theorem 4.29. Let f : S - T be a function from one metric space (S, ds) to another (T, dT). Assume that f is one-to-one on S, so that the inverse function f -1 exists. If S is compact and if f is continuous on S; then f -1 is continuous on f(S). Proof. By Theorem 4.24 (applied to f -1) we need only show that for every closed set X in S the image f(X) is closed in T. (Note that f(X) is the inverse image of 84 Limits and Continuity Def. 4.30 X under f -1.) Since X is closed and S is compact, X is compact (by Theorem 3.39), sof(X) is compact (by Theorem 4.25) and hencef(X) is closed (by Theorem 3.38). This completes the proof. Example. This example shows that compactness of S is an essential part of Theorem 4.29. Let S = [0, 1) with the usual metric of RI and consider the complex-valued function f defined by f(x) = e2nrx for 0 < x < 1. This is a one-to-one continuous mapping of the half-open interval [0, 1) onto the unit circle Iz I = I in the complex plane. However, f -1 is not continuous at the point f (O). For example, if x = 1 - 1 In, the sequence {f(x) } converges to f (O) but [x.) does not converge in S. 4.14 TOPOLOGICAL MAPPINGS (HOMEOMORPHISMS) Definition 4.30. Let f : S --+ T be a function from one metric space (S, ds) to another (T, dT). Assume also that f is one-to-one on S, so that the inverse function f -1 exists. If f is continuous on S and if f -1 is continuous on f(S), then f is called a topological mapping or a homeomorphism, and the metric spaces (S, ds) and (f(S), dT) are said to be homeomorphic. If f is a homeomorphism, then so is f -1. Theorem 4.23 shows that a homeo-morphism maps open subsets of S onto open subsets off(S). It also maps closed subsets of S onto closed subsets off(S). A property of a set which remains invariant under every topological mapping is called a topological property. Thus the properties of being open, closed, or compact are topological properties. An important example of a homeomorphism is an isometry. This is a function f : S --+ T which is one-to-one on S and which preserves the metric; that is, dT(f(x), f(y)) = ds(x, y) for all points x and y in S. If there is an isometry from (S, ds) to (f(S), dT) the two metric spaces are called isometric. Topological mappings are particularly important in the theory of space curves. For example, a simple arc is the topological image of an interval, and a simple closed curve is the topological image of a circle. 4.15 BOLZANO'S THEOREM This section is devoted to a famous theorem of Bolzano which concerns a global property of real-valued functions continuous on compact intervals [a, b] in R. If the graph of f lies above the x-axis at a and below the x-axis at b, Bolzano's theorem asserts that the graph must cross the axis somewhere in between. Our proof will be- based on a local property of continuous functions known as the sign preserving property. Th. 4.33 Bolzano's Theorem 85 Theorem 4.31. Let f be defined on an interval S in R. Assume that f is continuous at a point c in S and that f(c) 0. Then there is a 1-ball B(c; 6) such that f(x) has the same sign as f(c) in B(c; S) n S. Proof. Assume f(c) > 0. For every e > 0 there is a S > 0 such that f(c) - e < f(x) < f(c) + e whenever x e B(c; S) n S. Take the S corresponding to e = f(c)/2 (this E is positive). Then we have jf(c) < f(x) < 2f(c) whenever x e B(c; 6) n S, so f(x) has the same sign as f(c) in B(c; 6) r S. The proof is similar if f(c) < 0, except that we take e = - j f(c). Theorem 432 (Bolzano). Let f be real-valued and continuous on a compact interval [a, b] in R, and suppose that f(a) and f(b) have opposite signs; that is, assume f(a)f(b) < 0. Then there is at least one point c in the open interval (a, b) such that f(c) = 0. Proof. For definiteness, assume f(a) > 0 and f(b) < 0. Let A = {x : x e [a, b] and f(x) >- 0}. Then A is nonempty since a e A, and A is bounded above by b. Let c = sup A. Then a < c < b. We will prove that f(c) = 0. If f(c) # 0, there is a 1-ball B(c; 6) in which f has the same sign as f(c). If f(c) > 0, there are points x > c at which f(x) > 0, contradicting the definition of c. If f(c) < 0, then c - 8/2 is an upper bound for A, again contradicting the definition of c. Therefore we must have f(c) = 0. From Bolzano's theorem we can easily deduce the intermediate value theorem for continuous functions. Theorem 4.33. Assume f is real-valued and continuous on a compact interval S in R. Suppose there are two points or < /3 in S such that f(a) # f(8). Then f takes every value between f(a) and f(/3) in the interval (a, /3). Proof. Let k be a number between f(a) and f(8) and apply Bolzano's theorem to the function g defined on [a, /3] by the equation g(x) = f(x) - k. The intermediate value theorem, together with Theorem 4.28, implies that the continuous image of a compact interval S under a real-valued function is another compact interval, namely, [inf f(S), sup f(S)]. (If f is constant-on S, this will be a degenerate interval.) The next section extends this property to the more general setting of metric spaces. 86 4.16 CONNECTEDNESS Limits and Continuity Def. 4.34 This section describes the concept of connectedness and its relation to continuity. Definition 4.34. A metric space S is called disconnected if S = A U B, where A and B are disjoint nonempty open sets in S. We call S connected if it is not dis-connected. NOTE. A subset X of a metric space S is called connected if, when regarded as a metric subspace of S, it is a connected metric space. Examples 1. The metric space S = R - {0} with the usual Euclidean metric is disconnected, since it is the union of two disjoint nonempty open sets, the positive real numbers and the negative real numbers. 2. Every open interval in R is connected. This was proved in Section 3.4 as a conse-quence of Theorem 3.11. 3. The set Q of rational numbers, regarded as a metric subspace of Euclidean space R', is disconnected. In fact, Q = A u B, where A consists of all rational numbers < and B of all rational numbers > -d Similarly, every ball in Q is disconnected. 4. Every metric space S contains nonempty connected subsets. In fact, for each p in S the set {p} is connected. To relate connectedness with continuity we introduce the concept of a two-valued function. Definition 4.35. A real-valued function f which is continuous on a metric space S is said to be two-valued on S if f(S) c {0, 1}. In other words, a two-valued function is a continuous function whose only possible values are 0 and 1. This can be regarded as a continuous function from S to the metric space T = 10, 1}, where T has the discrete metric. We recall that every subset of a discrete metric space T is both open and closed in T. Theorem 4.36 A metric space S is connected if, and only if, every two-valued function on S is constant. Proof. Assume S is connected and let f be a two-valued function on S. We must show that f is constant. Let A = f -1({0}) and B = f -1({1}) be the inverse images of the subsets {0} and {1}. Since {0) and (1) are open subsets of the discrete metric space {0, 1 }, both A and B are open in S. Hence, S = A U B, where A and B are disjoint open sets. But since S is connected, either A is empty and B = S, or else B is empty and A = S. In either case, f is constant on S. Conversely, assume that S is disconnected, so that S = A u B, where A and B are disjoint nonempty open subsets of S. We will exhibit a two-valued function on S which is not constant. Let f(x) _ 10 ifxeA, 1 ifxeB. Th. 4.39 Components of a Metric Space 87 Since A and B are nonempty, f takes both values 0 and 1, so f is not constant. Also, f is continuous on S because the inverse image ofevery open subset of {0, 1} is open in S. Next we show that the continuous image of a connected set is connected. Theorem 4.37. Let f : S -+ M be a function from a metric space S to another metric space M. Let X be a connected subset of S. If f is continuous on X, then f(X) is a connected subset of M. Proof. Let g be a two-valued function on f(X). We will show that g is constant. Consider the composite function h defined on X by the equation h(x) = g(f(x)). Then h is continuous on X and can only take the values 0 and 1, so h is two-valued on X. Since X is connected, h is constant on X and this implies that g is constant on f(X). Therefore f(X) is connected. Example. Since an interval X in RI is connected, every continuous image f(X) is con-nected. If f has real values, the image f (X) is another interval. If f has values in R", the image f(X) is called a curve in W. Thus, every curve in R" is connected. As a corollary of Theorem 4.37 we have the following extension of Bolzano's theorem. Theorem 4.38 (Intermediate-value theorem for real continuous functions). Let f be real-valued and continuous on a connected subset S of R". If f takes on two different values in S, say a and b, then for each real c between a and b there exists a point x in S such that f(x) = c. Proof. The image f(S) is a connected subset of R1. Hence, f(S) is an interval containing a and b (see Exercise 4.38). If some value c between a and b were not in f(S), then f(S) would be disconnected. 4.17 COMPONENTS OF A METRIC SPACE This section shows that every metric space S can be expressed in a unique way as a union of connected "pieces" called components. First we prove the following: Theorem 4.39. Let F be a collection of connected subsets of a metric space S such that the intersection T = nAEF A is not empty. Then the union U = UAEF A is connected. Proof. Since T # 0, there is some t in T. Let f be a two-valued function on U. We will show that f is constant on U by showing that f(x) = f(t) for all x in U. If x e U, then x e A for some A in F. Since A is connected, f is constant on A and, since t e A, f(x) = f(t). Every point x in a metric space S belongs to at least one connected subset of S, namely {x}. By Theorem 4.39, the union of all the connected subsets which contain x is also, connected. We call this union a component of S, and we denote it by U(x). Thus, U(x) is the maximal connected subset of S which contains x. 88 Limits and Continuity Th. 4.40 Theorem 4.40. Every point of a metric space S belongs to a uniquely determined component of S. In other words, the components of S form a collection of disjoint sets whose union is S. Proof. Two distinct components cannot contain a point x; otherwise (by Theorem 4.39) their union would be a larger connected set containing x. 4.18 ARCWISE CONNECTEDNESS This section describes a special property, called arcwise connectedness, which is possessed by some (but not all) connected sets in Euclidean space R". Definition 4.41. A set S in R" is called arcwise connected if for any two points a and b in S there is a continuous function f : [0, 1] -> S such that f(0) = a and f(1) = b. NOTE. Such a function is called a path from a to b. If f(0) # f(l), the image of [0, 1] under f is called an arc joining a and b. Thus, S is arcwise connected if every pair of distinct points in S can be joined by an arc lying in S. Arcwise connected sets are also called pathwise connected. If f(t) = tb + (1 - t)a for 0 < t < 1, the curve joining a and b is called a line segment. Examples 1. Every convex set in R" is arcwise connected, since the line segment joining two points of such a set lies in the set. In particular, every n-ball is arcwise connected. 2. The set in Fig. 4.4 (a union of two tangent closed disks) is arcwise connected. Figure 4.4 3. The set in Fig. 4.5 consists of those points on the curve described by y = sin (1/x), 0 < x.:5 1, along with the points on the horizontal segment -1 < x < 0. This set is connected but not arcwise connected (Exercise 4.46). 1 Figure 4.5 Def. 4.45 Arcwise Connectedness The next theorem relates arcwise connectedness with connectedness. 89 Theorem 4.42. Every arcwise connected set S in R" is connected. Proof. Let g be two-valued on S. We will prove that g is constant on S. Choose a point a in S. If x e S, join a to x by an arc r lying in S. Since t is connected, g is constant on t so g(x) = g(a). But since x is an arbitrary point of S, this shows that g is constant on S, so S is connected. We have already noted that there are connected sets which are not arcwise connected. However, the concepts are equivalent for open sets. Theorem 4.43. Every open connected set in R" is arcwise connected. Proof. Let S be an open connected set in R" and assume x e S. We will show that x can be joined to every point y in S by an arc lying in S. Let A denote that subset of S which can be so joined to x, and let B = S - A. Then S = A u B, where A and B are disjoint. We will show that A and B are both open in R". Assume that a e A and join a to x by an arc, say r, lying in S. Since a e S and S is open, there is an n-ball B(a)s S. Every y in B(a) can be joined to a by a line segment (in S) and thence to x by r. Thus y e A if y e B(a). That is, B(a) c A, and hence A is open. To see that B is also open, assume that b e B. Then there is an n-ball B(b) S-= S, since S is open. But if a point y in B(b) could be joined to x by an arc, say t,, lying in S, the point b itself could also be so joined by first joining b to y (by a line segment in B(b)) and then using t'. But since b 0 A, no point of B(b) can be in A. That is, B(b) s B, so B is open. Therefore we have a decomposition S = A u B, where A and B are disjoint open sets in R". Moreover, A is not empty since x e A. Since S is connected, it follows that B must be empty, so S = A. Now A is clearly arcwise connected, because any two of its points can be suitably joined by first joining each of them to x. Therefore, S is arcwise connected and the proof is complete. NOTE. A path f : [0, 1] - S is said to be polygonal if the image of [0, 1] under f is the union of a finite number of line segments. The same argument used to prove Theorem 4.43 also shows that every open connected set in R" is polygonally con-nected. That is, every pair of points in the set can be joined by a polygonal arc lying in the set. Theorem 4.44. Every open set S in R" can be expressed in one and only one way as a countable disjoint union of open connected sets. Proof. By Theorem 4.40, the components of S form a collection of disjoint sets whose union is S. Each component T of S is open, because if x e T then there is an n-ball B(x) contained in S. Since B(x) is connected, B(x) E- T, so T is open. By the Lindelof theorem (Theorem 3.28), the components of S form a countable collection, and by Theorem 4.40 the decomposition into components is unique. Definition 4.45. A set in R" is called a region if it is the union of an open connected set with some, none, or all its boundary points. If none of the boundary points are 90 Limits and continuity Def. 4.46 included, the region is called an open region. If all the boundary points are included, the region is called a closed region. NOTE. Some authors use the term domain instead of open region, especially in the complex plane. 4.19 UNIFORM CONTINUITY Suppose f is defined on a metric space (S, ds), with values in another metric space (T, dT), and assume that f is continuous on a subset A of S. Then, given any point p in A and any e > 0, there is a 6 > 0 (depending on p and on s) such that, if x e A, then dT(f(x), f(p)) < e whenever ds(x, p) < 6. In general we cannot expect that for a fixed a the same value of 6 will serve equally well for every point p in A. This might happen, however. When it does, the function is called uniformly continuous on A. Definition 4.46. Let f : S -p T be a function from one metric space (S, ds) to another (T, dT). Then f is said to be uniformly continuous on a subset A of S if the following condition holds: For every s > 0 there exists a 6 > 0 (depending only on a) such that if x e A and peAthen dT(f(x), f(p)) < s whenever ds(x, p) < 6. (6) To emphasize the difference between continuity on A and uniform continuity on A we consider the following examples of real-valued functions. Examples 1. Let f(x) = 1/x for x > 0 and take A = (0, 1]. This function is continuous on A but not uniformly continuous on A. To prove this, let e = 10, and suppose we could find a 6, 0 < 6 < 1, to satisfy the condition of the definition. Taking x = S, p = 6/11, we obtain Ix - pI < 6 and 1f(X) - f(p)I = 11 - 8 = 8 > 10. Hence, for these two points we would always have If(x) - f(p)I > 10, contradicting the definition of uniform continuity. 2. Let f(x) = x2 if x e RI and take A = (0, 1 ] as above. This function is uniformly continuous on A. To prove this, observe that Jf(x) - f(p)l = 1x2 - p21 = I(x - p)(x + p)I < 21x - pl. If Ix - p1 < 6, then 1f(x) - f(p)l < 28. Hence, if a is given, we need only take 8 = e/2 to-guarantee that If(x) - f(p)I < e for every pair x, p with Ix - pl < 6. This shows that f is uniformly continuous on A. Th. 4.47 Uniform Continuity and Compact Sets .91 An instructive exercise is to show that the function in Example 2 is not uni-formly continuous on R'. 4.20 UNIFORM CONTINUITY AND COMPACT SETS Uniform continuity on a set A implies continuity on A. (The reader should verify this.) The converse is also true if A is compact. Theorem 4.47 (Heine). Let f : S -+ T be a function from one metric space (S, ds) to another (T, dT). Let A be a compact subset of S and assume that f is continuous on A. Then f is uniformly continuous on A. Proof. Let s > 0 be given. Then each point a in A has associated with it a ball Bs(a; r), with r depending on a, such that dT(f(x), f(a)) < 2 whenever x e Bs(a; r) n A. Consider the collection of balls Bs(a; r/2) each with radius r/2. These cover A and, since A is compact, a finite number of them also cover A, say A C U Bsak; rk/ . k=1 2 In any ball of twice the radius, B(ak; rk), we have dT(f(x), f(ak)) < 2 whenever x e Bs(ak; rk) n A. Let S be the smallest of the numbers r1/2, ... , rm/2. We shall show that this S works in the definition of uniform continuity. For this purpose, consider two points of A, say x and p with ds(x, p) < S. By the above discussion there is some ball Bs(ak; rk/2) containing x, so dT(f(x), f(ak)) < 2 By the triangle inequality we have ds(p, ak) < ds(p, x) + ds(x, ak) < S + 2 < r2 + z = rk. Hence, p e Bs(ak; rk) n S, so we also have dT(f(p), f(ak)) < e/2. Using the triangle inequality once more we find dT(f(x),1(p)) < dT(.f(x), f(ak)) + dT(f(ak),1'(r)) < 2 + 2 = E. This completes the proof. 92 Limits and Continuity Th. 4.48 4.21 FIXED-POINT THEOREM FOR CONTRACTIONS Let f : S -- S be a function from a metric space (S, d) into itself. A point p in S is called a fixed point of f if f(p) = p. The function f is called a contraction of S if there is a positive number a < I (called a contraction constant), such that d(f(x), f(y)) < ad(x, y) for all x, y in S. (7) Clearly, a contraction of any metric space is uniformly continuous on S. Theorem 4.48 (Fixed-point theorem). A contraction f of a complete metric space S has a unique fixed point p. Proof. If p and p' are two fixed points, (7) implies d(p, p') < ad(p, p'), so so d(p, p') = 0 and p = p'. Hence f has at most one fixed point. To prove it has one, take any point x in S and consider the sequence of iterates: x, f(x), f (f(x)), ... That is, define a sequence {p"} inductively as follows: Po = X, Pn+ 1 = f(Pn), n = 0, 1, 2, .. . We will prove that {p"} converges to a fixed point off. First we show that {pn} is a Cauchy sequence. From (7) we have d(Pn+1,Pn) = d(.f(Pn),f(Pn-1)) < ad(Pn,Pn-1), so, by induction, we find d(Pn+ 1, Pn) < a" d(P1, PO) = Can, where c = d(p1, po). Using the triangle inequality we find, for m > n, m-1 m-1 d(Pr,pn)< d(Pk + 1, Pk) < CEak=Can - am< c a". k=n k=n 1- a 1- a Since a" - 0 as n -+ oo, this inequality shows that {p"} is a Cauchy sequence. But S is complete so there is a point p in S such that pn -+ p. By continuity of f, f(p)=f(n-+limao p.)= nlimf(pn)= limPn+1 =P, n-, ao so p is a fixed point off. This completes the proof. Many important existence theorems in analysis are easy consequences of the fixed point theorem. Examples are given in Exercises 7.36 and 7.37. Reference 4.4 gives applications to numerical analysis. 4.22 DISCONTINUITIES OF REAL-VALUED FUNCTIONS The rest of this chapter is devoted to special properties of real-valued functions defined on subintervals of R. Def. 4.49 Discontinuities of Real-Valued Functions 93 Let f be defined on an interval (a, b). Assume c e [a, b). If f(x) -- A as x -+ c through values greater than c, we say that A is the righthand limit off at c and we indicate this by writing lim f(x) = A. x--c+ The righthand limit A is also denoted byf(c+). In the s, 6 terminology this means that for every e > 0 there is a 6 > 0 such that t f(x) - f(c+)I < e whenever c < x < c + 6 < b. Note that f need not be defined at the point c itself. If f is defined at c and if f(c+) = f(c), we say that f is continuous from the right at c. Lefthand limits and continuity from the left at c are similarly defined if c e (a, b]. If a < c < b, then f is continuous at c if, and only if, f(c) = .f(c+) = f(c-). We say c is a discontinuity of f if f is not continuous at c. In this case one of the following conditions is satisfied: a) Eitherf(c+) orf(c-) does not exist. b) Bothf(c+) andf(c-) exist but have different values. c) Both f(c +) and f(c -) exist and f(c +) = f(c -) 0 f(c). In case (c), the point c is called a removable discontinuity, since the discontinuity could be removed by redefining f at c to have the valuef(c+) = f(c-). In cases (a) and (b), we call c an irremovable discontinuity because the discontinuity cannot be removed by redefining f at c. Definition 4.49. Let f be defined on a closed interval [a, b]. If f(c+) and f(c-) both exist at some interior point c, then: a) f(c) - f(c-) is called the lefthand jump off at c, b) f(c+) - f(c) is called the righthand jump off at c, c) f(c +) - f(c -) is called the jump off at c. If any one of these three numbers is different from 0, then c is called a jump dis-continuity off. For the endpoints a and b, only one-sided jumps are considered, the righthand - ). jump at a, f (a +) - f (a), and the lefthand jump at b, f (b) - f (b Examples 1. The function f defined by f(x) = x/lxl if x ;4 0, f(0) = A, has a jump discontinuity at 0, regardless of the value of A. Here f (0+) = 1 and f (0 -) = - 1. (See Fig. 4.6.) 2. The function f defined by f(x) = 1 if x -A 0, f(0) = 0, has a removable jump dis-continuity at 0. In this case 1(0+) = f(0-) = 1. 94 Limits and Continuity Def. 4.50 Figure 4.6 Figure 4.7 3. The function f' defined by f(x) = I/x if x t- 0, f(0) = A, has an irremovable dis-continuity at 0. In this case neither f(0+) nor f(0-) exists. (See Fig. 4.7.) 4. The function f defined byf(x) = sin (1/x) if x -,6 0, f(0) = A, has an irremovable dis-continuity at 0 since neither f(0+) nor f(0-) exists. (See Fig. 4.8.) 5. The function f defined by f(x) = x sin (1/x) if x t- 0, f(0) = 1, has a removable jump discontinuity at 0, since f(0+) = f(0-) = 0. (See Fig. 4.9.) Figure 4.8 4.23 MONOTONIC FUNCTIONS Figure 4.9 Definition 4.50. Let f be a real-valued function defined on a subset S of R. Then f is said to be increasing (or nondecreasing) on S if for every pair of points x and y in S, x < y implies f(x) < f(y). If x < y impliesf(x) < f(y), then f is said to be strictly increasing on S. (Decreasing functions are similarly defined.) A function is called monotonic on S if it is increasing on S or decreasing on S. If f is an increasing function, then -f is a decreasing function. Because of this simple fact, in many situations involving monotonic functions it suffices to consider only the case of increasing functions. Th. 4.53 Monotonic Functions 95 We shall prove that functions which are monotonic on compact intervals always have finite right- and lefthand limits. Hence their discontinuities (if any) must be jump discontinuities. Theorem 4.51. If f is increasing on [a, b], then f(c+) and f(c-) both exist for each c in (a, b) and we have .f(c-) :9 f(c) < f(c +). At the endpoints we have f(a) < f(a +) and f(b -) < f(b). Proof. Let A = {f(x) : a < x < c}. Since f is increasing, this set is bounded above by f(c). Let a = sup A. Then a < f(c) and we shall prove that f(c-) exists and equals a. To do this we must show that for every e > 0 there is a 6 > 0 such that c - b < x < c implies I f(x) - al < e. But since a = sup A, there is an element f(xl) of A such that a - e < f(x,) < a. Since f is increasing, for every x in (x,, c) we also have a - e < f(x) < a, and hence If(x) - al < e. Therefore the number 6 = c - x, has the required property. (The proof that f(c+) exists and is >- f(c) is similar, and only trivial modifications are needed for the endpoints.) There is, of course, a corresponding theorem for decreasing functions which the reader can formulate for himself. Theorem 4.52. Let f be strictly increasing on a set S in R. Then f exists and is strictly increasing on f(S). Proof. Since f is strictly increasing it is one-to-one on S, so f -' exists. To see that f -' is strictly increasing, let y, < y2 be two points inf(S) and let x, = f -'(y,), x2 = f '(y2). We cannot have x, >- x2i for then we would also have y, >- y2. The only alternative is x, < x2, and this means that f -' is strictly increasing. Theorem 4.52, together with Theorem 4.29, now gives us: Theorem 4.53. Let f be strictly increasing and continuous on a compact interval [a, b]. Then f -' is continuous and strictly increasing on the interval [f(a), f(b)]. NOTE. Theorem 4.53 tells us that a continuous, strictly increasing function is a topological mapping. Conversely, every topological mapping of an interval [a, b] onto an interval [c, d] must be a strictly monotonic function. The verification of this fact will be an instructive exercise for the reader (Exercise 4.62). 96 EXERCISES Limits and Continuity Limits of sequences 4.1 Prove each of the following statements about sequences in C. a) z" -+ 0 if Iz I < 1; {z") diverges if 1z > 1. b) If z" -+ 0 and if {cn} is bounded, then {cnzn} -- 0. c) z"/n! --+ 0 for every complex z. d) If an=N/n2+2- n, then a, 0. 4.2 If an+2 = (an+1 + an)/2 for all n 1, show that an -+ (a1 + 2a2)/3. Hint. an+2 -an+1 = Yan - an+,) 4.3 If 0 < x1 < 1 and if xn+1 = I - 1 - Xn for all n >- 1, prove that {xn} is a decreasing sequence with limit 0. Prove also that xn+ 1 /xn -+ I-4.4 Two sequences of positive integers {an} and {bn} are defined recursively by taking a1 = b1 = I and equating rational and irrational parts in the equation an + bnV 2 = (an-1 + b.-,N/2)2 for n - 2. Prove that a2 - 2b2, = I for n >- 2. Deduce that an/bn -+ V2 through values > V2, and that 2bn/Qn -+ V2 through values < ,J2. 4.5 A real sequence {xn} satisfies 7xn+1 = xn + 6 for n 1. If x1 = }, prove that the sequence increases and find its limit. What happens if x1 = or if x1 = I? 2 2 +1 - and for all n >- 1, prove that {an} 4.6 If Ianl < 2 and Ian+2 - an+1l < BIan converges. 4.7 In a metric space (S, d), assume that xn -+ x and yn - y. Prove that d(xn, Yn) -d (x, y). 4.8 Prove that in a compact metric space (S, d), every sequence in S has a subsequence which converges in S. This property also implies that S is compact but you are not re-quired to prove this. (For a proof see either Reference 4.2 or. 4.3.) 4.9 Let A be a subset of a metric space S. If A is complete, prove that A is closed. Prove that the converse also holds if S is complete. Limits of functions NOTE. In Exercises 4.10 through 4.28, all functions are real-valued. 4.10 Let f be defined on an open interval (a, b) and assume x e (a, b). Consider the two statements a)limlf(x+h)-f(x)I=0; b) Jimlf(x+h)-f(x-h)I=0. h-+0 h-.0 Prove that (a) always implies (b), and give an example in which (b) holds but (a) does not. 4.11 Let f be defined on R2. If lim f(x, y) = L (x,y)-.(a,b) and if the one-dimensional limits f(x, y) and limy..b f(x, y) both exist, prove that lim [limf(x, y)] = lim [limf(x, y)] = L. x-.a y-.b y-.b x-a Exercises Now consider the functions f defined on R2 as follows: xy a) f(x, y) = x2 + y2 if (x, Y) t- (0, 0), f(0, 0) = 0. b) Ax, y) = 2 (xy)2 if (x, Y) ;4 (0, 0), f(0, 0) = 0. (xy) + (x - Y) c) f(x, y) = 1 sin (xy) X if x # 0, AO' y) = Y. d) f (x, y) = 1(x + y) sin (11x) sin (11Y) if x 0 and y # 0, 0 if x = Dory = 0. sin x - sin y e) f(x, y) = tan x - tan y cos3 x if tan x ?6 tan y, if tan x = tan Y. 97 In each of the preceding examples, determine whether the following limits exist and evaluate those limits that do exist: lim [lim f(x, y)] ; lim [lim f(x, y)] ; lim AX, y). x-'0 Y- O y-+0 x-.O (x,y)- (O.O) 4.12 If x c [0, 1 ] prove that the following limit exists, lira [lira cos2n (m! irx)] , M-00 n-ao and that its value is 0 or 1, according to whether x is irrational or rational. Continuity of real-valued functions 4.13 Let f be continuous on [a, b] and let f(x) = 0 when x is rational. Prove that f(x) = 0 for every x in [a, b]. 4.14 Let f be continuous at the point a = (a1, a2i ..., an) in R". Keep a2, a3, ... , an fixed and define a new function g of one real variable by the equation g(x) = f(x, a2, ... , an). Prove that g is continuous at the point x = al. (This is sometimes stated as follows: A continuous function of n variables is continuous in each variable separately.) 4.15 Show by an example that the converse of the statement in Exercise 4.14 is not true in general. 4.16 Let f, g, and h be defined on [0, 11 as follows: f(x) = g(x) = h(x) = 0, whenever x is irrational; f (x) = 1 and g(x) = x, whenever x is rational; h(x) = 1/n, if x is the rational number m/n (in lowest terms); h(0) = 1. Prove that f is not continuous anywhere in [0, 1 ], that g is continuous only at x = 0, and that h is continuous only at the irrational points in [0, 1 ]. 4.17 For each x in [0, 1 ], let f(x) = x if x is rational, and let f(x) = I - x if x is irrational. Prove that: a) f(f(x)) = x for all x in [0, 1 ]. b) f(x) + Al - x) = I for all x in [0, 1 ]. 98 Limits and Continuity c) f is continuous only at the point x = . d) f assumes every value between 0 and 1. e) f(x + y) - f(x) - f(y) is rational for all x and y in [0, 1 ]. 4.18 Let f be defined on R and assume that there exists at least one point x0 in Rat wh f is continuous. Suppose also that, for every x and y in R, f satisfies the equation .f(x + y) = f(x) + f(y). Prove that there exists a constant a such that f (x) = ax for all x. 4.19 Let f be continuous on [a, b] and define g as follows : g(a) = f (a) and, for a < x < let g(x) be the maximum value off in the subinterval [a, x ]. Show that g is continuous [a,b]. 4.20 Let fl, ... , be m real-valued functions defined on a set Sin R". Assume that e, fk is continuous at the point a of S. Define a new function f as follows: For each x in f(x) is the largest of them numbers f,(x), ... , fm(x). Discuss the continuity off at a. 4.21 Let f : S - R be continuous on an open set Sin R", assume that p e S, and assu that f(p) > 0. Prove that there is an n-ball B(p; r) such that f(x) > 0 for every x in ball. 4.22 Let f be defined and continuous on a closed set S in R. Let A = {x : x e S and f(x) = 0 }. Prove that A is a closed subset of R. 4.23 Given a function f : R -+ R, define two sets A and B in RZ as follows: A = {(x, y) : y < f(x) }, B = {(x, y) : y > fix)). Prove that f is continuous on R if, and only if, both A and B are open subsets of R2. 4.24 Let f be defined and bounded on a compact interval S in R. If T c S, the num nf(T) = sup {f(x) - fly) : x e T, y e T} is called the oscillation (or span) off on T. If x e S, the oscillation offat x is defines be the number cvf(x) = lim S2 f(B(x ; h) n S). h-+0+ Prove that this limit always exists and that cof(x) = 0 if, and only if, f is continuous a 4.25 Let f be continuous on a compact interval [a, b]. Suppose that f has a local rr imum at x, and a local maximum at x2. Show that there must be a third point betw x, and x2 where f has a local minimum. NOTE. To say that f has a local maximum at x, means that there is a 1-ball B(x,) s that f(x) 5 f(x,) for all x in B(x,) n [a, b]. Local minimum is similarly defined. 4.26 Let f be a real-valued function, continuous on [0, 1 ], with the following prope For every real y, either there is no x in [0, 1 ] for which f(x) = y or there is exactly such x. Prove that f is strictly monotonic on [0, 1 ]. 4.27 Let f be a function defined on [0, 1 ] with the following property: For every number y, either there is no x in [0, 1 ] for which f (x) = y or there are exactly two va of x in [0, 1 ] for which f(x) = y. Exercises 99 a) Prove that f cannot be continuous on [0, 1 ]. b) Construct a function f which has the above property. c) Prove that any function with this property has infinitely many discontinuities on 10111. 4.28 In each case, give an example of a function f, continuous on f(S) = T, or else explain why there can be no such f: a) S = (0, 1), b) S = (0, 1), c)S=R', d) S = [0, 1 ] v [2, 3 ], e) S = [0, 1 ] x [0, 1 ], f) S = [0, 1 ] x [0, 1 ], g) S = (0, 1) x (0, 1), T= (0, 1]. T (0, 1) u (1, 2). T = the set of rational numbers. T = {0, 1 }. T = R2. T = (0, 1) x (0, 1). T R2. S and such that Continuity in metric spaces In Exercises 4.29 through 4.33, we assume that f : S -T is a function from one metric space (S, ds) to another (T, dT). 4.29 Prove that f is continuous on S if, and only if, f -'(int B) S int f -'(B) for every subset B of T. 4.30 Prove that f is continuous on S if, and only if, f(A) E- 7) for every subset A of S. 4.31 Prove that f is continuous on S if, and only if, f is continuous on every compact subset of S. Hint. If x - p in S, the set {p, x1, x2, ... } is compact. 4.32 A function f : S T is called a closed mapping on S if the image f(A) is closed in T for every closed subset A of S. Prove that f is continuous and closed on S if, and only if, f(A) = f(A) for every subset A of S. 4.33 Give an example of a continuous f and a Cauchy sequence in some metric space S for which is not a Cauchy sequence in T. 4.34 Prove that the interval (- 1, 1) in R' is homeomorphic to R'. This shows that neither boundedness nor completeness is a topological property. 4.35 Section 9.7 contains an example of a function f, continuous on [0, 1 ], with f ([0, 1 ]) = [0, 1 ] x [0, 1 ]. Prove that no such f can be one-to-one on [0, 1 ]. Connectedness 4.36 Prove that a metric space S is disconnected if, and only if, there is a nonempty subset A of S, A 0 S, which is both open and closed in S. 4.37 Prove that a metric space S is connected if, and only if, the only subsets of S which are both open and closed in S are the empty set and S itself. 4.38 Prove that the only connected subsets of R are (a) the empty set, (b) sets consisting of a single point, and (c) intervals (open, closed, half-open, or infinite). 100 Limits and Continuity 4.39 Let X be a connected subset of a metric space S. Let Y be a subset of S such that X s Y c X, where X is the closure of X. Prove that Y is also connected. In particular, this shows that X is connected. 4.40 If x is a point in a metric space S, let U(x) be the component of S containing x. Prove that U(x) is closed in S. 4.41 Let S be an open subset of R. By Theorem 3.11, S is the union of a countable dis-joint collection of open intervals in R. Prove that each of these open intervals is a com-ponent of the metric subspace S. Explain why this does not contradict Exercise 4.40. 4.42 Given a compact set S in R"' with the following property: For every pair of points a and b in S and for every e > 0 there exists a finite set of points {x0, x1, ... , x"} in S with xo = a and x,, = b such that IJxk - xk_111 < fork = 1, 2,. .., n. Prove or disprove: S is connected. 4.43 Prove that a metric space S is connected if, and only if, every nonempty proper subset of S has a nonempty boundary. 4.44 Prove that every convex subset of R" is connected. 4.45 Given a function f : R" -- R' which is one-to-one and continuous on W. If A is open and disconnected in R", prove that f(A) is open and disconnected in f(R"). 4.46 Let A = {(x, y) : 0 < x < 1, y = sin 1/x), B = {(x, y) : y = 0, - 1 < x < 0}, and let S = A v B. Prove that S is connected but not arcwise connected. (See Fig. 4.5, Section 4.18.) 4.47 Let F = {F1, F2,... } be a countable collection of connected compact sets in R" such that Fk+1 Fk for each k >- 1. Prove that the intersection n, 1 Fk is connected and closed. 4.48 Let S be an open connected set in R". Let T be a component of R" - S. Prove that R" - T is connected. 4.49 Let (S, d) be a connected metric space which is not bounded. Prove that for every a in S and every r > 0, the set {x : d(x, a) = r ) is nonempty. Uniform continuity 4.50 Prove that a function which is uniformly continuous on S is also continuous on S. 4.51 If f(x) = x2 for x in R, prove that I. is not uniformly continuous on R. 4.52 Assume that f is uniformly continuous on a bounded set S in R". Prove that f must be bounded on S. 4.53 Let f be a function defined on a set S in R" and assume that f(S) S R'". Let g be defined on f(S) with value in Rk, and let h denote the composite function defined by h(x) = g[f(x)] if x e S. If f is uniformly continuous on S and if g is uniformly continuous on f(S), show that h is uniformly continuous on S. 4.54 Assume f : S -+ T is uniformly continuous on S, where S and T are metric spaces. If {x"} is any Cauchy sequence in S, prove that {f(x")} is a Cauchy sequence in T. (Com-pare with Exercise 4.33.) Exercises 101 4.55 Let f : S --> T be a function from a metric space S to another metric space T. Assume f is uniformly continuous on a subset A of S and that T is complete. Prove that there is a unique extension off to A which is uniformly continuous on A. 4.56 In a metric space (S, d), let A be a nonempty subset of S. Define a function fA : S - R by the equation fA(x) = inf (d(x, y) : y e A) for each x in S. The number fA(x) is called the distance from x to A. a) Prove that fA is uniformly continuous on S. b)Prove that A = (x : x e S and fA(x) = 0}. 4.57 In a metric space (S, d), let A and B be disjoint closed subsets of S. Prove that there exist disjoint open subsets U and V of S such that A s U and B 9 V. Hint. Let g(x) = fA(x) - fB(x), in the notation of Exercise 4.56, and consider g-'(- oo, 0) and g-'(0, +00). Discontinuities 4.58 Locate and classify the discontinuities of the functions f defined on R' by the follow-ing equations:. a) f (x) = (sin x)/x if x -A 0, f (O) = 0. b) f (x) = e"x if x : 0, f (O) = 0. c) f(x) = el/x + sin (1/x) if x t 0, f(0) = 0. d) f(x) = 1/(1 - e"x) if x 96 0, f(0) = 0. 4.59 Locate the points in RZ at which each of the functions in Exercise 4.11 is not con-tinuous. Monotonic functions 4.60 Let f be defined in the open interval (a, b) and assume that for each interior point x of (a, b) there exists a 1-ball B(x) in which f is increasing. Prove that f is an increasing function throughout (a, b). 4.61 Let f be continuous on a compact interval [a, b] and assume that f does not have a local maximum or a local minimum at any interior point. (See the NOTE following Exercise 4.25.) Prove that f must be monotonic on [a, b]. 4.62 If f is one-to-one and continuous on [a, b ], prove that f must be strictly monotonic on [a, b]. That is, prove that every topological mapping of [a, b] onto an interval [c, d] must be strictly monotonic. 4.63 Let f be an increasing function defined on [a, b] and let x1, ... , x be n points in the interior such that a < x1 < x2 < < x < b. a) Show that F_k=1 [f(xk+) - f(xk-)] <- f(b-) - f(a+). b) Deduce from part (a) that the set of discontinuities off is countable. c) Prove that f has points of continuity in every open subinterval of [a, b]. 4.64 Give an example of a function f, defined and strictly increasing on a set S in R, such that f -1 is not continuous on f(S). 102 Limits and Continuity 4.65 Let f be strictly increasing on a subset S of R. Assume that the image f(S) has one of the following properties: (a) f(S) is open; (b) f(S) is connected; (c) f (S) is closed. Prove that f must be continuous on S. Metric spaces and fixed points 4.66 Let B(S) denote the set of all real-valued functions which are defined and bounded on a nonempty set S. If f'e B(S), let 11111 = sup If(x)I. XES The number 11 f 11 is called the "sup norm" off a) Prove that the formula d(f, g) = 11f - g 1l defines a metric don B(S). b) Prove that the metric space (B(S), d) is complete. Hint. If (f") is a Cauchy sequence in B(S), show that {f(x) } is a Cauchy sequence of real numbers for each x in S. 4.67 Refer to Exercise 4.66 and let C(S) denote the subset of B(S) consisting of all func-tions continuous and bounded on S, where now S is a metric space. a) Prove that C(S) is a closed subset of B(S). b) Prove that the metric subspace C(S) is complete. 4.68 Refer to the proof of the fixed-point theorem (Theorem 4.48) for notation. a) Prove that d(p, p") < d(x, f(x))a"/(1 - a). This inequality, which is useful in numerical work, provides an estimate for the distance from p" to the fixed point p. An example is given in (b). b) Take f(x) = 4(x + 2/x), S = [1, +oo). Prove that f is a contraction of S with contraction constant a = } and fixed point p = %12. Form the sequence {p"} starting with x = po = I and show that Ip" - v/2I 5 2-" 4.69 Show by counterexamples that the fixed-point theorem for contractions need not hold if either (a) the underlying metric space is not complete, or (b) the contraction constant a >: 1. 4.70 Let f : S S be a function from a complete metric space (S, d) into itself. Assume there is a real sequence {a"} which converges to 0 such that d(f"(x), f"(y)) < a"d(x, y) for all n >- 1 and all x, y in S, where j'" is the nth iterate off, that is, f' (x) = f (x), f"+' (x) = f(f"(x)) for n >-1. Prove that f has a unique fixed point. Hint. Apply the fixed-point theorem to f' for a suitable m. 4.71 Let f : S S be a function from a metric space (S, d) into itself such that d(f(x),f(y)) < d(x, y) whenever x 0 y. a) Prove that f has at most one fixed point, and give an example of such an f with no fixed point. b) If S is compact, prove that f has exactly one fixed point. Hint. Show that g(x) = d(x, f(x)) attains its minimum on S. c) Give an example with S compact in which f is not a contraction. References 103 4.72 Assume that f satisfies the condition in Exercise 4.71. If x e S, let po = x, pn+ 1 = f(pn), and c = d(pn, Pn+ 1) for n >_ 0. a) Prove that {cn) is a decreasing sequence, and let c = lim c,,. b) Assume there is a subsequence which converges to a point q in S. Prove that c = d(q, f(q)) = d(f(q),.f[f(q)]) Deduce that q is a fixed point off and that p,, q. SUGGESTED REFERENCES FOR FURTHER STUDY 4.1 Boas, R. P., A Primer of Real Functions. Carus Monograph No. 13. Wiley, New York, 1960. 4.2 Gleason, A., Fundamentals of Abstract Analysis. Addison-Wesley, Reading, 1966. 4.3 Simmons, G. F., Introduction to Topology and Modern Analysis. McGraw-Hill, New York, 1963. 4.4 Todd, J., Survey of Numerical Analysis. McGraw-Hill, New York, 1962. CHAPTER 5 DERIVATIVI 5.1 INTRODUCTION This chapter treats the derivative, the central concept of differential calculus. T different types of problem-the physical problem of finding the instantanec velocity of a moving particle, and the geometrical problem of finding the tang, line to a curve at a given point-both lead quite naturally to the notion of deri tive. Here, we shall not be concerned with applications to mechanics and geomet but instead will confine our study to general properties of derivatives. This chapter deals primarily with derivatives of functions of one real varial specifically, real-valued functions defined on intervals in R. It also discus briefly derivatives of vector-valued functions of one real variable, and par derivatives, since these topics involve no new ideas. Much of this material sho be familiar to the reader from elementary calculus. A more detailed treatment derivative theory for functions of several variables involves significant chan and is dealt with in Chapter 12. The last part of the chapter discusses derivatives of complex-valued functi4 of a complex variable. 5.2 DEFINITION OF DERIVATIVE If f is defined on an open interval (a, b), then for two distinct points x and (a, b) we can form the difference quotient f(x) - f(c) x-c We keep c fixed and study the behavior of this quotient as x - c. Definition 5.1. Let f be defined on an open interval (a, b), and assume that c e (a, Then f is said to be differentiable at c whenever the limit lim f(x) - f(c) X-C x - c exists. The limit, denoted by f'(c), is called the derivative off at c. This limit process defines a new function f', whose domain consists of th points in (a, b) at which f is differentiable. The function f is called the j 104 Th. 5.3 Derivatives and Continuity 105 derivative off. Similarly, the nth derivative off, denoted by f("), is defined to be the first derivative of f(n-1), for n = 2, 3, .... (By our definition, we do not consider P) unless f " - 1) is defined on an open interval.) Other notations with which the reader may be familiar are f'(c) Df(c) (c) = I [where y = f(x)], ax ax x=c or similar notations. The function f itself is sometimes written f(O). The process which produces f' from f is called differentiation. 5.3 DERIVATIVES AND CONTINUITY The next theorem makes it possible to reduce some of the theorems on derivatives to theorems on continuity. Theorem 5.2. If f is defined on (a, b) and differentiable at a point c in (a, b), then there is a function f (depending on f and on c) which is continuous at c and which satisfies the equation f(x) - f(c) = (x - c)f(x), (1) for all x in (a, b), with f(c) = f'(c). Conversely, if there is a function f , con-tinuous at c, which satisfies (1), then f is differentiable at c andf'(c) = f(c). Proof .If f'(c) exists, let f be defined on (a, b) as follows : f(x) = f(x) - f(c) if x c, f (c) = f '(c) X - c Then f is continuous at c and (1) holds for all x in (a, b). Conversely, if (1) holds for some f continuous at c, then by dividing by x - c and letting x - c we see that f'(c) exists and equals f (c). - As an immediate consequence of (1) we obtain: Theorem 5.3. If f is differentiable at c, then f is continuous at c. Proof. Let x -+ c in (1). NOTE. Equation (1) has a geometric interpretation which helps us gain insight into its meaning. Since f is continuous at c, f (x) is nearly equal to f (c) = f'(c) if x is near c. Replacing f (x) by f'(c) in (1) we obtain the equation f(x) = f(c) + f'(c)(x - c), which should be approximately correct when x - c is small. In other words, if f is differentiable at c, then f is approximately a linear function near c. (See Fig. 5.1). Differential calculus continually exploits this geometric property of functions. 106 Derivatives Th. 5.4 (x f(x))Rr-----, Tan ent line g with slope f,(c) f(x) - f(c) f"(c)(X - c) (c, f(c)) -----L--- L Figure 5.1 C x 5.4 ALGEBRA OF DERIVATIVES The next theorem describes the usual formulas for differentiating the sum, differ-ence, product and quotient of two functions. Theorem 5.4. Assume f and g are defined on (a, b) and differentiable at c. Then f + g, f - g, and f - g are also differentiable at c. This is also true off/g if g(c) 96 0. The derivatives at c are given by the following formulas: a) (f ± g)'(c) = f'(c) ± g'(c), b) (.f' g)'(c) = f(c)g'(c) + f '(c)g(c), c) (f lg),(c) = g(c)f,(c9( f(c)g'(c) provided g(c) # 0. c) Proof We shall prove (b). Using Theorem 5.2 we write f(x) = f(c) + (x - c)f(x), g(x) = g(c) + (x - c)g(x). Thus, f(x)g(x) - f(c)g(c) = (x - c)[f(c)g(x) +f (x)g(c)] + (x - c)2 f (x)g(x). Dividing by x - c and letting x - c we obtain (b). Proofs of the other statements are similar. From the definition we see at once that if f is constant on (a, b) then f' = 0 on (a, b). Also, if f(x) = x, then f'(x) = 1 for all x. Repeated application of Theorem 5.4 tells us that if f(x) = x" (n a positive integer), then f'(x) = nx"-1 for all x. Applying Theorem 5.4 again, we see that every polynomial has a deriva-tive everywhere in R and every rational function has a derivative wherever it is defined. 5.5 THE CHAIN RULE A much deeper result is the so-called chain rule for differentiating composite func-tions. Def. 5.6 One-Sided Derivatives and Infinite Derivatives 107 Theorem 5.5 (Chain rule). Let f be defined on an open interval S, let g be defined on f(S), and consider the composite function g -f defined on S by the equation (g °f)(x) = g[f(x)]. Assume there is a point c in S such that f(c) is an interior point of f(S). If f is differentiable at c and if g is differentiable at f(c) then g -f is differentiable at c and we have (g °f)'(c) = 9'[f(c)]f'(c) Proof Using Theorem 5.2 we can write f(x) - f(c) = (x - c)f (x) for all x in S, where f is continuous at c and f (c) = f'(c). Similarly, 9(Y) - 9[.f(c)] = [Y - f(c)]9(Y), for all y in some open subinterval T off(S) containing f(c). Here g is continuous at f(c) and g[f(c)] = g'[f(c)]. Choosing x in S so that y = f(x) c T, we then have g[f(x)] - g[f(c)] = [f(x) - f(c)]g[f(x)] = (x - c)f(x)gU(x)] (2) By the continuity theorem for composite functions, 9[f(x)] -+ g[f(c)] = 9'U(c)] as x -, c. Therefore, if we divide by x - c in (2) and let x - c, we obtain lim g[.f(x)] - 9[f(c)] = g'U(c)]f'(c), as required. x-+c X - C 5.6 ONE-SIDED DERIVATIVES AND INFINT1E DERIVATIVES Up to this point, the statement that f has a derivative at c has meant that c was interior to an interval in which f was defined and that the limit defining f'(c) was finite. It is convenient to extend the scope of our ideas somewhat in order to discuss derivatives at endpoints of intervals. It is also desirable to introduce infinite derivatives, so that the usual geometric interpretation of a derivative as the slope of a tangent line will still be valid in case the tangent line happens to be vertical. In such a case we cannot prove that f is continuous at c. Therefore, we explicitly require it to be so. Definition 5.6. Let f be defined on a closed interval S and assume that f is continuous at the point c in S. Then f is said to have a righthand derivative at c if the righthand limit lim AX) - f (c) xyc+ x - c 108 Derivatives Th. exists as a finite value, or if the limit is + o0 or - oo. This limit will be denoted f +(c). Lefthand derivatives, denoted by f _ (c), are similarly defined. In additi if c is an interior point of S, then we say that f has the derivative f'(c) = + 00 both the right- and lefthand derivatives at c are + oo. (The derivative f'(c) is similarly defined.) It is clear that f has a derivative (finite or infinite) at an interior point c if, a only if, f+(c) = f-' (c), in which case f+(c) = f_(c) = f'(c). xl x2 x3 x4 X5 x7 X6 Figure 5.2 Figure 5.2 illustrates some of these concepts. At the point x, we have f+(x,) - oo. At x2 the lefthand derivative is 0 and the righthand derivative is -1. Al f'(x3) = - 00, f=(x4) = -1, f+(X4) = + 1, f'(x6) = + oo, and f_(x7) = There is no derivative (one-sided or otherwise) at x5, since f is not continuo there. 5.7 FUNCTIONS WITH NONZERO DERIVATIVE Theorem 5.7. Let f be defined on an open interval (a, b) and assume that for so, c in (a, b) we have f'(c) > 0 or f'(c) = +oo. Then there is a 1-ball B(c) c (a, in which f(x) > f(c) if x > c, and f(x) < f(c) if x < c. Proof. If f'(c) is finite and positive we can write f(x) - f(c) = (x - Of (x), where f is continuous at c and f (c) = f'(c) > 0. By the sign preserving prope of continuous functions there is a 1-ball B(c) s (a, b) in which f (x) has the sa sign as f(c), and this means that f(x) - f(c) has the same sign as x - c. Th. 5.9 Zero Derivatives and Local Extrema 109 If f'(c) = +oo, there is a 1-ball B(c) in which f(x) - f(c) > 1 x - c whenever x 0 c. In this ball the quotient is again positive and the conclusion follows as before. A result similar to Theorem 5.7 holds, of course, iff'(c) < 0 or iff'(c) 00 at some interior point c of (a, b). 5.8 ZERO DERIVATIVES AND LOCAL EXTREMA Definition 5.8. Let f be a real-valued function defined on a subset S of a metric space M, and assume a e S. Then f is said to have a local maximum at a if there is a ball B(a) such that f(x) < f(a) for all x in B(a) n S. If f(x) >- f(a) for all x in B(a) n S, then f is said to have a local minimum at a. NOTE. A local maximum at a is the absolute maximum off on the subset B(a) n S. If f has an absolute maximum at a, then a is also a local maximum. However, f can have local maxima at several points in S without having an absolute maximum on the whole set S. The next theorem shows a connection between zero derivatives and local extrema (maxima or minima) at interior points. Theorem 5.9. Let f be defined on an open interval (a, b) and assume that f has a local maximum or a local minimum at an interior point c of (a, b). If f has a derivative (finite or infinite) at c, then f'(c) must be 0. Proof. If f'(c) is positive or + oo, then f cannot have a local extremum at c because of Theorem 5.7. Similarly, f'(c) cannot be negative or - oo. However, because there is a derivative at c, the only other possibility isf'(c) = 0. The converse of Theorem 5.9 is not true. In general, knowing that f'(c) = 0 is not enough to determine whether f has an extremum at c. In fact, it may have neither, as can be verified by the example f(x) = x3 and c = 0. In this case, f'(0) = 0 but f is increasing in every neighborhood of 0. Furthermore, it should be emphasized that f can have a local extremum at c without f'(c) being zero. The example f(x) = IxI has a minimum at x = 0 but, of course, there is no derivative at 0. Theorem 5.9 assumes that f has a derivative (finite or infinite) at c. The theorem also assumes that c is an interior point of (a, b). In the example f(x) = x, where a < x S b, f takes on its maximum and minimum at the endpoints butf'(x) is never zero in [a, b]. 110 Derivatives Th. 5.10 5.9 ROLLE'S THEOREM It is geometrically evident that a sufficiently "smooth" curve which crosses the x-axis at both endpoints of an interval [a, b] must have a "turning point" some-where between a and b. The precise statement of this fact is known as Rolle's theorem. Theorem 5.10 (Rolle). Assume f has a derivative (finite or infinite) at each point of an open interval (a, b), and assume that f is continuous at both endpoints a and b. If f(a) = f(b) there is at least one interior point c at which f'(c) = 0. Proof. We assume f' is never 0 in (a, b) and obtain a contradiction. Since f is continuous on a compact set, it attains its maximum M and its minimum m some-where in [a, b]. Neither extreme value is attained at an interior point (otherwise f' would vanish there) so both are attained at the endpoints. Since f(a) = f(b), then m = M, and hence f is constant on [a, b]. This contradicts the assumption that f' is never 0 on (a, b). Therefore f'(c) = 0 for some c in (a, b). 5.10 THE MEAN-VALUE THEOREM FOR DERIVATIVES Theorem 5.11 (Mean- Value Theorem). Assume that f has a derivative (finite or infinite) at each point of an open interval (a, b), and assume also that f is continuous at both endpoints a and b. Then there is a point c in (a, b) such that f(b) - f(a) = f'(c)(b - a). Geometrically, this states that a sufficiently smooth curve joining two points A and B has a tangent line with the same slope as the chord AB. We will deduce Theorem 5.11 from a more general version which involves two functions f and g in a symmetric fashion. Theorem 5.12 (Generalized Mean- Value Theorem). Let f and g be two functions, each having a derivative (finite or infinite) at each point of an open interval (a, b) and each continuous at the endpoints a and b. Assume also that there is no interior point x at which both f'(x) and g'(x) are infinite. Then for some interior point c we have f'(c)[g(b) - g(a)] = g'(c)[f(b) - f(a)]. NOTE. When g(x) = x, this gives Theorem 5.11. Proof. Let h(x) = f(x)[g(b) - g(a)] - g(x)[f(b) - f(a)]. Then h'(x) is finite if both f'(x) and g'(x) are finite, and h'(x) is infinite if exactly one off'(x) or g'(x) is infinite. (The hypothesis excludes the case of both f'(x) and g'(x) being infinite.) Also, h is continuous at the endpoints, and h(a) = h(b) = f(a)g(b) - g(a)f(b). By Rolle's theorem we have h'(c) = 0 for some interior point, and this proves the assertion. Cor. 5.15 Intermediate-Value Theorem 111 NOTE. The reader should interpret Theorem 5.12 geometrically by referring to the curve in the xy-plane described by the parametric equations x = g(t), y = f(t), a<t<b. There is also an extension which does not require continuity at the endpoints. Theorem 5.13. Let f and g be two functions, each having a derivative (finite or infinite) at each point of (a, b). At the endpoints assume that the limits f(a+), g(a+), f(b-) and g(b-) exist as finite values. Assume further that there is no interior point x at which both f'(x) and g'(x) are infinite. Then for some interior point c we have f'(c)[g(b-) - g(a+)] = g'(c)[f(b-) - f(a+)] Proof. Define new functions F and G on [a, b] as follows: F(x) = f(x) and G(x) = g(x) if x e (a, b); F(a) = f(a+), G(a) = g(a+), F(b) = f(b - ), G(b) = g(b - ). Then F and G are continuous on [a, b] and we can apply Theorem 5.12 to F and G to obtain the desired conclusion. The next result is an immediate consequence of the Mean-Value Theorem. Theorem 5.14. Assume f has a derivative (finite or infinite) at each point of an open interval (a, b) and that f is continuous at the endpoints a and b. a) If f' takes only positive values (finite or infinite) in (a, b), then f is strictly increasing on [a, b]. b) If f' takes only negative values (finite or infinite) in (a, b), then f is strictly decreasing on [a; b], c) If f' is zero everywhere in (a, b) then f is constant on [a, b]. Proof Choose x < y and apply the Mean-Value Theorem to the subinterval [x, y] of [a, b] to obtain f(y) - f(x) = f'(c)(y - x) where c e (x, y). All the statements of the theorem follow at once from this equation. By applying Theorem 5.14 (c) to the difference f - g we obtain : Corollary 5.15..If f and g are continuous on [a, b] and have equal finite derivatives in (a, b), then f - g is constant on [a, b]. 5.11 INTERMEDIATE-VALUE THEOREM FOR DERIVATIVES In Theorem 4.33 we proved that a function f which is continuous on a compact interval [a, b] assumes every value between its maximum and its minimum on 112 Derivatives 1L. 5.16 the interval. In particular, f assumes every value betweenf(a) andf(b). A similar result will now be proved for functions which are derivatives. Theorem 5.16 (Intermediate-value theorem for derivatives). Assume that f is de-fined on a compact interval [a, b] and that f has a derivative (finite or infinite) at each interior point. Assume also that f has finite one-sided derivatives f+ (a) and f_(b) at the endpoints, with f+ (a) # f_(b). Then, if c is a real number between f+ (a) and f_(b), there exists at least one interior point x such that f'(x) = c. Proof. Define a new function g as follows: g(x) = f(x) - f(a) if x # a, g(a) = f+(a). x-a Then g is continuous on the closed interval [a, b]. By the intermediate-value theorem for continuous functions, g takes on every value between f+(a) and [f(b) - f(a)]/(b - a) in the interior (a, b). By the Mean-Value have g(x) = f'(k) for some k in (a, x) whenever x e (a, b). Therefore f' takes on every value between f+(a) and [f(b) - f(a)]/(b - a) in the interior (a, b). A similar argument applied to the function h, defined by h(x) = f(x) - f(b) if x # b, h(b) = f_(b), shows that f' takes on every value between [f(b) - f(a)]/(b - a) and f'_ (b) in the interior (a, b). Combining these results, we see that f' takes on every value between f+(a) and f_(b) in the interior (a, b), and this proves the theorem. NOTE. Theorem 5.16 is still valid if one or both of the one-sided derivatives f+(a), f_(b), is infinite. The proof in this case can be given by considering the auxiliary function g defined by the equation g(x) = f(x) - cx, if x e [a, b]. Details are left to the reader. The intermediate-value theorem shows that a derivative cannot change sign in an interval without taking the value 0. Therefore, we have the following strengthening of Theorem 5.14(a) and (b). Theorem 5.17. Assume f has a derivative (finite or infinite) on (a, b) and is con-tinuous at the endpoints a and b. If f'(x) # 0 for all x in (a, b) then f is strictly monotonic on [a, b]. The intermediate-value theorem also shows that monotonic derivatives are necessarily continuous. Theorem 5.18. Assume f' exists and is monotonic on an open interval (a, b). Then f' is continuous on (a, b). Proof. We assume f' has a discontinuity at some point c in (a, b) and arrive at a contradiction. Choose a closed subinterval [a, /3] of (a, b) which contains c in its interior. Since f' is monotonic on [a, /3] the discontinuity at c must be a jump Th. 5.20 Taylor's Formula with Remainder 113 discontinuity (by Theorem 4.51). Hence f' omits some value between f'(a) and f'(fl), contradicting the intermediate-value theorem. 5.12 TAYLOR'S FORMULA WITH REMAINDER As noted earlier, if f is differentiable at c, then f is approximately a linear function near c. That is, the equation f(x) = f(c) + f'(c)(x - c), is approximately correct when x - c is small. Taylor's theorem tells us that, more generally, f can be approximated by a polynomial of degree n - 1 if f has a deriva-tive of order n. Moreover, Taylor's theorem gives a useful expression for the error made by this approximation. Theorem 5.19 (Taylor). Let f be a function having finite nth derivative f11 every-where in an open interval (a, b) and assume that f( n-1) is continuous on the closed interval [a, b]. Assume that c e [a, b]. Then, for every x in [a, b], x # c, there exists a point x1 interior to the interval joining x and c such that n- I (k) a f(n)(xl f(x) = f(c) + E k=if k! ) (x - c)k + n) (x - Cr. Taylor's theorem will be obtained as a consequence of a more general result that is a direct extension of the generalized Mean-Value Theorem. Theorem 5.20. Let f and g be two functions having finite nth derivatives f O) and g(") in an open interval (a, b) ,and continuous (n - 1)st derivatives in the closed interval [a, b]. Assume that c e [a, b]. Then, for every x in [a, b], x # c, there exists a point xl interior to the interval joining x and c such that [f(x) - Ef(') c) (x - c)k gn°(x1) = 9(x) - E 9(k(c) (x - c)k . NoTE. For the special case in which g(x) = (x - c)r, we have g(k)(c) = 0 for 0 S k 5 n - 1 and g(")(x) = n!. This theorem then reduces to Taylor's theorem. Proof. For simplicity, assume that c < b and that x > c. Keep x fixed and define new functions F and G as follows : n- I (k) F(t)=f(t)+E- kit)(x-t)k, n-1 (k) G(t) = g(t) + (t) (x - t)k, kL=r1 k! . for each t in [e, x]. Then F and G are continuous on the closed interval [c, x] and have finite derivatives in the open interval (c, x). Therefore, Theorem 5.12 is 114 Derivatives applicable and we can write F'(x,)[G(x) - G(c)] = G'(x1)[F(x) - F(c)], where x, e (c, x). This reduces to the equation F'(x1)[g(x) - G(c)] = G'(x,)[f(x) - F(c)], (a) since G(x) = g(x) and F(x) = f (x). If, now, we compute the derivative of the sum defining F(t), keeping in mind that each term of the sum is a product, we find that all terms cancel but one, and we are left with F'(t) = (x - t)n- 1 f "°(t). (n - 1)! Similarly, we obtain G'(t) - (x -t)n-1 g0")(t). (n 1)! If we put t = x1 and substitute into (a), we obtain the formula of the theorem. 5.13 DERIVATIVES OF VECTOR-VALUED FUNCTIONS Let f : (a, b) - R" be a vector-valued function defined on an open interval (a, b) in R. Then f = (f1i ... , fn) where each component fk is a real-valued function defined on (a, b). We say that f is differentiable at a point c in (a, b) if each com-ponent fk is differentiable at c and we define f '(C) = (fl(c), ... , f.'(0). In other words, the derivative f'(c) is obtained by differentiating each component of f at c. In view of this definition, it is not surprising to find that many of the theorems on differentiation are also valid for vector-valued functions. For example, if f and g are vector-valued functions differentiable at c and if A is a real-valued function differentiable at c, then the sum f + g, the product )f, and the dot product f g are differentiable at c and we have (f + g)'(c) = U (C) + g'(c), (Af)'(c) = )'(c)f(c) + 2(c)f'(c), (f g)'(c) = f'(c) g(c) + f(c) - g'(c). The proofs follow easily by considering components. There is also a chain rule for differentiating composite functions which is proved in the same way. If f is vector-valued and if u is real-valued, then the composite function g given by g(x) _ f[u(x)] is vector-valued. The chain rule states that g'(c) = f'[u(c)]u'(c), if the domain of f contains a neighborhood of u(c) and if u'(c) and f'[u(c)] both exist. Partial Derivatives 115 The Mean-Value Theorem, as stated in Theorem 5.11, does not hold for vector-valued functions. For example, if f(t) = (cos t, sin t) for all real t, then f(27r) - f(O) = 0, but f'(t) is never zero. In fact, JIf'(t)fl = I for all t. A modified version of the Mean-Value Theorem for vector-valued functions is given in Chapter 12 (Theorem 12.8). 5.14 PARTIAL DERIVATIVES Let S be an open set in Euclidean space R", and let f : S -. R be a real-valued function defined on S. If x = (x1, ... , x") and c = (c1, ... , c.) are two points of S having corresponding coordinates equal except for the kth, that is, if x; = c; for i 0 k and if xk ck, then we can consider the limit lim f(x) - f(c) Xk-Ck Xk - Ck When this limit exists, it is called the partial derivative off with respect to the kth coordinate and is denoted by Dkf(c), fk(c), of (c), OXk or by a similar expression. We shall adhere to the notation Dkf(c). This process produces n further functions D1 f, D2 f, ... , defined at those points in S where the corresponding limits exist. Partial differentiation is not really a new concept. We are merely treating f(x1i ... , x") as a function of one variable at a time, holding the others fixed. That is, if we introduce a function g defined by 9(xk) = ./ (c1, ... , C1-1, Xk, Ck+1, ... , C.), then the partial derivative Dkf(c) is exactly the same as the ordinary derivative g'(ck). This is usually described by saying that we differentiate f with respect to the kth variable, holding the others fixed. In generalizing a concept from R1 to R, we seek to preserve the important properties in the one-dimensional case. For example, in the one-dimensional case, the existence of the derivative at c implies continuity at c. Therefore it seems desirable to have a concept of derivative for functions of several variables which will imply continuity. Partial derivatives do not do this. A function of n variables can have partial derivatives at a point with respect to each of the variables and yet not be continuous at the point. We illustrate with the following example of a function of two variables: .f(x, y) = x+y, ifx=Dory=O, 1, otherwise. 116 Derivatives The partial derivatives D1 f(0, 0) and D2 f(0, 0) both exist. In fact, D1 f(0, 0) = limf(X, 0) - f(0, 0) = lim x = 1, X-0 x - 0 x--'OX and, similarly, D2 f(0, 0) = 1. On the other hand, it is clear that this function is not continuous at (0, 0). The existence of the partial derivatives with respect to each variable separately implies continuity in each variable separately; but, as we have just seen, this does not necessarily imply continuity in all the variables simultaneously.-The difficulty with partial derivatives is that by their very definition we are forced to consider only one variable at a time. Partial derivatives give us the rate of change of a function in the direction of each coordinate axis. There is a more general concept of derivative which does not restrict our considerations to the special directions of the coordinate axes. This will be studied in detail in Chapter 12. The purpose of this section is merely to introduce the notation for partial derivatives, since we shall use them occasionally before we reach Chapter 12. If f has partial derivatives D1 f, ... , D"f on an open set S, then we can also consider their partial derivatives. These are called second-order partial derivatives. We write D,,kf for the partial derivative of Dkf with respect to the rth variable. Thus, D,,k.f = D,(Dk f ) Higher-order partial derivatives are similarly defined. Other notations are D f 02f D f = , a3f r,k aXr 0Xk P9.r axp aXq ax, 5.15 DIFFERENTIATION OF FUNCTIONS OF A COMPLEX VARIABLE In this section we shall discuss briefly derivatives of complex-valued functions defined on subsets of the complex plane. Such functions are, of course, vector-valued functions whose domain and range are subsets of R2. All the considerations of Chapter 4 concerning limits and continuity of vector-valued functions apply, in particular, to functions of a complex variable. There is, however, one essential difference between the set of complex numbers C and the set of n-dimensional vectors R" (when n > 2) that plays an important role here. In the complex number system we have the four algebraic operations of addition, subtraction, multiplica-tion, and division, and these operations satisfy most of the "usual" laws of algebra that told for the real number system. In particular, they satisfy the first five axioms for real numbers listed in Chapter 1. (Axioms 6 through 10 involve the ordering relation <, which cannot exist among the complex numbers.) Any algebraic system which satisfies Axioms I through 5 is called a field. (For a thorough discussion of fields, see Reference 1.4.) Multiplication and division, it turns out, cannot be introduced in R" (for n > 2) in such a way that R" will Def. 5.21 Functions of a Complex Variable 117 become a fieldt which includes C. Since division is possible in C, however, we can form the fundamental difference quotient [f(z) - f(c)]/(z - c) which was used to define the derivative in R, and it now becomes clear how the derivative should be defined in C. Definition 5.21. Let f be a complex-valued function defined on an open set S in C, and assume c e S. Then f is said to be differentiable at c if the limit limf(z) - f(c) = f'(c) Z_C z - c exists. By means of this limit process, a new complex-valued function f' is defined at those points z of S where f'(z) exists. Higher-order derivatives are, of course, similarly defined. The following statements can now be proved for complex-valued functions defined on an open set S by exactly the same proofs used in the real case: a) f is differentiable at c if, and only if, there is a function f , continuous at c, such that f(z) - f(c) = (z - Of (z), for all z in S, with f (c) = f'(c). NOTE. If we let g(z) = f(z) - f'(c) the equation in (a) can be put in the form f(z) = f(c) + f'(c)(z - c) + g(z)(z - c), where g(z) -> 0 as z -+ c. This is called a first-order Taylor formula for f. b) If f is differentiable at c, then f is continuous at c. c) If two functions f and g have derivatives at c, then their sum, difference, product, and quotient also have derivatives at c and are given by the usual formulas (as in Theorem 5.4). In the case off/g, we must assume g(c) 0. d) The chain rule is valid; that is to say, we have (g °f)'(c) = g'[.f(c)].f'(c)' if the domain of g contains a neighborhood of f(c) and if f'(c) and g'[f(c)] both exist. Whenf(z) = z, we findf'(z) = I for all z in C. Using (c) repeatedly, we find that f'(z) = nz"-1 when f(z) = z" (n is a positive integer). This also holds when t For example, if it were possible to define multiplication in R3 so as to make R3 a field including C, we could argue as follows: For every x in R3 the vectors 1, x, x2, x3 would be linearly dependent (see Reference 5.1, p. 558). Hence for each x in R3, a relation of the form ao + a,x + a2x2 + a3x3 = 0 would hold, where ao, a,, a2, a3 are real numbers. But every polynomial of degree three with real coefficients is a product of a linear polynomial and a quadratic polynomial with real coefficients. The only roots such polynomials can have are either real numbers or complex numbers. 118 Derivatives Th. 5.22 n is a negative integer, provided z 0. Therefore, we may compute derivatives of complex polynomials and complex rational functions by the same techniques used in elementary differential calculus. 5.16 THE CAUCHY-RIEMANN EQUATIONS If f is a complex-valued function of a complex variable, we can write each function value in the form f(z) = u(z) + iv(z), where u and v are real-valued functions of a complex variable. We can, of course, also consider u and v to be real-valued functions of two real variables and then we write f(z) = u(x, y) + iv(x, y), if z = x + iy. In either case, we write f = u + iv and we refer to u and v as the real and imag-inary parts off For example, in the case of the complex exponential function f, defined by f(z) = ez = ex cos y jj iex sin y, the real and imaginary parts are given by u(x, y) = ex cos y, v(x, y) = ex sin y. Similarly, when f(z) = z2 = (x + iy)2, we find u(x, y) = x2 - y2, v(x, y) = 2xy. In the next theorem we shall see that the existence of the derivative f' places a rather severe restriction on the real and imaginary parts u and v. Theorem 5.22. Let f = u + iv be defined on an open set S in C. If f'(c) exists for some c in S, then the partial derivatives D,u(c), D2u(c), D,v(c) and D2v(c) also exist and we have f'(c) = D1u(c) + i D,v(c), (3) and f'(c) = D2v(c) - i D2u(c). This implies, in particular, that D, u(c) = D2v(c) and D, v(c) = - D2u(c). (4) NOTE. These last two equations are known as the Cauchy-Riemann equations. They are usually seen in the form au av av au ax ay, ax ay Proof. Since f'(c) exists there is a function f defined on S such that f(z) - f(c) = (z - c)f(z), (5) 11. 5.23 Cauchy-Riemann Equations 119 where f is continuous at c and f (c) = f'(c). Write z=x+iy, c=a+ib, and f(z)=A(z)+iB(z), where A(z) and B(z) are real. Note that A(z) -' A(c) and B(z) --> B(c) as z -+ c. By considering only those z in S with y = b and taking real and imaginary parts of (5), we find u(x, b) - u(a, b) = (x - a)A(x + ib), v(x, b) - v(a, b) = (x - a)B(x + ib). Dividing by x - a and letting x - a we find D,u(c) = A(c) and Dlv(c) = B(c). Sincef'(c) = A(c) + iB(c), this proves (3). Similarly, by considering only those z in S with x = a we find D2v(c) = A(c) and D2u(c) = -B(c), which proves (4). Applications of the Cauchy-Riemann equations are given in the next theorem. Theorem 5.23. Let f = u + iv be a function with a derivative everywhere in an open disk D centered at (a, b). If any one of u, v, or If I is constant' on D, then f is constant on D. Also, f is constant if f'(z) = 0 for all z in D. Proof. Suppose u is constant on D. The Cauchy-Riemann equations show that D2v = D1v = 0 on D. Applying the one-dimensional Mean-Value Theorem twice we find, for some y' between b and y, v(x, y) - v(x, b) = (y - b)D2v(x, y') = 0, and, for some x' between a and x, v(x, b) - v(a, b) = (x - a)D, v(x', b) = 0. Therefore v(x, y) = v(a, b) for all (x, y) in D, so v is constant on D. A similar argument shows that if v is constant then u is constant. Now suppose If I is constant on D. Then. If I2 = u2 + v2 is constant on D. Taking partial derivatives we find uD1u+vD1v=0, uD2u+vD2v=0. By the Cauchy-Riemann equations the second equation can be written as vD,u-uD1v=0. Combining this with the first to eliminate D1v we find (u2 + v2)D,u = 0. If u2+v2=0, then u = v = 0,sof=0. Ifuz+v2 #0then D,u=0;hence u is constant, so f is constant. t Here Ifs denotes the function whose value at z is If(z)I. 120 Derivatives Finally, if f' = 0 on D, both partial derivatives D,v and D2v are zero on D. Again, as in the first part of the proof, we find f is constant on D. Theorem 5.22 tells us that a necessary condition for the function f = u + iv to have a derivative at c is that the four partials D,u, D2u, D,v, D2v, exist at c and satisfy the Cauchy-Riemann equations. This condition, however, is not sufficient, as we see by considering the following example. Example. Let u and v be defined as follows: x3 - y3 u(x, Y) = x2 + Y2 if (x, Y) 54 (0, 0), u(0, 0) = 0, v(x, Y) = X3 + Y3 if (x, Y) j4 (0, 0), v(o, 0) = 0. It is easily seen that D1u(0, 0) = Dlv(0, 0) = 1 and that D2u(0, 0) = - D2v(0, 0) = - 1, so that the Cauchy-Riemann equations hold at (0, 0). Nevertheless, the function f = u + iv cannot have a derivative at z = 0. In fact, for x = 0, the difference quotient becomes f(z) - f(0) = -Y + iy = 1 + i, z-0 iy whereas for x = y, it becomes f(z) - f(0) xi 1 + i z-0 x+ix 2 and hence f'(0) cannot exist. In Chapter 12 we shall prove that the Cauchy-Riemann equations do suffice to establish existence of the derivative off = u + iv at c if the partial derivatives of u and v are continuous in some neighborhood of c. To illustrate how this result is used in practice, we shall obtain the derivative of the exponential function. Let f(z)=e2=u+iv. Then and hence u(x, y) = ex cos y, v(x, y) = ex sin y, D,u(x, y) = ex cos y = D2v(x, y), D2u(x, y) = -ex sin y = -D,v(x, y). Since these partial derivatives are continuous everywhere in R2 and satisfy the Cauchy-Riemann equations, the derivativef'(z) exists for all z. To compute it we use Theorem 5.22 to obtain f'(z) = ex cos y + iex sin y = f(z). Thus, the exponential function is its own derivative (as in the real case). Exercises 121 EXERCISES Real-valued functions In the following exercises assume, where necessary, a knowledge of the formulas for differentiating the elementary trigonometric, exponential, and logarithmic functions. 5.1 A function f is said to satisfy a Lipschitz condition of order a at c if there exists a positive number M (which may depend on c) and a 1-ball B(c) such that (f(x) - f(c)1 < Mix - cI" whenever x e B(c), x c. a) Show that a function which satisfies a Lipschitz condition of order a is continuous at c if a > 0, and has a derivative at c if a > 1. b) Give an example of a function satisfying a Lipschitz condition of order 1 at c for which f'(c) does not exist. 5.2 In each of the following cases, determine the intervals in which the function f is increasing or decreasing and find the maxima and minima (if any) in the set where each f is defined. a)f(x)=x3+ax+b, xeR. b) f(x) = log (x2 - 9), IxI > 3. C) f(x) = x213(x - 1)4, 0 < x < 1. d) f (x) = (sin x)/x if x 96 0, f (O) = 1, 0 < x < n/2. 5.3 Find a polynomial f of lowest possible degree such that f(x1) = a1, f(x2) = a2, f'(xl) = b1, f'(x2) = b2, where x1 0 x2 and a1, a2, bl, b2 are given real numbers. 5.4 Define f as follows: f(x) = e-1/x2 if x : 0, f(0) = 0. Show that a) f is continuous for all x. b) f(") is continuous for all x, and that f (1)(0) = 0, (n = 1, 2, ... ). 5.5 Define f, g, and has follows: f(0) = g(0) = h(0) = 0 and, if x ;4 0, f(x) = sin (1/x), g(x) = x sin (1/x), h(x) = x2 sin (1/x). Show that a) f'(x) = -1/x2 cos (1/x), if x 54 0; f'(0) does not exist. b) g'(x) = sin (1/x) - l/x cos (1/x), if x j4 0; g'(0) does not exist. c) h'(x) = 2x sin (1/x) - cos (1/x), if x 0 0; h'(0) = 0; limx_o h'(x) does not exist. 5.6 Derive Leibnitz's formula for the nth derivative of the product h of two functions f and g : h(r)(x) = r (n/ f(k)(x)g(n-k)(x), where (n\ = n! kk=0 k kl k! (n - k)! 5.7 Let f and g be two functions defined and having finite third-order derivatives f "(x) and g"(x) for all x in R. If f(x)g(x) = 1 for all x, show that the relations in (a), (b), (c), 122 Derivatives and (d) hold at those points where the denominators are not zero: a) f'(x)/f(x) + g'(x)/g(x) = 0. b) f"(x)lf'(x) - 2f '(x)lf(x) - 9"(x)l9 (x) = 0. C) f,,,(x) - 3f '(x)9 (x) - 3f "(x) - 9 (x) = 0. f'(x) f (x)9'(x) f (x) 9'(x) d) fl(x) - 3 f (x) 2 g(x) -3 g"(x) 2 f'(x) 2 \.f'(x)/ -g '(x) 2 \9 (x)/ NOTE. The expression which appears on the left side of (d) is called the Schwarzian derivative off at x. e) Show that f and g have the same Schwarzian derivative if g(x) = [af(x) + b ]/ [cf (x) + d ], where ad - be j4 0. Hint. If c 34 0, write (af + b)l(cf + d) = (a/c) + (bc - ad)/[c(cf + d)], and apply part (d). 5.8 Let f1, f2, g1, g2 be four functions having derivatives in (a, b). Define F by means of the determinant F(x) = f1(x) f2(x) 91(x) 92(x) ifxe(a,b). a) Show that F'(x) exists for each x in (a, b) and that F'(x) = A(x) fi(x) 91(x) 92(x) fi(x) f2(x) 9i(x) 9s(x) b) State and prove a more general result for nth order determinants. 5.9 Given n functions f1, ... , f", each having nth order derivatives in (a, b). A function W, called the Wronskian of f1, ... , f", is defined as follows: For each x in (a, b), W(x) is the value of the determinant of order n whose element in the kth row and mth column is where k = 1, 2, ... , n and m = 1, 2, ... , n. [The expression is written for fm(x). ] a) Show that W'(x) can be obtained by replacing the last row of the determinant defining W(x) by the nth derivatives f (")(x), ... , b) Assuming the existence of n constants c1, ... , c", not all zero, such that c1f1(x) + + c"f"(x) = 0 for every x in (a, b), show that W(x) = 0 for each x in (a, b). NOTE. A set of functions satisfying such a relation is said to be a linearly dependent set on (a, b). c) The vanishing of the Wronskian throughout (a, b) is necessary, but not sufficient, for linear dependence of f1, . . . , f". Show that in the case of two functions, if the Wronskian vanishes throughout (a, b) and if one of the functions does not vanish in (a, b), then they form a linearly dependent set in (a, b). Exercises 123 Mean-Value Theorem 5.10 Given a function f defined and having a finite derivative in (a, b) and such that limx_,b_ f(x) = + oo. Prove that limX...b_ f'(x) either fails to exist or is infinite. 5.11 Show that the formula in the Mean-Value Theorem can be written as follows: f(x + h) - f(x) = f'(x + Oh), h where 0 < 0 < 1. Determine 9 as a function of x and h when a) f(x) = x2, b) f(x) = x3, c) f(x) = ex, d) f(x) = log x, x > 0. Keep x 34 0 fixed, and find limb,.o 0 in each case. 5.12 Take f (x) = 3x4 - 2x3 - x2 + 1 and g(x) = 4x3 - 3x2 - 2x in Theorem 5.20. Show that f'(x)/g'(x) is never equal to the quotient [f(1) - f(0)]/[g(1) - g(0)] if 0 < x <- 1. How do you reconcile this with the equation f(b) - f(a) = f''(xi) g(b) - g(a) 9 (xi) a < x < obtainable from Theorem 5.20 when n = I? 5.13 In each of the following special cases of Theorem 5.20, take n = 1, c = a, x = b, and show that xl = (a + b)/2. a) f(x) = sin x, g(x) = cos x; b) f(x) = ex, g(x) = e-X. Can you find a general class of such pairs of functions f and g for which xl will always be (a + b)/2 and such that both examples (a) and (b) are in this class? 5.14 Given a function f defined and having a finite derivative f' in the half-open interval 0 < x < 1 and such that I f'(x)I < 1. Define a = f(1/n) for n = 1, 2, 3, ... , and show that lime-,,, a exists. Hint. Cauchy condition. 5.15 Assume that f has a finite derivative at each point of the open interval (a, b). Assume also that limX.4 f'(x) exists and is finite for some interior point c. Prove that the value of this limit must be f'(c). 5.16 Let f be continuous on (a, b) with a finite derivative f' everywhere in (a, b), except possibly at c. If limX_, f'(x) exists and has the value A, show that f'(c) must also exist and have the value A. 5.17 Let f be continuous on [0, 1 ], f(0) = 0, f'(x) finite for each x in (0, 1). Prove that if f' is an increasing function on (0, 1), then so too is the function g defined by the equa-tion g(x) = f(x)/x. 5.18 Assume f has a finite derivative in (a, b) and is continuous on [a, b] with f(a) _ f(b) = 0. Prove that for every real .Z there is some c in (a, b) such that f'(c) = )f(c). Hint. Apply Rolle's theorem to g(x)f(x) for a suitable g depending on A. 5.19 Assume f is continuous on [a, b] and has a finite second derivative f" in the open interval (a, b). Assume that the line segment joining the points A = (a, f(a)) and B = (b, f(b)) intersects the graph off in a third point P different from A and B. Prove that f"(c) = 0 for some c in (a, b). 124 Derivatives 5.20 If f has a finite third derivative f A" in [a, b ] and if f (a) = f(a) = f (b) = f(b) = 0, prove that f A"(c) = 0 for some c in (a, b). 5.21 Assume f is nonnegative and has a finite third derivative f '" in the open interval (0, 1). If f (x) = 0 for at least two values of x in (0, 1), prove that f b"(c) = 0 for some c in (0, 1). 5.22 Assume f has a finite derivative in some interval (a, + oo). a) If f(x) - 1 and f'(x) -). c as x -). + oo, prove that c = 0. b) If f(x) -p 1 as x -> + oo, prove that f(x)lx 1 as x - + cc. c) If f'(x) -+ 0 as x -). + cc, prove that f(x)lx 0 as x --> + cc. 5.23 Let h be a fixed positive number. Show that there is no function f satisfying the following three conditions: f(x) exists for x ? 0, f'(0) = 0, f'(x) ? h for x > 0. 5.24 If h > 0 and if f(x) exists (and is finite) for every x in (a - h, a + h), and if f is continuous on [a - h, a + h], show that we have : a)f(a+h)-f(a- h) = f'a+Oh + f 'a - Oh0 < 0 < 1; h b) f(a + h) - 2f(a) + f(a - h) = f'(a + Ah - f'(a - ,h), 0 < A < 1. h ) c) If f '(a) exists, show that f "(a) = lim f (a + h) - 2f (a) + f (a - h) h-+O h2 d) Give an example where the limit of the quotient in (c) exists but wheref"(a) does not exist. 5.25 Let f have a finite derivative in (a, b) and assume that c e (a, b). Consider the following condition: For every c > 0 there exists a 1-ball B(c; 6), whose radius 6 depends only on c and not on c, such that if x e B(c; 6), and x t- c, then f(x) f(c) - f '(c) X c < E. Show that f' is continuous on (a, b) if this condition holds throughout (a, b). 5.26 Assume f has a finite derivative in (a, b) and is continuous on [a, b], with a _< f (x) -< b for all x in [a, b] and If'(x)I a< 1 for all x in (a, b). Prove that f has a unique fixed point in [a, b]. 5.27 Give an example of a pair of functions f and g having finite derivatives in (0, 1), such that lim (x) = 0, x -+ 0 g(x) but such that limx-+0 f'(x)/g'(x) does not exist, choosing g so that g'(x) is never zero. Exercises 125 5.28 Prove the following theorem : Let f and g be two functions having finite nth derivatives in (a, b). For some interior point c in (a, b), assume that f (c) = f'(c) = = f(")(c) = 0, and that g(c) = g'(c) _ = g(" -1)(c) = 0, but that g(")(x) is never zero in (a, b). Show that lim f(x) = f(")(c) x-+c g(x) g(")(C) NOTE. P") and g(") are not assumed to be continuous at c. Hint. Let (x - c)"-'f(n -1)(C) F(x) = f (X) -(n- 1)! define G similarly, and apply Theorem 5.20 to the functions F and G. 5.29 Show that the formula in Taylor's theorem can also be written as follows : n--1 f (k)(C) -k (x - C)(x - x1)"-1 (n) f (x) (x c) + f (X1)9 k=o k. (n where x1 is interior to the interval joining x and c. Let 1 - 0 = (x - x1)l(x - c). Show that 0 < 0 < 1 and deduce the following form of the remainder term (due to Cauchy) : 0 - 0)n-1(x - C)" (n) 8 1 8 Hint. Take G(t) = g(t) = t in the proof of Theorem 5.20. Vector-valued functions 5.30 If a vector-valued function f is differentiable at c, prove that f ,(c) = lim 1 [f(c + h) - f(c) . h-+ O h Conversely, if this limit exists, prove that f is differentiable at c. 5.31 A vector-valued function f is differentiable at each point of (a, b) and has constant norm 11 f 11. Prove that f (t) f'(t) = 0 on (a, b). 5.32 A vector-valued function f is never zero and has a derivative f' which exists and is continuous on R. If there is a real function A such that f'(t) = 4(t)f(t) for all t, prove that there is a positive real function u and a constant vector c such that f (t) = u(t)c for all t. Partial derivatives 5.33 Consider the function f defined on R2 by the following formulas : f(x, Y) = xy x2 + y2 if (x, y) : (0, 0) A09 0) = 0. Prove that the partial derivatives D1 f (x, y) and D2 f (x., y) exist for every (x, y) in R2 and evaluate these derivatives explicitly in terms of x and y. Also, show that f is not con-tinuous at (0, 0). 126 Derivatives 5.34 Let f be defined on R2 as follows : Ax, Y) = y x _ y2 2 2 if (x, Y) : (0, 0), f (0, 0) = 0. x +y Compute the first- and second-order partial derivatives off at the origin, when they exist. Complex-valued functions 5.35 Let S be an open set in C and let S be the set of complex conjugates 2, where z e S. -7 If f is defined on S, define g on S as follows: g(z) = J (z , the complex conjugate of f(z). If f is differentiable at c prove that g is differentiable at c and that g'(c) = 7'(c). 5.36 i) In each of the following examples write f = u + iv and find explicit formulas for u(x, y) and v(x, y) : a) f (z) = sin z, b) f (z) = cos z, c) f(Z) = IzI, d) f(z) = Z, e) f (z) = arg z (z 0 0), f) f (z) = Log z (z : 0), g) f(z) = ez2, h) f (z) = z°L (a complex, z : 0). (These functions are to be defined as indicated in Chapter 1.) ii) Show that u and v satisfy the Cauchy-Riemann equations for the following values of z : All z in (a), (b), (g) ; no z in (c), (d), (e) ; all z except real z < 0 in (f), (h). (In part (h), the Cauchy-Riemann equations hold for all z if a is a nonnegative integer, and they hold for all z : 0 if a is a negative integer.) iii) Compute the derivativef'(z) in (a), (b), (f), (g), (h), assuming it exists. 5.37 Write f = u + iv and assume that f has a derivative at each point of an open disk D centered at (0, 0). If au2 + bv2 is constant on D for some real a and b, not both 0, prove that f is constant on D. SUGGESTED REFERENCES FOR FURTHER STUDY 5.1 Apostol, T. M., Calculus, Vol. 1, 2nd ed. Xerox, Waltham, 1967. 5.2 Chaundy, T. W., The Differential Calculus. Clarendon Press, Oxford, 1935. CHAPTER 6 FUNCTIONS OF BOUNDED VARIATION AND RECTIFIABLE CURVES 6.1 INTRODUCTION Some of the basic properties of monotonic functions were derived in Chapter 4. This brief chapter discusses functions of bounded variation, a class of functions closely related to monotonic functions. We shall find that these functions are intimately connected with curves having finite arc length (rectifiable curves). They also play a role in the theory of Riemann-Stieltjes integration which is developed in the next chapter. 6.2 PROPERTIEOF MONOTONIC FUNCTIONS Theorem 6.1' Let f be an increasing function defined on [a, b] and let x0, x1, ... , x be n + 1 points such that a=x0 <x1 <x2 <<xn=b. Then we have the inequality n-1 L.: [f(x,+) - f(xk-)] < f(b) -' f(a). k=1 Proof. Assume thatyk e (xk, xk+l) For 1 < k < n - 1, we havef(xk+) <_ f(Yk) and f(Yk-1) < ftxk-), so that f(xk+) - f(Xk-) f(YJ - f(Yk- 0-If we add these inequalities, the sum on the right telescopes to f(Yn- 1) - f(y0). Since f(y -1) - f(Yo) <- f(b) - f(a), this completes the proof. The difference f(xk+) - f(xk-) is, of course, the jump of f at xk. The fore-going theorem tells us that for every finite collection of points xk in (a, b), the sum of the jumps at these points is always bounded by f(b) - f(a). This result can be used to prove the following theorem. Theorem 6.2. If f is monotonic on [a, b], then the set of discontinuities of f is countable. Proof. Assume that f is increasing and let S. be the set of points in (a, b) at which the jump off exceeds 1/m, m > 0. If xl < x2 < < are in Theorem 6.1 tells us that n I :!! f(b) - f(a). m 127 128 Functions of Bounded Variation and Rectifiable Curves Def. 6.3 This means that S. must be a finite set. But the set of discontinuities off in (a, b) is a subset of the union UM=1 S. and hence is countable. (If f is decreasing, the argument can be applied to -f.) 6.3 FUNCTIONS OF BOUNDED VARIATION Definition 6.3. If [a, b] is a compact interval, a set of points P = {x0, x1, ... , xn}, satisfying the inequalities a = x0 <xn=b, is called a partition of [a, b]. The interval [Xk..... 1, Xk] is called the kth subinterval of P and we write exk = Xk - Xk _ 1, so that F,k " =1 exk = b - a. The collection of all possible partitions of [a, b] will be denoted by t[a, b]. D e f i n i t i o n 6.4. Let f be d e f i n e d on [a, b]. If P = {xo, x1, ... , xn} is a partition of [a, b], writ 4 fk = f(xk) - f(xk_ 1), for k = 1, 2, ... , n. If there exists a positive nu er M such that n 1: k=1 for all partitions of [a, b], then f is said to be of bounded variation on [a, b]. Examples of functions of bounded variation are provided by the next two theorems. Theorem 6.5. If f is monotonic on [a, b], then f is of bounded variation on [a, b]. Proof. Let f be increasing. Then for every partition of [a, b] we have Ofk >_ 0 and hence n n n k=1 EAfk k=1 = E [f(Xk) - AXk- 01 =f(b) - f(a). Theorem 6.6. If f is continuous on [a, b] and if f' exists and is bounded in the interior, say If'(x)I < A for all x in (a, b), then f is of bounded variation on [a, b]. Proof. Applying the Mean-Value Theorem, we have Afk = f(xk) - f(xk- 1) _ f '(tk)(Xk - xk_ 1), where tk E (Xk_ 1, xk) This implies n n tt 1: VfkI = 1: If'(tk)I oXk 5 A 1, Oxk = A(b - a). k=1 k=1 k=1 Theorem 6.7. If f is of bounded variation on [a, b], say Y, JAfkj < M for all par-titions of [a, b], then f is bounded on [a, b]. In fact, 11(x)1 < I + M for all x in [a, b]. Def. 6.8 Total Variation 129 Proof. Assume that x c- (a, b). Using the special partition P = {a, x, b}, we find Iftx) - f(a)l + If(b) - f(x)l M. This implies lf(x) -f(a)d < M, Jf(x)j < I+ M. The same inequality holds ifx=aorx=b. Examples 1. It is easy to construct a continuous function which is not of bounded variation. For example, let f (x) = x cos (7r/(2x)) if x : 0, f (O) = 0. Then f is continuous on [0, 1 ], but if we consider the partition into 2n subintervals P 031 1 1 1 1 1 2n' 2n - 1'...,3' 2' an easy calculation shows that we have 1 1 1 1 2n 57 + + + -x=1 2n 2n 2n - 2 2n -his is not bounded for all n, since the se his is not bounded for all n, since the se +...+ 1 +. 1 = 1 + 1 +...+ I. 2 2 2 2 n ries YR'=, (11n) diverges. In this example variation on such an interval. the derivative f' exists in (0, 1) but f' is not bounded on (0, 1). However, f' is bounded on any compact interval not containing the origin and hence f will be of bounded 2. An example similar to the first is given by f (x) = x2 cos (11X) if x t- 0, f(0) = 0. This f is of bounded variation on [0, 1 ], since f' is bounded on [0, 1 ]. In fact, f'(0) = 0 and, for x : 0, f'(x) = sin (11x) + 2x cos (11x), so that If'(x)I < 3 for all x in [0, 1 ]. 3. Boundedness off' is not necessary for f to be of bounded variation. For example, let AX) = x1/3 . This function is monotonic (and hence of bounded variation) on every finite interval. However, f'(x) -' + oo as x -+ 0. 6.4 TOTAL VARIATION Definition 6.8. Let f be of bounded variation on [a, b], and let Y, (P) denote the sum Ekn [a, b]. The = 1 JAfkj corresponding to the partition P = {xo, xl, of number Vf(a, b) = sup (P) : P E 9[a, b]}, is called the total variation off on the interval [a, b]. NOTE. When there is no danger of misunderstanding, we will write Vf instead of Vf(a, b). Since f is of bounded variation on [a, b], the number Vf is finite. Also, Vf >_ 0, since each sum Y, (P) >_ 0. Moreover, Vf(a, b) = 0 if, and only if, f is constant on [a, b]. 130 Functions of Bounded Variation and Rectifiable Curves Th. 6.9 Theorem 6.9. Assume that f and g are each of bounded variation on [a, b]. Then so are their sum, difference, and product. Also, we have where Vf±g < V f + V 9 and Vf.9 < AVf + BV9, A = sup {Ig(x)l : x e [a, b]}, B = sup {I f(x)I : x e [a, b]}. Proof. Let h(x) = f(x)g(x). For every partition P of [a, b], we have IAhkl = If(xk)g(xk) -f(xk-1)g(xk-1)I = I[f(xk)g(xk) - J (xk-1)g(xk)] + [f(xk-1)g(xk) -f(xk-i)g(xk-1)]I < AlAfkl + BlAgkl This implies that h is of bounded variation and that Vh < AVf + BVg. The proofs for the sum and difference are simpler and will be omitted. NOTE. Quotients were not included in the foregoing theorem because the reciprocal of a function of bounded variation need not be of bounded variation. For example, if f(x) -+ 0 as x --> x0, then 1/f will not be bounded on any interval containing x0 and (by Theorem 6.7) 1/f cannot be of bounded variation on such an interval. To extend Theorem 6.9 to quotients, it suffices to exclude functions whose values become arbitrarily close to zero. Theorem 6.10. Let f be of bounded variation on [a, b]. and assume that f is bounded away from zero; that is, suppose that there exists a positive number m such that 0 < m < I f(x)I for all x in [a, b]. Then g = 1/f is also of bounded variation on [a, b], and V. < Vf/m2. Proof legki = 1 i f(xk) f(xk-1) Afk f(xk)f(xk-1) 6.5 ADDITIVE PROPERTY OF TOTAL VARIATION Iofki m2 In the last two theorems the interval [a, b] was kept fixed and Vf(a, b) was con-sidered as a function of f. If we keep f fixed and study the total variation as a function of the interval [a, b], we can prove the following additive property. Theorem 6.11. Let f be of bounded variation on [a, b], and assume that c e (a, b). Then f is of bounded variation on [a, c] and on [c, b] and we have Vf(a, b) = Vf(a, c) + Vf(c, b). Proof. We first prove that f is of bounded variation on [a, c] and on [c, b]. Let P1 be a partition of [a, c] and let P2 be a partition of [c, b]. Then P0 = P1 U P2 is a partition of [a, b]. If 7_ (P) denotes the sum Y_ IAfkl corresponding to the partition P (of the appropriate interval), we can write E (P1) + E (P2) = E (Po) < Vf(a, b). (1) Th. 6.12 Total Variation on [a, x] as a Function of x 131 This shows that each sum Y_ (PI) and Y_ (P2) is bounded by Vf(a, b) and this means that f is of bounded variation on [a, c] and on [c, b]. From (1) we also obtain the inequality Vf(a, c) + Vf(c, b) < Vf(a, b), because of Theorem 1.15. To obtain the reverse inequality, let P = {x0, x1, ... , a 9[a, b] and let P0 = P u {c}-be the (possibly new) partition obtained by adjoining the point c. If c e [xk_ 1, xk], then we have If(xk) - f(xk-1) I < If(xk) - .f(c)I + If(C) - f(xk-1)I , and hence (P) < Y, (Pa). Now the points of Po in [a, c] determine a partition PI of [a, c] and those in [c, b] determine a partition P2 of [c, b]. The corre-sponding sums for all these partitions are connected by the relation E (P) < E (Po) = E (PI) + E (P2) < Vf(a, C) + Vf(c, b). Therefore, Vf(a, c) + Vf(c, b) is an upper bound for every sum Y_ (P). Since this cannot be smaller than the least upper bound, we must have Vf(a, b) < Vf(a, c) + Vf(c, b), and this completes the proof. 6.6 TOTAL VARIATION ON [a, x] AS A FUNCTION OF x Now we keep the function f and the left endpoint of the interval fixed and study the total variation as a function of the right endpoint. The additive property implies important consequences for this function. Theorem 6.12. Let f be of bounded variation on [a, b]. Let V be defined on [a, b] as follows: V(x) = Vf(a, x) if a < x <- b, V(a) = 0. Then: i) V is an increasing function on [a, b]. ii) V-f is an increasing function on [a, b]. Proof. If a < x < y < b, we can write Vf(a, y) = Vf(a, x) + Vf(x, y). This implies V(y) - V(x) = V f(x, y) z 0. Hence V(x) < V(y), and (i) holds. To prove (ii), let D(x) = V(x) - f(x) if x e [a, b]. Then, if a 5 x < y < b, we have D(y) - D(x) = V(y) - V(x) - [f(y) - f(x)] = Vf(x, Y) - [f(Y) - f(x)]. But from the definition of Vf(x, y) it follows that we have f(Y) - f(x) <- Vf(x, Y) This means that D(y) - D(x) >- 0, and (ii) holds. NOTE. For some functions f, the total variation Vf(a, x) can be expressed as an integral. (See Exercise 7.20.) 132 Functions of Bounded Variation and Rectifiable Curves Th. 6.13 6.7 FUNCTIONS OF BOUNDED VARIATION EXPRESSED AS THE DIFFERENCE OF INCREASING FUNCTIONS The following simple and elegant characterization of functions of bounded varia-tion is a consequence of Theorem 6.12. Theorem 6.13. Let f be defined on [a, b]. Then f is of bounded variation on [a, b] if, and only if, f can be expressed as the difference of two increasing functions. Proof. If f is of bounded variation on [a, b], we can write f = V - D, where V is the function of Theorem 6.12 and D = V - f Both V and D are increasing functions on [a, b]. The converse follows at once from Theorems 6.5 and 6.9. The representation of a function of bounded variation as a difference of two increasing functions is by no means unique. If f = fl - f2, where fi and f2 are increasing, we also have f = (f1 + g) - (f2 + g), where g is an arbitrary in-creasing function, and we get a new representation off. If g is strictly increasing, the same will be true of fi + g and f2 + g. Therefore, Theorem 6.13 also holds if "increasing" is replaced by "strictly increasing." 6.8 CONTINUOUS FUNCTIONS OF BOUNDED VARIATION Theorem 6.14. Let f be of bounded variation on [a, b]. If x e (a, b], let V(x) = Vf(a, x) and put V(a) = 0. Then every point of continuity off is also a point of continuity of V. The converse is also true. Proof. Since V is monotonic, the right- and lefthand limits V(x+) and V(x-) exist for each point x in (a, b). Because of Theorem 6.13, the same is true of f(x+) andf(x-). If a < x < y < b, then we have [by definition of Vf(x, y)] 0 <- If(y) - f(x)1 <- V(y) - V(x). Letting y --> x, we find 0 < I f(x+) - f(x)1 < V(x+) - V(x). Similarly, 0 < If(x) - f(x-)I < V(x) - V(x-). These inequalities imply that a point of continuity of V is also a point of continuity off. To prove the converse, let f be continuous at the point c in (a, b). Then, given s > 0, there exists a S > 0 such that 0 < Ix - cl < S implies If(x) - f(c)I < c/2. For this same s, there also exists a partition P of [c, b], say P = {x0, x1i ... , x = b, such that n Vf(c, b) - 2 < E lofkf. Th. 6.15 Curves and Paths 133 Adding more points to P can only increase the sum Y_ I Afk I and hence we can assume that 0 < x1 - x0 < 6. This means that JAf1I = if(x1) - f(c)I < 2 , and the foregoing inequality now becomes n Vf(c,b)-2<2+ ElAfk1 <-2+Vf(x1,b), since {x1, x2, ... , x"} is a partition of [x1, b]. We therefore have Vf(c, b) - Vf(x1, b) < s. But 0 < V(x1) - V(c) = V1(a, x1) - Vf(a, c) = V1(c, x1) = Vf(c, b) - Vf(x1i b) < e. Hence we have shown that 0 < x1 - c < 6 implies 0 < V(x1) - V(c) < E. This proves that V(c+) = V(c). A similar argument yields V(c-) = V(c). The theorem is therefore proved for all interior points of [a, b]. (Trivial modifications are needed for the endpoints.) Combining Theorem 6.14 with 6.13, we can state Theorem 6.15. Let f be continuous on [a, b]. Then f is of bounded variation on [a, b] if, and only if, f can be expressed as the difference of two increasing continuous functions. NOTE. The theorem also holds if "increasing" is replaced by "strictly increasing." Of course, discontinuities (if any) of a function of bounded variation must be jump discontinuities because of Theorem 6.13. Moreover, Theorem 6.2 tells us that they form a countable set. 6.9 CURVES AND PATHS Let f : [a, b] -+ R" be a vector-valued function, continuous on a compact interval [a, b] in R. As t runs through [a, b], the function values f(t) trace out a set of points in R" called the graph of f or the curve described by f. A curve is a compact and connected subset of R" since it is the continuous image of a compact interval. The function f itself is called a path. It is often helpful to imagine a curve as being traced out by a moving particle. The interval [a, b] is thought of as a time interval and the vector f(t) specifies the position of the particle at time t. In this interpretation, the function f itself is called a motion. 134 Functions of Bounded Variation and Rectifiable Curves Def. 6.16 Different paths can trace out the same curve. For example, the two complex-valued functions f(t)=e2' , g(t)=a-2'1t, 0<t<1, each trace out the unit circle x2 + y2 = 1, but the points are visited in opposite directions. The same circle is traced out five times by the function h(t) = e"'11, 0 6.10 RECTIFIABLE PATHS AND ARC LENGTH Next we introduce the concept of arc length of a curve. The idea is to approximate the curve by inscribed polygons, a technique learned from ancient geometers. Our intuition tells us that the length of any inscribed polygon should not exceed that of the curve (since a straight line is the shortest path between two points), so the length of a curve should be an upper bound to the lengths of all inscribed polygons. Therefore, it seems natural to define the length of a curve to be the least upper bound of the lengths of all possible inscribed polygons. For most curves that arise in practice, this gives a useful definition of arc length. However, as we will see presently, there are curves for which there is no upper bound to the lengths of the inscribed polygons. Therefore, it becomes necessary to classify curves into two categories: those which have a length, and those which do not. The former are called rectifiable, the latter nonrectifiable. We now turn to a formal description of these ideas. Let f : [a, b] -+ R" be a path in R". For any partition of [a, b] given by P = {t0, t1, ... , tm}, the points f(to), f(t1), ... , are the vertices of an inscribed polygon. (An example is shown in Fig. 6.1.) The length of this polygon is denoted by Af(P) and is defined to be the sum M Af(P) = E, Ilf(tk) - f(tk-1)If. k=1 Definition 6.16. If the set of numbers Af(P) is bounded for all partitions P of [a, b], then the path f is said to be rectifiable and its arc length, denoted by Af(a, b), is a = tp t1 t2 t3 t4 t5 t$ = b Th. 6.18 Additive and Continuity Properties of Arc Length 135 defined by the equation Af(a, b) = sup {Af(P) : P e P'[a, b]}. If the set of numbers Af(P) is unbounded, f is called nonrectiflable. It is an easy matter to characterize all rectifiable curves. Theorem 6.17. Consider a path f : [a, b] -+ R" with components f = (f1, ... , f is rectifiable if, and only if, each component fk is of bounded variation on [a, b]. If f is rectifiable, we have the inequalities Vk(a, b) < Af(a, b) < V1(a, b) + ... + V"(a, b), (k = 1, 2, ... , n), (2) where Vk(a, b) denotes the total variation of fk on [a, b]. Proof If P = {to, t1, ... , t,"} is a partition of [a, b] we have m n I fk(t) - fk(ti-1)I < Af(P) < E E I fj(ti) - fj(ti-1)I , (3) i=1 i=1 j=1 for each k. All assertions of the theorem follow easily from (3). Examples 1. As noted earlier, the function given by R x) = x cos {nl(2x) } for x A 0, f (O) = 0, is continuous but not of bounded variation on [0, 1 ]. Therefore its graph is a non-rectifiable curve. 2. It can be shown (Exercise 7.21) that if f' is continuous on [a, b], then f is rectifiable and its arc length can be expressed as an integral, Ar(a, b) = Ilf'(t)II dt. b I 6.11 ADDITIVE AND CONTINUITY PROPERTIES OF ARC LENGTH Let f = (fl, ... , f") be a rectifiable path defined on [a, b]. Then each component fk is of bounded variation on every subinterval [x, y] of [a, b]. In this section we keep f fixed and study the arc length Af(x, y) as a function of the interval [x, y]. First we prove an additive property. Theorem 6.18. If c e (a, b) we have Af(a, b) = Af(a, c) + Af(c, b). Proof Adjoining the point c to a partition P of [a, b], we get a partition P1 of [a, c] and a partition P2 of [c, b] such that Af(P) < Af(P1) + Af(P2) <- Af(a, c) + Af(c, b). This implies Af(a, b) < Af(a, c) + Af(c, b). To obtain the reverse inequality, let P1 and P2 be-arbitrary partitions of [a, c] and [c, b], respectively. Then P= P1UP2, 136 Functions of Bounded Variation and Rectifiable Curves Th. 6.19 is a partition of [a, b] for which we have Af(P1) + Af(P2) = Af(P) <- Af(a, b). Since the supremum of all sums Af(P1) + Af(P2) is the sum Af(a, c) + Af(c, b) (see Theorem 1.15), the theorem follows. Theorem 6.19. Consider a rectifiable path f defined on [a, b]. If x e (a, b], let s(x) = Af(a, x) and let s(a) = 0. Then we have: i) The function s so defined is increasing and continuous on [a, b]. ii) If there is no subinterval of [a, b] on which f is constant, then s is strictly in-creasing on [a, b]. Proof. If a 5 x < y < b, Theorem 6.18 implies s(y) - s(x) = Af(x, y) >- 0. This proves that s is increasing on [a, b]. Furthermore, we have s(y) - s(x) > 0 unless Af(x, y) = 0. But, by inequality (2), Af(x, y) = 0 implies Vk(x, y) = 0 for each k and this, in turn, implies that f is constant on [x, y]. Hence (ii) holds. To prove that s is continuous, we use inequality (2) again to write n 0 _< s(y) - s(x) = Af(x, y) _< E Vk(x, y) k=1 If we let y -+ x, we find each term Vk(x, y) --+ 0 and hence s(x) = s(x+). Similarly, s(x) = s(x-) and the proof is complete. 6.12 EQUIVALENCE OF PATHS. CHANGE OF PARAMETER This section describes a class of paths having the same graph. Let f : [a, b] -+ R" be a path in R". Let u : [c, d] -+ [a, b] be a real-valued function, continuous and strictly monotonic on [c, d] with range [a,, b]. Then the composite function g = f o u given by g(t) = f[u(t)] for c < t < d, is a path having the same graph as f. Two paths f and g so related are called equivalent. They are said to provide different parametric representations of the same curve. The function u is said to define a change of parameter. Let C denote the common graph of two equivalent paths f and g. If u is strictly increasing, we say that f and g trace out C in the same direction. If u is strictly decreasing, we say that f and g trace out C in opposite directions. In the first case, u is said to be orientation preserving; in the second case, orientation-reversing. Theorem 6.20. Let f : [a, b] -+ R" and g : [c, d] -+ W be two paths in R", each of which is one-to-one on its domain. Then f and g are equivalent if, and only if, they have the same graph. Proof. Equivalent paths necessarily have the same graph. To prove the converse, assume that f and g have the same graph. Since f is one-to-one and continuous on Exercises 137 the compact set [a, b], Theorem 4.29 tells us that f -1 exists and is continuous on its graph. Define u(t) = f-i[g(t)] if t e [c, d]. Then u is continuous on [c, d] and g(t) = f[u(t)]. The reader can verify that u is strictly monotonic, and hence f and g are equivalent paths. EXERCISES Functions of bounded variation 6.1 Determine which of the following functions are of bounded variation on [0, 1 ]. a) f (x) = x2 sin (1 /x) if x : 0, f (O) = 0. b) f(x) = sin (1/x) if x - 0, f(0) = 0. 6.2 A function f, defined on [a, b], is said to satisfy a uniform Lipschitz condition of order a > 0 on [a, b] if there exists a constant M > 0 such that l f (x) - f(y) l < Mix - yl' for all x and y in [a, b]. (Compare with Exercise 5.1.) a) If f is such a function, show that a > I implies f is constant on [a, b], whereas a = 1 implies f is of bounded variation [a, b]. b) Give an example of a function f satisfying a uniform Lipschitz condition of order a < 1 on [a, b] such that f is not of bounded variation on [a, b]. c) Give an example of a function f which is of bounded variation on [a, b] but which satisfies no uniform Lipschitz condition on [a, b]. 6.3 Show that a polynomial f is of bounded variation on every compact interval [a, b ]. Describe a method for finding the total variation off on [a, b] if the zeros of the derivative f' are known. 6.4 A nonempty set S of real-valued functions defined on an interval [a, b] is called a linear space of functions if it has the following two properties: a) If f e $, then cf e S for every real number c. b)-If f e S and g e S, then f + geS. Theorem 6.9 shows that the set V of all functions of bounded variation on [a, b ] is a linear space. If S is any linear space which contains all monotonic functions on [a, b], prove that V s S. This can be described by saying that the functions of bounded variation form the smallest linear space containing all monotonic functions. 6.5 Let f be a real-valued function defined on [0, 1 ] such that f(0) > 0, f(x) 96 x for all x, and f (x) <- f (y) whenever x <- y. Let A = {x : f (x) > x). Prove that sup A e A and that f(1) > 1. 6.6 If f is defined everywhere in R1, then f is said to be of bounded variation on (- oo, + oo) if f is of bounded variation on every finite interval and if there exists a positive number M such that Vf(a, b) < M for all compact intervals [a, b ]. The total variation of f on (- co, + oo) is then defined to be the sup of all numbers Vf(a, b), - co < a < b < + oo, and is denoted by Vf(- oo, + oo). Similar definitions apply to half-open infinite intervals [a, +oo) and (-oo, b]. a) State and prove theorems for the infinite interval (- oo, + co) analogous to Theorems 6.7, 6.9, 6.10, 6.11, and 6.12. 138 Functions of Bounded Variation and Rectifiable Curves b) Show that Theorem 6.5 is true for (- oo, + oo) if "monotonic" is replaced by "bounded and monotonic." State and prove a similar modification of Theorem 6.13. 6.7 Assume that f is of bounded variation on [a, b ] and let P = As usual, write Afk = f(xk) - f(xk_1), k = 1, 2, ... , n. Define A(P) = {k : Afk > 0}, B(P) = {k : Afk < 0}. The numbers eY[a, b] pf(a, b) = sup (Afk:P and (nf(a, b) = sup { E IA.fkI : P e b]) are called, respectively, the positive and negative variations off on [a, b]. For each x in (a, b], let V(x) = Vf(a, x), p(x) = pf(a, x), n(x) = nf(a, x), and let V(a) = p(a) _ n(a) = 0. Show that we have: a) V(x) = p(x) + n(x). b) 0 < p(x) < V(x) and 0 < n(x) <- V(x). c) p and n are increasing on [a, b]. d) f(x) = f(a) + p(x) - n(x). Part (d) gives an alternative proof of Theorem 6.13. e) 2p(x) = V(x) + f (x) - f (a), 2n(x) = V(x) - f (x) + f (a). f) Every point of continuity off is also a point of continuity of p and of n. Curves 6.8 Let f and g be complex-valued functions defined as follows: f(t) = e2nu if t E [0, 1], g(t) = eel t if t e [0, 2]. a) Prove that f and g have the same graph but are not equivalent according to the definition in Section 6.12. b) Prove that the length of g is twice that of I. 6.9 Let f be a rectifiable path of length L defined on [a, b], and assume that f is not constant on any subinterval of [a, b]. Let s denote the arc-length function given by s(x) = Af(a, x) if a < x < b, s(a) = 0. a) Prove that s-1 exists and is continuous on [0, L]. b) Define g(t) = f [s-1(t)] if t e [0, L] and show that g is equivalent to f. Since f(t) = g[s(t)], the function g is said to provide a representation of the graph of f with arc length as parameter. 6.10 Let f and g be two real-valued continuous functions of bounded variation defined on [a, b], with 0 < f(x) < g(x) for each x in (a, b), f(a) = g(a), f(b) = g(b). Let h be the complex-valued function defined on the interval [a, 2b - a] as follows: h(t)=t+if(t), if a<_t<_b, h(t)=2b-t+ig(2b-t), ifb<-t<-2b-a. References 139 a) Show that h describes a rectifiable curve F. b) Explain, by means of a sketch, the geometric relationship between f, g, and h. c) Show that the set of points S = {(x, Y) : a <- x <- b, .f(x) 5 y <- g(x)} is a region in RI whose boundary is the curve F. d) Let H be the complex-valued function defined on [a, 2b - a] as follows: H(t) = t - 4i [g(t) - f(t) ], if a -< t 5 b, H(t)=t++i[g(2b-t)-f(2b-t)], if b:t52b-a. Show that H describes a rectifiable curve I'0 which is the boundary of the region So = {(x, y) : a 5 x -< b, f(x) - g(x) <- 2y <- g(x) - f(x)). e) Show that So has the x-axis as a line of symmetry. (The region So is called the symmetrization of S with respect to the x-axis.) f) Show that the length of I'o does not exceed the length of F. Absolutely continuous functions A real-valued function f defined on [a, b ] is said to be absolutely continuous on [a, b ] if for every c > 0 there is a S > 0 such that n E, If(bk) - f(ak)I < e k=1 for every n disjoint open subintervals (ak, bk) of [a, b], n = 1, 2, ... , the sum of whose lengths Y_k=1 (bk - ak) is less than S. Absolutely continuous functions occur in the Lebesgue theory of integration and differentiation. The following exercises give some of their elementary properties. 6.11 Prove that every absolutely continuous function on [a, b] is continuous and of bounded variation on [a, b]. NOTE. There exist functions which are continuous and of bounded variation but not absolutely continuous. 6.12 Prove that f is absolutely continuous if it satisfies a uniform Lipschitz condition of order 1 on [a, b]. (See Exercise 6.2.) 6.13 If f and g are absolutely continuous on [a, b], prove that each of the following is also: If I, cf (c constant), f + g, f g; also fig ifg is bounded away from zero. SUGGESTED REFERENCES FOR FURTHER STUDY 6.1 Apostol, T. M., Calculus, Vol. 1, 2nd ed. Xerox, Waltham, 1967. 6.2 Natanson, I. P., Theory of Functions of a Real Variable, Vol. 1, rev. ed. Leo F. Boron, translator. Ungar, New York, 1961. CHAPTER 7 THE RIEMANN-STIELTJES INTEGRAL 7.1 INTRODUCTION Calculus deals principally with two geometric problems: finding the tangent line to a curve, and finding the area of a region under a curve. The first is studied by a limit process known as differentiation; the second by another limit process-integration-to which we turn now. The reader will recall from elementary calculus that to find the area of the region under the graph of a positive function f defined on [a, b], we subdivide the interval [a, b] into a finite number of subintervals, say n, the kth subinterval having length Axk, and we consider sums of the form Ek=1 f(tk) Axk, where tk is some point in the kth subinterval. Such a sum is an approximation to the area by means of rectangles. If f is sufficiently well behaved in [a, b]-continuous, for example-then there is some hope that these sums will tend to a limit as we let n - oo, making the successive subdivisions finer and finer. This, roughly speaking, is what is involved in Riemann's definition of the definite integral f; f(x) dx. (A precise definition is given below.) The two concepts, derivative and integral, arise in entirely different ways and it is a remarkable fact indeed that the two are intimately connected. If we consider the definite integral of a continuous function f as a function of its upper limit, say we write F(x) = Jf(t) dt, a then F has a derivative and F'(x) = f(x). This important result shows that differentiation and integration are, in a sense, inverse operations. In this chapter we study the process of integration in some detail. Actually we consider a more general concept than that of Riemann: this is the Riemann-Stieltjes integral, which involves two functions f and a. The symbol for such an integral is f; f(x) da(x), or something similar, and the usual Riemann integral occurs as the special case in which a(x) = x. When a has a continuous derivative, the definition is such that the Stieltjes integral 1b. f(x) da(x) becomes the Riemann integral f; f(x) a'(x) dx. However, the Stieltjes integral still makes sense when a is not differentiable or even when a is discontinuous. In fact, it is in dealing with discontinuous a that the importance of the Stieltjes integral becomes apparent. By a suitable choice of a discontinuous a, any finite or infinite sum can be expressed as a Stieltjes integral, and summation and ordinary Riemann integration then 140 Def. 7.1 Definition of Riemann-Stieltjes Integral 141 become special cases of this more general process. Problems in physics which involve mass distributions that are partly discrete and partly continuous can also be treated by using Stieltjes integrals. In the mathematical theory of probability this integral is a very useful tool that makes possible the simultaneous treatment of continuous and discrete random variables. In Chapter 10 we discuss another generalization of the Riemann integral known as the Lebesgue integral. 7.2 NOTATION For brevity we make certain stipulations concerning notation and terminology to be used in this chapter. We shall be working with a compact interval [a, b] and, unless otherwise stated, all functions denoted by f, g, a, fi, etc., will be assumed to be real-valued functions defined and bounded on [a, b]. Complex-valued functions are dealt with in Section 7.27, and extensions to unbounded functions and infinite intervals will be discussed in Chapter 10. As in Chapter 6, a partition P of [a, b] is a finite set of points, say P = {x0, x1, ... , xn}, such that a = x0 < x1 < < xn_1 < x = b. A partition P' of [a, b] is said to be finer than P (or a refinement of P) if P c P', which we also write P' 2 P. The symbol Aak denotes the difference Aak = a(xk) - a(xk_ 1), so that n E Aak = a(b) - oc(a). k=1 The set of all possible partitions of [a, b] is denoted by 9[a, b]. The norm of a partition P is the length of the largest subinterval of P and is denoted by 11P11. Note that P' 2 P implies IIP'II < IIPII That is, refinement of a partition decreases its norm, but the converse does not necessarily hold. 7.3 THE DEFINITION OF THE RIEMANN-STIELTJES INTEGRAL Definition 7.1. Let P = {x0, x1, ... , xn} be a partition of [a, b] and let tk be a point in the subinterval [xk_ 1, xk]. A sum of the form S(P, f, a) = L J (tk) Aak k=1 is called a Riemann-Stieltjes sum off with respect to a. We say f is Riemann-integrable with respect to a on [a, b], and we write "f a R(a) on [a, b]" if there exists a number A having the following property: For every s > 0, there exists a partition Pa of [a, b] such that for every partition P finer than P. and for every choice of the points tk in [xk_ 1, xk], we have I S(P, f, a) - Al < e. 142 The Riemann-Stieltjes Integral Th. 7.2 When such a number A exists, it is uniquely determined and is denoted by f a f da or by f; f(x) da(x). We also say that the Riemann-Stieltjes integral f a f da exists. The functions f and a are referred to as the integrand and the integrator, respectively. In the special case when a(x) = x, we write S(P, f) instead of S(P, f, a), and f e R instead off e R(a). The integral is then called a Riemann integral and is denoted by fQ f dx or by fQ f(x) dx. The numerical value of f a f(x) da(x) depends only on f, a, a, and b, and does not depend on the symbol x. The letter x is a "dummy variable" and may be replaced by any other convenient symbol. NOTE. This is one of several accepted definitions of the Riemann-Stieltjes integral. An alternative (but not equivalent) definition is stated in Exercise 7.3. 7.4 LINEAR PROPERTIES It is an easy matter to prove that the integral operates in a linear fashion on both the integrand and the integrator. This is the context of the next two theorems. Theorem 7.2. If f e R(a) and if g e R(a) on [a, b], then c1 f + c2g a R(a) on [a, b] (for any two constants cl and c2) and we have b b b f (c1 f + c2g) da = c1 f f dot + c2 g da. Jo I Proof. Let h = c1 f + c2g. Given a partition P of [a, b], we can write n nn nn j S(P, h, a) = E h(tk) Dak = c1 E f(tk) eak + C2 E 9(tk) eak k=1 k=1 k=1 = c1S(P, f, a) + c2S(P, g, a). Given e > 0, choose PE so that P 2 PE implies I S(P, f, a) - fQ f dal < e, and choose PE so that P ? Pe implies I S(P, g, a) - fQ g dal < e. If we take PE = P' u P then, for P finer than PE, we have S(P, h, a) - c1 - c2 g daIdle + Ic2Ic, and this proves the theorem. Theorem 7.3. If f E R(a) and f e R(fl) on [a, b], then f E R(c1cc + on [a, b] (for any two constants cl and c2) and we have fd(ca+c2)=c1Jfda+c2Jbfdfi. Ja r a a The proof is similar to that of Theorem 7.2 and is left as an exercise. A result somewhat analogous to the previous two theorems tells us that the integral is also additive with respect to the interval of integration. Def. 7.5 Linear Properties 143 Theorem 7.4. Assume that c e (a, b). If two of the three integrals in (1) exist, then the third also exists and we have b b fda + I fda I = fda. (1) o c Proof. If P is a partition of [a, b] such that c e P, let P' = P n [a, c] and P" = P n [c, b], denote the corresponding partitions of [a, c] and [c, b], respectively. The Rie-mann-Stieltjes sums for these partitions are connected by the equation S(P, f, a) = S(P', f, a) + S(P", f, a). Assume that f a f da and f b f da exist. Then, given e > 0, there is a partition P' of [a, c] such that S(P', f, a) - f f da I < a whenever P' is finer than PE, a and a partition P" of [c, b] such that b S(P", f, a) - r f da < whenever P" is finer than P. 2 .J Then P. = P' u P' is a partition of [a, b] such that P finer than P. implies P' 2 P and P" 2 P. Hence, if P is finer than Pe, we can combine the foregoing results to obtain the inequality I S(P, f, a) -f c f doe - fb f da <e. This proves that fa f da exists and equals f; f da + f,, f da. The reader can easily verify that a similar argument proves the theorem in the remaining cases. Using mathematical induction, we can prove a similar result for a decomposi-tion of [a, b] into a finite number of subintervals. NOTE. The preceding type of argument cannot be used to prove that the integral J .c f da exists whenever J .b f da exists. The conclusion is correct, however. For integrators a of bounded variation, this fact will later be proved in Theorem 7.25. Definition 7.5. If a < b, we define f'bf da = - f; f da whenever 1b. f da exists. We also define P. f da = 0. The equation in Theorem 7.4 can now be written as follows : c a fda+ fda+ f fda=0. Sa' lb c 144 The Riemann-Stieltjes Integral 7.5 INTEGRATION BY PARTS Th. 7.6 A remarkable connection exists between the integrand and the integrator in a Riemann-Stieltjes integral. The existence of f a f doe implies the existence of I.' a df, and the converse is also true. Moreover, a very simple relation holds between the two integrals. Theorem 7.6. If f e R(a) on [a, b], then a e R(f) on [a, b] and we have da(x) + a(x) d f(x) f (x) = f (b)a(b) - f (a)a(a). f b b a NOTE. This equation, which provides a kind of reciprocity law for the integral, is known as the formula for integration by parts. Proof. Let e > 0 be given. Since fa f dot exists, there is a partition P. of [a, b] such that for every P' finer than PE, we have S(P', f, a) -J 6 f da a < a. (2) Consider an arbitrary Riemann-Stieltjes sum for the integral f; a df, say n // nn / n S(P, a, f) = E a(tk) Afk = E a(tk)f(xk) - E a(tk)f(xk-1), k=1 k=1 k=1 where P is finer than P. Writing A = f(b)a(b) - f(a)a(a), we have the identity A = Ef(xk)a(xk) - Ef(xk-1)a(xk-1) k=1 k=1 Subtracting the last two displayed equations, we find n n A - S(P, acf) = r L.rf(xk)[a(xk) - a(tk)] + Ef(xk-1)[a(tk) - a(xk-1)] k=1 k=1 The two sums on the right can be combined into a single sum of the form S(P',f, a), where P' is that partition of [a, b] obtained by taking the points xk and tk together. Then P' is finer than P and hence finer than P. Therefore the inequality (2) is valid and this means that we have A - S(P, at, f) -f b f doe a < a, whenever P is finer than P. But this is exactly the statement that fa a df exists and equals A - $a f doc. 7.6 CHANGE OF VARIABLE IN A RIEMANN-STIELTJES INTEGRAL Theorem 7.7. - Let f e R(a) on [a, b] and let g be a strictly monotonic continuous function defined on an interval S having endpoints c and d. Assume that a = g(c), Th. 7.7 Reduction to a Riemann Integral b = g(d). Let h and fl be the composite functions defined as follows: h(x) = f [g(x)], /3(x) = a[g(x)], if x e S. Then h e R(fl) on S and we have f .b f da = f d h dQ. That is, 145 f J g(c) f(t) da(t) = J df [9(x)] d{a[9(x)]}. 9(c) a Proof. For definiteness, assume that g is strictly increasing on S. (This implies c < d.) Then g is one-to-one and has a strictly increasing, continuous inverse g-1 defined on [a, b]. Therefore, for every partition P = {yo, ... , yn} of [c, d], there corresponds one and only one partition P' _ {xo, ... , of [a, b] with xk = g(yk). In fact, we can write P' = g(P) and P = g -'(P'). Furthermore, a refinement of P produces a corresponding refinement of P', and the converse also holds. If c > 0 is given, there is a partition PE of [a, b] such that P' finer than PE implies I S(P', f, a) - f a f daJ < c. Let P. = g 1(PE) be the corresponding par-tition of [c, d], and let P = {yo, ... , yn} be a partition of [c, d] finer than P. Form a Riemann-Stieltjes sum S(P, h, h(uk) NN., k=1 where uk e [yk- 1, yk] and AI3k = N(Yk) - /3(Yk- 1). If we put tk = g(uk) and xk = g(yk), then P' = {x0, ... , xn} is a partition of [a, b] finer than P. Moreover, we then have S(P, h, /3) _ f[9(uk)]{a[9(Yk)] - a[9(Yk-1)]} k=1 nn E f(tk){a(xk) - a(xk- 1)} = S(P', f, a), k=1 since tk a [xk_ 1, xk]. Therefore, (S(P, h, /3) - J. 'f dal < a and the theorem is proved. NOTE. This theorem applies, in particular, to Riemann integrals, that is, when a(x) = x. Another theorem of this type, in which g is not required to be mono-tonic, will later be proved for Riemann integrals. (See Theorem 7.36.) 7.7 REDUCTION TO A RIEMANN INTEGRAL The next theorem tells us that we are permitted to replace the symbol da(x) by a'(x) dx in the integral f; f(x) da(x) whenever a has a continuous derivative a'. 146 The Riemann-Stieltjes Integral Th. 7.8 Theorem 7.8. Assume f e R(a) on [a, b] and assume that a has a continuous derivative a' on [a, b]. Then the Riemann integral f; f(x)a'(x) dx exists and we have f b b f (x) doe(x) = f f (x)a'(x) dx. Ja a Proof. Let g(x) = f(x)a'(x) and consider a Riemann sum nn nn S(P, g) = E g(tk) Axk = L/ f(tk)a'(tk) Oxk. k=1 k=1 The same partition P and the same choice of the tk can be used to form the Riemann-Stieltjes sum S(P, f, a) = L/ f(tk) Dak. k=1 Applying the Mean-Value Theorem, we can write Dak = a'(Vk) Oxk, where vk a (xk_ 1, xk), and hence n S(P,f, a) - S(P, g) _ f(tk)[a'(vk) - a'(tk)] AXk. k=1 Since f is bounded, we have I f(x)I < M for all x in [a, b], where M > 0. Con-tinuity of a' on [a, b] implies uniform continuity on [a, b]. Hence, if a > 0 is given, there exists a S > 0 (depending only on s) such that 0 < Ix - yI < 6 implies Ia'(x) - a'(y)I < E 2M(b - a) If we take a partition P' with norm IIPEII < 6, then for any finer partition P we will have I a'(vk) - a'(tk)I < sl[2M(b - a)] in the preceding equation. For such P we therefore have I S(P, f, a) - S(P, g)I < 2 On the other hand, since f e R(a) on [a, b], there exists a partition P8 such that P finer than P E implies S(P,f,a) - Ja'f dal < 8. Combining the last two inequalities, we see that when P is finer than Pt = P' v P we will have I S(P, g) - $ f dal < a, and this proves the theorem. NOTE. A stronger result not requiring continuity of a' is proved in Theorem 7.35. Th. 7.9 Step Functions as Integrators 147 7.8 STEP FUNCTIONS AS INTEGRATORS If a is constant throughout [a, b], the integral f a f dot exists and has the value 0, since each sum S(P, f, a) = 0. However, if a is constant except for a jump dis-continuity at one point, the integral f o f da need not exist and, if it does exist, its value need not be zero. The situation is described more fully in the following theorem : Theorem 7.9. Given a < c < b. Define a on [a, b] as follows: The values a(a), a(c), a(b) are arbitrary; and a(x) = a(a) a(x) = a(b) if a < x < c, if c < x < b. Let f be defined on [a, b] in such a way that at least one of the functions for a is continuous from the left at c and at least one is continuous from the right at c. Then f E R(a) on [a, b] and we have fa f da f(c)[a(c+) - a(cNOTE. The result also holds if c = a, provided that we write a(c) for a(c-), and it holds for c = b if we write a(c) for a(c+). We will prove later (Theorem 7.29) that the integral does not exist if both f and a are discontinuous from the right or from the left at c. Proof. If c e P, every term in the sum S(P, f, a) is zero except the two terms arising from the subinterval separated by c, say S(P, f, a) = f(tk- i)[a(c) - a(c-)] + .f(tk)[a(c+) - a(c)], where tk-1 < c < tk. This equation can also be written as follows: A = [f(tk-1) - f(c)][a(c) - a(c-)] + [f(tk) - f(c)][a(c+) - a(c)], where A = S(P, f, a) - f(c)[a(c+) - a(c-)]. Hence we have IAI s If(tk-1) - f(c)I Ice(c) - a(c-)I + If(tk) - f(c)I I a(c+) - a(c)I. If f is continuous at c, for every s > 0 there is a S > 0 such that IIPII < S implies If(tk-1) - f(c) I < a and If(tk) - f(c)I < s. In this case, we obtain the inequality IDI < ela(c) - a(c-)I + sla(c+) - a(c)I. But this inequality holds whether or not f is continuous at c. For example, if f is discontinuous both from the right and from the left at c, then a(c) = a(c-) and a(c) = a(c+) and we get 0 = 0. On the other hand, if f is continuous from the left and discontinuous from the right at c, we must have a(c) = a(c+) and we get 148 The Riemann-Stieltjes Integral Def. 7.10 JAI < sIa(c) - a(c-)I. Similarly, if f is continuous from the right and discon-tinuous from the left at c, we have a(c) = a(c-) and JAI < ela(c+) - a(c)I. Hence the last displayed inequality holds in every case. This proves the theorem. Example. Theorem 7.9 tells us that the value of a Riemann-Stieltjes integral can be altered by changing the value off at a single point. The following example shows that the existence of the integral can also be affected by such a change. Let a(x) = 0, if x ;6 0, a(0) = - 1, f(x) = 1, if -1 5 x < +1. In this case Theorem 7.9 implies f'_ , f dot = 0. But if we re-define f so that f(0) = 2 and f(x) = 1 if x # 0, we can easily see that f 1 , f da will not exist. In fact, when P is a par-tition which includes 0 as a point of subdivision, we find S(P,11a) = f(tk) [0k) - a(0)1 + f(tk -1) [a(0) -a(xk-2) ] = f(tk) - f(tk-1), where xk_2 < tk-1 < 0 - tk < xk. The value of this sum is 0, 1, or -1, depending on the choice of tk and th_1. Hence, J1, f da does not exist in this case. However, in a Riemann integral fo f(x) dx, the values of f can be changed at a finite number of points without affecting either the existence or the value of the integral. To prove this, it suffices to consider the case where f(x) = 0 for all x in [a, b] except for one point, say x = c. But for such a function it is obvious that IS(P, f)I <- If(c)I IPII Since IIPII can be made arbitrarily small, it follows that fa f(x) dr = 0. 7.9 REDUCTION OF A RIEMANN-STIELTJES INTEGRAL TO A FINITE SUM The integrator a in Theorem 7.9 is a special case of an important class of functions known as step functions. These are functions which are constant throughout an interval except for a finite number of jump discontinuities. Definition 7.10 (Step function). A function a defined on [a, b] is called a step function if there is a partition a = x, <x2 b such that a is constant on each open subinterval (xk_1, xk). The number a(xk+) -a(xk-) is called the jump at Xk if 1 < k < n. The jump at x1 is a(xl+) - a(x1), and the jump at x,, is a(x - ). Step functions provide the connecting link between Riemann-Stieltjes integrals and finite sums: Theorem 7.11 (Reduction of a Riemann-Stieltjes integral to a finite sum). Let a be a step function d e f i n e d on [a, b] with j u m p ak at Xk, where x1, ... , x,, are as described in Definition 7.10. Let f be defined on [a, b] in such a way that not both f and a are Th. 7.13 Euler's Summation Formula 149 discontinuous from the right or from the left at each xk. Then f a f da exists and we have lab f(x) da(x) _ E {/ J (xk)ak k=1 Proof. By Theorem 7.4, J 'f f da can be written as a sum of integrals of the type considered in Theorem 7.9. One of the simplest step functions is the greatest-integer function. Its value at x is the greatest integer which is less than or equal to x and is denoted by [x]. Thus, [x] is the unique integer satisfying the inequalities [x] < x < [x] + 1. Theorem 7.12. Every finite sum can be written as a Riemann-Stieltjes integral. In fact, given a sum Ek=1 ak, define f on [0, n] as follows: f(x) = ak if k - 1 < x < k (k = 1, 2, ... , n), f(0) = 0. Then n n (' E ak = E f(k) = J f(x) d[x], k=1 k=1 0 where [x] is the greatest integer < x. Proof. The greatest-integer function is a step function, continuous from the right and having jump 1 at each integer. The function f is continuous from the left at 1, 2, ... , n. Now apply Theorem 7.11. 7.10 EULER'S SUMMATION FORMULA We shall illustrate the use of Riemann-Stieltjes integrals by deriving a remarkable formula known as Euler's summation formula, which relates the integral of a function over an interval [a, b] with the sum of the function values at the integers in [a, b]. It can sometimes be used to approximate integrals by sums or, conversely, to estimate the values of certain sums by means of integrals. Theorem 7.13 (Euler's summation formula). If f has a continuous derivative f' on [a, b], then we have E f(n) = $f(x) dx + ff'(x)((x)) dx + f (a)((a)) - f (b)((b)), a<n5b where ((x)) = x - [x]. When a and b are integers, this becomes b f(a) + f(b) = Sabf dx + ff'(x) (x - [x] -dx + \ J NOTE. >.<n_-5b means the sum from n = [a] + 1 to n = [b]. 150 The Riewann-Stieltjes Integral Th. 7.13 Proof. Applying Theorem 7.6 (integration by parts), we have fa f(x) d(x - [x]) + J b (x - [x]) df(x) = f(b)(b - [b]) - f(a)(a - [a]). a Since the greatest-integer function has unit jumps at the integers [a] + 1, [a] + 2, ... , [b], we can write ('b a f(x) d[x] = E f(n). a<n5b If we combine this with the previous equation, the theorem follows at once. 7.11 MONOTONICALLY INCREASING INTEGRATORS. UPPER AND LOWER INTEGRALS The further theory of Riemann-Stieltjes integration will now be developed for monotonically increasing integrators, and we shall see later (in Theorem 7.24) that for many purposes this is just as general as studying the theory for integrators which are of bounded variation. When a is increasing, the differences Dak which appear in the Riemann-Stieltjes sums are all nonnegative. This simple fact plays a vital role in the develop-ment of the theory. For brevity, we shall use the abbreviation "a i on [a, b]" to mean that "a is increasing on [a, b]." As stated earlier, to find the area of the region under the graph of a function f we consider Riemann sums E f(tk) Oxk as approximations to the area by means of rectangles. Such sums also arise quite naturally in certain physical problems requiring the use of integration for their solution. Another approach to these problems is by means of upper and lower Riemann sums. For example, in the case of areas, we can consider approximations from "above" and from "below" by means of the sums EMk Oxk and Emk Oxk, where Mk and Mk denote, respectively, the sup and inf of the function values in the kth subinterval. Our geometric intuition tells us that the upper sums are at least as big as the area we seek, whereas the lower sums cannot exceed this area. (See Fig. 7.1.) Therefore it seems natural a b Figure 7.1 Th. 7.15 Upper and Lower Integrals 151 to ask: What is the smallest possible value of the upper sums? This leads us to consider the inf of all upper sums, a number called the upper integral off. The lower integral is similarly defined to be the sup of all lower sums. For reasonable functions (for example, continuous functions) both these integrals will be equal to f a f(x) dx. However, in general, these integrals will be different and it becomes an important problem to find conditions on the function which will ensure that the upper and lower integrals will be the same. We now discuss this type of problem for Riemann-Stieltjes integrals. Definition 7.14. Let P be a partition of [a, b] and let Mk(f) = sup {f(x) : x e [Xk-1, Xk]}, mk(f) = inf {f(x) : x e [xk-1, Xk]}. The numbers n n U(P, f, a) _ Mk(f) Dak and L(P, f, a) = E mk(f) oak, k=1 k=1 are called, respectively, the upper and lower Stieltjes sums off with respect to a for the partition P. NOTE. We always have Mk(f) < Mk(f). If a, on [a, b], then Dak >_ 0 and we can also write Mk(f) Dak < Mk(f) Dak, from which it follows that the lower sums do not exceed the upper sums. Furthermore, if tk a [Xk_1, xk], then mk(f) <_ f(tk) <_ Mk(f). Therefore, when a T , we have the inequalities L(P, f, a) 5 S(P, f, a) < U(P, a) relating the upper and lower sums to the Riemann-Stieltjes sums. These inequali-ties, which are frequently used in the material that follows, do not necessarily hold when a is not an increasing function. The next theorem shows that, for increasing a, refinement of the partition increases the lower sums and decreases the upper sums. Theorem 7.15. Assume that a," on [a, b]. Then: i) If P' is finer than P, we have U(P',f, a) < U(P, f, a) and L(P', f, a) >_ L(P, f, a). ii) For any two partitions P1 and P2, we have L(P1, f, a) 5 U(P2i f, 4 152 The Riemann-Stieltjes Integral Def. 7.16 Proof It suffices to prove (i) when P' contains exactly one more point than P, say the point c. If c is in the ith subinterval of P, we can write U(P', f, a) = E Mk(f) Aak + M'[a(c) - a(xi_1)] + M"[a(xi) - a(c)], k=1 ki where M' and M" denote the sup off in [xi_ 1, c] and [c, x.]. But, since M' < M1(f) and M" < Mi(f), we have U(P',f, a) < U(P, f, a). (The inequality for lower sums is proved in a similar fashion.) To prove (ii), let P = P1 U P2. Then we have L(P1, f a) <_ L(P, f, a) < U(P, f, a) 5 U(P2, f, a) NOTE. It follows from this theorem that we also have (for increasing a) m[a(b) - a(a)] < L(P1, f a) < U(P2i.f, a) < M[a(b) - a(a)], where M and m denote the sup and inf off on [a, b]. Definition 7.16. Assume that aT on [a, b]. The upper Stieltjes integral off with respect to a is defined as follows: f. fda = inf {U(P, f, a) : P e .9[a, b]}. The lower Stieltjes integral is similarly defined: (b a fda = sup {L(P, f, a) : P e 9[a, b]}. NOTE. We sometimes write 1(f, a) and I(f a) for the upper and lower integrals. In the special case where a(x) = x, the upper and lower sums are denoted by U(P, f) and L(P, f) and are called upper and lower Riemann sums. The corre-sponding integrals, denoted by $f(x) dx and by f a f(x) dx, are called upper and lower Riemann integrals. They were first introduced by J. G. Darboux (1875). Theorem 7.17. Assume that a/ on [a, b]. Then I(f, a) < I(f, a). Proof. If e > 0 is given, there exists a partition P1 such that U(P1, f, a) < I(f, a) + e. By Theorem 7.15, it follows that I(f, a) + e is an upper bound to all lower sums L(P, f, a). Hence, I(f, a) < I(f, a) + e, and, since a is arbitrary, this implies I(.f, a) < 1(f a). Example. It is easy to give an example in which I(f, a) < I(f, a). Let a(x) = x and define f on [0, 1] as follows: f(x) = 1, if x is rational, f(x) = 0, if x is irrational. Th. 7.19 Riemann's Condition 153 Then for every partition P of [0, 1], we have Mk(f) = 1 and mk(f) = 0, since every subinterval contains both rational and irrational numbers. Therefore, U(P, f) = I and L(P, f) = 0 for all P. It follows that we have, for [a, b ] = 0, 11, I b b I f dx = 1 and f f dx = 0. a .Ja Observe that the same result holds if f(x) = 0 when x is rational, and f(x) = I when x is irrational. 7.12 ADDITIVE AND LINEARITY PROPERTIES OF UPPER AND LOWER INTEGRALS Upper and lower integrals share many of the properties of the integral. For ex-ample, we have bfda+ fb f da, Jb dcc = ra a a c if a < c < b, and the same equation holds for lower integrals. However, certain equations which hold for integrals must be replaced by inequalities when they are stated for upper and lower integrals. For example, we have f b (f + g) da < rbf dot + fb g da, a Ja and fa 6(f+g)da> fa' fda+ Jbgda. a T hese remarks can be easily verified by the reader. (See Exercise 7.11.) 7.13 RIEMANN'S CONDITION If we are to expect equality of the upper and lower integrals, then we must also expect the upper sums to become arbitrarily close to the lower sums. Hence it seems reasonable to seek those functions f for which the difference U(P, f, a) -L(P, f, a) can be made arbitrarily small. Definition 7.18. We say that f satisfies Riemann's condition with respect to of on [a, b] if, for every e > 0, there exists a partition P. such that P finer than PE implies 0 < U(P, f, a) - L(P, f, a) < e. Theorem 7.19. Assume that a ,;q on [a, b]. Then the following three statements are equivalent: _ i) f e R(a) on [a, b]. 154 The Riemann-Stieltjes Integral Th. 7.19 ii) f satisfies Riemann's condition with respect to a on [a, b]. iii) I(f, M) = I(f, a). Proof. We will prove that part (i) implies (ii), part (ii) implies (iii), and part (iii) implies (i). Assume that (i) holds. If a(b) = a(a), then (ii) holds trivially, so we can assume that a(a) < a(b). Given E > 0, choose PE so that for any finer P and all choices of tk and tk in [xk 1, xk], we have n E f(tk) eak - A k=1 < 3 and IkE f(tk) eak - Al < 3 , where A = f a f da. Combining these inequalities, we find nn E [f(tk) - f(tk)] eak k=1 2 <3a. Since Mk(f) - mk(f) = sup {f(x) - f(x') : x, x' in [xkxk]}, it follows that for every h > 0 we can choose tk and tk so that f(tk) - f(tk) > Mk(f) - mk(f) - h. Making a choice corresponding to h = 1E/[a(b) - a(a)], we can write n U(P, f, a) - L(P, f, a) = E [Mk(f) - mk(f )] eak k=1 Hence, (i) implies (ii). n n < E [f(tk) - f(tk)] Aak + h E eak < E. k=1 Next, assume that (ii) holds. If e > 0 is given, there exists a partition PE such that P finer than P. implies U(P, f a) < L(P, f, a) + E. Hence, for such P we have I(f,cc) < U(P,f a) <L(P,f a)+E <1(f cc)+E. That is, I(f, a) < I(f, a) + e for every e > 0. Therefore, I(f, a) < I(f a). But, by Theorem 7.17, we also have the, opposite inequality. Hence (ii) implies (iii). Finally, assume that 1(f a) = I(f, a) and let A denote their common value. We will prove that f 4 f da exists and equals A. Given e > 0, choose P' so that U(P, f, a) < 1(f a) + E for all P finer than F. Also choose P" such that L(P,f,cc)>I(f,cc)-E for all P finer than P". If PE = PE u P", we can write 1(f, a) - E < L(P, f, a) < S(P, f, a) < U(P, f, a) < 1(f, a) + E for every P finer than PE. But, since I(f a) = 1(f a) = A, this means that I S(P, f, a) - Al < E whenever P is finer than PE. This proves that f; f da exists and equals A, and the proof of the theorem is now complete. Th. 7.22 Comparison Theorems 155 7.14 COMPARISON THEOREMS Theorem 7.20. Assume that a.,- on [a, b]. If f E R(a) and g e R(a) on [a, b] and if f(x) < g(x) for all x in ['a, b], then we have J f(x) da(x) < rb g(x) da(x). a a Proof. For every partition P, the corresponding Riemann-Stieltjes sums satisfy n n S(P1 f, a) _ f(tk) Auk S g(tk) Aak = S(P, g, a), k=1 k=1 since a.- on [a, b]. From this the theorem follows easily. In particular, this theorem implies that f a g(x) da(x) >- 0 whenever g(x) > 0 and a i' on [a, b]. Theorem 7.21. Assume that a ,,x on [a, b]. If f e R(a) on [a, b], then If I e R(a) on [a, b] and we have the inequality f I ff(x) da(x)I < If(x)I da(x). Proof Using the notation of Definition 7.14, we can write Mk(f) - mk(f) = sup {f(x) - f(y) : x, y in [xk-1, xk]}. Since the inequality jIf(x)I - I.f(y)II < If(x) - f(y)I always holds, it follows that we have Mk(IfI) - mk(IfI) <_ Mk(f) - mk(f) Multiplying by Aak and summing on k, we obtain U(P, if I, a) - L(P, If I, cc) < U(P, f, a) - L(P, f, a), for every partition P of [a, b]. By applying Riemann's condition, we find that If I e R(a) on [a, b]. The inequality in the theorem follows by taking g = If I in Theorem 7.20. NOTE. The converse of Theorem 7.21 is not true. (See Exercise 7.12.) Theorem 7.22. Assume that aT on [a, b]. If f e R(a) on [a, b], then f2 e R(a) on [a, b]. Proof. Using the notation of Definition 7.14, we have Mk(f2) = [Mk(If I )]2 and mk(f 2) = [mk(I fl )]2. Hence we can write Mk(f2) - mk(f2) = [Mk(IfI) + mk(IfI)][Mk(IfI) - mk(IfI)] 2M[Mk(IfI) - mk(IfI)], 156 The Riemann-Stieltjes Integral Th. 7.23 where M is an upper bound for If I on [a, b]. By applying Riemann's condition, the conclusion follows. Theorem 7.23. Assume that a,- on [a, b]. If f e R(a) and g e R(a) on [a, b], then the product f- g e R(a) on [a, b]. Proof. We use Theorem 7.22 along with the identity 2f(x)g(x) = [f(x) + g(x)]2 - [f(x)]2 - [g(x)]2. 7.15 INTEGRATORS OF BOUNDED VARIATION In Theorem 6.13 we found that every function a of bounded variation on [a, b] can be expressed as the difference of two increasing functions. If a = al - a2 is such a decomposition and iff e R(a1) and f e R(a2) on [a, b], it follows by linearity that f e R(a) on [a, b]. However, the converse is not always true. If f e R(a) on [a, b], it is quite possible to choose increasing functions al and a2 such that or = al - a2, but such that neither integral J .'f dal, J 'f dal exists. The difficulty, of course, is due to the nonuniqueness of the decomposition a = al - a2. How-ever, we can prove that there is at least one decomposition for which the converse is true, namely, when al is the total variation of a and a2 = al - a. (Recall Definition 6.8.) Theorem 7.24. Assume that a is of bounded variation on [a, b]. Let V(x) denote the total variation of a on [a, x] if a < x < b, and let V(a) = 0. Let f be defined and bounded on [a, b]. If f e R(a) on [a, b], then f e R(V) on [a, b]. Proof If V(b) = 0, then V is contant and the result is trivial. Suppose therefore, that V(b) > 0. Suppose also that I f(x)I < M if x e [a, b]. Since V is increasing, we need only verify that f satisfies Riemann's condition with respect to V on [a, b]. Given e > 0, choose P, so that for any finer P and all choices of points tk and tk in [xk _ 1 i xk] we have . to [J (tk) - ! (tk)] Aak n < and V(b) < E Ioakl + e 4 k=1 4M For P finer than P. we will establish the two inequalities and E Mk(f) - mk(f) < 2 , nn E [Mk(f) - mk(f)] Ioakl < 2 which, by addition, yield U(P, f, V) - L(P, f, V) < e. Th. 7.25 Integrators of Bounded Variation To prove the first inequality, we note that AVk - IAakl >- 0 and hence nn r Lr ) - mk({ J) [Mk(J < 2M E (AVk - IOak!) k=1 k=1 = 214 CV(b) -IDakI) < e . k=1 2 157 To prove the second inequality, let A(P) = {k : oak >- 0}, B(P) = {k : Aak < 0}, and let h = c/V(b). If k e A(P), choose tk and tk so that f'(tk) - f(tk) > Mk(f) - mk(f) - h; but, if k e B(P), choose tk and tk so that f(tk) - f(tk) > Mk(f) - mk(f) - h. Then nn E [Mk({ J) - mk(f)] Ieakl < k=1 [f(tk) - f(tk)] Ieakl kEA(P) n + E [f(ek) - f(tk)] Ioakl + h E Ioakl keB(P) k=1 nn n E [f(tk) - f(tk)] eak + h : Ieakl k=1 k=1 < e+hV(b)= E+ E= E 4 4 4 2 It follows that f e R(V) on [a, b]. NOTE. This theorem (together with Theorem 6.12) enables us to reduce the theory of Riemann-Stieltjes integration for integrators of bounded variation to the case of increasing integrators. Riemann's condition then becomes available and it turns out to be a particularly useful tool in this work. As a first application we shall obtain a result which is closely related to Theorem 7.4. Theorem 7.25. Let a be of bounded variation on [a, b] and assume that f e R(a) on [a, b]. Then f e R(a) on every subinterval [c, d] of [a, b]. Proof Let V(x) denote the total variation of a on [a, x], with V(a) = 0. Then a = V - (V - a), where both V and V - a are increasing on [a, b] (Theorem 6.12). By Theorem 7.24, f e R(V), and hence f e R(V - a) on [a, b]. Therefore, if the theorem is true for increasing integrators, it follows that f e R(V) on [c, d] and f e R(V - a) on [c, d], so f e R(a) on [c, d]. 158 The Riemann-Stieltjes Integral Th. 7.26 Hence, it suffices to prove the theorem when a ,,x on [a, b]. By Theorem 7.4 it suffices to prove that each integral f a f da and f a f dot exists. Assume that a < c < b. If P is a partition of [a, x], let A(P, x) denote the difference 0(P, x) = U(P, f, a) - L(P, f, a), of the upper and lower sums associated with the interval [a, x]. Since f e R(a) on [a, b], Riemann's condition holds. Hence, if E > 0 is given, there exists a partition PE of [a, b] such that A(P, b) < E if P is finer than P. We can assume that c e P. The points of PE in [a, c] form a partition PE of [a, c]. If P' is a partition of [a, c] finer than PE, then P = P' u P. is a partition of [a, b] com-posed of the points of P' along with those points of PE in [c, b]. Now the sum defining 0(P', c) contains only part of the terms in the sum defining A(P, b). Since each term is > 0 and since P is finer than PE, we have A(P', c) < A(P, b) < e. That is, P' finer than Pe implies 0(P', c) < E. Hence, f satisfies Riemann's con-dition on [a, c] and fa f dot exists. The same argument, of course, shows that f a f da exists, and by Theorem 7.4 it follows that f " f dot exists. The next theorem is an application of Theorems 7.23, 7.21, and 7.25. Theorem 7.26. Assume f e R(a) and g e R(a) on [a, b], where o c,-on [a, b]. 'Define F(x) = xf(t) daft) Ix and a Then f e R(G), g e R(F), and the product f g e R(a) on [a, b], and we have f (x)g(x) da(x) = f (x) dG(x) b 6 6 = f g(x) dF(x). 0 Proof The integral f a f g da exists by Theorem 7.23. For every partition P of [a, b] we have ('xk ('xk S(P, f, G) f(tk) J g(t) da(t) J f(tk)g(t) da(t), k =1 xk -1 k= 1 xk-i G(x) = g(t) da(t) if x e [a, b]. fx and _ Jbf(x)g(x) da(x) _ E f(t)g(t) da(t). k=1 s:z t Th. 7.28 Sufficient Conditions for Existence 159 Therefore, if Mg jbf.g sup {Ig(x): x e [[a, b]}, we have S(P,f, G) -dal = Ik=1 Jxk 1 {f(tk) - f(t)}g(t) da(t) < Mg kE fXk"-xIf(tk) - f(t)I d«(t) < Mg rxk [M / da t 1 k=1 xk_1 = Mg{U(P, f, a) - L(P, f, a)}. Since f e R(a), for every E > 0 there is a partition PE such that P finer than P. implies U(P, f, a) - L(P, f, a) < e. This proves that f e R(G) on [a, b] and that f a f g da = J 'f f dG. A similar argument shows that g e R(F) on [a, b] and that f af g d o e = fa g dF. NOTE. Theorem 7.26 is also valid if a is of bounded variation on [a, b]. 7.16 SUFFICIENT CONDITIONS FOR EXISTENCE OF RIEMANN-STIELTJES INTEGRALS In most of the previous theorems we have assumed that certain integrals existed and then studied their properties. It is quite natural to ask : When does the integral exist? Two useful sufficient conditions will be obtained. Theorem 7.27. If f is continuous on [a, b] and if a is of bounded variation on [a, b], then f e R(a) on [a, b]. NOTE. By Theorem 7.6, a second sufficient condition can be obtained by inter-changing f and a in the hypothesis. Proof. It suffices to prove the theorem when a a with a(a) < a(b). Continuity of f on [a, b] implies uniform continuity, so that if e > 0 is given, we can find S > 0 (depending only on e) such that Ix - yI < S implies If(x) - f (y)I < E/A, where A = 2[a(b) - a(a)]. If P, is a partition with norm IIPEII < S, then for P finer than P. we must have Mk(f) - mk(f) <- E/A, since Mk(f) - mk(f) = sup {f(x) - f(y) : x, y in [xk xk]}. Multiplying the inequality by Dak and summing, we find n U(P, f, a) - L(P, f, a) < E Aak = E < E, A k=1 2 and we see that Riemann's condition holds. Hence, f e R(a) on [a, b]. For the special case in which a(x) = x, Theorems 7.27 and 7.6 give the following corollary : Theorem 7.28. Each of the following conditions is sufficient for the existence of the Riemann integral f; f(x) dx: a) f is continuous on [a, b]. b) f is of bounded variation on [a, b]. 160 The Riemann-Stieltjes Integral Th. 7.29 7.17 NECESSARY CONDITIONS FOR EXISTENCE OF RIEMANN-STIELTJES INTEGRALS When a is of bounded variation on [a, b], continuity off is sufficient for the exis-tence of Ja f da. Continuity off throughout [a, b] is by no means necessary, however. For example, in Theorem 7.9 we found that when cc is a step function, then f can be defined quite arbitrarily in [a, b] provided only that f is continuous at the discontinuities of a. The next theorem tells us that common discontinuities from the right or from the left must be avoided if the integral is to exist. Theorem 7.29. Assume that a,, on [a, b] and let a < c < b. Assume further that both a and f are discontinuous from the right at x = c; that is, assume that there exists an e > 0 such that for every S > 0 there are values of x and y in the interval (c, c + S) for which If(x) - f(c)1 >- e and la(y) - a(c)I > e. Then the integral f a f(x) da(x) cannot exist. The integral also fails to exist if a and f are discontinuous from the left at c. Proof. Let P be a partition of [a, b] containing c as a point of subdivision and form the difference n U(P,11 a) - L(P, f, a) = E [Mk(f) - mk(f)] Dak. k=1 If the ith subinterval has c as its left endpoint, then U(P, f, a) - L(P, f, a) >- [Mi(f) - m1(f)][a(x,) - a(c)], since each term of the sum is >- 0. If c is a common discontinuity from the right, we can assume that the point x; is chosen so that a(x,) - a(c) >- e. Furthermore, the hypothesis of the theorem implies M.(f) - mi(f) e. Hence, U(P, f, a) - L(P, f, a) >- e2, and Riemann's condition cannot be satisfied. (If c is a common discontinuity from the left, the argument is similar.) 7.18 MEAN-VALUE THEOREMS FOR RIEMANN-STIELTJES INTEGRALS Although integrals occur in a wide variety of problems, there are relatively few cases in which the explicit value of the integral can be obtained. However, it often suffices to have an estimate for the integral rather than its exact value. The Mean Value Theorems of this section are especially useful in making such estimates. Theorem 7.30 (First Mean- Value Theorem for Riemann-Stieltjes integrals). Assume that aT and let f e R(a) on [a, b]. Let M and m denote, respectively, the sup and inf of the set {f(x) : x e [a, b]}. Then there exists a real number c satisfying Th. 7.32 Integral as a Function of the Interval 161 m < c < M such that f b f(x) da(x) = c f 6 da(x) = c[a(b) - a(a)]. a Ja In particular, if f is continuous on [a, b], then c = f(xo) for some x0 in [a, b]. Proof. If a(a) = a(b), the theorem holds trivially, both sides being 0. Hence we can assume that a(a) < a(b). Since all upper and lower sums satisfy m[a(b) - a(a)] < L(P, f, a) < U(P, f, a) < M[a(b) - a(a)], the integral f 'f f da must lie between the same bounds. Therefore, the quotient c = (fa f da)/(fa doe) lies between m and M. When f is continuous on [a, b], the intermediate value theorem yields c = f(xo) for some x0 in [a, b]. A second theorem of this type can be obtained from the first by using integra-tion by parts. Theorem 7.31 (Second Mean- Value Theorem for Riemann-Stieltjes integrals). Assume that a is continuous and that f? on [a, b]. Then there exists a point xo in [a, b] such that f b b ). f (x) da(x) = f (a) f x0 da(x) + f (b) fX0 da(x Ja Proof. By Theorem 7.6, we have f b f (x) da(x) = f(b)a(b) - f (a)a(a) -f b a(x) df(x). a Applying Theorem 7.30 to the integral on the right, we find /'b a f(x) da(x) = f(a)[a(xo) - a(a)] + f(b)[a(b) - a(xo)], where x0 e [a, b], which is the statement we set out to prove. 7.19 THE INTEGRAL AS A FUNCTION OF THE INTERVAL If f e R(a) on [a, b] and if a is of bounded variation, then (by Theorem 7.25) the integral f a f da exists for each x in [a, b] and can be studied as a function of x. Some properties of this function will now be obtained. Theorem 7.32. Let a be of bounded variation on [a, b] and assume that f e R(a) on [a, b]. Define F by the equation ffdcc, F(x) = f x e [a, b]. 162 The Riemann-Stieltjes Integral Th. 7.33 Then we have: i) F is of bounded variation on [a, b]. ii) Every point of continuity of a is also a point of continuity of F. iii) If aT on [a, b], the derivative F'(x) exists at each point x in (a, b) where a'(x) exists and where f is continuous. For such x, we have F'(x) = f(x)a'(x). Proof. It suffices to assume that aT on [a, b]. If x # y, Theorem 7.30 implies that F(y) - F(x) = f yf dot = c[a(y) - a(x)], x where m < c < M (in the notation of Theorem 7.30). Statements (i) and (ii) follow at once from this equation. To prove (iii), we divide by y - x and observe that c -> f(x) as y - x. When Theorem 7.32 is used in conjunction with Theorem 7.26, we obtain the following theorem which converts a Riemann integral of a product f - g into a Riemann-Stieltjes integral f' f dG with a continuous integrator of bounded variation. Theorem 7.33. If f E R and g c- R on [a, b], let F(x) = J dt, G(x) = rx g(t) dt if x E [a, b]. a Then F and G are continuous functions of bounded variation on [a, b]. Also, f e R(G) and g e R(F) on [a, b], and we have f(x)g(x) dx = bf(x) dG(x) = f 6 g(x) dF(x). f a fa a Proof. Parts (i) and (ii) of Theorem 7.32 show that F and G are continuous func-tions of bounded variation on [a, b]. The existence of the integrals and the two formulas for f a f(x)g(x) dx follow by taking a(x) = x in Theorem 7.26. NOTE. When a(x) = x, part (iii) of Theorem 7.32 is sometimes called the first fundamental theorem of integral calculus. It states that F'(x) = f(x) at each point of continuity off. A companion result, called the second fundamental theorem, is given in the next section. 7.20 SECOND FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS The next theorem tells how to integrate a derivative. Theorem 7.34 (Second fundamental theorem of integral calculus). Assume that f E R on [a, b]. Let g be a function defined on [a, b] such that the derivative g' exists in Th. 7.35 Change of Variable (a, b) and has the value g'(x) = f(x) for every x in (a, b). At the endpoints assume that g(a +) and g(b -) exist and satisfy g(a) - g(a+) = g(b) - g(b-). Then we have J bf(x) dx = J g '(x) dx = g(b) - g(a). O A Proof. For every partition of [a, b] we can write n n g(b) - g(a) _ [g(xk) - g(xk- 1)] = E g'(tk) AXk = Ej f(tk) Axk, k=1 k=1 k=1 163 where tk is a point in (xk_1, xk) determined by the Mean-Value Theorem of differential calculus. But, for a given e > 0, the partition can be taken so fine that Ig(b) - g(a) - fbf(x) dxl = I kEf(tk) AXk -fa f(x) dxl < E, =1 a and this proves the theorem. The second fundamental theorem can be combined with Theorem 7.33 to give the following strengthening of Theorem 7.8. Theorem 7.35. Assume f e R on [a, b]. Let a be a function which is continuous on [a, b] and whose derivative a' is Riemann integrable on [a, b]. Then the following integrals exist and are equal: f f b f(x) da(x) = 6 f(x)a'(x) dx. a a Proof. By the second fundamental theorem we have, for each x in [a, b], a(x) - a(a) = f: a'(t) dt. Taking g = a' in Theorem 7.33 we obtain Theorem 7.35. NOTE. A related result is described in Exercise 7.34. 7.21 CHANGE OF VARIABLE IN A RIEMANN INTEGRAL The formula I.' f da = f l h dfi of Theorem 7.7 for changing the variable in an integral assumes the form f 8(a)f(x) dx = J df[g(t)]g'(t) dt, 9(c) c 164 The Riemann Stieltjes-Integral Th. 7.36 when a(x) = x and when g is a strictly monotonic function with a continuous derivative g'. It is valid if f E R on [a, b]. When f is continuous, we can use Theorem 7.32 to remove the restriction that g be monotonic. In fact, we have the following theorem: Theorem 7.36 (Change of variable in a Riemann integral). Assume that g has a continuous derivative g' on an interval [c, d]. Let f be continuous on g([c, d]) and define F by the equation ) dt if x E g([c, d]). F(x) = fg(,) X f(t Then, for each x in [c, d] the integral J' f[g(t)]g'(t) dt exists and has the value F[g(x)]. In particular, we have ' J 9(d) f(x) dx = J df[g(t)]9'(t) A g(c) c Proof. Since both g' and the composite function fo g are continuous on [c, d] the integral in question exists. Define G on [c, d] as follows: G(x) = $Xf[g(t)]gl(t) dt. Weare to show that G(x) = F[g(x)]. By Theorem 7.32, we have G' (x) = .f [g(x)]g' (x), and, by the chain rule, the derivative of F[g(x)] is also f [g(x)]g'(x), since F'(x) _ f(x). Hence, G(x) - F[g(x)] is constant. But, when x = c, we get G(c) = 0 and F[g(c)] = 0, so this constant must be 0. Hence, G(x) = F[g(x)] for all x in [c, d]. In particular, when x = d, we get G(d) = F[g(d)] and this is the last equation,in the theorem. NOTE. Some texts prove the preceding theorem under the added hypothesis that g' is never zero on [c, d], which, of course, implies monotonicity of g. The above proof shows that this is not needed. It should be noted that g is continuous on [c, d], so g([c, d]) is an interval which contains the interval joining g(c) and g(d). g(d) g(c) Th. 7.37 Second Mean-Value Theorem for Riemann Integrals 165 In particular, the result is valid if g(c) = g(d). This makes the theorem especially useful in the applications. (See Fig. 7.2 for a permissible g.) Actually, there is a more general version of Theorem 7.36 which does not require continuity off or of g', but the proof is considerably more difficult. Assume that h e R on [c, d] and, if x E [c, d], let g(x) = fq h(t) dt, where a is a fixed point in [c, d]. Then if f e R on g([c, d]) the integral fc f [g(t)] h(t) dt exists and we have f J g(d) f(x) dx = J f[g(t)]h(t) dt. g(c) c This appears to be the most general theorem on change of variable in a Riemann integral. (For a proof, see the article by H. Kestelman, Mathematical Gazette, 45 (1961), pp. 17-23.) Theorem 7.36 is the special case in which h is continuous on [c, d] and f is continuous on g([c, d]). 7.22 SECOND MEAN-VALUE THEOREM FOR RIEMANN INTEGRALS Theorem 7.37. Let g be continuous and assume that f,, on [a, b]. Let A and B be two real numbers satisfying the inequalities A 5 f(a+) and B >- f(b-). Then there exists a point xa in [a, b] such that b g(x) dx. i) rb f(x)g(x) dx = A rx0 g(x) dx + B fX0 .J a J In particular, iff(x) > 0 for all x in [a, b], we have ii) f n f(x)g(x) dx = B f n g(x) dx, where x0 E [a, b]. ,J a ,J xo NOTE. Part (ii) is known as Bonnet's theorem. Proof. If a(x) = Ix g(t) dt, then a' = g, Theorem 7.31 is applicable, and we get f 6 f(x)g(x) dx = f(a) fx0 g(x) dx + f (b) fb g(x) dx. J a a xo This proves (i) whenever A = f(a) and B = f(b). Now if A and B are any two real numbers satisfying A < f(a+) and B f(b-), we can redefine f at the end-points a and b to have the values f(a) = A and f(b) = B. The modified f is still increasing on [a, b] and, as we have remarked before, changing the value of f at a finite number of points does not affect the value of a Riemann integral. (Of course, the point x0 in (i) will depend on the choice of A and B.) By taking A = 0, part (ii) follows from part (i). 166 The Riemann-Stieltjes Integral Th. 7.38 7.23 RIEMANN-STIELTJES INTEGRALS DEPENDING ON A PARAMETER Theorem 7.38. Let f be continuous at each point (x, y) of a rectangle Q={(x,y):a<x<b, c 0, there exists a S > 0 (depending only on E) such that for every pair of points z = (x, y) and z' = (x', y') in Q with Iz - z'j < S, we have l f(x, y) - f(x', y')l < s. If fy - y'I < 6, we have IF(Y) - F(Y)I < J b If(x, y) - f(x, Y')I da(x) < E a b a This establishes the continuity of F on [c, d]. Of course, when a(x) = x, this becomes a continuity theorem for Riemann integrals involving a parameter. However, we can derive a much more useful result for Riemann integrals than that obtained by simply setting a(x) = x if we employ Theorem 7.26. Theorem 7.39. If f is continuous on the rectangle [a, b] x [c, d], and ifg e R on [a, b], then the function F defined by the equation F(y) = J b g(x)f(x, y) dx, a is continuous on [c, d]. That is, if yo e [c, d], we have 6 lim fa g(x)f(x, y) dx J b g(x).f(x, yo) dx. Y-Yo a Proof. If G(x) = $a g(t) dt, Theorem 7.26 shows that F(y) = fo f(x, y) dG(x). Now apply Theorem 7.38. Th. 7.41 Interchanging Order of Integration 7.24 DIFFERENTIATION UNDER THE INTEGRAL SIGN 167 Theorem 7.40. Let Q = {(x, y) : a < x < b, c < y < d}. Assume that a is of bounded variation on [a, b] and, for each fixed y in [c, d], assume that the integral b F(y) = J f(x, y) da(x), a exists. If the partial derivative D2f is continuous on Q, the derivative F(y) exists for each y in (c, d) and is given by F(y) = J D2f(x, y) da(x). 6 A NOTE. In particular, when g e R on [a, b] and a(x) = f a g(t) dt, we get F(y) = J g(x)f(x, y) dx and F(y) = J g(x) D2f(x, y) dx. 6 6 A a Proof. If yo e (c, d) and y # yo, we have F(y) - F(yo) = fl'f(x, Y) - f(x, Yo) da(x) = J b D2f(x, Y) da(x), Y - Yo JA Y - Yo A where y is between y and yo. Since D2 f is continuous on Q, we obtain the con-clusion by arguing as in the proof of Theorem 7.38. 7.25 INTERCHANGING THE ORDER OF INTEGRATION Theorem 7.41. Let Q = {(x, y) : a < x < b, c < y < d}. Assume that a is of bounded variation on [a, b], /3 is of bounded variation on [c, d], and f is continuous on Q. If (x, y) a Q, define F(Y) = J bf(x, y) da(x), G(x) = df(x, y) df(Y) fc Then F e R(fl) on [c, d], G e R(a) on [a, b], and we have f d F(y) df3(Y) = J 6 G(x) da(x). C A In other words, we may interchange the order of in as folllows: Jab LJf(x, y) dl(Y)] d(x) = f d [Sa" f(x, y) da(x)J df3(y). Proof. By Theorem 7.38, F is continuous on [c, d] and hence F e R(fl) on [c, d]. Similarly, G e R(a) on [a, b]. To prove the equality of the two integrals, it suffices to consider the case in which a,,,, on [a, b] and /3 r on [c, d]. 168 The Riemann-Stieltjes Integral Th. 7.41 By uniform continuity, given E > 0 there is a 6 > 0 such that for every pair of points z = (x, y) and z' = (x', y') in Q, with Iz - z'I < 6, we have If(x, Y) - f(x', Y')I < E. Let us now subdivide Q into n2 equal rectangles by subdividing [a, b] and [c, d] each into n equal parts, where n is chosen so that (b-a)< S and (d-c)< 6 Writing n n T2 =a+k(b-a) x and = c+k(d-c) for k = 0, k n 1, 2,... , n, we have Yk n f (fd f(x, y) df(Y)) da(x) = I kk+1 \Jy ryl+' f(x, y) df3(y) da(x). k=O j=0 J , We apply Theorem 7.30 twice on the right. The double sum becomes n-1 n-1 LIf(xk, Yj)[!'(Yj+l) - #(Yj)][a(xk+1) - 0001 k=0 j=0 where (xk, yj) is in the rectangle Qk, j having (xk, yj) and (xk+ , Yj+ ) as opposite vertices. Similarly, we find f a (fb f (x, y) da(x) J dfi(Y) n-1 n-i _ f(xk, Y';)[P(Yj+1) - f(Yj)][a(xk+1) -k=0 j=0 where (x', y 'j') a Qk,j. But I f(xk, y;) - f(x'k, y;)I < E and hence f b G(x) da(x) -I'd F(Y) dfl(Y) a n1 7 n-1 1 < ELI -1 [l3(Yj+1) - i3(Yj)] Lam! [a(xk+1) - a(xk)] j=0 k=0 = E[f(d) - f3(c)1[a(b) - a(a)]. Since a is arbitrary, this implies equality of the two integrals. Theorem 7.41 together with Theorem 7.26 gives the following result for Rie-mann integrals. Def. 7.43 Lebesgue's Criterion 169 Theorem 7.42. Let f be continuous on the rectangle [a, b] x [c, d]. If g E R on [a, b] and if h E R on [c, d], then we have jb [J1 g(x)h(y)f(x, y) dy] dx = J d LJ b g(x)h(y)f(x, y) dxl dy. J o JJ Proof Let a(x) = f a g(u) du and let fl(y) = f h(v) dv, and apply Theorems 7.26 and 7.41. 7.26 LEBESGUE'S CRITERION FOR EXISTENCE OF RIEMANN INTEGRALS Every continuous function is Riemann integrable. However, continuity is certainly not necessary, for we have seen that f e R when f is of bounded variation on [a, b]. In particular, f can be a monotonic function with a countable set of discontinuities and yet the integral J' f(x) dx will exist. Actually, there are Riemann-integrable functions whose discontinuities form a noncountable set. (See Exercise 7.32.) Therefore, it is natural to ask "how many" discontinuities a function can have and still be Riemann integrable. The definitive theorem on this question was dis-covered by Lebesgue and is proved in this section. The idea behind Lebesgue's theorem is revealed by examining Riemann's condition to see the kind of restriction it puts on the set of discontinuities off. The difference between the upper and lower Riemanil sums is given by E [Mk(f) - Mk(f) )] exk, k=1 and, roughly speaking, f will be integrable if, and only if, this sum can be made arbitrarily small. Split this sum into two parts, say S1 + S2, where S1 comes from subintervals containing only points of continuity of f, and S2 contains the re-maining terms. In S1, each difference Mk(f) - Mk(J)is small because of continuity and hence a large number of such terms can occur and still keep S1 small. In S21 however, the differences Mk(f) - mk(f) need not be small; but because they are bounded (say by M), we have IS21 < M Y_Lxk, so that S2 will be small if the sum of the lengths of the subintervals corresponding to S2 is small. Hence we may expect that the set of discontinuities of an integrable function can be covered by intervals whose total length is small. This is the central idea in Lebesgue's theorem. To formulate it more precisely we introduce sets of measure zero. Definition 7.43. A set S of real numbers is said to have measure zero if, for every e > 0, there is a countable covering of S by open intervals, the sum of whose lengths is less than If the intervals are denoted by (ak, bk), the definition requires that S g U (ak, bk) and E (bk - ak) < e. (3) k k 170 The Riemann-Stieltjes Integral Th. 7.44 If the collection of intervals is finite, the index k in (3) runs over a finite set. If the collection is countably infinite, then k goes from 1 to oo, and the sum of the lengths is the sum of an infinite series given by 00 N E (bk - ak) = lim E (bk - ak). k=1 N-.oo k=1 Besides the definition, we need one more result about sets of measure zero. Theorem 7.44. Let F be a countable collection ofsets in R, say F= {F1,F2,...}, each of which has measure zero. Then their union 00 S = U Fk, k=1 also has a measure zero. Proof. Given a > 0, there is a countable covering of Fk by open intervals, the sum of whose lengths is less than E/2k. The union of all these coverings is itself a countable covering of S by open intervals and the sum of the lengths of all the intervals is less than 00 = E. k = 1 2k Examples. Since a set consisting of just one point has measure zero, it follows that every countable subset of R has measure zero. In particular, the set of rational numbers has measure zero. However, there are uncountable sets which have measure zero. (See Exer-cise 7.32.) Next we introduce the concept of oscillation. Definition 7.45. Let f be defined and bounded on an interval S. If T c S, the number nf(T) = sup {f(x) - f(y) : x e T, y e T}, is called the oscillation off on T. The oscillation off at x is defined to be the number co f(x) = lim l2 f(B(x; h) r S). h-+o+ NOTE. This limit always exists, since S2 f(B(x; h) r S) is a decreasing function of h. In fact, T1 c T2 implies Of(TI) < Q f(T2). Also, w f(x) = 0 if, and only if, f is continuous at x (Exercise 4.24). The next theorem tells us that if cof(x) < E at each point of a compact interval [a, b], then Of(T) < e for all sufficiently small subintervals T. Theorem 7.46. - Let f be defined and bounded on [a, b], and let E > 0 be given. Assume that w f(x) < E for every x in [a, b]. Then there exists a 6 > 0 (depending Th. 7.48 Lebesgue's Criterion 171 only on ) such that for every closed subinterval T c [a, b], we have O f(T) < whenever the length of T is less than 8. Proof. For each x in [a, b] there exists a 1-ball Bx = B(x; 8x) such that D f(Bx n [a, b]) < co f(x) + [ - w1(x)] = . The set of all halfsize balls B(x; 6x/2) forms an open covering of [a, b]. By compactness, a finite number (say k) of these cover [a, b]. Let their radii be 8112, ..., 8k/2 and let 8 be the smallest of these k numbers. When the interval T has length <8, then T is partly covered by at least one of these balls, say by B(x p; 8 p/2). However, the ball B(x p; 8 p) completely covers T (since 8p >- 28). Moreover, in B(xp; 8 p) n [a, b] the oscillation off is less than s. This implies that O f(T) < and the theorem is proved. Theorem 7.47. Let f be defined and bounded on [a, b]. For each > 0 define the set J. as follows: JJ = {x: x c- [a, b], co f(x) }. Then JE is a closed set. Proof. Let x be an accumulation point of J. If x 0 J, we have (of (x) < . Hence there is a 1-ball B(x) such that El f(B(x) n [a, b]) < . Thus no points of B(x) can belong to JE, contradicting the statement that x is an accumulation point of JE. Therefore, x e J. and JE is closed. Theorem 7.48 (Lebesgue's criterion for Riemann-integrability). Let f be defined and bounded on [a, b] and let D denote the set of discontinuities off in [a, b]. Then f e R on [a, b] if, and only if, D has measure zero. Proof (Necessity). First we assume that D does not have measure zero and show that f is not integrable. We can write D as a countable union of sets 00 D = U D., r=1 where D, = x : oo f(x) > 11 . r If x e D, then co f(x) > 0, so D is the union of the sets D, for r = 1, 2, .. . Now if D does not have measure zero, then some set D, does not (by Theorem 7.44). Therefore, there is some > 0 such that every countable collection of open intervals covering D, has a sum of lengths >. For any partition P of [a, b] we have _ U(P,f) - L(P,f) = L.i [Mk(J) - mk(J )] Axk = S1 + S2 > S1, k=1 172 The Riemann-Stieltjes Integral Th. 7.48 where Sl contains those terms coming from subintervals containing points of D in their interior, and S2 contains the remaining terms. The open intervals from S1 cover D, except possibly for a finite subset of D which has measure 0, so the sum of their lengths is at least s. Moreover, in these intervals we have Mk(f) - Mk(f) > 1 and hence S1 > s r r This means that U(P,f) - L(P,f) ? E, r for every partition P, so Riemann's condition cannot be satisfied. Therefore f is not integrable. In other words, if f e R, then D has measure zero. (Sufficiency). Now we assume that D has measure zero and show that the Riemann condition is satisfied. Again we write D = Ur- 1, D,, where D, is the set of points x at which co f(x) >- 1 /r. Since D, s D, each D, has measure 0, so D, can be covered by open intervals, the sum of whose lengths is < 1 /r. Since D, is compact (Theorem 7.47), a finite number of these intervals cover D,. The union of these intervals is an open set which we denote by A,. The complement B, = [a, b] - A, is the union of a finite number of closed subintervals of [a, b]. Let I be a typical subinterval of B,. If x e I, then co f(x) < 1 /r so, by Theorem 7.46, there is a S > 0 (depending only on r) such that I can be further subdivided into a finite number of subintervals T of length <6 in which D f(T) < l/r. The endpoints of all these subintervals determine a partition P, of [a, b]. If P is finer than P, we can write U(P, f) - L(P, f) = E [Mk(f) - mk(f )] Axk = S 1 + S2, k=1 where S1 contains those terms coming from subintervals containing points of D,, and S2 contains the remaining terms. In the kth term of S2 we have Mk(f) - mk(f) < 1 and hence S2 < b - a r r Since A, covers all the intervals contributing to S1, we have Si < M - m r where M and m are the sup and inf off on [a, b]. Therefore U(P,f)-L(P,f) 1, we see that Riemann's condition holds, so f e R on [a, b]. NOTE. A property is said to hold almost everywhere on a subset S of Rl if it holds everywhere on S except for a set of measure 0. Thus, Lebesgue's theorem states 11. 7.50 Complex-valued Integrals 173 that a bounded function f on a compact interval [a, b] is Riemann integrable on [a, b] if, and only if, f is continuous almost everywhere on [a, b]. The following statements (some of which were proved earlier in the chapter) are immediate consequences of Lebesgue's theorem. Theorem 7.49. a) If f is of bounded variation on [a, b], then f e R on [a, b]. b) If f e R on [a, b], then f e R on [c, d] for every subinterval [c, d] c [a, b], If I E R and f 2 e R on [a, b]. Also, f- g e R on [a, b] whenever geR on [a, b]. c) If f e R and g e R on [a, b], then f/g e R on [a, b] whenever g is bounded away from 0. d) If f and g are bounded functions having the same discontinuities on [a, b], then f e R on [a, b] if, and only if, g e R on [a, b]. e) Let g e R on [a, b] and assume that m < g(x) < M for all x in [a, b]. If f is continuous on [m, M], the composite function h defined by h(x) = f [g(x)] is Riemann-integrable on [a, b]. NOTE. Statement (e) need not hold if we assume only that fe R on [m, M]. (See Exercise 7.29.) 7.27 COMPLEX-VALUED RIEMANN-STIELTJES INTEGRALS Riemann-Stieltjes integrals of the form f o f da, in which f and a are complex-valued functions defined and bounded on an interval [a, b], are of fundamental importance in the theory of functions of a complex variable. They can be intro-duced by exactly the same definition we have used in the real case. In fact, Definition 7.1 is meaningful when f and a are complex-valued. The sums of the products f(tk)[a(xk) - a(xk,)] which are used to form Riemann-Stieltjes sums need only be interpreted as sums of products of complex numbers. Since complex numbers satisfy the commutative, associative, and distributive laws which hold for real numbers, it is not surprising that complex-valued integrals share many of the properties of real-valued integrals. In particular, Theorems 7.2, 7.3, 7.4, 7.6, and 7.7 (as well as their proofs) are all valid (word for word) when f and a are complex-valued functions. (In Theorems 7.2 and 7.3, the constants c1 and c2 may now be complex numbers.) In addition, we have the following theorem which, in effect, reduces the theory of complex Stieltjes integrals to the real case. Theorem 7.50. Let f = f1 + if2 and a = al defined on an interval [a, b]. Then we have + ia2 be complex-valued functions dal - $/2 dal) + fabf da (fa" f, whenever all four integrals on the right exist. i (fbf2 dal + J bf1 da), 174 The Riemann-Stieltjes Integral The proof of Theorem 7.50 is immediate from the definition and is left to the reader. The use of this theorem permits us to extend most of the important properties of real integrals to the complex case. For example, the connection between differentiation and integration established in Theorem 7.32 remains valid for complex integrals if we simply define such notions as continuity, differentiability and bounded variation by components, as with vector-valued functions. Thus, we say that the complex-valued function a = al + ia2 is of bounded variation on [a, b] if each component al and a2 is of bounded variation on [a, b]. Similarly, the derivative a'(t) is defined by the equation a'(t) = ai(t) + k2(t) whenever the derivatives a'1(t) and a2(t) exist. (One-sided derivatives are defined in the same way.) With this understanding, Theorems 7.32 and 7.34 (the fundamental theorems of integral calculus) both remain valid when f and a are complex-valued. The proofs follow from the real case by using Theorem 7.50 in a straightforward manner. We shall return to complex-valued integrals in Chapter 16, when we study functions of a complex variable in more detail. EXERCISES Riemann-Stieltjes integrals 7.1 Prove that f .b da(x) = a(b) - a(a), directly from Definition 7.1. 7.2 If f e R(a) on [a, b ] and if f; f da = 0 for every f which is monotonic on [a, b ], prove that a must be constant on [a, b]. 7.3 The following definition of a Riemann-Stieltjes integral is often used in the literature: We say f is integrable with respect to a if there exists a real number A having the property that for every e > 0, there exists a b > 0 such that for every partition P of [a, b] with norm IIPII < 6 and for every choice of tk in [xk_1, xk], we have IS(P, f, a) - AI < S. a) Show that if S Q f da exists according to this definition, then it also exists according to Definition 7.1 and the two integrals are equal. b) Let f (x) = a(x) = 0 for a 5 x < c, f (x) = a(x) = I for c < x <- b, f (c) = 0, a(c) = 1. Show that f a f da exists according to Definition 7.1 but does not exist by this second definition. 7.4 If f e R according to Definition 7.1, prove that fa f(x) dx also exists according to the definition of Exercise 7.3. [Contrast with Exercise 7.3(b). ] Hint. Let I = fa f(x) dx, M = sup {I f(x)I : x e [a, b]}. Given e > 0, choose Pt so that U(P, f) < I + e/2 (notation of Section 7.11). Let N be the number of subdivision points in PE and let S = e/(2MN). If IIPII < 6, write U(P, f) _ Mk(f) Axk = S1 + S2, where S1 is the sum of terms arising from those subintervals of P containing no points of PE and S2 is the sum of the remaining terms. Then Sl < U(PE, f) < I + e/2 and S2 < NMIIPII < NM6 = e/2, Exercises 175 and hence U(P, f) < I + B. Similarly, L(P, f) > I - e if 11 P11 < b' for some S'. Hence I S(P, f) - II < e if 11P11 < min (S, 8'). 7.5 Let {an} be a sequence of real numbers. For x >- 0, define [x7 A(x) Ea"= Ea,,, nsx n=1 where [x] is the greatest integer in x and empty sums are interpreted as zero. Let f have a continuous derivative in the interval 1 <- x < a. Use Stieltjes integrals to derive the following formula : E a"f(n) fa A(x)f'(x) dx + A(a)f(a). nsa 7.6 Use Euler's summation formula, or integration by parts in a Stieltjes integral, to derive the following identities: n 1 1 J" [x] a) E s= n 1 s-1 I s s+1 d x ifs 0 1. k=1 k X b) - = log n -2 dx + 1. " 1 Tx-[x] x 7.7 Assume f' is continuous on [1, 2n] and use Euler's summation formula or integra-tion by parts to prove that 2n ( - 2[x/2J) dx. 1)kf(k) = flit 7.8 Let p1(x) = x - [x] - j if x # integer, and let VI(x) = 0 if x = integer. Also, let 'P2(x) = fo p1(t) dt. If f" is continuous on [1, n] prove that Euler's summation formula implies that n n Ef(k) = 1 f(x) dx - f i 'P2(x)f"(x) dx + f(1) 2 f(n) 7.9 Take f(x) = log x in Exercise 7.8 and prove that log n! = (n + 1) log n - n + I + f " (022t) dt. 1 t 7.10 If x >- 1, let ir(x) denote the number of primes <- x, that is, 7r(x) = E 1, psx where the sum is extended over all primes p <- x. The prime number theorem states that lira n(x) log x = 1. x-+OD x 176 The Riemann-Stieltjes Integral This is usually proved by studying a related function 9 given by 9(x) = E log P, p5x where again the sum is extended over all primes p <- x. Both functions n and 9 are step functions with jumps at the primes. This exercise shows how the Riemann-Stieltjes integral can be used to relate these two functions. a) If x >- 2, prove that ir(x) and 9(x) can be expressed as the following Riemann-Stieltjes integrals: 3(x) = f3X1 logt dir(t), 7r(x) = J I d9(t). 2 ,1312 log t NOTE. The lower limit can be replaced by any number in the open interval (1, 2). b) If x > 2, use integration by parts to show that 3(x) = 7r(x) log x -f x n(t) dt, 2 t 70) = 9(x) + 3(t) dt. log x f2x t loge t These equations can be used to prove that the prime number theorem is equivalent to the relation limx 3(x)/x = 1. 7.11 If a/ on [a, b], prove that we have b a) ffda= fcfda+ fbfda, (a<c<b), Ja Ja Jc fda+ fgda, b) f.' (f+g)da< fb 6 Ja Ja f c) b(f + g) doe >-fb f da + f6 g da. Ja a 7.12 Give an example of a bounded function f and an increasing function a defined on [a, b] such that if I E R(a) but for which fQ f da does not exist. 7.13 Let a be a continuous function of bounded variation on [a, b]. Assume g e R(a) on [a, b ] and define fl(x) = J.' g(t) da(t) if x e [a, b ]. Show that : a) If f, on [a, b], there exists a point xo in [a, b] such that J 6f d/ = f (a) f x0 g da + f (b) f 6 g da. a a ,Jxo o b) If, in addition, f is continuous on [a, b ], we also have fab f(x)g(x) dx(x) = f (a) fax g doe + f (b) 6 g da. Jxo Exercises 177 7.14 Assume f e R(a) on [a, b], where a is of bounded variation on [a, b]. Let V(x) denote the total variation of a on [a, x] for each x in (a, b], and let V(a) = 0. Show that fda f b f IfI dV < MV(b), where M is an upper bound for If I on [a, b]. In particular, when a(x) = x, the inequality becomes a f(x) dx < M(b - a). 7.15 Let {a,,} be a sequence of functions of bounded variation on [a, b]. Suppose there exists a function a defined on [a, b] such that the total variation of a - an on [a, b] tends to 0 as n -> co. Assume also that a(a) = aa(a) = 0 for each n = 1, 2,... If f is con-tinuous on [a, b], prove that Jim f f (X) da(x) b n -. c0 a a 7.16 If f e R(a), f 2 e R(a), g e R(a), and g2 e R(a) on [a, b], prove that I b b f(x) 9(x) 2 da(y) da(x) 2 1 LJa If(y) 9(Y)I = (f"f(x)2 da(x)) (f.' 9(x)2 da(x)) \J bf(x)9(x) da(x)) . a 1 When ax on [a, b], deduce the Cauchy-Schwarz inequality (fbf(x)g(x) da(x)) < (5a2 da(x)I (Jb 9(x)2 da(x)) 1 (Compare with Exercise 1.23.) 7.17 Assume that f e R(a), g e R(a), and f g e R(a) on [a, b]. Show that 1 b b 1 2 fa [fa (f(y) - .f(x))(9(Y) - 9(x)) da(Y)] da(x) b _ (a(b) - a(a)) b f f (x)g(x) da(x) - (f f (x) da(x)) (f g(x) da(x)) . a a b /(f / If a/ on [a, b], deduce the inequality (f"b f(x) da(x)) (Ia b g(x) da(x)) < (a(b) - a(a)) f f(x)g(x) da(x) / /J Ja b when both f and g are increasing (or both are decreasing) on [a, b ]. Show that the reverse inequality holds if f increases and g decreases on [a, b]. 178 The Riemann-Stieltjes Integral Riemann integrals 7.18 Assume f e R on [a, b]. Use Exercise 7.4 to prove that the limit limb -a n-00 n k=1 n exists and has the value fa f(x) dx. Deduce that lim n = n , lim (n2 + k2)-1I2 = log (1 + -F2). n Ej+ n2 4 n-,c k=1 7.19 Define 2 1 -x2(t2+1) f(x) = (f' e '2 dt) , g(x) = f o et2 + 1 dt. J a) Show that g'(x) + f'(x) = 0 for all x and deduce that g(x) + f(x) = n14. b) Use (a) to prove that lim Ixe t2dt= 11,17t. x-.00 0 2 7.20 Assume g e R on [a, b] and define f(x) = fa g(t) dt if x e [a, b]. Prove that the integral fa Jg(t)j dt gives the total variation off on [a, x]. 7.21 Let f = (fl, . . . ,f.) be a vector-valued function with a continuous derivative f' on [a, b]. Prove that the curve described by f has length b Af(a, b) = I IIf'(t) II dt. 7.22 If f(n+1) is continuous on [a, x], define X 4(x) = n, (x - t)"f (n+1)(t) dt. a a) Show that Ik-1(x) - Ik(x) _ " k)(Q)(x - Q)k k ! k = 1, 2, ... , n. b) Use (a) to express the remainder in Taylor's formula (Theorem 5.19) as an integral. 7.23 Let f be continuous on [0, a]. If x e [0, a], define fo(x) = f(x) and let fn+1(x)= 1 fx(x-t)"f(t)dt, n = 0, 1, 2,... n! o a) Show that the nth derivative of fn exists and. equals f. b) Prove the following theorem of M. Fekete: The number of changes in sign of f in [0, a] is not less than the number of changes in sign in the ordered set of numbers f(a), fl(a), ... , fn(a) Hint. Use mathematical induction. c) Use (b) to prove the following theorem of L. Fejbr: The number of changes in sign off in [0, a] is not less than the number of changes in sign in the ordered set f (O), f af (t) dt, a tf(t) dt, ..., t"f (t) dt. 0 0 foa 7.24 Let f be a positive continuous function in [a, b]. Let M denote the maximum value off on [a, b ]. Show that lim n-+ 00 (fbf(x)" )1/fl = M. 7.25 A function f of two real variables is defined for each point (x, y) in the unit square 0 -< x 5 1, 0 5 y< I as follows: f(x, y) _ (2y, 1, if x is rational, if x is irrational. a) Compute Jo f(x, y) dx and Jol f(x, y) dx in terms of y. b) Show that Jo f(x, y) dy exists for each fixed x and compute Jo f(x, y) dy in terms of xand tfor 0-x51,0-<<t5 1. c) Let F(x) = Jo f(x, y) dy. Show that Jo F(x) dx exists and find its value. 7.26 Let f be defined on [0, 1 ] as follows: f(0) = 0; if 2-R-1 < x 5 2'°, thenf(x) = 2'", for n = 0, 1, 2, .. . a) Give two reasons why Jo f(x) dx exists. b) Let F(x) = Jo f(t) dt. Show that for 0 < x _< 1 we have F(x) = xA(x) - 3A(x)2, where A(x) = 2-1-1o9x/1o82] and where [y] is the greatest integer in y. 7.27 Assume f has a derivative which is monotonic decreasing and satisfies f'(x) z m > 0 for all x in [a, b]. Prove that IJ6cosf(x)dxI -<2. a m Hint. Multiply and divide the integrand by f'(x) and use Theorem 7.37(ii). 7.28 Given a decreasing sequence of real numbers {G(n)} such that G(n) -- 0 as n - oo. Define a function f on [0, 1 ] in terms of {G(n)} as follows: f(0) = 1; if x is irrational, then f(x) = 0; if x is the rational m/n (in lowest terms), then f(m/n) = G(n). Compute the oscillation co f(x) at each x in [0, 1 ] and show that f e R on [0, 1 ]. 7.29 Let f be defined as in Exercise 7.28 with G(n) = 1/n. Let g(x) = 1 if 0 < x 5 1, g(0) = 0. Show that the composite function h defined by h(x) = g[f(x)] is not Riemann-integrable on [0, 1 ], although both f e R and g e R on [0, 1 ]. 7.30 Use Lebesgue's theorem to prove Theorem 7.49. 7.31 Use Lebesgue's theorem to prove that if f e R and g e R on [a, b] and if f(x) m > 0 for all x in [a, b ], then the function h defined by h(x) = f(x)a(x) is Riemann-integrable on [a, b ]. 180 The Riemann-Stieltjes Integral 7.32 Let I = [0, 1] and let Al = I - (;, 1) be that subset of I obtained by removing those points which lie in the open middle third of I; that is, Al = [0, 1] v [1, 1 J. Let A2 be that subset of Al obtained by removing the open middle third of [0, 3 ] and of [ , 11. Continue this process and define A3, A4,... The set C = nn,=1 A. is called the Cantor set. Prove that: a) C is a compact set having measure zero. b) x e C if, and only if, x = n 1 a"3-", where each a" is either 0 or 2. c) C is uncountable. d) Let f(x) = I if x e C, f(x) = 0 if x C. Prove that f e R on [0, 1 ]. 7.33 This exercise outlines a proof (due to Ivan Niven) that n2 is irrational. Let f(x) _ x"(1 - x)"/n!. Prove that: a)0<f(x)< 1/n! if0<x<1. b) Each kth derivative f (k)(0) and f (k)(1) is an integer. Now assume that R2 = a/b, where a and b are positive integers, and let n F(x) = b" E (- I)k f(2k)(x) x2n-2k k=0 Prove that : c) F(0) and F(1) are integers. d) n2a"f(x) sin nx = dx {F'(x) sin nx - xF(x) cos 7rx}. 1 e) F(1) + F(0) = na" f f (x) sin 7Cx dx. 0 f) Use (a) in (e) to deduce that 0 < F(1) + F(0) < 1 if n is sufficiently large. This contradicts (c) and shows that 7C2 (and hence n) is irrational. 7.34 Given a real-valued function a, continuous on the interval [a, b] and having a finite bounded derivative a' on (a, b). Let f be defined and bounded on [a, b] and assume that both integrals lb f (x) a'(x) dx f (x) da(x) and fa 6 exist. Prove that these integrals are equal. (It is not assumed that a' is continuous.) 7.35 Prove the following theorem, which implies that a function with a positive integral must itself be positive on some interval. Assume that f e R on [a, b] and that 0 5 f(x) < M on [a, b], where M > 0. Let I = f a f (x) dx, let h = +I/(M + b - a), and assume that I > 0. Then the set T = {x : f(x) >- h} contains a finite number of intervals, the sum of whose lengths is at least h. Hint. Let P be a partition of [a, b] such that every Riemann sum S(P, f) = Ek=1 f (tk) Axk satisfies S(P, f) > 1/2. Split S(P, f) into two parts, S(P, f) _ L A + LkEB, Where A = {k:[xk_1,xk]ST}, and B = {k:kOA}. If k e A, use the inequality f (tk) s M; if k e B, choose tk so that f(tk) < h. Deduce that L 4 AXk > h. Exercises 181 Existence theorems for integral and differential equations The following exercises illustrate how the fixed-point theorem for contractions (Theorem 4.48) is used to prove existence theorems for solutions of certain integral and differential equations. We denote by C [a, b] the metric space of all real continuous functions on [a, b] with the metric d(f, g) = Ilf - g11 = max If(x) - g(x)I, a5xsb and recall that C [a, b] is a complete metric space (Exercise 4.67). 7.36 Given a function g in C [a, b], and a function K continuous on the rectangle Q = [a, b ] x [a, b ], consider the function T defined on C [a, b ] by the equation T(rp)(x) = g(x) + A r K(x, t)rp(t) dt, b Ja where A is a given constant. a) Prove that T maps C [a, b ] into itself. b) If IK(x, y)I < M on Q, where M > 0, and if JAI < M-1(b - a)-1, prove that T is a contraction of C [a, b] and hence has a fixed point rp which is a solution of the integral equation rp(x) = g(x) + A fa K(x, t)rp(t) dt. 7.37 Assume f is continuous on a rectangle Q = [a - h, a + h] x [b - k, b + k], where h > 0, k > 0. a) Let rp be a function, continuous on [a - h, a + h], such that (x, (O(x)) e Q for all x in [a - h, a + h]. If 0 < c <- h, prove that c satisfies the differential equation y' = f(x, y) on (a - c, a + c) and the initial condition p(a) = b if, and only if, rp satisfies the integral equation ,p(x) = b + ff(t49(t )) dt on (a - c, a + c). b) Assume that I f (x, y)l < M on Q, where M > 0, and let c = min {h, k/M }. Let S denote the metric subspace of C [a - c, a + c] consisting of all rp such that I rp(x) - bI < Me on [a - c, a + c]. Prove that S is a closed subspace of C [a - c, a + c] and hence that S is itself a complete metric space. c) Prove that the function T defined on S by the equation T(rp)(x) = b + ff(t49(t)) dt maps S into itself. d) Now assume that f satisfies a Lipschitz condition of the form I f(x, y) - f(x, z)I < Al y - z l for every pair of points (x, y) and (x, z) in Q, where A > 0. Prove that T is a contraction of S if h < 1/A. Deduce that for h < 1/A the differential equation y' = f(x, y) has exactly one solution y = 49(x) on (a - c, a + c) such that p(a) = b. 182 The Riemann-Stieltjes Integral SUGGESTED REFERENCES FOR FURTHER STUDY 7.1 Hildebrandt, T. H., Introduction to the Theory of Integration. Academic Press, New York, 1963. 7.2 Kestelman, H., Modern Theories of Integration. Oxford University Press, Oxford, 1937. 7.3 Rankin, R. A., An Introduction to Mathematical Analysis. Pergamon Press, Oxford, 1963. 7.4 Rogosinski, W. W., Volume and Integral. Wiley, New York, 1952. 7.5 Shilov, G. E., and Gurevich, B. L., Integral, Measure and Derivative: A Unified Approach. R. Silverman, translator. Prentice-Hall, Englewood Cliffs, 1966. CHAPTER 8 INFINITE SERIES AND INFINITE PRODUCTS 8.1 INTRODUCTION This chapter gives a brief development of the theory of infinite series and infinite products. These are merely special infinite sequences whose terms are real or complex numbers. Convergent sequences were discussed in Chapter 4 in the setting of general metric spaces. We recall some of the concepts of Chapter 4 as they apply to sequences in C with the usual Euclidean metric. 8.2 CONVERGENT AND DIVERGENT SEQUENCES OF COMPLEX NUMBERS Definition 8.1. A sequence of points in C is said to converge if there is a point p in C with the following property: For every e > 0 there is an integer N (depending on e) such that Ian - PI < e whenever n > N. If converges to p, we write lim, a = p and call p the limit of the sequence. A sequence is called divergent if it is not convergent. A sequence in C is called a Cauchy sequence if it satisfies the Cauchy condition; that is, for every e > 0 there is an integer N such that Ian - aml < e whenever n > N and m > N. Since C is a complete metric space, we know from Chapter 4 that a sequence in C is convergent if, and only if, it is a Cauchy sequence. The Cauchy condition is particularly useful in establishing convergence when we do not know the actual value to which the sequence converges. Every convergent sequence is bounded (Theorem 4.3) and hence an unbounded sequence necessarily diverges. If a sequence {an} converges to p, then every subsequence {ak.} also converges to p (Theorem 4.5). A sequence whose terms are real numbers is said to diverge to + oo if, for every M > 0, there is an integer N (depending on M) such that a, > M whenever n >- N. In this case we write limn- a _ + oo. If (-an) = + oo, we write a oo and say that {an) diverges to - Co. Of course, there are divergent real-valued sequences which do not diverge 183 184 Infinite Series and Infinite Products Def. 8.2 to + oo or to - oo. For example, the sequence {(-1)n(1 + 1/n)} diverges but does not diverge to + o0 or to - co. 8.3 LIMIT SUPERIOR AND LIMIT INFERIOR OF A REAL-VALUED SEQUENCE Definition 8.2. Let {an} be a sequence of real numbers. Suppose there is a real number U satisfying the following two conditions: i) For every s > 0 there exists an integer N such that n > N implies an < U + s. ii) Givens > 0 and given m > 0, there exists an integer n > m such that an> U - s. Then U is called the limit superior (or upper limit) of {an} and we write U = lim sup an. Statement (i) implies that the set {al, a2, ... } is bounded above. If this set is not bounded above, we define lim sup an = + oo. n-00 If the set is bounded above but not bounded below and if {an} has no finite limit superior, then we say lim sup, an = - oo. The limit inferior (or lower limit) of {an} is defined as follows: lim inf an = -lim sup bn, where b = - an for n = 1, 2, .. . n- 00 n- 00 NOTE. Statement (i) means that ultimately all terms of the sequence lie to the left, of U + E. Statement (ii) means that infinitely many terms lie to the right of U - s. It is clear that there cannot be more than one U which satisfies both (i) and (ii). Every real sequence has a limit superior and a limit inferior in the extended real number system R. (See Exercise 8.1.) The reader should supply the proofs of the following theorems: Theorem 8.3. Let {an} be a sequence of real numbers. Then we have: a) lim info-. an < Jim sup.-. an. b) The sequence converges if, and only if, Jim sup.. an and lim inf,, an are both finite and equal, in which case Jim.-. an = lim inf.-.,, an = lim sup.-. an. c) The sequence diverges to + oo if, and only if, lim inf.-, an = lim sup.-,,,, an = +00. d) The sequence diverges to - oo if, and only if, lim inf, an = lim supn-,,, an = -00. Def. 8.7 Infinite Series 185 NOTE. A sequence for which lim info,,, an # lim supn,oo an is said to oscillate. Theorem 8.4. Assume that an < bn for each n = 1, 2.... Then we have: lim inf an < lim inf bn and lim sup an < lim sup bn. n-00 n_00 n-OD n_00 Examples 1. an = (-1)"(1 + 1/n), Jim info,,,) a" _ -1, lim sup an = + 1. 2. a" _ (-1)", lim info,,. an = -1, Jim supra-,o a,, = + 1. 3. an = (-1)" n, lim inf,-,, an = - oo, Jim suPn-.ao an = + oo. 4. an = n2 sin2 (Imr), Jim info.,, an = 0, lim sup an = + 00. 8.4 MONOTONIC SEQUENCES OF REAL NUMBERS Definition 8.5. Let {a,,} be a sequence of real numbers. We say the sequence is increasing and we write anT if an < an+1 for n = 1, 2.... If an > a,,+1 for all n, we say the sequence is decreasing and we write an ' . A sequence is called monotonic if it is increasing or if it is decreasing. The convergence or divergence of a monotonic sequence is particularly easy to determine. In fact, we have Theorem 8.6. A monotonic sequence converges if, and only if, it is bounded. P r o o f . If an / , limn," an = sup {an : n = 1, 2, ... }. If a" N, lim,, a" _ inf {an : n = 1, 2, ... }. 8.5 INFINITE SERIES Let {an} be a given sequence of real or complex numbers, and form a new sequence {sn} as follows: n sn = a1 + ... + an = E ak (n = 1, 2, ... ). (1) k=1 Definition 8.7. The ordered pair of sequences ({an), {sn}) is called an infinite series. The number s,, is called the nth partial sum of the series. The series is said to con-verge or to diverge according as {sn} is convergent or divergent. The following symbols are used to denote the series defined by (1): Go a1 + a2 + - + an + ... , a1 + a2 + a3 + ... , z ak. k=1 NOTE. The letter k used in F_k 1 ak is a "dummy variable" and may be replaced by any other convenient symbol. If p is an integer > 0, a symbol of the form E p bra is interpreted to mean En 1 an where a,, = bn+ p When there is no danger of misunderstanding, we write sbn instead of E ,'=p bra. 186 Infinite Series and Infinite Products Th. 8.8 If the sequence {sn} defined by (1) converges to s, the number s is called the sum of the series and we write Go S = E ak. k=1 Thus, for convergent series the symbol Eak is used to denote both the series and its sum. Example. If x has the infinite decimal expansion x = ao.ala2 (see Section 1.17), then the series Ek o akl0'k converges to x. Theorem 8.8. Let a = Lan and b = Ebn be convergent series. Then, for every pair of constants a and f, the series Y_(aan + fib.) converges to the sum as + fib. That is, 00 00 "0 E(aan+ fbn)=a1: b,. nQ=1 nQ/=1 n=1 Proof. -k=1 (aak + Nbk) = a Ek=1 ak + fi Ek=1 bk. Theorem 8.9. Assume that an Z 0 for each n = 1, 2, ... Then Lan converges if, and only if, the sequence of partial sums is bounded above. Proof. Let sn = a1 + + a. Then sn / and we can apply Theorem 8.6. Theorem 8.10 (Telescoping series). Let {an} and {bn} be two sequences such that an = bn+1 - bn for n = 1, 2,... Then Ean converges if, and only if, limn-. bn exists, in which case we have 00 E an = lim bn - b 1. n=1 n ao Proof. Jk=1 ak = Lk=1 (bk+l - bk) - bn+1 - bl Theorem 8.11 (Cauchy condition for series). The series ian converges if, and only if, for every e > 0 there exists an integer N such that n > N implies <e for each p = 1,2,... (2) Proof. Let sn = Y_,= 1 ak, write sn+ p - sn = an+ 1 + + an+ p, and apply Theorem 4.8 and Theorem 4.6. Taking p = I in (2), we find that limn,, an = 0 is a necessary condition for convergence of Ean. That this condition is not sufficient is seen by considering the example in which an = 1/n. When n = 2m and p = 2m in (2), we find 1 1 2m 1 an+1 +... F an+p= + ... } __ 2m + 1 2m + 2m _ 2m+2m 2 and hence the Cauchy condition cannot be satisfied when e < f. Therefore the series Eh 1 1 /n diverges. This series is called the harmonic series. Th. 8.14 Inserting and Removing Parentheses 8.6 INSERTING AND REMOVING PARENTHESES 187 Definition 8.12. Let p be a function whose domain is the set of positive integers and whose range is a subset of the positive integers such that i) p(n) < p(m), if n < m. Let Ea" and Ebn be two series related as follows: ii) bl=al+a2+...+ap(1), bn+1 = ap(n)+1 + ap(n)+2 + ... + ap(n+1) if n Then we say that Ebn is obtained from Ean by inserting parentheses, and that Ea" is obtained from Ebn by removing parentheses. Theorem 8.13. If Fan converges to s, every series Ebn obtained from Ea" by in-serting parentheses also converges to s. Proof. Let Ea" and Ebn be related by (ii) and write s" = Ek=1 ak, t" = En =1 bk. Then {t"} is a subsequence of {s"}. In fact, t" = sp(n). Therefore, convergence of {s"} to s implies convergence of {t,} to s. Removing parentheses may destroy convergence. To see this, consider the series Ebn in which each term is 0 (obviously convergent). Let p(n) = 2n and let a" = (-1)". Then (i) and (ii) hold but Ea" diverges. Parentheses can be removed if we further restrict Ea" and p. Theorem 8.14. Let Ea., Ebn be related as in Definition 8.12. Assume that there exists a constant M > 0 such that p(n + 1) - p(n) < M for all n, and assume that 1imn,,o a" = 0. Then Ean converges if, and only if, Eb" converges, in which case they have the same sum. Proof. If Ean converges, the result follows from Theorem 8.13. The whole difficulty lies in the converse deduction. Let sn=a1 to=b1 t=limtn. n-' ao Let e > 0 be given and choose N so that n > N implies stn - t I < 2 and Ia"I < 2M If n > p(N), we can find m > N so that N 5 p(m) < n < p(m + 1). [Why?] For such n, we have sn = al + ... + ap(m+l) - (an+1 + an+2 + ... + ap(m+1)) = tm+1 - (an+1 + an+2 + - - - + ap(m+1)), 188 Infinite Series and Infinite Products and hence ISn - tI < Itm+l - tI + Ian+1 + an+2 + ... + ap(m+1)I S Itm+1 - tI + Iap(m)+1I + Iap(m)+21 + + Iap(m+l)I <E+(Am+1)-p(m)) E 0 for each n, the series E0 1 1)"+' an is called an alternating series. Theorem 8.16. If {an} is a decreasing sequence converging to 0, the alternating series Y(-1)"+ 1 an converges. Ifs denotes its sum and sn its nth partial sum, we have the inequality 0 < (-1)"(s - sn) < an+1i for n = 1, 2, ... (3) NOTE. Inequality (3) tells us that when we "approximate" s by sn, the error made has the same sign as the first neglected term and is less than the absolute value of this term. Proof. We insert parentheses in Y_(-1)"+1 an, grouping together two terms at a time. That is, we take p(n) = 2n and form a new series Ebn according to Definition 8.12, with . b1 = a1 - a2, b2 = a3 - a4, ... , bn = a2n-1 - a2,-Since an -+ 0 and p(n + 1) - p(n) = 2, Theorem 8.14 tells us that 7-(-1)"+1 an converges if Ebn converges. But Ebn is a series of nonnegative terms (since an N and its partial sums are bounded above, since bk = a1 - (a2 - a3) - ... - (a2n-2 - a2n-1) - a2n < a1. k=1 Therefore Ebn converges, so F_(- 1)"' an also converges. Inequality (3) is a consequence of the following relations: 00 00 (-1)"(S - sn) = 1 (-1)k+lan+k = r (an+2k-I - an+2k) > 0, k=1 k=1 and 00 (-I)n(S - Sn) = an+l -(an+2k - an+2k+1) < an+1. k=1 Th. 8.19 Real and Imaginary Parts of a Complex Series 8.8 ABSOLUTE AND CONDITIONAL CONVERGENCE 189 Definition 8.17. A series Ea is called absolutely convergent if EIanI converges. It is called conditionally convergent if Y_a converges but EIanI diverges. Theorem 8.18. Absolute convergence of )a implies convergence. Proof. Apply the Cauchy condition to the inequality Ian+1 + ' ' ' + an+pI Ian+1I + - - - + Ian+pI To see that the converse is not true, consider the example E (-1)n+1 n=1 n This alternating series converges, by Theorem 8.16, but it does not converge absolutely. Theorem 8.19. Let Y _an be a given series with real-valued terms and define Pn = lanl + an , q° = lanl - an (n = 1, 2, ... ). (4) 2 2 Then: i) If Ean is conditionally convergent, both Y_pn and Eqn diverge. ii) If EIanI converges, both pn and sqn converge and we have Ej an =E Pn - E q,,. n=1 n=1 n=1 NOTE. P. = an and qn = 0 if an >- 0, whereas q = - an and p = 0 if an < 0. Proof. We have an = Pn - qn, Ianl = Pn + qn. To prove (i), assume that Fan converges and >Ianl diverges. If sq converges, then Y_p,, also converges (by Theorem 8.8), since pn = an + qn. Similarly, if Epn converges, then sqn also converges. Hence, if either Epn or >2q converges, both must converge and we deduce that EIa,I converges, since Ia,,I = Pn + This contradiction proves (i). To prove (ii), we simply use (4) in conjunction with Theorem 8.8. 8.9 REAL AND IMAGINARY PARTS OF A COMPLEX SERIES Let Ecn be a series with complex terms and write cn = an + ibn, where an and bn are real. The series Ea,, and Ebn are called, respectively, the real and imaginary parts of Y _c.. In situations involving complex series, it is often convenient to treat the real and imaginary parts separately. Of course, convergence of both Ean and Eb implies convergence of Ec implies con-vergence of both Ean and Ebn. The same remarks hold for absolute convergence. 190 Infinite Series and Infinite Products Th. 8.20 However, when Ec is conditionally convergent, one (but not both) of Y-an and Ebn might be absolutely convergent. (See Exercise 8.19.) If Y_c" converges absolutely, we can apply part (ii) of Theorem 8.19 to the real and imaginary parts separately, to obtain the decomposition. Ec = E(pn + E(9. + ivn), where F_pn, Y_gn, Eu", F_vn are convergent series of nonnegative terms. 8.10 TESTS FOR CONVERGENCE OF SERIES WITH POSITIVE TERMS Theorem 8.20 (Comparison test). If a" > 0 and b" > 0 for n = 1, 2, ... , and if there exist positive constants c and N such that an < cb" for n > N, then convergence of > b" implies convergence of Ea.. Proof. The partial sums ofY_a" are bounded if the partial sums of Ebn are bounded. By Theorem 8.9, this completes the proof. Theorem 8.21 (Limit comparison test). Assume that an > 0 and bn > 0 for n = 1, 2, ... , and suppose that a lim ° = I. ny00 bn Then Y_an converges if, and only if, Ebn converges. Proof There exists N such that n >- N implies # < anlbn < 2. The theorem fol-lows by applying Theorem 8.20 twice. NOTE. Theorem 8.21 also holds if limnyco anlb" = c, provided that c # 0. If limn.. an/bn = 0, we can only conclude that convergence of Ebn implies con-vergence of Fan. 8.11 THE GEOMETRIC SERIES To use comparison tests effectively, we must have at our disposal some examples of series of known behavior. One of the most important series for comparison purposes is the geometric series. Theorem 8.22. If Jxi < 1, the series 1 + x + x2 + converges and has sum 1/(1 - x). If Ixl > 1, the series diverges. Proof. (1 - x) yk=o xk = Ek=o (xk - xk+1) = 1 -x"". When jxj < 1, we find lim, x- +I = 0. If lxi > 1, the general term does not tend to zero and the series cannot converge. 11. 8.23 The Integral Test 8.12 THE INTEGRAL TEST 191 Further examples of series of known behavior can be obtained very simply by applying the integral test. Theorem 8.23 (Integral test). Let f be a positive decreasing function defined on [1, + oo) such that lim, +,, f(x) = 0. For n = 1, 2, ... , define n sn = E f(k), t = f f(x) dx, d = sn -k=1 1 Then we have: i) 0 < f(n + 1) < dn + 1 <d <f(1), forn= 1,2,... ii) d exists. iii) Y- 1 f(n) converges if, and only if, the sequence converges. iv) 0 < d k -d < f ( k ), f o r k Proof. To prove (i), write ('n+ 1 f(x) dx = n /k+ 1 f(x) dx < (k+ 1 f(k) dx J1 k=1 Jk k=1 Jk n _ > f(k) = k=1 This implies that f(n + 1) = sn+ 1 - sn <- sn+ 1 - tn+ 1 = dn+ 1, and we obtain But we also have n+1 do -to - (sn+1 - sn) = f(x) dx - f(n + 1) (5) nn + 1 J f(n+l)dx-f(n+l)=0, n and hence d < d1 = f(l). This proves (i). But now it is clear that (i) implies (ii) and that (ii) implies (iii). To prove part (iv), we use (5) again to write + 1 n f(n) dx - f(n + 1) = f(n) - f(n + 1). Summing on n, we get 00 Go 05 J ifk> 1. 1-1, n=k When we evaluate the sums of these telescoping series, we get (iv). 192 Infinite Series and Infinite Products Def. 8.24 NOTE. Let D = Then (i) implies 0 < D < f(l), whereas (iv) gives us 0 S k f(k) - J i f (x) dx - D <- f(n). (6) This inequality is extremely useful for approximating certain finite sums by integrals. 8.13 THE BIG OH AND LITTLE OH NOTATION Definition 8.24. Given two sequences and such that b > 0 for all n. We write a = (read: "a is big oh of if there exists a constant M > 0 such that lanl < Mb for all n. We write a = as n --oo (read: "a is little oh of if lim, 0. NOTE. An equation of the form a = c + means a - c = Sim-ilarly, a = c + means a - c = The advantage of this notation is that it allows us to replace certain inequalities by equations. For example, (6) implies E f (k) = Jf(x) dx + D + 0(f (n)). (7) Example 1. Let f(x) = 1/x in Theorem 8.23. We find t = log n and hence Y_1/n diverges. However, (ii) establishes the existence of the limit lim (± 1 - log n) , n-00 k=1 k a famous number known as Euler's constant, usually denoted by C (or by y). Equation (7) becomes k = log n + C + O(nl . (8) k=1 Example 2. Let f(x) = x-S, s ;4 1, in Theorem 8.23. We find that F_n_s converges if s > 1 and diverges if s < 1. For s > 1, this series defines an important function known as the Riemann zeta function: 00 1 C(s) _ s (s > 1). n=1 n For s > 0, s -A 1, we can apply (7) to write 1 _ n1-S - 1 (1 ks 1-s +C(s)+O where C(s) = lim (Y' k=1 k_3 - (hl--' - 1)/(1 - s)). 11. 8.26 Dirichlet's Test and Abel's Test 8.14 THE RATIO TEST AND ROOT TEST Theorem 8.25 (Ratio test). Given a series > an of nonzero complex terms, let r = lim inf n-oc an+1 an n - 00 an+1 an 193 a) The series Ean converges absolutely if R < 1. b) The series Ean diverges if r > 1. c) The test is inconclusive if r < 1 < R. Proof. Assume that R < I and choose x so that R < x < 1. The definition of R implies the existence of N such that Ian+lla.I < x if n > N. Since x = xn+1/x", this means that In+ 1 < 'an' < 1xN1 , if n > N, x and hence IanI < cx" if n > N, where c = IaNIx-N. Statement (a) now follows by applying the comparison test. To prove (b), we simply observe that r > 1 implies Ian+1 1 > IanJ for all n > N for some N and hence we cannot have limn-,, an = 0. To prove (c), consider the two examples En' 1 and Y-n - 2. In both cases, r = R = I but Y_n -1 diverges, whereas F_n - 2 converges. Theorem 8.26 (Root test). Given a series Y_an of complex terms, let p = lim sup V Ianl n-00 a) The series F a,, converges absolutely if p < 1. b) The series Ea diverges if p > 1. c) The test is inconclusive if p = 1. Proof. Assume that p < I and choose x so that p < x < 1. The definition of p implies the existence of N such that lanI < x" for n > N. Hence, ZIa,,I converges by the comparison test. This proves (a). To prove (b), we observe that p > 1 implies lan1 > 1 infinitely often and hence we cannot have limn an = 0. Finally, (c) is proved by using the same examples as in Theorem 8.25. NOTE. The root test is more "powerful" than the ratio test. That is, whenever the root test is inconclusive, so is the ratio test. But there are examples where the ratio test fails and the root test is conclusive. (See Exercise 8.4.) 8.15 DIRICHLET'S TEST AND ABEL'S TEST, R = lim sup All the tests in -the previous section help us to determine absolute convergence of a series with complex terms. It is also important to have tests for determining 194 Infinite Series and Infinite Products Th. 8.27 convergence when the series might not converge absolutely. The tests in this section are particularly useful for this purpose. They all depend on the partial summation formula of Abel (equation (9) in the next theorem). Theorem 8.27. If {an} and {bn} are two sequences of complex numbers, define An=a1 +...+an. Then we have the identity n n E akbk = Anbn+ 1 - E Ak(bk+ 1 - bk) (9) k=1 k=1 Therefore, Ek 1 akbk converges if both the series Ek 1 Ak(bk+ 1 - bk) and the sequence {Anbn+ 1 } converge. Proof. Writing AO = 0, we have n n nn rn E akbk = (Ak - Ak-l)bk = L Akbk - Lj Akbk+l k=1 k=1 k=1 k=1 + Anbn+ 1 The second assertion follows at once from this identity. NOTE. Formula (9) is analogous to the formula for integration by parts in a Riemann-Stieltjes integral. Theorem 8.28 (Dirichlet's test). Let Ian be a series of complex terms whose partial sums form a bounded sequence. Let {bn} be a decreasing sequence which converges to 0. Then Eanbn converges. Proof. Let A. = a1 + + an and assume that IAn1 < M for all n. Then lim Anbn+ 1 = 0. n - 00 Therefore, to establish convergence of we need only show that EAk(bk+ 1 - bk) is convergent. Since bn ' , we have IAk(bk+1 bk)I < M(bk - bk+1) But the series E(bk+ 1 - bk) is a convergent telescoping series. Hence the com-parison test implies absolute convergence of EAk(bk+1 - bk). Theorem 8.29 (Abel's test). The series Eanbn converges if La,, converges and if {bn} is a monotonic convergent sequence. Proof. Convergence of Lan and of {bn} establishes the existence of the limit lim, Anbn+1, where An = a1 + + an. Also, {An} is a bounded sequence. The remainder of the proof is similar to that of Theorem 8.28. (Two further tests, similar to the-above, are given in Exercise 8.27.) Th. 8.30 Geometric Series Y_z" 195 8.16 PARTIAL SUMS OF THE GEOMETRIC SERIES Ez" ON THE UNIT CIRCLE Izl = 1 To use Dirichlet's test effectively, we must be acquainted with a few series having bounded partial sums. Of course, all convergent series have this property. The next theorem gives an example of a divergent series whose partial sums are bounded. This is the geometric series E z" with IzI = 1, that is, with z = eix where x is real. The formula for the partial sums of this series is of fundamental importance in the theory of Fourier series. Theorem 8.30. For every real x 2mn (m is an integer), we have " inx eikx = eix 1 - e = sin (nx/2) ei(n+1)x/2 k=1 1 - e'x sin (x/2) NOTE. This identity yields the following estimate: 1 Isin (x/2)1 Proof. (1 - e") Y_k=1 eikx = Yk (eikx - ei(k+1)x) =eix - ei(n+l)x lishes the first equality in (10). The second follows from the identity inx 1nx12 -inx/2 eix 1 - e = e - e-ei(n+1)x/z 1 - eix eix/z _ e- 1x/2 NOTE. By considering the real and imaginary parts of (10), we obtain E cos kx = sin nx cos (n + 1) x sin x k=1 2 2/ 2 1 - 2 + 2 sin (2n + 1) 2 /sin -x , sin kx = sin nx sin (n + 1) x sin x . k=1 2 2/ 2 Using (10), we can also write n n r ei(2k-I)x = e-ix E eik(2x) = Sin nx einx' k=1 k=1 sin x (10) This estab-(12) (13) (14) an identity valid for every x 96 m7r (m is an integer). Taking real and imaginary 196 Infinite Series and Infinite Products Def. 8.31 parts of (14) we obtain cos (2k - 1)x = sin 2nx k=1 2sinx' sin (2k - 1)x = sine nx k=1 sin x Formulas (12) and (16) occur in the theory of Fourier series. (15) (16) 8.17 REARRANGEMENTS OF SERIES We recall that Z+ denotes the set of positive integers, Z+ = {1, 2, 3, ... }. Definition 8.31. Let f be a function whose domain is Z+ and whose range is Z+, and assume that f is one-to-one on V. Let Ea and >bn be two series such that bn = af(n) for n = 1, 2, ... (17) Then Y_bn is said to be a rearrangement of Ean. NOTE. Equation (17) implies a = bf-1(n) and hence Ean is also a rearrangement of Ebn. Theorem 8.32. Let Ean be an absolutely convergent series having sum s. Then every rearrangement of Ean also converges absolutely and has sum s. Proof. Let {bn} be defined by (17). Then OD Ib11 + ... + IbnI = Iaf(1)I + ... + Iaf(n)I -< E lakl, k=1 so Y, IbnI has bounded partial sums. Hence Ebn converges absolutely. To show thatEbn=s, let +a.. Given c > 0, choose N so that IsN - sI < e/2 and so that Ek 1 IaN+kI < E/2. Then Itn - SI < Itn - SNI+ISN - SI < Itn-SNl+2. Choose M so that {1, 2, ... , N} c {f (1), f (2), ... ,f (M)}. Then n > M implies f (n) > N, and for such n we have Itn-SNI=Ib1+...+bn-(a1+...+aN)I = I af(1) + ... + a f(.) - (a1 + ... + aN)I <IaN+1I+IaN+21+...<<25 since all the terms a1, ... , aN cancel out in the subtraction. Hence, n > M im-plies It,, - sl < s and this means Y_bn = s. Def. 8.34 Subseries 197 8.18 RIEMANN'S THEOREM ON CONDITIONALLY CONVERGENT SERIES The hypothesis of absolute convergence is essential in Theorem 8.32. Riemann discovered that any conditionally convergent series of real terms can be rearranged to yield a series which converges to any prescribed sum. This remarkable fact is a consequence of the following theorem: Theorem 8.33. Let Y_an be a conditionally convergent series with real-valued terms. Let x and y be given numbers in the closed interval [ - oo, + co], with x < y. Then there exists a rearrangement >.b of Ean such that lim inf to =-x and lim sup to = y, n - co n- co where to = b1 + + bn. Proof. Discarding those terms of a series which are zero does not affect its con-vergence or divergence. Hence we might as well assume that no terms of Ean are zero. Let pn denote the nth positive term of > an and let - qn denote its nth negative term. Then Y_pn and Y_gn are both divergent series of positive terms. [Why?] Next, construct two sequences of real numbers, say {xn} and {yn}, such that lim X. = x, lim Y. = y, with xn < yn, Yi > 0. n-ao n- oo The idea of the proof is now quite simple. We take just enough (say k1) positive terms so that P1 +"'+Pkl>Y1, followed by just enough (say r1) negative terms so that P1 +...+A, - q1 q Y2, followed by just enough further negative terms to satisfy the inequality PL + " ' + Pk, - q1 - " ' - qr, + Pk, + 1 + " + pk2 - q,.,+1 - "' - q.2 < x2. These steps are possible since Y_pn and Y_gn are both divergent series of positive terms. If the process is continued in this way, we obviously obtain a rearrangement of Ea.. We leave it to the reader to show that the partial sums of this rearrangement have limit superior y and limit inferior x. 8.19 SUBSERIES Definition 8.34. Let f be a function whose domain is Z + and whose range is an infinite subset of Z+, and assume that f is one-to-one on Z+. Let Y_an and Ebn be 198 Infinite Series and Infinite Products two series such that bn = af(n), ifnEZ+. Th. 8.35 Then Ebn is said to be a subseries of Ea.. Theorem 8.35. If Ean converges absolutely, every subseries Ebn also converges absolutely. Moreover, we have 00 E bn n=1 00 00 < Ibnl < IanI. n=1 n=1 Proof. Given n, let N be the largest integer in the set {f(l), ... , f(n)}. Then N aro bk E Ibkl E Iakl < IakI I k=1 k=1 k=1 lk=n The inequality Ek= 1 Ibkl -< Ek 1 IakI implies absolute convergence of Ebn. Theorem 8.36. Let { f1, f2, ... } be a countable collection of functions, each defined on Z+, having the following properties: a) Each fn is one-to-one on Z+. b) The range fn(Z+) is a subset Qn of Z+. c) {Q1, Q2, ... } is a collection of disjoint sets whose union is Z+. Let Ean be an absolutely convergent series and define bk(n) = afk(n), if n e Z+, k e Z+. Then: i) For each k, En 1 bk(n) is an absolutely convergent subseries of Ea.. ii) If sk = E 1 bk(n), the series Y_, ';O= sk converges absolutely and has the same sum as Ek 1 ak. Proof. Theorem 8.35 implies (i). To prove (ii), let tk = Is1I + .. + IskI. Then 00 00 OD tk 5 E I b1(n)I + ... + I bk(n)I = E (I b1(n)I + ... + I bk(n)I ) n=1 n=1 n=1 00 _ E (Ia.fi(n)I + ... + Iafk(n)I) n=1 But E00 1 (laf,(n)I + ... + Iajk(n)I) < En= 1 lanl This proves that EIskI has bounded partial sums and hence Esk converges absolutely. To find the sum of Esk, we proceed as follows : Let e > 0 be given and choose N so that n >_ N implies 00 nr IakI - L IakI < 2 (18) k=1 k=1 11. 8.39 Double Sequences 199 Choose enough functions f1 i ... , f, so that each term a1, a2, ... , aN will appear somewhere in the sum 00 E ati(n) +...+ n=1 00 E af,( ). n=1 The number r depends on N and hence on F. If n > r and n > N, we have IS1 + S2 + + S, -kI:akl < IaN+lI + IaN+2I +...< (19) =1 2 because the terms a1, a2, ... , ay cancel in the subtraction. Now (18) implies 00 IFak - Eak k=1 k=1 When this is combined with (19) we find s, +...+sn-ak k=1 if n > r, n > N. This completes the proof of (ii). < E, 8.20 DOUBLE SEQUENCES Definition 8.37. A .function f whose domain is Z+ x Z+ is called a double sequence. NOTE. We shall be interested only in real- or complex-valued double sequences. Definition 8.38. If a e C, we write limp,,- f(p, q) = a and we say that the double sequence f converges to a, provided that the following condition is satisfied: For every s > 0, there exists an N such that If(p, q) - al < E whenever both p>Nandq>N. Theorem 8.39. Assume that limp,q. , f(p, q) = a. For each fixed p, assume that the limit limq-,,f(p, q) exists. Then the limit limp-. (limq..,f(p, q)) also exists and has the value a. NOTE. To distinguish limp,q-,,,, f(p, q) from limp-,,,, (limq-. f(p, q)), the first is called a double limit, the second an iterated limit. Proof Let F(p) = limq-d. f(p, q). Given e > 0, choose N1 so that If(p,q)-aj <-, ifp> N1 and q > N1. (20) For each p we can choose N2 so that IF(p) -f(p, q)I < 2 , if q > N2. (21) 200 Infinite Series and Infinite Products Def. 8.40 (Note that N2 depends on p as well as on e.) For each p > N1 choose N2i and then choose a fixed q greater than both N1 and N2. Then both (20) and (21) hold and hence IF(p) - al < e, ifp>N1. Therefore, limp. F(p) = a. NOTE. A similar result holds if we interchange the roles of p and q. Thus the existence of the double limit f(p, q) and of limq-. f(p, q) implies the existence of the iterated limit lim(limf(p,q) P_ q-.ao The following example shows that the converse is not true. Example. Let f(p, q) = 2pq 2, (p = 1, 2,..., q = 1, 2,...). p + q Then limq_. f(p, q) = 0 and hence limp-,,, (limq-, f(p, q)) = 0. But f(p, q) _ when p = q and f(p, q) = 5 when p = 2q, and hence it is clear that the double limit cannot exist in this case. A suitable converse of Theorem 8.39 can be established by introducing the notion of uniform convergence. (This is done in the next chapter in Theorem 9.16.) Further examples illustrating the behavior of double sequences are given in Exercise 8.28. 8.21 DOUBLE SERIES Definition 8.40. Let f be a double sequence and let s be the double sequence defined by the equation p 9 s(p, q) =E E f(m, n). m=1 n=1 The pair (f, s) is called a double series and is denoted by the symbol Lm,n f(m, n) or, more briefly, by Ef(m, n). The double series is said to converge to the sum a if lim s(p, q) = a. p,9- oc Each number f(m, n) is called a term of the double series and each s(p, q) is a partial sum. If Y_f(m, n) has only positive terms, it is easy to show that it con-verges if, and only if, the set of partial sums is bounded. (See Exercise 8.29.) We say F_f(m, n) converges absolutely if El f(m, n)l converges. Theorem 8.18 is valid for double series. (See Exercise 8.29.) Th. 8.42 Rearrangement Theorem 8.22 REARRANGEMENT THEOREM FOR DOUBLE SERIES 201 Definition 8.41. Let f be a double sequence and let g be a one-to-one function defined on Z+ with range Z+ x Z+. Let G be the sequence defined by G(n) = f [g(n)] if n e Z. Then g is said to be an arrangement of the double sequence f into the sequence G. Theorem 8.42. Let Y-f(m, n) be a given double series and let g be an arrangement of the double sequence f into a sequence G. Then a) EG(n) converges absolutely if, and only if Ef(m, n) converges absolutely. Assuming that Ef(m, n) does converge absolutely, with sum S, we have further: b) Y- 1 G(n) = S. c) E 1 f(m, n) and Em=1 f(m, n) both converge absolutely. d) If A. = E 1 f(m, n) and B. = Em= 1 J '(m, n), both series EAm and FB converge absolutely and both have sum S. That is, 00 OD DD 00 E f(m, n) = F, E f(m, n) = S. m=1 n=1 n=1 m=1 Proof. Let T. = IG(1)I + + IG(k)I and let P 9 S(p, q) = E E If(m, n)I M=1 n=1 Then, for each k, there exists a pair (p, q) such that Tk < S(p, q) and, conversely, for each pair (p, q) there exists an integer r such that S(p, q) < T,. These in-equalities tell us that EIG(n)I has bounded partial sums if, and only if, EI f(m, n)) has bounded partial sums. This proves (a). Now assume that EI f(m, n)I converges. Before we prove (b), we will show that the sum of the series Y_G(n) is independent of the function g used to construct G from f. To see this, let h be another arrangement of the double sequence f into a sequence H. Then we have G(n) =.f [g(n)] and H(n) = f [h(n)]. But this means that G(n) = H[k(n)], where k(n) = h-1[g(n)]. Since k is a one-to-one mapping of Z+ onto Z+, the series EH(n) is a rearrangement of EG(n), and hence has the same sum. Let us denote this common sum by S. We will show later that S' = S. Now observe that each series in (c) is a subseries of EG(n). Hence (c) follows from (a). Applying Theorem 8.36, we conclude that EAm converges absolutely and has sum S'. The same thing is true of EBn. It remains to prove that S' = S. 202 Infinite Series and Infinite Products Th. 8.43 For this purpose let T = limp,q., S(p, q). Given e > 0, choose N so that 0 < T - S(p, q) < E/2 whenever p > N and q > N. Now write k p q tk = 1 G(n), s(p, q) = E E f(m, n). n=1 m=1 n=1 Choose M so that tM includes all terms f(m, n) with 1<m<N+1, 1<n<N+1. Then tM - s(N + 1, N + 1) is a sum of terms f(m, n) with either m > N or n > N. Therefore, if n > M, we have s(N+1,N+1)1 <T-S(N+1,N+1)<2. Similarly, IS - s(N+ 1, N + 1)1 <T-S(N+1,N+1)<2. Thus, given e > 0, we can always find M so that It. - SI < e whenever n >- M. Since Jimt = S', it follows that S' = S. NoTE. The series Em= En 1 f(m, n) and E 1 Ym=1 f(m, n) are called "iterated series". Convergence of both iterated series does not imply their equality. For example, suppose 1 , ifm = n + 1, n = 1, 2, ... , f(m,n)= -1, ifm=n-1,n=1,2,..., 0, otherwise. Then E E f(m, n) but E E f(m, n) = 1. M=1 n=1 n=1 m=1 8.23 A SUFFICIENT CONDITION FOR EQUALITY OF ITERATED SERIES Theorem 8.43. Let f be a complex-valued double sequence. Assume that En 1 f(m, n) converges absolutely for each fixed m and that E E If(m, n)I, m=1 n=1 converges. Then: a) The double series Em, f(m, n) converges absolutely. b) The series Y_m=1 f(m, n) converges absolutely for each n. Th. 8.44 Multiplication of Series 203 c) Both iterated series Y_,'= 1 Em=1 f(m, n) and Em=1 E' 1 f(m, n) converge absolutely and we have 00 00 00 00 E E f(m, n) = E E f(m, n) _ E f(m, n). m=1 n=1 n=1 m=1 m,n Proof. Let g be an arrangement of the double sequence f into a sequence G. Then EG(n) is absolutely convergent since all the partial sums of EIG(n)I are bounded by Em=1 En 1 I f(m, n)I. By Theorem 8.42(a), the double series Em,n f(m, n) converges absolutely, and statements (b) and (c) also follow from Theorem 8.42. As an application of Theorem 8.43 we prove the following theorem concerning double series Em,n f(m, n) whose terms can be factored into a function of m times a function of n. Theorem 8.44. Let Eam and Ebb be two absolutely convergent series with sums A and B, respectively. Let f be the double sequence defined by the equation f(m, n) = ambn, if (m, n) e Z+ X Z+. Then Em,n f(m, n) converges absolutely and has the sum AB. Proof We have OD OD '0 OD 00 OD E Iaml E Ibnl = E (iami E Ibnl) _ ± ± Iaml Ibni. M=1 n=1 m=1 n=1 m=1 n=1 Therefore, by Theorem 8.43, the double series Em,n ambn converges absolutely and has sum AB. 8.24 MULTIPLICATION OF SERIES Given two series Ean and Ebn, we can always form the double series Ef(m, n), where f(m, n) = ambn. For every arrangement g off into a sequence G, we are led to a further series EG(n). By analogy with finite sums, it seems natural to refer to Ef(m, n) or to EG(n) as the "product" of Ean and Ebb, and Theorem 8.44 justifies this terminology when the two given series Ean and Ebb are absolutely convergent. However, if either Ean or Ebb is conditionally convergent, we have no guarantee that either E f(m, n) or EG(n) will converge. Moreover, if one of them does converge, its sum need not be AB. The convergence and the sum will depend on the arrangement g. Different choices of g may yield different values of the product. There is one very important case in which the terms f(m, n) are arranged "diag-onally" to produce EG(n), and then parentheses are inserted by grouping together those terms ambn for which m + it has a fixed value. This product is called the Cauchy product and is defined as follows : 204 Infinite Series and Infinite Products Definition 8.45. Given two series Y_.'=0 an and E 0 bn, define Def. 8.45 n if Cn = r LI akbn-kg iJ n = 0, 1, 2, ... (22) k=0 The series F_.'= 0 c,, is called the Cauchy product of Ean and Ebn. NOTE. The Cauchy product arises in a natural way when we multiply two power series. (See Exercise 8.33.) Because of Theorems 8.44 and 8.13, absolute convergence of both Ean and Ebn implies convergence of the Cauchy product to the value 00 00 OD E cn an bn n=0 - (n=0 ) (n=0 ) (23) This equation may fail to hold if both Ean and Ebn are conditionally convergent. (See Exercise 8.32.) However, we can prove that (23) is valid if at least one of Ea,,, Ebn is absolutely convergent. Theorem 8.46 (Mertens). Assume that En=O a,, converges absolutely and has sum A, and suppose E 0 bn converges with sum B. Then the Cauchy product of these two series converges and has sum AB. Proof. Define An = Ek= O ak, B,, = Ek= o bk, Cn = Ek = 0 Ck, where ck is given by (22). Let d,, = B - Bn and en = Ek=0 akdn-k. Then P n P P CP E E akbn-k = 1 E fn(k), n=0 k=0 n=0 k=0 where fn(k) = abn-k, to, ifn>>-k, ifn < k. Then (24) becomes P P P P P P-k P Cp = E E fn// (k) = E E akbn-k = E ak E bm = E akBp-k k=0 n=0 k=0 n=k k=0 m=0 k=0 P E ak(B - dP-k) = APB - ep. k=0 (24) To complete the proof, it suffices to show that ep - 0 as p -+ oo. The sequence {dn} converges to 0, since B = Ebn. Choose M > 0 so that Idni < M for all n, and let K = F,, '=o lanl Given e > 0, choose Nso that n > Nimplies Idnl < e/(2K) and also so that 00 e n=N+1 2M Def. 8.47 Cesdro Summability Then, for p > 2N, we can write I epl <- E Iakdp-kI + L I akd p-kI -< E E Iakl + M E I akI k=0 k=N+1 2K k=0 k=N+1 00 00 <-E lakl +M E Iakl G E+ E= E. 2K k=0 k=N+1 2 2 205 This proves that ep -4 0 asp -+ co, and hence Cp - AB asp - co. A related theorem (due to Abel), in which no absolute convergence is assumed, will be proved in the next chapter. (See Theorem 9.32.) Another product, known as the Dirichlet product, is of particular importance in the Theory of Numbers. We take ao = bo = 0 and, instead of defining cn by (22), we use the formula cn = adb",d, (n = 1, 2, ... ), (25) din where Eden means a sum extended over all positive divisors of n (including 1 and n). For example, c6 = a1b6 + a2b3 + a3b2 + a6b1, and c7 = alb? + a7b1. The analog of Mertens' theorem holds also for this product. The Dirichlet product arises in a natural way when we multiply Dirichlet series. (See Exercise 8.34.) 8.25 CESARO SUMMABILITY Definition 8.47. Let sn denote the nth partial sum of the 'series an, and let (a,,) be the sequence of arithmetic means defined by an sl .} ... S" if n = 1, 2, .. . n (26) The series F_an is said to be Cesdro summable (or (C, 1) summable) if {an} converges. If lim,, vn = S, then S is called the Cesdro sum (or (C, 1) sum) of Y-an, and we write Ean = S (C, 1). Example 1. Let an = z", Iz I = 1, z 0 1. Then s -1 -zn and v -1 - 1 z(1 - z") " 1z z 1-z n(1-z)2 Therefore, zn-1 = (C,1) E 1 - z In particular, 00 E (-1)"-1 = n=1 1 (C, 1). 206 Infinite Series and Infinite Products Example 2. Let an = (-1)"+1n. In this case, 1im sup an = 1, lim inf an = 0, n-00 n-oo Th. 8.48 and hence E(-1)nt1n is not (C, 1) summable. Theorem 8.48. If a series is convergent with sum S, then it is also (C, 1) summable with Cesdro sum S. Proof. Let sn denote the nth partial sum of the series, define an by (26), and introduce to = sn - S, T. = Qn - S. Then we have t1 + ... + to Tn = n (27) and we must prove that Tn -+ 0 as n -+ oo. Choose A > 0 so that each It,, < A. Given e > 0, choose N so that n > N implies It, I < s. Taking n > N in (27), we obtain ITnI < IiI i+...+ItNI+ItN+II+...+It,, <NA+E. n n n Hence, lim sup, IT,I < E. Since s is arbitrary, it follows that limn-W ITnI = 0. NOTE. We have really proved that if a sequence {sn} converges, then the sequence {°n} of arithmetic means also converges and, in fact, to the same limit. Cesiiro summability is just one of a large class of "summability methods" which can be used to assign a "sum" to an infinite series. Theorem 8.48 and Example I (following Definition 8.47) show us that Cesiiro's method has a wider scope than ordinary convergence. The theory of summability methods is an important and fascinating subject, but one which we cannot enter into here. For an excellent account of the subject the reader is referred to Hardy's Divergent Series (Reference 8.1). We shall see later that (C, 1) summability plays an impor-tant role in the theory of Fourier series. (See Theorem 11.15.) 8.26 INFINITE PRODUCTS This section gives a brief introduction to the theory of infinite products. Definition 8.49. Given a sequence {un} of real or complex numbers, let pi = U1, n P2 = u1u2, Pn = u1u2 ... un = Id uk (28) k=1 The ordered-pair of sequences ({un}, { pn}) is called an infinite product (or simply, a product). The number pn is called the nth partial product and u,, is called the nth Th. 8.51 Infinite Products 207 factor of the product. The following symbols are used to denote the product defined by (28): 00 uiu2 ... un ... , ]a un n=1 (29) NOTE. The symbol jjn=N+ 1 un means 1I 1 uN+n We also write 11u,, when there is no danger of misunderstanding. By analogy with infinite series, it would seem natural to call the product (29) convergent if {pn} converges. However, this definition would be inconvenient since every product having one factor equal to zero would converge, regardless of the behavior of the remaining factors. The following definition turns out to be more useful: D e f i n i t i o n 8.50. Given an i n fi n i t e product fl 1 u,,, let p,, = l1k=1 uk. a) If infinitely many factors u are zero, we say the product diverges to zero. b) If no factor u is zero, we say the product converges if there exists a number p 0 such that {p,,} converges to p. In this case, p is called the value of the product and we write p = fl1 u,,. If J p.} converges to zero, we say the product diverges to zero. c) If there exists an N such that n > N implies un # 0, we say H 1 un converges, provided that fln N+1 un converges as described in (b). In this case, the value of the product 1J.'= 1 un is OD u1u2 ... UN un. n=N+1 d) Iln 1 un is called divergent if it does not converge as described in (b) or (c). Note that the value of a convergent infinite product can be zero. But this happens if, and only if, a finite number of factors are zero. The convergence of an infinite product is not affected by inserting or removing a finite number of factors, zero or not. It is this fact which makes Definition 8.50 very convenient. Example. [J 1 (1 + 1/n) and Iln' 2 (1 - 1/n) are both divergent. In the first case, pn = n + 1, and in the second case, pn = 1 /n. Theorem 8.51 (Cauchy condition for products). The infinite product jjun con-verges if, and only if, for every s > 0 there exists an N such that n > N implies Iun+lun+2 "' un+k - 11 < e, for k = 1, 2, 3, ... (30) Proof. Assume that the product IN converges. We can assume that no un is zero (discarding a few terms if necessary). Let pn = u1 un and p = limn-.. P,,. Then p : 0 and hence there exists an M > 0 such that IPnl > M. Now {Pn} satisfies the Cauchy condition for sequences. Hence, given s > 0, there is an N such that n > -N implies IPn+k - Pn1 < eM for k = 1, 2, ... Dividing by IPnI, we obtain (30). 208 Infinite Series and Infinite Products Th. 8.52 Now assume that condition (30) holds. Then n > N implies un # 0. [Why?] Take e = I in (30), let No be the corresponding N, and let qn = t2No+ 1UNo+2 ' ' ' un if n > No. Then (30) implies # < Ignl < 2. Therefore, if {qn} converges, it cannot converge to zero. To show that {qn} does converge, let s > 0 be arbitrary and write (30) as follows: < E. This gives us I qn+k - qnl < EIgnI < ze. Therefore, {qn} satisfies the Cauchy condition for sequences and hence is convergent. This means that the product jlun converges. NOTE. Taking k = 1 in (30), we find that convergence of Hun implies limn,, un = 1. For this reason, the factors of a product are written as un = I + an. Thus convergence of 1(1 + an) implies limn.cc an = 0. Theorem 8.52. Assume that each an > 0. Then the product f(1 + an) converges if, and only if, the series Y _a,, converges. Proof. Part of the proof is based on the following inequality: I + x < ex. (31) Although (31) holds for all real x, we need it only for x Z 0. When x > 0, (31) is a simple consequence of the Mean-Value Theorem, which gives us ex - I = xex°, where 0 < x0 < x. Since ex0 1, (31) follows at once from this equation. Now let sn = a, + a2 + -+ an, pn = (1 + a,)(1 + a2) (1 + an). Both sequences {sn} and {pn} are increasing, and hence to prove the theorem we need only show that {sn} is bounded if, and only if, {p,,} is bounded. First, the inequality pn > Sn is obvious. Next, taking x = ak in (31), where k = 1, 2, ... , n, and multiplying, we find pn < en. Hence, {sn} is bounded if, and only if, {pn} is bounded. Note that {pn} cannot converge to zero since each pn >- 1. Note also that Pn -'+ao if sn--+Co. Definition 8.53. The product jl(1 + an) is said to converge absolutely if Ij(1 + Ial) converges. Theorem 8.54. Absolute convergence of jl(1 + an) implies convergence. Proof. / Use the Cauchy condition along with the inequality l(1 + a.+,)(' + a,,+2) ... (1 + an+k) - 11 < (1 + Ian+il)(1 + Ian+2I) ... (1 + Ian+kI) - 1 Th. 8.56 Euler's Product 209 NOTE. Theorem 8.52 tells us that 11(1 + an) converges absolutely if, and only if, 'an converges absolutely. In Exercise 8.43 we give an example in which 11(1 + an) converges but Ea diverges. A result analogous to Theorem 8.52 is the following: Theorem 8.55. Assume that each an > 0. Then the product 11(1 -converges if, and only if, the series Ea converges. Proof. Convergence of Ea. implies absolute convergence (and hence convergence) of 11(1 - an). To prove.the converse, assume that Ea diverges. If does not converge to zero, then 11(1 - an) also diverges. Therefore we can assume that a,, -+ 0 as n --> oo. Discarding a few terms if necessary, we can assume that each an < 1. Then each factor 1 - a > I (and hence : 0). Let p, = (1 - a1)(1 - a2) ... (1 -q _ (1 + a1)(1 + a2) ... (1 + an)-Since we have (1 - ak)(1 + ak) = 1 - a' <_ 1, we can write p,< < But in the proof of Theorem 8.52 we observed that qn -> + oo if Ea diverges. Therefore, pn - 0 as n -- oo and, by part (b) of Definition 8.50, it follows that 11(1 -diverges to 0. 8.27 EULER'S PRODUCT FOR THE RIEMANN ZETA FUNCTION We conclude this chapter with a theorem of Euler which expresses the Riemann zeta function C(s) = El 1 n_s as an infinite product extended over all primes. Theorem 8.56. Let pk denote the kth prime number. Then ifs > 1 we have 0 0 1 r(S)= s= II n=1 n k=1 1- pk The product converges absolutely. Proof. We consider the partial product P. = j1k 1 (1 - pk s)-1 and show that Pm -+ C(s) as m -+ oo. Writing each factor as a geometric series we have pm= M I+ + +...1 k=1 A A'S a product of a finite number of absolutely, convergent series. When we multiply these series together and rearrange the terms according to increasing denominators, we get another absolutely convergent series, a typical term of which is 1 1 Pi's p22s ... pm s - ns where n = Pi' . - - Pm 210 Infinite Series and Infinite Products and each a; >- 0. Therefore we have P. = 1 , i n s where Y_1 is summed over those n having all their prime factors <pm. By the unique factorization theorem (Theorem 1.9), each such n occurs once and only once in Y_1. Subtracting Pm from c(s) we get °° 1 r 1 1 C(S) - P. = -m n=1 ns1' ns 2 ns where Z2 is summed over those n having at least one prime factor >pm. Since these n occur among the integers >pm, we have IC(S) - PmI < S n>p.n n As m - oo the last sum tends to 0 because Y n-' converges, so Pm -4 (s). To prove that the product converges absolutely we use Theorem 8.52. The product has the form 11(1 + ak), where 1 1 .. . ak = s + + 2s pk Pk The series >ak converges absolutely since it is dominated by En-s. Therefore 11(1 + ak) also converges absolutely. EXERCISES Sequences 8.1 a) Given a real-valued sequence {an) bounded above, let un = sup {ak : k >- n}. Then un and hence U = limn un is either finite or - oo. Prove that U = lim sup an = lim (sup {ak : k >- n}). n- CO n-00 b) Similarly, if {an} is bounded below, prove that V = lim inf an = lim (inf {a : k >- n}). n- ao n- o0 If U and V are finite, show that: c) There exists a subsequence of {an} which converges to U and a subsequence which converges to V. d) If U = V, every subsequence of {an} converges to U. 8.2 Given two real-valued sequences {an} and {bn} bounded below. Prove that a) lim sup,, (an + bn) 5 lim sup, an + lim sup,,-,,, bn. Exercises 211 b) lim sup... (lim sup.-,, an)(lim sup.-, bn) if an > 0, bn > 0 for all n, and if both lim sup,,,,, an and lim supny,,, b,, are finite or both are infinite. 8.3 Prove Theorems 8.3 and 8.4. 8.4 If each an > 0, prove that lim inf as +1 < lim inf Van <- lim sup Ian <- lim sup a" n-.ao an n-.co n-.oo n- 00 an 8.5 Let an = n"/n!. Show that limn-,,, an+1/an = e and use Exercise 8.4 to deduce that Jim n = e. n- oo (n!)1/" 8.6 Let {a,,} be a real-valued sequence and let an = (a1 + - - - + a,,)/n. Show that lim inf an < lim inf an < lim sup an <- lim sup an. n-Go n-ao n -.oo n -.w 8.7 Find lim sup,, a,, and lim infn_ an if an is given by a) cos n, b) I + 1 cos nn, c) n sin nir , n 3 d) sin 2 cos 2 , e) (-1)"n/(1 + n)", f) 3 - NOTE. In (f), [x] denotes the greatest integer <-x. 8.8 Let a,, = Y"=1 11V. Prove that the sequence {an} converges to a limit p in the interval 1 < p < 2. In each of Exercises 8.9 through 8.14, show that the real-valued sequence {an} is con-vergent. The given conditions are assumed to hold for all n >- 1. In Exercises 8.10 through 8.14, show that {an} has the limit L indicated. 8.9 lanl < 2, lan+2 - an+1l < lan+i - a2.1-8.10 a1 0, a2 0, a,,+2 = (anan+l)1/2, L = (a1a2)1/3-8.11 a1 = 2, a2 = 8, a2n+1 = Z a2n + a2n+1)+ a2n+2 -- a2na2n-1 L = 4. 11( a2n+ 1 8.12 a1 - -i-, 3an+1 = 2 + L = 1. Modify a1 to make L = -2. 8.13 a1 = 3, an+l - 3(1 + L V3. 3 + an 8.14 an = where b1 = b2 = 1, bnt2 = b,, + bn+1, L = 1 +'15 bn 2 Hint. Show that bn+2bn - b,+1 = (-1)"+i and deduce that la,, - an+,1 < n'2, if n > 4. 212 Series Infinite Series and Infinite Products 8.15 Test for convergence (p and q denote fixed real numbers). 00 a) E n3e-", n=1 00 c) E p"n° (P > 0), n=1 00 e) E n-n=1 g) i) I En log (1 + 1/n)' 00 nE n log n (log log n)°' k) 7(V1 + n2 - n), n=1 00 m) E (Jn - 1)", n=1 b) E (log n)°, n=2 00 d) E n° - nq n=2 f) 00 1 R n n=1 p - q h) J) CO 1 E (log n)logn (0 < q < P), (0- N. Hint. For sufficiency, show that 2 (n - 1)/n! is a tele-scoping series with sum 1. 8.18 Let p and q be fixed integers, p q >- 1, and let pn n (-1)k+1 X"= E k, s n = E k k=qn+1 k=1 a) Use formula (8) to prove that limn, xn = log (p/q). b) When q = 1, p = 2, show that stn = x,, and deduce that n (-1)n+1 n=1 = log 2. n c) Rearrange the series in (b), writing alternately p positive terms followed by q negative terms and use (a) to show that this rearrangement has sum log 2 + I log (p/q). n=1 (-1)"+1(1/(3n - 2) - 1/(3n - 1)). d) Find the sum of 1' I 8.19 Let cn = -an + ibn, where an = (-1)"/N/n, bn = 1/nz. Show that Y-c,, is conditionally convergent. Exercises 213 8.20 Use Theorem 8.23 to derive the following formulas: a) log k = I log2 n + A + O (log n1 n /1 (A is constant). =1 k 2 b), k2 k log k = log (log n) + B + O(n log n) (B is constant). 8.21 If 0 < a <- 1, s > 1, define C(s, a) _ r°=o (n + a)-s. a) Show that this series converges absolutely for s > 1 and prove that k k C(s, -/ = ksZ(s) if k = 1, 2, ... , h=1 where C(s) = C(s, 1) is the Riemann zeta function. b) Prove that En 1 (-1)n-'Ins = (I - 21 -%(s) if s > 1. 8.22 Given a convergent series Y_a,,, where each an >- 0. Prove that converges if p > 1. Give a counterexample for p = #. 8.23 Given that Ean diverges. Prove that Enan also diverges. 8.24 Given that F a. converges, where each an > 0. Prove that L(anan+ 1)1 /2 also converges. Show that the converse is also true if {an} is monotonic. 8.25 Given that Ean converges absolutely. Show that each of the following series also converges absolutely: a a) E an, b) E 1 +n a (if no a = -1), 2 C) n 1 + an 8.26 Determine all real values of x for which the following series converges: 1 sin I 1 n1 n nx 8.27 Prove the following statements: a) Y_anbn converges if Ean converges and if E(bn - bn+ 1) converges absolutely. b) Eanbn converges if Ean has bounded partial sums and if E(bn - bn+1) converges absolutely, provided that bn 0 as n -+ oo. Double sequences and double series 8.28 Investigate the existence of the two iterated limits and the double limit of the double sequence f defined by a) f(p, q) = I , b) f(p, q) = p P + q p + q 214 Infinite Series and Infinite Products c) f(P, q) _ (-1)°p p + q e)f(P,q)=(-1)a q g) f(P, q) = cos p q d) f(P, q) _ (-1)"+4 (1 + 1) p q f) f(p, q) = (-1)n+9, h)f(P,9)= P sin" 9 n n=1 P Answer. Double limit exists in (a), (d), (e), (g). Both iterated limits exist in (a), (b), (h). Only one iterated limit exists in (c), (e). Neither iterated limit exists in (d), (f). 8.29 Prove the following statements: a) A double series of positive terms converges if, and only if, the set of partial sums is bounded. b) A double series converges if it converges absolutely. c) m ne-cm2+"Z converges. 8.30 Assume that the double series a(n)x'" converges absolutely for jxj < 1. Call its sum S(x). Show that each of the following series also converges absolutely for lxi < I and has sum S(x) : 00 x" 00 1: a(n) E " , A(n)x", where A(n) = E a(d). n=1 I - x n=1 din 8.31 If a is real, show that the double series (m + in)converges absolutely if, and only if, a > 2. Hint. Let s(p, q). _ YP"=1 Eq1 Im + inI -°. The set {m+in:m= 1,2,...,p,n= 1,2,...,p} consists of p2 complex numbers of which one has absolute value 'J2, three satisfy 11 + 2i j <- I m + inj <- 2N/2, five satisfy 11 + Y J <- I m + inj 5 3'J2, etc. Verify this geometrically and deduce the inequality /2°2n-1 ° 2n- 1 2 E nn 5 s(p, P) : E (n2 + 1)a/2 n=1 n=1 8.32 a) Show that the Cauchy product of En (, (-1)"+1/.n + 1 with itself is a divergent series. b) Show that the Cauchy product of (-1)n+1/(n + 1) with itself is the series n1) 2E(+11 I12+...+n) n-1 Does this converge? Why? 8.33 Given two absolutely convergent power series, say Y_n o and Y_R o bnx", having sums A(x) and B(x), respectively, show that 10 cnx" = A(x)B(x) where C" = akbn-k k=0 Exercises 215 8.34 A series of the form Y,'=1 an/n-' is called a Dirichlet series. Given two absolutely convergent Dirichlet series, say an/ns and Y_', bn/ns, having sums A(s) and B(s), respectively, show that Y_,'=1 cn/ns = A(s)B(s) where cn = Ldjn adbnid 8.35 If C(s) = En 1 1/n, s > 1, show that C2(s) = d(n)/n?, where d(n) is the number of positive divisors of n (including 1 and n). Cesaro summability 8.36 Show that each of the following series has (C, 1) sum 0: a) 1 - 1 - 1 + I + 1 - 1 - l + I + 1 - - + + . b) -- 1 +f+I - 1 +f+ - 1 c) cos x + cos 3x + cos 5x + (x real, x j4 mrz). 8.37 Given a series Ea,,, let n n n sn= Eak, to= Ekak, vn=->sk. k=1 k=1 n k=1 Prove that a) to = (n + 1)sn - nvn. b) If Fan is (C, 1) summable, then >2an converges if, and only if, to = o(n) as n -1- 00. c) Y a. is (C, 1) summable if, and only if, Y_n 1 tn/n(n + 1) converges. 8.38 Given a monotonic sequence {an} of positive terms, such that limn, a,, = 0. Let n n n Sn = E ak, un = E (- 1)kak, V. = L.d (- 1)kSk. k=1 k=1 k=1 Prove that : a) vn = 4 un + (-1)"Sn/2. b) ins,, is (C, 1) summable and has Cesaro sum (-1)nan c) znw= (-1)"(1 + I + + i/n) _ -log 2 (C, 1). Infinite products 8.39 Determine whether or not the following infinite products converge. Find the value of each convergent product. a) 1 2 2 ( n(n + 1)) ' b) (1-n-2), ao n3 - 1 C) 11 n3+1, d) 1 00 Al 00 111 0 + z2") if IzI < 1. 8.40 If each partial sum sn of the convergent series >.an is not zero and if the sum itself is not zero, show that the infinite product a1 HnD= 2 (1 + an/s. 1) converges and has the value Y_n 1 an. 216 Infinite Series and Infinite Products 8.41 Find the values of the following products by establishing the following identities and summing the series: / (i+2"-2 =2E2-". b) ft 1+ n2 )=2v' 00 1) "=1 n=2 1 n=1 n(n + 8.42 Determine all real x for which the product lI' 1 cos (x/2") converges and find the value of the product when it does converge. 8.43 a) Let a,, _ (-1)"/,[n for n = 1, 2.... Show that II(1 + a") diverges but that Y_a,, converges. b) Let a2n_ 1 = -1/Jn, a2n = 1/,J + 1/n for n = 1, 2.... Show that Il(1 + an) converges but that Ean diverges. 8.44 Assume that a,, >: 0 for each n = 1, 2.... Assume further that a2n a2n+2 < a2,+1 < 1 + a2n for n = 1, 2, .. . Show that Ilk 1 (1 + (-1)kak) converges if, and only if, Y_k 1 (-1)kak converges. 8.45 A complex-valued sequence {f(n)} is called multiplicative if f(1) = 1 and if f(mn) _ f(m)f(n) whenever m and n are relatively prime. (See Section 1.7.) It is called com-pletely multiplicative if f(1) = I and f(mn) = f(m)f(n) for all m and n. a) If {,'(n)} is multiplicative and if the series >f(n) converges absolutely, prove that ,f(n)= II{1 +f(pk)+f(pk)+...}, n=1 k=1 where pk denotes the kth prime, the product being absolutely convergent. b) If, in addition, {f(n)} is completely multiplicative, prove that the formula in (a) becomes 00 W I R=1 k=1 I - f(pk) . Note that Euler's product for C(s) (Theorem 8.56) is the special case in which f(n) = n-s. 8.46 This exercise outlines a simple proof of the formula C(2) = n2/6. Start with the inequality sin x < x < tan x, valid for 0 < x < it/2, take reciprocals, and square each member to obtain cot2 x < 1 < 1 + cot2 X. x2 Now put x = k7r/(2m + 1), where k and m are integers, with 1 <- k <- m, and sum on k to obtain '" kn (2m + 1)2 m 1 cot22m+ 1 < n2k2 < m + k=1 k=1 E cot2 k=1 kn 2m+1' References 217 Use the formula of Exercise 1.49(c) to deduce the inequality m(2m - 1)n2 "I 2m(m + 1)n2 3(2m + 1)2 < E k2 < 3(2m + 1)2 Now let m -4 oo to obtain C(2) = n2/6. 8.47 Use an argument similar to that outlined in Exercise 8.46 to prove that C(4) = rr4/90. SUGGESTED REFERENCES FOR FURTHER STUDY 8.1 Hardy, G. H., Divergent Series. Oxford University Press, Oxford, 1949. 8.2 Hirschmann, I. I., Infinite Series. Holt, Rinehart and Winston, New York, 1962. 8.3 Knopp, K., Theory and Application of Infinite Series, 2nd ed. R. C. Young, trans-lator. Hafner, New York, 1948. CHAPTER 9 SEQUENCES OF FUNCTIONS 9.1 POINTWISE CONVERGENCE OF SEQUENCES OF FUNCTIONS This chapter deals with sequences { fn} whose terms are real- or complex-valued functions having a common domain on the real line R or in the complex plane C. For each x in the domain we can form another sequence { fn(x)} whose terms are the corresponding function values. Let S denote the set of x for which this second sequence converges. The function f defined by the equation f(x) = lim fn(x), if x e S, n- 00 is called the limit function of the sequence { fn}, and we say that { fn} converges pointwise to f on the set S. Our chief interest in this chapter is the following type of question : If each function of a sequence {f,,} has a certain property, such as continuity, differen-tiability, or integrability, to what extent is this property transferred to the limit function? For example, if each function fn is continuous at c, is the limit function f also continuous at c? We shall see that, in general, it is not. In fact, we shall find that pointwise convergence is usually not strong enough to transfer any of the properties mentioned above from the individual terms f to the limit function f Therefore we are led to study stronger methods of convergence that do preserve these properties. The most important of these is the notion of uniform convergence. Before we introduce uniform convergence, let us formulate one of our basic questions in another way. When we ask whether continuity of each fn at c implies continuity of the limit function fat c, we are really asking whether the equation lim fn(x) = .fn(c), implies the equation X--C lim f(x) = f(c). (1) X-C But (1) can also be written as follows: lim Jim fn(x) = Jim Jim fn(x). X-+C n-oo X-.c (2) Therefore our question about continuity amounts to this: Can we interchange the limit symbols in (2)? We shall see that, in general, we cannot. First of all, the limit in (1) may not exist. Secondly, even if it does exist, it need not be equal to 218 Sequences of Real-Valued Functions 219 f(c). We encountered a similar situation in Chapter 8 in connection with iterated series when we found that Em=1 Eh 1 f(m, n) is not necessarily equal to Ln 1 Lm= 1 f(m, n). L The general question of whether we can reverse the order of two limit pro-cesses arises again and again in mathematical analysis. We shall find that uniform convergence is a far-reaching sufficient condition for the validity of interchanging certain limits, but it does not provide the complete answer to the question. We shall encounter examples in which the order of two limits can be interchanged although the sequence is not uniformly convergent. 9.2 EXAMPLES OF SEQUENCES OF REAL-VALUED FUNCTIONS The following examples illustrate some of the possibilities that might arise when we form the limit function of a sequence of real-valued functions. fa (x) + X2n x2n , n=1,2,3. f (x) = lim f" (x) . n-M Figure 9.1 Example 1. A sequence of continuous functions with a discontinuous limit function. Let f"(x) = x2n/(1 + x2") if x e R, n = 1, 2,... The graphs of a few terms are shown in Fig. 9.1. In this case f"(x) exists for every real x, and the limit function f is given by Each f" is continuous on R, but f is discontinuous at x = landx= -1. Example 2. A sequence of functions for which f o f"(x) dx :0 f o lim,- w f"(x) dx. Let f"(x) = n2x(1 - x)" if x e R, n = 1, 2, ... If 0 < x 5 1 the limit f(x) f"(x) exists and equals 0. (See Fig. 9.2.) Hence fo f(x) dx = 0. But f 1 f"(x) dx = n2 f 1 x(1 - x)" dx Jo = n2 101(1 - t)tdt = n2 -n2 = n2 n+1 n+2 (n+1)(n+2) 220 Sequences of Functions Figure 9.2 n= so 5 f .(X) dx = 1. In other words, the limit of the integrals is not equal to the integral of the limit function. Therefore the operations of "limit" and "integration" cannot always be interchanged. Example 3. A sequence of differentiable functions (f.) with limit 0 for which (f.} diverges. Let f .(x) = (sin n = 1, 2.... Then f (x) = 0 for every x. But f,(x) = Vn cos nx, so limp.. f,,(x) does not exist for any x. (See Fig. 9.3.) Figure 9.3 9.3 DEFINITION OF UNIFORM CONVERGENCE Let { f } be a sequence of functions which converges pointwise on a set S to a limit function f. This means that for each point x in S and for each s > 0, there exists an N (depending on both x and c) such that n > N implies 1 f(x) - f (x)I < e. Th. 9.2 Uniform Convergence 221 If the same N works equally well for every point in S, the convergence is said to be uniform on S. That is, we have Definition 9.1. A sequence of functions {f.} is said to converge uniformly to f on a set S if, for every s > 0, there exists an N (depending only on E) such that n > N implies f (x) - f(x)J < 8, for every x in S. We denote this symbolically by writing f -+ f uniformly on S. When each term of the sequence {f.) is real-valued, there is a useful geometric interpretation of uniform convergence. The inequality If(x) - f(x)I < s is then equivalent to the two inequalities f(x) - s < A(x) < f(x) + S. (3) If (3) is to hold for all n > N and for all x in S, this means that the entire graph of f (that is, the set {(x, y) : y = f (x), x e S}) lies within a "band" of height 2E situated symmetrically about the graph off (See Fig. 9.4.) Figure 9.4 A sequence {f.) is said to be uniformly bounded on S if there exists a constant M > 0 such that I f,(x)I < M for all x in S and all n. The number M is called a uniform bound for f f.}. If each individual function is bounded and if f - f uniformly on S, then it is easy to prove that {f.} is uniformly bounded on S. (See Exercise 9.1.) This observation often enables us to conclude that a sequence is not uniformly convergent. For instance, a glance at Fig. 9.2 tells us at once that the sequence of Example 2 cannot converge uniformly on any subset containing a neighborhood of the origin. However, the convergence in this example is uniform on every compact subinterval not containing the origin. 9.4 UNIFORM CONVERGENCE AND CONTINUITY Theorem 9.2. Assume that f -+ f uniformly on S. If each f is continuous at a point c of S, then the limit function f is also continuous at c. NOTE. If c is an accumulation point of S, the conclusion implies that lim lim f (x) = lim lim f (x). x-c n-oo n-oo x-c 222 Sequences of Functions Th. 9.3 Proof. If c is an isolated point of S, then f is automatically continuous at c. Suppose, then, that c is an accumulation point of S. By hypothesis, for every e > 0 there is an M such that n >- M implies f(x)I < 3 for every x in S. Since fm is continuous at c, there is a neighborhood B(c) such that x e B(c) n S implies I fM(x) - fM(c)I < 3 But If(x) - f(c)I < If(x) - fM(x)I + I fM(x) - fM(C)I + I fM(c) - f(c)I. If x e B(c) n S, each term on the right is less than s/3 and hence If(x) - f(c)I < s. This proves the theorem. NOTE. Uniform convergence of {f.} is sufficient but not necessary to transmit continuity from the individual terms to the limit function. In Example 2 (Section 9.2), we have a nonuniformly convergent sequence of continuous functions with a continuous limit function. 9.5 THE CAUCHY CONDITION FOR UNIFORM CONVERGENCE Theorem 9.3. Let {f.) be a sequence of functions defined on a set S. There exists a function f such that f, -+ f uniformly on S if, and only if, the following condition (called the Cauchy condition) is satisfied: For every e > 0 there exists an N such that m > N and n > N implies Ifm(x) - f(x)I < e, for every x in S. Proof. Assume that f -+ f uniformly on S. Then, given e > 0, we can find N so that n > N implies If(x) - f(x)I < e/2 for all x in S. Taking m > N, we also have I fm(x) - f(x)I < e/2, and hence I fm(x) - f (x)I < e for every x in S. Conversely, suppose the Cauchy condition is satisfied. Then, for each x in S, the sequence converges. Let f(x) = if x e S. We must show that f -+ f uniformly on S. If e > 0 is given, we can choose N so that n > N implies I f (x) - f +k(x)I < e/2 for every k = 1, 2, ... , and every x in S. There-fore, limk-. I f,(x) - f +k(x)I = If(x) - f(x)I < e/2. Hence, n > N implies I f(x) e for every x in S. This proves that f - f uniformly on S. NOTE. Pointwise and uniform convergence can be formulated in the more general setting of metric spaces. If f. and f are functions from a nonempty set S to a metric space (T, dT), we say that f -+ f uniformly on S, if, for every s > 0, there is an N (depending only on e) such that n N implies f(x)) < c for all x in S. Th. 9.7 Uniform Convergence of Infinite Series 223 Theorem 9.3 is valid in this more general setting and, if S is a metric space, Theorem 9.2 is also valid. The same proofs go through, with the appropriate replacement of the Euclidean metric by the metrics ds and dT. Since we are primarily interested in real- or complex-valued functions defined on subsets of R or of C, we will not pursue this extension any further except to mention the following example. Example. Consider the metric space (B(S), d) of all bounded real-valued functions on a nonempty set S, with metric d(f g) = if - gll, where if II = supXEs If(x)I is the sup norm. (See Exercise 4.66.) Thenfn --+ f in the metric space (B(S), d) if and only if fn -> f uniformly on S. In other words, uniform convergence on S is the same as ordinary con-vergence in the metric space (B(S), d). 9.6 UNIFORM CONVERGENCE OF INFINITE SERIES OF FUNCTIONS Definition 9.4. Given a sequence { fn} of functions defined on a set S. For each x in S, let n Sn(x) _ E Jf(x) (n = 1, 2, ... ). (4) k=1 If there exists a function f such that sn -+ f uniformly on S, we say the series >fn(x) converges uniformly on S and we write co E fn(x) = f(x) (uniformly on S). n=1 Theorem 9.5 (Cauchy condition for uniform convergence of series). The infinite series Efn(x) converges uniformly on S if, and only if, for every E > 0 there is an N such that n > N implies n+p E fk(x)I < E, for each p = 1, 2, ... , and every x in S. k=n+1 Proof. Define sn by (4) and apply Theorem 9.3. Theorem 9.6 (Weierstrass M-test). Let {Mn} be a sequence of nonnegative numbers such that 0 < If,,(x)I < Mn, for n = 1, 2, ... , and for every x in S. Then >fn(x) converges uniformly on S if Y-Mn converges. Proof. Apply Theorems 8.11 and 9.5 in conjunction with the inequality n+p E fk(x) k=n+i + p n[ < G, Mk. k=n+1 Theorem 9.7. Assume that Yfn(x) = f(x) (uniformly on S). If eachfn is continuous at a point x0 of S, then f is also continuous at x0. 224 Sequences of Functions Proof. Define s" by (4). Continuity of each fn at xo implies continuity of s" at x0, and the conclusion follows at once from Theorem 9.2. NOTE. If xo is an accumulation point of S, this theorem permits us to interchange limits and infinite sums, as follows : lim E fn(x) = E lim fn(x). x-'xo n=1 n=1 x-'xo 9.7 A SPACE-FILLING CURVE We can apply Theorem 9.7 to construct a space-filling curve. This is a continuous curve in R2 that passes through every point of the unit square [0, 1] x [0, 1]. Peano (1890) was the first to give an example of such a curve. The example to be presented here is due to I. J. Schoenberg (Bulletin of the American Mathematical Society, 1938) and can be described as follows: Let 0 be defined on the interval [0, 2] by the following formulas : 0, if0<t< orif3 <t<2, 3t- 1, if1<t<, -3t+5, if3<t3. Extend the definition of ¢ to all of R by the equation q5(t + 2) = 0(t). This makes 0 periodic with period 2. (The graph of 0 is shown in Fig. 9.5.) Figure 9.5 Now define two functions f1 and f2 by the following equations : 00 {{ °D /i(t) = ± 0(32n 2t)' f2(t) = ± 0(3 2n2n- 1 t) Both series converge absolutely for each real t and they converge uniformly on R. In fact, since 10(t)l < 1 for all t, the Weierstrass M-test is applicable with Mn = 2-". Since 0 is continuous on R, Theorem 9.7 tells us that f, and f2 are also continuous on R. Let f = (fl, f2) and let F denote the image of the unit interval [0, 1] under f. We will show that F "fills" the unit square, i.e., that F=[0,1] x [0, 1]. First, it is clear that 0 < fl(t) < I and 0 < f2(t) < 1 for each t, since Y 1 2-" = 1. Hence, F is a subset of the unit square. Next, we must show that Th. 9.8 Uniform Convergence and Integration 225 (a, b) e F whenever (a, b) e [0, 1] x [0, 1]. For this purpose we write a and b in the binary system. That is, we write 00 00 _ E a" _ E b" a n=12"' b n=12"' where each a" and each bn is either 0 or 1. (See Exercise 1.22.) Now let 00 c = 2 E C" where c2n_1 = an and c2n = b", n = 1, 2, .. . n=1 3" Clearly, 0 < c < 1 since 2y_' 1 3-" = 1. We will show that f1(c) = a and that f2(c) = b. If we can prove that 0(3kc) = ek+1, for each k = 0, 1, 2, ... , (5) then we will have j(32n-2c) = c2,,_1 = a" and 0(32n-1c) = c2n = bn, and this will give us f1(c) = a, f2(c) = b. To prove (5), we write k ao 3kc = 2 E cn E c'. n=1 3"-k + 2 n-k+1 3n k = (an even integer) + dk, where dk = 2En=1 Cn+k/3". Since (p has period 2, it follows that 0(3kc) = 0(dk) If ck+ 1 = 0, then we have 0 < dk < 2Y00 2 3-m and hence 0(dk) = 0. Therefore, 0(3kc) = ck+ 1 in this case. The only other case to consider is ck+ 1 = 1. But then we get < dk < 1 and hence 0(dk) = 1. Therefore, 4 (3kc) = ck+ 1 in all cases and this proves that f1(c) = a, f2(c) = b. Hence, IF fills the unit square. 9.8 UNIFORM CONVERGENCE AND RIEMANN-STIELTJES INTEGRATION Theorem 9.8. Let a be of bounded variation on [a, b]. Assume that each term of the sequence { fn} is a real-valued function such that fn e R(a) on [a, b] for each n = 1, 2, ... Assume that fn -+ f uniformly on [a, b] and define g"(x) = f a fn(t) da(t) if x e [a, b], n = 1, 2, ... Then we have: a) f e R(a) on [a, b]. b) gn -+ g uniformly on [a, b], where g(x) = f; f(t) da(t). NOTE. The conclusion implies that, for each x in [a, b], we can write lim f nda(t) = f x Jim fn(t) do(t). n-ao J"f(t) a ,Ja This property is often described by saying that a uniformly convergent sequence can be integrated term by term. 226 Sequences of Functions Th. 9.9 Proof. We can assume that a is increasing with a(a) < a(b). To prove (a), we will show that f satisfies Riemann's condition with respect to a on [a, b]. (See Theorem 7.19.) Given is > 0, choose N so that I f(x) - fN(X) I < E , for all x in [a, b]. 3[a(b) - a(a)] Then, for every partition P of [a, b], we have I U(P, f - IN, a)I <- 3 and I L(P, f - IN, a)1 .5 3 (using the notation of Definition 7.14). For this N, choose Pe so that P finer than P. implies U(P, fN, a) - L(P, fN, a) < E/3. Then for such P we have U(P, f, a) - L(P, f, a) < U(P, f - IN, a) - L(P,f - fN, a) + U(P, fN, a) - L(P, IN, a) < I U(P, f - IN, a) I + I L(P, f - IN, 001 + 3 < This proves (a). To prove (b), let s > 0 be given and choose N so that If (t) - f(t)I < E 2[a(b) - a(a)] ' for all n > N and every tin [a, b]. If x e [a, b], we have E. I9n(x) - g(X)1 < E Ifn(t) - f(t)I daft) < a(x) - a(a) e < E < a(b) - a(a) 2 2 This proves that gn - g uniformly on [a, b]. Theorem 9.9. Let a be of bounded variation on [a, b] and assume that E fn(x) = f(x) (uniformly on [a, b]), where each f is a real-valued function such that fn e R(a) on [a, b]. Then we have: a) f E R(a) on [a, b]. b) fa 1 fn(t) da(t) = En f a fn(t) da(t) (uniformly on [a, b]). Proof. Apply Theorem 9.8 to the sequence of partial sums. NOTE. This theorem is described by saying that a uniformly convergent series can be integrated term by term. 9.9 NONUNIFORMLY CONVERGENT SEQUENCES THAT CAN BE INTEGRATED TERM BY TERM Uniform convergence is a sufficient but not a necessary condition for term-by-term integration, as is seen by the following example. Th. 9.11 Nonuniformly Convergent Sequences Figure 9.6 227 Example. Let f"(x) = x" if 0 <_ x < 1. (See Fig. 9.6.) The limit function f has the value 0 in [0, 1) and f(l) = 1. Since this is a sequence of continuous functions with discon-tinuous limit, the convergence is not uniform on [0, 1 ]. Nevertheless, term-by-term integration on [0, 1 ] leads to a correct result in this case. In fact, we have f f. (x) dx = I x" dx = 1 - 0 as n -- oo, 0 o n + 1 so lim, f o f"(x) dx = f f (x) dx = 0. The sequence in the foregoing example, although not uniformly convergen on [0, 1], is uniformly convergent on every closed subinterval of [0, 1] not con-taining 1. The next theorem is a general result which permits term-by-term inte-gration in examples of this type. The added ingredient is that we assume that {f.) is uniformly bounded on [a, b] and that the limit function f is integrable. Definition 9.10. A sequence of functions {f.) is said to be boundedly convergent on T if {f.) is pointwise convergent and uniformly bounded on T. Theorem 9.11. Let {f.} be a boundedly convergent sequence on [a, b]. Assume that each f" e R on [a, b], and that the limit function f e R on [a, b]. Assume also that there is a partition P of [a, b], say P = {xo, X1, ... , xm}, such that, on every subinterval [c, d] not containing any of the points xk, the sequence {f.} converges uniformly to f Then we have b lim f"(t) dt = r f(t) dt. Jim f f"(t) dt = f.' b a "" °D .J a (6) Proof. Since f is bounded and {f.) is uniformly bounded, there is a positive number M such that I f(x)I 5 M and i f"(x)I 5 M for all x in [a, b] and all n >- 1. Given a > 0 such that 2e < IIPII, let h = a/(2m), where m is the number of subintervals of P, and consider a new partition P' of [a, b] given by P' = {xo, xp + h, xl - h, xl + h, ... , xm_ 1 - h, xm_ 1 + h, xm - h, xm}. Since If - f"I is integrable on [a, b] and bounded by 2M, the sum of the integrals 228 Sequences of Functions Th. 9.12 of If -taken over the intervals [x0, x0 + h], [xl - h, x1 + h], ... , [xm- 1 - h, xm _ 1 + h], [xm - h, xm], is at most 2M(2mh) = 2Ms. The remaining portion of [a, b] (call it S) is the union of a finite number of closed intervals, in each of which {f.} is uniformly convergent to f. Therefore, there is an integer N (depending only on E) such that for all x in S we have I.f(x) - f(x)I < E whenever n >- N. Hence the sum of the integrals of If - fl over the intervals of S is at most E(b - a), so If(x) - f (x)I dx < (2M + b - a)e whenever n >- N. This proves that f;f (x) dx -+ J .a f(x) dx as n - oo. There is a stronger theorem due to Arzelii which makes no reference whatever to uniform convergence. Theorem 9.12 (Arzela). Assume that f f.) is boundedly convergent on [a,b] and sup-pose each fa is Riemann-integrable on [a, b]. Assume also that the limit function f is Riemann-integrable on [a, b]. Then b 6 lim f(x) dx f f b a The is Theorem and the a on Lebesgue integrals which includes Arzela's theorem as a special case. (See Theorem 10.29). NOTE. It is easy to give an example of a boundedly convergent sequence { f} of Riemann-integrable functions whose limit f is not Riemann-integrable. If {r1, r2, . . . } denotes the set of rational numbers in [0, 1], define fa(x) to have the value I if x = rk for all k = 1, 2, ... , n, and put f (x) = 0 otherwise. Then the integral f o f(x) dx = 0 for each n, but the pointwise limit function f is not Riemann-integrable on [0, 1]. 9.10 UNIFORM CONVERGENCE AND DIFFERENTIATION By analogy with Theorems 9.2 and 9.8, one might expect the following result to hold: If fa - f uniformly on [a, b] and if f exists for each n, then f' exists and f -> f' uniformly on [a, b]. However, Example 3 of Section 9.2 shows that this cannot be true. Although the sequence { fa} of Example 3 converges uniformly on R, the sequence {f,.} does not even converge pointwise on R. For example, { f,,(0)} diverges since f;,(0) = Jn. Therefore the analog of Theorems 9.2 and 9.8 for differentiation must take a different form. Th. 9.13 Uniform Convergence and Differentiation 229 Theorem 9.13. Assume that each term of is a real-valued function having a finite derivative at each point of an open interval (a, b). Assume that for at least one point x0 in (a, b) the sequence { converges. Assume further that there exists a function g such that f;, -+ g uniformly on (a, b). Then: a) There exists a function f such that f - f uniformly on (a, b). b) For each x in (a, b) the derivative f'(x) exists and equals g(x). Proof. Assume that c e (a, b) and define a new sequence as follows: .f (x) - 4(c) if x c, gn(x) x-c if x = c. (8) The sequence so formed depends on the choice of c. Convergence of {g (c)} follows from the hypothesis, since g (c) = f,,(c). We will prove next that converges uniformly on (a, b). If x c, we have g (x) - gm(x) = h(x) - h(c) , (9) x - c where h(x) = fm(x). Now h'(x) exists for each x in (a, b) and has the value f ;, (x) -Applying the Mean-Value Theorem in (9), we get gn(x) - gm(x) = fn(x1) - f,(x 1), (10) where x1 lies between x and c. Since f f,,') converges uniformly on (a, b) (by hy-pothesis), we can use (10), together with the Cauchy condition (Theorem 9.3), to deduce that {g} converges uniformly on (a, b). Now we can show that f f.} converges uniformly on (a, b). Let us form the particular sequence corresponding to the special point c = xo for which { f (x0)} is assumed to converge. From (8) we can write .(x) = f,,(xo) + (x - x0)g.(x), an equation which holds for every x, in (a, b). Hence we have .fn(x) - .fn,(x) = .fn(x0) - .fm(x0) + (x - x0)[gn(x) - gm(x0)] This equation, with the help of the Cauchy condition, establishes the uniform convergence of {f,,} on (a, b). This proves (a). To prove (b), return to the sequence {gn} defined by (8) for an arbitrary point c in (a, b) and let G(x) = The hypothesis that f exists means that lim, In other words, each g is continuous at c. Since g - G uniformly on (a, b), the limit function G is also continuous at c. This means that G(c) = lim G(x), (11) X-C 230 Sequences of Functions Th. 9.14 the existence of the limit being part of the conclusion. But, for x c, we have G(x) = lim 9n(x) = lim fn(x) - fn(c) = f(x) - f(c) n- 00 n- 00 x - C X- C Hence, (11) states that the derivative f'(c) exists and equals G(c). But G(c) = lim gn(c) = lim .(c) = g(c); n-00 M 00 hence f'(c) = g(c). Since c is an arbitrary point of (a, b), this proves (b). When we reformulate Theorem 9.13 in terms of series, we obtain Theorem 9.14. Assume that each fn is a real-valued function defined on (a, b) such that the derivative f (x) exists for each x in (a, b). Assume that, for at least one point x0 in (a, b), the series F_fn(xo) converges. Assume further that there exists a function g such that g(x) (uniformly on (a, b)). Then: a) There exists a function f such that > fn(x) = f(x) (uniformly on (a, b)). b) If x e (a, b), the derivative f'(x) exists and equals Ef(x). 9.11 SUFFICIENT CONDITIONS FOR UNIFORM CONVERGENCE OF A SERIES The importance of uniformly convergent series has been amply illustrated in some of the preceding theorems. Therefore it seems natural to seek some simple ways of testing a series for uniform convergence without resorting to the definition in each case. One such test, the Weierstrass M-test, was described in Theorem 9.6. There are other tests that may be useful when the M-test is not applicable. One of these is the analog of Theorem 8.28. Theorem 9.15 (Dirichlet's test for uniform convergence). Let F,,(x) denote the nth partial sum of the series E fn(x), where each fn is a complex-valued function defined on a set S. Assume that {Fn} is uniformly bounded on S. Let {gn} be a sequence of real-valued functions such that gn+1(x) < gn(x) for each x in S and for every n = 1, 2, ... , and assume that gn y 0 uniformly on S. Then the series F_fn(x)gn(x) converges uniformly on S. Proof. Let sn(x) = Ek=, fk(x)gk(x). By partial summation we have n Sn(x) = E F'k(x)(9k(x) - 9k+1(x)) + gn+1(X)Fn(X), k=1 and hence if n > m, we can write nn Sn(x) - Sm(X) =j Fk(x)(9k(x) - gk+1(x)) + 9n+1(x)Fn(x) - gm+1(X)Fm(x) -k=m+1 Th. 9.16 Uniform Convergence and Double Sequences Therefore, if M is a uniform bound for {F"}, we have 231 ISn(x) - Sm(x)I 5 M E (gk(x) - gk+1(x)) + Mgn+1(x) + Mgm+1(x) k=m+1 = M(gm+1(x) - gn+1(x)) + Mg"+1(x) + Mgm+1(x) = 2Mgm+1(x). Since g" - 0 uniformly on S, this inequality (together with the Cauchy condition) implies that Yfn(x)gn(x) converges uniformly on S. The reader should have no difficulty in extending Theorem 8.29 (Abel's test) in a similar way so that it yields a test for uniform convergence. (Exercise 9.13.) Example. Let F"(x) = Y_k=1 e'kx. In the last chapter (see Theorem 8.30), we derived the inequality IF,,(x)I 5 1/Isin (x/2)J, valid for every real x 0 2mn (m is an integer). There-fore, if 0 < 6 < n, we have the estimate I F"(x) l 5 I /sin (6/2) if 6 <- x < 2n - J. Hence, {F,,} is uniformly bounded on the interval [6, tic - b]. If {g"} satisfies the condi-tions of Theorem 9.15, we can conclude that the series >g"(x)e"x converges uniformly on [6, 27r - 6]. In particular, if we take gn(x) = 1/n, this establishes the uniform con-vergence of the series on [6, 2,r - 6] if 0 < 6 < n. Note that the Weierstrass M-test cannot be used to estab-lish uniform convergence in this case, since le'nxl = 1. 9.12 UNIFORM CONVERGENCE AND DOUBLE SEQUENCES As a different type of application of uniform convergence, we deduce the following theorem on double sequences which can be viewed as a converse to Theorem 8.39. Theorem 9.16. Let f be a double sequence and let Z+ denote the set of positive integers. For each n = 1, 2, ... , define a function gn on Z+ as follows: gn(m) = f(m, n), if m c- Z+. Assume that gn -+ g uniformly on Z+, where g(m) = limn, f(m, n). If the iterated limit lim,. (lim"-. f(m, n)) exists, then the double limit limm,. f(m, n) also exists and has the same value. Proof. Given E > 0, choose N1 so that n > N1 implies , for every m in Z+. If(m, n) = g(m)l < 2 232 Sequences of Functions Def. 9.17 Let a = limm (limn-, f(m, n)) = g(m). For the same s, choose N2 so that m > N2 implies l g(m) - al < e/2. Then, if N is the larger of N1 and N2, we have I f(m, n) - al < e whenever both m > N and n > N. In other words, limm,n oof(m, n) = a. 9.13 MEAN CONVERGENCE The functions in this section may be real- or complex-valued. Definition 9.17 Let { fn} be a sequence of Riemann-integrable functions defined on [a, b]. Assume that f e R on [a, b]. The sequence {f,,} is said to converge in the mean to f on [a, b], and we write l.i.m. fn = f on [a, b], n-ao if Jim fb Ifn(x) -f(x)IZ dx = 0. n- 00 If the inequality If(x) - f"(x)I < c holds for every x in [a, b], then we have J .b 1 f(x) - f" (x)12 dx < e2(b - a). Therefore, uniform convergence of {f.} to f on [a, b] implies mean convergence, provided that each fn is Riemann-integrable on [a, b]. A rather surprising fact is that convergence in the mean need not imply pointwise convergence at any point of the interval. This can be seen as follows: For each integer n Z 0, subdivide the interval [0, 1] into 2" equal subintervals and let 2-1k denote that subinterval whose right endpoint is (k + 1)/2", where k = 0, 1, 2, ... , 2" - 1. This yields a collection {11, I2, ... } of subintervals of [0, 1], of which the first few are: 11 = [0, 11, 12 = [0, 1], 13 = [11 1], 14 = [0, 1], 15 = [11 +], 16 = and so forth. Define fn on [0, 1] as follows : f"(x) _ 1. if x e I", 0 if x e [0, 1] - In. Then { fn} converges in the mean to 0, since $u I fn(x)12 dx is the length of In, and this approaches 0 as n .- oo. On the other hand, for each x in [0, 1] we have lim sup fn(x) = 1 and lim inf fn(x) = 0. n -co n- co [Why?] Hence, {fn(x)} does not converge for any x in [0, 1]. The next t-heorem illustrates the importance of mean convergence. Th. 9.19 Mean Convergence 233 Theorem 9.18. Assume that l.i.m.,,.. fa = f on [a, b]. If g e R on [a, b], define h(x) = J xf(t)g(t) dt, h.(x) = dt, a a if x e [a, b]. Then h -+ h uniformly on [a, b]. Proof. The proof is based on the inequality 0 < (J x If(t) - fn(t)I Ig(t)I dt)2 a I -< (J X If(t) - f(t)I2 dt)( f z Ig(t)12 dt), (12) / which is a direct application of the Cauchy-Schwarz inequality for integrals. (See Exercise 7.16 for the statement of the Cauchy-Schwarz inequality and a sketch of its proof.) Given E > 0, we can choose N so that n > N implies Sat' If(t) - ff(t)12 dt (13) where A = 1 + f .b I g(t)12 dt. Substituting (13) in (12), we find that n > N implies 0 < I h(x) - h (x)I < e for every x in [a, b]. This theorem is particularly useful in the theory of Fourier series. (See Theorem 11.16.) The following generalization is also of interest. Theorem 9.19. Assume that l.i.m.,,y,, fa = f and l.i.m.a.. ga = g on [a, b]. Define h(x) = $Xf(t)g(t) dt, h(x) = J fn(t)g(t) dt, a if x e [a, b]. Then h - h uniformly on [a, b]. Proof. We have ha(x) - h(x) = J x (f - fn)(g - gn) dt a + ($Xfg dt -$Xfg d) + (JXfg t 1 . n dt -$Xfg d// Applying the Cauchy-Schwarz inequality, we can write 0<(fxIf-f.1Ig-galdt)2 <(JabIf-f.12dtX fIg-gn12dt). a 'a The proof is now an easy consequence of Theorem 9.18. J l 234 Sequences of Functions 9.14 POWER SERIES An infinite series of the form written more briefly as 00 a0 + L.r a"(z - zo)", n=1 OD Th. 9.20 E a"(z - z0)", (14) n=0 is called a power series in z - zo. Here z, zo, and a" (n = 0, 1, 2, ...) are complex numbers. With every power series (14) there is associated a disk, called the disk of convergence, such that the series converges absolutely for every z interior to this disk and diverges for every z outside this disk. The center of the disk is at zo and its radius is called the radius of convergence of the power series. (The radius may be 0 or + oo in extreme cases.) The next theorem establishes the existence of the disk of convergence and provides us with a way of calculating its radius. Theorem 9.20. Given a power series Y_,=0 a"(z - zo)", let A=limsup"Ia,, r= i, A (where r = 0 if .. = + oo and r = + oo if A = 0). Then the series converges absolutely if Iz - zol < r and diverges if Iz - zol > r. Furthermore, the series converges uniformly on every compact subset interior to the disk of convergence. Proof. Applying the root test (Theorem 8.26), we have lim sup V Ian(z - zo)"I = Iz - zol nyao r and hence Ea"(z - zo)" converges absolutely if Iz - zol < r and diverges if Iz - zol>r. To prove the second assertion, we simply observe that if T is a compact subset of the disk of convergence, there is a point p in T such that z e T implies Iz-zol<Ip-zol<r. Hence, la"(z - zo)nl < Ia"(p - zo)"I for each z in T, and the Weierstrass M-test is applicable. NOTE. If the limit lim"y., l ala"+ 11 exists (or if this limit is + 00), its value is also equal to the radius of convergence of (14). (See Exercise 9.30.) Example 1. The two series F,,'=o z" and z"/n2 have the same radius of convergence, namely, r = 1. - On the boundary of the disk of convergence, the first converges nowhere, the second converges everywhere. Th. 9.22 Power Series 235 Example 2. The series Y_n 1 z"ln has radius of convergence r = 1, but it does not con-verge at z = 1. However, it does converge everywhere else on the boundary because of Dirichlet's test (Theorem 8.28). These examples illustrate why Theorem 9.20 makes no assertion about the be-havior of a power series on the boundary of the disk of convergence. Theorem 9.21. Assume that the power series Y_,'= 0 a"(z - zo)" converges for each z in B(zo; r). Then the function f defined by the equation 00 f(z) = an(z - zo)", if z e B(zo; r), (15) n=0 is continuous on B(zo; r). Proof. Since each point in B(zo; r) belongs to some compact subset of B(zo; r), the conclusion follows at once from Theorem 9.7. NOTE. The series in (15) is said to represent f in B(zo; r). It is also called a power series expansion of f about zo. Functions having power series expansions are continuous inside the disk of convergence. Much more than this is true, however. We will later prove that such functions have derivatives of every order inside the disk of convergence. The proof will make use of the following theorem: Theorem 9.22. Assume that Ea"(z - zo)" converges if z e B(zo; r). Suppose that the equation {' o0 J (z) = Lj an(z - zo)", n=0 is known 'to be valid for each z in some open subset S of B(zo; r). Then, for each point z1 in S, there exists a neighborhood B(z1; R) s S in which f has a power series expansion of the form where f(z) = Ej bk(z - z1)k, (16) 00 k=0 x bk = E (ul)an(zi -z0)"-k (k = 0, 1, 2, ... ). n=k Proof. If z e S, we have f(z) = E zo)" = F an(z - z1 + z1 - zo)" H=O n=0 ao n a" (n) n (z - z1)k(z1 - Z0)" -k n=0 k=o (17) = 1 (: c.(k), n=0 k=0 236 Sequences of Functions where Th. 9.23 (Jn an(Z - zl)k(z, -ZO)"-k, if k <_ n, cn(k) _ k) to, if k > n. Now choose R so that B(zl ; R) c S and assume that z e B(zl ; R). Then the iterated series En o Ek 0 cn(k) converges absolutely, since 00 00 00 00 E E Icn(k)I = E lanl(Iz - zll + Izl - zol)n = E Ianl(z2 - Zn)", (18) n=0 k=0 n=0 n=0 where z2 = zo + Iz - zll + lzl - zol. But Iz2-z01 <R+1z, -zol <r, and hence the series in (18) converges. Therefore, by Theorem 8.43, we can inter-change the order of summation to obtain 00 00 00 00 f(z) = E E c .(k) = E E (n ) an(z - z1)k(z1 -Z0)"-k k=0 n=0 k=0 n=k k 00 E bk(z - Z1)k, k=0 where bk is given by (17). This completes the proof. NOTE. In the course of the proof we have shown that we may use any R > 0 that satisfies the condition B(z1; R) S S. (19) Theorem 9.23. Assume that Ean(z - zo)" converges for each z in B(zo; r). Then the function f defined by the equation f(z) = E an(z - zo)", if z e B(zo; r), (20) 00 n=0 has a derivative f'(z) for each z in B(zo; r), given by f'(z) _ zo)"-1 n=1 NOTE. The series in (20) and (21) have the same radius of convergence. (21) Proof. Assume that z1 a B(zo; r) and expand fin a power series about z1, as indicated in (16). Then, if z e B(z1; R), z z1, we have z z Co f( ) - f( 1) = b1 + E bk+l(z - Z1)k. Z - Z1 k=1 (22) Th. 9.24 Multiplication of Power Series 237 By continuity, the right member of (22) tends to b, as z -+ z,. Hence, f'(zl) exists and equals b,. Using (17) to compute b,, we find 00 b, _ E na"(z, -zo)"-' n=1 Since z, is an arbitrary point of B(zo; r), this proves (21). The two series have the same radius of convergence because 1 as n - oo. NOTE. By repeated application of (21), we find that for each k = 1, 2, ... , the derivative f(k)(z) exists in B(zo; r) and is given by the series 00 t J (k)(Z) = E n. an(Z -Zo)"-k. (23) n = k (n - k)! If we put z = zo in (23), we obtain the important formula f(k)(Zo) = k!ak (k = 1, 2, ... ). (24) This equation tells us that if two power series >an(z - zo)" and Ybn(z - zo)" both represent the same function in a neighborhood B(zo; r), then a" = b" for every n. That is, the power series expansion of a function f about a given point zo is uniquely determined (if it exists at all), and it is given by the formula () f(z) = L..rf (z0) (z - z0)", n=0 n! valid for each z in the disk of convergence. 9.15 MULTIPLICATION OF POWER SERIES Theorem 9.24. Given two power series expansions about the origin, say 00 f(z) = E a"z", if z c- B(0; r), and n=0 g(z) = E b"z", if z e B(0; R). 00 n=0 Then where the product f(z)g(z) is given by the power series f(z)g(z) _ E CnZn, 00 n=0 nn Cn = akbn-k if z E B(0; r) n B(0; R), (n = 0, 1, 2, ... ). k=0 238 Sequences of Functions Proof. The Cauchy product of the two given series is OD 00 akzk bn-kzn-k n E CnZ , to k=0 n=0 and the conclusion follows from Theorem 8.46 (Mertens' Theorem). NOTE. If the two series are identical, we get Th. 9.25 00 f(z)2 = E c,z", n=0 where cn = Ek=0 akan-k = E., +.,,=. amiam2. The symbol Lmi+m2=n indicates that the summation is to be extended over all nonnegative integers ml and m2 whose sum is n. Similarly, for any integer p > 0, we have AZ)" = cn(p)Z", n=0 where cn(p) = E ami ... amp. mi+ +mp=n 9.16 THE SUBSTITUTION THEOREM Theorem 9.25. Given two power series expansions about the origin, say 00 f(z) = E anz", if z e B(0; r), and n=0 00 g(z) = E bnz", if z e B(0; R). n=0 If, for a fixed z in B(0; R), we have E o r, then for this z we can write 00 f[g(z)] = E CkZk, k=0 where the coefficients ek are obtained as follows: Define the numbers bk(n) by the equation n g(Z)" bkzk/ = E bk(n)zk. 00 00 k 0 k=0 Then ck = En 0 anbk(n) for k = 0, 1, 2, .. . NOTE. The series Ek 0 ckzk is the power series which arises formally by substituting the series for g(,z) in place of z in the expansion off and then rearranging terms in increasing powers of z. Th. 9.26 Reciprocal of a Power Series 239 Proof By hypothesis, we can choose z so that Y_ 0 Ibnznl < r. For this z we have Ig(z)I < r and hence we can write J [9(z)] = E an9(z)n = E E anbk(n)z". n=0 n=0 k=0 If we are allowed to interchange the order of summation, we obtain "0 00 00 .f [9(z )] = L I anbk(n) zk = ckzk , k=O n=0 k=0 which is the statement we set out to prove. To justify the interchange, we will establish the convergence of the series E E lanbk(n)zkl = E Ianl E Ibk(n)Zkl. (25) n=0 k=0 n=0 k=0 Now each number bk(n) is a finite sum of the form bk(n) _ E bm, ... bm,,, and hence Ibk(n)I < Y-.,+ Ibm,l ... Ibmnl On the other hand, we have r n (E Ibklzk) = Bk(n)Zk, k=0 k=0 where Bk(n) = Em,+ Ibm,I ... Returning to (25), we have 00 00 eo 00 OD 00 n E Ianl E Ibk(n)z"I < 1 Ianl E Bk(n)IZk9 = E Ian) (E Ibkzkl) , n=0 k=0 n=0 k=0 n=0 k-0 and this establishes the convergence of (25). 9.17 RECIPROCAL OF A POWER SERIES As an application of the substitution theorem, we will show that the reciprocal of a power series in z is again a power series in z, provided that the constant term is not 0. Theorem 9.26. Assume that we have p(Z) _ pnzn, if z E B(0; h), n=0 where p(O) # 0. Then there exists a neighborhood B(0; S) in which the reciprocal of p has a power series expansion of the form 1 = gnzn p(Z) n=0 Furthermore, q0 = l /po. 240 Sequences of Functions Th. 9.26 Proof. Without loss in generality we can assume that po = 1. [Why?] Then p(O) = 1. Let P(z) = 1 + Y_ 1 (Pnz"I if z e B(0; h). By continuity, there exists a neighborhood B(0; S) such that IP(z) - II < 1 if z e B(0; S). The conclusion follows by applying Theorem 9.25 with 00 00 f(z) = i = E z" and g(z) = 1 - p(z) = E paz" 1 z 9.18 REAL POWER SERIES If x, x0, and a" are real numbers, the series Y_a"(x - x0)" is called a real power series. Its disk of convergence intersects the real axis in an interval (xo - r, x0 + r) called the interval of convergence. Each real power series defines a real-valued sum function whose value at each x in the interval of convergence is given by f(x) = L an(x - x0)". n=0 The series is said to represent f in the interval of convergence, and it is called a power-series expansion off about x0. Two problems concern us here: 1) Given the series, to find properties of the sum function f. 2) Given a function f, to find whether or not it can be represented by a power series. It turns out that only rather special functions possess power-series expansions. Nevertheless, the class of such functions includes a large number of examples that arise in practice, so their study is of great importance. Question (1) is answered by the theorems we have already proved for complex power series. A power series converges absolutely for each x in the open subinterval (xo - r, x0 + r) of convergence, and it converges uniformly on every compact subset of this interval. Since each term of the power series is continuous on R, the sum function f is continuous on every compact subset of the interval of convergence and hence f is continuous on (xo - r, x0 + r). Because of uniform convergence, Theorem 9.9 tells us that we can integrate a power series term by term on every compact subinterval inside the interval of con-vergence. Thus, for every x in (x0 - r, x0 + r) we have S x x 00 + E (x - x0)n f(t) dt = a" J (t - X0)" dt = a xo o n=0 x0 n=0 n + 1 The integrated series has the same radius of convergence. The sum function has derivatives of every order in the interval of convergence and they can be obtained by differentiating the series term by term. Moreover, Def. 9.27 Taylor's Series f (")(x0) = n !an so the sum function is represented by the power series f(x) = E f(")(Xo) (X - Xo)". n=o n! 241 (26) We turn now to question (2). Suppose we are given a real-valued function f defined on some open interval (xo - r, x0 + r), and suppose f has derivatives of every order in this interval. Then we can certainly form the power series on the right of (26). Does this series converge for any x besides x = x0? If so, is its sum equal to f(x)?, In general, the answer to both questions is "No." (See Exercise 9.33 for a counter example.) A necessary and sufficient condition for answering both questions in the affirmative is given in the next section with the help of Taylor's formula (Theorem 5.19.) 9.19 THE TAYLOR'S SERIES GENERATED BY A FUNCTION Definition 9.27. Let f be a real-valued function defined on an interval I in R. If f has derivatives of every order at each point of I, we write f e C°° on I. If f e C°° on some neighborhood of a point c, the power series n=o n! 00 (n)(C Ef ) (x-c)", is called the Taylor's series about c generated by f To indicate that f generates this series, we write (n) f(x) ~ :f (x - c)". n! The The question we are interested in is this: When can we replace the symbol - by the symbol = ? Taylor's formula states that if f e C°° on the closed interval [a, b] and if c e [a, b], then, for every x in [a, b] and for every n, we have f(x) = Ef k( X - C)" + f(nn X') (x - c)", (27) where x, is some point between x and c. The point x, depends on x, c, and on n. Hence a necessary and sufficient condition for the Taylor's series to converge to f(x) is that (n) lim f i (x - c)" = 0. n_00 n! (28) In practice it may be quite difficult to deal with this limit because of the unknown position of x, : In some cases, however, a suitable upper bound can be obtained for f (")(x,) and the limit can be shown to be zero. Since An/n! -+ 0 as n -+ oc for 242 Sequences of Functions Th. 9.28 all A, equation (28) will certainly hold if there is a positive, constant M such that If(n)(x)I < Mn, for all x in [a, b]. In other words, the Taylor's series of a function f converges if the nth derivative f (") grows no faster than the nth power of some positive number. This is stated more formally in the next theorem. Theorem 9.28. Assume that f E C°° on [a, b] and let c e [a, b]. Assume that there is a neighborhood B(c) and a constant M (which might depend on c) such that If(x) I < M" for every x in B(c) n [a, b] and every n = 1, 2, ... Then, for each x in B(c) n [a, b], we have f(x) = E f () (x - c)". 00 n=0 n! 9.20 BERNSTEIN'S THEOREM Another sufficient condition for convergence of the Taylor's series off, formulated by S. Bernstein, will be proved in this section. To simplify the proof we first obtain another form of Taylor's formula in which the error term is expressed as an integral. Theorem 9.29. Assume f has a continuous derivative of order n + 1 in some open interval I containing c, and define E"(x) for x in I by the equation (k) C (x) ) = Ff () (x -)k + E f( (29) " . x c k=0 k! Then E"(x) is also given by the integral E"(x) = 1 x (x - t)n f(n+1)(t) dt. (30) n! J Proof The proof is by induction on n. For n = I we have E1(x) = f(x) - f(c) - f'(c)(x - c) = J x [f '(t) - f '(c)] dt = f x u(t) dv(t), where u(t) = f'(t) - f'(c) and v(t) = t - x. Integration by parts gives u(t) dv(t) = u(x)v(x) - u(c)v(c) - fCX v(t) du(t) = fCX (x - t)f "(t) dt. This proves (30) for n = 1. Now we assume (30) is true for n and prove it for n + 1. From (29) we have (n+1) C En+1(x) = En(x) - ( + 1)! (x - c)"+1 7%. 9.30 Bernstein's Theorem 243 We write E"(x) as an integral and note that (x - c)"' = (n + 1) fx (x - t)" dt to obtain En+1(x) = I J x (x - t)"f("+1)(t) dt -f("+1)(c ('x (x - t)" dt n. n! = 1 f (x - t)" [f + 1)(t) - f("+1)(c)] dt = 1 Ix u(t) dv(t), n where u(t) = f("+ 1)(t) - f(n+1)(c) and v(t) _ -(x - t)"+1/(n + 1). Integration by parts gives us E.+ 1(x) n! J v(t) du(t) (n + 1)! fx (x - t)"+1 f("+2)(t) A This proves (30). NOTE. The change of variable t = x + (c - x)u transforms the integral in (30) to the form -c"+1 1 E"(x) _ (x ) unf("+1)[x + (c - x)u] du. (31) n! o Theorem 9.30 (Bernstein). Assume f and all its derivatives are nonnegative on a compact interval [b, b + r]. Then, if b < x < b + r, the Taylor's series f(k) ! (x - b)k , k=o k! converges to fix). Proof. By a translation we can assume b = 0. The result is trivial if x = 0 so we assume 0 < x < r. We use Taylor's formula with remainder and write f(x) = E f (k) (o) xk + E"(x). (32) k=o k! We will prove that the error term satisfies the inequalities x +1 0 < f(r). r This implies that E"(x) - 0 as n -+ oo since (x/r)"+1 -, 0 if 0 < x < r. To prove (33) we use (31) with c = 0 and find x"+1 E"(x) = fo, u"f ("+ 1)(x - xu) du, n! for each x in [0, r]. If x # 0, let (F"(x) = .s- 1 = 1 1 1 unf("+ 1)(x - xu) du. x n! o l (33) Th. 9.30 The function f(n+1) is monotonic increasing on [0, r] since its derivative is non-negative. Therefore we have f(n+1)(X - xu) = f(n+1)[x(1 - u)] < f(n+1)[r(1 - u)], if 0 5 u < 1, and this implies F"(x) < F,(r) if 0 < x < r. In other words, En(x)lx"+1 < En(r)lrn+1, or E"(x) < X +1 E"(r). (34) r Putting x = r in (32), we see that En(r) < f(r) since each term in the sum is nonnegative. Using this in (34), we obtain (33) which, in turn, completes the proof. 9.21 THE BINOMIAL SERIES As an example illustrating the use of Bernstein's theorem, we will obtain the fol-lowing expansion, known as the binomial series: 00 (1 + x)° = E Ca x", if -1 < x < 1, (35) n=0 n/ where a is an arbitrary real number and a(a - 1) . (a - n + 1)ln!. Bernstein's theorem is not directly applicable in this case. However we can argue as follows : Let f (x) = (1 - x) -, where c > 0 and x < 1. Then f (")(x) = c(c + 1) (c + n - 1)(1 - X)-", and hence f (")(x) 0 for each n, provided that x < 1. Applying Bernstein's theorem with b = -1 and r = 2 we find that f (x) has a power series expansion about the point b = -1, convergent for -1 < x < 1. Therefore, by Theorem 9.22, f (x) also has a power series expansion about 0, f (x) = Ek ° f (k)(0)xklk!, convergent for -1 < x < 1. But f (k)(0) 1)k k!, so _ ( l(- 1)kxk, if -1 < x < 1. (1 1 x)k = ° \k c) Replacing c by -a and x by -x in (36) we find that (35) is valid for each a < 0. But now (35) can be extended to all real a by successive integration. Of course, if a is a positive integer, say a = m, then 0 for n > m, and (35) reduces to a finite sum (the Binomial Theorem). 9.22 ABEL'S LIMIT THEOREM If -1 < x < 1, integration of the geometric series Th. 9.31 Abel's Limit Theorem gives us the series expansion 00 log (1 - x) = - E xn n=1 n 245 (37) also valid for -1 < x < 1. If we put x = -1 in the righthand side of (37), we obtain a convergent alternating series, namely, E(- 1)n+1/n. Can we also put x = -1 in the lefthand side of (37)? The next theorem answers this question in the affirmative. Theorem 9.31 (Abel's limit theorem). Assume that we have 00 f(x) = E anx", if -r < x < r. (38) n=0 If the series also converges at x = r, then the limit f(x) exists and we have 00 lim f(x) = E anr". x-4r-n=0 Proof. For simplicity, assume that r = 1 (this amounts to a change in scale). Then we are given that f(x) = Y_anx" for -1 < x < 1 and that Y _a. converges. Let us write f(1) = E,01 0 an. We are to prove that limx..1- f(x) = f(l), or, in other words, that f is continuous from the left at x = 1. If we multiply the series for f(x) by the geometric series and use Theorem 9.24, we find 00 n 1 f(x) = E cnx", where cn = E ak. 1 - x n-0 k=0 Hence we have CO f(x) - f(1) = (1 - x) E [cn - ff(1)]x n=0 if -I < x < 1. (39) By hypothesis, limn cn = f(1). Therefore, given s > 0, we can find N such that n >- N implies Icn - f(1)1 < s/2. If we split the sum (39) into two parts, we get N-1 00 .f(x) - .f(1) x) 1: [cn - .f(1)]x" + (1 - x) 1: [cn - ff(1)]xn (40) n=0 n=N Let M denote the largest of the N numbers Icn - f(1)1, n = 0, 1, 2, ... , N - 1. If 0 < x < 1, (40) gives us 00 11(x)-1(1)1 :5 (1-x)NM+(1-x)8EXn 2 n=N N (1 -x)NM+(1 - x) --x <(1 -x)NM+2. 246 Sequences of Functions Th. 9.32 Now let 6 = e/2NM. Then 0 < 1 - x < 6 implies If(x) - f(1)l < e, which means lima- 1 - f(x) = f(1). This completes the proof. Example. We may put x = -1 in (37) to obtain 00 log 2 = 1)n+ n=1 n (See Exercise 8.18 for another derivation of this formula.) As an application of Abel's theorem we can derive the following result on multiplication of series: Theorem 9.32. Let F,,'=0 a" and F,.'=0 bn be two convergent series and let Y-,'=0 c" denote their Cauchy product. If En 0 cn converges, we have NOTE. This result is similar to Theorem 8.46 except that we do not assume absolute convergence of either of the two given series. However, we do assume convergence of their Cauchy product. Proof. The two power series and Eb,,.e both converge for x = 1, and hence they converge in the neighborhood B(0; 1). Keep IxI < I and write G r00 CnXn = (E a, n=0 "=0 using Theorem 9.24. Now let x --+ 1- and apply Abel's theorem. 9.23 TAUBER'S THEOREM The converse of Abel's limit theorem is false in general. That is, if f is given by (38), the limit f(r-) may exist but yet the series Fanr" may fail to converge. For example, take an = (-1)". Then f(x) = 1/(1 + x) if -1 < x < 1 and f(x) -+ I as x - 1-. However, F,(-1)" diverges. A. Tauber (1897) discovered that by placing further restrictions on the coefficients an, one can obtain a converse to Abel's theorem. A large number of such results are now known and they are referred to as Tauberian theorems. The simplest of these, sometimes called Tauber's first theorem, is the following: Theorem 9.33 (Tauber). Let f(x) = Y_ 0 anx" for -1 < x < 1, and assume that limn. nan = 0. If f(x) - S as x - 1-, then E 0 a" converges and has sum S. Proof Let nan = Fk=o klakl Then an - 0 as n --> oo. (See Note following Theorem 8.48) Also, lim,, f(xn) = S if x,, = 1 - 1/n. Hence, given c > 0, 11. 933 Exercises we can choose N so that n >- N implies IJ (xn) - SI < 3 , an < 3 , nlanl < 3 Now let sn = Ek=0 ak. Then, for -1 < x < 1, we can write n 00 Sn - S = f (X) - S + E a,(1 - xk) -E akxk. k=0 k=n+1 Now keep x in (0, 1). Then (1 - xk)= (1 -x)(1 +x+ +x"-1) <-k(1 -x), for each k. Therefore, if n >- N and 0 < x < 1, we have 247 Isn - SI <- I.f(x) - SI + (1 - x) klakl + s k=o 3n(1 - x) Taking x = xn = 1 - 1/n, we find Isn - SI < s/3 + E/3 + E/3 = s. This com-pletes the proof. NOTE. See Exercise 9.37 for another Tauberian theorem. EXERCISES Uniform convergence 9.1 Assume that fn - f uniformly on S and that each fn is bounded on S. Prove that {fn } is uniformly bounded on S. 9.2 Define two sequences {fn} and {gn} as follows: fn(x) = x 1 + 1 1 ifxeR, n = 1, 2,..., n I if x = 0 or if x is irrational, gn(x) = n b + n if x is rational, say x = a , b > 0. Let hn(x) = fn(x)gn(x) a) Prove that both {fn } and (gn ) converge uniformly on every bounded interval. b) Prove that {hn } does not converge uniformly on any bounded interval. 9.3 Assume that fn - f uniformly on S, gn - g uniformly on S. a) Prove that fn + gn - f + g uniformly on S. b) Let hn(x) = fn(x)gn(x), h(x) = f(x)g(x), if x e S. Exercise 9.2 shows that the assertion hn -+ h uniformly on S is, in general, incorrect. Prove that it is correct if each f and each gn is bounded on S. 248 Sequences of Functions 9.4 Assume that f" -+ f uniformly on S and suppose there is a constant M > 0 such that I f(x)1 < M for all x in S and all n. Let g be continuous on the closure of the disk B (O; M) and define h"(x) = g [ f"(x) ], h(x) = g [ f(x) J, if x e S. Prove that h" - h uniformly on S. 9.5 a) Let f"(x) = 1/(nx + 1) if 0 < x < 1, n = 1, 2, ... Prove that {f"} converges pointwise but not uniformly on (0, 1). b) Let g"(x) = x/(nx + 1) if 0 < x < 1, n = 1, 2,... Prove that g" -+ 0 uni-formly on (0, 1). 9.6 Let f"(x) = x". The sequence f f.} converges pointwise but not uniformly on [0, 1 ]. Let g be continuous on [0, 1 ] with g(l) = 0. Prove that the sequence {g(x)x"} converges uniformly on [0, 1 ]. 9.7 Assume that f" - f uniformly on S, and that each f" is continuous on S. If x e S, let {x"} be a sequence of points in S such that x" - x. Prove that f"(x") - fix). 9.8 Let {f"} be a sequence of continuous functions defined on a compact set S and assume that {f"} converges pointwise on S to a limit function f. Prove that f" - f uni-formly on S if, and only if, the following two conditions hold: i) The limit function f is continuous on S. ii) For every e > 0, there exists an m > 0 and a S > 0 such that n > m and I f k ( x ) - f (x)I < 8 implies I fk+"(x) - f (x)I < e f o r all x in S and all k = 1, 2, .. . Hint. To prove the sufficiency of (i) and (ii), show that for each xo in S there is a neigh-borhood B(xo) and an integer k (depending on xo) such that Ifk(x) - f(x)I < 6 if x e B(xo). By compactness, a finite set of integers, say A = {k1,..., kr}, has the property that, for each x in S, some k in A satisfies I fk(x) - f(x)I < 6. Uniform convergence is an easy consequence of this fact. 9.9 a) Use Exercise 9.8 to prove the following theorem of Dini : If {f} is a sequence of real-valued continuous functions converging pointwise to a continuous limit function f on a compact set S, and if f"(x) > f"+ 1(x) for each x in S andevery n = 1, 2, ... , then f" -+ f uniformly on S. b) Use the sequence in Exercise 9.5(a) to show that compactness of S is essential in Dini's theorem. 9.10 Let f"(x) = nx(1 - x2)" for x real and n >- 1. Prove that {f"} converges pointwise on [0, 1 ] for every real c. Determine those c for which the convergence is uniform on [0, 1 ] and those for which term-by-term integration on [0, 1 ] leads to a correct result. 9.11 Prove that Ex"(1 - x) converges pointwise but not uniformly on [0, 11, whereas F_(-1)"x"(1 - x) converges uniformly on 10, 1 ]. This illustrates that uniformconvergence of E f"(x) along with pointwise convergence of FI f"(x)I does not necessarily imply uniform convergence of EI f"(x)I. 9.12 Assume that gnt 1(x) 5 g"(x) for each x in T and each n = 1, 2, ... , and suppose that g" - 0 uniformly on T. Prove that F_(-1)n+1g"(x) converges uniformly on T. 9.13 Prove Abel's test for uniform convergence: Let {g"} be a sequence of real-valued functions such that g"+1(x) < g"(x) for each x in T and for every n = 1, 2, ... If {g"} Exercises 249 is uniformly bounded on T and if E fn(x) converges uniformly on T, then E fn(x)gn(x) also converges uniformly on T. 9.14 Let fn(x) = x/(1 + nx2) if x E R, n = 1, 2.... Find the limit function f of the sequence { fn } and the limit function g of the sequence ff.). a) Prove that f'(x) exists for every x but that f'(0) 0 g(0). For what values of x is f'(x) = g(x)? b) In what subintervals of R does fn - f uniformly? c) In what subintervals of R does f,' g uniformly? 9.15 Let fn(x) = (1/n)e_n2x2 if x e R, n = 1, 2,... Prove that fn - 0 uniformly on R, that f - 0 pointwise on R, but that the convergence of { f } is not uniform on any interval containing the origin. 9.16 Let { fn} be a sequence of real-valued continuous functions defined on [0, 1 ] and assume that fn f uniformly on [0, 11. Prove or disprove urn fo 1-1/n fn(x) dx f (x) A. o 9.17 Mathematicians from Slobbovia decided that the Riemann integral was too compli-cated so they replaced it by the Slobbovian integral, defined as follows: If f is a function defined on the set Q of rational numbers in [0, 1 ], the Slobbovian integral of f, denoted by S(f), is defined to be the limit 1 n n) ,.,o n k=1 n whenever this limit exists. Let {fn) be a sequence of functions such that S(fn) exists for each n and such that fn - f uniformly on Q. Prove that {S(fn)} converges, that S(f) exists, and that S(fn) - S(f) as n oo. 9.18 Let fn(x) = 1/(1 + n2x2) if 0 < x <- 1, n = 1, 2,... Prove that {J.) converges pointwise but not uniformly on [0, 1 ]. Is term-by-term integration permissible? 9.19 Prove that En 1 x/na(1 + nx2) converges uniformly on every finite interval in R if a > 1. Is the convergence uniform on R? 9.20 Prove that the series ER 1 sin (1 + (x/n)) converges uniformly on every compact subset of R. 9.21 Prove that the series Y _,'=o (x2"+1/(2n + 1) - x"+1/(2n + 2)) converges pointwise but not uniformly on [0, 1 ]. 9.22 Prove that En 1 an sin nx and F_,'=1 a,, cos nx are uniformly convergent on R if En 1 la"I converges. 9.23 Let (an) be a decreasing sequence of positive terms. Prove that the series Ean sin nx converges uniformly on R if, and only if, nan -' 0 as n - oo. 9.24 Given a convergent series F_', an. Prove that the Dirichlet series En 1 ann-s converges uniformly on the half-infinite interval 0 - s < + oo. Use this to prove that ao -s o0 limn..o+ En=1 ann = n=1 an. 9.25 Prove that the series C(s) = n_S converges uniformly on every half-infinite interval 1 + h < s < + oo, where h > 0. Show that the equation log n CI(S) n, n=1 is valid for each s > 1 and obtain a similar formula for the kth derivative Cmkl(s). Mean convergence 9.26 Let f"(x) = n312xe-"Zx2. Prove that If,,) converges pointwise to 0 on [ -1, I] but that I.i.m."y. f " 7,- 0 on [-1, 1 ]. 9.27 Assume that {f"} converges pointwise to f on [a, b] and that l.i.m.n-oo f" = g on [a, b]. Prove that f = g if both f and g are continuous on [a, b]. 9.28 Let f"(x) = cos" x if 0 < x 5 jr. a) Prove that l.i.m.n-. fn = 0 on [0, 7r] but that { f"(ir) } does not converge. b) Prove that ff.) converges pointwise but not uniformly on [0, 7r/2). 9.29 Let f (x) = 0 if 0 < x < lln or if 2/n < x < 1, and let f"(x) = n if 1/n < x < 2/n. Prove that {f.} converges pointwise to 0 on [0, 1 ] but that f" 76 0 on [0, 1 ]. Power series 9.30 If r is the radius of convergence of Ya"(z - zo)", where each an : 0, show that lim inf "-00 an an+ 1 < r 5 lim sup n-ao an an+1 9.31 Given that the power series Fn '=0 anz" has radius ofconvergence 2. Find the radius of convergence of each of the following series : 00 00 a) akz", b) azkn c) E az"2 "=0 n=0 n=0 In (a) and (b), k is a fixed positive integer. 9.32 Given a power series F_o whose coefficients are related by an equation of the form a " + Aa"_1 + Ba"_2 = 0 (n = 2, 3, ... ). Show that for any x for which the series converges, its sum is ao + (a1 + Aao)x 1 + Ax + Bx2 9.33 Let f(x) = e- /_V2 if x 7 0, f(0) = 0. a) Show that f()(0) exists for all n > 1. b) Show that the Taylor's series about 0 generated by f converges everywhere on R but that it represents f only at the origin. References 251 9.34 Show that the binomial series 1 + x)' = Y ( n°°=0 () x" exhibits the following be-havior at the points x = ± 1. n a) If x = -1, the series converges for a >_ 0 and diverges for a < 0. b) If x = 1, the series diverges for a < - 1, converges conditionally for a in the interval -1 < a < 0, and converges absolutely for a >_ 0. 9.35 Show that Eanx" converges uniformly on [0, 1 ] if Lan converges. Use this fact to give another proof of Abel's limit theorem. 9.36 If each an > 0 and if F_an diverges, show that Ea,,x" + oo as x 1- . (Assume Eanx" converges for jxI < 1.) 9.37 If each an >- 0 and if limx.,1_ Y_anx" exists and equals A, prove that Lan converges and has sum A. (Compare with Theorem 9.33.) 9.38 For each real t, define f ,(x) = xe" t/(ex - 1) if x e R, x t- 0, f'(0) = 1. a) Show that there is a disk B(0; b) in which f is represented by a power series in x. b) Define Po(t), P1(t), P2(t), ... , by the equation A(x)= ifxeB(0;6), n=0 and use the identity w x oo x" E PP(t) , = et: E P.(0) n_o n. n=0 n. to rove that P t n p () = Ek=o k Pk(0)t",t. This shows that each function P. is a polynomial. These are the Bernoulli polynomials. The numbers B. = P.(0) (n = 0, 1, 2, ...) are called the Bernoulli numbers. Derive the following further properties : n-1 C) Bo = 1, B1 1 k) B k = 0, if n = 2, 3, .. . k=o d) P (t) = n = 1, 2, .. . e) Pn(t + 1) - PP(t) = nt' if n = 1, 2, .. . f)Pn(1-t)_(-1)"Pp(t) g)B2n+1=0 ifn=1,2,... h) In + 2" + ... + (k - 1)" = Pn+1(k) - Pn+1(0) (n = 2, 3, ...). n + 1 SUGGESTED REFERENCES FOR FURTHER STUDY 9.1 Hardy, G. H., Divergent Series. Oxford Univ. Press, Oxford, 1949. 9.2 Hirschmann, I. I., Infinite Series. Holt, Rinehart and Winston, New York, 1962. 9.3 Knopp, K.,- Theory and Application of Infinite Series, 2nd ed. R. C. Young, trans-lator. Hafner, New York, 1948. CHAPTER 10 THE LEBESGUE INTEGRAL 10.1 INTRODUCTION The Riemann integral f a f(x) dx, as developed in Chapter 7, is well motivated, simple to describe, and serves all the needs of elementary calculus. However, this integral does not meet all the requirements of advanced analysis. An extension, called the Lebesgue integral, is discussed in this chapter. It permits more general functions as integrands, it treats bounded and unbounded functions simultaneously, and it enables us to replace the interval [a, b] by more general sets. The Lebesgue integral also gives more satisfying convergence theorems. If a sequence of functions { fa} converges pointwise to a limit function f on [a, b], it is desirable to conclude that b lim f b f(x) dx rroo a Ja with a minimum of additional hypotheses. The definitive result of this type is Lebesgue's dominated convergence theorem, which permits term-by-term integra-tion if each {f.} is Lebesgue-integrable and if the sequence is dominated by a Lebesgue-integrable function. (See Theorem 10.27.) Here Lebesgue integrals are essential. The theorem is false for Riemann integrals. In Riemann's approach the interval of integration is subdivided into a finite number of subintervals. In Lebesgue's approach the interval is subdivided into more general types of sets called measurable sets. In a classic memoir, Integrale, Iongueur, aire, published in 1902, Lebesgue gave a definition of measure for point sets and applied this to develop his new integral. Since Lebesgue's early work, both measure theory and integration theory have undergone many generalizations and modifications. The work of Young, Daniell, Riesz, Stone, and others has shown that the Lebesgue integral can be introduced by a method which does not depend on measure theory but which focuses directly on functions and their integrals. This chapter follows this approach, as outlined in Reference 10.10. The only concept required from measure theory is sets of measure zero, a simple idea introduced in Chapter 7. Later, we indicate briefly how measure theory can be developed with the help of the Lebesgue integral. 252 Def. 10.1 Integral of a Step Function 253 10.2 THE INTEGRAL OF A STEP FUNCTION The approach used here is to define the integral first for step functions, then for a larger class (called upper functions) which contains limits of certain increasing sequences of step functions, and finally for an even larger class, the Lebesgue-integrable functions. We recall that a function s, defined on a compact interval [a, b], is called a step function if there is a partition P = {x0, x1, ... , of [a, b] such that s is constant on every open subinterval, say s(x) = Ck 'f X E (xk_1, Xk). A step function is Riemann-integrable on each subinterval [xk_1, xk] and its integral over this subinterval is given by xk s(x) dx = Ck(Xk - Xk _ 1), xk-I regardless of the values of s at the endpoints. The Riemann integral of s over [a, b] is therefore equal to the sum b s(x) dx = E Ck(Xk - Xk-1) n k=1 (1) NOTE. Lebesgue theory can be developed without prior knowledge of Riemann integration by using equation (1) as the definition of the integral of a step function. It should be noted that the sum in (1) is independent of the choice of P as long as s is constant on the open subintervals of P. It is convenient to remove the restriction that the domain of a step function be compact. Definition 10.1. Let I denote a general interval (bounded, unbounded, open, closed, or half-open). A function s is called a step function on I if there is a compact subinterval [a, b] of I such that s is a step function on [a, b] and s(x) = 0 if x e I - [a, b]. The integral of s over 1, denoted by f, s(x) dx or by f'r s, is defined to be the integral of s over [a, b], as given by (1). There are, of course, many compact intervals [a, b] outside of which s vanishes, but the integral of s is independent of the choice of [a, b]. The sum and product of two step functions is also a step function. The follow-ing properties of the integral for step functions are easily deduced from the fore-going definition: Jr (s + t) = fr s + f I t, f r cs = c f r s for every constant c, Is < ft if s(x) < t(x) for all x in 1. r r 254 The Lebesgue Integral Th. 10.2 Also, if I is expressed as the union of a finite set of subintervals, say I = UP= 1 [a,., b,], where no two subintervals have interior points in common, then SI P br s(x) dx = f s(x) dx. r=1 ar 10.3 MONOTONIC SEQUENCES OF STEP FUNCTIONS A sequence of real-valued functions { fn} defined on a set S is said to be increasing on S if fn(x) 5fn+1(x) for all xin Sandalln. A decreasing sequence is one satisfying the reverse inequality. NOTE. We remind the reader that a subset T of R is said to be of measure 0 if, for every s > 0, T can be covered by a countable collection of intervals, the sum of whose lengths is less than e. A property is said to hold almost everywhere on a set S (written : a.e. on S) if it holds everywhere on S except for a set of measure 0. NOTATION. If If,} is an increasing sequence of functions on S such that f -> f almost everywhere on S, we indicate this by writing fn f a.e. on S. Similarly, the notation fn ' f a.e. on S means that {f.} is a decreasing sequence on S which converges to f almost everywhere on S. The next theorem is concerned with decreasing sequences of step functions on a general interval I. Theorem 10.2. Let {sn} be a decreasing sequence of nonnegative step functions such that sn N 0 a.e. on an interval I. Then lim f S. = 0. n- oo r Proof. The idea of the proof is to write Sn = fA Sn SB SI where each of A and B is a finite union of intervals. The set A is chosen so that in its intervals the integrand is small if n is sufficiently large. In B the integrand need not be small but the sum of the lengths of its intervals will be small. To carry out this idea we proceed as follows. There is a compact interval [a, b] outside of which sl vanishes. Since 0 < sn(x) < s1(x) for all x in I, each s,, vanishes outside [a, b]. Now sn is constant on each open subinterval of S" Th. 10.2 Monotonic Sequences of Step Functions 255 some partition of [a, b]. Let D. denote the set of endpoints of these subintervals, and let D = Un 1 D. Since each D. is a finite set, the union D is countable and therefore has measure 0. Let E denote the set of points in [a, b] at which the sequence {sn} does not converge to 0. By hypothesis, E has measure 0 so the set F=DVE also has measure 0. Therefore, if E > 0 is given we can cover F by a countable collection of open intervals F1, F2, . . . , the sum of whose lengths is less than E. Now suppose x e [a, b] - F. Then x E, so sn(x) -- 0 as n -> oo. Therefore there is an integer N = N(x) such that sN(x) < E. Also, x 0 D so x is interior to some interval of constancy of sN. Hence there is an open interval B(x) such that sN(t) < E for all t in B(x). Since {sn} is decreasing, we also have sn(t) < E for all n > N and all t in B(x). (2) The set of all intervals B(x) obtained as x ranges through [a, b] - F, together with the intervals F1, F2, . . . , form an open covering of [a, b]. Since [a, b] is compact there is a finite subcover, say P 9 [a, b] U B(xi) u U F,. i=1 r=1 Let N o denote the largest of the integers N(x1), ... , N(xp). From (2) we see that P sn(t) < E for all n > No and all tin U B(xi). (3) i=1 Now define A and B as follows : 9 B= U Fr, A=[a,b]-B. r=1 Then A is a finite union of disjoint intervals and we have Sn=fb Sn=J Sn+fD Sn. I A First we estimate the integral over B. Let M be an upper bound for s1 on [a, b]. Since {sn} is decreasing, we have sn(x) < s1(x) < M for all x in [a, b]. The sum of the lengths of the intervals in B is less than e, so we have SB S. < ME. Next we estimate the integral over A. Since A s U° 1 B(xi), the inequality in (3) shows that sn(x) < e if x e A and n -> No. The sum of the lengths of the intervals in A does not exceed b - a, so we have the estimate r sn<(b-a)e ifn - No. A 256 The Lebesgue Integral Th. 10.3 The two estimates together give us 11 s,, < (M + b - a)e if n > No, and this shows that limn- ,, 1, s = 0. Theorem 10.3. Let {t,,} be a sequence of step functions on an interval I such that: a) There is a function f such that t , f a.e. on I, and b) the sequence {f, tn} converges. Then for any step function t such that t(x) 5 f(x) a.e. on I, we have f t < lim f t,,. I n-.co I (4) Proof. Define a new sequence of nonnegative step functions {sn} on I as follows : _ to - tn(x) if t(x) > tn(x), sn(x) 10 if t(x) < Note that sn(x) = max {t(x) - tn(x), 0}. Now {sn} is decreasing on I since {tn} is increasing, and sn(x) -- max {t(x) - f(x), 0} a.e. on I. But t(x) < f(x) a.e. on I, and therefore s ,, 0 a.e. on I. Hence, by Theorem 10.2, limn., fI sn = 0. But sn(x) >- t(x) - tn(x) for all x in I, so f Sn > JI t - JI tn Now let n -+ oo to obtain (4). 10.4 UPPER FUNCTIONS AND THEIR INTEGRALS Let S(I) denote the set of all step functions on an interval I. The integral has been defined for all functions in S(I). Now we shall extend the definition to a larger class U(I) which contains limits of certain increasing sequences of step functions. The functions in this class are called upper functions and they are defined as follows : Definition 10.4. A real-valued function f defined on an interval I is called an upper function on I, and we write f e U(I), if there exists an increasing sequence of step functions {sn} such that a) sn T f a.e. on 1, and b) limns 11 sn is finite. The sequence {sn} is said to generate f. The integral off over I is defined by the equation $f=lirn$Sn. (5) I 'n-'c° r Th. 10.6 Upper Functions and Their Integrals 257 NOTE. Since { f r sn} is an increasing sequence of real numbers, condition (b) is equivalent to saying that If, is bounded above. The next theorem shows that the definition of the integral in (5) is unambiguous. Theorem 10.5. Assume f e U(I) and let {s} and {tm} be two sequences generating f Then n-'a0 m-a0 r j S. = lim f tn,. lim f Proof. The sequence {tm} satisfies hypotheses (a) and (b) of Theorem 10.3. Also, for every n we have sn(x) < f(x) a.e. on I, so (4) gives us II S. < lim f tm. m-r Since this holds for every n, we have n-a0 r m- 00 r lim f S. < lim f tm. The same argument, with the sequences {sn} and {tm) interchanged, gives the reverse inequality and completes the proof. It is easy to see that every step function is an upper function and that its integral, as given by (5), is the same as that given by the earlier definition in Section 10.2. Further properties of the integral for upper functions are described in the next theorem. Theorem 10.6. Assume f e U(I) and g c- U(I). Then: a) (f + g) E U(1) and I (f+g)=ff+f9. r r r b) cf e U(I) for every constant c >- 0, and f cf= c If r r Jr c) f, f S 11 g if f(x) < g(x) a.e. on I. NOTE. In part (b) the requirement c >- 0 is essential. There are examples for which f e U(I) but -f 0 U(I). (See Exercise 10.4.) However, if fE U(I) and if s e S(I), then f - s c- U(I) since f - s = f + (-s). Proof. Parts (a) and (b) are easy consequences of the corresponding properties for step functions. To prove (c), let {sm} be a sequence which generates f, and let 258 The Lebesgue Integral Th. 10.7 {tn} be a sequence which generates g. Then sm x f and t i' g a.e. on I, and lim f sm = r f , lim f to = f g. m-o0 JI n-'00 I r But for each m we have sm(x) < f(x) < g(x) = lim tn(x) a.e. on 1. n- 00 Hence, by Theorem 10.3, r n-00 r r Now, let m -> oo to obtain (c). The next theorem describes an important consequence of part (c). Theorem 10.7. If f E U(I) and g E U(I), and if f(x) = g(x) almost everywhere on I, then 11f=11g. Proof. We have both inequalities f(x) < g(x) and g(x) < f(x) almost everywhere on I, so Theorem 10.6 (c) gives frf < f 1 g and 1, g <_ I, f Definition 10.8. Let f and g be real-valued functions defined on I. We define max (f, g) and min (f, g) to be the functions whose values at each x in I are equal to max {f(x), g(x)} and min { f(x), g(x)}, respectively. The reader can easily verify the following properties of max and min : a) max (f, g) + min (f, g) = f + g, b) max (f + h, g + h) = max (f, g) + h, and min (f + h, g + h) = min (f, g) + h. Iffn , f a.e. on I, and if gn T g a.e. on I, then c) max (fn, g,) T max (f, g) a.e. on I, and min (fn, gn) / min (f, g) a.e. on I. Theorem 10.9. Iff E U(I) andg e U(I), then max (f, g) e U(I) and min (f g) E U(1). Proof. Let {sn} and {tn} be sequences of step functions which generate f and g, respectively, and let un = max (sn, tn), vn = min (sn, tn). Then un and vn are step functions such that un I' max (f, g) and vn T min (f, g) a.e. on I. To prove that min (f, g) E U(I), it suffices to show that the sequence {11 vn} is bounded above. But vn = min (sn, tn) < f a.e. on I, so 11 v,, < 11 f. Therefore the sequence {11 v,} converges. But the sequence {J, un} also converges since, by property (a), un = s,, + to - vn and hence l= funj'sn+$tn_$vnjf+$_$mmn(f9). The next theorem describes an additive property of the integral with respect to the interval of integration. Th. 10.11 Examples of Upper Functions 259 Theorem 10.10. Let I be an interval which is the union of two subintervals, say I = Il u I2, where I, and I2 have no interior points in common. a) If f e U(I) and if f > 0 a.e. on I, then f E U(I1), f c- U(I2), and ff=Jf+f f. (6) b) Assumef1 E U(I1), f2 E U(I2), and let f be defined on I as follows: f(x) = Jfl(x) if x E I1, f2(x) ifxeI-I1. Then f e U(1) and J1f= Jfi + Jf2. Proof. If {s"} is an increasing sequence of step functions which generates f on I, let s (x) = max {s"(x), 0} for each x in I. Then {s, } is an increasing sequence of nonnegative step functions which generates f on I (since f >- 0). Moreover, for every subinterval J of I we have f, s < f, s < f, f so Is. +} generates f on J. Also S" S" + f Sn , ! !1 .Ilz so we let n -> oo to obtain (a). The proof of (b) is left as an exercise. NOTE. There is a corresponding theorem (which can be proved by induction) for an interval which is expressed as the union of a finite number of subintervals, no two of which have interior points in common. 10.5 RIEMANN-INTEGRABLE FUNCTIONS AS EXAMPLES OF UPPER FUNCTIONS The next theorem shows that the class of upper functions includes all the Riemann-integrable functions. Theorem 10.11. Let f be defined and bounded on a compact interval [a, b], and assume that f is continuous almost everywhere on [a, b]. Then f e U([a, b]) and the integral off, as a function in U([a, b]), is equal to the Riemann integral f a f(x) dx. Proof. Let P. = {x0, x1, ... , x2"} be a partition of [a, b] into 2" equal sub-intervals of length (b - a)/2". The subintervals of Pn+1 are obtained by bisecting those of P". Let mk = inf {f(x) : x e [xk_ 1, xk]} for 1 <- k < 2", 260 The Lebesgue Integral Def. 10.12 and define a step function sn on [a, b] as follows: sn(x) = Mk if xk_ 1 < X G xk, sn(a) = m1. Then sn(x) < f(x) for all x in [a, b]. Also, {sn} is increasing because the inf of f in a subinterval of [xk_ 1, xk] cannot be less than that in [xk_ 1, xk]. Next, we prove that sn(x) -> f (x) at each interior point of continuity off. Since the set of discontinuities of f on [a, b] has measure 0, this will show that sn --> f almost everywhere on [a, b]. If f is continuous at x, then for every e > 0 there is a S (depending on x and on e) such that f (x) - e < f (y) < f (x) + s whenever x-S<y<x+S. Let m(S)=inf{f(y):ye(x-S,x+S)}. Then f(x) - e < m(S), so f(x) < m(S) + s. Some partition PN has a subinterval [xk_1, xk].containing x and lying within the interval (x - S, x + S). Therefore SN(x) = Mk < f (X) < m(S) + e < Mk + e = SN(X) + e. But sn(x) < f(x) for all n and sN(x) < sn(x) for all n >- N. Hence sn(x) <- f(x) < sn(x) + e if n >- N, which shows that sn(x) - f(x) as n -> oo. The sequence of integrals {1' a s,,} converges because it is an increasing sequence, bounded above by M(b - a), where M = sup {f(x) : x e [a, b]}. Moreover, I b Z. Sn = E mk(xk-xk-1)=L(Pn,{ J), Ja k=1 where L(Pn, f) is a lower Riemann sum. Since the limit of an increasing sequence is equal to its supremum, the sequence { f a sn} converges to the Riemann integral off over [a, b]. (The Riemann integral f a f(x) dx exists because of Lebesgue's criterion, Theorem 7.48.) NOTE. As already mentioned, there exist functions fin U(I) such that -f U(I). Therefore the class U(I) is,actually larger than the class of Riemann-integrable functions on I, since -f e R on I if f e R on I. 10.6 THE CLASS OF LEBESGUE-INTEGRABLE FUNCTIONS ON A GENERAL INTERVAL If u and v are upper functions, the difference u - v is not necessarily an upper function. We eliminate this undesirable property by enlarging the class of inte-grable functions. Definition 10.12. We denote by L(I) the set of all functions f of the form f = u - v, where u e U(I) and v e U(I). Each function f in L(1) is said to be Lebesgue-integrable on I, and its integral is defined by the equation frf= fu - fl v. (7) Def. 10.15 Basic Properties of the Lebesgue Integral 261 If f e L (I) it is possible to write f as a difference of two upper functions u - v in more than one way. The next theorem shows that the integral off is independent of the choice of u and v. Theorem 10.13. Let u, v, u1, and vl be functions in U(I) such that u - v = ul - v1. Then fu_fv=fui_fvi. (8) Proof. The functions u + vl and ul + v are in U(I) and u + vl = ul + v. Hence, by Theorem 10.6(a), we have fr u + f, vl = fr ul + fr-v, which proves (8). NOTE. If the interval I has endpoints a and b in the extended real number system R, where a < b, we also write a" or dx for the Lebesgue integral fr f We also define f b f = - f o f. If [a, b] is a compact interval, every function which is Riemann-integrable on [a, b] is in U([a, b]) and therefore also in L([a, b]). 10.7 BASIC PROPERTIES OF THE LEBESGUE INTEGRAL Theorem 10.14. Assume f e L(I) and g e L(I). Then we have: a) (af + bg) e L(1) for every real a and b, and 1, (af+bg)=a f, ff+b f r r 9. b) fr f - 0 if f(x) z 0 a.e. on I. c) f r f f r 9 ff(x) z g(x) a.e. on I. d) 11f = fr g if f(x) = g(x) a.e. on I. Proof. Part (a) follows easily from Theorem 10.6. To prove (b) we write f = u - v, where u e U(I) and v e U(I). Then u(x) > v(x) almost everywhere on I so, by Theorem 10.6(c), we have 11 u z f, v and hence I fu-rPart (c) follows by applying) (b) to f - g, and part (d) follows by applying (c) twice. Definition 10.15. If f is a real-valued function, its positive part, denoted by f +, and its negative part, denoted by f -, are defined by the equations f+ = max (f, 0), f = max (-f, 0). 262 The Lebesgue Integral 0 Th. 10.16 Figure 10.1 Note that f + and f - are nonnegative functions and that f=f+ -f', Ill =f+ +f-. Examples are shown in Fig. 10.1. Theorem 10.16. If f and g are in L (I), then so are the functions f +, f -, If 1, max (f, g) and min (f, g). Moreover, we have If fj (9) Proof. Write f = u - v, where u e U(I) and v e U(I). Then f + = max (u - v, 0) = max (u, v) - v. But max (u, v) a U(I), by Theorem 10.9, and v e U(I), so f+ e L(I). Since f - = f + - f, we see that f - e L(I). Finally, If I = f + + f -, so IfI e L(I). Since - I f(x)I < f(x) < If(x)I for all x in I we have - $ IfI < f f < f if 1, which proves (9). To complete the proof we use the relations max (f, g) = J(f + g + If - gi), min (f, g) = J(f + g - If - gl) The next theorem describes the behavior of a Lebesgue integral when the inter-val of integration is translated, expanded or contracted, or reflected through the origin. We use the following notation, where c denotes any real number: I + c = {x + c:xel}, cI = {cx:xeI). Theorem 10.17. Assume f e L(I). Then we have: a) Invariance under translation. Ifg(x) = f(x - c) for x in I + c, then g e L(I + c), and r+C = if ,J g f b) Behavior under expansion or contraction. If g(x) = f(x/c) for x in cI, where c > 0, then g e L(cI) and f"i g=cfl Th. 10.18 Basic Properties of the Lebesgue Integral c) Invariance under reflection. If g(x) = f(- x) for x in - I, then g e L(- I) and J r9 =f f NoTE. If I has endpoints a < b, where a and b are in the extended real number system R, the formula in (a) can also be written as follows : 6+c f(x - c) dx = f f(x) dx. Ja+c Ja Properties (b) and (c) can be combined into a single formula which includes both positive and negative values of c: b f(x) dx I ca f(x/c) dx = Icl a ,J ca ,J a if c 0. Proof. In proving a theorem of this type, the procedure is always the same. First, we verify the theorem for step functions, then for upper functions, and finally for Lebesgue-integrable functions. At each step the argument is straightforward, so we omit the details. Theorem 10.18. Let I be an interval which is the union of two subintervals, say I = I1 u I2, where I, and I2 have no interior points in common. a) If f E L(I), then f e L(I1), f e L(I2), and ff=5111+jf = b) Assume f, e L(I,), f2 e L(I2), and let f be defined on I as follows: f(x) = {.fl(x) f X E 11, lf2(x) if x e 1 - 11. Then f e L(I) and f, f = 11, fi + fl, f2 Proof. Write f = u - v where u e U(I) and v e U(I). Then u = u+ - u- and v = v+ - v-, so f = u+ + v- - (u- + v+). Now apply Theorem 10.10 to each of the nonnegative functions u+ + v- and u- + v+ to deduce part (a). The proof of part (b) is left to the reader. NOTE. There is an extension of Theorem 10.18 for an interval which can be expressed as the union of a finite number of subintervals, no two of which have interior points in common. The reader can formulate this for himself. We conclude this section with two approximation properties that will be needed later. The first tells us that every Lebesgue-integrable function f is equal to an upper function u minus a nonnegative upper function v with a small integral. The second tells us that f is equal to a step function s plus an integrable function 264 The Lebesgue Integral Th. 10.19 g with a small integral. More precisely, we have: Theorem 10.19. Assume f E L (I) and let c > 0 be given. Then: a) There exist functions u and v in U(I) such that f = u - v, where v is non-negative a.e. on I and f, v < E. b) There exists a step function s and a function g in L (I) such that f = s + g, where J, I I < E. Proof. Since f e L(I), we can write f = u1 - v1 where u1 and v1 are in U(I). Let be a sequence which generates v1. Since J, t - J, v1, we can choose N so that 0 < fI (vl - tN) < E. Now let v = vl - tN and u = ul - tN. Then both u and v are in U(I) and u - v = ul - v1 = f Also, v is nonnegative a.e. on I and f, v < s. This proves (a). To prove (b) we use (a) to choose u and v in U(I) so that v >- 0 a.e. on I, f=u-v and 0 J v<B. Now choose a step function s such that 0 < f, (u - s) < s/2. Then f=u-v=s+(u-s)-v=s+g, where g = (u - s) v. Hence g E L(I) and lgl f lu - sl + IVI < + E=E. I 2 2 10.8 LEBESGUE INTEGRATION AND SETS OF MEASURE ZERO The theorems in this section show that the behavior of a Lebesgue-integrable function on a set of measure zero does not affect its integral. Theorem 10.20. Let f be defined on I. If f = 0 almost everywhere on I, then fe L(1) and f1f = 0. Proof. Let s (x) = 0 for all x in I. Then is an increasing sequence of step functions which converges to 0 everywhere on I. Hence converges to f almost everywhere on L Since f, s = 0 the sequence {fI converges. Therefore f is an upper function, so f e L(I) and fI f = lim, f, s = 0. Theorem 10.21. Let f and g be defined on I. If f E L (I) and if f = g almost every-where on I, then g e L (I) and J, f = J, g. Proof. Apply Theorem 10.20 to f - g. Then f - g e L (I) and f, (f - g) = 0. Hence g=f-(f-g)EL(I)andflg=f'f-f'(f-g)=.1f Example. Define f on the interval [0, 1 ] as follows: f(x) = { 1 if x is rational 0 if x is irrational. Th. 10.22 The Levi Monotone Convergence Theorems 265 Then f = 0 almost everywhere on [0, 1 ] so f is Lebesgue-integrable on [0, 1 ] and its Lebesgue integral is 0. As noted in Chapter 7, this function is not Riemann-integrable on [0, 1]. NOTE. Theorem 10.21 suggests a definition of the integral for functions that are defined almost everywhere on I. If g is such a function and if g(x) = f (x) almost everywhere on I, where f e L(I), we say that g e L(1) and that 10.9 THE LEVI MONOTONE CONVERGENCE THEOREMS We turn next to convergence theorems concerning term-by-term integration of monotonic sequences of functions. We begin with three versions of a famous theorem of Beppo Levi. The first concerns sequences of step functions, the second sequences of upper functions, and the third sequences of Lebesgue-integrable functions. Although the theorems are stated for increasing sequences, there are corresponding results for decreasing sequences. Theorem 10.22 (Levi theorem for step functions). Let {sn} be a sequence of step functions such that a) increases on an interval I, and b) limn., f, s exists. Then {sn} converges almost everywhere on I to a limit function f in U(I), and f f = lim f sn. Ji n-co r Proof. We can assume, without loss of generality, that the step functions s are nonnegative. (If not, consider instead the sequence {sn - sl }. If the theorem is true for {sn - sl}, then it is also true for {sn}.) Let D be the set of x in I for which diverges, and let e > 0 be given. We will prove that D has measure 0 by showing that D can be covered by a countable collection of intervals, the sum of whose lengths is < a. Since the sequence {J, sn} converges it is bounded by some positive constant M. Let E Sn(x)] if x e I, tn(x) = [2M where [y] denotes the greatest integer 1 for infinitely many values of n. Let Dn = {x : x e I and to+1(x) - tn(x) > 1}. 266 The Lebesgue Integral Th. 10.23 Then Dn is the union of a finite number of intervals, the sum of whose lengths we denote by IDnJ. Now W D U D,,, n=1 so if we prove that Y_,'= 1 IDnI < s, this will show that D has measure 0. To do this we integrate the nonnegative step function tn+ 1 - to over I and obtain the inequalities f, (tn+1 - tn) > JD (t+1 - Q > 1 = I D. Hence for every m >- I we have IDnI (tn+1 - tn) t1+ n=1 n=1I- fI 1, E < 2M , jSm+i Therefore E' 1 IDni < e/2 < s, so D has measure 0. This proves that {sn} converges almost everywhere on I. Let .f(x) = limn-. sn(x) if x e I - D, to ifxeD. E 2 Then f is defined everywhere on I and sn -+ f almost everywhere on I. Therefore, f E U(I) and f1 f = limn, f, sn. Theorem 10.23 (Levi theorem for upper functions). Let { fn} be a sequence of upper functions such that a) { fn} increases almost everywhere on an interval I, and b) limn., f, fn exists. Then { fn} converges almost everywhere on I to a limit function fin U(I), and 1, f = lim J fn. n-.ao , Proof. For each k there is an increasing sequence of step functions {s,,,k} which generates fk. Define a new step function to on I by the equation tn(x) = max {sn,1(x), Sn,2(x), ... , Sn,n(x)}. Then {tn} is increasing on 7 because to+1(x) _ maX {Sn+1,1(x), sn+1,n+l(x)} max {sn,1(x), ... , sn,n+1(x)} > max {sn,1(x),... , ;,.(x)} = tn(x) Th. 10.24 The Levi Monotone Convergence Theorems But sn,k(x) < fk(x) and { fk} increases almost everywhere on I, so we have tn(x) < max {fi(x), ... , fn(x)} = fn(x) almost everywhere on I. Therefore, by Theorem 10.6(c) we obtain But, by (b), {J , fn} is bounded above so the increasing sequence If, tn} is also bounded above and hence converges. By the Levi theorem for step functions, {tn} converges almost everywhere on I to a limit function f in U(I), and f, f = limn, f, tn. We prove next that fn -+ f almost everywhere on I. The definition of tn(x) implies sn,k(x) < tn(x) for all k < n and all x in I. Letting n - oo we find fk(x) < f(x) almost everywhere on I. (12) Therefore the increasing sequence {fk(x)) is bounded above by f(x) almost every-where on I, so it converges almost everywhere on I to a limit function g satisfying g(x) < f(x) almost everywhere on I. But (10) states that tn(x) < fn(x) almost everywhere on I so, letting n - co, we find f(x) < g(x) almost everywhere on I. In other words, we have (11) 267 (10) lim fn(x) = f(x) almost everywhere on I. n-oo Finally, we show that Jr f = limn.,n f, fn. Letting n -+ co in (11) we obtain ii f < lim 1fn. n-+ao , (13) Now integrate (12), using Theorem 10.6(c) again, to get f, fk < f, f. Letting k -, oo we obtain limk., fI fk < f I f which, together with (13), completes the proof. NOTE. The class U(1) of upper functions was constructed from the class S(I) of step functions by a certain process which we can call P. Beppo Levi's theorem shows that when process P is applied to U(I) it again gives functions in U(I). The next theorem shows that when P is applied to L(I) it again gives functions in L(I). Theorem 10.24 (Levi theorem for sequences of Lebesgue-integrable functions). Let { fn} be a sequence of functions in L(I) such that a) { fn} increases almost everywhere on I, and b) limn. f, fn exists. 268 The Lebesgue Integral 11.10.25 Then {f.} converges almost everywhere on I to a limit function f in L(I), and 1, f = lim f". f n-+m r We shall deduce this theorem from an equivalent result stated for series of functions. Theorem 10.25 (Levi theorem for series of Lebesgue-integrable functions). Let {g"} be a sequence of functions in L(I) such that a) each g" is nonnegative almost everywhere on I, and b) the series E 1 fI g" converges. Then the series g" converges almost everywhere on I to a sum function g in L(I), and we have J 9 = fI 9n = J 9n. I n=1 n=1 I (14) Proof. Since gn e L(I), Theorem 10.19 tells us that for every e > 0 we can write gn = Un - vn, where u" e U(I), vn e U(I), vn > 0 a.e. on I, and J, v" < a. Choose u" and vn corresponding to a = (f)". Then U. = g" + v", where I V. < ()" The inequality on 11 v" assures us that the series E 1 fI v" converges. Now u" Z 0 almost everywhere on I, so the partial sums U5(x) = E Uk(x) k=1 form a sequence of upper functions {U.} which increases almost everywhere on I. Since 11 n UnJ uk= Uk-gk+J vk j k=1 k=1 I k=1 k=1 I the sequence of integrals {fI U"} converges because both series F-', fI gk and Y-' 1 f, vk converge. Therefore, by the Levi theorem for upper functions, the sequence {U"} converges almost everywhere on I to a limit function U in U(I), and fI U = limn-.. f, U. But Th. 10.26 The Levi Monotone Convergence Theorems so 1, 00 uk. U=E k=1 J Similarly, the sequence of partial sums {V.} given by nn Vn(X) = Lj Vk(X) k=1 converges almost everywhere on I to a limit function V in U(I) and f,Vf vk. k=1 I 269 Therefore U - V E L(I) and the sequence {Ek=1 gk} _ U. - Vn} converges almost everywhere on I to U - V. Let g = U - V. Then g c- L(I) and g I J=j'u_j'v=EJ(uk_vk)=Jk. This completes the proof of Theorem 10.25. Proof of Theorem 10.24. Assume { fn} satisfies the hypotheses of Theorem 10.24. Let gl = fl and let gn = fn - fn_ 1 for n > 2, so that n fn = E 9k k=1 Applying Theorem 10.25 to {gn}, we find that Ert 1 gn converges almost everywhere on I to a sum function g in L(I), and Equation (14) holds. Therefore fn -+ g almost everywhere on I and 119 = limn-. f f fn-In the following version of the Levi theorem for series, the terms of the series are not assumed to be nonnegative. Theorem 10.26. Let {gn} be a sequence of functions in L(I) such that the series JInI is convergent. Then the series E 1 gn converges almost everywhere on I to a sum function g in L (I) and we have ao E9. = E I n=1 n=1 Proof. Write gn = g - g; and apply Theorem 10.25 to the sequences {g } and {g, } separately. The following examples illustrate the use of the Levi theorem for sequences. 270 The Lebesgue Integral Th. 10.27 Example 1. Let f (x) = xS for x > 0, f (O) = 0. Prove that the Lebesgue integral fo f(x) dx exists and has the value 1/(s + 1) ifs > -1. Solution. If s >- 0, then f is bounded and Riemann-integrable on [0, 1 ] and its Riemann integral is equal to 11(s + 1). If s < 0, then f is not bounded and hence not Riemann-integrable on [0, 1 ]. Define a sequence of functions {fn} as follows: fn(X) _ {XS if x > 1/n, 0 if0-x<1/n. Then { fn} is increasing and fn -+ f everywhere on [0, 1 ]. Each fn is Riemann-integrable and hence Lebesgue-integrable on [0, 1 ] and $01 1 fn(x) dx. = f xs dx En S+ 1 If s + 1 > 0, the sequence {fo fn} converges to 1/(s + 1). Therefore, the Levi theorem for sequences shows that f o f exists and equals 1/(s + 1). Example 2. The same type of argument shows that the Lebesgue integral f10 a-"xy-1 dx exists for every real y > 0. This integral will be used later in discussing the Gamma function. 10.10 THE LEBESGUE DOMINATED CONVERGENCE THEOREM Levi's theorems have many important consequences. The first is Lebesgue's dominated convergence theorem, the cornerstone of Lebesgue's theory of inte-gration. Theorem 10.27 (Lebesgue dominated convergence theorem). Let { fn} be a sequence of Lebesgue-integrable functions on an interval I. Assume that a) { fn} converges almost everywhere on I to a limit function f, and b) there is a nonnegative function g in L (I) such that, for all n >_ 1, l fn(x)I <_ g(x) a.e. on I. Then the limit function f e L (I), the sequence If, fn} converges and SI f = lim fn. n-+oo i (15) NOTE. Property (b) is described by saying that the sequence { fn} is dominated by g almost everywhere on I. Proof. The idea of the proof is to obtain upper and lower bounds of the form gn(x) < fn(x) < Gn(x) (16) Th. 10.27 The Lebesgue Dominated Convergence Theorem 271 where increases and decreases almost everywhere on Ito the limit function f. Then we use the Levi theorem to show that f e L (I) and that f, f = lim,, f, g = f, G,,, from which we obtain (15). To construct and we make repeated use of the Levi theorem for sequences in L (I). First we define a sequence {G1} as follows : G,,,1(x) = max {f1(x),f2(x), ... Each function G,,,1 e L(I), by Theorem 10.16, and the sequence {G,,,1} is in-creasing on I. Since IG,,,1(x)I < g(x) almost everywhere on I, we have <- f IG,,.1I <-JI 9. 1 (17) Therefore the increasing sequence of numbers {f1 G,,,1} is bounded above by J, g, so 11 G,,,1 exists. By the Levi theorem, the sequence {G,,,1 } converges almost everywhere on I to a function G1 in L(I), and f G1 G. , 9 -I I n~0D I Because of (17) we also have the inequality -J, g < f, G1. Note that if x is a point in I for which G,,,1(x) -+ G1(x), then we also have G1(x) = sup {.fi(x),f2(x),... In the same way, for each fixed r -> I we let G,,,,(x) = max {f,(x), f,+ 1(x), ... , f (x)} for n >- r. Then the sequence {G,,,,} increases and converges almost everywhere on I to a limit function G, in L(I) with _f1g :!9 fG,<f1g Also, at those points for which G,(x) we have G,(x) = sup {f,(x), f,+ 1(x), . . . }, so f,(x) < G,(x) a.e. on I. Now we examine properties of the sequence {G (x)}. Since A 9 B implies sup A -< sup B, the sequence {G,(x)} decreases almost everywhere and hence converges almost everywhere on I. We show next that f(x) whenever lim P X). (18) If (18) holds, then for every s > 0 there is an integer N such that for alln>N. 272 The Lebesgue Integral Hence, if m > N we have f(x) - E < sup {fm(x), fm+ 1(x), ... } < f(x) + E. In other words, m>N and this implies that implies f(x) - E < Gm(x) < f(x) + E, lim Gm(x) = f(x) almost everywhere on I. M-00 (19) On the other hand, the decreasing sequence of numbers fl, is bounded below by -1, g, so it converges. By (19) and the Levi theorem, we see that f e L(I) and Jim n- 00 By applying the same type of argument to the sequence min {f.(x), f.+ 1(x), . . . , f (x)}, for n > r, we find that {g,,,,} decreases and converges almost everywhere to a limit function g, in L (I), where g,(x) = inf {f,(x), f,+ 1(x), ... } a.e. on I. Also, almost everywhere on I we have g,(x) 5 f,(x), {g,} increases, lime .. f (x), and n- 00 .Ji lim 19= If. Since (16) holds almost everywhere on I we have n -+ oo we find that { f, f } converges and that $1f=f1f fig. :9 J., fn < 11Gn. Th. 10.28 Letting 10.11 APPLICATIONS OF LEBESGUE'S DOMINATED CONVERGENCE THEOREM , The first application concerns term-by-term integration of series and is a companion result to Levi's theorem on series. Theorem 10.28. Let be a sequence of functions in L (I) such that: a) each g is nonnegative almost everywhere on I, and b) the series g converges almost everywhere on I to a function g which is bounded above by a function in L (I). 11. 10.30 Applications of Lebesgue's Theorem 273 Then g e L(I), the series En 1 fr gn converges, and we have E gn = E JIfgn-n=1 n=1 00 Proof. Let n fn(x) = 9k(x) 'f X E I. k=1 Then fn -+ g almost everywhere on I, and { fn} is dominated almost everywhere on I by the function in L(I) which bounds g from above. Therefore, by the Le-besgue dominated convergence theorem, g c- L (I), the sequence { fI fn} converges, and fI g = limn fI fn. This proves the theorem. The next application, sometimes called the Lebesgue bounded convergence theorem, refers to a bounded interval. Theorem 10.29. Let I be a bounded interval. Assume { fn} is a sequence of functions in L (I) which is boundedly convergent almost everywhere on I. That is, assume there is a limit function f and a positive constant M such that lim fn(x) = f(x) and I fn(x)I < M, almost everywhere on I. n-+w Then f E L(I) and limn..+ f rfn = f r f Proof. Apply Theorem 10.27 with g(x) = M for all x in I. Then g E L(I), since I is a bounded interval. NOTE. A special case of Theorem 10.29 is Arzela's theorem stated earlier (Theorem 9.12). If I fn} is a boundedly convergent sequence of Riemann-integrable functions on a compact interval [a, b], then each fn e L([a, b]), the limit function f e L([a, b]), and we have b lim f = n-00 f a f a f If the limit function f is Riemann-integrable (as assumed in Arzela's theorem), then the Lebesgue integral J1 f is the same as the Riemann integral fo f(x) dx. The next theorem is often used to show that functions are Lebesgue-integrable. Theorem 10.30. Let {fn} be a sequence offunctions in L (I) which converges almost everywhere on I to a limit function f. Assume that there is a nonnegative function g in L(I) such that f(x)l < g(x) a.e. on I. Then f E L(I). Proof. Define a new sequence of functions {g,,) on I as follows : g,, = max {min (fn, g), -g}. 274 The Lebesgue Integral Th. 10.31 Figure 10.2 Geometrically, the function gn is obtained from fn by cutting off the graph of fn from above by g and from below by -g, as shown by the example in Fig. 10.2. Then Ign(x)I < g(x) almost everywhere on I, and it is easy to verify that gn -+ f almost everywhere on I. Therefore, by the Lebesgue dominated convergence theorem, f e L(I). 10.12 LEBESGUE INTEGRALS ON UNBOUNDED INTERVALS AS LIMITS OF INTEGRALS ON BOUNDED INTERVALS Theorem 10.31. Let f be defined on the half-infinite interval I = [a, + co). Assume that f is Lebesgue-integrable on the compact interval [a, b] for each b >- a, and that there is a positive constant M such that If I < M for all b 2: a. (20) Then f e L(I), the limit limbs+a, f; f exists, and +OD a f = lim b-+00 ff (21) Proof. Let be any increasing sequence of real numbers with b >: a such that lima co b = + oo. Define a sequence { fn} on I as follows : A W = If (x) if a < x < bn, to otherwise. Each f e L(I) (by Theorem 10.18) and fa -+ f on I. Hence, Ifnl -+ IfI on I. But Ifnl is increasing and, by (20), the sequence {f, I fl} is bounded above by M. Therefore f, Ifal exists. By the Levi theorem, the limit function If I E L(I). Now each Ifal < If I and f -+ f on I, so by the Lebesgue dominated convergence theorem, f e L(I) and lim, f, fn = f, f. Therefore b + OD lim f = f n~00 a a for all sequences {bn} which increase to + oo. This completes the proof. Th. 10.31 Lebesgue Integrals on Unbounded Intervals 275 There is, of course, a corresponding theorem for the interval (- oo, a] which concludes that f:c= J'2 0 provided that $ If I < M for all c < a. If f'C If I < M for all real c and b with c < b, the two theorems together show that f e L(R) and that + a Pb J f = lim I f + lim c-.-00 by+1 f Example 1. Let f(x) = 1/(1 + x2) for all x in R. We shall prove that f e L(R) and that f R f = it. Now f is nonnegative, and if c <- b we have 6 b [fix f= f 2 = arctan b - arctan c <- it. J 1+x Therefore, f e L(R) and L 0 fc 0 dac b dx n it f = lim + li m + - = n. 1 + x2 b-.+OO fo 1 + x2 2 2 Example 2. In this example the limit on the right of (21) exists but f 0 L(I). Let I = [0, + oo) and define f on 1 as follows : f(x) = (_ 1) ifn - 1 5x< it, for n= 1,2,... n If b > 0, let m = [b], the greatest integer s b. Then 0 f bf = fo Mf + J bf = E (-1)n + (b - m)(-1)'"+1 m n=1 it m+ 1 As b - + oo the last term -- 0, and we find b lim f f = E (-1) _ -log 2. b-++ao o n=1 n Now we assume f e L(I) and obtain a contradiction. Let fn be defined by fn(x) _ (I f (x) I for 0 5 x 5 it, 0 for x > n. Then {fn } increases and fn(x) -+ I f (x) I everywhere on I. Since f e L(I) we also have Ifs e L(I). But Ifn(x)I <- If(x)I everywhere on I so by the Lebesgue dominated con-vergence theorem the sequence {fr converges. But this is a contradiction since asn -+ oo. 276 The Lebesgue Integral 10.13 IMPROPER RIEMANN INTEGRALS Def. 10.32 Definition 10.32. If f is Riemann-integrable on [a, b] for every b > a, and if the limit b Jim f(x) dx exists, b-+oo a then f is said to be improper Riemann-integrable on [a, + oo) and the improper Riemann integral of f, denoted by fa ao f(x) dx or fa f(x) dx, is defined by the equation b +00 f(x) dx = lim f(x) dx. a b-+oo Ja In Example 2 of the foregoing section the improper Riemann integral 10' °° f(x) dx exists but f is not Lebesgue-integrable on [0, + oo). That example should be contrasted with the following theorem. Theorem 10.33. Assume f is Riemann-integrable on [a, b] for every b >- a, and assume there is a positive constant M such that f b If(x)I dx < M for every b > a. (22) Then both f and If I are improper Riemann-integrable on [a, + oo). Also, f is Lebesgue-integrable on [a, + oo) and the Lebesgue integral off is equal to the im-proper Riemann integral off. Proof. Let F(b) = Ja If(x) I dx. Then F is an increasing function which is bounded above by M, so limb.. +,,o F(b) exists. Therefore If I is improper Riemann-integrable on [a, + oo). Since the limit 0 < If(x)I - f(x) < 21 f(x)I, lim f b {If(x)I - f(x)} dx b- +co a also exists; hence the limit limb-+. $ f(x) dx exists. This proves that f is improper Riemann-integrable on [a, + oo). Now we use inequality (22), along with Theorem 10.31, to deduce that f is Lebesgue-integrable on [a, + oo) and that the Lebesgue integral off is equal to the improper Riemann integral off. NOTE. There are corresponding results for improper Riemann integrals of the form b b f j _ f(x) dx = lim f f(x) dx, CO a - CO Ja f f (x) dx = lim f b f (x) dx, Ja Ja Th. 10.33 Improper Riemann Integrals and 6 f f(x) dx = lim f(x) dx, J c a-c+ Ja 277 which the reader can formulate for himself. If both integrals f_ f(x) dx and la' ' f(x) dx exist, we say that the integral +' f(x) dx exists, and its value is defined to be their sum, f+00 f(x) dx f(x) dx + f f(x) dx. +"O °° Ja If the integral f ±' f(x) dx exists, its value is also equal to the symmetric limit b f(x) dx. lim f-b b- + co However, it is important to realize that the symmetric limit might exist even when f +'f(x) dx does not exist (for example, take f(x) = x for all x). In this case the 00 symmetric limit is called the Cauchy principal value of f +' ,f(x) dx. Thus f +' x dx 00 -has Cauchy principal value 0, but the integral does not exist. Example 1. Let f(x) = e-xxy-', where y is a fixed real number. Since e-x/2xy-1 -, 0 as x - + oo, there is a constant M such that e-x/2xy-1 -< M for all x > 1. Then e-xxy-1 < Me-x12, so 6 b If(x)Idx<M f ex/2dx=2M(1-eb12)<2M. f 1 o Hence the integral f i °° a-xxy-1 dx exists for every real y, both as an improper Riemann integral and as a Lebesgue integral. Example 2. The Gamma function integral. Adding the integral of Example 1 to the integral fo a-xxy-1 dx of Example 2 of Section 10.9, we find that the Lebesgue integral +00 r(y) = f e-xx-1 dx 0 exists for each real y > 0. The function r so defined is called the Gamma function.-Example 4 below shows its relation to the Riemann zeta function. NOTE. Many of the theorems in Chapter 7 concerning Riemann integrals can be converted into theorems on improper Riemann integrals. To illustrate the straight-forward manner in which some of these extensions can be made, consider the formula for integration by parts : f(x)9'(x) dx = f(b)g(b) - f(a)g(a) g(x)f'(x) dx. - ja b b Since b appears in three terms of this equation, there are three limits to consider 278 The Lebesgue Integral as b -- + oo. If two of these limits exist, the third also exists and we get the formula f f(x)g'(x) dx = lim f(b)g(b) - f(a)g(a) - f g(x)f(x) dx. b-++oo Ja Other theorems on Riemann integrals can be extended in much the same way to improper Riemann integrals. However, it is not necessary to develop the details of these extensions any further, since in any particular example, it suffices to apply the required theorem to a compact interval [a, b] and then let b -> + oo. Example 3. The functional equation I'(y + 1) = yF(y). If 0 < a < b, integration by parts gives I b b e-xxy A = aye - ye-b + y f e-xxy-1 A. Ja Letting a --. 0+ and b --> + oo, we find r(y + 1) = yI'(y). Example 4. Integral representation for the Riemann zeta function. The Riemann zeta function C is defined for s > I by the equation 00 1 C(s) = E s . n=1 n This example shows how the Levi convergence theorem for series can be used to derive an integral representation, C(s)r(s) = Jo ex -11 dx. The integral exists as a Lebesgue integral. In the integral for r(s) we make the change of variable t = nx, n > 0, to obtain r(s) = J ets-1 dt = ns 0 a n"xs-1 dx. 0 Hence, if s > 0, we have n_sr(s) = f,0 e nxxs dx. 0 If s > 1, the series Y_n 1 n-s converges, so we have e nxxs-1 dx, C(s)r(s) = f,"o n=1 he series on the right being convergent. Since the integrand is nonnegative, Levi's con-t vergence theorem (Theorem 10.25) tells us that the series e-' xs-1 converges almost everywhere to a sum function which is Lebesgue-integrable on [0, + oo) and that C(s)j'(s) enxxs-1dx = fo, enxxs-1 dx. n=1 O nn=+1 Th. 10.36 Measurable Functions 279 But if x > 0, we have 0 < e-x < 1 and hence, e' e -X 1 - ex ex - 1' n=1 the series being a geometric series. Therefore we have 00 --S-1 e-"x-n=1 ex almost everywhere on [0, + oo), in fact everywhere except at 0, so C(s)r(s) = E e nxxs-1 dx = xs- 1 dx. 0n=1 00 f ex - I 10.14 MEASURABLE FUNCTIONS Every function f which is Lebesgue-integrable on an interval I is the limit, almost everywhere on I, of a certain sequence of step functions. However, the converse is not true. For example, the constant function f = 1 is a limit of step functions on the real line R, but this function is not in L(R). Therefore, the class of functions which are limits of step functions is larger than the class of Lebesgue-integrable functions. The functions in this larger class are called measurable functions. Definition 10.34. A function f defined on I is called measurable on I, and we write f e M(1), if there exists a sequence of step functions {sn} on I such that lim sn(x) = f(x) almost everywhere on I. n-i a0 NOTE. If f is measurable on I then f is measurable on every subinterval of I. As already noted, every function in L(I) is measurable on I, but the converse is not true. The next theorem provides a partial converse. Theorem 10.35. If f e M(I) and if I f(x)I < g(x) almost everywhere on I for some nonnegative g in L(I), then f e L(I). Proof. There is a sequence of step functions {sn} such that sn(x) - f(x) almost everywhere on I. Now apply Theorem 10.30 to deduce that f e L(I). Corollary 1. if f e M(1) and If I e L (l), then f e L (l). Corollary 2. If f is measurable and bounded on a bounded interval 1, then f e L (I). Further properties of measurable functions are given in the next theorem. Theorem 10.36. Let p be a real-valued function continuous on R2. If f e M(I) and g e M(1), define h on I by the equation h(x) = q[f(x), g(x)]. 280 The Lebesgue Integral Th. 10.37 Then h e M(I). In particular, f + g, f g, If I, max (f, g), and min (f g) are in M(I). Also, 1/f e M(I) if f(x) # 0 almost everywhere on I. Proof. Let {s"} and {t"} denote sequences of step functions such that s" -+ f and t" g almost everywhere on I. Then the function u" = (p(s,,, t") is a step function such that u" - h almost everywhere on I. Hence h c- M(I). The next theorem shows that the class M(I) cannot be enlarged by taking limits of functions in M(I). Theorem 10.37. Let f be defined on I and assume that {f.1 is a sequence of measur-able functions on I such that f"(x) -> f(x) almost everywhere on I. Then f is measur-able on I. Proof. Choose any positive function g in L(1), for example, g(x) = 1/(1 + x2) for all x in I. Let F"(x) = g(x) 1 +"( f x)I for x in I. Then F"(x) g(x)f(x) almost everywhere on I. 1 + If(x)I Let F(x) = g(x)f(x)l{I + If(x)I}. Since each F. is measurable on I and since IF"(x)I < g(x) for all x, Theorem 10.35 shows that each F. e L(I). Also, IF(x)I < g(x) for all x in I so, by Theorem 10.30, F E L(I) and hence F E M(I). Now we have f(x){g(x) - IF(x)I} = f(x)g(x) 1 -If(x)1 __ f(x)g(x) = F(x) I 1 + If(x)I .1 + If(x)1 for all x in I, so f(x) = F(x) g(x) - I F'(x)I Therefore f e M(I) since each of F, g, and Ill is in M(I) and g(x) - IF(x)I > 0 for all x in I. NOTE. There exist nonmeasurable functions, but the foregoing theorems show that it is not easy to construct an example. The usual operations of analysis, applied to measurable functions, produce measurable functions. Therefore, every function which occurs in practice is likely to be measurable. (See Exercise 10.37 for an example of a nonmeasurable function.) Th. 10.38 Functions Defined by Lebesgue Integrals 281 10.15 CONTINUITY OF FUNCTIONS DEFINED BY LEBESGUE INTEGRALS Let f be a real-valued function of two variables defined on a subset of R2 of the form X x Y, where each of X and Y is a general subinferval of R. Many functions in analysis appear as integrals of the form F(y) = fx f(x, y) dx. J We shall discuss three theorems which transmit continuity, differentiability, and integrability from the integrand f to the function F. The first theorem concerns continuity. Theorem 10.38. Let X and Y be two subintervals of R, and let f be a junction defined on X x Y and satisfying the following conditions: a) For each fixed y in Y, the function fY defined on X by the equation ff(x) = f(x, y) is measurable on X. b) There exists a nonnegative function g in L (X) such that, for each y in Y, I.f(x, y)I < g(x) a.e. on X. c) For each fixed y in Y, lim f(x, t) = f(x, y) a.e. on X. t-.Y Then the Lebesgue integral f x f (x, y) dx exists for each y in Y, and the function F defined by the equation F(y) = fx f(x, y) dx J is continuous on Y. That is, if y e Y we have lim f f(x, t) dx = fX lim f(x, t) dx. t-.y x tY Proof. Since fY is measurable on X and dominated almost everywhere on X by a nonnegative function g in L(X), Theorem 10.35 shows that fY e L(X). In other words, the Lebesgue integral $x f(x, y) dx exists for each y in Y. Now choose a fixed y in Y and let {y.} be any sequence of points in Y such that lim y = y. We will prove that lim F(y) Let f(x, Each G. e L(X) and (c) shows that f(x, y) almost everywhere on X. Note that $x GR(x) dx. Since (b) holds, the Lebesgue dominated convergence 282 The Lebesgue Integral theorem shows that the sequence {F(yn)} converges and that Jim f (yn) = J f(x, y) dx = F(y) n-00 x Example 1. Continuity of the Gamma function r(y) = f o 0D e -'x' -1 dx for y > 0. We apply Theorem 10.38 with X = [0, + oo), Y = (0, + oo). For each y > 0 the integrand, as a function of x, is continuous (hence measurable) almost everywhere on X, so (a) holds. For each fixed x > 0, the integrand, as a function of y, is continuous on Y, so (c) holds. Finally, we verify (b), not on Y but on every compact subinterval [a, b ], where 0 < a < b. For each y in [a, b] the integrand is dominated by the function xa-1 if 0 < x < 1, me-x12 if x > 1, where M is some positive constant. This g is Lebesgue-integrable on X, by Theorem 10.18, so Theorem 10.38 tells us that F is continuous on [a, b]. But since this is true for every subinterval [a, b], it follows that IF is continuous on Y = (0, + oo). Example 2. Continuity of F(y) = +00 ex' sin x dx f o x for y > 0. In this example it is understood that the quotient (sin x)/x is to be replaced by 1 when x = 0. Let X = [0, + oo), Y = (0, + oo). Conditions (a) and (c) of Theorem 10.38 are satisfied. As in Example 1, we verify (b) on each subinterval Y. = [a, + co), a > 0. Since (sin x)/xJ 5 1, the integrand is dominated on Y. by the function g(x) = e-ax for x > 0. Since g is Lebesgue-integrable on X, F is continuous on Ya for every a > 0; hence F is continuous on Y = (0, + oo). To illustrate another use of the Lebesgue dominated convergence theorem we shall prove that F(y) -+ 0 as y -+ + oo. Let {yn} be any increasing sequence of real numbers such that ya Z I and yn -+ + oo as n -+ oo. We will prove that F(ya) - 0 as n -+ oo. Let fn(x) = e-xyn sin x x for x _ 0. Then limn 0O fn(x) = 0 almost everywhere on [0, + co), in fact, for all x except 0. Now ya ;-> 1 implies I fn(x)I 5 e-x for all x > 0. Also, each fn is Riemann-integrable on [0, b] for every b > 0 and 6 b Ifnl fo dx < 1. 0 0 Th. 10.39 Differentiation under the Integral Sign 283 Therefore, by Theorem 10.33, f is Lebesgue-integrable on [0, + oo). Since the sequence f f.} is dominated by the function g(x) = e-" which is Lebesgue-inte-grable on [0, + oo), the Lebesgue dominated convergence theorem shows that the sequence {f o converges and that lim fn= f limfn=0. n_°o o o n_c But 10+ °° fn = F(y ), so F(yn) --> 0 as n --> oo. Hence, F(y) -1- 0 as y - + oo. NOTE. In much of the material that follows, we shall have occasion to deal with integrals involving the quotient (sin x)/x. It will be understood that this quotient is to be replaced by 1 when x = 0. Similarly, a quotient of the form (sin xy)/x is to be replaced by y, its limit as x -+ 0. More generally, if we are dealing with an integrand which has removable discontinuities at certain isolated points within the interval of integration, we will agree that these discontinuities are to be "re-moved" by redefining the integrand suitably at these exceptional points. At points where the integrand is not defined, we assign the value 0 to the integrand. 10.16 DIFFERENTIATION UNDER THE INTEGRAL SIGN Theorem 10.39. Let X and Y be two subintervals of R, and let f be a junction defined on X x Y and satisfying the following conditions: a) For each fixed y in Y, the function fy defined on X by the equation fy(x) = f(x, y) is measurable on X, and f, E L(X) for some a in Y. b) The partial derivative D2 f(x, y) exists for each interior point (x, y) of X x Y. c) There is a nonnegative function G in L(X) such that I D2 f(x, y)I < G(x) for all interior points of X x Y. Then the Lebesgue integral $x f(x, y) dx exists for every y in Y, and the function F defined by F(y) = fxf(x, y) dx J is differentiable at each interior point of Y. Moreover, its derivative is given by the formula F'(y) = f X y) dx. x NOTE. The derivative F'(y) is said to be obtained by differentiation under the integral sign. Proof. First we establish the inequality I fy(x)l < Ifa(x)I + l y - al G(x), (23) 284 The Lebesgue Integral for all interior points (x, y) of X x Y. The Mean-Value Theorem gives us f(x, y) - f(x, a) = (y - a) D2f(x, c), where c lies between a and y. Since I D2 f(x, c)I < G(x), this implies If(x,Y)I - If(x, a)I + Iy - at G(x), which proves (23). Since fy is measurable on X and dominated almost everywhere on X by a nonnegative function in L(X), Theorem 10.35 shows that fy e L(X). In other words, the integral Ix f(x, y) dx exists for each y in Y. Now choose any sequence {y.} of points in Y such that each y" y but lim y" = y. Define a sequence of functions {q"} on X by the equation q"(x) = f (X, Y") - f (X' Y) Y" - Y Then q" a L(X) and q"(x) -p D2f(x, y) at each interior point of X. By the Mean-Value Theorem we have q"(x) = D2 f(x, c"), where c" lies between y" and y. Hence, by (c) we have lq"(x)I < G(x) almost everywhere on X. Lebesgue's dominated convergence theorem shows that the sequence {j x q"} converges, the integral Ix D2f(x, y) dx exists, and lim fX q" = f 1im q" = f D2 f (x, y) dx. "-' 00 X "- 00 J x But Y" Y,Ix {f(x, y") - f(x, y)} dx = F(Yy) _ F(y) f q" = Since this last quotient tends to a limit for all sequences { y"}, it follows that F(y) exists and that F(y) = lim fx q" = J D2f(x, y) dx. " OD X Example 1. Derivative of the Gamma function. The derivative r"(y) exists for each y > 0 and is given by the integral log x dx, r"(Y) = f o Jo obtained by differentiating the integral for r'(y) under the integral sign. This is a conse-quence of Theorem 10.39 because for each y in [a, b], 0 < a < b, the partial deriva-tive D2(e'"xy-') is dominated a.e. by a function g which is integrable on [0, + oo). In fact, D2(e 'xy-') = ay (e-"xy-') = e-J°xy-' log x if x > 0, Differentiation under the Integral Sign 285 so if y >- a the partial derivative is dominated (except at 0) by the function x°-1 log xI if 0 < x < 1, g(x) = Me-x12 if x > 1, 0 ifx=0, where M is some positive constant. The reader can easily verify that g is Lebesgue-integrable on [0, + oo). Example 2. Evaluation of the integral F(y) _ Applying Theorem 10.39, we find e_x" sin x dx. x ao fo + F(y) = f o xy sin x dx if -Y > 0. 0 (As in Example 1, we prove the result on every interval Yo = [a, + oo), a > 0.) In this example, the Riemann integral f b a-x' sin x dx can be calculated by the methods of elementary calculus (using integration by parts twice). This gives us f b e-'' sin x dx = e b'(- y sin b - cos b) + 1 1 + y2 F+-;2 for all real y. Letting b - + oo we find +00 e x'sinxdx= ify>0. 0 1+y2 Therefore F'(y) _ -1/(1 + y2) if y > 0. Integration of this equation gives us (24) F(y) - F(b) b' 1 + dt t 2 = arctan b - arctan y, for y > 0, b > 0. Now let b -+ + oo. Then arctan b -+ x/2 and F(b) -+ 0 (see Example 2, Section 10.15), so F(y) = x/2 - arctan y. In other words, we have +00 a x ' si x dx = - arctan if > 0 25 2 y y . z Jo ( ) This equation is also valid if y = 0. That is, we have the formula "sin x d n x=-(26) o x 2 However, we cannot deduce this by putting y = 0 in (25) because we have not shown that F is continuous at 0. In fact, the integral in (26) exists as an improper Riemann integral. It does not exist as a Lebesgue integral. (See Exercise 10.9.) 286 The Lebesgue Integral Example 3. Proof of the formula + °0 sin x dx = lim ° sin x dx = n fo x b-.+. fo x 2 Let {gn} be the sequence of functions defined for all real y by the equation 9n(Y) Jf e_x,,sinx &C. 0 X First we note that g"(n) -+ 0 as n - oo since 9n(n)1 ex"dx1 f,,2 etdt fn o n n Now we differentiate (27) and use (24) to obtain 9 - f" e"' sin x dx e-"'(- y sin n - cos n) + I ;,(Y) = = -2 0 (27) an equation valid for all real y. This shows that g' .(y) -+ -1/(1 + y2) for all y and that ' gn(y) I < e '(y + 1) + 1 for all y >- 0. 1 + y2 Therefore the function fn defined by fn (Y) - (gn(Y) if 0 < y < n, 0 ify> n, is Lebesgue-integrable on [0, +oo) and is dominated by the nonnegative function g(y)=e''(y+1)+1 1+y2 Also, g is Lebesgue-integrable on [0, + oo). Since f"(y) - -1/(1 + y2) on [0, + oo), the Lebesgue dominated convergence theorem implies +00 +00 dy n lim f f" _ - f a-, 0 0 1+y22 But we have ('+ 00 n fn = J gn(Y) dy = gn(n) - gn(0). 0 0 f' Letting n - co, we find g"(0) - n/2. Now if b > 0 and if n = [b], we have 1+y o f b sin x dx = So" sin x + ns= g(0) + fb sin x dx. x n x n x Th. 10.40 Interchanging the Order of Integration Since 0< Ibsinxdxl < b l dx= b- n < asb -, + oo, I x I n n n we have b sin x n lim fo dx = lim b-,+w x 2 This formula will be needed in Chapter 11 in the study of Fourier series. 287 10.17 INTERCHANGING THE ORDER OF INTEGRATION Theorem 10.40. Let X and Y be two subintervals of R, and let k be a function which is defined, continuous, and bounded on X x Y, say jk(x, y)I < M for all (x, y) in X x Y. Assume f e L(X) and g e L(Y). Then we have: a) For each y in Y, the Lebesgue integral fx f(x)k(x, y) dx exists, and the function F defined on Y by the equation F (y) = f f(x)k(x, y) dx x is continuous on Y. b) For each x in X, the Lebesgue integral ly g(y)k(x, y) dy exists, and the function G defined on X by the equation G(x) = f g(y)k(x, y) dy y is continuous on X. c) The two Lebesgue integrals ly g(y)F(y) dy and f x f(x)G(x) dx exist and are equal. That is, , y ) dx] dy. (28) fX .f(x) C fy g(y)k(x, y) dy] dx = fy g(Y) [$f(x)k(x J Proof. For each fixed y in Y, let fy(x) = f(x)k(x, y). Then fy is measurable on X and satisfies the inequality Ify(x)I = If(x)k(x, y)l < Mlf(x)l for all x in X. Also, since k is continuous on X x Y we have lim f(x)k(x, t) = f(x)k(x, y) for all x in X. 1"Y Therefore, part (a) follows from Theorem 10.38. A similar argument proves (b). 288 The Lebesgue Integral Th. 10.40 Now the product f G is measurable on X and satisfies the inequality If(x)G(x)I < If(x)I f y Ik(x, y)l dy < M' If(x)I, Y where M' = M ly I g(y) I dy. By Theorem 10.35 we see that f - G E L(X). A similar argument shows that g - F E L(Y). Next we prove (28). First we note that (28) is true if each of f and g is a step function. In this case, each off and g vanishes outside a compact interval, so each is Riemann-integrable on that interval and (28) is an immediate consequence of Theorem 7.42. Now we use Theorem 10.19(b) to approximate each off and g by step functions. If e > 0 is given, there are step functions s and t such that Ix Therefore we have If - SI < e f, and 1. Ig - tl<e. f G = f s - G + A1, x where IA11 = L1-s) GI -< J If - SI fy Ig(y)I Ik(x, y)I dy < eM f IgI X x JY Also, we have where I Therefore where G(x) = fy g(y)k(x, y) dy f t(y)k(x, y) dy + A2, Y A21 = I fy - t)k(x, y) dy Y <M f Ig - tl<sM. Y t(y)k(x, y) d y] dx + A3, fX s G = fX s(x) r fy J IA3I = A2 f s(x) dx X < SM fX Isl J <eM f {Is - fI + IfI} <e2M+eM f Ifl, X x (29) Th. 10.42 Measurable Sets on the Real Line so (29) becomes 289 t(y)k(x, y) d y] dx + A l + A3. (30) fxf G = fx s(x) [ fy J Similarly, we find f g F = f t(y) [ 1 s(x)k(x, y) dxl dy + B1 + B3, (31) Y Y L X .f where IB1 I < eM f IfI and IB3I < EM r t i < e2M + eM fy IgI. x JY But the iterated integrals on the right of (30) and (31) are equal, so we have Jf.G - JYg.F --<IA11+IA3I+IBII+IB3I < 2e2M + 2cM f fx IfI + fy 19I . ) Since this holds for every e > 0 we have fx f G = l y g F, as required. NOTE. A more general version of Theorem 10.40 will be proved in Chapter 15 using double integrals. (See Theorem 15.6.) 10.18 MEASURABLE SETS ON THE REAL LINE Definition 10.41. Given any nonempty subset S of R. The function Xs defined by Xs(x) = J 1 if x E S, 0 if x e R - S, is called the characteristic function of S. If S is empty we define Xs(x) = 0 for all x. Theorem 10.42. Let R = (- oo, + oo). Then we have: a) If S has measure 0, then Xs e L(R) and SR Xs = 0-b) If Xs e L(R) and if fR Xs = 0, then S has measure 0. Proof. Part (a) follows by taking f = Xs in Theorem 10.20. To prove (b), let f,. = Xs for all n. Then IfI = Xs so I IfRI= Xs=0. n=1 J R n=1 1 R 290 The Lebesgue Integral Def. 10.43 By the Levi theorem for absolutely convergent series, it follows that the series En 1 f (x) converges everywhere on R except for a set T of measure 0. If x e S, the series cannot converge since each term is 1. If x 0 S, the series converges because each term is 0. Hence T = S, so S has measure 0. Definition 10.43. A subset S of R is called measurable if its characteristic function Xs is measurable. If, in addition, Xs is Lebesgue-integrable on R, then the measure µ(S) of the set S is defined by the equation u(S) = f R Xs-If Xs is measurable but not Lebesgue-integrable on R, we define µ(S) = + oo. The function it so defined is called Lebesgue measure. Examples 1. Theorem 10.42 shows that a set S of measure zero is measurable and that u(S) = 0. 2. Every interval I (bounded or unbounded) is measurable. If I is a bounded interval with endpoints a <- b, then u(I) = b - a. If I is an unbounded interval, then u(I) = + 00 . 3. If A and B are measurable and A S B, then u(A) < p(B). Theorem 10.44. a) If S and T are measurable, so is S - T. b) If S1, S2, , are measurable, so are 1 Si and n 1 S1. Proof. To prove (a) we note that the characteristic function of S - T is Xs - XSXT To prove (b), let n n ao Un = u Si, Vn = f l Si, U = U Si, i=1 i=1 i=1 Then we have Xu = max (Xs,, , and 00 V = f Si. =1 Xv = min (Xs,,... , so each of U. and Y. is measurable. Also, Xu = limn-,,,, Xu and Xv = limn-00 Xv,,, so U and V are measurable. Theorem 10.45. If A and B are disjoint measurable sets, then µ(A u B) = u(A) + µ(B). (32) Proof. Let S = A u B. Since A and B are disjoint we have Xs = XA + XB Suppose that Xs is integrable. Since both XA and XB are measurable and satisfy 0 < XA(x) < XS(x), 0 < XB(x) < XS(x) for all x, Def. 10.48 The Lebesgue Integral over Arbitrary Subsets of R 291 Theorem 10.35 shows that both XA(' and XB are integrable. Therefore P(S) = fit Xs = J XA + J XB = u(A) + p(B) R R In this case (32) holds with both members finite. If Xs is not integrable then at least one of XA or XB is not integrable, in which case (32) holds with both members infinite. The following extension of Theorem 10.45 can be proved by induction. Theorem 10.46. If {A1i ... , A^) is a finite disjoint collection of measurable sets, then u (c1 \ ^ A) = i=1 NOTE. This property is described by saying that Lebesgue measure is finitely additive. In the next theorem we prove that Lebesgue measure is countably additive. Theorem 10.47. If {A1i A2, ... } is a countable disjoint collection of measurable sets, then l 00 00 µ U A) = u(A1) (33) Proof. Let T. = U°=1 Ai, X. = XT.,, T = U i00=1 A i. Since p is finitely additive, we have AT.) = E p(A) for each n. i=1 We are to prove that p(T^) -+ µ(T) as n - oo. Note that p(T^) < p(Tn+1) so {p(T,a} is an increasing sequence. We consider two cases. If µ(T) is finite, then XT and each X. is integrable. Also, the sequence {p(T^)} is bounded above by p(T) so it converges. By the Lebesgue dominated convergence theorem, p(T^) -+ u(T). If p(T) = + oo, then XT is not integrable. Theorem 10.24 implies that either some X. is not integrable or else every X. is integrable but p(T^) - + op. In either case (33) holds with both members infinite. For a further study of measure theory and its relation to integration, the reader can consult the references at the end of this chapter. 10.19 THE LEBESGUE INTEGRAL OVER ARBITRARY SUBSETS OF R Definition 10.48. Let f be defined on a measurable subset S of it Define a new function f on R as follows: Ax) = Jf(x) if x e S, 0 ifxeR - S. 292 The Lebesgue Integral Th. 10.49 If f is Lebesgue-integrable on R, we say that f is Lebesgue-integrable on S and we write f E L(S). The integral off over S is defined by the equation 'SIR This definition immediately gives the following properties: If f E L (S), then f e L (T) for every subset of T of S. If S has finite measure, then µ(S) = Is 1. The following theorem describes a countably additive property of the Lebesgue integral. Its proof is left as an exercise for the reader. Theorem 10.49. Let {A1, A2,... } be a countable disjoint collection of sets in R, and let S = U g ?'= 1 A Let f be defined on S. a) If f E L (S), then f e L (A i) for each i and Jf=EJf Sb) If f E L (A 1) for each i and if the series in (a) converges, then f E L (S) and the equation in (a) holds. 10.20 LEBESGUE INTEGRALS OF COMPLEX-VALUED FUNCTIONS If f is a complex-valued function defined on an interval I, then f = u + iv, where u and v are real. We say f is Lebesgue-integrable on I if both u and v are Lebesgue-integrable on I, and we define Jf=$u + i V. Similarly, f is called measurable on I if both u and v are in M(I). It is easy to verify that sums and products of complex-valued measurable functions are also measurable. Moreover, since IfI = (u2 + v2)1/2, Theorem 10.36 shows that If I is measurable if f is. Many of the theorems concerning Lebesgue integrals of real-valued functions can be extended to complex-valued functions. However, we do not discuss these extensions since, in any particular case, it usually suffices to write f = u + iv and apply the theorems to u and v. The only result that needs to be formulated explicitly is -the following. Th. 10.52 Inner Products and Norms 293 Theorem 10.50. If a complex-valued function f is Lebesgue-integrable on I, then If i e L(I) and we have JfI < fIfl Proof. Write f = u + iv. Since f is measurable and If I < Jul + lvl, Theorem 10.35 shows that If I e L(I). Let Let a = $I f Then a = re°, where r = l al. We wish to prove that r < f if b e ie ifr > 0, 1 ifr=0. Then Ibl = I and r = ba = b fI f = 11 bf. Now write bf = U + iV, where U and V are real. Then 11 bf = fI U, since 11 bf is real. Hence r = f bf = fI U < f IUI << $ Ib'1 J Ifl 10.21 INNER PRODUCTS AND NORMS This section introduces inner products and norms, concepts which play an im-portant role in the theory of Fourier series, to be discussed in Chapter 11. Definition 10.51. Let f and g be two real-valued functions in L(I) whose product f g is in L(I). Then the integral fil f(x)g(x) dx (34) is called the inner product off and g, and is denoted by (f, g). If f2 a L(I), the nonnegative number (f, f)112, denoted by Il.f II, is called the L2-norm off. NoTE. The integral in (34) resembles the sum Ek=1 xkyk which defines the dot product of two vectors x = (x1, ... , and y = (y1, ... , The function values f(x) and g(x) in (34) play the role of the components xk and yk, and integra-tion takes the place of summation. The L2-norm off is analogous to the length of a vector. The first theorem gives a sufficient condition for a function in L(I) to have an L2-norm. Theorem 10.52. If f e L(I) and if f is bounded almost everywhere on I, then f2EL(I). Proof. Since f e L(1), f is measurable and hence f 2 is measurable on I and satisfies the inequality l f(x)12 < M l f(x)I almost everywhere on 1, where M is an upper bound for If 1.- By Theorem 10.35, f 2 e L(I). 294 The Lebesgue Integral Def. 10.53 10.22 THE SET L2(I) OF SQUARE-INTEGRABLE FUNCTIONS Definition 10.53. We denote by L2(I) the set of all real-valued measurable functions f on I such that f2 e L(I). The functions in L2(I) are said to be square-integrable. NOTE. The set L2(I) is neither larger than nor smaller than L(I). For example, the function given by f(x) = x-1'2 for 0 < x < 1, f(0) = 0, is in L([0, 1]) but not in L2([0, 1]). Similarly, the function g(x) = 1/x for x >_ 1 is in L2([1, + oo)) but not in L([1, + oo)). Theorem 10.54. If f e L2(I) and g e L2(I), then f g e L(1) and (af + bg) e L2(I) for every real a and b. Proof. Both f and g are measurable so f g e M(I). Since If(x)9(x)I < f 2(x) + g2(x) 2 Theorem 10.35 shows that f g e L(I). Also, (af + hg) a M(I) and (af + bg) 2 = a2f 2 + 2abf g + b2g2, so (af + bg) a L2(I). Thus, the inner product (f, g) is defined for every pair of functions f and g in L2(I). 'The basic properties of inner products and norms are described in the next theorem. Theorem 10.55. If f, g, and h are in L2 (I) and if c is real we have : a) (f g) = (g, f) (commutativity). b) (f + g, h) = (f, h) + (g, h) (linearity). c) (cf, g) = c(f g) (associativity). d) Ilcf II = Icl IIf II (homogeneity). e) I(.f, g)j <- 11.f II IIgII (Cauchy-Schwarz inequality). f) If + gAI < If II + IIgOI (triangle inequality). Proof. Parts (a) through (d) are immediate consequences of the definition. Part (e) follows at once from the inequality V(x)g(Y) - g(x).f(Y)I2 dy] dx >_ 0. To prove (f) we use (e) along with the relation If + gli2 = (f + g,.f + g) = (.f,.f) + 2(f, g) + (g, g) =IIf 112 + IIg1l2 + 2(f, g). Th. 10.55 Series of Functions in L2(I) 295 NoTE. The notion of inner product can be extended to complex-valued functions f such that If I c- L2(I). In this case, (f, g) is defined by the equation (f, g) = J .f(x)g(x) dx, r where the bar denotes the complex conjugate. The conjugate is introduced so that the inner product of f with itself will be a nonnegative quantity, namely, (f, f) = f r if 12 The L2-norm off is, as before, IIf 11 = (f, f)1"2. Theorem 10.55 is also valid for complex functions, except that part (a) must be modified by writing (f, g) = (g, f). This implies the following companion result to part (b) : (35) (f,g+h)=(g+h, )= G-J) +WT) = (f, g)+(f,h). In parts (c) and (d) the constant c can be complex. From (c) and (35) we obtain (f, cg) = c(f, g). The Cauchy-Schwarz inequality and the triangle inequality are also valid for complex functions. 10.23 THE SET L2(I) AS A SEMIMETRIC SPACE We recall (Definition 3.32) that a metric space is a set Ttogether with a nonnegative function d on T x T satisfying the following properties for all points x, y, z in T : 1. d(x, x) = 0. 2. d(x, y) > 0 if x # y. 3. d(x, y) = d(y, x). 4. d(x, y) < d(x, z) + d(z, y). We try to convert L2(1) into a metric space by defining the distance d(f, g) between any two complex-valued functions in L2(I) by the equation 1/2 d(f,g) = Ilf- gII = If - gI2 r This function satisfies properties 1, 3, and 4, but not 2. If f and g are functions in L2(1) which differ on a nonempty set of measure zero, then f # g but f - g = 0 almost everywhere on I, so d(f, g) = 0. A function d which satisfies 1, 3, and 4, but not 2, is called a semimetric. The set L2(I), together with the semimetric d, is called a semimetric space. 10.24 A CONVERGENCE THEOREM FOR SERIES OF FUNCTIONS IN L2(1) The following convergence theorem is analogous to the Levi theorem for series (Theorem 10.26). 296 The Lebesgue Integral Th. 10.56 Theorem 10.56. Let {gn} be a sequence of functions in L2(I) such that the series 00 E 119.11 n=1 converges. Then the series of functions F_,'= I gn converges almost everywhere on I to a function g in L2(I), and we have CO (36) R- 00 Proof. Let M = E"O 1 11g,II The triangle inequality, extended to finite sums, gives us n :!9 E119k11<_M. This implies f/ (k=1 If x e I, let I9k(x)I) dx = E 19"I k=1 < M2. (37) fn(x) = (I9k(x)I )2. / k=1 The sequence (f.) is increasing, each fn E L(I) (since each gk E L2(I)), and (37) shows that fI fn < M M. Therefore the sequence {11f.} converges. By the Levi theorem for sequences (Theorem 10.24), there is a function f in L(I) such that fn -+ f almost everywhere on I, and f f=lim f n~00 fn <M2. The refore the series Ek 1 gk(x) converges absolutely almost everywhere on I. Let 9(x) = lim 9k(x) M-00 k=1 at those points where the limit exists, and let Gn(x) = I Ej 9k(x)I2 k-1 Then each G. a L(I) and Gn(x) -+ Ig(x)12 almost everywhere on I. Also, G1(x) < f(x) on I. Therefore, by the Lebesgue dominated inated convergence theorem, IgI2 a L(I) and J IgI2 = lim f G. (38) I n- 00 I Th. 10.57 The Riesz-Fischer Theorem Since g is measurable, this shows that g e L2(I). Also, we have so (38) implies 2 -Igk112, 119112 = lim n-oo and this, in turn, implies (36). and 1G r r E 9k k=1 10.25 THE RIESZ-FISCHER THEOREM 2 < M2, 297 The convergence theorem which we have just proved can be used to prove that every Cauchy sequence in the semimetric space L2(I) converges to a function in L2(I). In other words, the semimetric space L2(I) is complete. This result, called the Riesz-Fischer theorem, plays an important role in the theory of Fourier series. Theorem 10.57. Let { fn} be a Cauchy sequence of complex-valued functions in L2(I). That is, assume that for every e > 0 there is an integer N such that 11f. - f.11 < s whenever m >- n >- N. (39) Then there exists a function fin L2(I) such that lim Ilfn - f II = 0. (40) n_00 Proof. By applying (39) repeatedly we can find an increasing sequence of integers n(1) < n(2) < such that IIfn - fn(k)II < I whenever m >- n(k). Let g1 = fn(1), and let gk = fn(k) - fn(k-1) for k >> 2. Then the series T_k 1 11gk1I converges, since it is dominated by 11f.(I)11 + E I1 fn(k)'-fn(k-1)11 < 11f1(l)11 + 12k = II4(1)II + I. k=2 k=1 Each gn is in L2(I). Hence, by Theorem 10.56, the series E gn converges almost everywhere on I to a function f in L2(I). To complete the proof we will show that hfm -f 11 -+0asm oo. For this purpose we use the triangle inequality to write 11fm _f11 <- 11fn, - fn(k)II + 11fn(k) _f11-(41) If m >- n(k), the first term on the right is < 1/2 k . To estimate the second term we note that 0 JJ f - fn(k) _ j lfn(r) - fn(r- I)}, r=k+ I 298 The Lebesgue Integral and that the series E' k+ 1 II fn(r) - .f"(r -1)11 converges. Therefore, we can use inequality (36) of Theorem 10.56 to write °D 1 1 if - fn(k)II < Ilfn(r) - fn(r-1)11 < Ej _ r=k+1 r=k+i 2r-1 2k-1 Hence, (41) becomes 11f. - f11 <- 2k 1 - 1 + 2 1 k = 2k 3 if m > n(k). Since n(k) -> oo as k -+ oo, this shows that Ilfn - f II -+ 0 as m -+ oo. NoTE. In the course of the proof we have shown that every Cauchy sequence of functions in L2(I) has a subsequence which converges pointwise almost everywhere on I to a limit function fin L2(I). However, it does not follow that the sequence f f.} itself converges pointwise almost everywhere to f on I. (A counterexample is described in Section 9.13.) Although {f.} converges to fin the semimetric space L2(I), this convergence is not the same as pointwise convergence. EXERCISES Upper functions 10.1 Prove that max (f, g) + min (f, g) = f + g, and that max (f+h,g+h)=max(f,g)+h, min (f+h,g+h)=min (f,g)+h. 10.2 Let {f"} and (g") be increasing sequences of functions on an interval I. Let u" _ max (fn, g,,) and v" = min (fn, g,). a) Prove that {u"} and {vn} are increasing on I. b) If fn ,c f a.e. on I and if g" w g a.e. on I, prove that u" / max (f, g) and v" / min (f, g) a.e. on I. 10.3 Let {s"} be an increasing sequence of step functions which converges pointwise on an interval I to a limit function f. If I is unbounded and if f(x) >- 1 almost everywhere on I, prove that the sequence {ft s"} diverges. 10.4 This exercise gives an example of an upper function f on the interval I = [0, 1 ] such that -f 0 U(I). Let {r1, r2,. .. } denote the set of rational numbers in [0, 1] and let I" = [r" - 4-", r" + 4-"] n I. Let f(x) = 1 if x e I. for some n, and let f(x) = 0 otherwise. a) Let f"(x) = I if x e I", f"(x) = 0 if x I", and let s" = max (fl, ... , Show that {s"} is an increasing sequence of step functions which generates f. This shows that f e U(I). b) Prove that fl f < 2/3. c) If a step function s satisfies s(x) <- -f(x) on I, show that s(x) <- -1 almost everywhere on I and hence f r s < -1. d) Assume that -f e U(I) and use (b) and (c) to obtain a contradiction. Exercises 299 NOTE. In the following exercises, the integrand is to be assigned the value 0 at points where it is undefined. Convergence theorems 10.5 If fn(x) = e-nx - 2e-'"'l show that 00 OD fx 0 f fn(x) dX. n=1 0 n=1 10.6 Justify the following equations: 1 1 1 00 ao 1 1 a) f log dx = f E dx = - f xndx = 1. 0 1- x o n=1 n n=1 n J0 b) log dz = E 2 (p > 0). fo 1 xP-1 00 1 1 -x x n=0(n+p) ) 10.7 Prove Tannery's convergence theorem for Riemann integrals: Given a sequence of functions { jn } and an increasing sequence {p,, } of real numbers such that pn -- + oo as n - co. Assume that a) fn - f uniformly on [a, b ] for every b >- a. b) fn is Riemann-integrable on [a, b]for every b z a. c) Ijn(x)I < g(x) almost everywhere on [a, + co), where g is nonnegative and im-proper Riemann-integrable on [a, + co). Then both f and If I are improper Riemann-integrable on [a, + co), the sequence {J°' fn} converges, and }0D Pn f f(x) dx = lim fn(x) dx. Ja o d) Use Tannery's theorem to prove that n n aD ) XP dx = f e-xxP dx, if p > -1. lim 0 (1 -n J Jo o 10.8 Prove Fatou's lemma: Given a sequence (fn) of nonnegative functions in L(I) such that (a) { fn } converges almost everywhere on I to a limit function f, and (b) L fn 5 A for some A > 0 and all n >- 1. Then the limit function f e L(I) and JI f 5 A. NOTE. It is not asserted that {JI fn} converges. (Compare with Theorem 10.24.) Hint. Let gn(x) = inf { fn(x), fn+ 1(x), ... }. Then gn W f a.e. on I and h gn -< Jr fn s A so limn, Jr gn exists and is s A. Now apply Theorem 10.24. Improper Riemann Integrals 10.9 a) If p > 1, prove that the integral 11' x-P sin x dx exists both as an improper Riemann integral and as a Lebesgue integral. Hint. Integration by parts. 300 The Lebesgue Integral b) If 0 < p <- 1, prove that the integral in (a) exists as an improper Riemann integral but not as a Lebesgue integral. Hint. Let 7r 37r 5x5 nn+ - forn= 1,2,..., if nn+-g(x) = 2x 4 4 0 otherwise, and show that (' 2 1 f n X-P sin xj dx >_ J nrz g(X) dx >_ -1 n 4 k=2 10.10 a) Use the trigonometric identity sin 2x = 2 sin x cos x, along with the formula JO' sin x/x dx = n/2, to show that f °° sin x cos x n dx = o x 4 b) Use integration by parts in (a) to derive the formula f I sing x n f dx = o x2 2 c) Use the identity sin2 x + cost x = 1, along with (b), to obtain fo, 0 sin4 x n x2dx = 4 d) Use the result of (c) to obtain . sin4 x n fdx = x4 3 10.11 If a > 1, prove that the integral Ja °D x° (log x)4 dx exists, both as an improper Riemann integral and as a Lebesgue integral for all q if p < -1, or for q < -1 if p = -1. 10.12 Prove that each of the following integrals exists, both as an improper Riemann integral and as a Lebesgue integral. a) I sin 2 1 dx, b) f x°e-x9 dx (p > 0, q > 0). 1 x o 10.13 Determine whether or not each of the following integrals exists, either as an improper Riemann integral or as a Lebesgue integral. f0OD e (t=+t-2) dt, a) 1 e) f o x sin dx, o x b) cos x dx fo 'x c) 00 log x dx, d) e_x sin 1 dx, f , x(x2 - 1)1/2 o x f) f0 e'x log (cost x) dx. Exercises 301 10.14 Determine those values of p and q for which the following Lebesgue integrals exist. 1 a) fo x°(1 x2)q dx, b) f xxe-x" dx, Jo c) xP-1 - xq-1 dx, d ('°° sin (xe) dx Jo 1 - x ) x° , 0 og x)l (sin x)-1'3 dx. (l e) J XP-1 dx, f) 100 o 1+x10.15 Prove that the following improper Riemann integrals have the values indicated (m and n denote positive integers). a) ('°° sine"+1 x dx = ic(2n)! b) log x dx = n-2 x 22n+1(n!)2 ' fl' c) J x"(l + dx = n!(m - 1)! I (m + n)! 10.16 Given that f is Riemann-integrable on [0, 1 ], that f is periodic with period 1, and that fo f(x) dx = 0. Prove that the improper Riemann integral fi 0° x-' f(x) dx exists ifs > 0. Hint. Let g(x) = fi f(t) dt and write f x_s f(x) dx = f, x-s dg(x). 10.17 Assume that f e R on [a, b] for every b > a > 0. Define g by the equation xg(x) = f f f (t) dt if x > 0, assume that the limit limx.., + . g(x) exists, and denote this limit by B. If a and b are fixed positive numbers, prove that a) fbf( x) dx = g(b) - g(a) + bg(x) dx. x a x b) lim T- + co f_)dx = BIog!. T c) J1 adf(ax) X f(bx) dx = B log b + bf(t) dt. t d) Assume that the limit limx-.o+ x J' If(j)t-2 dt exists, denote this limit by A, and prove that f'f(ax) - f(bx) b fbf(t) J dx = A log - J dt. o x a a t e) Combine (c) and (d) to deduce f °D f (ax) - f (bx) dx = (B - A) log a o x b and use this result to evaluate the following integrals: °D cos ax - cos bx dx f 00 e-ax - ebx dx. o x o X The Lebesgue Integral Lebesgue integrals 10.18 Prove that each of the following exists as a Lebesgue integral. a) fi xlogx dx, Jo (1 + x)2 i c) f log x log (1 + x) dx, 0 10.19 Assume that f is continuous on Lebesgue integral fl o f(x)x-31'2 dx exists. lx° - 1 b) dx (p > -1), Jo log x d) 1 log (1 - x) dx. fo (1 - x)112 [0, 1], f (O) = 0, f'(0) exists. Prove that the 10.20 Prove that the integrals in (a) and (c) exist as Lebesgue integrals but that those in (b) and (d) do not. a) x2e-xesin2x dx 0 J dx dx C) J , d) 1 1 + x° sin2 x b) x3e-sesinZS dx, 0 J 1 + x2 sin2 x f Hint. Obtain upper and lower bounds for the integrals over suitably chosen neighbor-hoods of the points nn (n = 1, 2, 3, ... ). Functions defined by integrals 10.21 Determine the set S of those real values of y for which each of the following integrals exists as a Lebesgue integral. a) fooo cos xy dx 1 + x2 x2 + y2)-1 dx, b) foo ( c) fo °° sin 2 e'"= cos 2xy dx. 2 Y dx, d) fo'o x 10.22 Let F(y) = fo e'"2 cos 2xy dx if y e R. Show that F satisfies the differential equation F(y) + 2y F(y) = 0 and deduce that F(y) = j/ne_ 2. (Use the result Ja e-X2 dx = 4'n, derived in Exercise 7.19.) 10.23 Let F(y) = Jo sin xy/x(x2 + 1) dx if y > 0. Show that F satisfies the differential equation F(y) - F(y) + nl2 = 0 and deduce that F(y) = 4n(1 - e-Y). Use this result to deduce the following equations, valid for y > 0 and a > 0: °° °° cos XY dx = ne-°Y' F sin xy dx = n (I - e °y), 2 2 + a Jo x(x2 + a2) 2a2 o x ' ( x it x sin xy dx = n e °Y you may use dx J = o sin x 2 x2 + a2 2 jo-10.24 Show that f ° [J° f(x, y) dx] dy :A I' [f' f(x, y) dy] dx if a) f(x, _ Y) -x - Y b) f(x, y) = x2 _ yz (x+Y)3 (x2+y2)2 Exercises 303 10.25 Show that the order of integration cannot be interchanged in the following integrals : a) fo [f0 (x + )3 dxl dy, b) fo [f: (e xv -2e 2xr) dy] dx. 10.26 Letf(x, y) = o dt/[(1 + x2t2)(1 + y2t2)] if (x, y) (0, 0). Show (by methods of elementary calculus) that f(x, y) = 4n(x + y)-1. Evaluate the iterated integral fo [fo f(x, y) dx] dy to derive the formula: ('°° (arctan x)2 dx = n log 2. J x2 0 10.27 Let f (y) = fo sin x cos xylx dx if y >- 0. Show (by methods of elementary calculus) that f(y) = it/2 if 0 < y < 1 and that f(y) = 0 if y > 1. Evaluate the integral f o f (y) dy to derive the formula na 0D sin ax sin x 2 o x2 7E 2 if 0<a<-1, if a>_ 1. 10.28 a) Ifs > 0 and a > 0, show that the series °O 1 r sin X n 2nrzx s "=l fao' converges and prove that lim 00 1 sin 2n7rx dx = 0 a - + + n f, XS b) Let f (x) sin (2mcx)/n. Show that fooo dx = (2x)s-1C(2 - s) fOD sin t dt, if 0 < s < 1, x+ t where C denotes the Riemann zeta function. 10.29 a) Derive the following formula for the nth derivative of the Gamma function: 00 V")(x) = f etex-1 (log t)" dt (x > 0). b) When x = 1, show that this can be written as follows: f 1 (t2 + (-1)"e-1/t)e-tt-2 (log t)" dt. 0 c) Use (b) to show that f°'(1) has the same sign as (- I)". In Exercises 10.30 and 10.31, F denotes the Gamma function. 10.30 Use the result Jo e-X2 dx = 4 to prove that F(4) = 1n. Prove that I'(n + 1) _ n! and that I'(n + 1) = (2n)! /4"n! if n = 0, 1, 2, .. . 304 The Lebesgue Integral 10.31 a) Show that for x > 0 we have the series representation (00 11'(X) = E -1)" 1 + E n=o n! n + x "=o where c" = (1/n!) f7 t-1e-(log t)" dt. Hint: Write f= fo + f' and use o an appropriate power series expansion in each integral. b) Show that the power series F,,'= 0 C"Z" converges for every complex z and that the series [(-1)"/n! ]/(n + z) converges for every complex z j4 0, -1, -2, .. 10.32 Assume that f is of bounded variation on [0, b] for every b > 0, and that lim, ,,, f(x) exists. Denote this limit by f(oo) and prove that lim Y J CO e xyf(x) dx = ftoo) Y-O+ o Hint. Use integration by parts. 10.33 Assume that f is of bounded variation on [0, 1 ]. Prove that lim y f x"-f(x) dx = f(0+). o Measurable functions 10.34 If f is Lebesgue-integrable on an open interval I and if f'(x) exists almost every-where on I, prove that f' is measurable on I. 10.35 a) Let {s"} be a sequence of step functions such that s" f everywhere on R. Prove that, for every real a, 00 00 f-1((a, +oo)) = U sk' \\a + 1 , +oo n=1 k=I In n b) If f is measurable on R, prove that for every open subset A of R the set f (A) is measurable. 10.36 This exercise describes an example of a nonmeasurable set in R. If x and y are real numbers in the interval [0, 11, we say that x and y are equivalent, written x - y, whenever x - y is rational. The relation - is an equivalence relation, and the interval [0, 11 can be expressed as a disjoint union of subsets (called equivalence classes) in each of which no two distinct points are equivalent. Choose a point from each equivalence class and let E be the set of points so chosen. We assume that E is measurable and obtain a contra-diction. Let A = {r1, r2, ...) denote the set of rational numbers in [-1, 11 and let E"= {r"+x:xeE}. a) Prove that each E" is measurable and that ,u(E") = p(E). b) Prove that {E1, E2, ... } is a disjoint collection of sets whose union contains [0, 1 ] and is contained in [-1, 2]. c) Use parts (a) and (b) along with the countable additivity of Lebesgue measure to obtain a contradiction. 10.37 Refer to Exercise 10.36 and prove that the characteristic function XE is not measur-able. Let f = XE - XI_E where I = [0, 1 ]. Prove that If I e L(I) but that f ¢ M(I). (Compare with Corollary 1 of Theorem 10.35.) References 305 Square-integrable functions In Exercises 10.38 through 10.42 all functions are assumed to be in L2(I). The L2-norm 11f II is defined by the formula, 11f II = (fi I fj2)112. 10.38 If 11f. - f II = 0, prove that 10.39 If limp.,, 11f. - f II = 0 and if lim,,_ f (x) = g(x) almost everywhere on I, prove that f(x) = g(x) almost everywhere on I. 10.40 If f - f uniformly on a compact interval I, and if each f is continuous on I, prove that Jim.,. 11f. - f II = 0-10.41 If 11f. - f II = 0, prove that fi f - g = fi f - g for every g in L2(I). 10.42 If f II = 0 and II g - g 1l = 0, prove that fi fn . gn = f, SUGGESTED REFERENCES FOR FURTHER STUDY 10.1 Asplund, E., and Bungart, L., A First Course in Integration. Holt, Rinehart and Winston, New York, 1966. 10.2 Bartle, R., The Elements of Integration. Wiley, New York, 1966. 10.3 Burkill, J. C., The Lebesgue Integral. Cambridge University Press, 1951. 10.4 Halmos, P., Measure Theory. Van Nostrand, New York, 1950. 10.5 Hawkins, T., Lebesgue's Theory of Integration: Its Origin and Development. University of Wisconsin Press, Madison, 1970. 10.6 Hildebrandt, T. H., Introduction to the Theory of Integration. Academic Press, New York, 1963. 10.7 Kestelman, H., Modern Theories of Integration. Oxford University Press, 1937. 10.8 Korevaar, J., Mathematical Methods, Vol. 1. Academic Press, New York, 1968. 10.9 Munroe, M. E., Measure and Integration, 2nd ed. Addison-Wesley, Reading, 1971. 10.10 Riesz, F., and Sz.-Nagy, B., Functional Analysis. L. Boron, translator. Ungar, New York, 1955. 10.11 Rudin, W., Principles of Mathematical Analysis, 2nd ed. McGraw-Hill, New York, 1964. 10.12 Shilov, G. E., and Gurevich, B. L., Integral, Measure and Derivative: A Unified Approach. Prentice-Hall, Englewood Cliffs, 1966. 10.13 Taylor, A. E., General Theory of Functions and Integration. Blaisdell, New York, 1965. 10.14 Zaanen, A. C., Integration. North-Holland, Amsterdam, 1967. CHAPTER 11 FOURIER SERIES AND FOURIER INTEGRALS 11.1 INTRODUCTION In 1807, Fourier astounded some of his contemporaries by asserting that an "arbitrary" function could be expressed as a linear combination of sines and co-sines. These linear combinations, now called Fourier series, have become an indispensable tool in the analysis of certain periodic phenomena (such as vibra-tions, and planetary and wave motion) which are studied in physics and engineering. Many important mathematical questions have also arisen in the study of Fourier series, and it is a remarkable historical fact that much of the development of modern mathematical analysis has been profoundly influenced by the search for answers to these questions. For a brief but excellent account of the history of this subject and its impact on the development of mathematics see Reference 11.1. 11.2 ORTHOGONAL SYSTEMS OF FUNCTIONS The basic problems in the theory of Fourier series are best described in the setting of a more general discipline known as the theory of orthogonal functions. There-fore we begin by introducing some terminology concerning orthogonal functions. NOTE. As in the previous chapter, we shall consider functions defined on a general subinterval I of R. The interval may be bounded, unbounded, open, closed, or half-open. We denote by L2(I) the set of all complex-valued functions f which are measurable on I and are such that If 12 e L(I). The inner product (f, g) of two such functions, defined by (f g) = J f(x)g(x) dx, r always exists. The nonnegative number 11f II = (f f)112 is the L2-norm off. Definition 11.1. Let S = {To, fpl, (p2, ... } be a collection of functions in L2(I). If (T.,. T.) = 0 whenever m n, the collection S is said to be an orthogonal system on I. If, in addition, each T. has norm 1, then S is said to be orthonormal on I. NOTE. Every orthogonal system for which each 11T.11 96 0 can be converted into an orthonormal system by dividing each cp,, by its norm. 306 Best Approximation 307 We shall be particularly interested in the special trigonometric system S = 1901 (PI, 92.... b where 1 W _ (P nx AX) =cos sin nx 1 , O / VLn , 92.-/ Vn (P2n(x) = V7C ( ) for n = 1, 2, ... It is a simple matter to verify that S is orthonormal on any interval of length 21r. (See Exercise 11.1.) The system in (1) consists of real-valued functions. An orthonormal system of complex-valued functions on every interval of length 2ir is given by e'"" cos nx + i sin nx cPn(x) _ _ n = 0, 1, 2, .. . -,/27r 11.3 THE THEOREM ON BEST APPROXIMATION One of the basic problems in the theory of orthogonal functions is to approximate a given function f in L2(I) as closely as possible by linear combinations of elements of an orthonormal system. More precisely, let S = IT O, T1, 92, ... } be ortho-normal on I and let n tn(x) = E bkTk(x), k=0 where bo, b1, ... , b" are arbitrary complex numbers. We use the norm I1f - tnll as a measure of the error made in approximating f by tn. The first task is to choose the constants bo, ... , bn so that this error will be as small as possible. The next theorem shows that there is a unique choice of the constants that minimizes this error. To motivate the results in the theorem we consider the most favorable case. If f is already a linear combination of coo, (p 1, ... , (p,, say f Ck(Pk, k=0 then the choice to = f will make If - tnll = 0. We can determine the constants c 0 , . . . , c" as follows. Form the inner product (f corn), where 0 < m 5 n. Using the properties of inner products we have n n (f (Pm) = C Ckcok, cm) = E Ck((Pk, (Pm) = Cm, since (cok, (pm) = 0 if k 0 m and (corn, cpm) = I. In other words, in the most favorable case we have cm = (f, corn) for m = 0, 1, ... , n. The next theorem shows that this choice of constants is best for all functions in L2(I). 308 Fourier Series and Fourier Integrals Th. 11.2 Theorem 11.2. Let {90, ipl, 92, ... } be orthonormal on I, and assume that f e L2(I). Define two sequences of functions {sn} and {tn} on I as follows: Sn(x) = 1 Ck(Pk(x), tn(x) = E bkcok(x), k=0 k=0 where Ck = (f, TO fork = 0, 1,2,..., and bo, bl, b2, ... , are arbitrary complex numbers. Then for each n we have IIf - SnII :5- If - tnll. Moreover, equality holds in (3) if, and only if, bk = ck for k = 0, 1, ... , n. Proof. We shall deduce (3) from the equation n n (2) (3) If - tnII2 = 11f112 - E ICk12 + 1 Ibk - CkI2. (4) k=0 k=0 It is clear that (4) implies (3) because the right member of (4) has its smallest value when bk = ck for each k. To prove (4), write IIf-toII2= f)-tn)-(t,f)+(tn,tn) Using the properties of inner products we find n n (tn, tn) = ( bk(pk, bm(Pm) k=0 m=0 and n n n (f, t,,) _ (f, E k(fi k) = E bkCk k=0 k=0 k=0 Also, (tn, f) = (f, tn) = Ek=o bkek, and hence nn nn nn If - tn!I2 = IIfII2 - 5kCk - E bkek +E IbkI2 k=0 k=0 k=0 n n 11f112 -ICk12 + L. (bk - Ck)(bk - Ck) k=0 k=0 nn nn r nn = E E bkbm(cok, (p.) = E IbkI2, k=0 m=0 k=0 n = IIf1I2-EICkI2+Ibk-CkI2. k=0 k=0 Th. 11.4 Properties of Fourier Coefficients 11.4 THE FOURIER SERIES OF A FUNCTION RELATIVE TO AN ORTHONORMAL SYSTEM 309 Definition 11.3. Let S = IT O, (pl, T2, ... } be orthonormal on I and assume that f e LZ(I). The notation 00 f(x) - E Cnq'n(x) n=0 will mean that the numbers co, c1, c2, ... are given by the formulas: (5) cn = (f, Tn) = J f(x)-T;(x) dx (n = 0, 1, 2,.. .). (6) I The series in (5) is called the Fourier series off relative to S, and the numbers CO, c1, c2, ... are called the Fourier coefficients off relative to S. NOTE. When I = [0, 2n] and S is the system of trigonometric functions described in (1), the series is called simply the Fourier series generated by f. We then write (5) in the form 00 f (x) - 2 + E (an cos nx + b,, sin nx), the coefficients being given by the following formulas: 1 2R 1 2a an = -f(t) cos nt dt, bn = - f f(t) sin nt dt. 7c 0 n J0 In this case the integrals for a and bn exist if f e L([0, 2n]). (7) 11.5 PROPERTIES OF THE FOURIER COEFFICIENTS Theorem 11.4. Let {q, TO, 91, 92, ... } be orthonormal on I, assume that f e L2(I), and suppose that 00 f(x) -cnQ (x) n=0 Then a) The series Y Icn12 converges and satisfies the inequality 00 I cn12 < 11f112 (Bessel's inequality). n=0 b) The equation co (8) cn12 = 11f112 (Parseval's formula) n=0 310 Fourier Series and Fourier Integrals holds if, and only if, we also have urn 11f - S.11 = 0, n- OD where {s"} is the sequence of partial sums defined by s .(x) _ Ckcok(x) k=0 Th. 11.4 Proof. We take bk = ck in (4) and observe that the left member is nonnegative. Therefore ICk12 << 11fI12 k=0 This establishes (a). To prove (b), we again put bk = ck in (4) to obtain n IIf - Sn112 = 11f112 - 1 ICkl2 k=0 Part (b) follows at once from this equation. As a further consequence of part (a) of Theorem 11.4 we observe that the Fourier coefficients c" tend to 0 as n - oo (since Y_ Ic"I2 converges). In particular, when Tn(x) = e'"/N[2-7r and I = [0, 2n] we find 2M lim f(x)e-'"" dx = 0, n-+ao o from which we obtain the important formulas lim fo ZR f (x) cos nx dx = lim f 2,, f (x) sin nx dx = 0. - °0 n- 00 ,J 0 (9) These formulas are also special cases of the Riemann-Lebesgue lemma (Theorem 11.6). NOTE. The Parseval formula 11f112=Ico12+IciI2+Ic212+ is analogous to the formula IIXII2=x;+x2+. +x2 for the length of a vector x = (x1, ... , x") in R". Each of these can be regarded as a generalization of the Pythagorean theorem for right triangles. Th. 11.5 The Riesz-Fischer Theorem 11.6 THE RIESZ-FISCHER THEOREM 311 The converse to part (a) of Theorem 11.4 is called the Riesz-Fischer theorem. Theorem 11.5. Assume {9o, Cpl, ... } is orthonormal on I. Let {cn} be any sequence of complex numbers such that Y_ Ick12 converges. Then there is a function f in L2(I) such that a) (f, Pk) = ck for each k >_ 0, and 00 b) 11fI12 = E ICk12. k=0 Proof Let s .(x) _ Ck cok(x) k=0 We will prove that there is a function f in L2(I) such that (f, 9k) = ck and such that lim IISn - .f II = 0. n- 00 Part (b) of Theorem 11.4 then implies part (b) of Theorem 11.5. First we note that {sn} is a Cauchy sequence in the semimetric space L2(I) because, if m > n we have m m IISn - S.112 = E E Ckcr(cok, (p,) k=n+1 r=n+1 M E ICk12, k=n+1 and the last sum can be made less than a if m and n are sufficiently large. By Theorem 10.57 there is a function f in L2(I) such that lim IISn - f II = 0. n-oD To show that (f, 9k) = Ck we note that (sn, p k) = ck if n Z k, and use the Cauchy-Schwarz inequality to obtain ICk - (,'P01 = I(Sn, (Pk) - (f,901 = I(S, - f, wk)I S IISn - fII. Since 11 s,, - f II 0 as n oo this proves (a). NOTE. The proof of this theorem depends on the fact that the semimetric space L2(I) is complete. There is no corresponding theorem for functions whose squares are Riemann-integrable. 312 Fourier Series and Fourier Integrals 11.7 THE CONVERGENCE AND REPRESENTATION PROBLEMS FOR TRIGONOMETRIC SERIES Consider the trigonometric Fourier series generated by a function f which is Lebesgue-integrable on the interval I = [0, 27r], say f(x) 2 + E (a cos nx + b sin nx). 00 Two questions arise. Does the series converge at some point x in I? If it does converge at x, is its sumf(x)? The first question is called the convergence problem; the second, the representation problem. In general, the answer to both questions is "No." In fact, there exist Lebesgue-integrable functions whose Fourier series diverge everywhere, and there exist continuous functions whose Fourier series diverge on an uncountable set. Ever since Fourier's time, an enormous literature has been published on these problems. The object of much of the research has been to find sufficient conditions to be satisfied by f in order that its Fourier series may converge, either throughout the interval or at particular points. We shall prove later that the convergence or divergence of the series at a particular point depends only on the behavior of the function in arbitrarily small neighborhoods of the point. (See Theorem 11.11, Riemann's localization theorem.) The efforts of Fourier and Dirichlet in the early nineteenth century, followed by the contributions of Riemann, Lipschitz, Heine, Cantor, Du Bois-Reymond, Dini, Jordan, and de la Vallbe-Poussin in the latter part of the century, led to the discovery of sufficient conditions of a wide scope for establishing convergence of the series, either at particular points, or generally, throughout the interval. After the discovery by Lebesgue, in 1902, of his general theory ofmeasure and integration, the field of investigation was considerably widened and the names chiefly associated with the subject since then are those of Fej6r, Hobson, W. H. Young, Hardy, and Littlewood. Fej6r showed, in 1903, that divergent Fourier series may be utilized by considering, instead of the sequence of partial sums the sequence of arithmetic means where 6n(x) = so(x) + s1(x) + ... + sn-1(x) n He established the remarkable theorem that the sequence {Q (x)} is convergent and its limit is j<[f(x+) + f(x -)] at every point in [0, 2n] where f(x +) and f(x-) exist, the only restriction on f being that it be Lebesgue-integrable on [0, 2n] (Theorem 11.15.). Fej6r also proved that every Fourier series, whether it converges or not, can be integrated term-by-term (Theorem 11.16.) The most striking result on Fourier series proved in recent times is that of Lennart Carleson, a Swedish mathematician, who proved that the Fourier series of a function in LZ(I) converges almost everywhere on I. (Acta Mathematica, 116 (1966), pp. 135-157.) Th. 11.6 The Riemann-Lebesgue Lemma 313 In this chapter we shall deduce some of the sufficient conditions for convergence of a Fourier series at a particular point. Then we shall prove Fejdr's theorems. The discussion rests on two fundamental limit formulas which will be discussed first. These limit formulas, which are also used in the theory of Fourier integrals, deal with integrals depending on a real parameter a, and we are interested in the behavior of these integrals as a - + oo. The first of these is a generalization of (9) and is known as the Riemann-Lebesgue lemma. 11.8 THE RIEMANN-LEBESGUE LEMMA Theorem 11.6. Assume that f e L(I). Then, for each real fi, we have lim f f(t) sin (at + fl) dt = 0. a-++OD , (10) Proof. If f is the characteristic function of a compact interval [a, b] the result is obvious since we have If sin (at + fl) dtl = cos (aa + f) - cos (ba + f) a ? < , ifa>0. a The result also holds if f is constant on the open interval (a, b) and zero outside [a, b], regardless of how we define f(a) and f(b). Therefore (10) is valid if f is a step function. But now it is easy to prove (10) for every Lebesgue-integrable function f If e > 0 is given, there exists a step function s such that S, If - sI < c/2 (by Theorem 10.19(b)). Since (10) holds for step functions, there is a positive M such that s(t) sin (at + /3) dtI < 2 if a >- M. Therefore, if a >- M we have f(t) sin (at + fi) dtl I f (f(t) - s(t)) sin (at + fi) dt J, + I r s(t) sin (at + fi) dtI I <JlIf(t)-s(t)l dt+2<2+2 This completes the proof of the Riemann-Lebesgue lemma. Example. Taking f = 0 and $ = n/2, we find, if f s L(I), 8. lim f(t) sin at dt = lim f f(t) cos at dt = 0. a-.+00 I a +00 r 314 Fourier Series and Fourier Integrals Th. 11.7 As an application of the Riemann-Lebesgue lemma we derive a result that will be needed in our discussion of Fourier integrals. Theorem 11.7. If f e L(- oo, + oo), we have lim f f(t) 1 - cos at dt = f(- t) dt, a +Go t Jo t (11) whenever the Lebesgue integral on the right exists. Proof. For each fixed a, the integral on the left of (11) exists as a Lebesgue integral since the quotient (1 - cos at)lt is continuous and bounded on (- oo, + oo). (At t = 0 the quotient is to be replaced by 0, its limit as t -+ 0.) Hence we can write ('°' f(t) 1 - cos at dt fo, f(t) 1 - cos at dt + f o f(t) cos at dt t t 00 t When a = f - [f(t) - f(-t)] 1 - Cos at dt o t = ff(t) - f(- t) dt -ft) f ( -. o t f o °° t + oo, the last integral tends to 0, by the Riemann-Lebesgue lemma. 11.9 THE DIRICHLET INTEGRALS Integrals of the form f u g(t)(sin at )l t dt (called Dirichlet integrals) play an im-portant role in the theory of Fourier series and also in the theory of Fourier integrals. The function g in the integrand is assumed to have a finite right-hand limit g(0+) = lim,...o+ g(t) and we are interested in formulating further con-ditions on g which will guarantee the validity of the following equation : a lim ? g(t) sin at dt = g(0+). (12) a-+ CO 76 0 t To get an idea why we might expect a formula like (12) to hold, let us first consider the case when g is constant (g(t) = g(0+)) on [0, 6]. Then (12) is a trivial con-sequence of the equation fo (sin t)lt dt = 7r/2 (see Example 3, Section 10.16), since a sin at as sin t 7c Jo t dt = f o t dt -+ 2 as a -+ + oo. More generally, if g e L([0, 6]), and if 0 < s < 6, we have lim ?f g(t) !t dt = 0, a-++oo 7C c t Th. 11.8 The Dirichlet Integrals 315 by the Riemann-Lebesgue lemma. Hence the validity of (12) is governed entirely by the local behavior of g near 0. Since g(t) is nearly g(0+) when t is near 0, there is some hope of proving (12) without placing too many additional restrictions on g. It would seem that continuity of g at 0 should certainly be enough to insure the existence of the limit in (12). Dirichlet showed that continuity of g on [0, 6] is sufficient to prove (12), if, in addition, g has only a finite number of maxima or minima on [0, S]. Jordan later proved (12) under the less restrictive condition that g be of bounded variation on [0, S]. However, all attempts to prove (12) under the sole hypothesis that g is continuous on [0, 6] have resulted in failure. In fact, Du Bois-Reymond discovered an example of a continuous function g for which the limit in (12) fails to exist. Jordan's result, and a related theorem due to Dini, will be discussed here. Theorem 11.8 (Jordan). If g is of bounded variation on [0, S], then lim 2 fo g(t) sin at dt = g(0+). a+ o0 7C t (13) Proof. It suffices to consider the case in which g is increasing on [0, S]. If a > 0 and if 0 < h < 6, we have f ('a g(t) sin at dt = f h [g(t) - g(0+)] sin at dt Jo t Jo t + g(0+) fo sin at dt + f h g(t) sin at dt = I1(a, h) + I2(a, h) + 13(aa, h), (14) let us say. We can apply the Riemann-Lebesgue lemma to I3(a, h) (since the integral fa g(t)lt dt exists) and we find 13(a, h) - 0 as a -+ + oo. Also, I2(a, h) = g(0+) f sin at dt o t J o ha sin t It = g(0+) t dt -' 2 g(0+) as at -+ + oo. 0 Next, choose M > 0 so that I fa (sin t)lt dt I < M for every b >- a >- 0. It follows that IS a (sin at )l t dt I < M for every b > a Z 0 if a > 0. Now let E > 0 be given and choose h in (0, 6) so that f g(h) - g(0+)I < e/(3M). Since g(t)-g(0+)z0 if 0 5t --.9 h, we can apply Bonnet's theorem (Theorem 7.37) in Il(a, h) to write I1(a, h) = fo a [g(t) - g(0+)] sin at dt -h sin at t f t 316 Fourier Series and Fourier Integrals Th. 11.9 where c e [0, h]. The definition of h gives us II i(a, h)I = I g(h) - g(0+)I I j'S1flXtdt t < E M= E. (15) 3M 3 For the same h we can choose A so that a >- A implies II3(a, h)I < 3 and I2(a, h) - 2 g(0+) < 3 . (16) Then, for a -> A, we can combine (14), (15), and (16) to get jg(t)mctdt - 2 g(0+) < E. This proves (13). A different kind of condition for the validity of (13) was found by Dini and can be described as follows: Theorem 11.9 (Dini). Assume that g(0+) exists and suppose that for some 6 > 0 the Lebesgue integral ra g(t) - g(0+) dt o t exists. Then we have ali m 2 a g(t) sin Oct dt = g(0+). 0 t Proof. Write a sin at a g(t) - g(0+) "a sin t g(t) dt = f sin at dt + g(0+) f dt. o t .J o t f o t When a -> + oo, the first term on the right tends to 0 (by the Riemann-Lebesgue lemma) and the second term tends to 1ng(0+). NOTE. If g e L([a, 6]) for every positive a < S, it is easy to show that Dini's condition is satisfied whenever g satisfies a "right-handed" Lipschitz condition at 0; that is, whenever there exist two positive constants M and p such that Ig(t) - g(0+)I < Mt", for every tin (0, 6]. (See Exercise 11.21.) In particular, the Lipschitz condition holds with p = 1 whenever g has a righthand derivative at 0. It is of interest to note that there exist functions which satisfy Dini's condition but which do not satisfy Jordan's con-dition. Similarly, there are functions which satisfy Jordan's condition but not Dini's. (See Reference 11.10.) Th. 11.10 Partial Sums of Fourier Series 317 11.10 AN INTEGRAL REPRESENTATION FOR THE PARTIAL SUMS OF A FOURIER SERIES A function f is said to be periodic with period p 0 if f is defined on R and if f (x + p) = f (x) for all x. The next theorem expresses the partial sums of a Fourier series in terms of the function sin (n ± -)t if t 2mn (m an integer), Cos kt = 2 sin t/2 (17) k=1 n + I if t = 2mir (m an integer). This formula was discussed in Section 8.16 in connection with the partial sums of the geometric series. The function D. is called Dirichlet's kernel. Theorem 11.10. Assume that f e L([0, 2nc]) and suppose that f is periodic with period 2Rc. Let denote the sequence of partial sums of the Fourier series generated byf,say 2 + E (ak cos kx + bk sin kx), (n = 1, 2, ...). (18) Then we have the integral representation 2 f'f(x + t) 2f(x -) dt. (19) n o Proof. The Fourier coefficients off are given by the integrals in (7). Substituting these integrals in (18) we find 2a n 1 f(t) {1 + E (cos kt cos kx + sin kt sin kx)} dt 7[ p 2 k=1 ('2a 2a fo f(t) + cos k(t - x) dt = f f x) dt. k=1 7T 0 Since both f and D. are periodic with period 2n, we can replace the interval of integration by [x - 7C, x + iv] and then make a translation u = t - x to get 1 x+n Sn(x) = -f(t)D,,(t - x) dt 7r x-a 1 f = -f (x + u)D (u) du. 7 R Using the equation D.(-u) = D (u), we obtain (19). 318 Fourier Series and Fourier Integrals 11.11 RIEMANN'S LOCALIZATION THEOREM Th. 11.11 Formula (19) tells us that the Fourier series generated by f will converge at a point x if, and only if, the following limit exists : lim 2 f 'f(x + t) + f(x - t) sin (n +4)t dt, (20) n-+ao 7c o 2 2 sin it in which case the value of this limit will be the sum of the series. This integral is essentially a Dirichlet integral of the type discussed in the previous section, except that 2 sin it appears in the denominator rather than t. However, the Riemann-Lebesgue lemma allows us to replace 2 sin it by t in (20) without affecting either the existence or the value of the limit. More precisely, the Riemann-Lebesgue lemma implies 2 f (1 -1 l f(x + t) + f (x - t) lim sin (n + J)t dt = 0 n-+oo 7C o t 2 sin 3t) 2 because the function F defined by the equation 1 1 F(t) = t 2 sin it if 0<t<7[, 0 ift=0, is continuous on [0, 7c]. Therefore the convergence problem for Fourier series amounts to finding conditions on f which will guarantee the existence of the following limit : lim 2 f nf(x + t) + f(x - t) sin (n + 4)t dt. (21) n- oo 7C o 2 t Using the Riemann-Lebesgue lemma once more, we need only consider the limit in (21) when the integral f o is replaced by f lo, where S is any positive number <76, because the integral f ,x tends to 0 as n -> oo. Therefore we can sum up the results of the previous section in the following theorem : Theorem 11.11. Assume that f e L([0, 27c]) and suppose f has period 27c. Then the Fourier series generated by f will converge for a given value of x if, and only if, for some positive S < 7C the following limit exists: lim 2 faf(x+t)+f(x-t)sin(n+.)tdt, (22) n 76 o 2 t in which case the value of this limit is the sum of the Fourier series. This theorem is known as Riemann's localization theorem. It tells us that the convergence or divergence of a Fourier series at a particular point is governed entirely by the behavior off in an arbitrarily small neighborhood of the point. This is rather surprising in view of the fact that the coefficients of the Fourier Th. 11.14 Cesaro Smnmability 319 series depend on the values which the function assumes throughout the entire interval [0, 2n]. 11.12 SUFFICIENT CONDITIONS FOR CONVERGENCE OF A FOURIER SERIES AT A PARTICULAR POINT Assume that f e L([0, 2ir]) and suppose that f has period 2Rc. Consider a fixed x in [0, 2ir] and a positive S < n. Let g(t)_f(x+t)+f(x-t) and let 2 ifte[0,s], s(x)=g(0+)= lim f(x + t) + f(x - t) t-o+ 2 whenever this limit exists. Note that s(x) = f(x) if f is continuous at x. By combining Theorem 11.11 with Theorems 11.8 and 11.9, respectively, we obtain the following sufficient conditions for convergence of a Fourier series. Theorem 11.12 (Jordan's test). If f is of bounded variation on the compact interval [x - S, x + S] for some S < iv, then the limit s(x) exists and the Fourier series generated by f converges to s(x). Theorem 11.13 (Din's test). If the limit s(x) exists and if the Lebesgue integral fo g(t) - s(x) dt t exists for some S < iv, then the Fourier series generated by f converges to s(x). 11.13 CESARO SUMMABILITY OF FOURIER SERIES Continuity of a function f is not a very fruitful hypothesis when it comes to studying convergence of the Fourier series generated by f. In 1873, Du Bois-Reymond gave an example of a function, continuous throughout the interval [0, 2n], whose Fourier series fails to converge on an uncountable subset of [0, 2n]. On the other hand, continuity does suffice to establish Cesaro summability of the series. This result (due to Fejr r) and some of its consequences will be discussed next. Our first task is to obtain an integral representation for the arithmetic means of the partial sums of a Fourier series. Theorem 11.14. Assume that f e L([0, 2n]) and suppose that f is periodic with period 2n. Let s denote the nth partial sum of the Fourier series generated by f and let an(x) = So(x) + s1(x) + ... + (n = 1, 2, ...). (23) n 320 Fourier Series and Fourier Integrals Th. 11.14 Then we have the integral representation o,,(x) = 1 fo R f(x + t) + f (x - t) sin' int dt. nit 2 sin' it (24) Proof. If we use the integral representation for given in (19) and form the sum defining we immediately obtain the required result because of formula (16), Section 8.16. NOTE. If we apply Theorem 11.14 to the constant function whose value is 1 at each point we find a.(x) = s,,(x) = I for each n and hence (24) becomes ['sin 2 1 sin nt dt = 1. (25) n7G o sine It Therefore, given any number s, we can combine (25) with (24) to write s = 1 f R ff(x + t) + f (x - t) - sl sin' in t d t. (26) na Jo 2 j sine it If we can choose a value of s such that the integral on the right of (26) tends to 0 as n - oo, it will follow that oR(x) - s as n -+ oo. The next theorem shows that it suffices to take s = [f(x+) + f(x-)]/2. Theorem 11.15 (Fejdr). Assume that f e L([0, 27c]) and suppose that f is periodic with period 27r. Define a function s by the following equation: s(x) = Jim f (x + t) + f (x - t) (27) 1-o+ 2 whenever the limit exists. Then, for each x for which s(x) is defined, the Fourier series generated by f is Cesdro summable and has (C, 1) sum s(x). That is, we have lim s(x), R - a0 where is the sequence of arithmetic means defined by (23). If, in addition, f is continuous on [0, 27r], then the sequence {o.} converges uniformly to f on [0, 27[]. Proof. Let gx(t) = [f(x + t) + f(x - t)]/2 - s(x), whenever s(x) is defined. Then gx(t) -+ 0 as t -> 0+. Therefore, given s > 0, there is a positive S < 76 such that 1gx(t)j < e/2 whenever 0 < t < 6. Note that 6 depends on x as well as on e. However, if f is continuous on [0, 2n], then f is uniformly continuous on [0, 27r], and there exists a 6 which serves equally well for every x in [0, 27c]. Now we use (26) and divide the interval of integration into two subintervals [0, 6] and [6, n]. On [0, S] we have 1 Ia gx(t) sin' +nt dt < e R sine int dt = e , nit sin 2 it 2n7r 0 sin 2 It 2 Th. 11.16 Consequences of Fear's Theorem because of (25). On [6, ir] we have 1 fa" gx(t) 'in' nt dt nic sin it 1 1(x) < nn sine S Igx(t)) dt < , a n7c sin' 16 321 where I(x) = f o Igx(t)I dt. Now choose N so that I(x)/(N is sin2 18) < E/2. Then n > N implies I a.(x) - s(x)I = 1 n gx(t) sin dt < E. nn 0 sin2 It In other words, an(x) -+ s(x) as n -+ oo. If f is continuous on [0, 2ic],- then, by periodicity, f is bounded on R and there is an M such that Igx(t)I < M for all x and t, and we may replace I(x) by nM in the above argument. The resulting N is then independent ofx and hence an -+ s = f uniformly on [0, 27c]. 11.14 CONSEQUENCES OF FEJER'S THEOREM Theorem 11.16. Let f be continuous on [0, 2a] and periodic with period 27r. Let {sn} denote the sequence of partial sums of the Fourier series generatedby f, say f (x) 2 + (an cos nx + b sin nx). (28) Then we have: a) s = f on [0, 27r]. 2 b) -fo If(x)12 dx = 2 + E (a 2 + b2.) (Parseval's formula). c) The Fourier series can be integrated term by term. That is, for all x we have fox f(t) dt = aOx + , f2, (a cos nt + b sin nt) dt, n=1 0 the integrated series being uniformly convergent on every interval, even if the Fourier series in (28) diverges. d) If the Fourier series in (28) converges for some x, then it converges to f(x). Proof. Applying formula (3) of Theorem 11.2, with tn(x) = a .(x) = (1/n) Ek = o sk(x), we obtain the inequality f 2x I1(x) - sn(x)I2 dx < ('2n I1(x) - an(x)I2 dx. (29) o Jo But, since Qn -+ f uniformly on [0, 2ir], it follows that an = f on [0, 2zc], and (29) implies (a). Part (b) follows from (a) because of Theorem 11.4. Part (c) 322 Fourier Series and Fourier Integrals Th. 11.17 also follows from (a), by Theorem 9.18. Finally, if {sn(x)) converges for some x, then {Q"(x)} must converge to the same limit. But since a(x) -- f(x) it follows that s"(x) -+ f(x), which proves (d). 11.15 THE WEIERSTRASS APPROXIMATION THEOREM Fejer's theorem can also be used to prove a famous theorem of Weierstrass which states that every continuous function on a compact interval can be uniformly approximated by a polynomial. More precisely, we have: Theorem 11.17. Let f be real-valued and continuous on a compact interval [a, b]. Then for every s > 0 there is a polynomial p (which may depend on c) such that f(x) - p(x)t < E for every x in [a, b]. (30) Proof If t s [0, iv), let g(t) = f[a + t(b - a)/7c]; if t c- [iv, 2n], let g(t) = f [a + (2xc - t)(b - a)/7r] and define g outside [0, 2x] so that g has period 21r. For the E given in the theorem, we can apply Fejt is theorem to find a function a defined by an equation of the form .N a(t) = Ao + E (Ak cos kt + Bk sin kt) k=1 such that Sg(t) - a(t)I < 6/2 for every t in [0, 2ic]. (Note that N, and hence a, depends on e.) Since a is a finite sum of trigonometric functions, it generates a power series expansion about the origin which converges uniformly on every finite interval. The partial sums of this power series expansion constitute a sequence of polynomials, say {pn}, such that p" --> a uniformly on [0, 27c]. Hence, for the same a, there exists an m such that I Pm(t) - a(t)I < 2 for every t in [0, 21C]. Therefore we have pm(t) - g(t )j < e, for every tin [0, 21r]. (31) Now define the polynomial p by the formula p(x) = pm[76(x - a)/(b - a)]. Then inequality (31) becomes (30) when we put t = ir(x - a)l(b - a). 11.16 OTHER FORMS OF FOURIER SERIES Using the formulas 2 cos nx = e"" + e-"" and 2i sin nx = e"" - e-"X the Fourier series generated by f can be expressed in terms of complex exponentials as follows : 00 00 f(x) N a + E (an cos nx + bn sin nx) = ao + E (ane" + fne-1nx), 2 n=1 2 n=1 Fourier Integral Theorem 323 where an = (an - ibn)/2 and P. = (an + ib,J/2. If we put a° = ao/2 and a-n = Yn, we can write the exponential form more briefly as follows: f(x) N E aneinx n=00 The formulas (7) for the coefficients now become 2x 1 an = - ° f(t)e-int dt (n = 0, ±1, ±2,...). If f has period 2ir, the interval of integration can be replaced by any other interval of length 27[. More generally, if f c- L([0, p]) and if f has period p, we write f(x) - a0 + 1 (an cos 2nnx + bn sin 27rnx \ 2 n=1 P P to mean that the coefficients are given by the formulas a,, = 2 Pf(t) cos.-2-cnt dt, P o P P 2 f bn=P of(t)sin2Ptdt (n=0,1,2,...). In exponential form we can write f(x) . -ane2Rinx1P, 00 n=-oo where P o ifn=0,±1,±2,.... All the convergence theorems for Fourier series of period 27r can also be applied to the case of a general period p by making a suitable change of scale. 11.17 THE FOURIER INTEGRAL. THEOREM The hypothesis of periodicity, which appears in all the convergence theorems dealing with Fourier series, is not as serious a restriction as it may appear to be at first sight. If a function is initially defined on a finite interval, say [a, b], we can always extend the definition off outside [a, b] by imposing some sort of periodicity condition. For example, iff(a) = f(b), we can define f everywhere on (- co, + co) by requiring the equation f(x + p) = f(x) to hold for every x, where p = b - a. (The condition f(a) = f(b) can always be brought about by changing the value off at one of the endpoints if necessary. This does not affect the existence or the values of the integrals which are used to compute the Fourier coefficients of f.) However, if the given function is already defined everywhere on (- co, + co) and 1 rP «n = J f(t)e-zainr1p dt, 324 Fourier Series and Fourier Integrals is not periodic, then there is no hope of obtaining a Fourier series which represents the function everywhere on (- oo, + oo). Nevertheless, in such a case the function can sometimes be represented by an infinite integral rather than by an infinite series. These integrals, which are in many ways analogous to Fourier series, are known as Fourier integrals, and the theorem which gives sufficient conditions for representing a function by such an integral is known as the Fourier integral theorem. The basic tools used in the theory are, as in the case of Fourier series, the Dirichlet integrals and the Riemann-Lebesgue lemma. Theorem 11.18 (Fourier integral theorem). Assume that f e L(- oo, + oo). Suppose there is a point x in R and an interval [x - S, x + S] about x such that either a) f is of bounded variation on [x - b, x + 8], or else b) both limits f(x +) and f(x-) exist and both Lebesgue integrals f d f(x + t) - f(x+) dt and f af(x - t) - f(x-) dt o t Jo t exist. Then we have the formula f(x +) + f(x -) = 1 °° f(u) cos v(u - x) du dv, (32) 2 7C fooo [ - co the integral Jo being an improper Riemann integral. Proof. The first step in the proof is to establish the following formula: lim 1 f(x + t) sin at dt = f(x+) + f(x-) -(33) a-++ 00 16 -0o t 2 For this purpose we write I f(x + t) sinatdt = f-a + f o + fa+ 1, itt .3 cc .3 a ,10 When a -> + oo, the first and fourth integrals on the right tend to 0, because of the Riemann-Lebesgue lemma. In the third integral, we can apply either Theorem 11.8 or Theorem 11.9 (depending on whether (a) or (b) is satisfied) to get lim fo df(x + t) sin at dt = f(x+) a+m 7rt Similarly, we have J o f(x + t) sin at dt = fo f(x - t) sin at dt -+ .f(x-) a itt ?[t 2 as a -+ + oo. Th. 11.19 Exponential Form of the Fourier Integral Theorem Thus we have established (33). If we make a translation, we get F f (X + t) sin at dt = f(u) sin a(u - x) du, 00 t f-u-x and if we use the elementary formula sin a(u - x) = cos v(u - x) dv, u - x o the limit relation in (33) becomes lim a-+ + Co 1 f"O. f(u) r f a cos v(u - x) dvl du =.f(x+) 2+ f(xL J o J -325 (34) But the formula we seek to prove is (34) with only the order of integration reversed. By Theorem 10.40 we have f oa r f - -f(u) cos v(u - x) du] dv = f [rfU)0cos v(u - x) Al du for every a > 0, since the cosine function is everywhere continuous and bounded. Since the limit in (34) exists, this proves that -.+cO7r JoJ J 2 a-f f(u) cos v(u - x) du] dv = f(x+) + f(x-) Jim 1 f I By Theorem 10.40, the integral f f(u) cos v(u - x) du is a continuous function of v on [0, a], so the integral f o' in (32) exists as an improper Riemann integral. It need not exist as a Lebesgue integral. 11.18 THE EXPONENTIAL FORM OF THE FOURIER INTEGRAL THEOREM Theorem 11.19. If f satisfies the hypotheses of the Fourier integral theorem, then we have f(x+) + f(x-) 1 - Jim f(u)e°(°-") dul dv. (35) 2 276 a++w J -w Proof. Let F(v) = f f(u) cos v(u - x) A. Then F is continuous on (- oo, + oo), F(v) = F(- v) and hence f °_a F(v) dv = f o F(- v) dv = f o F(v) A. Therefore (32) becomes f(x+) + f(x-) = lim 1 f F(v) dv = lim ra F(v) A. (36) a-+oD 76 Jo a-+ 00 27r J a 326 Fourier Series and Fourier Integrals Now define G on (- oo, + oo) by the equation G(v) = f f(u) sin v(u - x) du. Then G is everywhere continuous and G(v) = - G(- v). Hence f a a G(v) dv = 0 for every at, so lima + J'. G(v) dv = 0. Combining this with (36) we find + f(x-) = lim 1 {F(v) + iG(v)} A. 2 a +o0 2n a This is formula (35). 11.19 INTEGRAL TRANSFORMS Many functions in analysis can be expressed as Lebesgue integrals or improper Riemann integrals of the form g(y) = f K(x, y)f(x) dx. (37) A function g defined by an equation of this sort (in which y may be either real or complex) is called an integral transform off. The function K which appears in the integrand is referred to as the kernel of the transform. Integral transforms are employed very extensively in both pure and applied mathematics. They are especially useful in solving certain boundary value prob-lems and certain types of integral equations. Some of the more commonly used transforms are listed below: Exponential Fourier transform: f-e-xyf(x) dx. J Fourier cosine transform : Fourier sine transform : Laplace transform : Mellin transform : fo"o fooo fo"o f0'0 cos xy f (x) dx. sin xyf(x) dx. e-xyf(x) dx. xy"'f(x) dx. Since e- ix, = cos xy - i sin xy, the sine and cosine transforms are merely special cases of the exponential Fourier transform in which the function f vanishes on the negative real axis. The Laplace transform is also related to the exponential Fourier transform. If we consider a complex value of y, say y = u + iv, where Def. 11.20 Convolutions u and v are real, we can write 327 fo"o e-xyf(x) dx = f0,0 e-ixoe-'"f(x) dx = f00 e-rsocu(x) dx, \ where 4"(x) = e-x"f(x). Therefore the Laplace transform can also be regarded as a special case of the exponential Fourier transform. NOTE. An equation such as (37) is sometimes written more briefly in the form g = . ''(f) or g = . ''f, where Jr denotes the "operator" which converts f into g. Since integration is involved in this equation, the operator Y is referred to as an integral operator. It is clear that X' is also a linear operator. That is, Jr(a1.f1 + a2f2) = a1iff1 + a2-V f2, if a1 and a2 are constants. The operator defined by the Fourier transform is ofte denoted by F and that defined by the Laplace transform is denoted by Y. The exponential form of the Fourier integral theorem can be expressed in terms of Fourier transforms as follows. Let g denote the Fourier transform off, so that g(u) = f f(t)e dt. (38) J Then, at points of continuity off, formula (35) becomes " f(x) = lira 1 g(u)e"" du, (39) a-++ao 2n -a and this is called the inversion formula for Fourier transforms. It tells us that a continuous function f satisfying the conditions of the Fourier integral theorem is uniquely determined by its Fourier transform g. NOTE. If F denotes the operator defined by (38), it is customary to denote by the operator defined by (39). Equations (38) and (39) can be expressed symbolically by writing g = 9f and f = F-1g. The inversion formula tells us how to solve the equation g = 9f for f in terms of g. Before we pursue the study of Fourier transforms any further, we introduce a new notion, the convolution of two functions. This can be interpreted as a special kind of integral transform in which the kernel K(x, y) depends only on the difference x - y. 11.20 CONVOLUTIONS Definition 11.20. Given two functions f and g, both Lebesgue integrable on (- oo, + oo), let S denote the set of x for which the Lebesgue integral h(x) = J f(t)g(x - t) dt (40) 00 328 Fourier Series and Fourier Integrals Th. 11.21 exists. This integral defines a function h on S called the convolution off andg. We also write h = f g to denote this function. NOTE. It is easy to see (by a translation) that f g = g f whenever the integral exists. An important special case occurs when both f and g vanish on the negative real axis. In this case, g(x - t) = 0 if t > x, and (40) becomes h(x) = J:f(t)9(x - t) dt. (41) It is clear that, in this case, the convolution will be defined at each point of an interval [a, b] if both f and g are Riemann-integrable on [a, b]. However, this need not be so if we assume only that f and g are Lebesgue integrable on [a, b]. For example, let f(t) = 1/ and g(t) = 1 if 0 < t < 1, 1/ 1 - t and letf(t) = g(t) = 0 if t < 0 or if t >- 1. Then f has an infinite discontinuity at t = 0. Nevertheless, the Lebesgue integral f °°. f(t) dt = f o t dt exists. Similarly, the Lebesgue integral f °° , g(t) dt = f o (1 -dt exists, although g has an infinite discontinuity at t = 1. However, when we form the convolution integral in (40) corresponding to x = 1, we find t -' dt. f f(t)g(1 - t) dt = J Observe that the two discontinuities off and g have "coalesced" into one dis-continuity of such nature that the convolution integral does not exist. This example shows that there may be certain points on the real axis at which the integral in (40) fails to exist, even though both f and g are Lebesgue-integrable on (- oo, + oo). Let us refer to such points as "singularities" of h. It is easy to show that such singularities cannot occur unless both f and g have infinite dis-continuities. More precisely, we have the following theorem : Theorem 11.21. Let R = (- oo, + oo): Assume that f e L(R), g e L(R), and that either for g is bounded on R. Then the convolution integral h(x) = f f(t)g(x - t) dt (42) exists for every x in R, and the function h so defined is bounded on R. If, in addition, the bounded function f or g is continuous on R, then h is also continuous on R and h e L(R). Th. 11.23 Convolution Theorem for Fourier Transforms 329 Proof. Since f g = g f, it suffices to consider the case in which g is bounded. Suppose IgI < M. Then If(t)g(x - t)I <- MIf(t)I (43) The reader can verify that for each x, the product f(t)g(x - t) is a measurable function of t on R, so Theorem 10.35 shows that the integral for h(x) exists. The inequality (43) also shows that Ih(x)I < M I If I, so h is bounded on R. Now if g is also continuous on R, then Theorem 10.40 shows that h is continuous on R. Now for every compact interval [a, b] we have f 6Ih(x)I dx < f 6 r f If(t)I Ig(x - t)I dtl dx Ja LJ J f If(t)I [fix - t)Idx]dt f-00. If(t)1 [f b tt I g(Y)I dy] dt -< f If(t)I dt Ig(Y)I dy, so, by Theorem 10.31, h e L(R). Theorem 11.22. Let R = (- oo, + oo). Assume that f e L2(R) and g e L2(R). Then the convolution integral (42) exists for each x in R and the function h is bounded on R. Proof. For fixed x, let gs(t) = g(x - t). Then g,, is measurable on R and gx e L2(R), so Theorem 10.54 implies that the product f gx E L(R). In other words, the convolution integral h(x) exists. Now h(x) is an inner product, h(x) = (f gx), hence the Cauchy-Schwarz inequality shows that Ih(x)I <- IIfII Ilgxli = Ilfll IIgII, so h is bounded on R. 11.21 THE CONVOLUTION THEOREM FOR FOURIER TRANSFORMS The next theorem shows that the Fourier transform of a convolution f g is the product of the Fourier transforms off and of g. In operator notation, F(.f g) = F(f) - -17(g) Theorem 11.23. Let R = (- co, + oo). Assume that f c- L(R), g c- L(R), and that at least one off or g is continuous and bounded on R. Let h denote the convolution, 330 Fourier Series and Fourier Integrals h = f g. Then for every real u we have h(x)e-"" dx = (f,f(t)e u d) \J-g(Y)e-1y" dy) . f-O". ( f Th. 11.23 (44) The integral on the left exists both as a Lebesgue integral and as an improper Riemann integral. Proof. Assume that g is continuous and bounded on R. Let {aa} and {ba} be two increasing sequences of positive real numbers such that a --> + oo and b - + oo. Define a sequence of functions If.) on R as follows: fMO = b e-'"' g(x - t) dx. a Since fb le-i"" g(x - t)1 dt < f '0 IgI a -for all compact intervals [a, b], Theorem 10.31 shows that lim f"(t) = e "" g(x - t) dx for every real t. (45) atioo -00 J 00 The translation y = x - t gives us J e-i"" g(x - t) dx = e-wt f-e-'u" g(y) dY, 00 and (45) shows that urn f(t)f"(t) = f(t)e "t ( fe-'uy g(y) dY 1 for all t. Now fa is continuous on R (by Theorem 10.38), so the product f fa is measurable on R. Since I f(t)fn(t )1 5 jf(01 f-00 I gI, the product f f is Lebesgue-integrable on R, and the Lebesgue dominated con-vergence theorem shows that But 9(Y) dY) (46) lim J f(t)fn(t) dt = \J f(t)e-i"t dt! (J a-r"y ao f_1 5f(t)[$fle_I4xg(x f(t)fa(t) dt = - t) dx] dt. Convolution Theorem for Fourier Transforms 331 Since the function k defined by k(x, t) = g(x - t) is continuous and bounded on R2 and since the integral fa e" dx exists for every compact interval [a, b], Theorem 10.40 permits us to reverse the order of integration and we obtain J f(t).f.(t) dt = f b e-t"" r f "0 f(t)g(x - t) dtl dx a LJ J fi = e-fuxh(x) dx. J - a Therefore, (46) shows that li m a h(x)e-tux dx = (r f(t)e-fat dt) (f g(y)e "" dy which proves (44). The integral on the left also exists as an improper Riemann integral because the integrand is continuous and bounded on R and fa Ih(x)e-t"xl dx < f °°Q Jhi for every compact interval [a, b]. As an application of the convolution theorem we shall derive the following property of the Gamma function. Example. If p > 0 and q > 0, we have the formula f xp-1(1 -x)Q-1 dx = r(p)r(g) (47) Jo r(p + q) The integral on the left is called the Beta function and is usually denoted by B(p, q). To prove (47) we let fp(t) = tp-1e t if t > 0, 0 if t50. Then fp e L(R) and J°-°,, fp(t) dt = 100 tp-le-t dt = r(p). Let h denote the convolution, h = fp fQ. Taking u = 0 in the convolution formula (44) we find, if p > 1 or q > 1, h(x) dx = J fp(t) dt f-fq(Y) dy = r(p)r(q) (48) 00 00 Now we calculate the integral on the left in another way. Since both fp and fq vanish on the negative real axis, we have h(x) = fo xfp(t) fq(x - t) dt = e-x f tp-1(x t)q-1 dt ifx > 0, 0 0 if x<_0. The change of variable t = ux gives us, for x > 0, 1 h(x) = e-xxp+q-1 fo up-1(1 -u)a-1 du = e-xp+v-1B(p, q). Therefore f°_° h(x) dx = B(p, q) Jo e'"xp+a-1 dx = B(p, q)r(p + q) which, when used in (48), proves (47) if p > 1 or q > 1. To obtain the result for p > 0, q > 0 use the relation pB(p, q) = (p + q)B(p + 1, q). 332 Fourier Series and Fourier Integrals 11.22 THE POISSON SUMMATION FORMULA Th. 11.24 We conclude this chapter with a discussion of an important formula, called Poisson's summation formula, which has many applications. The formula can be expressed in different ways. For the applications we have in mind, the following form is convenient. Theorem 11.24. Let f be a nonnegative function such that the integral f °_°. f(x) dx exists as an improper Riemann integral. Assume also that f increases on (- oo, 0] and decreases on [0, + oo). Then we have f-. f(m+) + f(m"0 .r(t)e-2at't dt, 2 n=-0o each series being absolutely convergent. Proof. The proof makes use of the Fourier expansion of the function F defined by the series +00 F(x) _ f(m + x). (50) m=-ao First we show that this series converges absolutely for each real x and that the convergence is uniform on the interval [0, 1]. Since f decreases on [0, + co) we have, for x >t 0, E f(m + x) < f(O) + E f(m) < f(O) +f f(t) dt. M=0 m=1 X10 Therefore, by the Weierstrass M-test (Theorem 9.6), the series EM=o f(m + x) converges uniformly on [0, + co). A similar argument shows that the series Em= _ f(m + x) converges uniformly on (- oo, 1]. Therefore the series in (50) converges for all x and the convergence is uniform on the intersection (-oo,1]n[0,+o0)=[0,1]. The sum function F is periodic with period 1. In fact, we have F(x + 1) _ Em f(m + x + 1), and this series is merely a rearrangement of that in (50). Since all its terms are nonnegative, it converges to the same sum. Hence F(x + 1) = F(x). Next we show that F is of bounded variation on every compact interval. If 0 < x < 1, then f(m + x) is a decreasing function of x if m >- 0, and an in-creasing function of x if m < 0. Therefore we have 00 - 1 F(x) Ef(m+x)- E {-f(m+x)}, m=0 m=-00 so F is the difference of two decreasing functions. Therefore F is of bounded Th. 11.24 The Poisson Summation Formula 333 variation on [0, 1]. A similar argument shows that F is also of bounded variation on [-- , 0]. By periodicity, F is of bounded variation on every compact interval. Now consider the Fourier series (in exponential form) generated by F, say +00 F(x) a"e2ainx n=-oo Since F is of bounded variation on [0, 1] it is Riemann-integrable on [0, 1], and the Fourier coefficients are given by the formula 1 an = foF(x)e-2a'nx dx. (51) Also, since F is of bounded variation on every compact interval, Jordan's test shows that the Fourier series converges for every x and that F(x+) + F(x-) _ E a e2ainx (52) 2 n= - 00 " To obtain the Poisson summation formula we express the coefficients an in another form. We use (50) in (51) and integrate term by term (justified by uniform convergence) to obtain +00 an = E J f(m + x)e-2n"x dx. m=- oo 0 The change of variable t = m + x gives us a" = ±0D fM+1 f(t)e-taint dt = f o0 0 f(t)e-taint dt, ,, since e1. Using this in (52) we obtain F(x+) 2 F(X-) f(t)e-2ainr dte2'inx(53) When x = 0 this reduces to (49). NOTE. In Theorem 11.24 there are no continuity requirements on f. However, if f is continuous at each integer, then each term f(m + x) in the series (50) is con-tinuous at x = 0 and hence, because of uniform convergence, the sum function F is also continuous at 0. In this case, (49) becomes +,0 +"o E f(m) _ E f(t)e-2ainr dt. (54) m= - ao n=-,0 The monotonicity requirements on f can be relaxed. For example, since each member of (49) depends linearly on f, if the theorem is true for f1 and for f2 then it is also true for any linear combination a1 f1 + a2 f2. In particular, the formula holds for a complex-valued function f = u + iv if it holds for u and v separately. 334 Fourier Series and Fourier Integrals Example 1. Transformation formula for the theta function. The theta function 0 is defined for all x > 0 by the equation +"0 0(x) _ e-xn2x. OW n=-oo We shall use Poisson's formula to derive the transformation equation 0(x) e1 for x > 0. (55) x (X) For fixed a > 0, let f(x) = ex2 for all real x. This function satisfies all the hypoth-esis of Theorem 11.24 and is continuous everywhere. Therefore, Poisson's formula implies =-ao n=-oo m - OD (56) The left member is 6(a/n). The integral on the right is equal to f eat e2xini dt = 2 e2 cos 271nt dt = ? f -e-X2 cos 27rnx dx = 2 F(nn fo Tat a where -x2 cos 2xy dx. e F(y) = Jo0 But F(y) _ ne_r2 (see Exercise 10.22), so fOD eat e2xini dt = (a Using this in (56) and taking a = nx we obtain (55). Example 2. Partial fraction decomposition of coth x. The hyperbolic cotangent, coth x, is defined for x # 0 by the equation coth x = e2x + 1 e2x - 1 We shall use Poisson's formula to derive the so-called partial-fraction decomposition 1 cothx= +2xE I n x n-1 x2 + 71 2 2 (57) for x > 0. For fixed a > 0, let f(x) = (e-ctx if x > 0, to ifx<0. Then f clearly satisfies the hypotheses of Theorem 11.24. Also, f is continuous everywhere except at 0, where f(0+) = I and f(0-) = 0. Therefore, the Poisson formula implies OD +00 ao + r e-ma = [: f e-a'-2xinr dt. (58) M=1 n=-00 0 E e2 = 2 J f e-02 2xinr dt. 1/2 / ex2n2/a Exercises 335 The sum on the left is a geometric series with sum 1/(ea - 1), and the integral on the right is equal to 1/(a + 2nin). Therefore (58) becomes 1+ 1 1+ 1 + 1 1 2 e" - 1 a E a + 2nin a - 2nin/f ' and this gives (57) when a is replaced by 2x. EXERCISES Orthogonal systems 11.1 Verify that the trigonometric system in (1) is orthonormal on [0, 2n]. 11.2 A finite collection of functions {rpo, ip,,... , rpm} is said to be linearly independent on [a, b] if the equation M E ckrpk(x) = 0 for all x in [a, b] k=0 implies co = cl = = cM = 0. An infinite collection is called linearly independent on [a, b ] if every finite subset is linearly independent on [a, b ]. Prove that every orthonormal system on [a, b] is linearly independent on [a, b]. 11.3 This exercise describes the Gram-Schmidt process for converting any linearly inde-pendent system to an orthogonal system. Let { fo, fl, ... } be a linearly independent system on [a, b] (as defined in Exercise 11.2). Define a new system {go, g,, ... } recur-sively as follows : go = Jo, gr+ 1 = f + 1 - E, akgk, k=1 where ak = (f,+ 1, gk)/(gk, gk) if II gk II : 0, and ak = 0 if II gk II = 0. Prove that is orthogonal to each of go, g,, ... , g for every n z 0. 11.4 Refer to Exercise 11.3. Let (f, g) = f '-I f(t)g(t) dt. Apply the Gram-Schmidt process to the system of polynomials {1, t, t2, ... } on the interval [-1, 1 ] and show that lp g1(t)=t, g2(t)=t2-4, g3(t)=t3- it, 94(t)=t4- 6t2+ 11.5 a) Assume f e R on [0, 2n], where f is real and has period 2n. Prove that for every e > 0 there is a continuous function g of period 2n such that If - g I I < e. Hint. Choose a partition P, of [0, 2n] for which f satisfies Riemann's condition U(P, f) - L(P, f) < e and construct a piecewise linear g which agrees with f at the points of P. b) Use part (a) to show that Theorem 11.16(a), (b) and (c) holds if f is Riemann integrable on [0, 2n]. 11.6 In this exercise all functions are assumed to be continuous on a compact interval [a, b]. Let {rpo, p,.... } be an orthonormal system on [a, b]. a) Prove that the following three statements are equivalent. 336 Fourier Series and Fourier Integrals 1) (f, p,,) = (g, rpn) for all n implies f = g. (Two distinct continuous functions cannot have the same Fourier coefficients.) 2) (f, (") = 0 for all n implies f = 0. (The only continuous function orthogonal to every rp,, is the zero function.) 3) If T is an orthonormal set on [a, b] such that {rpo, (pl,... ) T, then {rpo, rpi, ... } = T. (We cannot enlarge the orthonormal set.) This property is described by saying that {rpo, rp1, ... } is maximal or complete. b) Let rp(x) = ei' /-2n for n an integer, and verify that the set {rp,,: n e Z) is com-plete on every interval of length 2;r. 11.7 If x e R and n = 1, 2, ... , let f"(x) = (x2 - 1)" and define Q0(X) = 1, cn(x) = n ! J n)(x). It is clear that 0" is a polynomial. This is called the Legendre polynomial of order n. The first few are q51(x) = x, 02(X) = Tx2 - i, 03(X) = x3 - ix, q54(X) = 4x4 - 4X2 + . Derive the following properties of Legendre polynomials : a) O,(x) = x0n-1(x) + nOn-1(x). b) rbn(x) = x¢"-i(x) + X2 4-1(x). n c) (n + 1)qn+1(x) = (2n + 1)xon(x) - non-1(x). d) ¢" satisfies the differential equation [(1 - x2) y' ]' + n(n + I) y = 0. e) [(1 - x 2) A(x) ]' + [m(m + 1) - n(n + I) ] q$m(x) ¢n(x) = 0, where A = 0" Om - 0m 0n. f) The set {00, 01, 02, ... } is orthogonal on [-1, 1 ]. 2n - 1 i g) ¢. A = _i 2n+1, i h) r 02dx 2 = ,1 i i 2n+I NOTE. The polynomials 0n2-1 dx. 2"(n !)2 gn(t) = _____ 0n0) arise by applying the Gram-Schmidt process to the system {1, t, t2, ... } on the interval [-1, 1 ]. (See Exercise 11.4.) Exercises 337 Trigonometric Fourier series 11.8 Assume that f e L( [ - it, 7t ]) and that f has period 2n. Show that the Fourier series generated by f assumes the following special forms under the conditions stated: a) If f(- x) = f (x) when 0 < x < it, then 00 a f (X) ^ 2 + a. cos nx, n=1 where a" = f(t) cos nt dt. 0 X b) If fl- x) = -f(x) when 0 < x < it, then R X) - E bn sin nx, n=1 where b" = 2 f(t) sin nt dt. 0 In Exercises 11.9 through 11.15, show that each of the expansions is valid in the range indicated. Suggestion. Use Exercise 11.8 and Theorem 11.16(c) when possible. 00 11.9 a)x=7r-2Esinnx if0<x<27r. n=1 n x2 7r2 00 cos nx b)= 7rx-+ 2 2 2 3 n=1 n xoTE. When x = 0 this gives C(2) = 7x2/6. it 00 sin (2n - 1)x 11.10 a)4-1: 5 1 2n- 1 it 4 00 cos (2n - 1)x b) x = -2 it (!2n-12 n=1 (-1)"-' sin nx 00 11.11 a) x = 2 F. n=1 n b) x2 = n2 + 4 E E (- 1)" cos nx 3 "=1 n2 00 cos x 4 2 ( t 11.12 x = 7r + 4 "=1 8 0D n sin 2nx 11.13 a)cosx= -7r =1 4n 1 ' 2 n=1 -if0<-x<<-27r. 2 4 0O cos 2nx z _ b)sinx=X-n"4n 1' 1 00 11.14 a) x cos x = - } sin x + 2 -1)2 "n sin nx n=2 n - 1 b) x sin x = 1 - + cos x - 2 E (-1)" cos nx z - I if0 < x< it. if 0<-x<-n. if-7r<x<7r. if -7r 5 x < it. if0<x<27r. if0<x<7r. if0<x<7r. if-7r<x<7r. if - 7r < x < it. n=2 n Fourier Series and Fourier Integrals 00 x cos nx 11.15 a) log sin _ - log 2 -if x 0 2kn (k an integer). 2 R=1 n x (-1)"cosnx 00 b) lo cos - _ -lo 2 - E 2k 1 if g c) log X tan -g , n=1 n cos (2n - 1)x -2 x: ( )ir. + if x :A kn. L.r n=1 2n - 1 11.16 a) Find a continuous function on [-n, n] which generates the Fourier series Y_n1 (-1)"n-3 sin nx. Then use Parseval's formula to prove that C(6) _ X6/945. b) Use an appropriate Fourier series in conjunction with Parseval's formula to show that C(4) = n4/90. 11.17 Assume that f has a continuous derivative on [0, 2n], that f(0) = f(2ir), and that f 2X f(t) dt = 0. Prove that Il f' 1I >: If 11, with equality if and only if f (x) = a cos x + b sin x. Hint. Use Parseval's formula. 11.18 A sequence {Bn} of periodic functions (of period 1) is defined on R as follows: B2n(x) _ (-1)n+' 2(2n)! r cos 2irkx L, (n = 2n / 2n 2rz) , k=1 k 9en+1(x) 2(2n + 1)! sin 27rkx (2n)2n+ 1 IS _j211+1 (Bn is called the Bernoulli function of order n.) Show that: (n=0,1,2,...). a) B1(x) = x - [x] - # if x is not an integer. ([x] is the greatest integer <_x.) b) f a B.(x) dx = 0 if n I and nBn-I (x) if n >: 2. c) Bl(x) = PP(x) if 0 < x < 1, where P. is the nth Bernoulli polynomial. (See Exercise 9.38 for the definition of Pn.) ' 00 e2nikx ). d) BB(x) = -E (n = 1, 2.... (2xi)n k_ _. k" k#0 11.19 Let f be the function of period 2rr whose values on [ - it, 7r ] are f(x)= 1 if0<x<Jr, f(x) = 0 if x = 0 or x = 1r. a) Show that f(x)= -1 if-7r<x<0, 4 °° sin (2n - 1)x f(x) = -- , for every x. nn=1 2n - 1 This is one example of a class of Fourier series which have a curious property known as Gibbs' phenomenon. This exercise is designed to illustrate this phenomenon. In that which follows, sn(x) denotes the nth partial sum of the series in part (a). Exercises 339 b) Show that 2 f sin 2nt sn(x _ - J dt. 7r o sin t c) Show that, in (0, 7r), sn has local maxima at x1, x3, ... , x2,,_ 1 and local minima at x2, x4, .... x2n_2, where xn = Jm7r/n (m = 1, 2, ... , 2n - 1). d) Show that sn(J7r/n) is the largest of the numbers sn(xm) (m = 1, 2, ... , 2n - 1). e) Interpret sn(J7r/n) as a Riemann sum and prove that lim sn n) = 2 o sin t dt. n-.CO 2n 7r t The value of the limit in (e) is about 1.179. Thus, although f has a jump equal to 2 at the origin, the graphs of the approximating curves sn tend to approximate a vertical segment of length 2.358 in the vicinity of the origin. This is the Gibbs phenomenon. 11.20 If f(x) - ao/2 + YR 1 (an cos nx + b sin nx) and if f is of bounded variation on [0, 27r], show that an = 0(1/n) and bn = 0(1/n). Hint. Write f = g - h, where g and h are increasing on [0, 27r]. Then 2 2 an = n f n g(x) d(sin nx) - 1 f h(x) d(sin nx). r o n7r o Now apply Theorem 7.31. 11.21 Suppose g e L( [a, 8]) for every a in (0, 6) and assume that g satisfies a "right-handed" Lipschitz condition at 0. (See the Note following Theorem 11.9.) Show that the Lebesgue integral f o Sg(t) - g(0+)Ilt dt exists. 11.22 Use Exercise 11.21 to prove that differentiability off at a point implies convergence of its Fourier series at the point. 11.23 Let g be continuous on [0, 1 ] and assume that f o tng(t) dt = 0 for n = 0, 1, 2,.... Show that: a) f o g(t)2 dt = f o g(t)(g(t) - P(t)) dt for every polynomial P. b) f o g(t)2 dt = 0. Hint. Use Theorem 11.17. c) g(t) = 0 for every t in [0, 1 ]. 11.24 Use the Weierstrass approximation theorem to prove each of the following state-ments. a) If f is continuous on [1, + oo) and if f(x) -+ a as x -+ + oo, then f can be uni-formly approximated on [1, + oc) by a function g of the form g(x) = p(1/x), where p is a polynomial. b) If f is continuous on [0, + oo) and if f(x) -> a as x --+ + oo, then f can be uniformly approximated on [0, + oo) by a function g of the form g(x) = p(e1, where p is a polynomial. 11.25 Assume that f(x) - a0/2 + F_n 1 (an cos nx + bn sin nx) and let {an} be the sequence of arithmetic means of the partial sums of this series, as it was given in (23). 340 Fourier Series and Fourier Integrals Show that : n-1 a) an(x) = ao + kl (ak cos kx + bk sin kx). 2 k=1\ n 2. b) fo I f(x) a (x)l2 dx If(x)12 &C 0 n-1 n-1 -2 ao - n F (ak + bk) + n2 k2(ak + bk). k=1 n k=1 c) If f is continuous on [0, 271 ] and has period 2ir, then n lim 2 k2(ak + bk) = 0. n-.oo n k=1 11.26 Consider the Fourier series (in exponential form) generated by a function f which is continuous on [0, 27v] and periodic with period 2n, say + CO f(x) E aena n=-ao Assume also that the derivative f e. R on [0, 2n ]. a) Prove that the series n2lan12 converges; then use the Cauchy-Schwarz inequality to deduce that Ian converges. b) From (a), deduce that the series 0D_ ae'n" converges uniformly to a con-tinuous sum function g on [0, 2n ]. Then prove that f = g. Fourier integrals 11.27 If f satisfies the hypotheses of the Fourier integral theorem, show that : a) If f is even, that is, if f(- t) = f(t) for every t, then f(x+) + f(x-) = 2 Urn f a cos vx f(u) cos vu du] dv. 2 n Jo f fo J b) If f is odd, that is, if ft- t) = -f(t) for every t, then f(x+) + f(x-) = 2 lim f sin vx Lfo u) sin vu dul dv. f( 2 n JO LJ Use the Fourier integral theorem to evaluate the improper integrals in Exercises 11.28 through 11.30. Suggestion. Use Exercise 11.27 when possible. 2 sin v cos vx 1 if -1 < x < 1, 11.28- f dv= 0 iflxI> 1, n o V # if 1xI = 1. 11.29 cos axe dx = 26 e- l °l b, if b > 0. 0 Hint. Apply Exercise 11.27 withf(u) = e-binl, Exercises 341 11.30 f'xsinaxdx- a ite-lat, if a 00. Jo 1 +x2 jai 2 11.31 a) Prove that r(p)r(p) = 112 2 xv-1(1 -x)v-1 dx. r(2p) Jo b) Make a suitable change of variable in (a) and derive the duplication formula for the Gamma function: r(2p)r(j) = 22p-lr(p)r(p + 1). NOTE. In Exercise 10.30 it is shown that r(f) = V7C. 11.32 If f(x) = e-x2/2 and g(x) = xf(x) for all x, prove that f(y) f (x) cos xy dx and g(y) f g(x) sin xy dx. o f o 11.33 This exercise describes another form of Poisson's summation formula. Assume that f is nonnegative, decreasing, and continuous on [0, + oo) and that f o' f(x) dx exists as an improper Riemann integral. Let g(y) = 2 f 00 f(x) cos xy dx. n o If a and ft are positive numbers such that aft = 2n, prove that Va (+1(0) + f(ma)} = {+(0) + E g(np)} . m=1 R=1 1 11.34 Prove that the transformation formula (55) for 0(x) can be put in the form + e-.W/2) + p..2/2) e , l t M=1 n=1 where aft = 2n, a > 0. 11.35 Ifs > 1, prove that 7r-a/2 I+I Zln-8 = JO e-Ra2xxs/2-1 dx and derive the formula (x)x/2-1 dx, n-8/2 r(2)C(s) = fo(,o w 342 Fourier Series and Fourier Integrals where 2V(x) = 6(x) - 1. Use this and the transformation formula for 0(x) to prove that n-S/2 l C(s) = As 1) + J1 (xsl2-1 + xu-s)12-1)1//(x) dx. Laplace transforms Let c be a positive number such that the integral fo e-"If(t)j dt exists as an improper Riemann integral. Let z = x + iy, where x > c. It is easy to show that the integral F(z) = f e-Zf(t) dt 0 exists both as an improper Riemann integral and as a Lebesgue integral. The function F so defined is called the Laplace transform of f, denoted by .P(f). The following exercises describe some properties of Laplace transforms. 11.36 Verify the entries in the following table of Laplace transforms. f(t) ear cos at sin at theat F(z) = fo e-Zf(t) dt z = x + iy (z - a)-1 (x > a) z/(z2 + a2) (x > 0) a/(z2 + a2) (x > 0) r(p + 1)/(z - a)p+1 (x > a, p > 0) 11.37 Show that the convolution h = f g assumes the form h(t) = J'tf(x)(t - x) dx 0 when both f and g vanish on the negative real axis. Use the convolution theorem for Fourier transforms to prove that 2'(f g) _ 9(f) -W(g). 11.38 Assume f is continuous on (0, + oo) and let F(z) = fo ef (t) dt for z = x + iy, x > c > 0. Ifs > c and a > 0 prove that : a) F(s + a) = a fo g(t)e-° dt, where g(x) = f o e-sf(t) A b) If F(s + na) = 0 for n = 0, 1, 2, ... , then f(t) = 0 for t > 0. Hint. Use Exercise 11.23. c) If h is continuous on (0, + oo) and if f and h have the same Laplace transform, then f(t) = h(t) for every t > 0. 11.39 Let F(z) = f o e-" f(t) dt for z = x + iy, x > c > 0. Let t be a point at which f satisfies one of the "local" conditions (a) or (b) of the Fourier integral theorem (Theorem 11.18). Prove that for each a > c we have f(t+) + f(t-) = 1 Jim fT e(a+l )tF(a + iv) dv. 2 2n T-++OD T This is called the inversion formula for Laplace transforms. The limit on the right is usually evaluated with the help of residue calculus, as described in Section 16.26. Hint. Let g(t) = e-af(t) fort >- 0, g(t) = 0 fort < 0, and apply Theorem 11.19 tog. References 343 SUGGESTED REFERENCES FOR FURTHER STUDY 11.1 Carslaw, H. S., Introduction to the Theory of Fourier's Series and Integrals, 3rd ed. Macmillan, London, 1930. 11.2 Edwards, R. E., Fourier Series, A Modern Introduction, Vol. 1. Holt, Rinehart and Winston, New York, 1967. 11.3 Hardy, G. H., and Rogosinski, W. W., Fourier Series. Cambridge University Press, 1950. 11.4 Hobson, E. W., The Theory of Functions of a Real Variable and the Theory of Fourier's Series, Vol. 1, 3rd ed. Cambridge University Press, 1927. 11.5 Indritz, J., Methods in Analysis. Macmillan, New York, 1963. 11.6 Jackson, D., Fourier Series and Orthogonal Polynomials. Carus Monograph No. 6. Open Court, New York, 1941. 11.7 Rogosinski, W. W., Fourier Series. H. Cohn and F. Steinhardt, translators. Chelsea, New York, 1950. 11.8 Titchmarsh, E. C., Theory of Fourier Integrals. Oxford University Press, 1937. 11.9 Wiener, N., The Fourier Integral. Cambridge University Press, 1933. 11.10 Zygmund, A., Trigonometrical Series, 2nd ed. Cambridge University Press, 1968. CHAPTER 12 MULTIVARIABLE DIFFERENTIAL CALCULUS 12.1 INTRODUCTION Partial derivatives of functions from R" to R' were discussed briefly in Chapter 5. We also introduced derivatives of vector-valued functions from Rl to R". This chapter extends derivative theory to functions from R" to R'. As noted in Section 5.14, the partial derivative is a somewhat unsatisfactory generalization of the usual derivative because existence of all the partial derivatives Dl f, ... , D"f at a particular point does not necessarily imply continuity of f at that point. The trouble with partial derivatives is that they treat a function of several variables as a function of one variable at a time. The partial derivative describes the rate of change of a function in the direction of each coordinate axis. There is a slight generalization, called the directional derivative, which studies the rate of change of a function in an arbitrary direction. It applies to both real- and vector-valued functions. 12.2 THE DIRECTIONAL DERIVATIVE Let S be a subset of R", and let f : S -+ R'° be a function defined on S with values in R'". We wish to study how f changes as we move from a point c in S along a line segment to a nearby point c + u, where u # 0. Each point on the segment can be expressed as c + hu, where h is real. The vector u describes the direction of the line segment. We assume that c is an interior point of S. Then there is an n-ball B(c; r) lying in S, and, if h is small enough, the line segment joining c to c + ho will lie in B(c; r) and hence in S. Definition 12.1. The directional derivative of f at c in the direction u, denoted by the symbol f'(c; u), is defined by the equation f'(c; u) = lim f(c + hu) - f(c) (1) h-0 h whenever the limit on the right exists. NOTE. Some authors require that Hull = 1, but this is not assumed here. Examples 1. The definition in (1) is meaningful if u = 0. In this case f'(c; 0) exists and equals 0 for every c in S. 344 Directional Derivatives 345 2. If u = uk, the kth unit coordinate vector, then f'(c; uk) is called a partial derivative and is denoted by Dkf(c). When f is real-valued this agrees with the definition given in Chapter 5. 3. If f = (ft, ... , fm), then f'(c; u) exists if and only if fk(c; u) exists for each k = 1, 2, ... , m, in which case f'(c; u) = (fi(c; u), ... , f,, (c; u)). In particular, when u = uk we find Dkf(c) = (Dkf1(c), ... , Dkfm(c)). (2) 4. If F(t) = f(c + tu), then F'(0) = f'(c; u). More generally, F'(t) = f'(c + tu; u) if either derivative exists. 5. If f(x) = 11x112, then F(t) = f(c + tu) = (c + tu) (c + tu) = 11e11' + etc u + t211u112, so F'(t) = 2c u + 2t11u112; hence F'(0) = f'(c; u) = 2c u. 6. Linear functions. A function f : R" - Rm is called linear if flax + by) = af(x) + bf(y) for every x and y in R" and every pair of scalars a and b. If f is linear, the quotient on the right of (1) simplifies to f(u), so f'(c; u) = f(u) for every c and ever' u. 12.3 DIRECTIONAL DERIVATIVES AND CONTINUITY If f'(c; u) exists in every direction u, then in particular all the partial derivatives Dkf(c),... , D"f(c) exist. However, the converse is not true. For example, consider the real-valued function f : R2 -+ Rt given by _ x+y ifx=Dory=O, f(x' Y) 1 otherwise. Then Dt f(0, 0) = D2f(O, 0) = 1. Nevertheless, if we consider any other direction u = (a1, a2), where at 0 and a2 0, then f(0 + hu) - f(0) _ f(hu) _ 1 h h h' and this does not tend to a limit as h -1- 0. A rather surprising fact is that a function can have a finite directional derivative f'(c; u) for every. u but may fail to be continuous at c. For example, let .f(x, Y) _ toY2I(x2 + Y4) ifx 0, ifx=0. Let u = (at, a2) be any vector in R2. Then we have -f(0 + hu) - f(0) _ f(hal, ha2) -alai h h a; + h 2a 4 ' 346 Multivariable Differential Calculus and hence Def. 12.2 P o; u) = a2/ai if al A 0, to if al = 0. Thus, f'(0; u) exists for all u. On the other hand, the function f takes the value I at each point of the parabola x = y2 (except at the origin), so f is not continuous at (0, 0), since f(0, 0) = 0. Thus we see that even the existence of all directional derivatives at a point fails to imply continuity at that point. For this reason, directional derivatives, like partial derivatives, are a somewhat unsatisfactory extension of the one-dimensional concept of derivative. We turn now to a more suitable generalization which implies continuity and, at the same time, extends the principal theorems of one-dimensional derivative theory to functions of several variables. This is called the total derivative. 12.4 THE TOTAL DERIVATIVE In the one-dimensional case, a function f with a derivative at c can be approximated near c by a linear polynomial. In fact, iff'(c) exists, let Ec(h) denote the difference f(c + h) - f (c) - f'(c) if h # 0, (3) h and let EJ0) = 0. Then we have f(c + h) = f(c) + f'(c)h + hEE(h), (4) an equation which holds also for h = 0. This is called the first-order Taylor formula for approximating f(c + h) - f(c) by f'(c)h. The error committed is hEE(h). From (3) we see that EE(h) -+ 0 as h -+ 0. The error hEE(h) is said to be of smaller order than h as h -+ 0. We focus attention on two properties of formula (4). First, the quantity f'(c)h is a linear function of h. That is, if we write Tc(h) = f.'(c)h, then Tc(ahl + bh2) = aT,(hl) + bTc(h2). Second, the error term hEc(h) is of smaller order than h as h -+ 0. The total derivative of a function f from R" to R' will now be defined in such a way that it preserves these two properties. Let f : S -+ R' be a function defined on a set S in W with values in R'°. Let c be an interior point of S, and let B(c; r) be an n-ball lying in S. Let v be a point in R" with Ilvll < r, so that c + v e B(c; r). Definition 12.2. The function f is said to be differentiable at c if there exists a linear function T, : R" -+ R' such that f(c + v) = f(c) + T.(v) + llvll E.(v), (5) where E,,(v) -+ 0 as v - 0. Th. 12.5 The Total Derivative 347 NOTE. Equation (5) is called a first-order Taylor formula. It is to hold for all v in R" with Ilvll < r. The linear function T,, is called the total derivative of fat c. We also write (5) in the form f(c + v) = f(c) + Tc(v) + o(llvll) as v - 0. The next theorem shows that if the total derivative exists, it is unique. It also relates the total derivative to directional derivatives. Theorem 12.3. Assume f is differentiable at c with total derivative Tc. Then the directional derivative f'(c; u) exists for every u in R" and we have T,,(u) = f'(c; u). (6) Proof. If v = 0 then f'(c; 0) = 0 and Tr (0) = 0. Therefore we can assume that v # 0. Take v = hu in Taylor's formula (5), with h 0, to get f(c + hu) - f(c) = Tc(hu) + Ilhull E,,(v) = hTju) + IhI (lull Ejv) Now divide by h and let h - 0 to obtain (6). Theorem 12.4. If f is differentiable at c, then f is continuous at c. Proof. Let v -- 0 in the Taylor formula (5). The error term IIvii E,,(v) -- 0; the linear term T,,(v) also tends to 0 because if v = v1u1 + - - + v"u", where u1, ... , u" are the unit coordinate vectors, then by linearity we have T.(u) = v1Tc(ul) + ... + and each term on the right tends to 0 as v -+ 0. NOTE. The total derivative T, is also written as f'(c) to resemble the notation used in the one-dimensional theory. With this notation, the Taylor formula (5) takes the form f(c + v) = f(c) + f'(c)(v) + llvfl E.(v), (7) where E.(v) -+ 0 as v -+ 0. However, it should be realized that f'(c) is a linear function, not a number. It is defined everywhere on R"; the vector f'(c)(v) is the value of U(c) at v. Example. If f is itself a linear function, then f(c + v) = f(c) + f(v), so the derivative f'(c) exists for every c and equals f. In other words, the total derivative of a linear function is the function itself. 12.5 THE TOTAL DERIVATIVE EXPRESSED IN TERMS OF PARTIAL DERIVATIVES The next theorem shows that the vector f'(c)(v) is a linear combination of the partial derivatives of f. Theorem 12.5. -Let f : S -+ R' be differentiable at an interior point c of S, where S S R. If v = v1u1 + -- - + vu", where u1, ... , u" are the unit coordinate 348 Multivariable Differential Calculus vectors in R", then n f'(c)(v) = E vkDk f (c). k=1 In particular, if f is real-valued (m = 1) we have f'(c)(v) = Vf(c) - v, the dot product of v with the vector Vf(c) = (D1 f(c), ... , Dnf(c)). Proof. We use the linearity of f'(c) to write n n f '(C)(V) _ f '(C)(vkuk) _ vk f '(C)(uk) k=1 k=1 nn / n = E vk f '(c; uk) = E vkDk f (c) k=1 k=1 Th. 12.6 (8) NOTE. The vector Vf(c) in (8) is called the gradient vector off at c. It. is defined at each point where the partials D1 f, ... , D"f exist. The Taylor formula for real-valued f now takes the form f(c+v)=f(c)+ Vf(c)-v+o(Ilvll) asv-+0. 12.6 AN APPLICATION TO COMPLEX-VALUED FUNCTIONS Let f = u + iv be a complex-valued function of a complex variable. Theorem 5.22 showed that a necessary condition for f to have a derivative at a point c is that the four partials D1u, D2u, D1v, D2v exist at c and satisfy the Cauchy-Riemann equations : D1u(c) = D2v(e), D1v(c) = -D2u(c). Also, an example showed that the equations by themselves are not sufficient for existence off '(c). The next theorem shows that the Cauchy-Riemann equations, along with differentiability of u and v, imply existence of f'(c). Theorem 12.6. Let u and v be-two real-valued functions defined on a subset S of the complex plane. Assume also that u and v are differentiable at an interior point c of S and that the partial derivatives satisfy the Cauchy-Riemann equations at c. Then the function f = u + iv has a derivative at c. Moreover, f'(c) = D1u(c) + iD1v(c). Proof. We have f(z) - f(c) = u(z) - u(c) + i{v(z) - v(c)} for each z in S. Since each of u and v is differentiable at c, for z sufficiently near to c we have u(z) - u(c) = Vu(c) (z - c) + 0(11Z -- CID and -v(z) - v(c) = Vv(c) - (z - c) + o(IIz - cll). Matrix of a Linear Function 349 Here we use vector notation and consider complex numbers as vectors in R2. We then have Writing z = x + iy and c = a + ib, we find {VU(C) + i Vv(c)} (z - c) = D1u(c)(x - a) + D2u(c)(y - b) + i {Dlv(c)(x - a) + D2v(c)(y - b)} = D1u(c){(x - a) + i(y - b)} + iD1v(c){(x - a) + i(y - b)}, because of the Cauchy-Riemann equations. Hence f(z) - f(c) = {D1u(c) + iD1v(c)} (z - c) + o(Ilz - cll) Dividing by z - c and letting z - c we see that f(c) exists and is equal to D1u(c) + iD1v(c). 12.7 THE MATRIX OF A LINEAR FUNCTION In this section we digress briefly to record some elementary facts from linear algebra that are useful in certain calculations with derivatives. Let T:-IV -i- Rm be a linear function. (In our applications, T will be the total derivative of a function f.) We will show that T determines an m x n matrix of scalars (see (9) below) which is obtained as follows : Let u1, ... , u" denote the unit coordinate vectors in R". If x e R" we have x = x1u1 + + x"u" so, by linearity, T(x) = E xkT(uk). k=1 Therefore T is completely determined by its action on the coordinate vectors u1, , u". Now let e1, ... , em denote the unit coordinate vectors in R. Since T(uk) a R'", we can write T(uk) as a linear combination of e1, ... , em, say T(uk) _ tike,-The scalars tlk, , tmk are the coordinates of T(uk). We display these scalars vertically as follows : t1k t2k tmk 350 Multivariable Differential Calculus This array is called a column vector. We form the column vector for each of T(u1), ... , T(u") and place them side by side to obtain the rectangular array (9) This is called the matrix of T and is denoted by m(T). It consists of m rows and n columns. The numbers going down the kth column are the components of T(uk). We also use the notation m(T) = [tik]inkn 1 or m(T) = (tik) to denote the matrix in (9). Now let T : R" - Rm and S : Rm -+ RP be two linear functions, with the domain of S containing the range of T. Then we can form the composition S o T defined by (S o T)(x) = S[T(x)] for all x in W. The composition S o T is also linear and it maps R" into RP. Let us calculate the matrix m(S d T). Denote the unit coordinate vectors in R", Rm, and RP, respectively, by ul, ... , u", e1, ... , em, and w1, , wP. Suppose that S and T have matrices (sij) and (tij), respectively. This means that P S(ek) = E sikWi fork = 1, 2, ... , m and i=1 M T(uj) = E tkjek for j = 1, 2,..., n. Then k=1 mr mr PP (S o T)(uj) = S[T(uj)] = Lj tk jS(ek) = Lj tkj SikWi k=1 k=1 i=1 P m Siktkj Wi i=1 k=1 so M P.n m(SoT) = C Siktkj J k=1 ,j=1 In other words, m(S o T) is a p x n matrix whose entry in the ith row and jth More precisely, the matrix of T relative to the given bases u1, ... , u,, of R" and el,... , em of Rm. The Jacobian Matrix 351 column is Siktkj, k=1 the dot product of the ith row of m(S) with thejth column of m(T). This matrix is also called the product m(S)m(T). Thus, m(S o T) = m(S)m(T). 12.8 THE JACOBIAN MATRIX Next we show how matrices arise in connection with total derivatives. Let f be a function with values in Rm which is differentiable at a point c in R", and let T = f'(c) be the total derivative of f at c. To find the matrix of T we consider its action on the unit coordinate vectors u1, ... , u,,. By Theorem 12.3 we have T(uk) = f'(c; Uk) = Dkf(c). To express this as a linear combination of the unit coordinate vectors e1, ... , em of Rm we write f = (f1, ... , fm) so that Dkf = (Dkf1, ... , Dkfm), and hence T(uk) = Dkf (c) _ Dkfi(c)ei i=1 Therefore the matrix of T is m(T) = (Dk fi(c)). This is called the Jacobian matrix of f at c and is denoted by Df(c). That is, D1f1(c) D2f1(c) ... D"f1(c) Df(c) = D1f2(c) D2f2(c) ... D"f2(c) (10) Dlfm(C) D2fm(C) ... D,,fm(C) The entry in the ith row and kth column is Dkfi(c). Thus, to get the entries in the kth column, differentiate the components of f with respect to the kth coordinate vector. The Jacobian matrix Df(c) is defined at each point c in R" where all the partial derivatives Dk fi(c) exist. The kth row of the Jacobian matrix (10) is a vector in R" called the gradient vector of fk, denoted by Vfk(c). That is,/ Vfk(c) = (Dlfk(c), ... , Dnfk(c)). In the special case when f is real-valued (m = 1), the Jacobian matrix consists of only one row. In this case Df(c) = Vf(c), and Equation (8) of Theorem 12.5 shows that the directional derivative f'(c; v) is the dot product of the gradient vector Vf(c) with the direction v. For a vector-valued function f = (f1, ... , fm) we have m 1 f'(C)(V) = f'(C; V) = E fk(c; v)ek = r L f {Vfk(C) . VJek, (11) k=1 k=1 352 Multivariable Differential Calculus Th. 12.7 so the vector f'(c)(v) has components (Qf1(C) - V, ... , Vfm(C) - V). Thus, the components of f'(c)(v) are obtained by taking the dot product of the successive rows of the Jacobian matrix with the vector v. If we regard f'(c)(v) as an m x 1 matrix, or column vector, then f'(c)(v) is equal to the matrix product Df(c)v, where Df(c) is the m x n Jacobian matrix and v is regarded as an n x 1 matrix, or column vector. NOTE. Equation (11), used in conjunction with the triangle inequality and the Cauchy-Schwarz inequality, gives us II f'(c)(v)II = Therefore we have m m m E {VA(C) v}ekll < E IVfk(C) . VI s Ilvll E IIVfk(C)II k=1 k=1 k=1 Ilf'(c)(v)II < MllvMI, (12) where M = Ek=1 II Vfk(c)II This inequality will be used in the proof of the chain rule. It also shows that f'(c)(v) - 0 as v - 0. 12.9 THE CHAIN RULE Let f and g be functions such that the composition h = f o g is defined in a neighborhood of a point a. The chain rule tells us how to compute the total derivative of h in terms of total derivatives of f and of g. Theorem 12.7. Assume that g is differentiable at a, with total derivative g'(a). Let b = g(a) and assume that f is differentiable at b, with total derivative f'(b). Then the composite function h = f o g is differentiable at a, and the total derivative h'(a) is given by h'(a) = f'(b) o g'(a), the composition of the linear functions f'(b) and g'(a). Proof. We consider the difference h(a + y) -,h(a) for small Ilyll, and show that we have a first-order Taylor formula. We have h(a + y) - h(a) = f [g(a + y)] - f [g(a)] = f(b + v) - f(b), (13) where b = g(a) and v = g(a + y) - b. The Taylor formula for g(a + y) implies v = g'(a)(y) + Ilyll E,(y), where EQ(y) - 0 as y -> 0. (14) The Taylor formula for f(b + v) implies f(b + v) - f(b) = f'(b)(v) + IIvii Eb(v), where Eb(v) - 0 as v -+ 0. (15) Matrix Form of the Chain Rule 353 Using (14) in (15) we find f(b + v) - f(b) = f'(b)[g'(a)(y)] + f'(b)[IIYIl E.(Y)] + Ilvll Eb(v) = f'(b)[g'(a)(y)] + IIYII E(y), where E(0) = 0 and (16) E(y) = f'(b)[E.(Y)] + IIII Vll Eb(v) if Y # 0. (17) IIYII To complete the proof we need to show that E(y) -' 0 as y -' 0. The first term on the right of (17) tends to 0 as y -- 0 because E.(y) -+ 0. In the second term, the factor Eb(v) -+ 0 because v - 0 as y -+ 0. Now we show that the quotient Ilvll/IIYII remains bounded as y - 0. Using (14) and (12) to estimate the numerator we find IIYII <- Ilg'(a)(Y)II + IIYII IIE.(Y)II <- IIYII{M + IIE.(Y)II}, where M = Ek-1 II Vgk(a)II. Hence II VII < M + IIE.(Y)il, IIYII so Ilvll/IIYII remains bounded as y -> 0. Using (13) and (16) we obtain the Taylor formula h(a + y) - h(a) = f'(b)[g'(a)(y)] + IIYII E(y), where E(y) -+ 0 as y -+ 0. This proves that h is differentiable at a and that its total derivative at a is the composition f'(b) o g'(a). 12.10 MATRIX FORM OF THE CHAIN RULE The chain rule states that h'(a) = f'(b) ° g'(a), (18) where h = f 0 g and b = g(a). Since the matrix of a composition is the product of the corresponding matrices, (18) implies the following relation for Jacobian matrices : Dh(a) = Df(b)Dg(a). (19) This is called the matrix form of the chain rule. It can also be written as a set of scalar equations by expressing each matrix in terms of its entries. Specifically, suppose that a e RP, b = g(a) a R", and f (b) a Rm. Then h(a) a Rm and we can write g = (91, ... , 9.), f = (.f1, . . . ,.fm), h = (h1,. .. , hm). Then Dh(a) is an m x p matrix, Df (b) is an m x n matrix, and Dg(a) is an n x p 354 Multivariable Differential Calculus matrix, given by Dh(a) = [Djhi(a)]m;-P 1, Th. 12.8 Df(b) = [Dkff (b)]mkn=1, Dg(a) = [Djgk(a)]k;1= 1 The matrix equation (19) is equivalent to the mp scalar equations n D;hi(a) _ E Dk gk(a), for i = 1, 2, ... , m and j = 1, 2, ... , p. (20) k=1 These equations express the partial derivatives of the components of h in terms of the partial derivatives of the components of f and g. The equations in (20) can be put in a form that is easier to remember. Write y = f(x) and x = g(t). Then y = f[g(t)] = h(t), and (20) becomes aYi aYiaXk (21) at k=1 axk at; where ay' = D h, ayi = DkJi, and axk = Djgk at; ' axk at; Examples. Suppose m = 1. Then both f and h = f o g are real-valued and there are p equations in (20), one for each of the partial derivatives of h: n Djh(a) = E Dkf(b)Djgk(a), p. k=1 The right member is the dot product of the two vectors Vf(b) and Djg(a). In this case Equation (21) takes the form ay " ay axk atj = axk atj 2,...,p. In particular, if p = 1 we get only one equation, h'(a) = E Dkf(b)gk(a) = Vf(b) Dg(a), k=1 where the Jacobian matrix Dg(a) is a column vector. The chain rule can be used to give a simple proof of the following theorem for differentiating an integral with respect to a parameter which appears both in the integrand and in the limits of integration. Theorem 12.8. Let f and D2f be continuous on a rectangle [a, b] x [c, d]. Let p and q be differentiable on [e, d], where p(y) a [a, b] and q(y) e [a, b] for each y in [c, d]. Define F by the equation 9(Y) F(y) = f(x, y) dx, if y e [c, d]. P(y) Th. 12.9 Mean-Value Theorem for Differentiable Functions 355 Then F(y) exists for each y in (c, d) and is given by q(Y) F(y) = D2.f(X, Y) dx + .f(q(Y), y)q'(y) - .f(p(Y), y)p'(y). v( y) Proof. Let G(xl, x2, x3) = f x; f (t, x3) dt whenever x, and x2 are in [a, b] and x3 E [c, d]. Then F is the composite function given by F(y) = G(p(y), q(y), y). The chain rule implies F(y) = D1G(p(Y), q(y), y)p'(y) + D2G(p(y), q(y), y)q'(y) + D3G(p(y), q(y), y). By Theorem 7.32, we have D,G(x,, x2, x3) = -f(x,, x3) and D2G(xl, x2, x3) _ f (x2i x3). By Theorem 7.40, we also have xz D3G(x,, X2, X3) = D2.f(t, X3) dt. xl Using these results in the formula for F(y) we obtain the theorem. 12.11 THE MEAN-VALUE THEOREM FOR DIFFERENTIABLE FUNCTIONS The Mean-Value Theorem for functions from R' to R1 states that f(y) - f(x) = f'(z)(Y - x), (22) where z lies between x and y. This equation is false, in general, for vector-valued functions from R" to R', when m > 1. (See Exercise 12.19.) However, we will show that a correct equation is obtained by taking the dot product of each member of (22) with any vector in R, provided z is suitably chosen. This gives a useful generalization of the Mean-Value Theorem for vector-valued functions. In the statement of the theorem we use the notation L(x, y) to denote the line segment joining two points x and y in R". That is, L(x, y) = {tx + (I -t)y:0<t<1}. Theorem 12.9 (Mean-Value Theorem.) Let S be an open subset of R" and assume that f : S -+ R' is differentiable at each point of S. Let x and y be two points in S such that L(x, y) c S. Then for every vector a in R' there is a point z in L(x, y) such that a {f(y) - f(x)} = a -If '(z)(Y - x)}. (23) Proof. Let u = y - x. Since S is open and L(x, y) c S, there is a 6 > 0 such that x + toe S for all real tin the interval (-6, 1 + S). Let a be a fixed vector in R' and let F be the real-valued function defined on (-S, 1 + a) by the equation F(t) = a f(x + tu). Then F is differentiable on (-S, 1 + S) and its derivative is given by F(t) = a f'(x + tu; u) = a {f'(x + tu)(u)}. 356 Multivariable Differential Calculus Th. 12.10 By the usual Mean-Value Theorem we have F(1) - F(O) = F'(0), where 0 < 0 < 1. Now F'(0) = a - {f'(x + Ou)(u)} = a - {f'(z)(y - x)), where z = x + Ou e L(x, y). But F(1) - F(0) = a - {f(y) - f(x)}, so we obtain (23). Of course, the point z depends on F, and hence on a. NoTE. If S is convex, then L(x, y) S for all x, y in S so (23) holds for all x and y in S. Examples 1. If f is real-valued (m = 1) we can take a = 1 in (23) to obtain f(Y) - f(x) = f'(z)(Y - x) = Vf(z) (Y - x). (24) 2. If f is vector-valued and if a is a unit vector in R'", IIaII = 1, Eq. (23) and the Cauchy-Schwarz inequality give us IIf(Y) - f(x)II < IIf'(z)(Y - x)II. Using (12) we obtain the inequality IIf(Y)-f(x)II <MIIY - xli. where M = k 1 II Vfk(z) II . Note that M depends on z and hence on x and y. 3. If S is convex and if all the partial derivatives D f fk are bounded on S, then there is a constant A > 0 such that IIf(Y) - f(x)II < Ally - xli In other words, f satisfies a Lipschitz condition on S. The Mean-Value Theorem gives a simple proof of the following result concern-ing functions with zero total derivative. Theorem 12.10. Let S be an open connected subset of R", and let f : S -+ R' be differentiable at each point of S. If f'(c) = 0 for each c in S, then f is constant on S. Proof. Since S is open and connected, it is polygonally connected . (See Section 4.18.) Therefore, every pair of points x and y in S can be joined by a polygonal arc lying in S. Denote the vertices of this arc by pl, ... , p,, where pl = x and p, = y. Since each segment L(pi+ 1, p) c S, the Mean-Value Theorem shows that a - {f(pr+1) - f(p3} = 0, for every vector a. Adding these equations for i = 1 , 2, ... , r - 1, we find a - {f(y) - f(x)} = 0, for every a. Taking a = f(y) - f(x) we find f(x) = f(y), so f is constant on S. Th. 12.11 Sufficient Condition for Differentiability 357 12.12 A SUFFICIENT CONDITION FOR DIFFERENTIABILITY Up to now we have been deriving consequences of the hypothesis that a function is differentiable. We have also seen that neither the existence of all partial deriv-atives nor the existence of all directional derivatives suffices to establish differ-entiability (since neither implies continuity). The next theorem shows that continuity of all but one of the partials does imply differentiability. Theorem 12.11. Assume that one of the partial derivatives D1f, ... , Dnf exists at c and that the remaining n - 1 partial derivatives exist in some n-ball B(c) and are continuous at c. Then f is differentiable at c. Proof. First we note that a vector-valued function f = (f1, ... , fn) is differentiable at c if, and only if, each componentfk is differentiable at c. (The proof of this is an easy exercise.) Therefore, it suffices to prove the theorem when f is real-valued. For the proof we suppose that D1 f(c) exists and that the continuous partials The only candidate forf'(c) is the gradient vector Vf(c). We will prove that f(c + v) - f(c) = Vf(c) v + o(IlvIj) as v - 0, and this will prove the theorem. The idea is to express the differencef(c + v) - f(c) as a sum of n terms, where the kth term is an approximation to Dkf(c)vk. For this purpose we write v = Ay, where flyjj = 1 and A = 1(vUU. We keep A small enough so that c + v lies in the ball B(c) in which the partial derivatives D2 f, ... , Dn f exist. Expressing y in terms of its components we have Y = y1u1 + ...+ ynun, where uk is the kth unit coordinate vector. Now we write the differencef(c + v) -f(c) as a telescoping sum, 0 f(c + v) - f(c) = f(c + 1y) - f(c) = E {f(c + Avk) - f(c + Avk-1)}, (25) where YO = 0, V1 = y1u1, V2 = y1u1 + Y2U2, ... , Vn = y1111 + ... + ynun The first term in the sum is f(c + Ay1u1) - f(c). Since the two points c and c + Ay1u1 differ only in their first component, and since D1 f(c) exists, we can write f({/ c + Ay1u1) - f(c) = (c) + Ay1E1(A), where E1(A) -+ 0 as A -+ 0. For k //> 2, the kth term in the sum is {/ f (c + Avk-I + Aykuk) - f (c + Avk- 1) = J (bk + Aykuk) - f (bk), where bk = c +-Avk-1. The two points bk and bk + Aykuk differ only in their kth component, and we can apply the one-dimensional Mean-Value Theorem for 358 derivatives to write Multivariable Differential Calculus f(bk + AYkuk) - f(bk) = AYkDkf(ak), (26) where ak lies on the line segment joining bk to bk + A.ykuk. Note that bk -+ c and hence ak -+ c as A -+ 0. Since each Dk f is continuous at c for k z 2 we can write Dkf(ak) = Dkf(C) + Ek('O, where Ek(2) -), 0 as X Using this in (26) we find that (25) becomes where n n f(C + v) - f(c) = A E Dkf(C)Yk + A E YAW k=1 k=1 = IIv1IE(A), n E (A) _ YkEkW -+ 0 as II v II - 0-k=1 0. This completes the proof. NOTE. Continuity of at least n - 1 of the partials D1f, ... , D"f at c, although sufficient, is by no means necessary for differentiability of f at c. (See Exercises 12.5 and 12.6.) 12.13 A SUFFICIENT CONDITION FOR EQUALITY OF MIXED PARTIAL DERIVATIVES The partial derivatives D1f, ... , DJ of a function from R" to R' are themselves functions from R" to R' and they, in turn, can have partial derivatives. These are called second-order partial derivatives. We use the notation introduced in Chapter 5 for real-valued functions: 82 D, kf = D,(Dkf) = f ax,axk Higher-order partial derivatives are similarly defined. The example f(x, Y) = xY(x2 - YZ)I(x2 + Y2) if (x, y) 0 (0, 0), t0 if (x, y) _ (0, 0), shows that D1,2f(x, y) is not necessarily the same as D2,1f(x, y). In fact, in this example we have P1f(x, y) = Y(x4 + 4xzYz -Ya) 2 2 2 0) y) (0 if (x , , , ) + y (x Th. 12.12 Sufficient Condition for Equality of Mixed Partials 359 and D1 f(0, 0) = 0. Hence, D1 f(0, y) _ -y for all y and therefore D2, 1f(0, Y) _ -1, D2,1f(0, 0) _ -1. On the other hand, we have D2f(x, Y) = x(x4 x (x2 2y2)2 Y4) if (x, Y) 91 (0, 0), and D2 f(0, 0) = 0, so that D2 f(x, 0) = x for all x. Therefore, D1,2 f(x, 0) = 1, D1,2 f(0, 0) = 1, and we see that D2,1 f(0, 0) D1,2f(0, 0). The next theorem gives us a criterion for determining when the two mixed partials D1,2f and D2 1f will be equal. Theorem 12.12. If both partial derivatives D,f and Dkf exist in an n-ball B(c; b) and if both are differentiable at c, then Dr,kf(c) = Dk,,f(c). (27) P r o o f . If f = (fl, ... ,f m) , then Dkf = (Dkfl, ... , Dkfm). Therefore it suffices to prove the theorem for real-valued f Also, since only two components are involved in (27), it suffices to consider the case n = 2. For simplicity, we assume that c = (0, 0). We shall prove that D1,2f(0, 0) = D2,1f(0, 0) Choose h # 0 so that the square with vertices (0, 0), (h, 0), (h, h), and (0, h) lies in the 2-ball B(0; a). Consider the quantity 0(h) = f(h, h) - f(h, 0) - f(0, h) + f(0, 0). We will show that 0(h)lh2 tends to both D2,1f(0, 0) and D1,2f(0, 0) as h - 0. Let G(x) = f(x, h) - f(x, 0) and note that 0(h) = G(h) - G(0). (28) By the one-dimensional Mean-Value Theorem we have G(h) - G(0) = hG'(xl) = h{Dlf(xl, h) - D1f(xl, 0)), (29) where x1 lies between 0 and h. Since DI f is differentiable at (0, 0), we have the first-order Taylor formulas D1f(xl, h) = D1f(0, 0) + D1,1f(0, 0)x1 + D2,l.f(0, 0)h + (xi + h2)''2E1(h), and D1f(x1, 0) = D1f(0, 0) + D1,1f(0, 0)x1 + Ix11 E2(h), where E1(h) and E2(h) - 0 as h - 0. Using these in (29) and (28) we find 0(h) = D2,1f(0, 0)h2 + E(h), 360 Multivariable Differential Calculus Th. 12.13 where E(h) = h(xi + h2)1'2E1(h) + hlx1I E2(h). Since Ix,I < Ihl, we have 0 < IE(h)I S -,/2 h2 IE1(h)I + h2 IE2(h)I, so lim 0(h) = D2,1f(0, 0) h-+o h2 Applying the same procedure to the function H(y) = f(h, y) - f(0, y) in place of G(x), we find that lim e(2) = D1 2f(0, 0), h which completes the proof. As a consequence of Theorems 12.11 and 12.12 we have: Theorem 12.13. If both partial derivatives Drf and Dkf exist in an n-ball B(c) and if both Dr kf and Dk rf are continuous at c, then Dr,kf(C) = Dk,rf(C) NOTE. We mention (without proof) another result which states that if Drf, Dkf and Dk,,f are continuous in an n-ball B(c), then D,kf(c) exists and equals Dk,rf(c). If f is a real-valued function of two variables, there are four second-order partial derivatives to consider; namely, D1,1f, D1,2f, D2,1f, and D2,2f. We have just shown that only three of these are distinct if f is suitably restricted. The number of partial derivatives of order k which can be formed is 2k. If all these derivatives are continuous in a neighborhood of the point (x, y), then certpin of the mixed partials will be equal. Each mixed partial is of the form D,1, ... , rkf, where each rr is either 1 or 2. If we have two such mixed partials, Dr1, ... , rkf and Dp1, ... , pkf, where the k-tuple (r1, ... , rk) is a permutation of the k-tuple (pl, ... , pk), then the two partials will be equal at (x, y) if all 2k partials are continuous in a neighborhood of (x, y). This statement can be easily proved by mathematical induction, using Theorem 12.13 (which is the case k = 2). We omit the proof for general k. From this it follows that among the 2k partial derivatives of order k, there are only k + 1 distinct partials in general, namely, those of the form Dr,, ... , rkf where the k-tuple (r1, ... , rk) assumes the following k + I forms : (2,2,...,2), (1,2,2,...,2), (1, 1,2,...,2),..., (1, 1, ..., 1, 2), (1, ... , 1). Similar statements hold, of course, for functions of n variables. In this case, there are nk partial derivatives of order k that can be formed. Continuity of all these partials at a point x implies that D,1, ... , ,J(X) is unchanged when the indices r1, . .. , rk are permuted. Each r, is now a positive integer <n. Th. 12.14 Taylor's Formula 361 12.14 TAYLOR'S FORMULA FOR FUNCTIONS FROM R" TO R1 Taylor's formula (Theorem 5.19) can be extended to real-valued functions f defined on subsets of R". In order to state the general theorem in a form which resembles the one-dimensional case, we introduce special symbols f"(x; t, f,,,(x; ), ... , J"(x; t), for certain sums that arise in Taylor's formula. These play the role of higher-order directional derivatives, and they are defined as follows : If x is a point in R" where all second-order partial derivatives off exist, and if t = (t1, ... , t") is an arbitrary point in R", we write n n f"(x; t) = E E Di.jf(x)tjt1. i=1 j=1 We also define' nn n n f"'(x; t) = E L/ r E Di,j.kf(x)tktjti i=1 j=1 k=1 if all third-order partial derivatives exist at x. The symbol f ('")(x; t) is similarly defined if all mth-order partials exist. These sums are analogous to the formula f'(x; t) _ Dif (x)ti for the directional derivative of a function which is differentiable at x. Theorem 12.14 (Taylor's formula). Assume that f and all its partial derivatives of order 0 such that a + t(b - a) e S for all real t in the interval -S < t < I + S. Define g on (-S, 1 + S) by the equation g(t) = f [a + t(b - a)]. Then f(b) - f(a) = g(1) - g(0). We will prove the theorem by applying the one-dimensional Taylor formula to g, writing m-1 g(1) - g(0) = E 'i g(k)(0) + i g(m)(9), where 0 < 0 < 1. (30) Nowg is a composite function given byg(t) = f [p(t)], where p(t) = a + t(b - a). The kth component of p has derivative pk(t) = bk - ak. Applying the chain rule, 362 Multivariable Differential Calculus we see that g'(t) exists in the interval (-S, 1 + S) and is given by the formula n ge(t) = E Djf p(t) = f '(p(t); b - a). j=1 Again applying the chain rule, we obtain n n g"(t) = E E Dt.Jfp(t)(b1 - ai) =f"(p(t); b - a). i=1 j=1 Similarly, we find that g(m)(t) = f(ml(p(t); b - a). When these are used in (30) we obtain the theorem, since the point z = a + 9(b - a) e L(a, b). EXERCISES Differentiable functions 12.1 Let S be an open subset of R", and let f : S - R be a real-valued function with finite partial derivatives D1 f, ... , Dn f on S. If f has a local maximum or a local minimum at a point c in S, prove that Dk f (c) = 0 for each k. 12.2 Calculate all first-order partial derivatives and the directional derivative f'(x; u) for each of the real-valued functions defined on R" as follows: a) f(x) = a x, where a is a fixed vector in W. b) f(x) = IjxII4. c) f(x) = x L(x), where L : R" - R" is a linear function. n n d) f (x) = E E ai jxix j, where ai j = a ji. t=1 J=1 12.3 Let f and g be functions with values in R' such that the directional derivatives f'(c; u) and g'(c; u) exist. Prove that the sum f + g and dot product f g have directional derivatives given by and (f + g)'(c; u) = f'(c; u) + g'(c; u) (f g)'(c; u) = f(c) g'(c; u) + g(c) f'(c; u). 12.4 If S S R", let f : S - R'bea function with values in R, and write f = (fi, ... ,fn) Prove that f is differentiable at an interior point c of S if, and only if, each ft is differentiable at c. 12.5 Given n real-valued functions fi, ... ,1,, each differentiable on an open interval (a, b) in R. For each x = (x1, ... , x") in the n-dimensional open interval S = {(xi, ... , x,.): a < xk < b, k = 1, 2, ... , n}, define f(x) = f1(x1) + + f"(x"). Prove that f is differentiable at each point of S and that n f'(x)(u) = E fi(xt)ut, where u = (u1, ... , un). i=1 Exercises 363 12.6 Given n real-valued functions fl, . . . , f" defined on an open set S in R". For each x in S, define f (x) = ft (x) + + f"(x). Assume that for each k = 1, 2, ... , n, the following limit exists: lim My) - fk(x) Y_X Yk#Xk Call this limit ak(x). Prove that f is differentiable at x and that n f'(x)(u) _ E ak(X)Uk if u = (u1, ... , U"). k=1 12.7 Let f and g be functions from R" to R. Assume that f is differentiable at c, that f(c) = 0, and that g is continuous at c. Let h(x) = g(x) f(x). Prove that h is differen-tiable at c and that h'(c)(u) = g(c) {f'(c)(u) } if u e W. 12.8 Let f : R2 -+ R3 be defined by the equation f(x, y) = (sin x cos y, sin x sin y, cos x cos y). Determine the Jacobian matrix Df(x, y). 12.9 Prove that there is no real-valued function f such that f'(c; u) > 0 for a fixed point c in W and every nonzero vector u in W. Give an example such that f'(c; u) > 0 for a fixed direction u and every c in W. 12.10 Let f = u + iv be a complex-valued function such that the derivative f(c) exists for some complex c. Write z = c + re" (where a is real and fixed) and let r -+ 0 in the difference quotient [f(z) - f(c)]/(z - c) to obtain f'(c) = e-ta[u'(c; a) + iv'(c; a)], where a = (cos a, sin a), and u'(c; a) and v'(c; a) are directional derivatives. Let b = (cos f, sin f), where f = a + 1n, and show by a similar argument that f(c) = e-ta[v'(c; b) - iu'(c; b)]. Deduce that u'(c; a) = v'(c; b) and v'(c; a) u'(c; b). The Cauchy-Riemann equa-tions (Theorem 5.22) are a special case. Gradients and the chain rule 12.11 Let f be real-valued and differentiable at a point c in R", and assume that 11 Vf (c)11 0 0. Prove that there is one and only one unit vector u in W such that If'(c; u)l = 11 Vf (c) 11, and that this is the unit vector for which I f'(c; u)j has its maximum value. 12.12 Compute the gradient vector Vf(x, y) at those points (x, y) in R2 where it exists: a) f(x, y) = x2y2 log (x2 + y2) if (x, y) (0, 0), f(0, 0) = 0. b) f(x, y) = xy sin 1 if (x, y) (0, 0), f(0, 0) = 0. x2 + y2 364 Multivariable Differential Calculus 12.13 Let f and g be real-valued functions defined on R1 with continuous second deriva-tives f" and g". Define F(x, y) = f [x + g(y)] for each (x, y) in R2. Find formulas for all partials of F of first and second order in terms of the derivatives of f and g. Verify the relation (D1F)(D1.2F) = (D2F)(D1,1F) 12.14 Given a function f defined in R2. Let F(r, 0) = f(r cos 0, r sin 6). a) Assume appropriate differentiability properties off and show that D1F(r, 6) = cos 0 Dlf(x, y) + sin 0 D2f(x, y), D1,1F(r, 6) = cost 9D1,1f(x, y) + 2 sin 6cos 9 D1,2f(x, y) + sin 2 9D2,2f(x, y), where x = r cos 0, y = r sin 6. b) Find similar formulas for D2F, D1,2F, and D2,2F. c) Verify the formula IIVf(r cos 6, r sin 6)112 = [D1F(r, 9)]2 + [D2F(r, 9)]2. s r 12.15 If f and g have gradient vectors Vf(x) and Vg(x) at a point x in R" show that the product function h defined by h(x) = f(x)g(x) also has a gradient vector at x and that Vh(x) = f(x)Vg(x) + g(x)Vf(x). State and prove a similar result for the quotient f/g. 12.16 Let f be a function having a derivative f' at each point in R1 and let g be defined on R3 by the equation 9(x,Y,z)=x2+y2+ z2. If h denotes the composite function h = f o g, show that II Vh(x, y, z)112 = 49(x, y, z){f'[9(x, y, z)]}2 12.17 Assume f is differentiable at each point (x, y) in R2. Let g1 and g2 be defined on R3 by the equations 91(x, Y, z)=x2+Y2+ z2, 92(x, Y,z)=x+y+z, and let g be the vector-valued function whose values (in R2) are given by g(x, Y, z) _ (91(x, Y, z), 92(x, Y, Z))' Let h be the composite function h = f o g and show that IIohII2 = 4(D1f)291 + 4(D1f)(D2f)92 + 3(D2f)2. 12.18 Let f be defined on an open set S in R". We say that f is homogeneous of degree p over S if f(Ax) = 2°f(x) for every real A and for every x in S for which Ax e S. If such a Exercises 365 function is differentiable at x, show that x - Vf(x) = pf(x). NOTE. This is known as Euler's theorem for homogeneous functions. Hint. For fixed x, define g(A) = f(Ax) and compute g'(1). Also prove the converse. That is, show that if x - Vf(x) = pf(x) for all x in an open set S, then f must be homogeneous of degree p over S. Mean-Value theorems 12.19 Let f : R -+ R2 be defined by the equation f(t) = (cos t, sin t). Then f'(t)(u) _ u(- sin t, cos t) for every real u. The Mean-Value formula f(y) - f(x) = f'(z)(y - x) cannot hold when x = 0, y = 2ir, since the left member is zero and the right member is a vector of length 2n. Nevertheless, Theorem 12.9 states that for every vector a in R2 there is a z in the interval (0, 2n) such that a - {f(y) - f(x)} = a - {f'(z)(y - x)}. Determine z in terms of a when x = 0 and y = 2n. 12.20 Let f be a real-valued function differentiable on a 2-ball B(x). By considering the function g(t) = f[tyl + (1 - t)x1, Y21 + f[x1, tY2 + (1 - t)x2] prove that f(y) - f(x) = (yl - x1)D1f(z1, y2) + (Y2 - x2)D2f(x1, z2), where zl a L(xl, yl) and z2 E L(x2, Y2)-12.21 State and prove a generalization of the result in Exercise 12.20 for a real-valued function differentiable on an n-ball B(x). 12.22 Let f be real-valued and assume that the directional derivative f'(c + tu; u) exists for each tin the interval '0 < t < 1. Prove that for some 0 in the open interval (0, 1) we have f(c + u) - f(c) = f'(c + 9u; u). 12.23 a) If f is real-valued and if the directional derivativef'(x; u) = 0 for every x in an n-ball B(c) and every direction u, prove that f is constant on B(c). b) What can you conclude about f if f'(x; u) = 0 for a fixed direction u and every x in B(c)? Derivatives of higher order and Taylor's formula 12.24 For each of the following functions, verify that the mixed partial derivatives D1,2f and D2,1 If are equal. a) f(x, y) = x4 + y4 - 4x2y2. b) f(x, y) = log (x2 + y2), (x, y) t (0, 0). c) f (x, y) = tan (x2/y), y # 0. 366 Multivariable Differential Calculus 12.25 Let f be a function of two variables. Use induction and Theorem 12.13 to prove that if the 2k partial derivatives off of order k are continuous in a neighborhood of a point (x, y), then all mixed partials of the form Drl and DPI. will be equal at (x, y) if the k-tuple (r1, ... , r r) contains the same number of ones as the k-tuple ( P1 , . . . , pk). 12.26 If f is a function of two variables having continuous partials of order k on some open set S in R2, show that k (k) f(k)(X; t) _ tit2-rD {' D1, ... , Pk (X), if X E S, t = (t1, t2), r=O r where in the rth term we have pl = = pr = I and pr+1 = . = pk = 2. Use this result to give an alternative expression for Taylor's formula (Theorem 12.14) in the case when n = 2. The symbol (kr) is the binomial coefficient k!/[r! (k - r)!]. 12.27 Use Taylor's formula to express the following in powers of (x - 1) and (y - 2): a) f(x, y) = x3 + y3 + xy2, b) f(x, y) = x2 + xy + y2. SUGGESTED REFERENCES FOR FURTHER STUDY 12.1 Apostol, T. M., Calculus, Vol. 2, 2nd ed. Xerox, Waltham, 1969. 12.2 Chaundy, T. W., The Differential Calculus. Clarendon Press, Oxford, 1935. 12.3 Woll, J. W., Functions of Several Variables. Harcourt Brace and World, New York, 1966. CHAPTER 13 IMPLICIT FUNCTIONS AND EXTREMUM PROBLEMS 13.1 INTRODUCTION This chapter consists of two principal parts. The first part discusses an important theorem of analysis called the implicit function theorem; the second part treats extremum problems. Both parts use the theorems developed in Chapter 12. The implicit function theorem in its simplest form deals with an equation of the form f(x, t) = 0. (1) The problem is to decide whether this equation determines x as a function of t. If so, we have x = g(t), for some function g. We say that g is defined "implicitly" by (1). The problem assumes a more general form when we have a system of several equations involving several variables and we ask whether we can solve these equations for some of the variables in terms of the remaining variables. This is the same type of problem as above, except that x and t are replaced by vectors, and f and g are replaced by vector-valued functions. Under rather general con-ditions, a solution always exists. The implicit function theorem gives a description of these conditions and some conclusions about the solution. An important special case is the familiar problem in algebra of solving n linear equations of the form n E aijxj t; (i = 1, 2, ... , n), (2) j=1 where the ai j and ti are considered as given numbers and x1, ... , x represent unknowns. In linear algebra it is shown that such a system has a unique solution if, and only if, the determinant of the coefficient matrix A = [ai j] is nonzero. NOTE. The determinant of a square matrix A = [aij] is denoted by det A or det [ai j]. If det [ai j] # 0, the solution of (2) can be obtained by Cramer's rule which expresses each xk as a quotient of two determinants, say xk = Ak/D, where D = det [ai j] and A. is the determinant of the matrix obtained by replacing the kth column of [ai j] by t 1, ... , tn. (For a proof of Cramer's rule, see Reference 13.1, Theorem 3.14.) In particular, if each t, = 0, then each xk = 0. 367 368 Implicit Functions and Extremum Problems Th. 13.1 Next we show that the system (2) can be written in the form (1). Each equation in (2) has the form fi(x, t) = 0 where x = (xl, ... , t = (tl, and / i(X, t) _ ai jx j - ti. j=1 Therefore the system in (2) can be expressed as one vector equation f(x, t) = 0, where f = (f1i ... , f ). If D jfi denotes the partial derivative off; with respect to the j th coordinate xj, then D j f i(x, t) = ai j. Thus the coefficient matrix A = [ai j] in (2) is a Jacobian matrix. Linear algebra tells us that (2) has a unique solution if the determinant of this Jacobian matrix is nonzero. In the general implicit function theorem, the nonvanishing of the determinant of a Jacobian matrix also plays a role. This comes about by approximating f by a linear function. The equation f(x, t) = 0 gets replaced by a system of linear equations whose coefficient matrix is the Jacobian matrix of f. NOTATION. If f = (fl, ... , and x = (x1, ... , x.), the Jacobian matrix Df(x) = [D j fi(x)] is an n x n matrix. Its determinant is called a Jacobian determinant and is denoted by Jf(x). Thus, Jf(x) = det Df(x) = det [Djfi(x)]. The notation a(fl, ... I A) a(x1, ... , x,.) is also used to denote the Jacobian determinant Jf(x). The next theorem relates the Jacobian determinant of a complex-valued function with its derivative. Theorem 13.1. If f = u + iv is a complex-valued function with a derivative at a point z in C, then Jf(z) _ I f'(z)12. Proof We have f'(z) = Dlu + iD1v, so I f'(z)j2 = (Dlu)2 + (Dlv)2. Also, J f(z) = det Dlu D2ul = Dlu D2v - Dlv D2u = (Dlu)2 + (Dlv)2, 1Dly D2vJ by the Cauchy-Riemann equations. 13.2 FUNCTIONS WITH NONZERO JACOBIAN DETERMINANT This section gives some properties of functions with nonzero Jacobian determinant at certain points. These results will be used later in the proof of the implicit function theorem. Th. 13.2 Nonzero Jacobian Determinant f f(B) f(a) Figure 13.1 369 Theorem 13.2. Let B = B(a; r) be an n-ball in R", let 8B denote its boundary, 8B= {x:llx-all=r}, and let B = B u 8B denote its closure. Let f = (f1, ... , f") be continuous on B, and assume that all the partial derivatives Dj f;(x) exist if x e B. Assume further that f(x) 0 f(a) if x e 8B and that the Jacobian determinant JJ(x) : 0 for each x in B. Then f(B), the image of B under f, contains an n-ball with center at f(a). Proof. Define a real-valued function g on 8B as follows: g(x) = Ilf(x) - f(a)ll ifxeaB. Then g(x) > 0 for each x in 8B because f(x) # f(a) if x e 8B. Also, g is continuous on 8B since f is continuous on B. Since 8B is compact, g takes on its absolute minimum (call it m) somewhere on 8B. Note that m > 0 since g is positive on 8B. Let T denote the n-ball T = B(f(a); 2) . We will prove that T c f(B) and this will prove the theorem. (See Fig. 13.1.) To do this we show that y e T implies y e f(B). Choose a point y in T, keep y fixed, and define a new real-valued function h on B as follows : h(x) = IIf(x) - Yll ifxeB. Then h is continuous on the compact set B and hence attains its absolute minimum on B. We will show that h attains its minimum somewhere in the open n-ball B. At the center we have h(a) = Ilf(a) - yll < m/2 since y e T. Hence the minimum value of h in B must also be - llf(x) - f(a)ll - Ilf(a) - YII > g(x) - 2 >- 2 so the minimum of h cannot occur on the boundary B. Hence there is an interior point c in B at which h attains its minimum. At this point the square of h also has 370 Implicit Functions and Extremum Problems a minimum. Since '(X) - y,.]2, h2(x) = Ilf(x) - YIIZ = [f Th. 13.3 =1 and since each partial derivative Dk(h2) must be zero at c, we must have n E [f,(c) - y,]Dkf,(c) = 0 for k = 1, 2, ... , n. r=1 But this is a system of linear equations whose determinant Jf(c) is not zero, since c e B. Therefore f,(c) = y, for each r, or f(c) = y. That is, y ef(B). Hence T s f(B) and the proof is complete. A function f : S -+ T from one metric space (S, ds) to another (T, dT) is called an open mapping if, for every open set A in S, the image f(A) is open in T. The next theorem gives a sufficient condition for a mapping to carry open sets onto open sets. (See also Theorem 13.5.) Theorem 13.3. Let A be an open subset of R" and assume that f : A R" is con-tinuous and has finite partial derivatives D3 f on A. If f is one-to-one on A and if Jf(x) 0 for each x in A, then f(A) is open. Proof. If b e f(A), then b = f(a) for some a in A. There is an n-ball B(a; r) c A on which f satisfies the hypotheses of Theorem 13.2, so f(B) contains an n-ball with center at b. Therefore, b is an interior point of f(A), so f(A) is open. The next theorem shows that a function with continuous partial derivatives is locally one-to-one near a point where the Jacobian determinant does not vanish. Theorem 13.4. Assume that f = (fl, ... , f") has continuous partial derivatives Dj f, on an open set S in R", and that the Jacobian determinant Jf(a) 0 0 for some point a in S. Then there is an n-ball B(a) on which f is one-to-one. Proof. Let Z1, ... , Z. be n points in S and let Z = (Z1; ... ; Z") denote that point in R"Z whose first n components are the components of Z1, whose next n components are the components of Z2, and so on. Define a real-valued function h as follows : h(Z) = det [D; f (Z)]. This function is continuous at those points Z in R"2 where h(Z) is defined because each D3 f is continuous on S and a determinant is a polynomial in its n2 entries. Let Z be the special point in R"2 obtained by putting Z1 = Z2 = ... = Z" = a. Then h(Z) = Jf(a) # 0 and hence, by continuity, there is some n-ball B(a) such that det [D j f (Z)] # 0 if each Z, e B(a). We will prove that f is one-to-one on B(a). Th. 13.5 Nonzero Jacobian Determinant 371 Assume the contrary. That is, assume that f(x) = f(y) for some pair of points x # y in B(a). Since B(a) is convex, the line segment L(x, y) c B(a) and we can apply the Mean-Value Theorem to each component of f to write 0 = .fi(y) - .fi(x) = Vfi(Z) - (y - x) for i = 1, 2, ... , n, where each Zi a L(x, y) and hence Zi a B(a). (The Mean-Value Theorem is applicable because f is differentiable on S.) But this is a system of linear equations of the form " r (Yk - xk)aik = 0 with aik = Dkf(Zi) The determinant of this system is not zero, since Zi e B(a). Hence yk - xk = 0 for each, k, and this contradicts the assumption that x # y. We have shown, therefore, that x # y implies f(x) # f(y) and hence that f is one-to-one on B(a). NOTE. The reader should be cautioned that Theorem 13.4 is a local theorem and not a global theorem. The nonvanishing of Jf(a) guarantees that f is one-to-one on a neighborhood of a. It does not follow that f is one-to-one on S, even when Jf(x) # 0 for every x in S. The following example illustrates this point. Let f be the complex-valued function defined byf(z) = eZ if z e C. If z = x + iy we have Jf(z) = If'(Z)12 = 1e12 = e2x. Thus Jf(z) # 0 for every z in C. However, f is not one-to-one on C because f(zl) = f(z2) for every pair of points z, and z2 which differ by 27ri. The next theorem gives a global property of functions with nonzero Jacobian determinant. Theorem 13.5. Let A be an open subset of R" and assume that f : A - R" has continuous partial derivatives Dj fi on A. If Jf(x) # 0 for all x in A, then f is an open mapping. Proof Let S be any open subset of A. If x e S there is an n-ball B(x) in which f is one-to-one (by Theorem 13.4). Therefore, by Theorem 13.3, the image f(B(x)) is open in R". But we can write S = U.s B(x). Applying f we find f(S) _ UxEs f(B(x)), so f(S) is open. NOTE. If a function f = (fl, ... ,f") has continuous partial derivatives on a set S, we say that f is continuously differentiable on S, and we write f e C' on S. In view of Theorem 12.11, continuous differentiability at a point implies differentiability at that point. Theorem 13.4 shows that a continuously differentiable function with a non-vanishing Jacobian at a point a has a local inverse in a neighborhood of a. The next theorem gives some local differentiability properties of this local inverse function. -372 Implicit Functions and Extremum Problems Th. 13.6 13.3 THE INVERSE FUNCTION THEOREM Theorem 13.6. Assume f = (fl, ... , f") a C' on an open set S in R", and let T = f(S). If the Jacobian determinant Jf(a) # 0 for some point a in S, then there are two open sets X S S and Y E- T and a uniquely determined function g such that a) a e X and f(a) a Y, b) Y = f(X), c) f is one-to-one on X, d) g is defined on Y, g(Y) = X, and g[f(x)] = x for every x in X, e) g e C' on Y. Proof. The function Jf is continuous on S and, since Jf(a) # 0, there is an n-ball B1(a) such that Jf(x) # 0 for all x in B1(a). By Theorem 13.4, there is an n-ball B(a) g B1(a) on which f is one-to-one. Let B be an n-ball with center at a and radius smaller than that of B(a). Then, by Theorem 13.2, f(B) contains an n-ball with center at f(a). Denote this by Y and let X = f -1(Y) n B. Then X is open since both f -1(Y) and B are open. (See Fig. 13.2.) Figure 13.2 The set B (the closure of B) is compact and f is one-to-one and continuous on B. Hence, by Theorem 4.29, there exists a function g (the inverse function f-1 of Theorem 4.29) defined on f(B) such that g[f(x)] = x for all x in B. Moreover, g is continuous on f(B). Since X c B and Y c f(B), this proves parts (a), (b), (c) and (d). The uniqueness of g follows from (d). Next we prove (e). For this purpose, define a real-valued function h by the equation h(Z) = det [Dj fi(Zl)], where Z1, ... , Z. are n points in S, and Z = (Zr;... ... ; Z") is the corresponding point in R"2. Then, arguing as in the proof of Theorem 13.4, there is an n-ball B2(a) such that h(Z) # 0 if each Zi e B2(a). We can now assume that, in the earlier part of the proof, the n-ball B(a) was chosen so that B(a) c B2(a). Then B c B2(a) and h(Z) # 0 if each Zi e B. To prove (e), write g = (g1, ... , g"). We will show that each gk e C' on Y. To prove that D,gk exists on Y, assume y e Y and consider the difference quotient [9k(Y + tu,) - gk(y)]/t, where u, is the rth unit coordinate vector. (Since Y is Implicit Function Theorem 0 373 open, y + tur e Y if t is sufficiently small.) Let x = g(y) and let x' = g(y + tu,). Then both x and x' are in X and f(x') - f(x) = tu,. Hence f;(x') - f,(x) is 0 if i r, and is t if i = r. By the Mean-Value Theorem we have f (x') - AX) = Vf (Z1) - x' - x for i = 1, 2, ... , n, t t where each Zt is on the line segment joining x and x'; hence Zi a B. The expression on the left is 1 or 0, according to whether i = r or i r. This is a system of n linear equations in n unknowns (x; - xj)lt and has a unique solution, since det [Djf1(Zi)] = h(Z) : 0. Solving for the kth unknown by Cramer's rule, we obtain an expression for [gk(y + tUr) - gk(y)]/t as a quotient of determinants. As t -+ 0, the point x -> x, since g is continuous, and hence each Z; -+ x, since Zi is on the segment joining x to V. The determinant which appears in the denominator has for its limit the number det [Djff(x)] = Jf(x), and this is nonzero, since x e X. Therefore, the following limit exists : lim t 9k(Y + tur) - 9k(Y) = Dr9k(Y)-This establishes the existence of Drgk(y) for each y in Y and each r = 1, 2, ... , n. Moreover, this limit is a quotient of two determinants involving the derivatives D j fi(x). Continuity of the Dj f, implies continuity of each partial D,gk. This completes the proof of (e). NOTE. The foregoing proof also provides a method for computing D,gk(y). In practice, the derivatives Dgk can be obtained more easily (without recourse to a limiting process) by using the fact that, if y = f(x), the product of the two Jacobian matrices Df(x) and Dg(y) is the identity matrix. When this is written out in detail it gives the following system of n2 equations: n E = Dk91(Y)Djjk(x) = 1 if i j, {0 if i j. For each fixed i, we obtain n linear equations as j runs through the values 1, 2, ... , n. These can then be solved for the n unknowns, D1gj(y),... , D g1(y), by Cramer's rule, or by some other method. 13.4 THE IMPLICIT FUNCTION THEOREM The reader knows that the equation of a curve in the xy-plane can be expressed either in an "explicit" form, such as y = f(x), or in an "implicit" form, such as F(x, y) = 0. However, if we are given an equation of the form F(x, y) = 0, this does not necessarily represent a function. (Take, for example, x2 + y2 - 5 = 0.) The equation F(x, y) = 0 does always represent a relation, namely, that set of all 374 Implicit Functions and Extremum Problems Th. 13.7 pairs (x, y) which satisfy the equation. The following question therefore presents itself quite naturally: When is the relation defined by F(x, y) = 0 also a function? In other words, when can the equation F(x, y) = 0 be solved explicitly for y in terms of x, yielding a unique- solution? The implicit function theorem deals with this question locally. It tells us that, give a point (xo, yo) such that F(xo, yo) = 0, under certain conditions there will be a neighborhood of (xo, yo) such that in this neighborhood the relation defined by F(x, y) = 0 is also a function. The conditions are that F and D2F be continuous in some neighborhood of (xo, yo) and that D2F(xo, yo) # 0. In its more general form, the theorem treats, instead of one equation in two variables, a system of n equations in n + k variables: f.(x1, ... , xn; t1, .. - , tk) = 0 (r = 1, 2, ... , n). This system can be solved for x1, .. . , x" in terms of t1, ... , tk, provided that certain partial derivatives are continuous and provided that the n x n Jacobian determinant 8(fl, ... , fn)/8(x1, ... , xn) is not zero. For brevity, we shall adopt the following notation in this theorem: Points in (n + k)-dimensional space R"+k will be written in the form (x; t), where x = (x1,...,xn)ER" and t = (t1,...,tk)eRk. Theorem 13.7 (Implicit function theorem). Let f = (f 1, ... , f") be a vector-valued function defined on an open set S in R"+k with values in R". Suppose f e C' on S. Let (xo; to) be a point in S for which f(xo; to) = 0 and for which the n x n determi-nant det [Djfi(xo; to)] # 0. Then there exists a k-dimensional open set To con-taining to and one, and only one, vector-valued function g, defined on To and having values in R", such that a) g e C' on To, b) g(to) = xo, c) f(g(t); t) = 0 for every t in To. Proof. We shall apply the inverse function theorem to a certain vector-valued function F = (F1, ... , Fn; Fn+1, ... , Fn+k) defined on S and having values in R"+k The function F is defined as follows: For 1 < m < n, let Fm(x; t) = fn(x; t), and for 1 < m < k, let Fn+m(x; t) = We can then write F = (f; I), where f = (fl, ... ,fn) and where I is the identity function defined by I(t) = t for each t in Rk. The Jacobian JF(x; t) then has the same value as the n x n determinant det [Djfi(x; t)] because the terms which appear in the last k rows and also in the last k columns of JF(x; t) form a k x k determinant with ones along the main diagonal and zeros elsewhere; the intersection of the first n rows and n columns consists of the determinant det [Djfi(x; t)], and DiF,+ j(x; t) = 0 for 1 5 i < n, 1 < j 5 k. Hence the Jacobian JF(xo; to) .96 0. Also, F(xo; to) _ (0; to). Therefore, by Theorem 1-3.6, there exist open sets X and Y containing (xo; to) and (0; to), respectively, such that F is one-to-one on X, and X = F-1(Y). Also, there exists Extrema of Real-Valued Functions 375 a local inverse function G, defined on Y and having values in X, such that G[F(x; t)] _ (x; t), and such that G e C' on Y. Now G can be reduced to components as follows: G = (v; w) where v = (v1, ... , v") is a vector-valued function defined on Y with values in R" and w = (w1, ... , wk) is also defined on Y but has values in R'. We can now determine v and w explicitly. The equation G[F(x; t)] = (x; t), when written in terms of the components v and w, gives us the two equations v[F(x; t)] = x and w[F(x; t)] = t. But now, every point (x; t) in Ycan be written uniquely in the form (x; t) = F(x'; t') for some (x'; t') in X, because F is one-to-one on X and the inverse image F-'(Y) contains X. Furthermore, by the manner in which F was defined, when we write (x; t) = F(x'; t'), we must have t' = t. Therefore, v(x; t) = v[F(x'; t)] = x' and w(x; t) = w[F(x'; t)] = t. Hence the function G can be described as follows: Given a point (x; t) in Y, we have G(x; t) = (x'; t), where x' is that point in R" such that (x; t) = F(x'; t). This statement implies that F[v(x; t); t] = (x; t) for every (x; t) in Y. Now we are ready to define the set To and the function g in the theorem. Let To = {t : t e Rk, (0; t) a Y}, and for each tin To define g(t) = v(0; t). The set To is open in Rk. Moreover, g e C' on To because G e C' on Y and the components of g are taken from the components of G. Also, g(to) = v(0; to) = xo because (0; to) = F(xo; to). Finally, the equation F[v(x; t); t] = (x; t), which holds for every (x; t) in Y, yields (by considering the components in R") the equation f[v(x; t); t] = x. Taking x = 0, we see that for every tin To, we have f[g(t); t] = 0, and this completes the proof of statements (a), (b), and (c). It remains to prove that there is only one such function g. But this follows at once from the one-to-one character of f. If there were another function, say h, which satisfied (c), then we would have f[g(t); t] = f[h(t); t], and this would imply (g(t); t) = (h(t); t), or g(t) = h(t) for every tin To. 13.5 EXTREMA OF REAL-VALUED FUNCTIONS OF ONE VARIABLE In the remainder of this chapter we shall consider real-valued functions f with a view toward determining those points (if any) at which f has a local extremum, that is, either a local maximum or a local minimum. 376 Implicit Functions and Extremum Problems Th. 13.8 We have already obtained one result in this connection for functions of one variable (Theorem 5.9). In that theorem we found that a necessary condition for a function f to have a local extremum at an interior point c of an interval is that f'(c) = 0, provided thatf'(c) exists. This condition, however, is not sufficient, as we can see by taking f(x) = x3, c = 0. We now derive a sufficient condition. Theorem 13.8. For some integer n > 1, let f have a continuous nth derivative in the open interval (a, b',. Suppose also that for some interior point c in (a, b) we have f '(c) = f "(c) = ... = f("-1)(c) = 0, but f (")(c) # 0. Then for n even, f has a local minimum at c if f (")(c) > 0, and a local maximum a c if f (")(c) < 0. If n is odd, there is neither a local maximum nor a local minimum at c. Proof. Since f (")(c) # 0, there exists an interval B(c) such that for every x in B(c), the derivative f (")(x) will have the same sign as f (")(c). Now by Taylor's formula (Theorem 5.19), for every x in B(c) we have f(x) - f(c) = f(")(X 1) (x - c)", where x1 a B(c). n ' . If n is even, this equation implies f(x) f(c) when f(")(c) > 0, and f(x) < f(c) when P ")(c) < 0. If n is odd and f (")(c) > 0, then f(x) > f(c) when x > c, but f(x) < f(c) when x < c, and there can be no extremum at c. A similar statement holds if n is odd and f (")(c) < 0. This proves the theorem. 13.6 EXTREMA OF REAL-VALUED FUNCTIONS OF SEVERAL VARIABLES We turn now to functions of several variables. Exercise 12.1 gives a necessary condition for a function to have a local maximum or a local minimum at an interior point a of an open set. The condition is that each partial derivative Dkf(a) must be zero at that point. We can also state this in terms of directional derivatives by saying that f'(&; u) must be zero for every direction u. The converse of this statement is not 'true, however. Consider the following example of a function of two real variables : f(x, y) = (y - x2)(y - 2x2). Here we have D1 f(0, 0) = D2 f(0, 0) = 0. Now f(0, 0) = 0, but the function assumes' both positive and negative values in every neighborhood of (0, 0), so there is neither a local maximum nor a local minimum at (0, 0). (See Fig. 13.3.) This example illustrates another interesting phenomenon. If we take a fixed straight line through the origin and restrict the point (x, y) to move along this line toward (0, 0), then the point will finally enter the region above the parabola y = 2x2 (or below the parabola y = x2) in which f(x, y) becomes and stays positive for every (x, y) # (0, 0). Therefore, along every such line, f has a minimum at (0, 0), but the origin is not a local minimum in any two-dimensional neighbor-hood of (0, 0). Def. 13.9 Extrema of Fwictioas of Several Variables Figure 13.3 377 Definition 13.9. If f is differentiable at a and if Vf(a) = 0, the point a is called a stationary point of f. A stationary point is called a saddle point if every n-ball B(a) contains points x such that f(x) > f(a) and other points such that f(x) < f(a). In the foregoing example, the origin is a saddle point of the function. To determine whether a function of n variables has a local maximum, a local minimum, or a saddle point at a stationary point a, we must determine the algebraic sign of f(x) - f(a) for all x in a neighborhood of a. As in the one-dimensional case, this is done with the help of Taylor's formula (Theorem 12.14). Take m = 2 and y = a + tin Theorem 12.14. If the partial derivatives off are differentiable on an n-ball B(a) then f(a + t) - f(a) = Vf(a) t + If "(z; 0, (3) where z lies on the line segment joining a and a + t, and n f"(z; t) _ E D;,;f(z)tit;. i=1 j=1 At a stationary point we have Vf(a) = 0 so (3) becomes f(a + t) - f(a) = If "(z; 0. Therefore, as a + t ranges over B(a), the algebraic sign of f(a + t) - f(a) is determined by that of f"(z; t). We can write (3) in the form f(a + t) - f(a) = 4f"(a; t) + IItII2E(t), (4) where The inequality II t112E(t) = - f"(z; t) -If'"(a; t). IIt112 IE(t)1.< i E E IDr.;f(z) - Di.if(a)I 11tl12, 21=i j=i shows that E(t) --p 0 as t --p 0 if the second-order partial derivatives of f are continuous at-a. Since 11 t I12E(t) tends to zero faster than II t1l 2, it seems reasonable to expect that the algebraic sign of f(a + t) - f(a) should be determined by that off"(a; t). This is what is proved in the next theorem. 378 Implicit Functions and Extremum Problems Th. 13.10 Theorem 13.10 (Second-derivative test for extrema). Assume that the second-order partial derivatives Di,j f exist in an n-ball B(a) and are continuous at a, where a is a stationary point off. Let 1 Q(t) = if"(a, t) = E E Di.jf(a)titj 2 i=1 j=1 a) If Q(t) > 0 for all t # 0, f has a relative minimum at a. b) If Q(t) < 0 for all t # 0, f has a relative maximum at a. (5) c) If Q(t) takes both positive and negative values, then f has a saddle point at a. Proof The function Q is continuous at each point tin R". Let S = {t : II t1l = 1 } denote the boundary of the n-ball B(0; 1). If Q(t) > 0 for all t # 0, then Q(t) is positive on S. Since S is compact, Q has a minimum on S (call it m), and m > 0. Now Q(ct) = c2Q(t) for every real c. Taking c = 1/11t11 where t # 0 we see that ct E S and hence c2Q(t) >- m, so Q(t) >- m II tIi 2 Using this in (4) we find f(a + t) - f(a) = Q(t) + IIt1l 2E(t) > m 11 t1l 2 + II t1l 2E(t). Since E(t) -+ 0 as t -+ 0, there is a positive number r such that IE(t)I < -m whenever 0 < 11 t1l < r. For such t we have 0 < 11 t112 IE(t)I < 4m11 t112, so f(a+t)-f(a)> mIItfl2 -4m11t112 =4m1It112>0. Therefore f has a relative minimum at a, which proves (a). To prove (b) we use a. similar argument, or simply apply part (a) to -f. Finally, we prove (c). For each A > 0 we have, from (4), f(a + At) - f(a) = Q(At) + A211tII2E(At) = A2{Q(t) + IIt112E(At)}. Suppose Q(t) 0 0 for some t. Since E(y) -+ 0 as y -+ 0, there is a positive r such that IItOI2E(At) < 4IQ(t)I if 0 < A < r. Therefore, for each such A the quantity .2{Q(t) + IItfl2E(At)} has the same sign as Q(t). Therefore, if 0 < A < r, the difference f(a + At) - f(a) has the same sign as Q(t). Hence, if Q(t) takes both positive and negative values, it follows that f has a saddle point at a. NOTE. A real-valued function Q defined on R" by an equation of the type nn Q(X) = r LI L. r aijxixj, i=1 j=1 where x = (x1, ... , xn) and the ai j are real is called a quadratic form. The form is called symmetric if aij = aji for all i and j, positive definite if x # 0 implies Q(x) > 0, and negative definite if x 0 0 implies Q(x) < 0. In general, it is not easy to determine whether a quadratic form is positive or negative definite. One criterion, involving eigenvalues, is described in Reference Th. 13.11 Extreme of Functions of Several Variables 379 13.1, Theorem 9.5. Another, involving determinants, can be described as follows. Let A = det [a;;] and let Ak denote the determinant of the k x k matrix obtained by deleting the last (n - k) rows and columns of [aij]. Also, put AO = 1. From the theory of quadratic forms it is known that a necessary and sufficient condition for a symmetric form to be positive definite is that the n + 1 numbers Ao, A1, ... , A. be positive. The, form is negative definite if, and only if, the same n + 1 numbers are alternately positive and negative. (See Reference, 13.2, pp. 304-308.) The quadratic form which appears in (5) is symmetric because the mixed partials D;,j f(a) and DD,; f(a) are equal. Therefore, under the conditions of Theorem 13.10, we see that f has a local minimum at a if the (n + 1) numbers Ao, A1, ... , A are all positive, and a local maximum if these numbers are alternately positive and negative. The case n = 2 can be handled directly and gives the following criterion. Theorem 13.11. Let f be a real-valued function with continuous second-order partial derivatives at a stationary point a in R2. Let A = D1,1f(a), B = D1,2f(a), C = D2,2.f(a), and let A=det[AB B1 =AC-B2. Then we have: a) If A > 0 and A > 0, f has a relative minimum at a. b) If A > 0 and A < 0, f has a relative maximum at a. c) If A < 0, f has a saddle point at a. Proof. In the two-dimensional case we can write the quadratic form in (5) as follows : Q(x, y) = +{Ax2 + 2Bxy + Cy2}. If A 0 0, this can also be written as Q(x, Y) = ZA {(Ax + By)2 + Aye}. If A > 0, the expression in brackets is the sum of two squares, so Q(x, y) has the same sign as A. Therefore, statements (a) and (b) follow at once from parts (a) and (b) of Theorem 13.10. If A < 0, the quadratic form is the product of two linear factors. Therefore, the set of points (x, y) such that Q(x, y) = 0 consists of two lines in the xy-plane intersecting at (0, 0). These lines divide the plane into four regions; Q(x, y) is positive in two of these regions and negative in the other two. Therefore f has a saddle point at a. NOTE. If A = 0, there may be a local maximum, a local minimum, or a saddle point at a. 380 Implicit Functions and Extremum Problems 13.7 EXTREMUM PROBLEMS WITH SIDE CONDITIONS Consider the following type of extremum problem. Suppose that f(x, y, z) represents the temperature at the point (x, y, z) in space and we ask for the maxi-mum or minimum value of the temperature on a certain surface. If the equation of the surface is given explicitly in the form z = h(x, y), then in the expression f(x, y, z) we can replace z by h(x, y) to obtain the temperature on the surface as a function of x and y alone, say F(x, y) = f [x, y, h(x, y)]. The problem is then reduced to finding the extreme values of F. However, in practice, certain difficulties arise. The equation of the surface might be given in an implicit form, say g(x, y, z) = 0, and it may be impossible, in practice, to solve this equation explicitly for z in terms of x and y, or even for x or y in terms of the remaining variables. The problem might be further complicated by asking for the extreme values of the temperature at those points which lie on a given curve in space. Such a curve is the intersection of two surfaces, say g1(x, y, z) = 0 and g2(x, y, z) = 0. If we could solve these two equations simultaneously, say for x and y in terms of z, then we could introduce these expressions into f and obtain a new function of z alone, whose extrema we would then seek. In general, however, this procedure cannot be carried out and a more practicable method must be sought. A very elegant and useful method for attacking such problems was developed by Lagrange. Lagrange's method provides a necessary condition for an extremum and can be described as follows. Let f(x1, ... , be an expression whose extreme values are sought when the variables are restricted by a certain number of side conditions, say g1(x1, ... , xn) = 0, ... , gm(xl, ... , xn) = 0. We then form the linear combination 0(X1, ... , xn) = f(X1, ... , xn) + 2191(x1, ... , Xn) + ....+ 2.9.(x1, ... , xn), where 21, ... , A. are m constants. We then differentiate 0 with respect to each coordinate and consider the following system of n + m equations : Dr4(xl, ... , xn) = 0, r = 1, 2, ... , n, 9k(xl,...,xn) = 0, k = 1,2,...,m. Lagrange discovered that if the point (x1, ... , xn) is a solution of the extremum problem, then it will also satisfy this system of n + m equations. In practice, one attempts to solve this system for the n + m "unknowns," 21, ... , and x1, ... , xn. The points (x1, ... , xn) so obtained must then be tested to determine whether they yield a maximum, a minimum, or neither. The numbers 21, ... , 2m, which are introduced only to help solve the system for x1, ... , xn, are known as Lagrange's multipliers. One multiplier is introduced for each side condition. A complicated analytic criterion exists for distinguishing between maxima and minima in such problems. (See, for example, Reference 13.3.) However, this criterion is not very useful in practice and in any particular prolem it is usually easier to rely- on some other means (for example, physical or geometrical consider-ations) to make this distinction. Th. 13.12 Extremum Problems with Side Conditions 381 The following theorem establishes the validity of Lagrange's method: Theorem 13.12. Let f be a real-valued function such that f E C' on an open set S in W. Let g1, ... , gm be m real-valued functions such that g = (g1, ... , gm) E C' on S, and assume that m < n. Let X0 be that subset of S on which g vanishes, that is, Xo = {x : x E S, g(x) = 0}. Assume that xo e Xo and assume that there exists an n-ball B(xo) such that f(x) < f(xo) for all x in Xo r B(xo) or such that f(x) > f(xo) for all x in X0 n B(xo). Assume also that the m-rowed determinant det [D1g1(xo)] 0 0. Then there exist m real numbers A1i ... 2 Am such that the following n equations are satisfied: m Dr.f(xo) + E 'ZkD,gk(xo) = 0 (r = 1, 2,... , n). (6) k=1 NOTE. The n equations in (6) are equivalent to the following vector equation: Vf(xo) + Al V91(xo) + ... + Am Vg n(xo) = 0-Proof. Consider the following system of m linear equations in the m unknowns m L, 2kDr9k(xo) D, f(xo) (r = 1, 2, ... , m). k=1 This system has a unique solution since, by hypothesis, the determinant of the system is not zero. Therefore, the first m equations in (6) are satisfied. We must now verify that for this choice of A1, ... , Am, the remaining n - m equations in (6) are also satisfied. To do this, we apply the implicit function theorem. Since m < n, every point x in S can be written in the form x = (x'; t), say, where x' E Rm and t e R". In the remainder of this proof we will write x' for (x1, ... , xm) and t for (xm+ 1, ... , x"), so that tk = xm+k. In terms of the vector-valued function g = (g1, -, gm), we can now write g(xo; to) = 0 if xo = (xo; to). Since g c- C' on S, and since the determinant det [Djg;(xo; to)] 0, all the conditions of the implicit function theorem. are satisfied. Therefore, there exists an (n - m)-dimensional neighborhood To of to and a unique vector-valued function h = (h1, ... , hm), defined on To and having values in Rm such that h e C' on To, h(to) = xo, and for every t in To, we have g[h(t); t] = 0. This amounts to saying that the system of m equations 91(x1,...,x") = 0,...,9m(x1,...,x") = 0, can be solved for x1, ... , xm in terms of xm+ 1, ... , x,,, giving the solutions in the form x, = hr(xm+1, ... , x"), r = 1, 2, ... , m. We shall now substitute these expressions for x1, ... , xm into the expression f(x1, ... , x") and also into each 382 Implicit Functions and Extremum Problems expression gp(x1, ... , xn). That is to say, we define a new function F as follows: F(Xm+1) ... , xe) = J [hl(xm+l, ... , X . ) ,-.. , hm(Xm+l, ... , xn); Xm+1'... , xn]; and we define m new functions Gl,... , Gm as follows: Gp(Xm+ 19 ... , xn) = gp[hl(xm+ 1, ... , xn), ... , hm(Xm+ 1, ... , xn); Xm+ 19 ... , Xn]. More briefly, we can write F(t) = f [H(t)] and Gp(t) = gp[H(t)], where H(t) _ (h(t); t). Here t is restricted to lie in the set To. Each function Gp so defined is identically zero on the set To by the implicit function theorem. Therefore, each derivative D,Gp is also identically zero on To and, in particular, D,Gp(to) = 0. But by the chain rule (Eq. 12.20), we can com-pute these derivatives as follows : DrGp(to) = E Dkgp(xo)D,Hk(to) (r = 1, 2, ... , n - m). k=1 But Hk(t) = hk(t) if 1 < k 5 m, and Hk(t) = xk if m + 1 < k < n. Therefore, when m + 1 < k < n, we have D,Hk(t) - 0 if m + r # k and D,Hm+r(t) = 1 for every t. Hence the above set of equations becomes E Dkgp(xo)Drhk(to) + Dm+rgp(xo) = 0P 1, 2, ... , m, (7) k=1 Ir= 1, 2,... , n - m. By continuity of h, there is an (n - m)-ball B(to) c To such that t e B(to) implies (h(t); t) a B(xo), where B(xo) is the n-ball in the statement of the theorem. Hence, t e B(to) implies (h(t); t) a Xo n B(xo) and therefore, by hypothesis, we have either F(t) < F(to) for all t in B(to) or else we have F(t) >- F(to) for all t in B(to). That is, F has a local maximum or a local minimum at the interior point to. Each partial derivative D,F(to) must therefore be zero. If we use the chain rule to compute these derivatives, we find D,F(to) = L, Dkf(xo)DPHk(to) (r = 1, ... , n - m), k=1 and hence we can write m E Dkf(xo)Drhk(to) + Dm+r.f(xo) = 0 (r = 1, ... , n - m). (8) k=1 If we now multiply (7) by Ap, sum on p, and add the result to (8), we find E'" m [Dkf(xo) + )PDkp(xo) Drh k(to) + Dm+rf(xo) + i.pDm+rgp(xo) = 0, p=1 p=1 for r = 1, ... -n - m. In the sum over k, the expression in square brackets Extremum. Problems with Side Conditions 383 vanishes because of the way 21, ... , A,,, were defined. Thus we are left with m Dm+rf(xo) + E 2pDm+rgp(xo) = 0 (r = 1, 2, ... , n - m), p=1 and these are exactly the equations needed to complete the proof. NOTE. In attempting the solution of a particular extremum problem by Lagrange's method, it is usually very easy to determine the system of equations (6) but, in general, it is not a simple matter to actually solve the system. Special devices can often be employed to obtain the extreme values off directly from (6) without first finding the particular points where these extremes are taken on. The following example illustrates some of these devices : Example. A quadric surface with center at the origin has the equation Axe + Bye + Cz2 + 2Dyz + 2Ezx + 2Fxy = 1. Find the lengths of its semi-axes. Solution. Let us write (x1, x2, x3) instead of (x, y, z), and introduce the quadratic form 3 3 q(x) = E E a1jxixi, (9) J=1 i=1 where x = (x1, x2, x3) and the air = aji are chosen so that the equation of the surface becomes q(x) = 1. (Hence the quadratic form is symmetric and positive definite.) The problem is equivalent to finding the extreme values of f(x) = IIx1I2 = x1 + x2 + x3 subject to the side condition g(x) = 0, where g(x) = q (x) - 1. Using Lagrange's method, we introduce one multiplier and consider the vector equation Vf(x) + 2 Vq (x) = 0 (10) (since Vg = Vq). In this particular case, both f and q are homogeneous functions of degree 2 and we can apply Euler's theorem (see Exercise 12.18) in (10) to obtain x Vf(x) + Ax Vq(x) = 2f(x) + 2Aq(x) = 0. Since q(x) = I on the surface we find 2 = -f(x), and (10) becomes t Vf(x) - Vq(x) = 0, (11) where t = 1/f(x). (We cannot havef(x) = 0 in this problem.) The vector equation (11) then leads to the following three equations for x1, x2, x3: (a11-t)x1+ a12x2 + a13X3 = 0, a21x1 + (a22 - t)x2 + a23x3 = 0, a31x1 + a32x2 + (a33 - t)x3 = 0. Since x = 0 cannot yield a solution to our problem, the determinant of this system must 384 Implicit Functions and Extremum Problems vanish. That is, we must have all - t a12 a13 a21 a22 - t a23 a31 a32 a33 - t Equation (12) is called the characteristic equation of the quadratic form in (9). In this case, the geometrical nature of the problem assures us that the three roots tl, t2, t3 of this cubic must be real and positive. [Since q(x) is symmetric and positive definite, the general theory of quadratic forms also guarantees that the roots of (12) are all real and positive. (See Reference 13.1, Theorem 9.5.)] The semi-axes of the quadric surface are ti 1/2, t2 1/2, t3 -1/2 EXERCISES Jacobians 13.1 Let f be the complex-valued function defined for each complex z 0 by the equation f(z) = 1/z. Show that Jf(z) Iz I'4. Show that f is one-to-one and compute f -1 explicitly. 13.2 Let f = (fl,f2,f3) be the vector-valued function defined (for every point (x1, x2, x3) in R3 for which xl + x2 + x3 96 -1) as follows: fk(x1, x2, x3) = k + x3 (k = 1, 2, 3). 1 + xl + x2 Show that Jf(x1, x2, x3) = (1 + x1 + x2 + x3)'4. Show that f is one-to-one and compute f'1 explicitly. 13.3 Let f = (fi..... f") be a vector-valued function defined in R", suppose f e C' on R", and let J f ( x ) denote the Jacobian determinant. L e t g 1 , . .. , g" be n real-valued functions defined on R1 and having 'continuous derivatives g', . . . , g,;. Let hk(x) _ fk[gl(x1), . . . , g"(x,.], k = 1, 2, ... , n, and put h = (hl, ... , h"). Show that J4(x) = if [91(x1), ... , g"(x")1g'1(x1) ... gg(x") 13.4 a) If x(r, 0) = r cos 0, y(r, 0) = r sin 0, show that a(x, Y) = r. a(r, 0) b) If x(r, 0, q$) = r cos 0 sin q, y(r, 0, 0) = r sin 0 sin 0, z = r cos 0, show that a(x, Y, z) r2 sin q$. a(r, 0, 0) 13.5 a) State conditions on f and g which will ensure that the equations x = f (u, v), y = g(u, v) can be solved for u and v in a neighborhood of (xo, yo). If the solu-tions are u = F(x, y), v = G(x, y), and if J = a(f, g)/a(u, v), show that aF- 1ag a F Iaf aG= - lag . 2G_ laf ax j av ' ay , J av ' ax j au ' ay J au Exercises 385 b) Compute J and the partial derivatives of F and G at (xo, yo) = (1, 1) when flu, v) = u2 - v2, g(u, v) = 2uv. 13.6 Let f and g be related as in Theorem 13.6. Consider the case n = 3 and show that we have JE(x)D1 gi(y) = ai,l ai,2 ai,3 D1f2(x) D1f3(x) D2f2(x) D2f3(x) D3f2(x) D3f3(x) (i = 1, 2, 3), where y = f(x) and 8i, j = 0 or I according as i A j or i j. Use this to deduce the formula a(f2,f3) / a( 1,f2,f3) Dl gl = (X2, x3) a(x1, x2, x3) There are similar expressions for the other eight derivatives Dkgi. 13.7 Let f = u + iv be a complex-valued function satisfying the following conditions: u e C' and v e C' on the open disk A = {z : Iz < 11; f is continuous on the closed disk A = {z : Iz < 11; u(x, y) = x and v(x, y) = y whenever x2 + y2 = 1; the Jacobian Jf(z) > 0 if z e A. Let B = f(A) denote the image of A under f and prove that: a) If X is an open subset of A, then f(X) is an open subset of B. b) B is an open disk of radius 1. c) For each point uo + ivo in B, there is only a finite number of points z in A such that f(z) = uo + ivo. Extremum problems 13.8 Find and classify the extreme values (if any) of the functions defined by the following equations: a) f(x, y) = y2 + x2y + x4, b)f(x,y)=x2+y2+x+y+xy, c) f(x, y) = (x - 1)4 + (x - y)4, d) f(x, y) = y2 - x3. 13.9 Find the shortest distance from the point (0, b) on the y-axis to the parabola x2 - 4y = 0. Solve this problem using Lagrange's method and also without using Lagrange's method. 13.10 Solve the following geometric problems by Lagrange's method: a) Find the shortest distance from the point (a1, a2, a3) in R3 to the plane whose equation is 61x1 + 62x2 + 63x3 + bo = 0. b) Find the point on the line of intersection of the two planes and a1x1 + a2X2 + a3x3 + ao = 0 61x1 + 62x2 + 63x3 + bo = 0 which is nearest the origin. 386 Implicit Functions and Exreemum Problems 13.11 Find the maximum value of lEk=1 akxkl, if k=1 xk = 1, by using a) the Cauchy-Schwarz inequality. b) Lagrange's method. 13.12 Find the maximum of (x1x2 ... x,,)2 under the restriction xi+ +x.=1. Use the result to derive the following inequality, valid for positive real numbers al, ... , an : 1/n a1 +...+ an (a1 ... an) n 13.13 If f(x) = xi + + 4, x = (x1, ... , xn), show that a local extreme of f, subject to the condition x1 + + xn = a, is aknl'k 13.14 Show that all points (x1i x2, x3, x4) where x1 + x2 has a local extremum subject to the two side conditions x1 + x3 + x4 = 4, x2 + 24 + 3x4 = 9, are found among (0, 0, ±J3, ±1), (0, ±1, +2, 0), (±1, 0, 0, ± /3), (±2, ±3, 0, 0). Which of these yield a local maximum and which yield a local minimum? Give reasons for your conclusions. 13.15 Show that the extreme values of f(x1, x2, x3) = xi + x2 + x3, subject to the two side conditions 33 33 E E aiJxixJ = 1 (a1J = aji) J=1 i=1 and 61x1 + 62x2 + 63x3 = 0, (bl, b2, b3) 9 (0, 0, 0), are tl 1, t2 1, where tl and t2 are the roots of the equation bl b2 b3 0 a12 a13 b1 a22 - t a23 b2 a32 a33 - t b3 = 0. Show that this is a quadratic equation in t and give a geometric argument to explain why the roots t1, t2 are real and positive. 13.16 Let A = det [xi; ] and let Xi = (xil, ... , xi ). A famous theorem of Hadamard states that JAI 5 dl ... d,,, if dl, ... , do are n positive constants such that IJX1112 = dt (i = 1, 2, ... , n). Prove this by treating A as a function of n2 variables subject to n constraints, using Lagrange's method to show that, when A has an extreme under these conditions, we must have dl 0 0 .. 0 A2 = 0 d2 0 0 0 0 0 ... d. SUGGESTED REFERENCES FOR FURTHER STUDY 13.1 Apostol, T. M., Calculus, Vol. 2, 2nd ed. Xerox, Waltham, 1969. 13.2 Gantmacher, F. R., The Theory of Matrices, Vol. 1. K. A. Hirsch, translator. Chelsea, New York, 1959. 13.3 Hancock, H., Theory of Maxima and Minima. Ginn, Boston, 1917. CHAPTER 14 MULTIPLE RIEMANN INTEGRALS 14.1 INTRODUCTION The Riemann integral f .b f(x) dx can be generalized by replacing the interval [a, b] by an n-dimensional region in which f is defined and bounded. The simplest regions in R" suitable for this purpose are n-dimensional intervals. For example, in R2 we take a rectangle I partitioned into subrectangles Ik and consider Riemann sums of the form Y_f(xk, yk)A(Ik), where (xk, yk) E IA, and A(Ik) denotes the area of Ik. This leads us to the concept of a double integral. Similarly, in R3 we use rectangular parallelepipeds subdivided into smaller parallelepipeds IA; and, by considering sums of the form Y_ f(xk, Yk, zk)V(Ik), where (xk, Yk, zk) E Ik and V(1k) is the volume of Ik, we are led to the concept of a triple integral. It is just as easy to discuss multiple integrals in R", provided that we have a suitable generalization of the notions of area and volume. This "generalized volume" is called measure or content and is defined in the next section. 14.2 THE MEASURE OF A BOUNDED INTERVAL IN R" Let A 1, ... , A. denote n general intervals in R'; that is, each At may be bounded, unbounded, open, closed, or half-open in R'. A set A in R" of the form A = Al x x A. = {(x1i...,x"):XkEAt fork = 1, 2,..., n), is called a general n-dimensional interval. We also allow the degenerate case in which one or more of the intervals Ak consists of a single point. If each Ak is open, closed, or bounded in R1, then A has the corresponding property in R". If each Ak is bounded, the n-dimensional measure (or n-measure) of A, denoted by µ(A), is defined by the equation µ(A) = µ(A1) ... µ(A"), where µ(Ak) is the one-dimensional measure (length) of Ak. When n = 2, this is called the area of A, and when n = 3, it is called the volume of A. Note that p(A) = 0 if µ(Ak) = 0 for some k. We turn next to a discussion of Riemann integration in R". The only essential difference between the case n = 1 and the case n > I is that the quantity Exk = xk - xk_ 1 which was used to measure the length of the subinterval 388 Def. 14.2 Riemann Integral of a Bounded Function 389 [xk_ 1, xk] is replaced by the measure / (Ik) of an n-dimensional subinterval. Since the work proceeds on exactly the same lines as the one-dimensional case, we shall omit many of the details in the discussions that follow. 14.3 THE RIEMANN INTEGRAL OF A BOUNDED FUNCTION DEFINED ON A COMPACT INTERVAL IN R" Definition 14.1. Let A = Al x x A. be a compact interval in W. If Pk is a partition of Ak, the Cartesian product P = P1 x x P", is said to be a partition of A. If Pk divides Ak into Mk one-dimensional subintervals, then P determines a decomposition of A as a union of m1 m" n-dimensional intervals (called subintervals of P). A partition P' of A is said to be finer than P if P c P'. The set of all partitions of A will be denoted by /(A). Figure 14.1 illustrates partitions of intervals in RZ and in R3. Figure 14.1 Definition 14.2. Let f be defined and bounded on a compact interval I in W. If P is a partition of I into m subintervals I,, .. , 1. and if tk e Ik, a sum of the form M S(P, f) = E f(tk)µ(Ik)r k=1 is called a Riemann sum. We say f is Riemann-integrable on land we write f e R on I, whenever there exists a real number A having the following property: For every e > 0 there exists a partition Pe of I such that P finer than Pe implies IS(P,f) - Al < E, for all Riemann sums S(P, f). When such a number A exists, it is uniquely 390 Multiple Rkn m Integrals determined and is denoted by SI f dx, fr f(x) dx, or by f f(x1, ... , x") d(x1, ... , x"). r r Def. 143 NOTE. For n > 1 the integral is called a multiple or n -fold integral. When n = 2 and 3, the terms double and triple integral are used. As in R1, the symbol x in j', f(x) dx is a "dummy variable" and may be replaced by any other convenient symbol. The notation f,f(x1, ... , x") dx1 .. dx" is also used instead of f, f(xl, ... , x") d(xl, ... , x"). Double integrals are sometimes written with two integral signs and triple integrals with three such signs, thus: JJf(x, y) dx dy, z) dx dy 555 f(x, y, Definition 14.3. Let f be defined and bounded on a compact interval I in R". If P is a partition of I into m subintervals I1, ... , Im, let mk(f) = inf {f(x) : x e Ik}, Mk(f) = sup {f(x) : x e Ik}. The numbers m m U(P, f) = E Mk(f),u(Ik) and L(P, f) = E mk(f)u(Ik), k=1 k=1 are called upper and lower Riemann sums. The upper and lower Riemann integrals off over I are defined as follows: SI SI f dx = inf {U(P, f) : P e &(I)}, f dx = sup {L(P, f) : P E g(I)}. The function f is said to satisfy Riemann's condition on I if, for every s > 0, there exists a partition Pa of I such that P finer than Pe implies U(P, f) - L(P, f) < e. NOTE. As in the one-dimensional case, upper and lower integrals have the following properties : a) f (f+g)dx5 fdx+ f, g dx, J r(f+g)dxZ rfdx+ rgdx. Jr r , Th. 14.5 Evaluation of a Multiple Integral 391 b) If an interval I is decomposed into a union of two nonoverlapping intervals I1, 12, then we have f dx = Jfdx + ffdx and f f dx = ffdx + $fdx. I , z J! i Iz The proof of the following theorem is essentially the same as that of Theorem 7.19 and will be omitted. Theorem 14.4. Let f be defined and bounded on a compact interval I in R. Then the following statements are equivalent: i) feRon I. ii) f satisfies Riemann's condition on I. iii) fI f dx = f, f dx. 14.4 SETS OF MEASURE ZERO AND LEBESGUE'S CRITERION FOR EXISTENCE OF A MULTIPLE RIEMANN INTEGRAL A subset T of R" is said to be of n-measure zero if, for every e > 0, T can be covered by a countable collection of n-dimensional intervals, the sum of whose n-measures is <.-. As in the one-dimensional case, the union of a countable collection of sets of n-measure 0 is itself of n-measure 0. If m < n, every subset of R, when considered as a subset of R", has n-measure 0. A property is said to hold almost everywhere on a set S in R" if it holds every-where on S except for a subset of n-measure 0. Lebesgue's criterion for the existence of a Riemann integral in R1 has a straightforward extension to multiple integrals. The proof is analogous to that of Theorem 7.48. Theorem 14.5. Let f be defined and bounded on a compact interval I in R". Then f e R on I if, and only if, the set of discontinuities off in I has n-measure zero. 14.5 EVALUATION OF A MULTIPLE INTEGRAL BY ITERATED INTEGRATION From elementary calculus the reader has learned to evaluate certain double and triple integrals by successive integration with respect to each variable. For example, if f is a function of two variables continuous on a compact rectangle Q in the xy-plane, say Q = {(x, y) : a 5 x 5 b, c 5 y 5 d}, then for each fixed y in [c, d] the function F defined by the equation F(x) = f(x, y) is continuous (and hence integrable) on [a, b]. The value of the integral fl. F(x) dx depends on y and 392 Multiple Riemann Integrals Th. 14.6 defines a new function G, where G(y) = f a f(x, y) dx. This function G is con-tinuous (by Theorem 7.38), and hence integrable, on [c, d]. The integral f" G(y) dy turns out to have the same value as the double integral IQ f(x, y) d(x, y). That is, we have the equation (' f f(x, y) d(x, y) = I d CJ bf(x, y) dxl dy. (l) Jc o (This formula will be proved later.) The question now arises as to whether a similar result holds when f is merely integrable (and not necessarily continuous) on Q. We can see at once that certain difficulties are inevitable. For example, the inner integral fa f(x, y) dx may not exist for certain values of y even though the double integral exists. In fact, if f is discontinuous at every point of the line segment y = yo, a < x < b, then f a f(x, yo) dx will fail to exist. However, this line segment is a set whose 2-measure is zero and therefore does not affect the integrability of f on the whole rectangle Q. In a case of this kind we must use upper and lower integrals to obtain a suitable generalization of (1). Theorem 14.6. Let f be defined and bounded on a compact rectangle Q = [a, b] x [c, d] in R2. Then we have: 1) IQ f d(x, y) 5 fa [p, f(x, y) dy] dx < Ja [ f " f(x, y) dy] A < f Q fd(x, y). ii) Statement (i) holds with Jd replaced by f" throughout. iii) IQfd(x, y) 5 ,a [jaf(x, y) dx] dy 5 P [5of(x, y) dx] dy 5 JQfd(x, y). iv) Statement (iii) holds with lab replaced by f .b throughout. v) When f Q f(x, y) d(x, y) exists, we have J f(x, y) d(x, y) = faa' [J d f(xy) dyl dx = lb [fdf(xy) dy] dx c J ('c c = d [la' y) dx] dy =d [fbf(x, y) dxl dy. Proof. To prove (i), define F by the equation d F(x) = f f(x, y) dy, if x e [a, b]. Then IF(x)I 5 M(d - c), where M = sup {If(x, y)j : (x, y) e Q}, and we can consider 1= f b F(x) dx = fb [fdf(x, y) dy] dx. Th.14.6 Evaluation of a Multiple Integral Similarly, we define 393 I = f b F(x) dx = f b [Jdf(x , y) d y] dx. a J Let P1 = {xo, x1, ... , xn} be a partition of [a, b] and let P2 = {Yo, Y1, ... , Ym}, be a partition of [c, d]. Then P = P1 X P2 is a partition of Q into mn sub-rectangles Qi, and we define z; r (' r (' Ii, Lf Y.r f(x, y) dy] dx, I; = I s L J YJ f(x, y) dyJ dx. xt-t YJ-1 xt-t YJ-1 J Since we have m jdf(x, y) dy = E f YJ f(x, y) dy, j=1 YJ-1 we can write fbLJCdf(x,y)dyl dx < E I IfYJ,f(x,y)dyl dx J J J m n fxi 1 [fYiJj t j=1 i=1 That is, we have the inequality Similarly, we find If we write f(x, y) dy] dx. m n I s E 37 Iii. j=1 i=1 m n I> ± EIi,. j=1 i=1 mi, = inf { f(x, y) : (x, y) a Qi,}, and Mi, = sup {f(x,y):(x,y)eQi;}, then from the inequality mi, < f(x, y) 5 Mi,, (x, y) e Qi,, we obtain YJ ( J YJ-I mi,(Y; - Yj -1) f(x, y) dy < Mii(Yi - Y.i-1) 394 Multiple Riemann Integrals This, in turn, implies mjiµ(Q,j) < f.,-. [rJ f(x, y) dyl dx 7J-, J x, 7J < CJ f(x, y) dyI dx < M1jy(Q1j) x+- i J-, Summing on i and j and using the above inequalities, we get L(P,f) S I S 15 U(P,f). Since this holds for all partitions P of Q, we must have rQf d(x, y) < I< 1< f d(x, y). J Q This proves statement (i). It is clear that the preceding proof could also be carried out if the function F were originally defined by the formula F(x) = f d f(x, y) dy, C and hence (ii) follows by the same argument. Statements (iii) and (iv) can be similarly proved by interchanging the roles of x and y. Finally, statement (v) is an immediate consequence of statements (i) through (iv). As a corollary, we have the formula mentioned earlier: fb[fC d Q f(X, Y) d(x, Y) , y) d.] dx = J [fbf(x,y) dxI dy, which is valid when f is continuous on Q. Ibis is often called Fubini's theorem. NOTE. The existence of the iterated integrals Jb [f df(x, y) dy] dx and fd [Jb f(x, y) dxl dy, does not imply the existence of JQ f(x, y) d(x, y). A counter example is given in Exercise 14.7. Before commenting on the analog of Theorem 14.6 in R", we first introduce some further notation and terminology. If k 5 n, the set of x in R". for which xk = 0 is called the coordinate hyperplane f jk. Given a set S in R", the projection Sk of S on r 1k- is defined to be the image of S under that mapping whose value at each point (x1, x2,... , x") in S is (x1, ... , Xk-1, 0, Xk+ 1, ... , x"). It is easy to Evaluation of a Multiple Integral 395 x3 x2 Figure 14.2 show that such a mapping is continuous on S. It follows that if S is compact, each projection Sk is compact. Also, if S is connected, each S. is connected. Projections in R3 are illustrated in Fig. 14.2. A theorem entirely analogous to Theorem 14.6 holds for n-fold integrals. It will suffice to indicate how the extension goes when n = 3. In this case, f is defined and bounded on a compact interval Q = [al, b1] x [a2, b2] x [a3, b3] in R3 and statement (i) of Theorem 14.6 is replaced by r rbi r r C J 61 [J f d(x2, x3)] dx1 < J f dx, (2) a Q Q where Q1 is the projection of Q on the coordinate plane 111. When f Q f(x) dx exists, the analog of part (v) of Theorem 14.6 is the formula 2, x3)] dxl = fe, Ub, 1 J d(x f (x) dx = bl 1fQI f Q f al J i f dXlJ d(x2, x3) (3) As in Theorem 14.6, similar statements hold with appropriate replacements of upper integrals by lower integrals, and there are also analogous formulas for the projections Q2 and Q3. The reader should have no difficulty in stating analogous results for n-fold integrals (they can be proved by the method used in Theorem 14.6). The special case in which the n-fold integral !Q f(x) dx exists is of particular importance and 396 Multiple Riemann Integrals can be stated as follows : Theorem 14.7. Let f be defined and bounded on a compact interval Q = [al, b1] x ... x [a", b"], in R". Assume that f Q f(x) dx exists. Then Th. 14.7 f d(x2, x")] dxl = J b, d(x2, x"). f f 61 f f dx = Q Q Similar formulas hold with upper integrals replaced by lower integrals and with Q1 replaced by Qk, the projection of Q on r 1k. 14.6 JORDAN-MEASURABLE SETS IN R" Up to this point the multiple integral fI f(x) dx has been defined only for intervals I. This, of course, is too restrictive for the applications of integration. It is not difficult to extend the definition to, encompass more general sets called Jordan-measurable sets. These are discussed in this section. The definition makes use of the boundary of a set S in R. We recall that a point x in R" is called a boundary point of S if every n-ball B(x) contains a point in S and also a point not in S. The set of all boundary points of S is called the boundary of S and is denoted by S. (See Section 3.16.) Definition 14.8. Let S be a subset of a compact interval I in W. For every partition P of I define J(P, S) to be the sum of the measures of those subintervals of P which contain only interior points of S and let J(P, S) be the sum of the measures of those subintervals of P which contain points of S u 8S. The numbers c(S) = sup {J(P, S) : P e 9(I)}, c(S) = inf {J(P, S) : P e .9(I)}, are called, respectively, the (n-dimensional) inner and outer Jordan content of S. The set S is said to be Jordan-measurable if c(S) = e(S), in which case this common value is called the Jordan content of S, denoted by c(S). It is easy to verify that c(S) and e(S) depend only on S and not on the interval I which contains S. Also, 0 5 c(S) 5 e(S). If S has content zero, then c(S) = e(S) = 0. Hence, for every e > 0, S can be covered by a finite collection of intervals, the sum of whose measures is <e. Note that content zero is described in terms of finite coverings, whereas measure zero is described in terms of countable coverings. Any set with content zero also has measure zero, but the converse is not necessarily true. Every compact interval Q is Jordan-measurable and its content, c(Q), is equal to its measure, p(Q). If k < n,'the n-dimensional content of every bounded set in Rk is zero. -Jordan=measurable sets S in R2 are also said to have area c(S). In this case, the sums J(P, S) and J(P, S) represent approximations to the area from the "inside" Def. 14.10 Integration over Jordan-Measurable Sets Figure 143 397 and the "outside" of S, respectively. This is illustrated in Fig. 14.3, where the lightly shaded rectangles are counted in J(P, S), the heavily shaded rectangles in J(P, S). For sets in R3, c(S) is also called the volume of S. The next theorem shows that a bounded set has Jordan content if, and only if, its boundary isn't too "thick." Theorem 14.9. Let S be a bounded set in R" and let DS denote its boundary. Then we have WS) = c(S) - c(S). Hence, S is Jordan-measurable if, and only if, OS has content zero. Proof. Let I be a compact interval containing S and 3S. Then for every partition P of I we have J(P, as) = J(P, S) J(P, S). Therefore, J(P, 3S) >- c(S) - c(S) and hence c(aS) >- c(S) - c(S). To obtain the reverse inequality, let e > 0 be given, choose Pl so that J(P1, S) < c(S) + e/2 and choose P2 so that J(P2, S) > c(S) _ e/2. Let P = Pl u P2. Since refine-ment increases the inner sums J and decreases the outer sums j, we find c(aS) 5 J(P, as) = J(P, S) - J(P, S) < J(P1, S) - J(P2, S) < c(S) - c(S) + e. Since a is arbitrary, this means that c(aS) 5 c(S) - c(S). Therefore, c(aS) _ c(S) - c(S) and the proof is complete. 14.7 MULTIPLE INTEGRATION OVER JORDAN-MEASURABLE SETS Definition 14.10. Let f be defined and bounded on a bounded Jordan-measurable set S in R". Let I be a compact interval containing S and define g 'on I as follows: 9(X) = to f (X) if X e S, ifXEl - S. 398 Multiple Rlemann Integrals Th. 14.11 Then f is said to be Riemann-integrable on S and we write f e R on S, whenever the integral J, g(x) dx exists. We also write f(x) dx = f g(x) dx. fS JI J The upper and lower integrals Is f(x) dx and Is f(x) dx are similarly defined. NOTE. By considering the Riemann sums which approximate fI g(x) dx, it is easy to see that the integral f s f(x) dx does not depend on the choice of the interval I used to enclose S. A necessary and sufficient condition for the existence of f s f(x) dx can now be given. Theorem 14.11. Let S be a Jordan-measurable set in R", and let f be defined and bounded on S. Then f e R on S if, and only if, the discontinuities off in S form a set of measure zero. Proof. Let I be a compact interval containing S and let g(x) = f(x) when x e S, g(x) = 0 when x e I - S. The discontinuities of f will be discontinuities of g. However, g may also have discontinuities at some or all of the boundary points of S. Since S is Jordan measurable, Theorem 14.9 tells us that c(8S) = 0. Therefore, g e R on I if, and only if, the discontinuities of f form a set of measure zero. 14.8 JORDAN CONTENT EXPRESSED AS A RIEMANN INTEGRAL Theorem 14.12. Let S be a compact Jordan-measurable set in R". Then the integral Is 1 exists and we have c(S) = f I. S Proof. Let I be a compact interval containing S and let Xs denote the characteristic function of S. That is, 1 ifxES, xssx) = 0 if xel - S. The discontinuities of Xs in I are the boundary points of S and these form a set of content zero, so the integral I., Xs exists, and hence Is I exists. Let P be a partition of I into subintervals I,, ... , I" and let A = {k : Ik n S is nonempty}. If k e A, we have Mk(Xs) = SUP {Xs(x) : x E Ik) = 1, Th. 14.13 Additive Property of the Riemann Integral and Mk(Xs) = 0 if k 0 A, so m U(P, Xs) = E Mk(Xs)u(Ik) = E AID = J(P, Xs). k=1 keA Since this holds for all partitions, we have f, Xs = E(S) = c(S). But Ji Xs = JI Xs so c(S) = f, Xs = f s 1. 399 14.9 ADDITIVE PROPERTY OF THE RIEMANN INTEGRAL The next theorem shows that the integral is additive with respect to sets having Jordan content. Theorem 14.13. Assume f e R on a Jordan-measurable set S in R". Suppose S = A v B, where A and B are Jordan-measurable but have no interior points in common. Then f e R on A, f e R on B, and we have f(x) dx = f f(x) dx + f f(x) dx. (4) JS A B Proof. Let I be a compact interval containing S and define g as follows : (f(x) ifxeS, 10 ifxel - S. The existence of fA f(x) dx and $B f(x) dx is an easy consequence of Theorem 14.11. To prove (4), let P be a partition of I into m subintervals I1, ... , I. and form a Riemann sum rm S(P, g) = L.r g(tk)P(Ik) k=1 If SA denotes that part of the sum arising from those subintervals containing points of A, and if S. is similarly defined, we can write S(P,9)=SA+SB-Sc, where Sc contains those terms coming from subintervals which contain both points of A and points of B. In particular, all points common to the two boundaries 8A and aB will fall in this third class. But now SA is a Riemann sum approximating the integral JA f(x) dx, and SB is a Riemann sum approximating JB f(x) A. Since c(aA n 8B) = 0, it follows that IS 1 can be made arbitrarily small when P is sufficiently fine. The equation in the theorem is an easy consequence of these remarks. NOTE. Formula (4) also holds for upper and lower integrals. 400 Multiple Riemann Integrals Th. 14.14 For sets S whose structure is relatively simple, Theorem 14.6 can be used to obtain formulas for evaluating double integrals by iterated integration. These formulas are given in the next theorem. Theorem 14.14. Let 01 and 02 be two continuous functions defined on [a, b] such that 01(x) < (/ 2(x) for each x in [a, b]. Let S be the compact set in R2 given by S={(x,y):a<x<b,01(x)<y<<02(x)}. If f e R on S, we have f f(x, y) d(x, y) = J 6 LJ m2(=) f(x, y) dyl dx. s a mi(x) J NOTE. The set S is Jordan-measurable because its boundary has content zero. (See Exercise 14.9.) Analogous statements hold for n-fold integrals. The extensions are too obvious to require further comment. a -Figure 14.4 Figure 14.4 illustrates the type of region described in the theorem. For sets which can be decomposed into a finite number of Jordan-measurable regions of this type, we can apply iterated integration to each separate part and add the results in accordance with Theorem 14.13. 14.10 MEAN-VALUE THEOREM FOR MULTIPLE INTEGRALS As in the one-dimensional case, multiple integrals satisfy a mean value property. This can be obtained as an easy consequence of the following theorem, the proof of which is left as an exercise. Theorem 14.15. Assume f e R and g e R on a Jordan-measurable set S in R". If f(x) < g(x) for each x in S, then we have f f(x) dx < f g(x) dx. s ,Js Th. 14.17 Mean-Value Theorem for Multiple Integrals 401 Theorem 14.16 (Mean- Value Theorem for multiple integrals). Assume that g e R and f e R on a Jordan-measurable set S in R" and suppose that g(x) >- 0 for each x in S. Let m = inf f(S), M = sup f(S). Then there exists a real number I in the interval m < A < M such that Is f(x)g(x) dx = 1 fs g(x) dx. (5) In particular, we have mc(S) < fs f(x) dx < Mc(S). (6) J NOTE. If, in addition, S is connected and f is continuous on S, then A = f(xa) for some xo in S (by Theorem 4.38.) and (5) becomes 1, f(x)g(x) dx = f(xo) f g(x) dx. s (7) In particular, (7) implies fs f(x) dx = f(xo)c(S), where xo a S. Proof. Since g(x) >- 0, we have mg(x) < f(x) g(x) < Mg(x) for each x in S. By Theorem 14.15, we can write g(x) dx. m f g(x) dx < f f(x)g(x) dx < M fs s If Is g(x) dx = 0, (5) holds for every A. If Is g(x) dx > 0, (5) holds with 1 = Is f(x)g(x) dx/ fs g(x) dx. Taking g(x) - 1, we obtain (6). We can use (6) to prove that the integrandf can be disturbed on a set of content zero without affecting the value of the integral. In fact, we have the following theorem : Theorem 14.17. Assume that f e R on a Jordan-measurable set S in R". Let T be a subset of S having n-dimensional Jordan content zero. Let g be a function, defined and bounded on S, such that g(x) = f(x) when x e S - T. Then g e R on S and J. f(x) dx = f s dx. s P r o o f. Let h = f - g. Then fs h(x) d x = IT h(x) d x + IS T h(x) dx. However, IT h(x) dx = 0 because of (6), and Is_T h(x) dx = 0 since h(x) = 0 for each x in S - T. NOTE. This theorem suggests a way of extending the definition of the Riemann integral fs f(x) dx for functions which may not be defined and bounded on the whole of S. In fact, let S be a bounded set in R" having Jordan content and let T be a subset of S having content zero. If f is defined and bounded on S - T and 402 Multiple Riemann Integrals if IS_ T f(x) dx exists, we agree to write f f(x) dx = f _ f(x) dx, S JS T and to say that f is Riemann-integrable on S. In view of the theorem just proved, this is essentially the same as extending the domain of definition off to the whole of S by defining f on Tin such a way that it remains bounded. EXERCISES Multiple integrals 14.1 If fl e R on [al, bl e R on [a., prove that bl (J fl(xl) dxl) ... (f:fn(xII)dxn), a where S = [al, bl ] x x [a., 14.2 Let f be defined and bounded on a compact rectangle Q = [a, b ] x [c, d ] in R2. Assume that for each fixed y in [c, d ], f (x, y) is an increasing function of x, and that for each fixed x in [a, b], f(x, y) is an increasing function of y. Prove that f e R on Q. 14.3 Evaluate each of the following double integrals. a) ffsin2 x sine y dx dy, where Q = [0, n] x [0, n]. Q b) 55 I cos (x + y) J dx dy, where Q = [0, n ] x [0, n ]. Q c) ff [x + y ] dx dy, where Q = [0, 2] x [0, 2), and [t] is the greatest Q integer < t. 14.4 Let Q = [0, 1 ] x [0, 1 ] and calculate f f Q f (x, y) dx dy in each case. a) f(x, y) = I - x - y if x + y <- 1, f(x, y) = 0 otherwise. b) f(x, y) = x2 + y2 if x2 + y2 < 1, f(x, y) = 0 otherwise. c) f(x, y) = x + y if x2 <- y <- 2x2, f(x, y) = 0 otherwise. 14.5 Define f on the square Q = [0, 1 ] x [0, 1 ] as follows: _ 1 if x is rational, f (x' y) 2y if x is irrational. a) Prove that f 'O f(x, y) dy exists for 0 < t <- 1 and that Ji [Jo f(x, y) dy] dx = t2, and Jo [f of(x, y) dyI dx = t. This shows that f o'[f o f(x, y) dy] dx exists and equals 1. b) Prove that fo [Jo f(x, y) dx] dy exists and find its value. c) Prove that the double integral JQ f(x, y) d(x, y) does not exist. 14.6 Define f on the square Q = [0, 1 ] x [0, 1 ] as follows: f.(x, y) = (0 if at least one of x, y is irrational, 1/n if y is rational and x = m/n, where in and n are relatively prime integers, n > 0. Prove that 1 1 1 f(x, y) dx] dy = f f(x, y) d(x, y) = 0 f0f(xY)dx f [o Q d but that fo f(x, y) dy does not exist for rational x. 14.7 If pk denotes the kth prime number, let , 1 :n= 1,2,...,Pk- 1, m= 1,2,...,Pk- 1}, S(Pk)= {\Pkn in t Pk/f 1111 let S = Uk 1 S(pk), and let Q = [0, 1 ] x [0, 1 ]. a) Prove that S is dense in Q (that is, the closure of S contains Q) but that any line parallel to the coordinate axes contains at most a finite subset of S. b) Define f on Q as follows : fix, y) = 0 if (x, y) a S, AX, y) = 1 if (x, y) e Q - S. Prove that fo [fo f(x, y) dy] dx = fo [fof(x, y) dx] dy = 1, but that the double integral fQ f(x, y) d(x, y) does not exist. Jordan content 14.8 Let S be a bounded set in W having at most a finite number of accumulation points. Prove that c(S) = 0. 14.9 Let f be a continuous real-valued function defined on [a, b]. Let S denote the graph off, that is, S = {(x, y) : y = f(x), a s x < b}. Prove that S has two-dimensional Jordan content zero. 14.10 Let I- be a rectifiable curve in R". Prove that IF has n-dimensional Jordan content zero. 14.11 Let f be a nonnegative function defined on a set S in W. The ordinate set offover S is defined to be the following subset of R"+ 1: {(xi, ... , x", x"+1) : (x1, ... , xn) e S, 0 s x"+1 5 f(x1, ... , x")). If S is a Jordan-measurable region in R" and if f is continuous on S, prove that the ordinate set off over S has (n + 1)-dimensional Jordan content whose value is j f(x1,... , x,) d(x1,... , x"). S Interpret this problem geometrically when n = 1 and n = 2. 404 Multiple Riemann Integrals 14.12 Assume that f e R on S and suppose Js f(x) dx = 0. (S is a subset of R"). Let A = {x : x e S, f(x) < 0} and assume that c(A) = 0. Prove that there exists a set B of measure zero such that f (x) = 0 for each x in S - B. 14.13 Assume that f e R on S, where S is a region in R" and f is continuous on S. Prove that there exists an interior point x0 of S such that 1 f(x) dx = f(xo)c(S) s 14.14 Let f be continuous on a rectangle Q = [a, b] x [c, d ]. For each interior point (x1, x2) in Q, define F(x1, X2) = I X, (f2 f(x, y) dy) dx. a c J Prove that D1,2F(xl, X2) = D2,1F(xl, x2) = f(x1, X2)-14.15 Let T denote the following triangular region in the plane: T (x, y):0<X+y<1 , where a > 0, b > 0. { a b Assume that f has a continuous second-order partial derivative D1, 2f on T. Prove that there is a point (xo, yo) on the segment joining (a, 0) and (0, b) such that fT D1,2f(x, y) d(x, y) = f(0, 0) - f(a, 0 )+ aD1f(xo, yo). SUGGESTED REFERENCES FOR FURTHER STUDY 14.1 Apostol, T. M., Calculus, Vol. 2, 2nd ed. Xerox, Waltham, 1969. 14.2 Kestelman, H., Modern Theories of Integration. Oxford University Press, 1937. 14.3 Rogosinski, W. W., Volume and Integral. Wiley, New York, 1952. CHAPTER 15 MULTIPLE LEBESGUE INTEGRALS 15.1 INTRODUCTION The Lebesgue integral was described in Chapter 10 for functions defined on subsets of R1. The method used there can be generalized to provide a theory of Lebesgue integration for functions defined on subsets of n-dimensional space W. The resulting integrals are called multiple integrals. When n = 2 they are called double integrals, and when n = 3 they are called triple integrals. As in the one-dimensional case, multiple Lebesgue integration is an extension of multiple Riemann integration. It permits more general functions as integrands, it treats unbounded as well as bounded functions, and it encompasses more general sets as regions of integration. The basic definitions and the principal convergence theorems are completely analogous to the one-dimensional case. However, there is one new feature that does not appear in R'. A multiple integral in R" can be evaluated by calculating a succession of n one-dimensional integrals. This result, called Fubini's Theorem, is one of the principal concerns of this chapter. As in the one-dimensional case we define the integral first for step functions, then for a larger class (called upper functions) which contains limits of certain increasing sequences of step functions, and finally for an even larger class, the Lebesgue-integrable functions. Since the development proceeds on exactly the same lines as in the one-dimensional case, we shall omit most of the details of the proofs. We recall some of the concepts introduced in Chapter 14. If I = Il x ... x I. is a bounded interval in R", the n-measure of I is defined by the equation YV) = u(Ii) ... u(I"), where y(Ik) is the one-dimensional measure, or length, of Ik. A subset T of R" is said to be of n-measure 0 if, for every e > 0, T can be covered by a countable collection of n-dimensional intervals, the sum of whose n-measures is 0, then g e L(cI) and Jf 9=c"Jf.. r f, In other words, expansion of the interval by a positive factor c has the effect of multiplying the integral by c", where n is the dimension of the space. The Levi convergence theorems (Theorems 10.22 through 10.26), and the Lebesgue dominated convergence theorem (Theorem 10.27) and its consequences (Theorems 10.28, 10.29, and 10.30) are also valid for multiple integrals. NOTATION. The integral J r f is also denoted by f f(x) dx r or f(xl, ... , x")d(xl, ... , x"). SI The notation 11 f(x1, ... , x") dx1 A. is also used. Double integrals are sometimes written with two integral signs, and triple integrals with three such signs, thus : y ) dx dy, j'$f(x r 15.4 MEASURABLE FUNCTIONS AND MEASURABLE SETS IN R" A real-valued function f defined on an interval I in R" is called measurable on I, and we write f e M(I), if there exists a sequence of step functions {s"} on I such that lim s"(x) = f(x) a.e. on I. n- CO The properties of measurable functions described in Theorems 10.35, 10.36, and 10.37 are also valid in this more general setting. f$$f(x y, z) dx dy dz. 408 Multiple Lebesgue Integrals Th. 15.1 A subset S of R" is called measurable if its characteristic function xs is measur-able. If, in addition, Xs is Lebesgue-integrable on R", then the n-measure µ(S) of the set S is defined by the equation Xs. p(S) = f JR" If Xs is measurable but not in L(R"), we define µ(S) = + oo. The function µ so defined is called n-dimensional Lebesgue measure. The properties of measure described in Theorems 10.44 through 10.47 are also valid for n-dimensional Lebesgue measure. Also, the Lebesgue integral can be defined for arbitrary subsets of R" by the method used in Section 10.19. We emphasize in particular -the countably additive property of Lebesgue measure described in Theorem 10.47: If (A,, A2, ... } is a countable disjoint collection of measurable sets in R", then the union 1 A, is measurable and 00 i (U A, I = p(A). The next theorem shows that every open subset of R" is measurable. Theorem 15.1. Every open set S in R" can be expressed as the union of a countable disjoint collection of bounded cubes whose closure is contained in S. Therefore S is measurable. Moreover, if S is bounded, then u(S) is finite. Proof. Fix an integer m Z 1 and consider all half-open intervals in R1 of the form (k k+1] fork=0,±1,±2,... 2m 2m All the intervals are of length 2-m, and they form a countable disjoint collection whose union is R'. The cartesian product of n such intervals is an n-dimensional cube of edge-length 2-m. Let Fm denote the collection of all these cubes. Then F. is a countable disjoint collection whose union is R". Note that the cubes in Fm+ 1 are obtained by bisecting the edges of those in Fm. Therefore, if Q. is a cube in Fm and if Qm+ 1 is a cube in Fm+ 1, then either Q.,1 q Qm, or Qm+ 1 and Q. are disjoint. Now we extract a subcollection G. from Fm as follows. If m = 1, G1 consists of all cubes in F1 whose closure lies in S. If m = 2, G2 consists of all cubes in F2 whose closure lies in S but not in any of the cubes in G1. Ifm = 3, G3 consists of all cubes in F3 whose closure lies in S but not in any of the cubes in G1 or G2, and so on. The construction is illustrated in Fig. 15.1 where S is a quarter of an open disk in R2. The blank square is in G1, the lightly shaded ones are in G2, and the darker ones are in G3. Now let 00 T= U U Q. m= 1 QEG,,, Th. 15.1 Fubini's Reduction Theorem 409 It C ' - s' lm_ Figure 15.1 That is, T is the union of all the cubes in G1, G2, ... We will prove that S = T and this will prove the theorem because T is a countable disjoint collection of cubes whose closure lies in S. Now T c S because each Q in G. is a subset of S. Hence we need only show that S s T. Let p = (pl, ... , p") be a point in S. Since S is open, there is a cube with center p and edge-length S > 0, which lies in S. Choose m so that 2_m < 6/2. Then for each i we have S 1 1 S Now choose k;, so that kj < k,+ 1 2" pi < _ 2" and let Q be the Cartesian product of the intervals (k12-'", (k, + 1)2-'n] for i = 1, 2, ... , n. Then p E Q for some cube Q in F",. If m is the smallest integer with this property, then Q E G,", so p e T. Hence S s T. The statements about the measurability of S follow at once from the countably additive property of Lebesgue measure. NOTE. If S is measurable, so is R" - S because Xin-s = I - Xs Therefore, every closed subset of R" is measurable. 15.5 FUBINI'S REDUCTION THEOREM FOR THE DOUBLE INTEGRAL OF A STEP FUNCTION Up to this point, Lebesgue theory in R" is completely analogous to the one-dimensional case. New ideas are required when we come to Fubini's theorem for calculating a multiple integral in R" by iterated lower-dimensional integrals. To better understand what is needed, we consider first the two-dimensional case. Let us recall the corresponding result for multiple Riemann integrals. If I = [a, b]- x [c, d] is a compact interval in R2 and if f is Riemann-integrable 410 Multiple Lebee Integrals Th.15.2 on I, then we have the following reduction formula (from part (v) of Theorem 14.6) : J f(x, y) d(x, y) = J d y) dxI dy. (3) I e a There is a companion formula with the lower integral 1b. replaced by the upper integral J .b, and there are two similar formulas with the order of integration re-versed. The upper and lower integrals are needed here because the hypothesis of Riemann-integrability on I is not strong enough to ensure the existence of the one-dimensional Riemann integral f a f(x, y) dx. This difficulty does not arise in the Lebesgue theory. Fubini's theorem for double Lebesgue integrals gives us the reduction formulas J f(x, y) d(x, y) = fd [S:fx (, y) dxdy = J b d.f(x, y) dy] dx, o und er the sole hypothesis that f is Lebesgue-integrable on I. We will show that the inner integrals always exist as Lebesgue integrals. This is another example illus-trating how Lebesgue theory overcomes difficulties inherent in the Riemann theory. In this section we prove Fubini's theorem for step functions, and in a later section we extend it to arbitrary Lebesgue integrable functions. Theorem 15.2. (Fubini's theorem for step functions). Let s be a step function on R2. Then for each fixed y in R1 the integral IR1 s(x, y) dx exists and, as a function of y, is Lebesgue-integrable on R1. Moreover, we have s(x, y) dxl dy. (4) f s(x, y) d(x, y) = SRI [Ski J J R Similarly, for each fixed x in R1 the integral 1.1 s(x, y) dy exists and, as a function of x, is Lebesgue-integrable on R1. Also, we have ff s(x, y) d(x, y) = R2 fR. UR1 s(x, y) dy] dx. (5) Proof This theorem can be derived from the reduction formula (3) for Riemann integrals, but we prefer to give a direct proof independent of the Riemann theory. There is a compact interval I = [a, b] x [c, d] such that s is a step function on I and s(x, y) = 0 if (x, y) a R2 - I. There is a partition of I into mn sub-rectangles I,j = [x1_1, x,] x [ y j ,, y j] such that s is constant on the interior of Ii j, say s(x, y) = c,j if (x, y) a int I,j. Def. 15.4 Some Properties of Sets of Measure Zero Then 411 S(x, y) dx1 dy. jj s(x, y) d(x, y) = c,j(x; - xi-1)(y; - y;-i) = J YJ i x JJ 1 J Y Ijj Summing on i and j we find J J s(x, y) d(x, y) = f d [Jab s(x, y) dxl dy. I J Since s vanishes outside I, this proves (4), and a similar argument proves (5). To extend Fubini's theorem to Lebesgue-integrable functions we need some further results concerning sets of measure zero. These are discussed in the next section. 15.6 SOME PROPERTIES OF SETS OF MEASURE ZERO Theorem 15.3. Let S be a subset of R". Then S has n-measure 0 if, and only if, there exists a countable collection of n-dimensional intervals {J1, J2, ... }, the sum of whose n-measures is finite, such that each point in S belongs to Jk for infinitely many k. Proof. Assume first that S has n-measure 0. Then, for every m Z 1, S can be covered by a countable collection of n-dimensional intervals {Im,l, Im,2i ... }, the sum of whose n-measures is <2-m. The set A consisting of all intervals Im,k for m = 1, 2, ... , and k = 1, 2, ... , is a countable collection which covers S, and the sum of the n-measures of all these intervals is < Em= 1 2-' = 1. Moreover, if a e S then, for each m, a e Im,k for some k. Therefore if we write A = {J1, J2, ... }, we see that a belongs to Jk for infinitely many k. Conversely, assume that there is a countable collection of n-dimensional intervals {Jl, J2.... } such that the series Ek 1 i (Jk) converges and such that each point in S belongs to Jk for infinitely many k. Given s > 0, there is an integer N such that 00 1: µ(Jk) < s. k=N Each point of S lies in the set U fk N Jk, so S C- U f k N JA;. Thus, S has been covered by a countable collection of intervals, the sum of whose n-measures is <c, so S has n-measure 0. Definition 15.4. If S is an arbitrary subset of R2, and if (x, y) e R2, we denote by SY and S" the following subsets of R': SY = {x : x E Rl and (x, y) E S}, S" = {y:yeR1 and '(x,y)ES}. 412 Multiple Lebesgue Integrals y Figure 15.2 X 171. 15.5 Examples are shown in Fig. 15.2. Geometrically, Sy is the projection on the x-axis of a horizontal cross section of S; and S" is the projection on the y-axis of a vertical cross section of S. Theorem 15.5. If S is a subset of R2 with 2-measure 0, then Sy has 1-measure O for almost all y in R1, and S" has 1-measure 0 for almost all x in R'. Proof. We will prove that Sy has 1-measure 0 for almost all y in R'. The proof makes use of Theorem 15.3. Since S has 2-measure 0, by Theorem 15.3 there is a countable collection of rectangles {Ik} such that the series 00 k=1 E µ(1k) converges, (6) and such that every point (x, y) of S belongs to Ik for infinitely many k. Write Ik = Xk x Yk, where Xk and Yk are subintervals of R'. Then P(1 k) = U(X k)U(yk) = p(Xk) XYk = J iu(Xk) XYk, RI where XYk is the characteristic function of the interval Yk. Let 9k = U(Xk)XYk. Then (6) implies that the series 00 gk converges. k=1 Ri Now {9k} is a sequence of nonnegative functions in L(R') such that the series F_k 1 J.1 9k converges. Therefore, by the Levi theorem (Theorem 10.25), the series Ex 1 9k converges almost everywhere on R'. In other words, there is a subset T of R' of 1-measure 0 such that the series -00 E u(Xk)XYk(y) converges for all y in R' - T. (7) k=1 Th. 15.6 Fubini's Reduction Theorem for Double Integrals 413 Take a point y in R' - T, keep y fixed and consider the set Sy. We will prove that Sy has 1-measure zero. We can assume that Sy is nonempty; otherwise the result is trivial. Let A(y)= {Xk:yEYk, k= 1,2,...}. Then A(y) is a countable collection of one-dimensional intervals which we relabel as {J1, J2, ... }. The sum of the lengths of all the intervals Jk converges because of (7). If x e Sy, then (x, y) E S so (x, y) E Ik = Xk x Yk for infinitely many k, and hence x e J. for infinitely many k. By the one-dimensional version of Theorem 15.3 it follows that Sy has 1-measure zero. This shows that Sy has 1-measure zero for almost all y in R', and a similar argument proves that S" has 1-measure zero for almost all x in R'. 15.7 FUBINI'S REDUCTION THEOREM FOR DOUBLE INTEGRALS Theorem 15.6. Assume f is Lebesgue-integrable on R2. Then we have: a) There is a set T of 1-measure 0 such that the Lebesgue integral !RI f(x, y) dx exsits for ally in R' - T. b) The function G defined on R1 by the equation f G(Y) f(x, y) dx ifyER' - T, -RI 0 ifyET, is Lebesgue-integrable on R'. c) f f = f G(y) dy. That is, RJ R1 ff f(x, y) d(x, y) = SRI [fRI f(x, y) dxl dy. RZ J NOTE. There is a corresponding result which concludes that y) dy] dx. J f(x, y) d(x, Y) = SRI [SRI' JJ Proof We have already proved the theorem for step functions. We prove it next for upper functions. If f e U(R2) there is an increasing sequence of step functions such that s (x, y) - f(x, y) for all (x, y) in R2 - S, where S is a set of 2-measure 0; also, lim f y) d(x, y) = Jf f(x, y) d(x, y). R_ 00 RZ 414 Multiple Lebesgue Integrals Th. 15.6 Now (x, y) a R2 - S if, and only if, x E R' - Sy. Hence sn(x, y) -+ f(x, y) if x c- R' - Sy. (8) Let tn(y) = $RI sn(x, y) dx. This integral exists for each real y and is an integrable function of y. Moreover, by Theorem 15.2 we have fRl tn(Y) dy = JR1JRI Sn(x, y) dx] dy = Sf s .(x, y) d(x, y) fff J R2 Since the sequence {tn} is increasing, the last inequality shows that limn-J R. tn(y) dy exists. Therefore, by the Levi theorem (Theorem 10.24) there is a function t in L(R') such that t -+ t almost everywhere on R'. In other words, there is a set T1 of 1-measure 0 such that tn(y) --> t(y) if yf e R' - T1. Moreover, t(y) dy = lim tn(y) dy. JR1 n-+CO Rl l Again, since {tn} is increasing, we have tn(Y) = J ;(x, y) dx <_ t(y) if y e R' - T1. l R Applying the Levi theorem to {sn} we find that if y E R' - T1 there is a function g in L(R') such that sn(x, y) -+ g(x, y) for x in R1 - A, where A is a set of 1-measure 0. (The set A depends on y.) Comparing this with (8) we see that if y e R1 - T1 then g(x, y) = f(x, y) if x e R' - (A u Sy). (9) But A has 1-measure 0 and Sy has 1-measure 0 for almost all y, say for all y in R' - T2, where T2 has 1-measure 0. Let T = T1 u T2. Then T has 1-measure 0. If y e R1 - T, the set A v Sy has 1-measure 0 and (9) holds. Since the integral SRI g(x, y) dx exists if y E R1 - T it follows that the integral $R, f(x, y) dx also exists if y E R' - T. This proves (a). Also, if y E R' - T we have Sal f(x, y) dx = fR g (x, y) dx = lim sn(x, y) dx = t(y). (10) 1 n-ao J al Since t e L(R'), this proves (b). Finally, we have f t(y) d y= f lim tn(y) d y= lim f tn(y) d y n1 RI n- oo n-+ao JRI = lim f [JR. sn(x, y) dxl d y = lim f sn(x, y) d(x, y) R= J n~0D ,J R j'j'f(x, y) d(x, y). R2 Th. 15.8 Tonelli-Hobson Test for Integrability 415 Comparing this with (10) we obtain (c). This proves Fubini's theorem for upper functions. To prove it for Lebesgue-integrable functions we write f = u - v, where u e L(R2) and v e L(R2) and we obtain U=U u -J v = $ [f.1 u(x, y) dx] dy - 5 [JRI v(x, y) dx] dY R = SRI [fRi {u(x, y) - v(x, y)) dx] dy = f R r f Rf(x, y) dx] dy. As an immediate corollary of Theorem 15.6 and the two-dimensional analog of Theorem 10.11 we obtain : Theorem 15.7. Assume that f is defined and bounded on a compact rectangle I = [a, b] x [c, d], and that f is continuous almost everywhere on I. Then f E L(I) and we have f J J f (x, y) d(x, y) = fd [fa"' y) dx I dy = fb [$df(xy) dy] dx. J J NOTE. The one-dimensional integral f a f(x, y) dx exists for almost all y in [c, d] as a Lebesgue integral. It need not exist as a Riemann integral. A similar remark applies to the integral f' f(x, y) dy. In the Riemann theory, the inner integrals in the reduction formula must be replaced by upper or lower integrals. (See Theorem 14.6, part (v).) There is, of course, an extension of Fubini's theorem to higher-dimensional integrals. If f is Lebesgue-integrable on R"'+k the analog of Theorem 15.6 concludes that y) dx] dy = f [SRk f(x; y) dyl dx. Rm+kf = IRk [i"1m f(x; Here we have written a point in Rm .k as (x; y), where x e R' and y e W. This can be proved by an extension of the method used to prove the two-dimensional case, but we shall omit the details. 15.8 THE TONELLI-HOBSON TEST FOR INTEGRABILITY Which functions are Lebesgue-integrable on R2? The next theorem gives a useful sufficient condition for integrability. Its proof makes use of Fubini's theorem. Theorem 15.8. Assume that f is measurable on R2 and assume that at least one of the two iterated integrals SRI [5 RI lf(x, Y)l dx] dy or J'Ri [SRI Jf(x, Y)I dy] dx, 416 Multiple Lebesgue Integrals exists. Then we have: Th. 15.8 a) f e L(R2). y) dxJ dy _ SRI [SRI f(x, y) dy] dx. b) ff f = L [SRI f(x, RZ J Proof. Part (b) follows from part (a) because of Fubini's theorem. We will also use Fubini's theorem to prove part (a). Assume that the iterated integral JR. [fRI If(x, y)I dx] dy exists. Let {sn} denote the increasing sequence of nonneg-ative step functions defined as follows: sn(x, y) _ n if IxJ < n and I yI <_ n, 0 otherwise. Let fn(x, y) = min {sn(x, y), I f(x, y)I }. Both s and If I are measurable so fn is measurable. Also, we have 0 < ,(x, y) < sn(x, y), so fn is dominated by a Lebesgue-integrable function. Therefore, fn e L(R2). Hence we can apply Fubini's theorem to fn along with the inequality 0 < fn(x, y) _< I f(x, y)I to obtain y) dx] dy f [S I f(x, y)I dxl dy. J'ffn = SRI [SRl' J RZ Since { fn} is increasing, this shows that the limit limn $.1R2 fn exists. By the Levi theorem (the two-dimensional analog of Theorem 10.24), { fn} converges almost everywhere on R2 to a limit function in L(R2). But fn(x, y) -+ I f(x, y)I as n -+ oo, so if I e L(R2). Since f is measurable, it follows that f e L(R2). This proves (a). The proof is similar if the other iterated integral exists. 15.9 COORDINATE TRANSFORMATIONS One of the most important results in the theory of multiple integration is the formula for making a change of variables. This is an extension of the formula f g(d)f(x) dx = J df[g(t)]g'(t) dt, g(c) which was proved in Theorem 7.36 for Riemann integrals under the assumption that g has a continuous derivative g' on an interval T = [c, d] and that f is continuous on the image g(T). Consider the special case in which g' is never zero (hence of constant sign) on T. If g' is positive on T, then g is increasing, so g(c) < g(d), g(T) = [g(c), g(d)], and the above formula can be written as follows: [g(t)]g'(t) dt. f f(x) dx = fTf 9(T) Th. 15.10 Coordinate Transformations 417 On the other hand, if g' is negative on T, then g(T) = [g(d), g(c)] and the above formula becomes (t)]g'(t) dt. f f(x) dx = - fT f[g 9(T) Both cases are included, therefore, in the single formula J f(x) dx = f T Ig'(t)I dt. (11) g(T) T Equation (11) is also valid when c > d, and it is in this form that the result will be generalized to multiple integrals. The function g which transforms the variables must be replaced by a vector-valued function called a coordinate transformation which is defined as follows. Definition 15.9. Let T be an open subset of W. A vector-valued function g : T --> R" is called a coordinate transformation on T if it has the following three properties: a) g e C' on T. b) g is one-to-one on T. c) The Jacobian determinant Jg(t) = det Dg(t) : 0 for all t in T. NOTE. A coordinate transformation is sometimes called a diffeomorphism. Property (a) states that g is continuously differentiable on T. From Theorem 13.4 we know that a continuously differentiable function is locally one-to-one near each point where its Jacobian determinant does not vanish. Property (b) assumes that g is globally one-to-one on T. This guarantees the existence of a global inverse g-1 which is defined and one-to-one on the image g(T). Properties (a) and (c) together imply that g is an open mapping (by Theorem 13.5). Also, g-1 is continuously differentiable on g(T) (by Theorem 13.6). Further properties of coordinate transformations will be deduced from the following multiplicative property of Jacobian determinants. Theorem 15.10 (Multiplication theorem for Jacobian determinants). Assume that g is differentiable on an open set T in R" and that h is differentiable on the image g(T). Then the composition k = h o g is differentiable on T, and for every t in T we have Jk(t) = Je[g(t)]J,(t) (12) Proof. The chain rule (Theorem 12.7) tells us that the composition k is differen-tiable on T, and the matrix form of the chain rule tells us that the corresponding Jacobian matrices are related as follows : Dk(t) = Dh[g(t)]Dg(t). (13) From the theory of determinants we know that det (AB) = det A det B, so (13) implies (12). 418 Multiple Lebesgue Integrals This theorem shows that if g is a coordinate transformation on T and if h is a coordinate transformation on g(T), then the composition k is a coordinate trans-formation on T. Also, if h = g-1, then k(t) = t for all tin T, and Jk(t) = 1, so Jh[g(t)]J1(t) = 1 and g-1 is a coordinate transformation on g(T). A coordinate transformation g and its inverse g-1 set up a one-to-one corre-spondence between the open subsets of T and the open subsets of g(T), and also between the compact subsets of T and the compact subsets of g(T). The following examples are commonly used coordinate transformations. Example 1. Polar coordinates in R2. In this case we take T = {(t1, t2) : t1 > 0, 0 < t2 < 27r}, and we let g = (g1, g2) be the function defined on T as follows: g1(t) = t1 cos t2, g2(t) = t1 sin t2. It is customary to denote the components of t by (r, 0) rather than (t1, t2). The co-ordinate transformation g maps each point (r, 0) in T onto the point (x, y) in g(T) given by the familiar formulas x = r cos 0, y = r sin 0. The image g(T) is the set R2 - {(x, 0) : x >- 0}, and the Jacobian determinant is j5(t) = cos 0 sin 0 _ r I- r sin o r cos B Example 2. Cylindrical coordinates in R3. Here we write t = (r, 0, z) and we take T = {(r, o, z) : r > 0, 0 < 0 < 2n, - oo < z < + oo }. The coordinate transformation g maps each point (r, o, z) in T onto the point (x, y, z) in g(T) given by the equations x = r cos o, y = r sin 0, z = Z. T (x, Y, Z) I r Figure 15.3 x Coordinate Transformations 419 The image g(T) is the set R3 - {(x, 0, 0) : x >- 0}, and the Jacobian determinant is given by cos 0 sin 0 0 - r sin 0 r cos 0 0 0 0 1 = r. The geometric significance of r, 0, and z is shown in Fig. 15.3. Example 3. Spherical coordinates in R3. In this case. we write t = (p, 0, gyp) and we take T= {(p,0,rp):p>0, 0<0<2,r, 0< 9P 0) u {(0, 0, z) : z e R}), and the Jacobian determinant is cos 0 sin (p sin 0 sin q cos - p sin 0 sin ip p cos 0 sin rp 0 p cos 0 cos ip p sin 0 cos ip - p sin ip = - p2 sin (p. The geometric significance of p, 0, and (p is shown in Fig. 15.4. P .4 (x,y,Z) P COs q y Figure 15.4 Psinq x Example 4. Linear transformations in R. Let g : R" - R", be a linear transformation represented by a matrix (a1J) = m(g), so that " n g(t) aljtj,... , aits J=1 Then g = (gi, ... , g") where gi(t) _ 1 a1Jt1, and the Jacobian matrix is Dg(t) = (DJgi(t)) = (ai,). Thus the Jacobian determinant J,(t) is constant, and equals det (a, j), the determinant of the matrix (ai J). We also call this the determinant of g and we write det g = det (at J). 420 Multiple Lebesgue Integrals A linear transformation g which is one-to-one on R" is called nonsingular. We shall use the following elementary facts concerning nonsingular transforma-tions from R" to R". (Proofs can be found in any text on linear algebra; see also Reference 14.1.) A linear transformation g is nonsingular if, and only if, its matrix A = m(g) has an inverse Asuch that AA-1 = I, where I is the identity matrix (the matrix of the identity transformation), in which case A is also called nonsingular. An n x n matrix A is nonsingular if, and only if, det A : 0. Thus, a linear function g is a coordinate transformation if, and only if, det g # 0. Every nonsingular g can be expressed as a composition of three special types of nonsingular transformations called elementary transformations, which we refer to as types a, b, and c. They are defined as follows : Type a: ga(t1, ... , tk, ... , t") _ (t1, ... , ttk, ... , t"), where A.: 0. In other words, ga multiplies one component of t by a nonzero scalar )L. In particular, ga maps the unit coordinate vectors as follows : ga(Uk) = ).Uk for some k, ga(ui) = u, for all i # k. The matrix of ga can be obtained by multiplying the entries in the kth row of the identity matrix by A. Also, det ga = A. Type b: gb(ti, ... , tk, ... , t") = (t1, ... , tk + t1, ... , t"), where j # k. Thus, gb replaces one component of t by itself plus another. In particular, gb maps the coordinate vectors as follows: gb(uk) = Uk + uj for some fixed k and j, k # j, gb(ui) = ui for all i # k. The matrix gb can be obtained from the identity matrix by replacing the kth row of I by the kth row of I plus the jth row of I. Also, det gb = 1. Type c: ge(d, ... , ti, ... , ti, ... , tn) = (t 1, ... , t;) ... , ti, ... , te), where i # j. That is, gc interchanges the ith and jth components of t for some i and j with i # j. In particular, g(ui) = u1, g(u;) = ui, and g(uk) = uk for all k # i, k j. The matrix of g, is the identity matrix with the ith and jth rows interchanged. In this case det gc = - 1. The inverse of an elementary transformation is another of the same type. The matrix of an elementary transformation is called an elementary matrix. Every nonsingular matrix A can be transformed to the identity matrix I by multiplying A on the left by a succession of elementary matrices. (This is the familiar Gauss-Jordan process of linear algebra.) Thus, I= T1T2...T,A, where each Tk is an elementary matrix, Hence, A = V 1 ... TZ 1 T1 1. Th. 15.12 Transformation Formula for Linear Coordinate Transformations 421 If A = m(g), this gives a corresponding factorization of g as a composition of elementary transformations. 15.10 THE TRANSFORMATION FORMULA FOR MULTIPLE INTEGRALS The rest of this chapter is devoted to a proof of the following transformation formula for multiple integrals. Theorem 15.11. Let T be an open subset of R" and let g be a coordinate transfor-mation on T. Let f be a real-valued function defined on the image g(T) and assume that the Lebesgue integral cg(T)f(x) dx exists. Then the Lebesgue integral 1T f [g(t)] IJg(t)I dt also exists and we have f f(x) dx = r f[g(t)] IJJ(t)I d t. (14) g(T) T The proof of Theorem 15.11 is divided into three parts. Part 1 shows that the formula holds for every linear coordinate transformation a. As a corollary we obtain the relation p[a(A)] = Idet al µ(A), for every subset A of R" with finite Lebesgue measure. In part 2 we consider a general coordinate transformation g and show that (14) holds when f is the characteristic function of a compact cube. This gives us µ(K) = IJg(t)I dt, (15) s '(K) for every compact cube K in g(T). This is the lengthiest part of the proof. In part 3 we use Equation (15) to deduce (14) in its general form. 15.11 PROOF OF THE TRANSFORMATION FORMULA FOR LINEAR COORDINATE TRANSFORMATIONS Theorem 15.12. Let a : R" -+ R" be a linear coordinate transformation. If the Lebesgue integral fR" f(x) dx exists, then the Lebesgue integral f R"f[a(t)] IJJ(t)I dt also exists, and the two integrals are equal. Proof. First we note that if the theorem is true for a and ft, then it is also true for the composition y = a o ft because f f(x) dx = $Rflf1111 IJ(t)I dt = fa.f(a[fl(t)]) J[/S(t)]I J,(t)I " _ .f'[y(t)] IJY(t)I dt, R" since J7(t) = Jj[ft(t)] JJ(t). 422 Multiple L,ebesgue Integrals Th. 15.13 Therefore, since every nonsingular linear transformation a is a composition of elementary transformations, it suffices to prove the theorem for every elemen-tary transformation. It also suffices to assume f >- 0. Suppose a is of type a. For simplicity, assume that a multiplies the last component of t by a nonzero scalar A, say a(tl, ... , t") = (t1, . . . , to-1, At"). Then IJJ(t)I = (det al = 1).J. We apply Fubini's theorem to write the integral off over R" as the iteration of an (n - 1)-dimensional integral over R"-1 and a one-dimensional integral over R1. For the integral over R1 we use Theorem 10.17(b) and (c), and we obtain , ... , x") dx" dx1 ... dx"-1 f fdx = J [Jf(xi fR.1 [lei Jf(xi. $Rn_1 [Jf[x(t)I IJ.(t)l dtn] dt1 ... dt"-1 jf[(t)] IJ.(t)l dt, " 1 where in the last step we use the Tonelli-Hobson theorem. This proves the theorem if a is of type a. If a is of type b, the proof is similar except that we use Theorem 10.17(a) in the one-dimensional integral. In this case IJJ(t)I = 1. Finally, if a is of type c we simply use Fubini's theorem to interchange the order of integration over the ith and jth coordinates. Again, IJ.(t)l = 1 in this case. As an immediate corollary we have: Theorem 15.13. If a : R" -+ R" is a linear coordinate transformation and if A is any subset of R" such that the Lebesgue integral f(x) dx exists, then the Lebesgue integral JA f [a(t)] IJ.(t)l dt also exists, and the two are equal. Proof. Let J(x) = f(x) if x e a(A), and let J(x) = 0 otherwise. Then LA) f(x) dx = SRn .fi(x) dx = $Rfl f[a(t)] IJJ(t)I dt = IA f[a(t)] IJ.(t)I dt. s a corollary of Theorem 15.13 we have the following relation between th A measure of A_and the measure of a(A). Th. 15.15 Transformation Formula for a Compact Cube 423 Theorem 15.14. Let a : R" -+ R" be a linear coordinate transformation. If A is a subset of R" with finite Lebesgue measure µ(A), then a(A) also has finite Lebesgue measure and p[a(A)] = Idet al p(A). (16) Proof. Write A = a-1(B), where B = a(A). Since a-1 is also a coordinate transformation, we find µ(A) = f A = dx = J Idet a-' I d t= Idet a-' I p(B). A JB This proves (16) since B = a(A) and det (a-1) = (det a)-1. Theorem 15.15. If A is a compact Jordan-measurable subset of R", then for any linear coordinate transformation a : R" -+ R" the image a(A) is a compact Jordan-measurable set and its content is given by c[a(A)] = Idet al c(A). Proof. The set a(A) is compact because a is continuous on A. To prove the theorem we argue as in the proof of Theorem 15.14. In this case, however, all the integrals exist both as Lebesgue integrals and as Riemann integrals. 15.12 PROOF OF THE TRANSFORMATION FORMULA FOR THE CHARACTERISTIC FUNCTION OF A COMPACT CUBE This section contains part 2 of the proof of Theorem 15.11. Throughout the section we assume that g is a coordinate transformation on an open set T in R. Our purpose is to prove that u(K) = l J.(T)l d t, ft-I(K) for every compact cube K in T. The auxiliary results needed to prove this formula are labelled as lemmas. To help simplify the details, we introduce some convenient notation. Instead of the usual Euclidean metric for R" we shall use the metric d given by d(x, y) = max Ixi - yii. Isis" This metric was introduced in Example 9, Section 3.13. In this section only we shall write Ilx - yll for d(x, y). With this metric, a ball B(a; r) with center a and radius r is an n-dimensional cube with center a and edge-length 2r; that is, B(a; r) is the cartesian product of n one-dimensional intervals, each of length 2r. The measure of such a cube is (2r)", the product of the edge-lengths. 424 Multiple Lebesgue Integrals Lemma 1 If a : R" -+ R" is a linear transformation represented by a matrix (ai j), so that n n 11 a(X) = (a,jxj, ... , E a"jxj), j=1 j=1 then II a(x) ll = max 15i<n We also define aijxjl IIXII max E laijl. j=1 15i5nj=1 (17) n Il ll = max E laijl. (18) 1<_i<_n j=1 This defines a metric Ila - P11 on the space of all linear transformations from R" to R". The first lemma gives some properties of this metric. Lemma 1. Let a and J denote linear transformations from R" to R". Then we have: a) Ilall = Ila(x)II for some x with 11x11 = 1. b) 11a(x)11 <- hall Ilx11 for all x in R". c) 11a ° P11 <- hall IIill. d) IIIII = 1, where I is the identity transformation. Proof. Suppose that max, s i5n E;=1 Iaijl is attained for i = p. Take xp = 1 if apj>-- 0,xp= -1 ifapj <0,andxj =0ifj p. Then 11x11 = I and 11211 = Ila(x)ll, which proves (a). Part (b) follows at once from (17) and (18). To prove (c) we use (b) to write 11((Z° P)(x)ll = Ila(P(x))II < II«II IIP(x)II < Ilall IIPII Ilxll Taking x with Ilxll = 1 so that 11(a ° P)(x)ll = Ila ° P11, we obtain (c). Finally, if I is the identity transformation, then each sum E1=1 IaijI = 1 in (18)so11111=1. The coordinate transformation g is differentiable on T, so for each t in T the total derivative g'(t) is a linear transformation from R" to R" represented by the Jacobian matrix Dg(t) = (Djgi(t)). Therefore, taking a = g'(t) in (18), we find n Ilg'(t)II = max E IDjgi(t)I 1<i5n j=1 We note that llg'(t)II is a continuous function of t since all the partial derivatives Djgi are continuous on T. If Q is a compact subset of T, each function Djgi is bounded on Q; hence llg'(t)II is also bounded on Q, and we define teQ tGQ 15i_n j=1 ' g(Q) = sup Ilg'(t)II = sup { max IDjgi(t)I} (19) Lemma 3 Transformation Formula for a Compact Cube 425 The next lemma states that the image g(Q) of a cube Q of edge-length 2r lies in another cube of edge-length 2r t g(Q). Lemma 2. Let Q = {x : 1l x - all < r} be a compact cube of edge-length 2r lying in T. Then for each x in Q we have Ilg(x) - g(a)ll < r),g(Q) Therefore g(Q) lies in a cube of edge-length 2r).g(Q). Proof. By the Mean-Value theorem for real-valued functions we have g.(x) - g1(a) = Og,(z,) - (x - a) _ D;g1(zj)(x - a;), j=1 (20) where z1 lies on the line segment joining x and a. Therefore n lg1(x) - g.(a)l < ID;g1(zi)l lxi - a,l <- Ilx - all E lD,g1(z1)I :5 rl.(Q), i=1 i=1 and this implies (20). NOTE. Inequality (20) shows that g(Q) lies inside a cube of content (2r),g(Q))" = {1g(Q)}"c(Q) Lemma 3. If A is any compact Jordan-measurable subset of T, then g(A) is a com-pact Jordan-measurable subset of g(T). Proof. The compactness of g(A) follows from the continuity of g. Since A is Jordan-measurable, its boundary OA has content zero. Also, 8(g(A)) = g(8A), since g is one-to-one and continuous. Therefore, to complete the proof, it suffices to show that g(8A) has content zero. Given e > 0, there is a finite number of open intervals A1, ... , Am lying in T, the sum of whose measures is < e, such that 8A S U"'1 A;. By Theorem 15.1, this union can also be expressed as a union U(e) of a countable disjoint collection of cubes, the sum of whose measures is < e. If e < 1 we can assume that each cube in U(e) is contained in U(1). (If not, intersect the cubes in U(e) with U(l) and apply Theorem 15.1 again.) Since 8A is compact, a finite subcollection of the cubes in U(e) covers 8A, say Q1, ... , Qk. By Lemma 2, the image g(Q1) lies in a cube of measure {).g(Q;)}"c(Q;). Let J. = ).g(U(1)). Then ),g(Q;) < A since Q; c U(1). Thus g(8A) is covered by a finite number of cubes, the sum of whose measures does not exceed A" Y_k=, c(Q,) < ei". Since this holds for every e < 1, it follows that g(8A) has Jordan content 0, so g(A) is Jordan-measurable. The next lemma relates the content of a cube Q with that of its image g(Q). 426 Multiple Lebesgue Integrals Lemma 4 Lemma 4. Let Q be a compact cube in T and let h = a o g, where a : R" - R" is any nonsingular linear transformation. Then c[g(Q)] < Idet al -1{Ah(Q)}"c(Q) (21) Proof. From Lemma 2 we have c[g(Q)] < { .(Q)}"c(Q). Applying this inequality to the coordinate transformation h, we find c[h(Q)] < {Ah(Q)}"c(Q) But by Theorem 15.15 we have c[h(Q)] = c[a(g(Q))] = Idet al c[g(Q)], so c[g(Q)] = Idet al -1 c[h(Q)] < Idet al -1 {Ae(Q)}"c(Q) Lemma 5. Let Q be a compact cube in T. Then for every e > 0, there is a b > 0 such that ifteQand aeQwe have 11g'(a)-1 o g'(t) 11 < 1 + e whenever lit - all < 6. (22) Proof. The function Ilg'(t)-111 is continuous and hence bounded on Q, say Ilg'(t)-1il < M for all tin Q where M > 0. By the continuity of Ilg'(t)ll, there is a & > 0 such that II g'(t) - g'(a) ll < M whenever lit - all < 5. If I denotes the identity transformation, then g'(a)-1 o g'(t) - I(t) = g'(a)-1 o {g'(t) - g'(a)}, so if lit - all < S we have I191(a)-1 o g'(t) - I(t)II < Ilg'(a)-' II Ilg'(t) - g'(a)ll < M M = a. The triangle inequality gives us Ilall IIPII + Ila - P11. Taking a = g'(a)-1 o g'(t) and we obtain (22). Lemma 6. Let Q be a compact cube in T. Then we have z(t)I dt. c[g(Q)] < fo I Proof. The integral on the right exists as a Riemann integral because the in grand is continuous and bounded on Q. Therefore, given s > 0, there is a partition PP of Q such that for every Riemann sum S(P, IJ,I) with P finer than Pe we have IS(P, J dt Q < a. Take such a partition P into a finite number of cubes Q1, ... , Q,", each of which Lemma 7 Transformation Formula for a Compact Cube 427 has edge-length <S, where S is the number (depending on s) given by Lemma 5. Let ai denote the center of Qi and apply Lemma 4 to Qi with a = g'(ai)-' to obtain the inequality c[g(Qi)] 5 Idet gr(ai)l {Ah(Qi)}" c(Qi), (23) where h = a o g. By the chain rule we have h'(t) = a'(x) o g'(t), where x = g(t). But a'(x) = a since a is a linear function, so hi(t) = a o g'(t) = g'(ai)-1 o g'(t). But by Lemma 5 we have IIh'(t)ll < 1 + s if t e Qi, so /tb(Q1) = sup 11b'(011 <_ 1 + S. teQ; Thus (23) gives us c[g(Qi)] <_ Idet g'(ai)1 (1 + s)" c(Q). Summing over all i, we find m c[g(Q)] < (1 + e)" E Idet g'(ai)I c(Q). i=1 Since det g'(ai) = Jg(a), the sum on the right is a Riemann sum S(P, IJg1 ), and since S(P, IJgi) < fe IJg(t)I dt + e, we find c[g(Q)] < (1 + e)" JQ IJ,(t)I dt + sl But a is arbitrary, so this implies c[g(Q)] < fe IJ,(t)I dt. Lemma 7. Let K be a compact cube in g(T). Then µ(K) < IJg(t)I dt. (24) g-'(K) Proof. ' The integral exists as a Riemann integral because the integrand is con-tinuous on the compact set g-1(K). Also, by Lemma 3, the integral over g-'(K) is equal to that over the interior of g- '(K). By Theorem 15.1 we can write OD int g-1(K) = U Ai, i=1 where {A1i A2, ... } is a countable disjoint collection of cubes whose closure lies in the interior of g-'(K). Thus, int g- 1(K) = UJ' 1 Qi where each Qi is the closure of A i. Since the integral in (24) is also a Lebesgue integral, we can use countable additivity along with Lemma 6 to write 00 IJ (t)I dt =r (' IJ (t)I dt t LI J g P[g(Qt)] = ul U g(Qi) I = u(K) g-1(K) i=1 Qi i=1 i=1 JJJ 428 Multiple Lebesgue Integrals Lemma 8 Lemma 8. Let K be a compact cube in g(T). Then for any nonnegative upper function f which is bounded on K, the integral 1. (K)f[g(t)] IJ1(t)I dt exists, and we have the inequality f f(x) dx < f f [g(t)] IJg(t)I dt. (25) K g (K) Proof. Let s be any nonnegative step function on K. Then there is a partition of K into a finite number of cubes K1, ... , K, such that s is constant on the interior of each Ki, say s(x) = ai >_ 0 if x e int Ki. Apply (24) to each cube Ki, multiply by ai and add, to obtain f s(x) dx < f s[g(t)] IJg(t)I dt. K g '(K) (26) Now let {sk} be an increasing sequence of nonnegative step functions which converges almost everywhere on K to the upper function f. Then (26) holds for each sk, and we let k -+ oo to obtain (25). The existence of the integral on the right follows from the Lebesgue bounded convergence theorem since both f [g(t)] and IJg(t)I are bounded on the compact set g-(K). Theorem 15.16. Let K be a compact cube in g(T). Then we have µ(K) = IJg(t)I dt. g-'(K) Proof. In view of Lemma 7, it suffices to prove the inequality IJg(t)I dt < µ(K) As in the proof of Lemma 7, we write (27) (28) int g-1(K) = U Ai = U Qi' i=1 i=1 where {A 1, A2, ... } is a countable, disjoint collection of cubes and Qi is the closure of A i. Then f IJg(t)) dt = f IJg(t)I dt. (29) Jg '(K) '-1 Qi Now we apply Lemma 8 to each integral f., IJg(t)I dt, taking f = IJgi and using the coordinate transformation h = g-1. This gives us the inequality J IJg(t)I dt < JS(Qi) IJg[h(u)]I IJh(u)) du = J du = µ[g(Qi)], QI g(Q,) which, when used in (29) gives (28). Th. 15.17 Completion of the Proof of the Transformation Formula 15.13 COMPLETION OF THE PROOF OF THE TRANSFORMATION FORMULA 429 Now it is relatively easy to complete the proof of the formula f [g(t)] IJg(t)I dt, (30) f f(x) dx = fT g(T) under the conditions stated in Theorem 15.11. That is, we assume that T is an open subset of R", that g is a coordinate transformation on T, and that the integral on the left of (30) exists. We are to prove that the integral on the right also exists and that the two are equal. This will be deduced from the special case in which the integral on the left is extended over a cube K. Theorem 15.17. Let K be a compact cube in g(T) and assume the Lebesgue integral $K f(x) dx exists. Then the Lebesgue integral $g -,(K) f [g(t)] IJg(t)I dt also exists, and the two are equal. Proof. It suffices to prove the theorem when f is an upper function on K. Then there is an increasing sequence of step functions {sk} such that sk -> f almost everywhere on K. By Theorem 15.16 we have fK Sk(x) dx = ft-sk[g(t)] IJg(t)) dt, '(K) for each step function sk. When k --, oo, we have f K sk(x) dx -+ f K f(x) dx. Now let fk(t) _ to IJg(t)) if t e g-'(K), 0 if teR" - g-'(K). Then I fk(t) d t = J Sk[g(t)] )Jg(t)I d t= J Sk(x) dx, R" g (K) K so lim f fk(t) d t = lim f sk(x) dx = f f(x) dx. k cc R" k-ao JK JK By the Levi theorem (the analog of Theorem 10.24), the sequence { fk} converges almost everywhere on R" to a function in L(R"). Since we have I'M fk(t) = to f[g(t)] IJg(t)I if t e g- '(K), k-ao if t o R" - g-'(K), almost everywhere on R", it follows that the integral f g-1(K) f [g(t)] )Jg(t)I dt exists and equals 1K f(x) A. This completes the proof of Theorem 15.17. 430 Multiple Lebesgue Integrals Proof of Theorem 15.11. Now assume that the integral 1 S(T) f(x) dx exists. Since g(T) is open, we can write g(T) = U A;, =1 where {A1, A2, ... } is a countable disjoint collection of cubes whose closure lies in g(T). Let K; denote the closure of A;. Using countable additivity and Theorem 15.17 we have f f(x) dx = ff(x) dx (T) i-1 K 00 J .f [g(t)] IJs(t)I dt i= 1 8 I(R[) = ff[(t)] IJ5(t)I d t. EXERCISES 15.1 If f e L(T), where T is the triangular region in R2 with vertices at (0, 0), (1, 0), and (0, 1), prove that f(x, y) d(x, y) = 1 s f rI f(x, y) dy I dx = 1 1 I II f(x, y) dx I dy. f r JT o o .J 0 L .J 15.2 For fixed c, 0 < c < 1, define f on R2 as follows: f(x, y) = {(1 - y)c/(x - y)if 0 < y < x, 0 < x < 1, 0 otherwise. Prove that f e L(R2) and calculate the double integral f12 f(x, y) d(x, y). 15.3 Let S be a measurable subset of R2 with finite measure u(S). Using the notation of Definition 15.4, prove that AS) = J p(SX) dx = J u(S,) dy. 00 15.4 Let f (x, y) = e-x' sin x sin y if x >- 0, y > 0, and let f (x, y) = 0 otherwise. Prove that both iterated integrals f [f11, y) dx] dy and f [Jaf(x, y) dy] dx exist and are equal, but that the double integral off over R2 does not exist. Also, explain why this does not contradict the Tonelli-Hobson test (Theorem 15.8). Exercises 431 15.5 Let f(x, y) = (x2 - y2)/(x2 + y2)2 for 0 < x < 1, 0 < y < 1, and let f(0, 0) _ 0. Prove that both iterated integrals fl [fo R X, y) dy] dx and 101 [fe' f(x, y) dx] dY exist but are not equal. This shows that f is not Lebesgue-integrable on [0, 1 ] x [0, 1 ]. 15.6 Let I = [0, 1 ] x [0, 1 ], let f(x, y) = (x - y)/(x + y)3 if (x, y) a I, (x, y) (0, 0), and let f(0, 0) = 0. Prove that f 0 L(I) by considering the iterated integrals f(x, Y) dx] dy. f [f f(x, Y) dy] dx and ji Ifo' 15.7 Let I = [0, 11 x [1, + oo) and let f(x, y) = e-I -2e_2xy if (x, y) a I. Prove that f 0 L(I) by considering the iterated integrals Jo I floo f(x, y) dy] dx and f i Ifo, f(x, y) dx] dy. 15.8 The following formulas for transforming double and triple integrals occur in ele-mentary calculus. Obtain them as consequences of Theorem 15.11 and give restrictions on T and T' for validity of these formulas. a) fff(x, y) dx dy = fff(r cos 0, r sin 0)r dr dB. T T' b) ffff(x, y, z) dx dy dz = fffi(r cos 0, r sin 0, z)r dr dO dz. T T' c) ffff(x y, z) dx dy dz T = ffff(P cos 0 sin ip, p sin 0 sin gyp, p cos rp) p2 sin p dp d9 dip. T' 15.9 a) Prove that fR2 e- 2+Y2) d(x, y) = x by transforming the integral to polar coordinates. b) Use part (a) to prove that f--. e_x2 dx =tin. c) Use part (b) to prove that JR e- II it2 d(xl,... , x") = e2. d) Use part (b) to calculate J°_° a-"2 dx and f°_° x2 e-t"2 dx, t > 0. 15.10 Let V"(a) denote the n-measure of the n-ball B(0; a) of radius a. This exercise outlines a proof of the formula nn/tan V"(a) = r(}n + 1) a) Use a linear change of variable to prove that Vn(a) = a"V"(1). 432 Multiple Lebesgue Integrals b) Assume n >- 3, express the integral for V"(1) as the iteration of an (n - 2)-fold integral and a double integral, and use part (a) for an (n - 2)-ball to obtain the formula 2" 1 V"(1) = V"-2(1) fo r f (1 -r2)"12-1r dr] dB = V"_2(1) 2n o J J n c) From the recursion formula in (b) deduce that V"(1) nn/2 r(+n + 1) 15.11 Refer to Exercise 15.10 and prove that f X2 d(x1,... , xn) = V"(1) 9(071) n + 2 for each k = 1, 2, ... , n. 15.12 Refer to Exercise 15.10 and express the integral for V"(1) as the iteration of an (n - 1)-fold integral and a one-dimensional integral, to obtain the recursion formula 1 V"(1) = 2V"_1(1) r (1 -x2)(,,-1)12 dx. 0 Put x = cos tin the integral, and use the formula of Exercise 15.10 to deduce that fo /2 cos" t dt = 2 r(In + 1) ,J 15.13 If a > 0, let S"(a) = {(x1,. .. , x.): jxl i + -- - + Ix"i <- a}, and let V"(a) denote the n-measure of S"(a). This exercise outlines a proof of the formula V"(a) = 21a"/n!. a) Use a linear change of variable to prove that V"(a) = a"V"(1). b) Assume n >- 2, express the integral for V"(1) as an iteration of a one-dimensional integral and an (n - 1)-fold integral, use (a) to show that ('1 V"(1) = V"-1(1) i (1 -lxl)"-1 dx = 2V"-1(1)ln, JJ 1 and deduce that V"(1) = 2"/n!. 15.14 If a > 0 and n >- 2, let S"(a) denote the following set in R": S"(a) = {(x1, ... , x") : Ixti + Ix"i < a for each i = 1, ... , n - 1 }. Let V"(a) denote the n-measure of S"(a). Use a method suggested by Exercise 15.13 to prove that V"(a) = 2"a"/n. 15.15 Let Q"(a) denote the "first quadrant" of the n-ball B(0: a) given by Q"(a) = {(xl,... , x") : jlxDD s a and 0 <- xi <- a for each i = 1, 2, ... , n}. Let f(x) = x1... x" and prove that f f(x) dx = a2" J Qn(a) SUGGESTED REFERENCES FOR FURTHER STUDY 15.1 Asplund, E., and Bungart, L., A First Course in Integration. Holt, Rinehart, and Winston, New York, 1966. 15.2 Bartle, R., The Elements of Integration. Wiley, New York, 1966. 15.3 Kestelman, H., Modern Theories of Integration. Oxford University Press, 1937. 15.4 Korevaar, J., Mathematical Methods, Vol. 1. Academic Press, New York, 1968. 15.5 Riesz, F., and Sz.-Nagy, B., Functional Analysis. L. Boron, translator. Ungar, New York, 1955. CHAPTER 16 CAUCHY'S THEOREM AND THE RESIDUE CALCULUS 16.1 ANALYTIC FUNCTIONS The concept of derivative for functions of a complex variable was introduced in Chapter 5 (Section 5.15). The most important functions in complex variable theory are those which possess a continuous derivative at each point of an open set. These are called analytic functions. Definition 16.1. Let f = u + iv be a complex-valued function defined on an open set S in the complex plane C. Then f is said to be analytic on S if the derivative f' exists and is continuous at every point of S. NOTE. If T is an arbitrary subset of C (not necessarily open), the terminology '!f is analytic on T" is used to mean that f is analytic on some open set containing T. In particular, f is analytic at a point z if there is an open disk about z on which f is analytic. It is possible for a function to have a derivative at a point without being analytic at the point. For example, if f(z) = Iz I2, then f has a derivative at 0 but at no other point of C. Examples of analytic functions were encountered in Chapter 5. If f(z) = z" (where n is a positive integer), then f is analytic everywhere in C and its derivative is f'(z) = nz"-1. When n is a negative integer, the equation f(z) = z" if z # 0 defines a function analytic everywhere except at 0. Polynomials are analytic everywhere in C, and rational functions are analytic everywhere except at points where the denominator vanishes. The exponential function, defined by the formula e= = ?(cos y + i sin y), where z = x + iy, is analytic everywhere in C and is equal to its derivative. The complex sine and cosine functions (being linear combinations of exponentials) are also analytic everywhere in C. Let f(z) = Log z if z # 0, where Log z denotes the principal logarithm of z (see Definition 1.53). Then f is analytic everywhere in C except at those points z = x + iy for which x S 0 and y = 0. At these points, the principal logarithm fails to be continuous. Analyticity at the other points is easily shown by verifying It can be shown that the existence of f' on S automatically implies continuity of f' on S (a fact di9overed by Goursat in 1900). Hence an analytic function can be defined as one which merely possesses a derivative everywhere on S. However, we shall include continuity off' as part of the definition of analyticity, since this allows some of the proofs to run more smoothly. 434 Paths and Curves in the Complex Plane 435 that the real and imaginary parts of f satisfy the Cauchy-Riemann equations (Theorem 12.6). We shall see later that analyticity at a point z puts severe restrictions on a function. It implies the existence of all higher derivatives in a neighborhood ofz and also guarantees the existence of a convergent power series which represents the function in a neighborhood of z. This is in marked contrast to the behavior of real-valued functions, where it is possible to have existence and continuity of the first derivative without existence of the second derivative. 16.2 PATHS AND CURVES IN THE COMPLEX PLANE Many fundamental properties of analytic functions are most easily deduced with the help of integrals taken along curves in the complex plane. These are called contour integrals (or complex line integrals) and they are discussed in the next section. This section lists some terminology used for different types of curves, such as those in Fig. 16.1. 1,-a w 0 are Jordan are closed curve Jordan curve Figure 16.1 We recall that a path in the complex plane is a complex-valued function y, continuous on a compact interval [a, b]. The image of [a, b] under y (the graph of y) is said to be a curve described by y and it is said to join the points y(a) and y(b). If y(a) y(b), the curve is called an arc with endpoints y(a) and y(b). If y is one-to-one on [a, b], the curve is called a simple arc or a Jordan arc. If y(a) = y(b), the curve is called a closed curve. If y(a) = y(b) and if y is one-to-one on the half-open interval [a, b), the curve is called a simple closed curve, or a Jordan curve. The path y is called rectifiable if it has finite arc length, as defined in Section 6.10. We recall that y is rectifiable if, and only if, y is of bounded variation on [a, b]. (See Section 7.27 and Theorem 6.17.) A path y is called piecewise smooth if it has a bounded derivative y' which is continuous everywhere on [a, b] except (possibly) at a finite number of points. At these exceptional points it is required that both right- and left-hand derivatives exist. Every piecewise smooth path is rectifiable and its arc length is given by the integral f; by'(t)I dt. A piecewise smooth closed path will be called a circuit. 436 Cauchy's Theorem and the Residue Calculus Def. 16.2 Definition 16.2. If a e C and r > 0, the path y defined by the equation y(0)=a+re'°, 0<0<27r, is called a positively oriented circle with center at a and radius r. NOTE. The geometric meaning of y(O) is shown in Fig. 16.2. As 0 varies from 0 to 2n, the point y(0) moves counterclockwise around the circle. 16.3 CONTOUR INTEGRALS Contour integrals will be defined in terms of complex Riemann-Stieltjes integrals, discussed in Section 7.27. Definition 16.3. Let y be a path in the complex plane with domain [a, b], and let f be a complex-valued function defined on the graph of y. The contour integral off along y, denoted by J .Yf, is defined by the equation f f = J f[y(t)] dy(t), y a whenever the Riemann-Stieltjes integral on the right exists. NOTATION. We also write J. Y(b) f(z) dz or fY(a) f(z) dz, J for the integral. The dummy symbol z can be replaced by any other convenient symbol. For example, J, f(z) dz = 1, f(w) dw. If y is rectifiable, then a sufficient condition for the existence of f y f is that f be continuous on the graph of y (Theorem 7.27). The effect of replacing y by an equivalent path (as defined in Section 6.12) is, at worst, a change in sign. In fact, we have: Theorem 16.4. Let y and 6 be equivalent paths describing the same curve IF. If fy f exists, then f j f also exists. Moreover, we have Th. 16.6 Contour Integrals if y and S trace out IF in the same direction, whereas J. 437 if y and S trace out IF in opposite directions. Proof. Suppose S(t) = y[u(t)] where u is strictly monotonic on [c, d]. From the change-of-variable formula for Riemann-Stieltjes integrals (Theorem 7.7) we have fu(d)f[y(t)] dy(t) = J f[S(t)] d6(t) = fa v(c) If u is increasing then u(c) = a, u(d) = b and (1) becomes f }, f = f j f. (1) If u is decreasing then u(c) = b, u(d) = a and (1) becomes - Jy f = jj f. The reader can easily verify the following additive properties of contour integrals. Theorem 16.5. Let y be a path with domain [a, b]. i) If the integrals f .,f and f Y g exist, then the integral fY (af + fig) exists for every pair of complex numbers a, fi, and we have f7 (af+fig) =aff+ fi f y v f, ii) Let yl and y2 denote the restrictions of y to [a, c] and [c, b], respectively, where a < c < b. If two of the three integrals in (2) exist, then the third also exists and we have 1. f+ Yif (2) In practice, most paths of integration are rectifiable. For such paths the following theorem is often used to estimate the absolute value of a contour integral. Theorem 16.6. Let y be a rectifiable path of length A(y). If the integral f Y f exists, and if I f(z)l 5 M for all z on the graph of y, then we have the inequality JYf Proof. We simply observe that all Riemann-Stieltjes sums which occur in the definition of f f [y(t)] dy(t) have absolute value not exceeding MA(y). Contour integrals taken over piecewise smooth curves can be expressed as Riemann integrals. The following theorem is an easy consequence of Theorem 7.8. f= J f 438 Cauchy's Theorem and die Residue Calculus Th. 16.7 Theorem 16.7. Let y be a piecewise smooth path with domain [a, b]. If the contour integral fY f exists, we have J f = J bf[y(t)] y'(t) dt. Y c 16.4 THE INTEGRAL ALONG A CIRCULAR PATH AS A FUNCTION OF THE RADIUS Consider a circular path y of radius r >: 0 and center a, given by y(9)=a+ret°, 050<2ir. In this section we study the integral 1. f as a function of the radius r. Let 9(r) = JY f. Since y'(9) = ireiB, Theorem 16.7 gives us pp(r) = fo ZR f(a + reB)ire'B d9. (3) As r varies over an interval [r1, r2], where 0 < r1 < r2, the points y(9) trace out an annulus which we denote by A(a; r1, r2). (See Fig. 16.3.) Thus, A(a; r1, r2) = {z: r1 5 lz - al < r2}. If r1 = 0 the annulus is a closed disk of radius r2. If f is continuous on the annulus, then (p is continuous on the interval [r1, r2]. If f is analytic on the annulus, then (P is differentiable on [r1, r2]. The next theorem shows that ap is constant on [r1, r2] if f is analytic everywhere on the annulus except possibly on a finite subset, pro-vided that f is continuous on this subset. Figure 16.3 Theorem 16.8. Assume f is analytic on the annulus A(a; r1, r2), except possibly at a finite number of points. At these exceptional points assume that f is continuous. Then the function (p defined by (3) is constant on the interval [r1, r2]. Moreover, if r1 = 0 the constant is 0. Proof. Let z1, ... , z denote the exceptional points where f fails to be analytic. Label these points according to increasing distances from the center, say Iz1-al SIz2-al and let R. = Izk - al. Also, let R0 = r1, 1 = r2. Th. 16.9 Homotopic Curves The union of the intervals [Rk, Rk+l] fo r k = 0, 1 , 2, ... , n is the interval [r1, r2]. We will show that qp is constant on each interval [Rk, Rk+l]. We write (3) in the form pp(r) = f2x g(r, B) dB, where g(r, 0) = f(a + reBireiB. o An easy application of the chain rule shows that we have 09 = it a9 . (4) ae Or (The reader should verify this formula.) Continuity off' implies continuity of the partial derivatives aglar and ag/a9. Therefore, on each open interval (Rk, Rk+i), we can calculate (p'(r) by differentiation under the integral sign (Theorem 7.40) and then use (4) and the second fundamental theorem of calculus (Theorem 7.34) to obtain (p'(r) = Jo fn or d9 it Jo a9 d9 it {g(r, 2ir) - 9(r, 0)} = 0. Applying Theorem 12.10, we see that (p is constant on each open subinterval (Rk, Rk+ 1). By continuity, 9 is constant on each closed subinterval [Rk, Rk+ 1] and hence on their union [r1, r2]. From (3) we see that (p(r) - 0 as r -+ 0 so the constant value of qp is 0 if rl = 0. 16.5 CAUCHY'S INTEGRAL THEOREM FOR A CIRCLE The following special case of Theorem 16.8 is of particular importance. Theorem 16.9 (Cauchy's integral theorem for a circle). If f is analytic on a disk B(a; R) except possibly for a finite number of points at which it is continuous, then for every circular path y with center at a and radius r < R. Proof. Choose r2 so that r < r2 < R and apply Theorem 16.8 with rl = 0. NOTE. There is a more general form of Cauchy's integral theorem in which the circular path y is replaced by a more general closed path. These more general paths will be introduced through the concept of homotopy. 16.6 HOMOTOPIC CURVES Figure 16.4 shows three arcs having the same endpoints A and B and lying in an open region D. - Arc I can be continuously deformed into arc 2 through a collection of intermediate arcs, each of which lies in D. Two arcs with this property are said 440 Cauchy's Theorem and the Residue Calculus Figure 16.4 Def. 16.10 to be homotopic in D. Arc 1 cannot be so deformed into arc 3 (because of the hole separating them) so they are not homotopic in D. In this section we give a formal definition of homotopy. Then we show that, if f is analytic in D, the contour integral off from A to B has the same value along any two homotopic paths in D. In other words, the value of a contour integral f A f is unaltered under a continuous deformation of the path, provided the in-termediate contours remain within the region of analyticity of f. This property of contour integrals is of utmost importance in the applications of complex integration. Definition 16.10. Let yo and yi be two paths with a common domain [a, b]. Assume that either a) yo and yi have the same endpoints: yo(a) = yi(a) and yo(b) = yi(b), or b) yo and yi are both closed paths: yo(a) = yo(b) and y, (a) = yi(b). Let D be a subset of C containing the graphs of yo and yi. Then yo and yi are said to be homotopic in D if there exists a function h, continuous on the rectangle [0, 1] x [a, b], and with values in D, such that 1) h(0, t) = yo(t) if t e [a, b], 2) h(1, t) = y1(t) if t e [a, b]. In addition we require that for each s in [0, 1] we have 3a) h(s, a) = yo(a) and h(s, b) = yo(b), in case (a); or 3b) h(s, a) = h(s, b), in case (b). The function h is called a homotopy. The concept of homotopy has a simple geometric interpretation. For each fixed s in [0, 1], let ys(t) = h(s, t). Then ys can be regarded as an intermediate moving path which starts from yo when s = 0 and ends at yi when s = 1. Example 1. Homotopy to a point. If yi is a constant function, so that its graph is a single point, and if yo is homotopic to yi in D, we say that yo is homotopic to a point in D. Example 2. Linear homotopy. If, for each t in [a, b], the line segment joining yo(t) and yi(t) lies in D, then yo and yi are homotopic in D because the function h(s, t) = syi(t) + (1 - s)yo(t) Th. 16.11 Homotopic Curves 441 serves as a homotopy. In this case we say that yo and 71 are linearly homotopic in D. In particular, any two paths with domain [a, b] are linearly homotopic in C (the complex plane) or, more generally, in any convex set containing their graphs. NOTE. Homotopy is an equivalence relation. The next theorem shows that between any two homotopic paths we can inter-polate a finite number of intermediate polygonal paths, each,of which is linearly homotopic to its neighbor. Theorem 16.11 (Polygonal interpolation theorem). Let yo and yl be homotopic paths in an open set D. Then there exist a finite number of paths ao, al, ... , an such that: a) ao = yo and a = y1, b) a1 is a polygonal path for 1 < j < n - 1, c) a1 is linearly homotopic in D to a1+ 1 for 0 < j < n - 1. Proof. Since yo and yl are homotopic in D, there is a homotopy h satisfying the conditions in Definition 16.10. Consider partitions {so, sl, ... , of [0, 1] and {to, t1i ... , of [a, b], into n equal parts, choosing n so large that the image of each rectangle [si, s1+ 11 x [tk, tk+1] under h is contained in an open disk D1k contained in D. (The reader should verify that this is possible because of uniform continuity of h.) On the intermediate path y f given by y,J(t) = h(s1, t) for 0 < j < n, we inscribe a polygonal path a1 with vertices at the points h(s1, tk). That is, a1(tk) = h(s1, tk) for k = 0, 1, ... , n, and a1 is linear on each subinterval [tk, tk+ 1] for 0 < k < n - 1. We also define ao = yo and an = yl. (An example is shown in Fig. 16.5.) The four vertices a1(tk), a1(tk+1), aJ+1(tk), and a1+1(tk+1) all lie in the disk D1k. Since D1k is convex, the line segments joining them also lie in D1k and hence the points saJ+1(t) + (1 - s)a1(t), (5) Figure 16.5 442 Cauchy's Theorem and the Residue Calculus Th. 16.12 lie in DJk for each (s, t) in [0, 1] x [tk, tk+1]. Therefore the points (5) lie in D for all (s, t) in [0, 1] x [a, b], so aj+1 is linearly homotopic to a} in D. 16.7 INVARIANCE OF CONTOUR INTEGRALS UNDER HOMOTOPY Theorem 16.12. Assume f is analytic on an open set D, except possibly for a finite number of points where it is continuous. If yo and yl are piecewise smooth paths which are homotopic in D we have ff=If 7oJYiProof. First we consider the case in which yo and yl are linearly homotopic. For each s in [0, 1] let ys(t) = sy1(t) + (1 - s)yo(t) if t E [a, b]. Then ys is piecewise smooth and its graph lies in D. Write ys(t) = yo(t) + sa(t), where a(t) = y1(t) - yo(t), and define 1V(s) = J f = J bf[ys(t)] dyo(t) + s J bf[y:(t)] da(t), r. a a for 0 5 s < 1. We wish to prove that q,(0) = lp(l). We will in fact prove that qp is constant on [0, 1]. We use Theorem 7.40 to calculate (p'(s) by differentiation under the integral sign. Since a ys(t) = a(t), this gives us V(S) = ' ( ( J b a(t)f'[ys(t)]ys(t) dt + J bf[ys(t)] do(t) a a b f'[y:(t)]a(t) dyo(t) + S J f'[y3(t)]a(t) da(t) + J bf[y.(t)] da(t) J f a a a 6 6 a a(t)f'[y3(t)] dy.(t) + Ja f[ys(t)] da(t) f 6 a(t) d{f[y3(t)]} + J bf[y.(t)] da(t) I a(b)f [y3(b)] - a(a)f [ys(a)], by the formula for integration by parts (Theorem 7.6). But, as the reader can eas Th. 16.15 Cauchy's Integral Formula 443 verify, the last expression vanishes because yo and yl are homotopic, so cp'(s) = 0 for all s in [0, 1]. Therefore (p is constant on [0, 1]. This proves the theorem when yo and y, are linearly homotopic in D. If they are homotopic in D under a general homotopy h, we interpolate poly-gonal paths ai as described in Theorem 16.11. Since each polygonal path is piece-wise smooth, we can repeatedly apply the result just proved to obtain Jf=ffjf=...=$f=ff 70 The general form of Cauchy's theorem referred to earlier can now be easily deduced from Theorems 16.9 and 16.12. We remind the reader that a circuit is a piecewise smooth closed path. Theorem 16.13 (Cauchy's integral theorem for circuits homotopic to a point). Assume f is analytic on an open set D, except possibly for a finite number of points at which we assume f is continuous. Then for every circuit y which is homotopic to a point in D we have J.f=0. Proof. Since y is homotopic to a point in D, y is also homotopic to a circular path S in D with arbitrarily small radius. Therefore fr f = f af, and 16f = 0 by Theorem 16.9. Definition 16.14. An open connected set D is called simply connected if every closed path in D is homotopic to a point in D. Geometrically, a simply connected region is one without holes, Cauchy's theorem shows that, in a simply connected region D the integral of an analytic function is zero around any circuit in D. 16.9 CAUCHY'S INTEGRAL FORMULA The next theorem reveals a remarkable property of analytic functions. It relates the value of an analytic function at a point with the values on a closed curve not containing the point. Theorem 16.15 (Cauchy's integral formula). Assume f is analytic on an open set D, and let y be any circuit which is homotopic to a point in D. Then for any point z in D which is not on the graph of y we have f f(w) dw = f(z) f dw. (6) V w- Z Y w- z 444 Cauchy's Theorem and the Residue Calculus Th. 16.16 Proof. Define a new function g on D as follows : .f(w) -.f(z) if w z g(w) = w - z f'(z) if w = Z. Then g is analytic at each point w # z in D and, at the point z itself, g is continuous. Applying Cauchy's integral theorem to g we have J,, g = 0 for every circuit y homotopic to a point in D. But if z is not on the graph of y we can write f g = f. f(w) - f(z) dw = f f(w) dw - f(z) f 1 dw, r wz ,Jyw - z yw - z which proves (6). NOTE. The same proof shows that (6) is also valid if there is a finite subset T of D on which f is not analytic, provided that f is continuous on T and z is not in T. The integral f), (w - z)-1 dw which appears in (6) plays an important role in complex integration theory and is discussed further in the next section. We can easily calculate its value for a circular path. Example. If y is a positively oriented circular path with center at z and radius r, we can write y(O) = z + rei°, 0 <- 0 <- 2n. Then y'(0) = ireie = i {y(0) - z }, and we find fy dwfo y,(B) I d9 J i d6 = 2ni. w- z Y(0) - z o NOTE. In this case Cauchy's integral formula (6) takes the form 2nif(z) = fy f(w) dw. w - z Again writing y(O) = z + reie, we can put this in the form 1 2m .f (z) = 2n f o f(z + re") d8. (7) This can be interpreted as a Mean- Value Theorem expressing the value off at the center of a disk as an average of its values at the boundary of the disk. The function f is assumed to be analytic on the closure of the disk, except possibly for a finite subset on which it is continuous. 16.10 THE WINDING NUMBER OF A CIRCUIT WITH RESPECT TO A POINT Theorem 16.16. Let y be a circuit and let z be a point not on the graph of y. Then there is an integer n (depending on y and on z) such that fy = 2nin. (8) W - z dw Def. 16.17 The Winding Number of a Circuit 445 Proof. Suppose y has domain [a, b]. By Theorem 16.7 we can express the integral in (8) as a Riemann integral, f dw = ('b y'(t) dt JYw - z JaY(t) - z Define a complex-valued function on the interval [a, b] by the equation F(x) = r" y'(t) dt if a < x < b. Ja y(t) - z To prove the theorem we must show that F(b) = 2itin for some integer n. Now F is continuous on [a, b] and has a derivative Ax) F '(x) _ y(x) - z at each point of continuity of y'. Therefore the function G defined by G(t) = e-F(t){y(t) - z} if t e [a, b], is also continuous on [a, b]. Moreover, at each point of continuity of y' we have G'(t) = e-F(t)y,(t) - F,(t)e-F(t){y(t) - z} = 0. Therefore G'(t) = 0 for each t in [a, b] except (possibly) for a finite number of points. By continuity, G is constant throughout [a, b]. Hence, G(b) = G(a). In other words, we have e-F(b) {y(b) - z} = y(a) - z. Since y(b) = y(a) # z we find e-F(b) = 1, which implies F(b) = 2irin, where n is an integer. This completes the proof. Definition 16.17. If y is a circuit whose graph does not contain z, then the integer n defined by (8) is called the winding number (or index) of y with respect to z, and is denoted by n(y, z). Thus, n (y, 1 (' dw (y, z 2ni Y w - z NOTE. Cauchy's integral formula (6) can now be restated in the form n(y, z)f(z) = 2ni f, w (w)z dw. v The term "winding number" is used because n(y, z) gives a. mathematically precise way of counting the number of times the point y(t) "winds around" the point z as t varies over the interval [a, b]. For example, if y is a positively oriented 446 Cauchy's Theorem and the Residue Calculus 11. 16.18 circle given by y(O) = z + re°, where 0 < 0 S 2ir, we have already seen that the winding number is 1. This is in accord with the physical interpretation of the point y(O) moving once around a circle in the positive direction as 0 varies from 0 to 2n. If 0 varies over the interval [0, 2nn], the point y(O) moves n times around the circle in the positive direction and an easy calculation shows that the winding number is n. On the other hand, if 6(0) = z + re-iB for 0 < 0 < 2irn, then 6(0) moves n times around the circle in the opposite direction and the winding number is -n. Such a path 6 is said to be negatively oriented. 16.11 THE UNBOUNDEDNESS OF THE SET OF POINTS WITH WINDING NUMBER ZERO Let F denote the graph of a circuit y. Since IF is a compact set, its complement C - IF is an open set which, by Theorem 4.44, is a countable union of disjoint open regions (the components of C - I,). If we consider the components as subsets of the extended plane C, exactly one of these contains the ideal point oo. In other words, one and only one of the components of C - IF is,unbounded. The next theorem shows that the winding number n(y, z) is 0 for each z in the unbounded component. Theorem 16.18. Let y be a circuit with graph F. Divide the set C - F into two subsets: E = {z : n(y, z) = 0} and I = {z : n(y, z) # 0}. Then both E and I are open. Moreover, E is unbounded and I is bounded. Proof. Define a function g on C - IF by the formula g(z) = n(y, z) = 1 J dw 2ni w - z By Theorem 7.38, g is continuous on C - r and, since g(z) is always an integer, it follows that g is constant on each component of C - t. Therefore both E and I are open since each is a union of components of C - r. Let U denote the unbounded component of C - r. If we prove that E con-tains U this will show that E is unbounded and that I is bounded. Let K be a constant such that ly(t)I < K for all t in the domain of y, and let c be a point in U such that Icl > K + A(y) where A(y) is the length of y. Then we have 1 < 1 ICI - ly(t)I icl - K Estimating the integral for n(y, c) by Theorem 16.6 we find 0 <_ Ig(c)I <_ A(y) < 1. Icl - K 71. 16.19 Analytic Functions Defined by Contour Integrals 447 Since g(c) is an integer we must have g(c) = 0, so g has the constant value 0 on U. Hence E contains the point c, so E contains all of U. There is a general theorem, called the Jordan curve theorem, which states that if IF is a Jordan curve (simple closed curve) described by y, then each of the sets E and I in Theorem 16.18 is connected. In other words, a Jordan curve F divides C - F into exactly two components E and I having IF as their common boundary. The set I is called the inner (or interior) region of IF, and its points are said to be inside IF. The set E is called the outer (or exterior) region of F, and its points are said to be outside F. Although the Jordan curve theorem is intuitively evident and easy to prove for certain familiar Jordan curves such as circles, triangles, and rectangles, the proof for an arbitrary Jordan curve is by no means simple. (Proofs can be found in References 16.3 and 16.5.) We shall not need the Jordan curve theorem to prove any of the theorems in this chapter. However, the reader should realize that the Jordan curves occurring in the ordinary applications of complex integration theory are usually made up of a finite number of line segments and circular arcs, and for such examples it is usually quite obvious that C - F consists of exactly two components. For points z inside such curves the winding number n(y, z) is + I or -1 because y is homo-topic in I to some circular path 6 with center z, so n(y, z) = n(S, z), and n(8, z) is + I or -1 depending on whether the circular path S is positively or negatively oriented. For this reason we say that a Jordan circuit y is positively oriented if, for some z inside F we have n(y, z) = + 1, and negatively oriented if n(y, z) = - 1. 16.12 ANALYTIC FUNCTIONS DEFINED BY CONTOUR INTEGRALS Cauchy's integral formula, which states that(' n(y, z)f(z) = 1 J f(w) dw, 2ni Yw-z has many important consequences. Some of these follow from the next theorem which treats integrals of a slightly more general type in which the integrand f(w)l(w - z) is replaced by (p(w)/(w - z), where qp is merely continuous and not necessarily analytic, and y is any rectifiable path, not necessarily a circuit. Theorem 16.19. Let y be a rectifiable path with graph F. Let 9 be a complex-valued /'unction which is continuous on I', and let f be defined on C - F by the equation f(z) = f w(w) dw if z 0 F. r w - Z Then f has the following properties: a) For each point a in C - I', f has a power-series representation f(Z) _ cn(z - a)n, (9) n=0 448 Cauchy's Theorem and the Residue Calculus where Th. 16.19 cn = f" (w T(a)r+ 1 dw for n = 0, 1, 2,... (10) b) The series in (a) has a positive radius of convergence R, where R=inf{jw-al :wEF}. (11) c) The function f has a derivative of every order n on C - IF given by f(")(z) = n! f y (w q(z)n+1 dw if z 0 F. (12) Proof. First we note that the number R defined by (11) is positive because the function g(w) = Iw - al has a minimum on the compact set IF, and this minimum is not zero since a 0 F. Thus, R is the distance from a to the nearest point of F. (See Fig. 16.6.) Figure 16.6 To prove (a) we begin with the identity 1 k t" + tk+1 1-t n=0 1-t' (13) valid for all t # 1. We take t = (z - a)/(w - a) where Iz - al < R and w e IF. Then 1/(1 - t) = (w - a)l(w - z): Multiplying (13) by (p(w)l(w - a) and integrating along y, we find f(z) = f T(w) dw J r w - z E(z-a)" f T(w) dw+ n=O JY (w - a)n+1 n=0 k E cn(z - a)" + Ek, 1, (p(w) (z - a w - z a where cn is given by (10) and Ek is given by _ fy (Z - a\k+l w - Z a Ek J dw. dw (14) Th. 16.20 Power-Series Expansions for Analytic Functions 449 Now we show that Ek -> 0 as k -+ oo by estimating the integrand in (14). We have <Iz - al and 1 -1 < 1 R Iw - zI Iw - a+a - zI R- la - zl Let M = max {I(p(w)I : w e F), and let A(y) denote the length of y. Then (14) gives us IEkI <_ MA(y) (y) Clz - allk+1 R- a-zI R J Since Iz - at < R we find that Ek -> 0 as k - co. This proves (a) and (b). Applying Theorem 9.23 to (9) we find that f has derivatives of every order on the disk B(a; R) and that f(")(a) = n!c". Since a is an arbitrary point of C - F this proves (c). NOTE. The series in (9) may have a radius of convergence greater than R, in which case it may or may not represent fat more distant points. 16.13 POWER-SERIES EXPANSIONS FOR ANALYTIC FUNCTIONS A combination of Cauchy's integral formula with Theorem 16.19 gives us : Theorem 16.20. Assume f is analytic on an open set S in C, and let a be any point of S. Then all derivatives f (")(a) exist, and f can be represented by the convergent power series f(z) = E f(")(a) (z - a)", "=o n! (15) in every disk B(a; R) whose closure lies in S. Moreover, for every n > 0 we have (a)"+ dw, (16) f (")(a) = 2ni f Y (w -where y is any positively oriented circular path with center at a and radius r < R. NOTE. The series in (15) is known as the Taylor expansion off about a. Equation (16) is called Cauchy's integral formula for f(")(a). Proof. Let y be a circuit homotopic to a point in S, and let F be the graph of y. Define g on C - F by the equation g(z) = f f(w) dw if z 0 F. yw-z If z e B(a; R), Cauchy's integral formula tells us that g(z) = 2nin(y, z)f(z). Hence, - n(y, z)f(z) = 2ni f w (W)z dw if Iz - al < R. r 450 Cauchy's Theorem and the Residue Calculus Now let y(O) = a + re'°, where Iz - al < r < R and 0 5 0 5 tic. Then n(y, z) = 1, so by applying Theorem 16.19 to (p(w) = f(w)/(2ai) we find a series representation 1(z) = E c"(z - a)", 00 n=0 convergent for Iz - al < R, where c" = f(")(a)/n!. Also, part (c) of Theorem 16.19 gives (16). Theorems 16.20 and 9.23 together tell us that a necessary and sufficient con-dition for a complex-valued function f to be analytic at a point a is that f be representable by a power series in some neighborhood of a. When such a power series exists, its radius of convergence is at least as large as the radius of any disk B(a) which lies in the region of analyticity off. Since the circle of convergence cannot contain any points in its interior where f fails to be analytic, it follows that the radius of convergence is exactly equal to the distance from a to the nearest point at which f fails to be analytic. This observation gives us a deeper insight concerning power-series expansions for real-valued functions of a real variable. For example, letf(x) = 1/(1 + x2) if x is real. This function is defined everywhere in R1 and has derivatives of every order at each point in R1. Also, it has a power-series expansion about the origin, namely, 1 2 4 6 1 + x2 =1-x +x -x +... However, this representation is valid only in the open interval (- 1, 1). From the standpoint of real-variable theory, there is nothing in the behavior off which explains this. But when we examine the situation in the complex plane, we see at once that the function f(z) = 1/(1 + z2) is analytic everywhere in C except at the points z = ± i. Therefore the radius of convergence of the power-series expansion about 0 must equal 1, the distance from 0 to i and to -i. Examples. The following power series expansions are valid for all z in C: w n OD (_ i)nz2n+1 a) e= = E sin z = F (2n + 1)! n=0 n! n=O z R c) Cos z = r (-i)nz2n n=0 (2n)! 16.14 CAUCHY'S INEQUALITIES. LIOUVILLE'S THEOREM If f is analytic on a closed disk B(a; R), Cauchy's integral formula (16) shows that f(n)(a) 2ni Jr (w f (a)"' d"', where y is any positively oriented circular path with center a and radius r < R. Th. 16.22 Isolation of the Zeros of an Analytic Function 451 We can write y(9) = a + reie, 0 < 0 < 27v, and put this in the form f(")(a) 2ar" L rzn f(a + re'°) a-'no dB. (17) Jo This formula expresses the nth derivative at a as a weighted average of the values of f on a circle with center at a. The special case n = 0 was obtained earlier in Section 16.9. Now, let M(r) denote the maximum value of If I on the graph of y. Estimating the integral in (17), we immediately obtain Cauchy's inequalities: If(n)(a)I < M(r)n! (n = 0, 1, 2, ... ). (18) r ' The next theorem is an easy consequence of the case n = 1. Theorem 16.21(Liouville's theorem). If f is analytic everywhere on C and bounded on C, then f is constant. Proof. Suppose If(z)I S M for all z in C. Then Cauchy's inequality with n = 1 gives us I f'(a) I < M/r for every r > 0. Letting r -> + co, we find f'(a) = 0 for every a in C and hence, by Theorem 5.23, f is constant. NOTE. A function analytic everywhere on C is called an entire function. Examples are polynomials, the sine and cosine, and the exponential. Liouville's theorem states that every bounded entire function is constant. Liouville's theorem leads to a simple proof of the Fundamental Theorem of Algebra. Theorem 16.22 (Fundamental Theorem of Algebra). Every polynomial of degree n >- 1 has a zero. Proof. Let P(z) = ao + a1z + + where n z I and an # 0. We assume that P has no zero and prove that P is constant. Let f(z) = 1/P(z). Then f is analytic everywhere on C since P is never zero. Also, since P(z)=z"(an +za1i+...+azl+a) we see that IP(z)I - + oo as Iz I - + oo, so f(z) - 0 as Iz I -' + co. Therefore f is bounded on C so, by Liouville's theorem, f and hence P is constant. 16.15 ISOLATION OF THE ZEROS OF AN ANALYTIC FUNCTION If f is analytic at a and iff(a) = 0, the Taylor expansion off about a has constant term zero and hence assumes the following form : f(Z) = E Cn(Z - a)". 00 E n=1 452 Cauchy's Theorem and the Residue Calculus Th. 16.23 This is valid for each z in some disk B(a). If f is identically zero on this disk [that is, if f(z) = 0 for every z in B(a)], then each c = 0, since c = f(°)(a)/n!. If f is not identically zero on this neighborhood, there will be a first nonzero coefficient ck in the expansion, in which case the point a is said to be a zero of order k. We will prove next that there is a neighborhood of a which contains no further zeros off This property is described by saying that the zeros of an analytic function are isolated. Theorem 16.23. Assume that f is analytic on an open set S in C. Suppose f(a) = 0 for some point a in S and assume that f is not identically zero on any neighborhood of a. Then there exists a disk B(a) in which f has no further zeros. Proof. The Taylor expansion about a becomesf(z) = (z - a)kg(z), where k > 1, g(z)=ck+ck+,(z-a)+..., and g(a)=ck96 0. Since g is continuous at a, there is a disk B(a) c S on which g does not vanish. Therefore, f(z) 0 0 for all z a in B(a). This theorem has several important consequences. For example, we can use it to show that a function which is analytic on an open region S cannot be zero on any nonempty open subset of S without being identically zero throughout S. We recall that an open region is an open connected set. (See Definitions 4.34 and 4.45.) Theorem 16.24. Assume that f is analytic on an open region S in C. Let A denote the set of those points z in S for which there exists a disk B(z) on which f is identically zero, and let B = S - A. Then one of the two sets A or B is empty and the other one is S itself. Proof. We have S = A u B, where A and B are disjoint sets. The set A is open by its very definition. If we prove that B is also open, it will follow from the connectedness of S that at least one of the two sets A or B is empty. To prove B is open, let a be a point of B and consider the two possibilities: f(a) # 0, f(a) = 0. If f(a) # 0, there is a disk B(a) S on which f does not vanish. Each point of this disk must therefore belong to B. Hence, a is an interior point of B if f(a) # 0. But, if f(a) = 0, Theorem 16.23 provides us with a disk B(a) containing no further zeros off. This means that B(a) c B. Hence, in either case, a is an interior point of B. Therefore, B is open and one of the two sets A or B must be empty. 16.16 THE IDENTITY THEOREM FOR ANALYTIC FUNCTIONS Theorem 16.25. Assume that f is analytic on an open region S in C. Let T be a subset of S having an accumulation point a in S. If f(z) = 0 for every z in T, then f(z) = 0 for every z in S. Proof. There exists an infinite sequence whose terms are points of T, such that lim .. z = a. By continuity, f(a) = 0. We will prove T1.16.27 Maximum and Minimum Modulus 453 next that there is a neighborhood of a on which f is identically zero. Suppose there is no such neighborhood. Then Theorem 16.23 tells us that there must be a disk B(a) on whichf(z) 0 if z a. But this is impossible, since every disk B(a) contains points of T other than a. Therefore there must be a neighborhood of a on which f vanishes identically. Hence the set A of Theorem 16.24 cannot be empty. Therefore, A = S, and this means f(z) = 0 for every z in S. As a corollary we have the following important result, sometimes referred to as the identity theorem for analytic functions: Theorem 16.26. Let f and g be analytic on an open region S in C. If T is a subset of S having an accumulation point a in S, and if f(z) = g(z) for every z in T, then f(z) = g(z) for every z in S. Proof. Apply Theorem 16.25 to f - g. 16.17 THE MAXIMUM AND MINIMUM MODULUS OF AN ANALYTIC FUNCTION The absolute value or modulus If I of an analytic function f is a real-valued non-negative function. The theorems of this section refer to maxima and minima of Ifi. Theorem 16.27 (Local maximum modulus principle). Assume f is analytic and not constant on an open region S. Then If I has no local maxima in S. That is, every disk B(a; R) in S contains points z such that If(z)I > If(a)j. Proof. We assume there is a disk B(a; R) in S in which If(z)I < If(a)I and prove that f is constant on S. Consider the concentric disk B(a; r) with 0 < r < R. From Cauchy's integral formula, as expressed in (7), we have 2a If(a)I s 1 2n I.f(a + reB)I d0. (19) 0 Now I f(a + ret°)I < I f(a)I for all 0. We show next that we cannot have strict inequality I f(a + reB)I < If(a)I for any 0. Otherwise, by continuity we would have I f(a + reB)I < I f(a)I - e for some e > 0 and all 0 in some subinterval I of [0, 2n] of positive length h, say. Let J = [0, 2n] - I. Then J has measure 2n - h, and (19) gives us f(a + re'B)I dO 2irjf(a)I <- 5 I f(a + reB)I dO + fj I I < h{If(a)I - e} + (2n - h) If(a)I = 27r If(a)I - he < 2ir If(a)I Thus we get the contradiction I f(a)I < If(a)I. This shows that if r < R, we cannot have strict inequality I f(a + reie)I < If(a)I forany 0. Hence I f(z)I = If(a)I for every z in B(a; R). Therefore If I is constant on this disk so, by Theorem 5.23, f itself is constant on this disk. By the identity theorem, f is constant on S. 454 Cauchy's Theorem and the Residue Calculus Th. 16.28 Theorem 16.28 (Absolute maximum modulus principle). Let T be a compact subset of the complex plane C. Assume f is continuous on T and analytic on the interior of T. Then the absolute maximum of If I on T is attained on 8T, the boundary of T. Proof. Since T is compact, If I attains its absolute maximum somewhere on T, say at a. If a e OT there is nothing to prove. If a e int T, let S be the component of int T containing a. Since If I has a local maximum at a, Theorem 16.27 implies that f is constant on S. By continuity, f is constant on 8S s T, so the maximum value, If(a)I, is attained on 8S. But 8S c 8T (Why?) so the maximum is attained on 8T. Theorem 16.29 (Minimum modulus principle). Assume f is analytic and not constant on an open region S. If If I has a local minimum in S at a, thenf(a) = 0. Proof. If f(a) # 0 we apply Theorem 16.27 tog = 1/f Then g is analytic in some open disk B(a; R) and I g I has a local maximum at a. Therefore g and hence f is constant on this disk and therefore on S, contradicting the hypothesis. 16.18 THE OPEN MAPPING THEOREM Nonconstant analytic functions are open mappings; that is, they map open sets onto open sets. We prove this as an application of the minimum modulus principle. Theorem 16.30 (Open mapping theorem). If f is analytic and not constant on an open region S, then f is open. Proof. Let A be any open subset of S. We are to prove that f(A) is open. Take any b in f(A) and write b = f(a), where a e A. First we note that a is an isolated point of the inverse-image f -1({b}). (If not, by the identity theorem f would be constant on S.) Hence there is some disk B = B(a; r) whose closure B lies in A and contains no point off -1({b}) except a. Since f(B) s f(A) the proof will be complete if we show that f(B) contains a disk with center at b. Let 8B denote the boundary of B, 3B = {z : Iz - al = r}. Then f(8B) is a compact set which does not contain b. Hence the number m defined by m = inf {If(z) - bI : z e 8B}, is positive. We will show that f(B) contains the disk B(b; m/2). To do this, we take any w in B(b; m/2) and show that w = f(zo) for some zo in B. Let g(z) = f(z) - w if z e B. We will prove that g(zo) = 0 for some zo in B. Now IgI is continuous on B and, since B is compact, there is a point zo in B at which IgI attains its minimum. Since a e B, this implies Ig(z0)I :5 Ig(a)I = If(a) - wl = Ib - wI < But if z e 8B, we have I9(z)l=I.f(z)-b+b-wl?I.f(z)-bl-Iw-bl>m-=2. Th. 16.30 Laureat Expansions Hence, zo 0 8B so zo is an interior point of B. In other words, I9I has a local minimum at z0. Since g is analytic and not constant on B, the minimum modulus principle shows that g(zo) = 0 and the proof is complete. 16.19 LAURENT EXPANSIONS FOR FUNCTIONS ANALYTIC IN AN ANNULUS Consider two functionsf1 and g1, both analytic at a point a, with gl(a) = 0. Then we have power-series expansions 91(z) bn(z - a)", for Iz - al < Ti, and n=1 00 fi(z) = F; cn(z - a)", for Iz - al < r2. (20) n=0 Letf2 denote the composite function given by f2(z) = 91 (z 1 + a) . Thenf2 is defined and analytic in the region Iz - al > r1 and is represented there by the convergent series f2(z) _ bn(z - a)-", for Iz - aI > ri. (21) n=1 Now if r1 < r2, the series in (20) and (21) will have a region of convergence in common, namely the set of z for which r1 < Iz - al < Ti. In this region, the interior of the annulus A(a; r1, r2), both f1 andf2 are analytic and their sum fi + f2 is given by 00 00 f1(z) + f2(z) = E c"(z - a)" + E bn(Z -a)-n. n=0 n=1 The sum on the right is written more briefly as 00 E cn(z - a)", n=-00 where c_" = bn for n = 1, 2, ... A series of this type, consisting of both positive and negative powers of z - a, is called a Laurent series. We say it converges if both parts converge separately. Every convergent Laurent series represents an analytic function in the interior of the annulus A(a; r1, r2). Now we will prove that, conversely, every function f which is analytic on an annulus can be represented in the interior of the annulus by a convergent Laurent series. 456 Cauchy's Theorem and the Residue Calculus Th. 16.31 Theorem 16.31. Assume that f is analytic on an annulus A(a; r1, r2). Then for every interior point z of this annulus we have f(z) = f1(z) + .f2(z), (22) where f1(z) _ cn(z - a)" and n=0 OD f2(z) = E c-.(z -a)-". n=1 The coefficients are given by the formulas c" = 1 f (w) dw (n = 0, ± 1 , ±2, ... ), (23) 27r1 7 (w - a)n+1 where y is any positively oriented circular path with center at a and radius r, with r1 < r < r2. The function f1 (called the regular part off at a) is analytic on the disk B(a; r2). The function f2 (called the principal part off at a) is analytic outside the closure of the disk B(a; r1). Proof Choose an interior point z of the annulus, keep z fixed, and define a function g on A(a; r1, r2) as follows: .f(w) - f(z) if w z 9(w) w-z f'(z) if w = Z. Then g is analytic at w if w # z and g is continuous at z. Let cp(r) = f g(w) dw, .J rr where y, is a positively oriented circular path with center a and radius r, with r1 < r 5 r2. By Theorem 16.8, cp(r1) = cp(r2) so fyi g(w) dw = f72 where y1 = Yr1 and 72 = y,2. Since z is not on the graph of y1 or of y2, in each of these integrals we can write 9(w) = .f(w) -f(Z) w - z w - z Substituting this in (24) and transposing terms, we find f(z) 1 dw dw. rI f(w) - z ) l -Jr2 w f (-w)z dw - f I r2 w - z - fy w 1 dw w- z g(w) dw, (24) (25) But $71 (w - z)-1 dw = 0 since the integrand is analytic on the disk B(a; r1), 11. 16.31 Isolated Singularities 457 and 172 (w - z)-1 dw = 27ri since n(y2, z) = 1. Therefore, (25) gives us the equation where f1(z) f(z) = fi(z) + f2(z), _ 1 f(w) dw 27ri f72 w - z and f2(z) _ - 2ni f, w (w)z dw. Yt By Theorem 16.19, f1 is analytic on the disk B(a; r2) and hence we have a Taylor expansion f1(z) _ E cn(z - a)" for Iz - al < r2, where n=0 1 f(w) c"= -dw. 27ri 12 (W - a)n+1 (26) Moreover, by Theorem 16.8, the path y2 can be replaced by y, for any r in the interval rl S r 5 r2. To find a series expansion forf2(z), we argue as in the proof of Theorem 16.19, using the identity (13) with t = (w - a)/(z - a). This gives us 1 = CW -a\"+(w-a k+l Z-a (27) 1 -(w-a)/(z-a) n=0 Z -a z -a) (z-W 6 If w is on the graph of y1, we have Iw - al = rl < Iz - aj, so ItI < 1. Now we multiply (27) by -f(w)l(z - a), integrate along y1, and let k -> oo to obtain 00 where f2(z)_ Ebn(z-a)-" forIz - aj>rl n=1 1 f(w) dw. (28) 27rr Y, (w - a)1-n By Theorem 16.8, the path yl can be replaced by y, for any r in [rl, r2]. If we take the same path y, in both (28) and (26) and if we write c_,, for bn, both formulas can be combined into one as indicated in (23). Since z was an arbitrary interior point of the annulus, this completes the proof. NOTE. Formula (23) shows that a function can have at most one Laurent ex-pansion in a given annulus. 16.20 ISOLATED SINGULARITIES A disk B(a; r) minus its center, that is, the set B(a; r) - {a}, is called a deleted neighborhood of a and is denoted by B'(a; r) or B'(a). 458 Cauchy's Theorem and the Residue Calculus Def. 1632 Definition 16.32. A point a is called an isolated singularity off if a) f is analytic on a deleted neighborhood of a, and b) f is not analytic at a. NOTE. f need not be defined at a. If a is an isolated singularity off, there is an annulus A(a; r1, r2) on which f is analytic. Hence f has a uniquely determined Laurent expansion, say f(z) _ cn(z - a)" + c-n(Z -a)-n, (29) n=0 n=1 Since the inner radius r1 can be arbitrarily small, (29) is valid in the deleted neighborhood B'(a; r2). The singularity a is classified into one of three types (depending on the form of the principal part) as follows: If no negative powers appear in (29), that is, if c_,, = 0 for every n = 1 , 2, ... , the point a is called a removable singularity. In this case, f(z) -+ co as z -+ a and the singularity can be removed by defining f at a to have the value f(a) = co. (See Example I below.) If only a finite number of negative powers appear, that is, if c_,, 96 0 for some n but c_, = 0 for every m > n, the point a is said to be a pole of order n. In this case, the principal part is simply a finite sum, namely, C-1 + C-2 + ... +. C-n z-a (z-a)2 (z-a)"' A pole of order 1 is usually called a simple pole. If there is a pole at a, then I.f(z)I - oo as z - a. Finally, if c_" # 0 for infinitely many values of n, the point a is called an essential singularity. In this case, f(z) does not tend to a limit as z - a. Example 1. Removable singularity. Let f(z) _ (sin z)/z if z 0 0, f(O) = 0. This func-tion is analytic everywhere except at 0. (It is discontinuous at 0, since (sin z)/z -+ 1 as z - 0.) The Laurent expansion about 0 has the form sin z z2 z4 - = 1 -+ - + z 3! 5! Since no negative powers of z appear, the point 0 is a removable singularity. If we re-define f to have the value 1 at 0, the modified function becomes analytic at 0. Example 2. Pole. Let f(z) = (sin z)/zs if z # 0. The Laurent expansion about 0 is sin z _4 1 _2 1 1 2 zs 3! 5! 7! - z -- z +---z + In this case, the point 0 is a pole of order 4. Note that nothing has been said about the value of fat 0. 1b. 16.33 Residue at an Isolated Singular Point 459 Example 3. Essential singularity. Let f(z) = e' if z # 0. The point 0 is an essential singularity, since el/z = l + Z-1 + 1 Z-2 + ... + 1 Z-n + ... 2! n! Theorem 16.33. Assume that f is analytic on an open region Sin C and define g by the equation g(z) = 1/f(z) if f(z) # 0. Then f has a zero of order k at a point a in S if, and only if, g has a pole of order k at a. Proof. If f has a zero of order k at a, there is a deleted neighborhood B'(a) in which f does not vanish. In the neighborhood B(a) we have f(z) = (z - a)h(z), where h(z) # 0 if z e B(a). Hence, 1/h is analytic in B(a) and has an expansion 1 1 h(z) b0 + b1(z - a) + , where bo = h a # 0. () Therefore, if z e B'(a), we have 1 _ bo b1 g(Z) _ -(z - a)kh(z) (z - a)k + (z -a)k-1 + ... and hence a is a pole of order k for g. The converse is similarly proved. 16.21 THE RESIDUE OF A FUNCTION AT AN ISOLATED SINGULAR POINT If a is an isolated singular point of f, there is a deleted neighborhood B'(a) on which f has a Laurent expansion, say 00 00 f(Z) = E cn(z - a)n + E c-.(z -a)-n. (30) n=0 n=1 The coefficient c_ 1 which multiplies (z - a)-1 is called the residue off at a and is denoted by the symbol c-1 = Res f(z). :=a Formula (23) tells us that Si f(z) dz = 2ai Res f(z), z=a (31) if y is any positively oriented circular path with center at a whose graph lies in the disk B(a). In many cases it is relatively easy to evaluate the residue at a point without the use of integration. For example, if a is a simple pole, we can use formula (30) to obtain Resf(z) = lim (z - a)f(z). (32) z=a z-a 460 Cauchy's Theorem and the Residue Calculus Similarly, if a is a pole of order 2, it is easy to show that Res f(z) = g'(a), where g(z) = (z - a)2f(z). z=a Th. 16.34 In cases like this, where the residue can be computed very easily, (31) gives us a simple method for evaluating contour integrals around circuits. Cauchy was the first to exploit this idea and he developed it into a powerful method known as the residue calculus. It is based on the Cauchy residue theorem which is a generalization of (31). 16.22 THE CAUCHY RESIDUE THEOREM Theorem 16.34. Let f be analytic on an open region S except for a finite number of isolated singularities z1, ... , z" in S. Let y be a circuit which is homotopic to a point in S, and assume that none of the singularities lies on the graph of y. Then we have 17 n f(z) dz = 2iri E n(y, zk) Res f(z), (33) k=1 z=zk where n(y, zk) is the winding number of y with respect to zk. Proof. The proof is based on the following formula, where m denotes an integer (positive, negative, or zero) : f (z - zk)' dz = 27rin(y, zk) if m = -1, y 0 ifm#-1. (34) The formula for m = -1 is just the definition of the winding number n(y, zk). Let [a, b] denote the domain of y. If m # -1, let g(t) _ {y(t) - zk}'"+ 1 for tin [a, b]. Then we have fr (z - zkm dz = J b {y(t) -7 zk}my'(t) dt = m + 1 f b g'(t) dt 1 {g(b) - g(a)} = 0, m + 1, since g(b) = g(a). This proves (34). To prove the residue theorem, letfk denote the principal part off at the point Zk. By Theorem 16.31, fk is analytic everywhere in C except at zk. Therefore f - fl is analytic in S except at z2,.... , zn. Similarly, f - f1 - f2 is analytic in S except at z3, ... , z" and, by induction, we find that f - Ek= 1 fk is analytic everywhere in S. Therefore, by Cauchy's integral theorem, fy (f - Ek=, fk) = 0, or f A f f k=1 y k=1 Now we express fk as a Laurent series about zk and integrate this series term by term, using (34) and the definition of residue to obtain (33). Th. 16.35 Counting Zeros and Poles in a Region 461 NOTE. If y is a positively oriented Jordan curve with graph I', then n(y, zk) = I for each zk inside I', and n(y, zk) = 0 for each zk outside F. In this case, the integral off along y is 2iri times the sum of the residues at those singularities lying inside F. Some of the applications of the Cauchy residue theorem are given in the next few sections. 16.23 COUNTING ZEROS AND POLES IN A REGION If f is analytic or has a pole at a, and if f is not identically 0, the Laurent expansion about a has the form f(z) = E cn(z - a)n, n=m where cm 0 0. If m > 0 there is a zero at a of order m; if m < 0 there is a pole at a of order -m, and if m = 0 there is neither a zero nor a pole at a. NOTE. We also write m(f; a) for m to emphasize that m depends on both f and a. Theorem 16.35. Let f be a function, not identically zero, which is analytic on an open region S, except possibly for a finite number of poles. Let y be a circuit which is homotopic to a point in S and whose graph contains no zero or pole off. Then we have 1 f f'(z) J dz n(y, a)m(f; a), (35) 27ri Y f(z) aeS where the sum on the right contains only a finite number of nonzero terms. NOTE. If y is a positively oriented Jordan curve with graph r, then n(y, a)- I for each a inside IF and (35) is usually written in the form f ( dz = N - P, (36) 2ni fy f(z) ) where N denotes the number of zeros and P the number of poles of f inside r, each counted as often as its order indicates. Proof Suppose that in a deleted neighborhood of a point a we have f(z) _ (z - a)mg(z), where g is analytic at a and g(a) # 0, m being an integer (positive or negative). Then there is a deleted neighborhood of a on which we can write f'(z) = In + g'(z) f(z) z-a g(z) the quotient g'/g being analytic at a. This equation tells us that a zero off of order m is a simple pole off'/f with residue m. Similarly, a pole off of order m is a simple pole of f'lf with residue -m. This fact, used in conjunction with Cauchy's residue theorem, yields (35). 462 Cauchy's Theorem and the Residue Calculus Th. 16.36 16.24 EVALUATION OF REAL-VALUED INTEGRALS BY MEANS OF RESIDUES Cauchy's residue theorem can sometimes be used to evaluate real-valued Riemann integrals. There are several techniques available, depending on the form of the integral. We shall describe briefly two of these methods. The first method deals with integrals of the form 10' R(sin 6, cos 6) d8, where R is a rational function of two variables. Theorem 16.36. Let R be a rational function of two variables and let f(z) = R(22 - 1 z2 + 1'\ 2iz , 2z whenever the expression on the right is finite. Let y denote the positively oriented unit circle with center at 0. Then 2x R(sin 6, cos 6) d9 = f f(?) dz, (37) 0 y 1Z provided that f has no poles on the graph of y. Proof. Since y(O) = e'° with 0 < 0 < 2ir, we have Y'(0) = iY(0), y(9)2 - 1 = sin 0, Y(e)2 + 1 = cos 0, 2iy(0) 2y(O) and (37) follows at once from Theorem 16.7. NoTE. To evaluate the integral on the right of (37), we need only compute the residues of the integrand at those poles which lie inside the unit circle. Example. Evaluate I = f' dOl(a + cos 6), where a is real, dal > 1. Applying (37), we find dz f z2 + 2az + 1 The integrand has simple poles at the roots of the equation z2 + 2az + 1 = 0. These are the points z1 = -a + a2 - 1, Z2 = -a - Va2 - 1. A function P defined on C x C by an equation of the form P 4 P(z1, Z2) = E E am,nzlz2 M=0 n=0 is called a polynomial in two variables. The coefficients am,,, may be real or complex. The quotient of two such polynomials is called a rational function of two variables. Th. 16.37 Evaluation of Real-Valued Integrals The corresponding residues R1 and R2 are given by R1 = lim -z - zi = 1 z-.z, z2 + 2az + 1 zi - z2 463 R2= lim z-Z2 = 1 z-.z2 z2 + 2az + 1 z2 - Z1 If a > 1, zl is inside the unit circle, z2 is outside, and I = 41r/(z, - z2) = 1. If a < -1, z2 is inside, z1 is outside, and we get I = -2n/Va2 - 1. Many improper integrals can be dealt with by means of the following theorem : Theorem 16.37. Let T = {x + iy : y >- 01 denote the upper half-plane. Let S be an open region in C which contains T and suppose f is analytic on S, except, possibly, for a finite number of poles. Suppose further that none of these poles is on the real axis. If lim then fo, f(Re'B) Re'° dO = 0, (38) lim JR f(x) dx = 2ni Res f(z). (39) R-.+R k=1 z=zk where z1, ... , zn are the poles off which lie in T. Proof. Let y be the positively oriented path formed by taking a portion of the real axis from - R to R and a semicircle in T having [ - R, R] as its diameter, where R is taken large enough to enclose all the poles z1, ... , zn. Then 2ni fR f(Re'B) Ret° dB. E Resf(z) = fy f(z) dz f(x) dx + i fox k=1 z=zk ,J R When R - + oo, the last integral tends to zero by (38) and we obtain (39). NOTE. Equation (38) is automatically satisfied if f is the quotient of two poly-nomials, say f = P/Q, provided that the degree of Q exceeds the degree of P by at least 2. (See Exercise 16.36.) Example. To evaluate f-'. dxl(1 + x4), let f(z) = 1/(z4 + 1). Then P(z) = 1, Q(z) = 1 + z4, and hence (38) holds. The poles of f are the roots of the equation 1 + z4 = 0. These are zl, z2, z3, z4i where zk = e(2k-1)at/4 (k = 1, 2, 3, 4). Of these, only zl and z2 lie in the upper half-plane. The residue at zl is 1 Res f(z) = lim (z - z1)f(z) e-z=zt z-z, 41 - Z2)(Z1 - Z3)(Z1 - z4) 4i 464 Cauchy's Theorem and the Residue Calculus Similarly, we find Rest=.2 f(z) = (1/4i)ext/4. Therefore, Th.16.38 dx _ 2nf -xt/4 t/4 n x = = + ( 0 n2 _ n cos 4 e e I + x4 4i F. 2 . 16.25 EVALUATION OF GAUSS'S SUM BY RESIDUE CALCULUS The residue theorem is often used to evaluate sums by integration. We illustrate with a famous example called Gauss's sum G(n), defined by the formula n-1 G(n) _ E e2 a1r2/n r=0 (40) where n >_ 1. This sum occurs in various parts of the Theory of Numbers. For small values of n it can easily be computed from its definition. For example, we have G(l) = 1, G(2) = 0, G(3) = i3, G(4) = 2(1 + 1). Although each term of the sum has absolute value 1, the sum itself has absolute value 0, -Vn, or 2n. In fact, Gauss proved the remarkable formula G(n) = 2 ,ln(1 + i)(1 + "e- ntn/2), (41) for every n >_ 1. A number of different proofs of (41) are known. We will deduce (41) by considering a more general sum S(a, n) introduced by Dirichlet, n-1 S(a, n) _ extar2/n r= O where n and a are positive integers. If a = 2, then S(2, n) = G(n). Dirichlet proved (41) as a corollary of a reciprocity law for S(a, n) which can be stated as follows : Theorem 16.38. If the product na is even, we have 1 + it S(n, a), (42) S(a, n) =f \ / a where the bar denotes the complex conjugate. NOTE. To deduce Gauss's formula (41), we take a = 2 in (42), and observe that S (n, 2) = 1 + e-xtn/2. Proof. The proof given here is particularly instructive because it illustrates several techniques used in complex analysis. Some minor computational details are left as exercises for the reader. Let g be the function defined by the equation n-1 g(z) = E ex1a(z+r)2/n (43) r=0 Th. 16.38 Evaluation of Gauss's Sum Then g is analytic everywhere, and g(O) = S(a, n). Since na is even we find a-1 g(z + 1) - g(z) = e,iaz=1n(e2ziaz - 1) = eaiaz2/n(e2xiz - 1) E e2ximz l , m=0 466 (Exercise 16.41). Now define f by the equation f(z) = g(z)/(e2xiz - 1). Then f is analytic everywhere except for a first-order pole at each integer, and f satisfies the equation where f(z + 1) = f(z) + (P(z), a-1 (p(z) = exiozz/n 2 e2ximz. M=0 The function (P is analytic everywhere. At z = 0 the residue off is g(0)/(2iri) (Exercise 16.41), and hence (44) (45) S(a, n) = g(0) = 2ni Resf(z) = I f(z) dz, (46) z=0 y where y is any positively oriented simple closed path whose graph contains only the pole z = 0 in its interior region. We will choose y so that it describes a paral-lelogram with vertices A, A + 1, B + 1, B, where A = -I - Rexi14 and B = -I + Rexti4, Figure 16.7 as shown in Fig. 16.7. Integrating f along y we have fy. A+1 B+1 B A .f= f+ r f+ f f+ f f. A A+1 B+1 B In the integral f A+ i f we make the change of variable w = z + I and then use (44) to get f f(w) dw = f A f(z + 1) dz = fA Bf(z) dz + f B (p(z) dz. A+1 A JA 466 Cauchy's Theorem and the Residue Calculus Therefore (46) becomes A+1 B+1 S(a, n) = fA co(z) dz + f f(z) dz - f f(z) dz. (47) A B Now we show that the integrals along the horizontal segments from A to A + I and from B to B + I tend to 0 as R - + oo. To do this we estimate the integrand on these segments. We write f(z)I = 19(Z)I Ie2xiz - 11' and estimate the numerator and denominator separately. On the segment joining B to B + 1 we let y(t) = t + Rexi/4, where -I < t < I. From (43) we find (48) 19[y(t)1I Iexp {lria(t + Rexi/4 + r)2) I (49) r=O - ira(,l2tR + R2 + V2rR)/n. Since I?+iYJ = e and exp {-iraN/2rR/n} < 1, each term in (49) has absolute value not exceeding exp { - 7raR2/n} exp { - ./2natR/n}. But -I < t < 1, so we obtain the estimate I9[y(t)]I < n e-xaR21n. For the denominator in (48) we use the triangle inequality in the form Ie2xiz - 1I Z (Ie2xizl Since lexp {2niy(t )} I = exp { - 21R sin (n/4)} = exp { - 'l27rR}, we find e2xiyu> Therefore on the line segment joining B to B + 1 we have the estimate e-xaR2/n I.f(z)I < 1 -o(l) as R - +oo. 2sR Here o(1) denotes a function of R which tends to 0 as R - + oo. A similar argument shows that the integrand tends to 0 on the segment joining A to A + I as R - + oo. Since the length of the path of integration is I in each case, this shows that the second and third integrals on the right of (47) tend to 0 Evaluation of Gauss's Sum 467 Figure 16.8 as R -+ + oo. Therefore we can write (47) in the form S(a, n) = fA B T(z) dz + o(1) as R -+ + oo. (50) J To deal with the integral f A .p we apply Cauchy's theorem, integrating cp around the parallelogram with vertices A, B, a, -a, where a = B + -- = Re"'/4. (See Fig. 16.8.) Since qp is analytic everywhere, its integral around this parallelogram is 0, so cp+ fA. q'=O. (51) f Bp+ fB a(p+ I-" A Because of the exponential factor axiaZ2/" in (45), an argument similar to that given above shows that the integral of (p along each horizontal segment -+0 as R - + oo. Therefore (51) gives us B a A(p= f app+o(1) a = Re"i/4. Using (45) we find f a-1 fat a-1 9(z) dz = E eniazz/n e2 aimz dz = F e- ainm=/a I (a 1, a m=0 m=0 where S(a, n) = (a (p(z) dz + o(l) as R - +oo, (52) I(a, m, n, R) = f as exp na (z + na121 dz. m, n, R), Applying Cauchy's theorem again to the parallelogram with vertices -a, a, a - nm/a, -a - nm/a, we find as before that the integrals along the horizontal 468 Cauchy's Theorem and the Residue Calculus T h. 16.39 segments -+0 as R -+ + oo, so a-mn/a iia nm 2 I(a, m, n, R) = f-exp f- (z + -1 1 dz + o(1) as R - + oo. J a-nm/a n a ) The change of variable w = -v/a/n(z + nm/a) puts this into the form an I(a, m, n, R) = J n e""'2 dw + o(1) as R -+ + oo. l fcol-a/n a a-1 S(a, n) = E e ainm2/an lim m=0 a R-.+co /e e1t 2 dw. (53) By writing T = Ja/nR, we see that the last limit is equal to f Te"i/4 lim eni"Z dw = I. T-. too - Te-114 say, where I is a number independent of a and n. Therefore (53) gives us S(a, n) = Jn IS(n, a). (54) a To evaluate I we take a = 1 and n = 2 in (54). Then S(1, 2) = 1 + i and S(2, 1) = 1, so (54) implies I = (1 + i)lf, and (54) reduces to (42). 16.26 APPLICATION OF THE RESIDUE THEOREM TO THE INVERSION FORMULA FOR LAPLACE TRANSFORMS The following theorem is, in many cases, the easiest method for evaluating the limit which appears in the inversion formula for Laplace transforms. (See Exercise 11.38.) Theorem 16.39. Let F be a function analytic everywhere in C except, possibly, for a finite number of poles. Suppose there exist three positive constants M, b, c such that IF(z)I < M whenever IzI b. z Let a be a positive number such that the vertical line x = a contains no poles of F and let z1, . . ., zn denote the poles of F which lie to the left of this line. Then, for each real t > 0, we have lim f e(a+i°)F(a + iv) dv = 2iv E Res {ezF(z)}. (55) T- +oo -T k=1 z=zk Inversion Formula for Laplace Transforms 469 Figure 16.9 Proof. We apply Cauchy's residue theorem to the positively oriented path r shown in Fig. 16.9, where the radius T of the circular part is taken large enough to enclose all the poles of F which lie to the left of the line x = a, and also T > b. The residue theorem gives us Sr Now write ?t F(z) dz = 27[i ERes {e2tF(z)}. (56) k=1 z=zk B E -JA +f +JCD+.ID + f E where A, B, C, D, E are the points indicated in Fig. 16.9, and denote these integrals by I1, I2, 13, 14, I5. We will prove that It -1, 0 as T - + oo when k > 1. First, we have 1121 < M' f l z etT cos a T d0 < Me"t ( a1 = Meet T aresin a Tc Tc-1 12 -/ Tc 1.J Since T aresin (a/T) - a as T -+ + oo, it follows /that I2 -- 0 as T - + oo. In the same way we prove I5 - 0 as T -± + oo. Next, consider 13. We have 1131 < M etT cos B A = M x/2 e-tT sin 4 d c-1 1 (P T E/2 Tc -o But sin ip -> 2q,/ic if 0 < 9 < ir/2, and hence M x/2 ItM 1131 < T-1 fo e-ztTq,/x d(p = (1 - etT) _- 0 as T -i +oo. 2tT Similarly, we find 14 -1- 0 as T - + co. But as T -- + oo the righthand side of 470 Cauchy's Theorem and the Residue Calculus (56) remains unchanged. Hence "MT-+co Ii exists and we have T lim Ii = lim f-T e(°+°)t F(a + iv) i dv = 21ri E Res {eF(z)}. T-+oo k=1 z=zk Example. Let F(z) = zl (z + a2), where a is real. Then F has simple poles at ± ia. Since z/(z2 + a2) _ 1[1/(z + ia) + 1/(z - ia)], we find Res {e tF(z)} = I e", Res {etF(t)} = # e tat z=ta z=-ia Therefore the limit in (55) has the value 2ni cos at. From Exercise 11.38 we see that the function f, continuous on (0, + oo), whose Laplace transform is F, is given by f(t) _ cos at. 16.27 CONFORMAL MAPPINGS An analytic function f will map two line segments, intersecting at a point c, into two curves intersecting at f(c). In this section we show that the tangent lines to these curves intersect at the same angle as the given line segments if f'(c) # 0. This property is geometrically obvious for linear functions. For example, suppose f(z) = z + b. This represents a translation which moves every line parallel to itself, and it is clear that angles are preserved. Another example is f(z) = az, where a 0. If jal = 1, then a = ea and this represents a rotation about the origin through an angle a. If JaJ 1, then a = Re" and f represents a rotation composed with a stretching (if R > 1) or a contraction (if R < 1). Again, angles are preserved. A general linear function f(z) = az + b with a # 0 is a composition of these types and hence also preserves angles. In the general case, differentiability at c means that we have a linear approx-imation near c, say f(z) = f(c) + f'(c)(z - c) + o(z - c), and if f'(c) # 0 we can expect angles to be preserved near c. To formalize these ideas, let yi and Y2 be two piecewise smooth paths with respective graphs r, and I'2, intersecting at c. Suppose that yi is one-to-one on an interval containing ti, and that Y2 is one-to-one on an interval containing t2, where y1(t1) = y2(t2) = c. Assume also that y'1(t1) # 0 and y2(t2) # 0. The difference arg [YZ(t2)] - arg [y,(ti)], is called the angle from I'1 to r2 at c. Now assume that f'(c) # 0. Then (by Theorem 13.4) there is a disk B(c) on which f is one-to-one. Hence the composite functions w1(t) =f[Y1(t)] and w2(t) =f[Y2(t)], will be locally one-to-one near ti and t2, respectively, and will describe arcs C1 and C2 intersecting at f(c). (See Fig. 16.10.) By the chain rule we have w'1(t1) = f'(c)Yi(t1) # 0 and w2(12) = f'(c)YZ(t2) # 0 Conformal Mappings 471 C2 Figure 16.10 Therefore, by Theorem 1.48 there exist integers nl and n2 such that arg [w'1(t1)] = arg [f'(c)] + arg [yi(tl)] + 2nn1, arg [wz(t2)] = arg [f'(c)] + arg [y2(t2)] + 2nn2, so the angle from Cl to C2 at f(c) is equal to the angle from f1 to f2 at c plus an integer multiple of 2ic. For this reason we say that f preserves angles at c. Such a function is also said to be conformal at c. Angles are not preserved at points where the derivative is zero. For example, iff(z) = z2, a straight line through the origin making an angle a with the real axis is mapped by f onto a straight line making an angle 2a with the real axis. In general, when f'(c) = 0, the Taylor expansion off assumes the form f(Z) - f(c) = (z - c)k[ak + ak+l(Z - C) + ... ], where k z 2. Using this equation, it is easy to see that angles between curves intersecting at c are multiplied by a factor k under the mapping f. Among the important examples of conformal mappings are the Mobius transformations. These are functions f defined as follows: If a, b, c, d are four complex numbers such that ad - be # 0, we define f(z) = az + b (57) cz + d whenever cz + d 0. It is convenient to define f everywhere on the extended plane C by setting f(-d/c) = oo and f(oo) = a/c. (If c = 0, these last two equations are to be replaced by the single equation f(oo) = oo.) Now (57) can be solved for z in terms of f(z) to get - df(z) + b z = cf(z) - a This means that the inverse function f -1 exists and is given by f_1(z) -dz + b cz - a 472 Cauchy's Theorem and the Residue Calculus with the understanding that f -'(a/c) = oo and f -1(oo) = -d/c. Thus we see that Mobius transformations are one-to-one mappings of C onto itself. They are also conformal at each finite z # - d/c, since .f'(z)= be - ad # 0. (cz + d)' One of the most important properties of these mappings is that they map circles onto circles (including straight lines as special cases of circles). The proof of this is sketched in Exercise 16.46. Further properties of Mobius transformations are also described in the exercises near the end of the chapter. EXERCISES Complex integration; Cauchy's integral formulas 16.1 Let y be a piecewise smooth path with domain [a, b] and graph r. Assume that the integral JY f exists. Let S be an open region containing r and let g be a function such that g'(z) exists and equals f(z) for each z on 11". Prove that fi= f g' = g(B) - g(A), where A = y(a) and B = y(b). Y In particular, if y is a circuit, then A = B and the integral is 0. Hint. Apply Theorem 7.34 to each interval of continuity of y'. 16.2 Let y be a positively oriented circular path with center 0 and radius 2. Verify each of the following by using one of Cauchy's integral formulas. a) f ez dz = 2xi. b) f e3 dz = .in . z Yz yz c) f ez dz = 31 . d) fy z ez 1 dz = 2nie. e) ez dz = 2ni(e - 1). f) ez dz = 2ni(e - 2). fy z(z - 1) y z2(z - 1) 16.3 Let f = u + iv be analytic on a disk B(a; R). If 0 < r < R, prove that i 2A f'(a) = - f u(a + reie)eie d9. 7rr Jo 16.4 a) Prove the following stronger version of Liouville's theorem: If f is an entire function such that lime... l f (z )/z I = 0, then f is a constant. b) What-can you conclude about an entire function which satisfies an inequality of the form Jf(z)I s MIzI' for every complex z, where c > 0? Exercises 473 16.5 Assume that f is analytic on B(0; R). Let y denote the positively oriented circle with center at 0 and radius r, where 0 < r < R. If a is inside y, show that f(a) = 27ri f f(z) i z 1 a z -lra } dz. r 2/ If a = Aea, show that this reduces to the formula f(a) = 1 2" (r2 - A2)f(ree) dB. ticf o r2 - 2rA cos (a - 6) + A2 By equating the real parts of this equation we obtain an expression known as Poisson's integral formula. 16.6 Assume that f is analytic on the closure of the disk B(0; 1). If jai < 1, show that f(z) z 1 - a dz, (1 - Jal2)f(a) = 2Ici fy where y is the positively oriented unit circle with center at 0. Deduce the inequality " Jf(eie)I dB. (1 - lal2)1f(a)! 5 2 2n fo 16.7 Letf(z) = Et o 2"z"l3" if IzI < 3/2, and let g(z) = E o (2z)-" if 1zI > 1. Let y be the positively oriented circular path of radius 1 and center 0, and define h(a) for jai 96 1 as follows: h(a) = 1 f \ f(z) + a2g(z) ) dz. 21ri y z - a z2-az Prove that if Jai > 1. Taylor expansions 16.8 Define f on the disk B(0; 1) by the equation f(z) _ o z". Find the Taylor expansion off about the point a = 4 and also about the point a Determine the radius of convergence in each case. 16.9 Assume that f has the Taylor expansion f(z) = En 0 a(n)z", valid in B(0; R). Let 1 D-1 g(z) = - E f(ze2"tk/D). P k=O Prove that the Taylor expansion ofg consists of every pth term in that of f That is, if z e B(0; R) we have 00 g(z) = E a(pn)z°". M=O 474 Cauchy's Theorem and the Residue Calculus 16.10 Assume that f has the Taylor expansion f(z) = En o anz", valid in B(0; R). Let sn(z) = Ek=o akz". If 0 < r < R and if Iz I < r, show that +1 -+1 ss(z) = 1f f(w) w" z° dw, "+1 2nt w w - z where y is the positively oriented circle with center at 0 and radius r. 16.11 Given the Taylor expansions f(z) = Ln o anz" and g(z) = En o bnz", valid for IzI <_ R1 and IzI < R2, respectively. Prove that if IzI < R1R2 we have 1 f(w) g ( z/ dw = 0° --a"b"Z", E 27ri y w w n_0 where y is the positively oriented circle of radius R1 with center at 0. 16.12 Assume that f has the Taylor expansion f(z) _ Y_n=0 an(z - a)n, valid in B(a; R). a) If 0 <- r < R, deduce Parseval's identity: 2n °° J If(a + reie)I2 dB = r Iaal2 r2" 2n 0 n=O b) Use (a) to deduce the inequality 0 Iant2 r2n <- M(r)2, where M(r) is the maximum of If I on the circle Iz - al = r. c) Use (b) to give another proof of the local maximum modulus principle (Theorem 16.27). 16.13 Prove Schwarz's lemma: Let f be analytic on the disk B(0; 1). Suppose that f(0) = 0 and I f(z)I 5 1 if Iz I < 1. Then If'(0)I <- 1 and If(z)I <- IzI, if IzI < 1. If I f'(0)I = 1 or if If(zo)I = Izol for at least one z0 in B'(0; 1), then f(z) = e"az, where a is real. Hint. Apply the maximum-modulus theorem to g, where g(0) = f'(0) and g(z) = f(z)lz if z# 0. Laurent expansions, singularities, residues 16.14 Let f and g be analytic on an open region S. Let y be a Jordan circuit with graph t such that both I' and its inner region lie within S. Suppose that Ig(z)I < j f(z)I for every zonI'. a). Show that 1 fe(z) + g'(z) dz = 1 L(z-) dz. 2nif y, f(z) + g(z) 2iri Y f(z) Hint. Let m = inf {If(z)I - Ig(z)I : z e t}. Then m > 0 and hence If(z)+tg(z)I?m>0 for each t in [0, 1 ] and each z on F. Now let fi(t) = 1 f'(z) + tg'(z) dz, if 0 < t < 1. 2nt r f(z) + tg(z) Then 0 is continuous, and hence constant, on [0, 11. Thus, 0(0) = 0(1). Exercises 475 b) Use (a) to prove that f and f + g have the same number of zeros inside IF (Rouche's theorem). 16.15 Let p be a polynomial of degree n, say p(z) = ao + a1z + + anz", where an 76 0. Take f(z) = a"z", g(z) = p(z) - f(z) in Rouche's theorem, and prove that p has exactly n zeros in C. 16.16 Let f be analytic on the closure of the disk B(0; 1) and suppose If(z)I < 1 if Iz I = 1. Show that there is one, and only one, point zo in B(0; 1) such that f(zo) = zo. Hint. Use Rouch6's theorem. 16.17 Let pn(z) denote the nth partial sum of the Taylor expansion ez = Y_,,° o z"/n!. Using Rouche's theorem (or otherwise), prove that for every r > 0 there exists an N (depending on r) such that n >- N implies pn(z) ;4 0 for every z in B(0; r). 16.18 If a > e, find the number of zeros of the function f(z) = ez - az" which lie inside the circle Iz I = 1. 16.19 Give an example of a function which has all the following properties, or else explain why there is no such function: f is analytic everywhere in C except for a pole of order 2 at 0 and simple poles at i and -i; f(z) = f(-z) for all z; f(1) = 1; the function g(z) = f(1/z) has a zero of order 2 at z = 0; and Rest=t f(z) = 2i. 16.20 Show that each of the following Laurent expansions is valid in the region indicated: 1 _ 000 Z" CO 1 + Z" if 1 < IzI < 2. a) (z - 1)(2 - z) = n=0 2nt1 n=1 b) 1 =X1-2"-1 (z - 1)(2 - z) n==2 Zn if IzI > 2. 16.21 For each fixed tin C, define Jn(t) to be the coefficient of z" in the Laurent expansion OD e(z-1/z)t/2 = r Jn(t)Zn n=-OD Show that for n >- 0 we have n o and that J_n(t) _ (-1)"JJ(t). Deduce the power series expansion JJ(t) = LI (-1)k(lt) n+2k (n > 0). kL=o0 k! (n Z+ k)! The function Jn is called the Bessel function of order n. 16.22 Prove Riemann's theorem: If zo is an isolated singularity off and if If is bounded on some deleted neighborhood B'(zo), then zo is a removable singularity. Hint. Estimate the integrals for the coefficients an in the Laurent expansion off and show that an = 0 for each n < 0. 16.23 Prove the Casorati-Weierstrass theorem: Assume that zo is an essential singularity of f and let c be an arbitrary complex number. Then, for every e > 0 and every disk B(zo), there exists a point z in B(zo) such that If(z) - cl < e. Hint. Assume that the theorem is false and arrive at a contradiction by applying Exercise 16.22 to g, where g(z) _ 1/[f(z) - c]. J,(t) = 1 cos (t sin 8 - nO) dB 476 Cauchy's Theorem and the Residue Calculus 16.24 The point at infinity. A function f is said to be analytic at oo if the function g defined by the equation g(z) = f(1/z) is analytic at the origin. Similarly, we say that f has a zero, a pole, a removable singularity, or an essential singularity at oo if g has a zero, a pole, etc., at 0. Liouville's theorem states that a function which is analytic everywhere in C must be a constant. Prove that a) f is a polynomial if, and only if, the only singularity of fin C is a pole at oo, in which case the order of the pole is equal to the degree of the polynomial. b) f is a rational function if, and only if, f has no singularities in C other than poles. 16.25 Derive the following "short cuts" for computing residues: a) If a is a first order pole for f, then Res f(z) = lim (z - a)f(z). z=a z-+a b) If a is a pole of order 2 for f, then Res f(z) = g'(a), where g(z) = (z - a)2f(z). z=a c) Suppose f and g are both analytic at a, with f (a) 76 0 and a a first-order zero for g. Show that f(z) = f(a) Res f(z) _ f(a)g '(a) - f(a)g"(a) Res -z=a 9(z) 9'(a) z=a [9(Z)]2 [9'(a)]3 d) If f and g are as in (c), except that a is a second-order zero for g, then Resf(z) 6f'(a)g"(a) - 2.f(a)9(a) z=a g(z) 3[g"(a)]2 16.26 Compute the residues at the poles off if zez a) f(z) = z2 - 1 ' smz c)f(z)= e) .f(z) = z cos z 1 I - Z" b) f(z) = ex z(z - 1)2 d) .f(z) = 1 1 - eZ' (where n is a positive integer). 16.27 If y(a; r) denotes the positively oriented circle with center at a and radius r, show that a) f,(0;4) 3z - 1 dz = 6ni, (z + 1)(z - 3) Z c) -dz = 2ni, Z - 1 y(0;2) Z4 b) 22z dz = 4ni, fyo;2) z z + 1 d) fy2i) eZ 2 dz = 2ie2. (z - 2) Exercises Evaluate the integrals in Exercises 16.28 through 16.35 by means of residues. 2a dt 16 28 2na . fo (a + b cos t)2 (a2 - b2)3/2 16.29 f' ft cos 2t dt = 2na2 1 - 2a cost + a2 1 - a2 2, (1 + cos 3t) dt 16 30 n(a2 a + . 1-2acost+a2 fo 1-a if0<b<a. if a2 < 1. if 0 < a < 1. 16.31 f 21t sin 2 t dt - 27c(a -a2 - 62) if 0 < b < a. o a + b cost b2 16.32 I 1 dx = 2n 3 J x2+x+1 3 16.33 f'4 x6 dx -3n' ,J J - (1 + x4)2 16 16.34 x2 dx = ft fo (x2 + 4)2(x2 + 9) 200 477 16.35 a) f OD X dx = -/sin 25 . 0 Hint. Integrate z/(1 + z5) around the boundary of the circular sector S = (rei° : 0 S r <- R, 0 <- 9 <- 2x/5), and let R -). oo. x2m 2m + 1 b) 1 + x2n dx = 2n/ sin ( 2n' x) , , m, n integers, 0 < m < n. fO'O 16.36 Prove that formula (38) holds if f is the quotient of two polynomials, say f = P/Q, where the degree of Q exceeds that of P by 2 or more. 16.37 Prove that formula (38) holds if f(z) = eimzP(z)/Q(z), where m > 0 and P and Q are polynomials such that the degree of Q exceeds that of P by 1 or more. This makes it possible to evaluate integrals of the form f a0 eimx P(x)dx -. . Q(x) by the method described in Theorem 16.37. 16.38 Use the method suggested in Exercise 16.37 to evaluate the following integrals : a) Jo x(a2 x2) dx = 2x2 (1 - e ") if m 0, a > 0. x4 ifm>0,a>0. b) fo, `7 1 478 Cauchy's Theorem and the Residue Calculus 16.39 Let w = e2' 113 and let y be a positively oriented circle whose graph does not pass through 1, w, or w2. (The numbers 1, w, w2 are the cube roots of 1.) Prove that the integral f (z + dz z3 - 1 is equal to 2ni(m + nw)/3, where m and n are integers. Determine the possible values of m and n and describe how they depend on y. 16.40 Let y be a positively oriented circle with center 0 and radius < 2n. If a is complex and n is an integer, let 1 zn-leaz I(n, a) = -dz. 2ni 7 1 - eZ Prove that 1(0, a) _ 4 - a, I (l, a) _ -1, and 1(n, a) = 0 if n > 1. Calculate I(-n, a) in terms of Bernoulli polynomials when n >- 1 (see Exercise 9.38). 16.41 This exercise requests some of the details of the proof of Theorem 16.38. Let n-1 g(z) = E e 1a(z+r)2/n' f (Z) (z) = 9(z)/(e2aiz - 1), r=0 where a and n are positive integers with na even. Prove that: a) g(z + 1) - g(z) = eniaz2In(e2niz - 1) -a-1 e2rzimz 1 m=0 b) Res.=Of(z) = g(0)/(27ri). c) The real part of i(t + Reni14 + r)2 is R2 + V2rR). One-to-one analytic functions 16.42 Let S be an open subset of C and assume that f is analytic and one-to-one on S. Prove that: a) f'(z) # 0 for each z in S. (Hence f is conformal at each point of S.) b) If g is the inverse of f, then g is analytic on f(S) and g'(w) = 1/f'(g(w)) if WEf(S). 16.43 Let f : C -+ C be analytic and one-to-one on C. Prove that f(z) = az + b, where a 0 0. What can you conclude if f is one-to-one on C and analytic on C except possibly for a finite number of poles? 16.44 If f and g are Mobius transformations, show that the composition f o g is also a Mobius transformation. 16.45 Describe geometrically what happens to a point z when it is carried into f(z) by the following special Mobius transformations: a) f(z) = z + b (Translation). b) f(z) = az, where a > 0 (Stretching or contraction). c) f(z) = ei"z, where a is real (Rotation). d) f(z) = 11z (Inversion). Exercises 479 16.46 If c t- 0, we have az+ba+ be - ad cz+d c c(cz+d) Hence every Mobius transformation can be expressed as a composition of the special cases described in Exercise 16.45. Use this fact to show that Mobius transformations carry circles into circles (where straight lines are considered as special cases of circles). 16.47 a) Show that all Mobius transformations which map the upper half-plane T = {x + iy : y >- 0} onto the closure of the disk B(0; 1) can be expressed in the form f(z) = ei8(z - a)l(z - a), where a is real and a e T. b) Show that a and a can always be chosen to map any three given points of the real axis onto any three given points on the unit circle. 16.48 Find all Mobius transformations which map the right half-plane S = {x+iy:x>-0} onto the closure of B(0; 1). 16.49 Find all Mobius transformations which map the closure of B(0; 1) onto itself. 16.50 The fixed points of a Mobius transformation f (z) = az + b (ad - be ;4 0) cz + d are those points z for which f(z) = z. Let D = (d - a)2 + 4bc. a) Determine all fixed points when c = 0. b) If c 0 0 and D 0 0, prove that f has exactly 2 fixed points z1 and z2 (both finite) and that they satisfy the equation f (z) - Zl = Re'° z - z1, where R > 0 and 8 is real. f(z) - z2 z - z2 c) If c 0 0 and D = 0, prove that f has exactly one fixed point z1 and that it satisfies the equation 1 = 1 + C for some C : 0. f(z) - z1 z - z1 d) Given any Mobius transformation, investigate the successive images of a given point w. That is, let w1 = f(w), w2 = .f (w1), ... , W. = f(wn-1), .... and study the behavior of the sequence {wn}. Consider the special case a, b, c, d real, ad - be = 1. MISCELLANEOUS EXERCISES 16.51 Determine all complex z such that OD n z = E E e2aikz/n. n=2 k=1 480 Cauchy's Theorem and the Residue Calculus 16.52 If f(z) = E o a,,z" is an entire function such that I f(reie)I < Me'k for all r > 0, where M > 0 and k > 0, prove that m a. 1 < for n >- 1. (n/k)n1k 16.53 Assume f is analytic on a deleted neighborhood B'(0; a). Prove that lim.,0 f(z) exists (possibly infinite) if, and only if, there exists an integer n and a function g, analytic on B(0; a), with g(0) ? 0, such that f(z) = z"g(z) in B'(0; a). 16.54 Let p(z) = Ek_ o akzk be a polynomial of degree n with real coefficients satisfying ao > a1 >...> an-1 > an > 0. Prove that p(z) = 0 implies jzi > 1. Hint. Consider (1 - z)p(z). 16.55 A function f, defined on a disk B(a; r), is said to have a zero of infinite order at a if, for every integer k > 0, there is a function gk, analytic at a, such that f (z) = (z - a)kgk(z) on B(a; r). If f has a zero of infinite order at a, prove that f = 0 everywhere in B(a; r). 16.56 Prove Morera's theorem : If f is continuous on an open region S in C and if f y f = 0 for every polygonal circuit y in S, then f is analytic on S. SUGGESTED REFERENCES FOR FURTHER STUDY 16.1 Ahlfors, L. V., Complex Analysis, 2nd ed. McGraw-Hill, New York, 1966. 16.2 Caratheodory, C., Theory of Functions of a Complex Variable, 2 vols. F. Steinhardt, translator. Chelsea, New York, 1954. 16.3 Estermann, T., Complex Numbers and Functions. Athlone Press, London, 1962. 16.4 Heins, M., Complex Function Theory. Academic Press, New York, 1968. 16.5 Heins, M., Selected Topics in the Classical Theory of Functions of a Complex Vari-able. Holt, Rinehart, and Winston, New York, 1962. 16.6 Knopp, K., Theory of Functions, 2 vols. F. Bagemihl, translator. Dover, New York, 1945. 16.7 Saks, S., and Zygmund, A., Analytic Functions, 2nd ed. E. J. Scott, translator. Monografie Matematyczne 28, Warsaw, 1965. 16.8 Sansone, G., and Gerretsen, J., Lectures on the Theory of Functions of a Complex Variable, 2 vols. P. Noordhoff, Grbningen, 1960. 16.9 Titchmarsh, E. C., Theory of Functions, 2nd ed. Oxford University Press, 1939. INDEX OF SPECIAL SYMBOLS e, 0, belongs to (does not belong to), 1, 32 c, is a subset of, 1, 33 R, set of real numbers, 1 R+, R', set of positive (negative) numbers, 2 {x: x satisfies P}, the set of x which satisfy property P, 3, 32 (a, b), [a, b], open (closed) interval with endpoints a and b, 4 [a, b), (a, b], half-open intervals, 4 (a, + oo), [a, + oo), (- oo, a), (- oo, a], infinite intervals, 4 Z+, set of positive integers, 4 Z, set of all integers (positive, negative, and zero), 4 Q, set of rational numbers, 6 max S, min S, largest (smallest) element of S, 8 sup, inf, supremum, (infimum), 9 [x], greatest integer 5 x, 11 R, extended real-number system, 14 C, the set of complex numbers, the complex plane, 16 C , extended complex-number system, 24 A x B, cartesian product of A and B, 33 F(S), image of S under F, 35 F: S -+ T, function from S to T, 35 {F"}, sequence whose nth term is F", 37 U, u, union, 40, 41 n, r), intersection, 41 B - A, the set of points in B but not in A, 41 f -'(Y), inverse image of Y under f, 44 (Ex. 2.7), 81 R", n-dimensional Euclidean space, 47 (x1, . . . , x"), point in R", 47 II x II, norm or length of a vector, 48 uk, kth-unit coordinate vector, 49 B(a), B(a; r), open n-ball with center a, (radius r), 49 int S, interior of S, 49, 61 (a, b), [a, b], n-dimensional open (closed) interval, 50, 52 S, closure of S, 53, 62 S', set of accumulation points of S, 54, 62 (M, d), metric space M with metric d, 60 481 482 Index of Special Symbols d(x,y), distance from x to y in metric space, 60 BM(a; r), ball in metric space M, 61 8S, boundary of a set S, 64 lim , lim , right- (left-)hand limit, 93 X -C+ X- C-f (c+ ), f (c - ), right- (left-)hand limit off at c, 93 O f(T), oscillation off on a set T, 98 (Ex. 4.24), 170 cof(x), oscillation off at a point x, 98 (Ex. 4.24), 170 f'(c), derivative off at c, 104, 114, 117 Dk f, partial derivative off with respect to the kth coordinate, 115 D,,k f, second-order partial derivative, 116 Y[a, b], set of all partitions of [a, b], 128, 141 Vf, total variation off, 129 Af, length of a rectifiable path f, 134 S(P, f, a), Riemann-Stieltjes sum, 141 f e R(a) on [a, b], f is Riemann-integrable with respect to a on [a, b], 141 f e R on [a, b], f is Riemann-integrable on [a, b], 142 a / on [a, b], a is increasing on [a, b], 150 U(P, f, a), L(P, f, a), upper (lower) Stieltjes sums, 151 Jim sup, limit superior (upper limit), 184 lim inf, limit inferior (lower limit), 184 a = 0(b.), a = o(b ), big oh (little oh) notation, 192 l.i.m. f = f, {f.) converges in the mean to f, 232 11-ao f e C °°, f has derivatives of every order, 241 a.e., almost everywhere, 172 f / f a.e. on S, sequence { fq} increases on S and converges to f a.e. on S, 254 S(I), set of step functions on an interval 1, 256 U(1), set of upper functions on an interval I, 256 L(I), set of Lebesgue-integrable functions on an interval 1, 260 f + f - positive (negative) part of a function f, 261 M(I), set of measurable functions on an interval 1, 279 Xs, characteristic function of S, 289 µ(S), Lebesgue measure of S, 290 (f, g), inner product of functions f and g, in L2(I), 294, 295 11f 11, L2-norm off, 294, 295 L2(I), set of square-integrable functions on 1, 294 f g, convolution off and g, 328 f'(c; u), directional derivative off at c in the direction u, 344 T,,, f'(c), total derivative, 347 Vf, gradient vector off, 348 m(T), matrix of a linear function T, 350 Df(c), Jacobian matrix off at c, 351 L(x, y), line segment joining x and y, 355 Index of Special Symbols det [aj j], determinant of matrix [a; j], 367 JJ, Jacobian determinant of f, 368 f e C, the components off have continuous first-order partials, 371 SI f(x) dx, multiple integral, 389, 407 c(S), c(S), inner (outer) Jordan content of S, 396 c(S), Jordan content of S, 396 Si f, contour integral off along y, 436 A(a; rl, r2), annulus with center a, 438 n(y, z), winding number of a circuit y with respect to z, 445 B'(a), B'(a; r), deleted neighborhood of a, 457 Res f(z), residue off at a, 459 483 z=a Abel, Neils Henrik, (1802-1829), 194, 245, 248 Abel, limit theorem, 245 partial summation formula, 194 test for convergence of series, 194, 248 (Ex. 9.13) Absolute convergence, of products, 208 of series, 189 Absolute value, 13, 18 Absolutely continuous function, 139 Accumulation point, 52, 62 Additive function, 45 (Ex. 2.22) Additivity of Lebesgue measure, 291 Adherent point, 52, 62 Algebraic number, 45 (Ex. 2.15) Almost everywhere, 172, 391 Analytic function, 434 Annulus, 438 Approximation theorem of Weierstrass, 322 Arc, 88, 435 Archimedean property of real numbers, 10 Arc length, 134 Arcwise connected set, 88 Area (content) of a plane region, 396 Argand, Jean-Robert (1768-1822), 17 Argument of complex number, 21 Arithmetic mean, 205 Arzela, Cesare (1847-1912), 228, 273 Arzela's theorem, 228, 273 Associative law, 2, 16 Axioms for real numbers, 1, 2, 9 Ball, in a metric space, 61 in R-, 49 Basis vectors, 49 Bernoulli, James (1654-1705), 251, 338, 478 Bernoulli, numbers, 251 (Ex. 9.38) periodic functions, 338 (Ex. 11.18) polynomials, 251 (Ex. 9.38), 478 (Ex. 16.40) INDEX Bernstein, Sergei Natanovic (1880-), 242 Bernstein's theorem, 242 Bessel, Friedrich Wilhelm (1784-1846), 309,475 Bessel function, 475 (Ex. 16.21) Bessel inequality, 309 Beta function, 331 Binary system, 225 Binomial series, 244 Bolzano, Bernard (1781-1848),54,85 Bolzano's theorem, 85 Bolzano-Weierstrass theorem, 54 Bonnet, Ossian (1819-1892), 165 Bonnet's theorem, 165 Borel, Emile (1871-1938), 58 Bound, greatest lower, 9 least upper, 9 lower, 8 uniform, 221 upper, 8 Boundary, of a set, 64 point, 64 Bounded, away from zero, 130 convergence, 227, 273 function, 83 set, 54, 63 variation, 128 Cantor, Georg (1845-1918), 8, 32, 56, 67, 180, 312 Cantor intersection theorem, 56 Cantor-Bendixon theorem, 67 (Ex. 3.25) Cantor set, 180 (Ex. 7.32) Cardinal number, 38 Carleson, Lennart, 312 Cartesian product, 33 Casorati-Weierstrass theorem, 475 (Ex. 16.23) Cauchy, Augustin-Louis (1789-1857), 14, 73, 118, 177, 183, 207, 222 485 486 Index Cauchy condition, for products, 207 for sequences, 73, 183 for series, 186 for uniform convergence, 222, 223 Cauchy, inequalities, 451 integral formula, 443 integral theorem, 439 principal value, 277 product, 204 residue theorem, 460 sequence, 73 Cauchy-Riemann equations, 118 Cauchy-Schwarz inequality, for inner products, 294 for integrals, 177 (Ex. 7.16), 294 for sums, 14, 27 (Ex. 1.23), 30 (Ex. 1.48) Cesaro, Ernesto (1859-1906), 205, 320 Cesiro, sum, 205 summability of Fourier series, 320 Chain rule, complex functions, 117 real functions, 107 matrix form of, 353 vector-valued functions, 114 Change of variables, in a Lebesgue integral, 262 in a multiple Lebesgue integral, 421 in a Riemann integral, 164 in a Riemann-Stieltjes integral, 144 Characteristic function, 289 Circuit, 435 Closed, ball, 67 (Ex. 3.31) curve, 435 interval, 4, 52 mapping, 99 (Ex. 4.32) region, 90 set, 53, 62 Closure of a set, 53 Commutative law, 2, 16 Compact set, 59, 63 Comparison test, 190 Complement, 41 Complete metric space, 74 Complete orthonormal set, 336 (Ex. 11.6) Completeness axiom, 9 Complex number, 15 Complex plane, 17 Component, interval, 51 of a metric space, 87 of a vector, 47 Composite function, 37 Condensation point, 67 (Ex. 3.23) Conditional convergent series, 189 rearrangement of, 197 Conformal mapping, 471 Conjugate complex number, 28 (Ex. 1.29) Connected, metric space, 86 set, 86 Content, 396 Continuity, 78 uniform, 90 Continuously differentiable function, 371 Contour integral, 436 Contraction, constant, 92 fixed-point theorem, 92 mapping, 92 Convergence, absolute, 189 bounded, 227 conditional, 189 in a metric space, 70 mean, 232 of a product, 207 of a sequence, 183 of a series, 185 pointwise, 218 uniform, 221 Converse of a relation, 36 Convex set, 66 (Ex. 3.14) Convolution integral, 328 Convolution theorem, for Fourier trans-forms, 329 for Laplace transforms, 342 (Ex. 11.36) Coordinate transformation, 417 Countable additivity, 291 Countable set, 39 Covering of a set, 56 Covering theorem, Heine-Borel, 58 Lindelof, 57 Cramer's rule, 367 Curve, closed, 435 Jordan, 435 piecewise-smooth, 435 rectifiable, 134 Daniell, P. J. (1889-1946), 252 Darboux, Gaston (1842-1917), 152 Decimals, 11, 12, 27 (Ex. 1.22) Dedekind, Richard (1831-1916), 8 Deleted neighborhood, 457 De Moivre, Ham (1667-1754), 29 De Moivre's theorem, 29 (Ex. 1.44) Index Dense set, 68 (Ex. 3.32) Denumerable set, 39 Derivative(s), of complex functions, 117 directional, 344 partial, 115 of real-valued functions, 104 total, 347 of vector-valued functions, 114 Derived set, 54, 62 Determinant, 367 Difference of two sets, 41 Differentiation, of integrals, 162, 167 of sequences, 229 of series, 230 Dini, Ulisse (1845-1918), 248, 312, 319 Dini's theorem, on Fourier series, 319 on uniform convergence, 248 (Ex. 9.9) Directional derivative, 344 Dirichlet, Peter Gustav Lejeune (1805-1859), 194, 205, 215, 230, 317, 464 Dirichlet, integrals, 314 kernel, 317 product, 205 series, 215 (Ex. 8.34) Dirichlet's test, for convergence of series, 194 for uniform convergence of series, 230 Disconnected set, 86 Discontinuity, 93 Discrete metric space, 61 Disjoint sets, 41 collection of, 42 Disk, 49 of convergence, 234 Distance function (metric), 60 Distributive law, 2, 16 Divergent, product, 207 sequence, 183 series, 185 Divisor, ,4 greatest common, 5 Domain (open region), 90 Domain of a function, 34 Dominated convergence theorem, 270 Dot product, 48 Double, integral, 390, 407 Double sequence, 199 Double series, 200 Du Bois-Reymond, Paul (1831-1889), 312 Duplication formula for the Gamma function, 341 (Ex. 11.31) 487 e, irrationality of, 7 Element of a set, 32 Empty set, 33 Equivalence, of paths, 136 relation, 43 (Ex. 2.2) Essential singularity, 458 Euclidean, metric, 48, 61 space R", 47 Euclid's lemma, 5 Euler, Leonard (1707-1783), 149, 192, 209, 365 Euler's, constant, 192 product for C(s), 209 summation formula, 149 theorem on homogeneous functions, 365 (Ex. 12.18) Exponential form, of Fourier integral theorem, 325 of Fourier series, 323 Exponential function, 7, 19 Extended complex plane, 25 Extended real-number system, 14 Extension of a function, 35 Exterior (or outer region) of a Jordan curve, 447 Extremum problems, 375 Fatou, Pierre (1878-1929), 299 Fatou's lemma, 299 (Ex. 10.8) Fej6r, Leopold (1880-1959),179,312,320 Fej6r's theorem, 179 (Ex. 7.23), 320 Fekete, Michel, 178 Field, of complex numbers, 116 of real numbers, 2 Finite set, 38 Fischer, Emst (1875-1954),297,311 Fixed point, of a function, 92 Fixed-point theorem, 92 Fourier, Joseph (1758-1830), 306, 309, 312, 324, 326 Fourier coefficient, 309 Fourier integral theorem, 324 Fourier series, 309 Fourier transform, 326 Fubini, Guido (1879-1943),405,410,413 Fubini's theorem, 410, 413 Function, definition of, 34 Fundamental theorem, of algebra, 15, 451, 475 (Ex. 16.15) of integral calculus, 162 488 Gamma function, continuity of, 282 definition of, 277 derivative of, 284, 303 (Ex. 10.29) duplication formula for, 341 (Ex. 11.31) functional equation for, 278 series for, 304 (Ex. 10.31) Gauss, Karl Friedrich (1777-1855), 17, 464 Gaussian sum, 464 Geometric series, 190, 195 Gibbs' phenomenon, 338 (Ex. 11.19) Global property, 79 Goursat, Ldouard (1858-1936), 434 Gradient, 348 Gram, Jorgen Pedersen (1850-1916), 335 Gram-Schmidt process, 335 (Ex. 11.3) Greatest lower bound, 9 Hadamard, Jacques (1865-1963), 386 Hadamard determinant theorem, 386 (Ex. 13.16) Half-open interval, 4 Hardy, Godfrey Harold (1877-1947), 30, 206, 217, 251, 312 Harmonic series, 186 Heine, Eduard (1821-1881), 58, 91, 312 Heine-Borel covering theorem, 58 Heine's theorem, 91 Hobson, Ernest William (1856-1933), 312, 415 Homeomorphism, 84 Homogeneous function, 364 (Ex. 12.18) Homotopic paths, 440 Hyperplane, 394 Identity theorem for analytic functions, 452 Image, 35 Imaginary part, 15 Imaginary unit, 18 Implicit-function theorem, 374 Improper Riemann integral, 276 Increasing function, 94, 150 Increasing sequence, of functions, 254 of numbers, 71, 185 Independent set of functions, 335 (Ex. 11.2) Induction principle, 4 Inductive set, 4 Inequality, Bessel, 309 Cauchy-Schwarz, 14, 177 (Ex. 7.16), 294 Minkowski, 27 (Ex. 1.25) triangle, 13, 294 Infimum, 9 Infinite, derivative, 108 product, 206 series, 185 set, 38 Infinity, in C, 24 in R, 14 Inner Jordan content, 396 Inner product, 48, 294 Integers, 4 Integrable function, Lebesgue, 260, 407 Riemann, 141, 389 Integral, equation, 181 test, 191 transform, 326 Integration by parts, 144, 278 Integrator, 142 Interior (or inner region) of a Jordan curve, 447 Interior, of a set, 49, 61 Interior point, 49, 61 Intermediate-value theorem, for continuous functions, 85 for derivatives, 112 Intersection of sets, 41 Interval, in R, 4 in R°, 50, 52 Inverse function, 36 Inverse-function theorem, 372 Inverse image, 44 (Ex. 2.7), 81 Inversion formula, for Fourier transforms, 327 for Laplace transforms, 342 (Ex. 11.38), 468 Irrational numbers, 7 Isolated point, 53 Isolated singularity, 458 Isolated zero, 452 Isometry, 84 Iterated integral, 167, 287 Iterated limit, 199 Iterated series, 202 Jacobi, Carl Gustav Jacob (1804-1851), 351, 368 1 Jacobian, determinant, 368 matrix, 351 Jordan, Camille (1838-1922), 312, 319, 396, 435, 447 Jordan, arc, 435 content, 396 489 curve, 435 curve theorem, 447 theorem on Fourier series, 319 Jordan-measurable set, 396 Jump, discontinuity, 93 of a function, 93 Kestelman, Hyman, 165, 182 Kronecker delta, 8;1, 385 (Ex. 13.6) Ls-norm, 293, 295 Lagrange, Joseph Louis (1736-1813), 27, 30, 380 Lagrange, identity, 27 (Ex. 1.23), 30 (Ex. 1.48), 380 multipliers, 380 Landau, Edmund (1877-1938), 31 Laplace, Pierre Simon (1749-1827), 326, 342,468 Laplace transform, 326, 342, 468 Laurent, Pierre Alphonse (1813-1854), 455 Laurent expansion, 455 Least upper bound, 9 Lebesgue, Henri (1875-1941), 141, 171, 260, 270, 273, 290, 292, 312, 391, 405 bounded convergence theorem, 273 criterion for Riemann integrability, 171, 391 dominated-convergence theorem, 270 integral of complex functions, 292 integral of real functions, 260, 407 measure, 290, 408 Legendre, Adrien-Marie (1752-1833), 336 Legendre polynomials, 336 (Ex. 11.7) Leibniz, Gottfried Wilhelm (1646-1716), 121 Leibniz' formula, 121 (Ex. 5.6) Length of a path, 134 Levi, Beppo (1875-1961), 265, 267, 268, 407 Levi monotone convergence theorem, for sequences, 267 for series, 268 for step functions, 265 Limit, inferior, 184 in a metric space, 71 superior, 184 Limit function, 218 Limit theorem of Abel, 245 Lindelof, Ernst - (1870-1946), 56 Lindelof covering theorem, 57 Linear function, 345 Linear space, 48 of functions, 137 (Ex. 6.4) Line segment in R", 88 Linearly dependent set of functions, 122 (Ex. 5.9) Liouville, Joseph (1809-1882), 451 Liouville's theorem, 451 Lipschitz, Rudolph (1831-1904),121,137, 312, 316 Lipschitz condition, 121 (Ex. 5.1), 137 (Ex. 6.2), 316 Littlewood, John Edensor (1885-312 Local extremum, 98 (Ex. 4.25) Local property, 79 Localization theorem, 318 Logarithm, 23 Lower bound, 8 Lower integral, 152 Lower limit, 184 Mapping, 35 Matrix, 350 product, 351 Maximum and minimum, 83, 375 Maximum-modulus principle, 453, 454 Mean convergence, 232 Mean-Value Theorem for derivatives, of real-valued functions, 110 of vector-valued functions, 355 Mean-Value Theorem for integrals, multiple integrals, 401 Riemann integrals, 160, 165 Riemann-Stieltjes integrals, 160 Measurable function, 279, 407 Measurable set, 290, 408 Measure, of a set, 290, 408 zero, 169, 290, 391, 405 Mertens, Franz (1840-1927), 204 Mertens' theorem, 204 Metric, 60 Metric space, 60 Minimum-modulus principle, 454 Minkowski, Hermann (1864-1909), 27 Minkowski's inequality, 27 (Ex. 1.25) MSbius, Augustus Ferdinand (1790-1868), 471 Mobius transformation, 471 Modulus of a complex number, 18 Monotonic function, 94 490 Monotonic sequence, 185 Multiple integral, 389, 407 Multiplicative function, 216 (Ex. 8.45) Index Neighborhood, 49 of infinity, 15, 25 Niven, Ivan M. (1915-), 180 (Ex. 7.33) n-measure, 408 Nonempty set, I Nonmeasurable function, 304 (Ex. 10.37) Nonmeasurable set, 304 (Ex. 10.36) Nonnegative, 3 Norm, of a function, 102 (Ex. 4.66) of a partition, 141 of a vector, 48 0, o, oh notation, 192 One-to-one function, 36 Onto, 35 Operator, 327 Open, covering, 56, 63 interval in R, 4 interval in R", 50 mapping, 370, 454 mapping theorem, 371, 454 set in a metric space, 62 set in R", 49 Order, of pole, 458 of zero, 452 Ordered n-tuple, 47 Ordered pair, 33 Order-preserving function, 38 Ordinate set, 403 (Ex. 14.11) Orientation of a circuit, 447 Orthogonal system of functions, 306 Orthonormal set of functions, 306 Oscillation of a function, 98 (Ex. 4.24), 170 Outer Jordan content, 396 Parallelogram law, 17 Parseval, Mark-Antoine (circa 1776-1836), 309, 474 Parseval's formula, 309, 474 (Ex. 16.12) Partial derivative, 115 of higher order, 116 Partial sum, 185 Partial summation formula, 194 Partition of an interval, 128, 141 Path, 88, 133, 435 Peano, Giuseppe (1858-1932), 224 Perfect set, 67 (Ex. 3.25) Periodic function, 224, 317 Pi, a, irrationality of, 180 (Ex. 7.33) Piecewise-smooth path, 435 Point, in a metric space, 60 in R", 47 Pointwise convergence, 218 Poisson, Sim6on Denis (1781-1840), 332, 473 Poisson, integral formula, 473 (Ex. 16.5) summation formula, 332 Polar coordinates, 20, 418 Polygonal curve, 89 Polygonally connected set, 89 Polynomial, 80 in two variables, 462 zeros of, 451, 475 (Ex. 16.15) Power series, 234 Powers of complex numbers, 21, 23 Prime number, 5 Prime-number theorem, 175 (Ex. 7.10) Principal part, 456 Projection, 394 Quadratic form, 378 Quadric surface, 383 Quotient, of complex numbers, 16 of real numbers, 2 Radius of convergence, 234 Range of a function, 34 Ratio test, 193 Rational function, 81, 462 Rational number, 6 Real number, 1 Real part, 15 Rearrangement of series, 196 Reciprocity law for Gauss sums, 464 Rectifiable path, 134 Reflexive relation, 43 (Ex. 2.2) Region, 89 Relation, 34 Removable discontinuity, 93 Removable singularity, 458 Residue, 459 Residue theorem, 460 Restriction of a function, 35 Riemann, Georg Friedrich Bernard (1826-1866), 17, 142, 153, 192, 209, 312, 313, 318, 389, 475 condition, 153 Index integral, 142, 389 localization theorem, 318 sphere, 17 theorem on singularities, 475 (Ex. 16.22) zeta function, 192, 209 Riemann-Lebesgue lemma, 313 Riesz, Frigyes (1880-1956), 252, 297, 305, 311 Riesz-Fischer theorem, 297, 311 Righthand derivative, 108 Righthand limit, 93 Rolle, Michel (1652-1719), 110, Rolle's theorem, 110 Root test, 193 Roots of complex numbers, 22 Rouch6, Eugene (1832-1910), 475 Rouch6's theorem, 475 (Ex. 16.14) Saddle point, 377 Scalar, 48 Schmidt, Erhard (1876-1959), 335 Schoenberg, Isaac J., (1903-), 224 Schwarz, Hermann Amandus (1843-1921), 14, 27, 30, 122, 177, 294 Schwarzian derivative, 122 (Ex. 5.7) Schwarz's lemma, 474 (Ex. 16.13) Second-derivative test for extrema, 378 Second Mean-Value Theorem for Riemann integrals, 165 Semimetric space, 295 Separable metric space, 68 (Ex. 3.33) Sequence, definition of, 37 Set algebra, 40 Similar (equinumerous) sets, 38 Simple curve, 435 Simply connected region, 443 Singularity, 458 essential, 459 pole, 458 removable, 458 Slobbovian integral, 249 (Ex. 9.17) Space-filling curve, 224 Spherical coordinates, 419 Square-integrable functions, 294 Stationary point, 377 Step function, 148, 406 Stereographic projection, 17 Stieltjes, Thomas Jan (1856-1894), 140 Stieltjes integral, 140 Stone, Marshall H. (1903-), 252 Strictly increasing function, 94 491 Subsequence, 38 Subset, 1, 32 Substitution theorem for power series, 238 Sup norm, 102 (Ex. 4.66) Supremum, 9 Symmetric quadratic form, 378 Symmetric relation, 43 (Ex. 2.2) Tannery, Jules (1848-1910), 299 Tannery's theorem, 299 (Ex. 10.7) Tauber, Alfred (1866-circa 1947), 246 Tauberian theorem, 246, 251 (Ex. 9.37) Taylor, Brook (1685-1731), 113, 241, 361, 449 Taylor's formula with remainder, 113 for functions of several variables, 361 Taylor's series, 241, 449 Telescoping series, 186 Theta function, 334 Tonelli, Leonida (1885-1946), 415 Tonelli-Hobson test, 415 Topological, mapping, 84 property, 84 Topology, point set, 47 Total variation, 129, 178 (Ex. 7.20) Transformation, 35, 417 Transitive relation, 43 (Ex. 2.2) Triangle inequality, 13, 19, 48, 60, 294 Trigonometric series, 312 Two-valued function, 86 Uncountable set, 39 Uniform bound, 221 Uniform continuity, 90 Uniform convergence, of sequences, 221 of series, 223 Uniformly bounded sequence, 201 Union of sets, 41 Unique factorization theorem, 6 Unit coordinate vectors, 49 Upper bound, 8 Upper half-plane, 463 Upper function, 256, 406 Upper integral, 152 Upper limit, 184 Vall6e-Poussin, C. J. de la (1866-1962), 312 Value of a function, 34 492 Variation, bounded, 128 total, 129 Vector, 47 Vector-valued function, 77 Volume, 388, 397 Index Well-ordering principle, 25 (Ex. 1.6) Weierstrass, Karl (1815-1897), 8, 54, 223, 322, 475 approximation theorem, 322 M-test, 223 Winding number, 445 Wronski, J. M. H. (1778-1853), 122 Wronskian, 122 (Ex. 5.9) Young, William Henry (1863-1942), 252, 312 Zero measure, 169, 391, 405 Zero of an analytic function, 452 Zero vector, 48 Zeta function, Euler product for, 209 integral representation, 278 series representation, 192
15656
http://physics.bu.edu/~duffy/semester2/c11_RC.html
| | | --- | | J2S._Canvas2D (circuit.CircuitApplet) "CircuitApp"[x] | J2S._Canvas2D (dataGraph.DataGraph) "DataGraphApp"[x] | An RC Circuit: Charging Circuits with resistors and batteries have time-independent solutions: the current doesn't change as time goes by. Adding one or more capacitors changes this. The solution is then time-dependent: the current is a function of time. Consider a series RC circuit with a battery, resistor, and capacitor in series. The capacitor is initially uncharged, but starts to charge when the switch is closed. Initially the potential difference across the resistor is the battery emf, but that steadily drops (as does the current) as the potential difference across the capacitor increases. Applying Kirchoff's loop rule: ε - IR - Q/C = 0 As Q increases I decreases, but Q changes because there is a current I. As the current decreases Q changes more slowly. I = dQ/dt, so the equation can be written: ε - R (dQ/dt) - Q/C = 0 This is a differential equation that can be solved for Q as a function of time. The solution (derived in the text) is: Q(t) = Qo [ 1 - e-t/τ ] where Qo = C ε and the time constant τ = RC. Differentiating this expression to get the current as a function of time gives: I(t) = (Qo/RC) e-t/τ = Io e-t/τ where Io = ε/R is the maximum current possible in the circuit. The time constant τ = RC determines how quickly the capacitor charges. If RC is small the capacitor charges quickly; if RC is large the capacitor charges more slowly. | time | current | --- | | 0 | Io | | 1τ | Io/e = 0.368 Io | | 2τ | Io/e2 = 0.135 Io | | 3τ | Io/e3 = 0.050 Io | An RC Circuit: Discharging What happens if the capacitor is now fully charged and is then discharged through the resistor? Now the potential difference across the resistor is the capacitor voltage, but that decreases (as does the current) as time goes by. Applying Kirchoff's loop rule: -IR - Q/C = 0 I = dQ/dt, so the equation can be written: R (dQ/dt) = -Q/C This is a differential equation that can be solved for Q as a function of time. The solution is: Q(t) = Qo e-t/τ where Qo is the initial charge on the capacitor and the time constant t = RC. Differentiating this expression to get the current as a function of time gives: I(t) = -(Qo/RC) e-t/τ = -Io e-t/τ where Io = Qo/RC Note that, except for the minus sign, this is the same expression for current we had when the capacitor was charging. The minus sign simply indicates that the charge flows in the opposite direction. Here the time constant τ = RC determines how quickly the capacitor discharges. If RC is small the capacitor discharges quickly; if RC is large the capacitor discharges more slowly. DataGraphApp ready
15657
https://www.gabormelli.com/RKB/Monotonic_Function
Monotonic Function From GM-RKB Jump to navigation Jump to search A Monotonic Function is a function that has an ordered function output. AKA: Monotone Function. Context: It can range from being a Monotone Increasing Function to being a Monotone Nondecreasing Function to being a Monotone Increasing Function to being a Monotone Increasing Function. Example(s): a Cumulative Distribution Function. a Monotonic Boolean Function. … Counter-Example(s): a Non-Monotonic Function. See: Monotonic, Modular Function, Partially Ordered Set. References 2017 (Wikipedia, 2017) ⇒ Retrieved:2017-8-14. In mathematics, a monotonic function (or monotone function) is a function between ordered sets that preserves or reverses the given order. This concept first arose in calculus, and was later generalized to the more abstract setting of order theory. 2010 QUOTE: 2009 (Wikipedia, 2009) ⇒ … In calculus, a function [math]\displaystyle{ f }[/math] defined on a subset of the real numbers with real values is called monotonic (also monotonically increasing or non-decreasing), if for all [math]\displaystyle{ x }[/math] and [math]\displaystyle{ y }[/math] such that [math]\displaystyle{ x }[/math] ≤ [math]\displaystyle{ y }[/math] one has [math]\displaystyle{ f }/math ≤ [math]\displaystyle{ f }/math, so [math]\displaystyle{ f }[/math] preserves the order (see Figure 1). Likewise, a function is called monotonically decreasing (non-increasing) if, whenever [math]\displaystyle{ x }[/math] ≤ [math]\displaystyle{ y }[/math], then [math]\displaystyle{ f }/math ≥ [math]\displaystyle{ f }/math, so it reverses the order (see Figure 2). The following properties are true for a monotonic function [math]\displaystyle{ f }[/math] : R → R: f has limits from the right and from the left at every point of its domain; f has a limit at infinity (either ∞ or −∞) of either a real number, ∞, or −∞. f can only have jump discontinuities; f can only have countably many discontinuities in its domain. These properties are the reason why monotonic functions are useful in technical work in analysis. Two facts about these functions are: if [math]\displaystyle{ f }[/math] is a monotonic function defined on an interval $I$, then [math]\displaystyle{ f }[/math] is differentiable almost everywhere on $I$, i.e. the set of numbers [math]\displaystyle{ x }[/math] in $I$ such that [math]\displaystyle{ f }[/math] is not differentiable in [math]\displaystyle{ x }[/math] has Lebesgue measure zero. if [math]\displaystyle{ f }[/math] is a monotonic function defined on an interval [a, b], then [math]\displaystyle{ f }[/math] is Riemann integrable. An important application of monotonic functions is in probability theory. If [math]\displaystyle{ X }[/math] is a random variable, its cumulative distribution function : FX(x) = Prob(X ≤ x) is a monotonically increasing function. A function is unimodal if it is monotonically increasing up to some point (the mode) and then monotonically decreasing. 2006 QUOTE: Monotonic function. A function that either never decreases or never increases. A non-decreasing, or isotonic, function satisfies: f(x') >= f(x) whenever x' >= x (it is strictly increasing if f(x') > f(x) for x' not= x). A non-increasing, or anatonic, function satisfies: f(x') <= f(x) whenever x' >= x (it is strictly decreasing if f(x') < f(x) for x' not= x). This extends to a vector function, where range(f) is in Rn: (f(x)-f(y))t (x-y) >= 0. 2002 Retrieved from " Category: Concept Navigation menu
15658
https://oercommons.org/curated-collections/390
Skip to Content Skip to Search Results Updating search results... CK-12 STEM Flexbooks Show More Show Less 64 affiliated resources Filter Resources Education Standards Subject Area Education Level Material Type License Types Content Source Primary User Media Format Educational Use Language Providers Open filters Close filters Conditional Remix & Share Permitted CC BY-NC-SA Nanoleap: Student Journal Rating 0.0 stars This is a Physical Science journal that supports student investgation of forces … This is a Physical Science journal that supports student investgation of forces of change and nanoscale. More Less Subject: : Applied Science : Engineering Material Type: : Activity/Lab : Homework/Assignment : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : MCREL : Morrow, Christine : Ristvey, John Date Added: : 10/13/2009 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Modeling and Simulation for High School Teachers: Principles, Problems, and Lesson Plans Rating 0.0 stars A collaboration between the National Aeronautics and Space Administration (NASA) and the … A collaboration between the National Aeronautics and Space Administration (NASA) and the CK-12 Foundation, this book provides high school mathematics and physics teachers with an introduction to the main principles of modeling and simulation used in science and engineering. An appendix of lesson plans is included. More Less Subject: : Applied Science : Engineering : Physical Science : Physics Material Type: : Lesson Plan : Teaching/Learning Strategy : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Date Added: : 10/24/2012 MoreLess Conditional Remix & Share Permitted CC BY-NC The Pythagorean Formula Mini-Flexbook Rating 0.0 stars This Flexbook is community contributed through ck12.org. It covers three lessons on … This Flexbook is community contributed through ck12.org. It covers three lessons on the Pythagorean Theorem. 1) Introduction and Determining if the Triangle is a Right Triangle, 2) Finding the Hypotenuse, and 3) Finding a leg. It includes step by step instructions, application problems, and answers (at the end of each lesson). Ck12.org material is downloadable, editable, and accessible offline and online. More Less Subject: : Algebra : Geometry : Mathematics Material Type: : Lecture Notes : Lesson : Reading Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Shana Friend Date Added: : 05/25/2018 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Human Biology - Digestion and Nutrition (Student's Edition) Rating 0.0 stars The Digestion and Nutrition Student Edition book is one of ten volumes … The Digestion and Nutrition Student Edition book is one of ten volumes making up the Human Biology curriculum, an interdisciplinary and inquiry-based approach to the study of life science. More Less Subject: : Biology : Life Science : Nutrition Material Type: : Activity/Lab : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Program in Human Biology, Stanford University Date Added: : 02/03/2011 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA People's Physics Book, Version 3 (with Videos) Rating 0.0 stars The People's Physics Book v3 is intended to be used as one … The People's Physics Book v3 is intended to be used as one small part of a multifaceted strategy to teach physics conceptually and mathematically More Less Subject: : Physical Science : Physics Material Type: : Activity/Lab : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Dann, James Date Added: : 08/09/2011 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Chemistry (Teacher's Edition) Rating 0.0 stars A work in progress, CK-12 Chemistry Teacher's Edition supports its Chemistry book … A work in progress, CK-12 Chemistry Teacher's Edition supports its Chemistry book covering: Matter; Atomic Structure; The Elements; Stoichiometry; Chemical Kinetics; Physical States of Matter; Thermodynamics; Nuclear and Organic Chemistry. More Less Subject: : Chemistry : Physical Science Material Type: : Activity/Lab : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Parsons, Richard : Robinson, Shonna Date Added: : 02/12/2010 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Calculus (Student's Edition) Rating 0.0 stars CK-12 Foundation's Single Variable Calculus FlexBook introduces high school students to the … CK-12 Foundation's Single Variable Calculus FlexBook introduces high school students to the topics covered in the Calculus AB course. Topics include: Limits, Derivatives, and Integration. More Less Subject: : Calculus : Functions : Mathematics Material Type: : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Date Added: : 10/06/2009 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Trigonometry - Second Edition (Student's Edition) Rating 0.0 stars CK-12 Foundation's Trigonometry FlexBook is an introduction to trigonometry for the high … CK-12 Foundation's Trigonometry FlexBook is an introduction to trigonometry for the high school student. Topics include: Trigonometric Identities & Equations, Circular Functions, and Polar Equations & Complex Numbers. More Less Subject: : Mathematics : Trigonometry Material Type: : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Fortgang, Art : Hayes, Andrea : Jordan, Lori : Landers, Mara : Meery, Brenda : Ottman, Larry Date Added: : 08/12/2010 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Geometría - Edición Espanola (Student's Edition) Rating 0.0 stars CK-12 Foundation's Geometry FlexBook is a clear presentation of the essentials of … CK-12 Foundation's Geometry FlexBook is a clear presentation of the essentials of geometry for the high school student. Topics include: Proof, Congruent Triangles, Quadrilaterals, Similarity, Perimeter & Area, Volume, and Transformations. More Less Subject: : Geometry : Mathematics Material Type: : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Date Added: : 11/23/2010 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Geometry - Basic (Student's Edition) Rating 0.0 stars CK-12's Basic Geometry FlexBook is designed to present students with geometric principles … CK-12's Basic Geometry FlexBook is designed to present students with geometric principles in a simpler, more graphics-oriented course. Students will explore geometry at a slower pace with an emphasis placed on visual aids and approachability. More Less Subject: : Geometry : Mathematics Material Type: : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Cifarelli, Victor : Farbizio, Annamaria (Editor) : Gloag, Andrew : Greenberg, Dan : Jordan, Lori : Sconyers, Jim : Zahner, Bill Date Added: : 10/11/2010 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Probability & Statistics - Advanced Second Edition (Student's Edition) Rating 0.0 stars CK-12 Advanced Probability and Statistics introduces students to basic topics in statistics … CK-12 Advanced Probability and Statistics introduces students to basic topics in statistics and probability but finishes with the rigorous topics an advanced placement course requires. Includes visualizations of data, introduction to probability, discrete probability distribution, normal distribution, planning and conducting a study, sampling distributions, hypothesis testing, regression and correlation, Chi-Square, analysis of variance, and non-parametric statistics. More Less Subject: : Mathematics : Statistics and Probability Material Type: : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Almukkahal, Raja : DeLancey, Danielle : Meery, Brenda : Ottman, Larry Date Added: : 10/01/2010 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Algebra Explorations, Pre-K through Grade 7 Rating 0.0 stars CK-12 Algebra Explorations is a hands-on series of activities that guides students … CK-12 Algebra Explorations is a hands-on series of activities that guides students from Pre-K to Grade 7 through algebraic concepts. More Less Subject: : Algebra : Mathematics Material Type: : Activity/Lab : Teaching/Learning Strategy Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Carol ­ Findell, Mary ­Cavanagh, Carole ­Greenes Date Added: : 02/23/2012 MoreLess Conditional Remix & Share Permitted CC BY-NC Trigonometry Concepts Rating 0.0 stars CK-12 Foundation's Trigonometry FlexBook is an introduction to trigonometry for the high … CK-12 Foundation's Trigonometry FlexBook is an introduction to trigonometry for the high school student. It includes chapters on graphs of trigonometric functions, trigonometric identities, inverse trigonometric functions, triangles and vectors, and the polar system. More Less Subject: : Mathematics : Trigonometry Material Type: : Reading : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Date Added: : 05/25/2014 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Geometry - Second Edition (Student's Edition) Rating 0.0 stars CK-12's Geometry - Second Edition is a clear presentation of the essentials … CK-12's Geometry - Second Edition is a clear presentation of the essentials of geometry for the high school student. Topics include: Proofs, Triangles, Quadrilaterals, Similarity, Perimeter & Area, Volume, and Transformations. More Less Subject: : Geometry : Mathematics Material Type: : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Dirga, Kathryn : Jordan, Lori Date Added: : 11/16/2010 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Calculus - TI Activities (Teacher's Edition) Rating 0.0 stars CK-12's Texas Instruments Calculus Teacher's Edition is a useful companion to a … CK-12's Texas Instruments Calculus Teacher's Edition is a useful companion to a Calculus course, offering extra assignments and opportunities for students to understand course material through their graphing calculator. More Less Subject: : Calculus : Functions : Mathematics Material Type: : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Jordan, Lori Date Added: : 12/17/2010 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Nanotechnology - Nano Sense (Student's Edition) Rating 0.0 stars These NanoSense Student Materials have been designed to help high school students … These NanoSense Student Materials have been designed to help high school students understand science concepts that account for nanoscale phenomena, and the principles, applications, and implications of nanoscale science. More Less Subject: : Applied Science : Engineering Material Type: : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Schank, Patricia : Stanford, Tina Date Added: : 10/09/2009 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA CK-12 Physical Science Concepts for Middle School Rating 0.0 stars CK-12 Physical Science Concepts covers the study of physical science for middle … CK-12 Physical Science Concepts covers the study of physical science for middle school students. The 5 chapters provide an introduction to physical science, matter, states of matter, chemical interactions and bonds, chemical reactions, motion and forces, and the types and characteristics of energy. More Less Subject: : Chemistry : Physical Science Material Type: : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Jean Brainard, Ph.D. Date Added: : 11/01/2012 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Organic Chemistry - Applications From Vitamins to Baked Goods Rating 0.0 stars This stand-alone module intends to provide some motivation for studying organic chemistry. … This stand-alone module intends to provide some motivation for studying organic chemistry. The topics touch briefly on some basic organic chemistry topics and focus on various organic compounds that readers would encounter in everyday life. More Less Subject: : Chemistry : Physical Science Material Type: : Activity/Lab : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Fan, Lizhi : Gallagher, Kieran : Sheng, Joy Date Added: : 11/08/2010 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Human Biology - Your Changing Body (Student's Edition) Rating 0.0 stars The Your Changing Body Student Edition book is one of ten volumes … The Your Changing Body Student Edition book is one of ten volumes making up the Human Biology curriculum, an interdisciplinary and inquiry-based approach to the study of life science. More Less Subject: : Biology : Life Science Material Type: : Activity/Lab : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Program in Human Biology, Stanford University Date Added: : 02/04/2011 MoreLess Conditional Remix & Share Permitted CC BY-NC-SA Life Science for Middle School (Teacher's Edition) Rating 0.0 stars CK-12 Life Science Honors For Middle School covers seven units: Understanding Living … CK-12 Life Science Honors For Middle School covers seven units: Understanding Living Things; Cells: The Building Blocks of Life; Genetics and Evolution; Prokaryotes, Protists, Fungi, and Plants; The Animal Kingdom; The Human Body; and Ecology. More Less Subject: : Biology : Life Science Material Type: : Activity/Lab : Textbook Provider: : CK-12 Foundation Provider Set: : CK-12 FlexBook Author: : Wilkin, Douglas (Editor) Date Added: : 04/21/2010 MoreLess ` No restrictions on your remixing, redistributing, or making derivative works. Give credit to the author, as required. Most restrictive license type. Prohibits most uses, sharing, and any changes. Copyrighted materials, available under Fair Use and the TEACH Act for US-based educators, or other custom arrangements. Go to the resource provider to see their individual restrictions.
15659
https://en.wikipedia.org/wiki/Rate_of_heat_flow
Rate of heat flow - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 See also 2 References Rate of heat flow [x] 17 languages العربية Български Boarisch Deutsch Eesti Español Esperanto فارسی Français Къарачай-малкъар Polski Português Русский Slovenščina Suomi Українська Tiếng Việt Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Thermodynamic property The rate of heat flow is the amount of heat that is transferred per unit of time in some material, usually measured in watts (joules per second). Heat is the flow of thermal energy driven by thermal non-equilibrium, so the term 'heat flow' is a redundancy (i.e. a pleonasm). Heat must not be confused with stored thermal energy, and moving a hot object from one place to another must not be called heat transfer. However, it is common to say ‘heat flow’ to mean ‘heat content’. The equation of heat flow is given by Fourier's law of heat conduction. Rate of heat flow = - (heat transfer coefficient) (area of the body) (variation of the temperature) / (length of the material) The formula for the rate of heat flow is: Q Δ t=−k A Δ T Δ x{\displaystyle {\frac {Q}{\Delta t}}=-kA{\frac {\Delta T}{\Delta x}}} where Q{\displaystyle Q} is the net heat (energy) transfer, Δ t{\displaystyle \Delta t} is the time taken, Δ T{\displaystyle \Delta T} is the difference in temperature between the cold and hot sides, Δ x{\displaystyle \Delta x} is the thickness of the material conducting heat (distance between hot and cold sides), k{\displaystyle k} is the thermal conductivity of the material conducting heat, and A{\displaystyle A} is the surface area of the surface emitting heat. If a piece of material whose cross-sectional area is A{\displaystyle A} and thickness is Δ x{\displaystyle \Delta x} with a temperature difference Δ T{\displaystyle \Delta T} between its faces is observed, heat flows between the two faces in a direction perpendicular to the faces. The time rate of heat flow, Q Δ t{\displaystyle {\frac {Q}{\Delta t}}}, for small Q{\displaystyle Q} and small Δ t{\displaystyle \Delta t}, is proportional to A×Δ T Δ x{\displaystyle A\times {\frac {\Delta T}{\Delta x}}}. In the limit of infinitesimal thickness Δ x{\displaystyle \Delta x}, with temperature difference Δ T{\displaystyle \Delta T}, this becomes H=−k A(Δ T Δ x){\displaystyle H=-kA({\frac {\Delta T}{\Delta x}})}, where H=(Q Δ t){\displaystyle H=({\frac {Q}{\Delta t}})} is the time rate of heat flow through the area A{\displaystyle A}, Δ T Δ x{\displaystyle {\frac {\Delta T}{\Delta x}}} is the temperature gradient across the material, and k{\displaystyle k}, the proportionality constant, is the thermal conductivity of the material. People often use k{\displaystyle k}, λ{\displaystyle \lambda }, or the Greek letter κ{\displaystyle \kappa } to represent this constant.[citation needed] The minus sign is there because the rate of heat flow is always negative—heat flows from the side at higher temperature to the one at lower temperature, not the other way around. See also [edit] energy portal Heat transfer coefficient Heat transfer Thermal conduction Thermal conductivity Heat flux Watt Flux References [edit] ^"Heat and mass transfer"(PDF). Archived from the original(PDF) on 2012-12-24. ^"52.09 -- Heat conduction in various metal rods". web.physics.ucsb.edu. Retrieved 2019-05-07. ^"Unit Operations in Food Processing - R. L. Earle". nzifst.org.nz. Retrieved 2019-05-07. | Authority control databases | GND | This thermodynamics-related article is a stub. You can help Wikipedia by expanding it. v t e Retrieved from " Categories: Thermodynamic properties Temporal rates Thermodynamics stubs Hidden categories: Articles with short description Short description is different from Wikidata All articles with unsourced statements Articles with unsourced statements from April 2023 All stub articles This page was last edited on 16 February 2025, at 18:23(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Rate of heat flow 17 languagesAdd topic
15660
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/01%3A_Basic_Concepts-_Atoms/1.09%3A_The_Aufbau_Principle/1.9B%3A_Valence_and_Core_Electrons
Skip to main content 1.9B: Valence and Core Electrons Last updated : Jan 15, 2023 Save as PDF 1.9A: Ground State Electronic Configurations 1.9C: Diagrammatic Representations of Electronic Configurations Page ID : 2606 ( \newcommand{\kernel}{\mathrm{null}\,}) The electrons of an atom are typically divided into two categories: valence and core electrons. Valence electrons occupy the outermost shell or highest energy level of an atom while core electrons are those occupying the innermost shell or lowest energy levels. This difference greatly influences the role of the two types of electrons in a chemical reaction. Generally, valence electrons can participate in the formation of chemical bonding, but core electrons cannot. While core electrons are not involved in bonding, they influence the chemical reactivity of an atom. The electron configuration of a oxygen atom is O:1s22s22p4(1.9B.1) which may be shorted O:[He]2s22p4(1.9B.2) where the [He] stands for the configuration of helium (1s2). Similarly, the configuration of calcium with 20 electrons can be written Ca:[Ar]4s2(1.9B.3) where the [Ar] stands for the configuration of argon (1s22s22p63s23p6). Electronic configurations that are the same as noble gases are very stable since they have a full octet (except helium with a full 1s orbital). The 1s electrons in oxygen do not participate in bonding (i.e., chemistry) and are called core electrons. The valence electrons (i.e., the 2s22p4 part) are valence electrons, which do participate in the making and breaking of bonds. Similarly, in calcium (Equation 1.9B.3), the electrons in the argon-like closed shell are the core electrons and the the two electrons in the 4s orbital are valence electrons. Example 1.9B.1: Cobalt What are the core and valence electrons in cobalt? Solution Start by writing the electron configuration of cobalt with 27 electrons: 1s22s22p63s23p64s23d7 However, argon has the electronic structure 1s22s22p23s23p6, so we can rewrite the configuration as [Ar]4s23d7 The two electrons in the 4s orbital and the seven electrons in the 3d are the valence electrons: all others are core electrons. The periodicity of valance electrons can be seen in the Periodic Table. Basically, the periodicity is only applied to the main group elements, while in transition metals, rules are complex. The core electrons remain the same in the increase of group numbers in the main group elements. On the other hand, the valance electrons increase by one from left to right of a main period, and remain the same down the column of a main group. This evolution gives periodical change in property of a period, and similar chemical property of a group, which is called periodical trend. The number of valence electrons in a main period is the same as its group number. The table below shows this rule clearly. Under construction Figure 1: 1A + 2A are metals. 3A to 8A are non-metals. However, this periodicity cannot be applied to the transition group, which is more complicated than that of the main group. Although the outermost electrons can be easily determined, the apparent valence electrons considered in chemical reactivity are complex and fluctuated. Electrons going into d sublevel can play either a role of valence electrons or shielding electrons. So there is not always a certain number of apparent valence electrons. The number of apparent valence electrons for the first transition metal period is shown in the table below. Under construction Figure 2: Valence electrons for transition metals. Relationship with Chemical Reactivity The chemical reactivity of an atom is mainly determined by valence electrons. Atoms which have a complete shell of valence electrons tend to be chemically inert. Atoms with one or two valence electrons are highly reactive. This phenomenon can be explained by Hund's rule, which states that orbitals that are empty, half-full, or full are more stable than those that are not. For example, Ne is chemically inert because it has two valence electrons that fill its outermost shell which makes it stable compared to atoms such as Al, which has three valence electrons, but its valence electrons does not fill its outermost shell. Although core electrons do not take part in chemical bonding, they play a role in determining the chemical reactivity of an atom. This influence is generally due to the effect it has on valence electrons. The effect can be observed from the gradual change of chemical reactivity in a group. As you go down a group, more shells are occupied by electrons, which increases the size of the atom. The more core electron shells an atom has, the larger the size of the atom, and the farther the valence electrons are from the nucleus, thus the valence electrons will experience less effective nuclear charge and will be easily lost. For example, Na and K can both react with water, but K has a more radical reaction because it has more shells of core electrons which makes the valence electron in its outermost orbital much easier to lose than the valence electron of Na. References Miessler, Gary L., and Donald A. Tarr. Inorganic Chemistry. Upper Saddle River, NJ: Pearson Prentice Hall, 2010. Print. Brown, Ian David. The Chemical Bond in Inorganic Chemistry the Bond Valence Model. Oxford: Oxford UP, 2006. Print. 1.9A: Ground State Electronic Configurations 1.9C: Diagrammatic Representations of Electronic Configurations
15661
https://www.youtube.com/watch?v=kK_pGcJZavQ
How to determine if two vectors are parallel, orthogonal or neither Brian McLogan 1600000 subscribers 1089 likes Description 205430 views Posted: 6 Feb 2016 Learn how to determine if two vectors are orthogonal, parallel or neither. You can setermine whether two vectors are parallel, orthogonal, or neither uxsing the dot/cross product or using the slope formula. Recall that two lines are parallel if they both have the same slope and are perpemdicular/othorgonal is the slope of one of the lines is the negative of the reciprocal of the slope of the other line. Also, if the scalar/dot product of two vectors gives zero then the two vectors are perpendicular/orthogonal but if the magnitude of the cross/vector product of two vectors is zero, then the two vectors are perpendicular. trigonometry#vectors vectors Timestamps: 0:00 Intro 0:45 Start of Problem Corrections: 1:00 Made a mistake, m is equal to - 5/4 47 comments Transcript: Intro so if you're going to do this basically what the question is asking is determine if these two vectors are parallel orthogonal or neither so the main important thing guys is I just want to figure out first thing I'm going to do is see if they're parallel so if I have two vectors and let's just look at a vector in component form if I have a vector in component form I can find the slope right which is the rise over the Run basically what I want to do is if here's my Vector V1 comma V2 the rise is V 2 and the run is V1 correct so therefore we could technically say the slope is V2 over V1 and if the slopes are exactly the same then the vectors are parallel Start of Problem correct so let's do V1 over V2 and I'll do U1 over U2 because I'll call those two different vectors so M = -12 15 and M = 5 over 4 well you can reduce this to -4 over 5 and this one to 5 4 is this the same problem 5-15 -124 five okay do these work I mean are these the same slope no so the next thing is so they're not the same slope the next thing if we do the dotproduct u. V equals z then they are orthogonal so remember the dot product u. V is equal to U1 V1 plus U2 V2 so therefore you can go ahead and show here um you could do 15 -4 plus -12 5 and therefore this gives you 15 -4 so therefore since it does not equal zero it is neither right I'm kind of jeez [Music]
15662
https://www.pdesas.org/ContentWeb/Content/Content/20968/Lesson
Multiplication Using Arrays - SAS Skip to Main Content You are impersonating . Stop Impersonating SAS Standards Search Standards View Standards Vertical Viewer English Language Development Standards (2020) Download Standards Download PSSA Anchors and Eligible Content Download Keystone Anchors Download PA Core Implementation Download PASA DLM Essential Elements Conceptual Vertical Alignment Viewer Assessments Assessment Center Project Based Assessment Keystone Exams Classroom Diagnostic Tools PSSA Analyzing Root Cause Assessment Literacy Early Learning: Pre-Kindergarten to Grade 3 PASA Online Assessment Resources Curriculum Framework Search Curriculum Framework Download Curriculum Framework Curriculum Maps PA Standards Instructional Frameworks: ELA PA Standards Instructional Frameworks: Math PA Standards Instructional Frameworks: Personal Finance Early Learning: Pre-Kindergarten to Grade 3 Instruction Educator Effectiveness Instructional Toolkits Teacher of the Year Pennsylvania Learns Early Learning: Pre-Kindergarten to Grade 3 PA Roadmap: Focus on Effective Instruction Pennsylvania High School Graduation Requirements Seal of Biliteracy Toolkit STEELS Hub Educator Professional Development Resource PDE English Learner (EL) Portal PA Disability Inclusive Curriculum Pilot Program Pennsylvania Inspired Leadership (PIL) Materials & Resources Search My Lessons Voluntary Model Curriculum (sample unit and lesson plans) Learning Progressions Pennsylvania Literacy Early Learning: Pre-Kindergarten to Grade 3 POWER Library Career Readiness TDA Toolkit Career Ready Toolkit Organ and Tissue Donation Awareness Toolkit Act 35 Civics Toolkit PBS Toolkit Content Collections CPR Toolkit Holocaust, Genocide, and Human Rights Toolkit Information and Media Literacy Toolkit September 11 Resource Toolkit Personal Finance Toolkit STEM & CS in PA USA's 250th Toolkit Safe & Supportive Schools Safe & Supportive Schools Menu Standards Assessments Curriculum Framework Instruction Materials & Resources Safe & Supportive Schools Quick Links 's SAS Tools Login Username Password Log In RegisterForgot Password? Profile ePortfolio My Assessments My Curriculum Maps My Lessons My Library Communities PD Center Workshop Evaluation Logout Quick Links Share Suggestion Subject Please include a subject for your suggestion. Suggestion Please enter information about your suggestion. Please check the "I'm not a robot" checkbox. Send Close Multiplication Using Arrays Lesson Multiplication Using Arrays Options Printer Friendly Version Email Grade Levels 4th Grade Related Academic StandardsCC.2.1.4.B.2 Use place value understanding and properties of operations to perform multi-digit arithmetic. CC.2.2.4.A.1 Represent and solve problems involving the four operations. Assessment AnchorsM04.A-T.2 Use place-value understanding and properties of operations to perform multi-digit arithmetic. M04.B-O.1 Use the four operations with whole numbers to solve problems. Eligible ContentM04.A-T.2.1.2 Multiply a whole number of up to four digits by a one-digit whole number and multiply 2 two-digit numbers. M04.B-O.1.1.1 Interpret a multiplication equation as a comparison. Represent verbal statements of multiplicative comparisons as multiplication equations. Example 1: Interpret 35 = 5 × 7 as a statement that 35 is 5 times as many as 7 and 7 times as many as 5. Example 2: Know that the statement 24 is 3 times as many as 8 can be represented by the equation 24 = 3 × 8 or 24 = 8 × 3. M04.B-O.1.1.2 Multiply or divide to solve word problems involving multiplicative comparison, distinguishing multiplicative comparison from additive comparison. Example: Know that 3 × 4 can be used to represent that Student A has 4 objects and Student B has 3 times as many objects not just 3 more objects. Big Ideas Concepts Competencies Objectives Students will find products of one-digit and two-digit factors using a dot array. They will: break apart arrays (modeling the distributive property of multiplication over addition) to compute a two-digit factor multiplied by a one-digit factor. find several ways to split an array to calculate the product of two factors. Essential Questions How is mathematics used to quantify, compare, represent, and model numbers? How can patterns be used to describe relationships in mathematical situations? Vocabulary Equivalence:The relationship between expressions which have an equal value. Multiple:The product of a given number and a whole number. Duration 45–60 minutes Prerequisite Skills Prerequisite Skills haven't been entered into the lesson plan. Materials Base-ten grid paper (M-4-2-2_Base-Ten Grid.docx) Array Pictures sheet (M-4-2-2_Array Pictures.docx) Array Practice worksheet and Array Practice KEY (M-4-2-2_Array Practice.docx and M-4-2-2_Array Practice KEY.docx) Related Unit and Lesson Plans Using Models and Algorithms to Solve Multiplication Problems Using Multiplication to Solve Word Problems Partial Products Related Materials & Resources The possible inclusion of commercial websites below is not an implied endorsement of their products, which are not free, and are not required for this lesson plan. Additional practice with multiplication using arrays: Formative Assessment View Use the Array Practice worksheet and Array Practice KEY (M-4-2-2_Array Practice.docx and M-4-2-2_Array Practice KEY.docx) to find the degree of student understanding. Use the think-pair-share activity in the Routine section of the Extension to help establish the level of student knowledge. Observe students solving multiplication problems using arrays to evaluate student comprehension of the materials. Suggested Instructional Supports View Scaffolding, Active Engagement, Modeling, Explicit Instruction W:Introduce students to arrays. Having students find the total number of objects in the arrays without giving any instruction will lead them to discover ways of breaking up the task. H:The visual nature of the arrays should hook students and help them to conceptualize these problems in several ways, including writing the corresponding number sentences. E:Discuss how to solve an array problem by decomposing it into rectangles based on place value (i.e., tens by tens, tens by ones, etc.). R:Have students practice solving arrays by working on problems in the problem set. E:To evaluate whether students are making the connection between arrays and number sentences, have students solve the same problems by drawing the arrays on grid paper, using decomposition, and labeling each part. T:The lesson may be adjusted as needed using strategies introduced in the Extension section. O:The goal of the lesson is for students to learn how to make an array and break it into smaller parts in order to multiply.In doing so, they create a pictorial model of distributing multiplication over addition and are introduced to the partial-products concept. At this stage, students are learning a variety of methods for solving multiplication problems. By discussing each strategy, students gain a better understanding of the multiplication process, especially as it applies to multidigit numbers. Instructional Procedures View “Today, we are going to talk about arrays. An array is defined as a rectangular arrangement of objects in equal rows or columns. Let’s look at some examples.” Show students the Array Pictures (M-4-2-2_Array Pictures.docx). Have them make observations about how the arrays are arranged. “How could you determine the number of objects shown in each array?” (repeated addition, multiplying rows by the number in each row, counting, grouping together, skip counting) “Believe it or not, we probably see arrays every day. For the next few minutes, you will find some arrays right here in our classroom (or hallway).” Give students a few minutes to walk around the classroom or hallway and find real-life arrays. Construct a class list of arrays they found. (Examples might be crayons in a box, chalk in a box, chairs in a row, monthly wall calendar, etc.) Provide additional contexts that facilitate thinking about arrays: There are three dozen eggs. How many eggs is three dozen? There are five 12-packs of soda. How many cans of soda are in five 12-packs? Ask students to draw a picture of the problem. Students can use pictures, dots, stars, or any shape to represent the objects in their pictures. “How many different number sentences can we write to represent this problem?”Here students should generate equivalent equations like those they practiced in Lesson 1. For example, a student might write (5 × 6) + (5 × 6) = 5 × 12. Point out that the sentence also can be written as (6 × 5) + (6 × 5) = 12 × 5 by the commutative property. (10 × 3) + (2 × 3) = 12 × 3. Look for opportunities to point out the properties in the students’ solutions. Present the following problem: Four girls saw a strip of unusual stickers for $0.45 at the school store. Each girl had a dime and two pennies left over from buying lunch. The girls combined their coins on the counter of the store. Did they have enough to buy the strip of stickers? Encourage students to explain how they would approach this problem. Try to get them to focus on the idea of putting like coins together. For instance, putting the 4 dimes together gives $0.40 (4 × 10), and putting all the pennies together gives 8 pennies (4 × 2). Encourage students to represent the grouping of the 10s and singles using pictures (four rows of 10 dots each and four rows of two dots each) displayed right next to the previous amount. Use the base-ten grid paper (M-4-2-2_Base-Ten Grid.docx) to color in the amount (4 × 10) + (4 × 2). Ask students to label the parts of the rectangular grid. Use “4” to label the width and “10 + 2” to label the length of the rectangle, as shown here. “Can someone explain what the first section represents?”(4 groups of 10, 4 girls each had a dime.)“And the second section?”(4 groups of 2, 4 girls each had 2 cents.) Probe with questions that encourage thinking about number relationships and properties (such as the distributive property). For example, “What if the four girls each had two dimes and two pennies? What would we have to change in our rectangular array and why? How is this answer related to $0.48?”One example might look like this: Notice that the context of the problem encourages students to think about 10s, decomposing the problem in a way that emphasizes place value. Present the following problem: Four children saw a used kite in the window of a thrift shop. Each child had a quarter and three pennies. The price marked on the kite was $1.15. Can the children combine their coins to buy the kite? Explain. This problem context encourages the children to combine all the quarters together to make $1.00 and combine the pennies together to make 12 cents. Use base-ten grid paper to outline the rectangular array that will represent the problem $0.28 × 4. (four rows of 28 squares) The base-ten grid paper points out the groups of 10 in 28 with the 8 added on. Use additional similar multiplication problems to reinforce understanding if necessary. Provide problem sets (like the following) for students to gain additional practice. Encourage them to use any strategy they have learned so far in the unit to solve these problems. Have students explain the relationship among the problems in each set. Have students draw rectangular arrays on the base-ten grid paper to match the problem sets above. Students should label the rectangular array, making the connection to the related problems. Students should be asked to explain one of the drawings or expressions to demonstrate an understanding of multiplication by decomposition using arrays. Remind students to focus on 10s in these problems. Extension: Routine:As a warm-up for the next day’s lesson, post a one-digit by two-digit multiplication problem on the board. Have students think about using an array to solve the problem and then draw the array and solve the problem. (They may use grid paper, draw freehand, draw lines of dots, or write Xs to make the array). Engage students in a think-pair-share to share their solutions with a partner and quickly discuss methodology. Then call on students to explain how they arrived at the solution. Expansion:Students who have mastered the concept quickly or easily can think about how to solve a two-digit by two-digit multiplication problem using arrays. Have them test their theory. Students could write word problems to solve and exchange their problems with a partner. This will provide additional practice and the added complexity of translating a story into symbols to reach a solution. Technology Connection: For students who have not yet mastered the concept, go to for additional practice using arrays for multiplication without the complexity of two-digit numbers. This Web site offers practice with multiplication arrays using one-digit by one-digit problems, so students can master the concept before moving on to two-digit multiplication. Related Instructional Videos Note: Video playback may not work on all devices. Instructional videos haven't been assigned to the lesson plan. Final 04/12/13 Copyright © 2025 Commonwealth of Pennsylvania HelpContact UsTerms of UseFAQ Please wait... Insert Template Cancel Insert Information Close
15663
https://education.ti.com/en-AU/Activity/activity-detail?id=1D2E6532323D4A14A50C65E56EB0A7F5
Classroom Activities: Dot Product - Texas Instruments - content Education Technology » Products TI-Nspire™ Family Graphing Calculators Scientific Calculators Computer Software TI-Innovator™ Hub TI-Nspire™ Navigator Data Collection Financial Calculators Teachers Australian Curriculum Nspired Senior Curriculum Inspirations IB® Curriculum Resources STEM Learning Centre TI Codes TI Recognition (rewards) Program Professional Development eNewsletter Subscription State QLD VIC Students On-line TI Technology Courses Student Resources TI Codes Support Downloads All Guidebooks All Software, OS and Apps TI-84 CE Apps and Updates TI-Nspire™ CX Updates Classroom Activities Where to Buy SiteAustralia and New Zealand Activities Software, OS Updates and Apps Guidebooks Activities Downloads Home Download Download All Files (ZIP) ##### Lesson Files TI-Nspire Activity Document Student Activity Teacher Notes and Answers Subject Area Author Texas Instruments Level Aust Senior Activity Time 30 Minutes Device TI-Nspire™ CX II TI-Nspire™ CX CAS Software TI-Nspire™ TI-Nspire™ CAS TI-Nspire Version 4.5 Report an Issue Dot Product Activity Overview Students manipulate two vectors as the corresponding dot product of the vectors is displayed graphically on the same page. By exploration students are able to explore the factors that affect the dot product, the conditions for which it is positive, zero and negative. Objectives This activity provides a dynamic and visual representation of the dot product where students can manipulate the two vectors to explore the conditions under which the dot product is positive, zero and negative in addition to understanding the formation of the dot product. Vocabulary Vector Scalar Magnitude Dot Product About the Lesson Students are provided with a number of functions that help illustrate the geometric transformations associated with the absolute value function. Particular focus is drawn to notation such as f(|x|) and |f(x)|. Students can see the original function and the resulting absolute value function making it easier to draw comparisons. © Copyright 1995-2025 Texas Instruments Incorporated. All rights reserved. TI-Nspire™ OSTI-84 OSWhere to BuyTrademarksPrivacy PolicyLink Policy Contact TI YouTube@ticalculatorsau TI websites use cookies to optimize site functionality and improve your experience. To find out more or to change your preferences, see our cookie policy page. Click Agree and Proceed to accept cookies and enter the site. Manage preferencesAgree and Proceed Control your cookie preferences You can control your preferences for how we use cookies to collect and use information while you're on TI websites by adjusting the status of these categories. | Category | Description | Allow | --- | Analytics and performance cookies | These cookies, including cookies from Google Analytics, allow us to recognize and count the number of visitors on TI sites and see how visitors navigate our sites. This helps us improve the way TI sites work (for example, by making it easier for you to find information on the site). | - [x] | | Advertising and marketing cookies | These cookies enable interest-based advertising on TI sites and third-party websites using information you make available to us when you interact with our sites. Interest-based ads are displayed to you based on cookies linked to your online activities, such as viewing products on our sites. We may also share this information with third parties for these purposes. These cookies help us tailor advertisements to better match your interests, manage the frequency with which you see an advertisement, and understand the effectiveness of our advertising. | - [x] | | Functional cookies | These cookies help identify who you are and store your activity and account information in order to deliver enhanced functionality, including a more personalized and relevant experience on our sites. If you do not allow these cookies, some or all site features and services may not function properly. If you do not allow these cookies, some or all of the site features and services may not function properly. | - [x] | | Social media cookies | These cookies allow identification of users and content connected to online social media, such as Facebook, Twitter and other social media platforms, and help TI improve its social media outreach. | - [x] | | Strictly necessary | These cookies are necessary for the operation of TI sites or to fulfill your requests (for example, to track what items you have placed into your cart on the TI.com, to access secure areas of the TI site, or to manage your configured cookie preferences). | Always On | Accept all Save and Close End User License Agreement Close Close Close Loading Close Notice Close Close
15664
https://medlineplus.gov/ency/article/000636.htm
Skip navigation An official website of the United States government Here’s how you know Official websites use .govA .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPSA lock ( ) or https:// means you’ve safely connected to the .gov website. Share sensitive information only on official, secure websites. National Library of Medicine The navigation menu has been collapsed. Health Topics Drugs & Supplements Genetics Medical Tests Medical Encyclopedia About MedlinePlus About MedlinePlus What's New Site Map Customer Support Health Topics Drugs & Supplements Genetics Medical Tests Medical Encyclopedia Español You Are Here: Home → Medical Encyclopedia → Donovanosis (granuloma inguinale) URL of this page: //medlineplus.gov/ency/article/000636.htm Donovanosis (granuloma inguinale) Donovanosis (granuloma inguinale) is a sexually transmitted disease (STD) that is rarely seen in the United States. Causes Donovanosis is caused by the bacterium Klebsiella granulomatis. The disease is commonly found in tropical and subtropical areas such as southeast India, Guyana, and New Guinea. There are about 100 cases reported per year in the United States. Most of these cases occur in people who have traveled to or are from places where the disease is common. The disease spreads mostly through vaginal or anal intercourse. Very rarely, it spreads during oral sex. Most infections occur in people ages 20 to 40. Symptoms Symptoms can occur 1 to 12 weeks after coming in contact with the bacteria. Symptoms may include: Sores in the anal area (in about half of cases) Small, beefy-red bumps on the genitals or around the anus The skin gradually wears away, and the bumps turn into raised, beefy-red, velvety nodules called granulation tissue. They are often painless, but they bleed easily if injured. The disease slowly spreads and destroys genital tissue. Tissue damage may spread to the groin. The genitals and the skin around them lose skin color. In its early stages, it may be hard to tell the difference between donovanosis and chancroid. Exams and Tests Tests may include: Culture of tissue sample (hard to do and not routinely available) Scrapings or biopsy of lesion Laboratory tests for diagnosing donovanosis are available only on a research basis. Treatment Antibiotics are used to treat donovanosis. These may include: Azithromycin Doxycycline Ciprofloxacin Erythromycin Trimethoprim-sulfamethoxazole To cure the condition, long-term treatment is needed. Most treatment courses require 3 weeks or until the sores have completely healed. A follow-up exam is needed because the disease can reappear after it seems to be cured. Outlook (Prognosis) Early treatment helps prevent tissue damage or scarring. Untreated, the disease leads to damage of the genital tissue. Possible Complications Health problems that may result from this disease include: Genital damage and scarring Loss of skin color in the genital area Permanent genital swelling due to scarring When to Contact a Medical Professional Contact your health care provider for an appointment if: You have had sexual contact with a person who is known to have donovanosis You develop symptoms of donovanosis You develop an ulcer in the genital area Prevention Avoiding all sexual activity is the only absolute way to prevent an STD such as donovanosis. However, safer sex behaviors may reduce your risk. The proper use of condoms, either the male or female type, greatly decreases the risk of catching an STD. You need to wear the condom from the beginning to the end of each sexual activity. Alternative Names Granuloma inguinale; Sexually transmitted disease - donovanosis; STD - donovanosis; Sexually transmitted infection - donovanosis; STI - donovanosis Images Skin layers References Belda W. Granuloma inguinale (Donovanosis). In: Goldman L, Cooney KA, eds. Goldman-Cecil Medicine. 27th ed. Philadelphia, PA: Elsevier; 2024:chap 292. Eckert LO, Lentz GM. Genital tract infections: vulva, vagina, cervix, toxic shock syndrome, endometritis, and salpingitis. In: Gershenson DM, Lentz GM, Valea FA, Lobo RA, eds. Comprehensive Gynecology. 8th ed. Philadelphia, PA: Elsevier; 2022:chap 23. Stoner BP, Reno HEL. Klebsiella granulomatis (donovanosis, granuloma inguinale). In: Bennett JE, Dolin R, Blaser MJ, eds. Mandell, Douglas, and Bennett's Principles and Practice of Infectious Diseases. 9th ed. Philadelphia, PA: Elsevier; 2020:chap 235. Review Date 8/26/2023 Updated by: Jatin M. Vyas, MD, PhD, Roy and Diana Vagelos Professor in Medicine, Columbia University Vagelos College of Physicians and Surgeons, Division of Infectious Diseases, Department of Medicine, New York, NY. Also reviewed by David C. Dugdale, MD, Medical Director, Brenda Conaway, Editorial Director, and the A.D.A.M. Editorial team. Related MedlinePlus Health Topics Sexually Transmitted Infections Skin Infections
15665
https://www.freemathhelp.com/forum/threads/finding-and-area-of-a-triangle.56972/
New posts Search forums Menu Log in Register Install the app Forums Free Math Help Beginning Algebra finding and area of a triangle Thread starter diana46 Start date D diana46 New member Joined : Aug 3, 2008 Messages : 3 #1 Hi all, I am so lost with my algebra class; I am taking this online and would appreciate any help, suggestions, feedback. The problem I am to do is this: Find an area of a triangle with vertices that have coordinates of (0,9), (0, -4), and (5, -4). Thanks if you can help. Diana L Loren Senior Member Joined : Aug 28, 2007 Messages : 1,299 #2 Here's one approach (a long one). The distance between two points is given by the formula... d=(x2​−x1​)2+(y2​−y1​)2​ So, the distance between (0,9) and (0, -4) is... (0−0)2+(−4−9)2​=02+(−13)2​=169​=13 units. In like manner you can find length of other two sides. Then use Hero's (Heron's) formula... A=s(s−a)(s−b)(s−c)​ where s=(1/2)(a+b+c) and a,b and c are the lengths of the sides of a triangle. T TchrWill Full Member Joined : Jul 7, 2005 Messages : 856 #3 diana46 said: Hi all, I am so lost with my algebra class; I am taking this online and would appreciate any help, suggestions, feedback. The problem I am to do is this: Find an area of a triangle with vertices that have coordinates of (0,9), (0, -4), and (5, -4). Thanks if you can help. Given the rectangular coordinates of the three vertices, the area of a triangle whose vertices are P1(x1,y1), P2(x2,y2), P3(x3,y3) is given by Area triangle P1P2P3 = 1/2[(x1y2 - x2y1) + (x2y3 - x3y2) + (x3y1 - x1y3)]. Method: First step-Write down the vertices in two columns, abscissas in one, ordinates in the other, repeating the coordinates of the first vertex. They must be in order around the triangle, clockwise preferably. Second step-Multiply each abscissa by the ordinate of the next row, and add the results. This gives x1y2+x2y3+x3y1. Third step-Multiply each ordinate by the abscissa of the next row, and add the results. This gives y1x2+y2x3+y3x1. Fourth step-Subtract the result of the third step from that of the second step, and divide by 2. This gives the required area. (If you list the vertices in clockwise order the result will be positive. If you list them is counterclockwise order the result will be negative.) For example, consider the coordinates (0, 4), (4, 2), (-4, -6) from which our table looks like: x1 y1 0 +4 x2 y2 +4 +2 x3 y3 -4 -6 x1 y2 0 +4 For the second step--0x3 + 4x(-6) + (-4)x4 = 0-24-16 = -40 For the third step------4x4 + 2x(-4) + (-6)x0 = 16-8+0 = 8 For the fourth step--[8-(-40)]/2 = 24 sq units. That was easy. This method applies to any multiple sided figure that a set of vertices produces. Click to expand... skeeter Elite Member Joined : Dec 15, 2005 Messages : 3,216 #4 diana46 said: Hi all, I am so lost with my algebra class; I am taking this online and would appreciate any help, suggestions, feedback. The problem I am to do is this: Find an area of a triangle with vertices that have coordinates of (0,9), (0, -4), and (5, -4). Thanks if you can help. Diana Click to expand... Sketch a diagram of the three points connected on a Cartesian coordinate grid. You'll see that the three points form a right triangle with base = 5 and height = 13. Now use the area formula for a triangle. Work smart ... not hard. D diana46 New member Joined : Aug 3, 2008 Messages : 3 #5 Thanks for the replies; it just makes no sense to me...I don't know what to do ... M Mrspi Senior Member Joined : Dec 17, 2005 Messages : 2,128 #6 diana46 said: Hi all, I am so lost with my algebra class; I am taking this online and would appreciate any help, suggestions, feedback. The problem I am to do is this: Find an area of a triangle with vertices that have coordinates of (0,9), (0, -4), and (5, -4). Thanks if you can help. Diana Click to expand... Skeeter gave you the BEST advice... plot the points given and draw the triangle. Once you do this, you should see that one side of the triangle is on a horizontal line...the segment connecting (0, -4) and (5, -4). Call that side the "base" of the triangle...and you should readily be able to determine the length of that side. The third vertex of the triangle is at (0, 9). You should be able to determine the perpendicular distance between this point and the base (which is known as the HEIGHT of the triangle)..after all, if you draw this out, you should see that the triangle is a RIGHT triangle! Now... For any triangle, Area = (1/2)base height If all of this is still confusing to you, I'd suggest that you have a face-to-face meeting with your instructor. D Denis Senior Member Joined : Feb 17, 2004 Messages : 1,707 #7 Diana, are you sure of coordinates (0,9)? Perhaps a typo and should be (0,8)? Anyhow, if you don't know basics like (as example) a 3-4-5 triangle is a right triangle with area = 34/2 = 6, then you definitely need to follow Mrspi's advice. D diana46 New member Joined : Aug 3, 2008 Messages : 3 #8 Hi Denis, I checked the numbers and what I posted is correct; I am hoping to find a tutor today, as I have never had algebra in high school. I wish I could have taken Algebra in a classroom, but I will my best. I am working towards my Medical Coding certificate and will graduate Sept 2008! Yeah for me. Thanks for the help and tips. Diane D Denis Senior Member Joined : Feb 17, 2004 Messages : 1,707 #9 Try googling "area of triangle with given coordinates". If nothing else, at least you'll see why it's almost "impossible" to "teach" that here You must log in or register to reply here. Share: Facebook Twitter Reddit Pinterest Tumblr WhatsApp Email Share Link Forums Free Math Help Beginning Algebra This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register. By continuing to use this site, you are consenting to our use of cookies. Accept Learn more…
15666
https://www.lessonplanet.com/teachers/paul-s-online-notes-differential-equations-first-order-differential-equations
Paul Dawkins Resource Details Curator Rating Educator Rating : Not yet Rated Grade : 9th - 10th Subjects : Math : 2 more... Resource Type : Activities & Projects Audiences : For Administrator Use : 2 more... Lexile Measures : 0L Activity Paul's Online Notes: Differential Equations: First Order Differential Equations Curated by ACT The tutorial explores first order differential equations. Topics examined are linear equations, separable equations, Bernoulli differential equations, equilibrium solutions, Euler?s method, and intervals of validity. Class notes, definitions, and examples with detailed solutions are included. The notes are available in pdf format. 3 Views 0 Downloads Concepts linear equations Show More Show Less Additional Tags equilibrium solutions, intervals of validity, separable equations, bernoulli differential equations, euler’s method, first order differential equations Show More Show Less Classroom Considerations Knovation Readability Score: 5 (1 low difficulty, 5 high difficulty) This resource is only available on an unencrypted HTTP website.It should be fine for general use, but don’t use it to share any personally identifiable information See similar resources: AP Test Prep Calculus AB/BC - Verifying Solutions for Differential Equations Flipped Math Find out if it is truly the solution. An informative presentation provides viewers with multiple examples of finding the particular solution to a differential equation. Learners find out how to verify the solution by differentiating the... 15 mins 10th - 12th Math AP Test Prep Calculus AB/BC - Exponential Models with Differential Equations Flipped Math Exponential models come with a shortcut. Pupils watch an example of how to determine if a differential equation is an exponential model. Learners realize the solution is in the same formula, allowing for a quick way to determine the... 17 mins 10th - 12th Math Activity Paul's Online Notes: Differential Equations Paul Dawkins The tutorial explores differential equations. Topics examined are first order differential equations, second order differential equations, Laplace transforms, Fourier series, higher order differential equations, and systems of... 9th - 10th Math Activity Paul's Online Notes: Differential Equations Second Order Differential Equation Paul Dawkins The tutorial explores second order differential equations. Topics examined are mechanical vibrations, the reduction of order, fundamental sets of solutions, nonhomogeneous differential equations, and undetermined coefficients. Class... 9th - 10th Math Instructional Video Finding the Equation Passing Through Two Points Corbett Maths What's the quickest way to get from point A to point B? A straight line, of course! Young scholars learn to write the equation of this line as they watch a video lesson. They then have the opportunity to deepen their understanding using... 4 mins 8th - 10th Math CCSS: Adaptable [Instructional Video Khan Academy: Differential Equations: First Order Homogeneous Equations 2 Khan Academy Video shows an example of using substitution to solve a first order homogeneous differential equation. [8:22] 9th - 10th Math](/teachers/khan-academy-differential-equations-first-order-homogeneous-equations-2 "Khan Academy: Differential Equations: First Order Homogeneous Equations 2 Instructional video") [Instructional Video Khan Academy: Differential Equations: First Order Homogeneous Equations Khan Academy Video first shows what homogeneous means in terms of differential equations, then shows how to use substitution to find solutions to the homogeneous differential equation. [7:21] 9th - 10th Math](/teachers/khan-academy-differential-equations-first-order-homogeneous-equations "Khan Academy: Differential Equations: First Order Homogeneous Equations Instructional video") Instructional Video Khan Academy: Differential Equations: 2nd Order Linear Homogeneous Equations 1 Khan Academy Video first explains what 2nd order linear differential equations are, then shows that a constant times a known solution to a 2nd order linear homogenous differential equation is also a solution and that the sum of two known solutions is... 9th - 10th Math Instructional Video Khan Academy: Differential Equations: 2nd Order Linear Homogeneous Equations 2 Khan Academy Video shows an example of solving a second order linear homogeneous differential equation. Includes how to write the solution in a general form using the properties of homogeneous equations shown in the previous video. Briefly discusses... 9th - 10th Math Handout Paul's Online Notes: Algebra: Linear Equations Paul Dawkins This math tutorial features notes and examples that illustrate the process for solving linear equations including equations with radical expressions. 9th - 10th Math CCSS: Adaptable Try It Free
15667
https://blog.prepscholar.com/lines-slopes-sat-math-geometry-strategies-and-practice
Published Time: Mon, 21 Jul 2025 22:03:47 GMT Lines and Slopes in SAT Math: Geometry Strategies · PrepScholar CALL NOW:+1 (866) 811-5546 Choose Your Test SAT Prep ACT Prep PrepScholar Advice Blog ☰ Search Blogs By Category SAT ACT College Admissions AP and IB Exams PSAT TOEFL GPA and Coursework Lines and Slopes in SAT Math: Geometry Strategies Posted byCourtney Montgomery SAT Math In our SAT guide to lines and angles, we dealt with parallel lines, perpendiculars, and the many different ways to find angle measures with two or more lines. Now, we’ll look at the other aspect of lines, namely their slopes and equations. This will be your complete guide to lines and slopes—what slopes mean, how to find them, and how to solve the many types of slope and line equation questions you’ll see on the SAT. What Are Lines and Slopes? Before we start, you may want to take a moment to familiarize yourself with our guide to SAT coordinate points in order to refresh yourself on the basics of coordinate geometry. Basically, coordinate geometry takes place in the space where the x-axis and the y-axis meet. Any place on this space is given a coordinate point—written as (x,y)—that indicates where the point is along each axis. A line (or line segment) is a completely straight marker with no curvature. It is made up of (and connects) a series of points together. A slope is the measure of the slant (steepness) of a line. A slope is found by finding the change in distance along the y axis over the change in distance along the x axis. You are probably most familiar with this concept by finding the “rise over run” to find the slope of a line. change in y change in x Here is a typical line, presented on the coordinate grid. To find our slope, first start by marking the points where the line hits the grid at perfect integer coordinates. This will make life simpler when we go to find the slope. Wherever the grid meets at a corner, we will have coordinates that are integers. We can see here that our line hits the coordinates: (−3,4), (0,2), and (3,0). Now let us find the rise and run of the line. Our "rise" will be -2, as we must move down 2 units to reach the next coordinate point in our line. Our run will be +3, as we must move 3 units to the right to reach the next coordinate point in our line. So our final slope will be: rise run −2 3 Properties of Slopes A slope can either be positive or negative. A positive slope rises from left to right. A negative slope falls from left to right. A straight horizontal line has a slope of zero. It will be defined by one axis only. y=2 A straight vertical line has an undefined slope (because run will always be 0, and you can't divide by 0). It will be defined by one axis only. x=2.5 The steeper the line, the larger the slope. _The red line is steepest, with a slope of 4 1, or 4. The blue line is not as steep, with a slope of 4 9_ San Francisco knows a little bit about steep slopes. Line and Slope Formulas Finding the Slope y 2−y 1 x 2−x 1 In order to find the slope of a line that connects two points, you must find the change in the y-values over the change in the x-values. (Note: It does not matter which points you assign as (x 1,y 1) and (x 2,y 2), so long as you are consistent.) Given the coordinates (2,2) and (−1,0), find the slope of the line. Now, we can solve this question in one of two ways—by drawing a graph and counting, or by using our formula. Since we already saw earlier how it's possible to count out our slope on a graph, let us use our formula to see how it's done. Of the two coordinate points, we must assign one set to be x 1 and y 1 and the other to be x 2 and y 2. Let us pick (2, 2) to be our x 1 and y 1 and (-1, 0) to be our x 2 and y 2. y 2−y 1 x 2−x 1 (0−2)(−1−2) −2−3 2 3 But what would have happened had we assigned (−1,0) to be our x 1 and y 1 and (2,2) to be our x 2 and y 2? We would have gotten the same results either way! y 2−y 1 x 2−x 1 (2−0)(2−−1) 2 3 No matter which coordinates we assign to be the first or second values for x and y, we will end up with the same slope so long as we are consistent. Equation of a Line y=m x+b This is called the “equation of a line” or a line in "slope-intercept form" and it shows exactly how a line is positioned along the x- and y-axes as well as how steep it is. This is the most important formula you’ll need when it comes to lines and slopes, so let’s break it into pieces. y is your y-coordinate value for any particular value of x. x is your x-coordinate value for any particular value of y. m is the measure of your slope. b is the y-intercept value of your line. This means that it is the value where the line hits the y-axis (remember, a straight line will only hit each axis a maximum of one time). Find the equation of the line from the graph. We're using the same graph from above, and we can see that the line intersects with the y-axis at around y=0.5. We also determined earlier that the slope is 2 3. So when we put those two pieces of information together, the equation of our line would be: y=2 3 x+0.5 If the question wants you to be more specific than "about a half" about what fraction of a number the line hits x=0 or y=0 at, it will have a more detailed drawing. In this question, while the actual intercept for a line with coordinates (2,2) and (0,−1) would actually be 2 3 or 0.66, not 0.5, the graph is not at a scale where you could reasonably visually estimate that. Remember to always re-write any line equations you are given into the proper form! Often, the test will try to trip you up by giving you an equation not written in proper form and asking for the slope of the line. This is to get people to make a mistake if they are going too quickly through the test. t x+12 y=−3 The equation above is the equation of a line in the x y-plane, and t is a constant. If the slope of the line is -10, what is the value of t? First, let us put this equation into the proper equation of a line. t x+12 y=−3 12 y=−t x−3 y=−t x 12−3 12 Now, we only care about finding the slope, so let’s find the value of t (our slope) by using our given. −t 12=−10 −t=−120 t=120 Our final answer is t=120. Always remember to set up your equation as your first step and you’ll be able to solve most any slope problem quickly and easily. Perpendicular Lines When two lines meet at right angles, the lines are called “perpendicular.” Perpendicular lines will always have slopes that are negative reciprocals of one another. This means that you must reverse both the sign of the slope as well as the fraction. For example, if a two lines are perpendicular to one another and one has a slope of 3 4 the other line will have a slope of −4 3. Perpendicular lines with slopes of 3 4 and −4 3 And if a line has a slope of −5, the line that meets it perpendicularly will have a slope of 1 5. Parallel Lines When two lines will never meet, no matter how infinitely long they extend, the lines are said to be parallel. This means that they are continuously equidistant. If two lines are parallel, they will also have the same slope. You can see why this makes sense, since the rise over run will always have to be the same in order to make sure that the lines will never touch. Parallel lines with slopes of 4 3 When lines turn devious. Typical Line and Slope Questions Most line and slope questions on the SAT are quite basic at their core.You’ll generally see two questions on slopes per test and almost all of them will simply ask you to find the slope of a line or the equation of a line. The test may attempt to complicate the question by using other shapes or figures, but the questions always boil down to these simple concepts. Just remember to re-write any given equations into the proper slope-intercept form and keep in mind your rules for finding slopes (and your rules for parallel or perpendicular lines). With this knowledge in hand, you’ll be able to solve these types of problems quickly and easily. From the graph, we can see that the y-intercept of the line is (0,1). The line also passes through the point (1,2). This is enough information for us to figure out the slope of the line, which we know is change in y change in x. 2−1 1−0=1 1=1 Now we know that the slope of the line is 1. In slope-intercept form, the equation for line l is y=x+1 or choice D. Our final answer is D. We know that slope is the change in y change in x. The equation y=k x+4 is already in slope-intercept form, so we know the slope of the line is k. We also know that the line contains the point (c,d) which means we can substitute those variables for (x,y) in the equation. This gives us d=k c+4 Solving for this equation for the slope, k, gives us k=d−4 c Our final answer is A,d−4 c How to Solve a Line and Slope Problem As you go through your line and slope problems, keep in mind these tips: #1: Always rearrange your equation into y=m x+b The test makers will often present you with an equation of a line that is not in proper form, for example: 4 y+3 x=12. If you are going too quickly through the test or if you forget to rearrange the given equation into proper slope-intercept form, you will misidentify the slope of the line. So always remember to rearrange your equation into proper form as your first step. 4 y+3 x=12 =>y=−3 4 x+3 #2: Always remember your rise run It can be easy to make a mistake try to find the change in x before finding the change in y, as our brains are used to doing things "in order." Keep careful track of your variables in order to reduce careless mistakes of this nature. Remember the mantra of "rise over run" and this will help you always know to find your change in y over your change in x. #3: Make your own graph and/or count to find your slope Because the slope is always "rise over run," you can always find the slope with a graph—whether from a given graph or from your own. It's never a bad idea to take a second and make your own graph if you are not provided with one. This will help you better visualize the problem and avoid errors. If you forget your formulas (or simply don't want to use them), simply count how the line rises (or falls) and trace its "run," and you will always find your slope. Oh, for the days when all we needed to know about lines was how to color in (or out of) them.... Test Your Knowledge Now that we’ve walked through the typical slope questions you’ll see on the test (and a few basic tips you’ll need to solve them), let’s put your knowledge to the test. 1. 2. 3. 4. Answers: D, A, B, D Answer Explanations: 1. We know that slope is the change in y change in x. We are also told that our slope is -2, which means it must be −2 1. This means that, for every time our y value decreases by 2, our x value increases by 1. And every time our y value increases by 2, our x value will decrease by 1. If we use our rectangle, we also have one point of reference on the line. We can see that the rectangle has a length of 3 (because it spans horizontally from x=0 to x=3) and a length of 4 (because it spans vertically from y=0 to y=4). This means that the rectangle hits the line at the top right corner at coordinates (3, 4). Now we can simply count where the line will hit the y-axis. Because the slope increases horizontally (along the x-axis) one unit at a time, we can see that there will be 3 1=3 even points along the line needed to find the y-intercept. Basically, this means that we are taking the slope, −2 1, and multiplying it by 3 to get −6 3. In other words, we are adding 6 to the change in y and subtracting 3 from the change in x because we are taking the slope backwards. So now we can find our new point by saying that we increased our y-value from 4 to 4+6=10, and we have decreased our x-value from 3 to 3−3=0, which would give us the new slope of: (4+6)(3−3)=10 0 So our new coordinate point is (0, 10), which means that our y-intercept is 10. Our final answer is D, 10. 2. We know that we can find the slope of the line using y 2−y 1 x 2−x 1, so let's plug in our coordinates of (0,r) and (s,0) for these values. y 2−y 1 x 2−x 1 (0−r)(s−0) r s Our s will remain unchanged, but our r value will become negative, as it is being subtracted from zero. Our final answer is A. 3. If you count to the point at which the line crosses the y-intercept, you can see it hits at y=3 In the equation y=m x+b, the b is the y-intercept. That means our b will be 3. We can therefore cross off answer choices A and D, leaving us down to B, C, and E. We can also see that our line falls from left to right, so our slope will be negative. This means we can eliminate answer choice E, leaving us between choices B and C. Now let’s take the two points at which the line hits the axes. We already saw that the line hits the y-intercept at y=3 and we can also see that the line hits the x-axis at x=2. This means our line hits coordinates (0, 3) and (2, 0). This means that the change in our y value is -3 and the change in our x value is +2. (Why? Because we decreased our y value by 3 and we increased our x value by 2.) So our slope must be −3 2, which means that our final equation will be: y=−3 2 x+3 Our final answer is B. 4. We're told that the equation of the line is y=2 x−5. This means that the slope of the line is 2 and the y-intercept is at (0,-5). A slope of 2 means that, for each increase in x of 1, y increases by 2. Looking at each of the graphs, choices C and D are the only lines with a slope of 2. (Don't be fooled by choice B which has a slope of -2). Next we can look at y-intercepts. The line in choice C has a y-intercept at (0,5), which doesn't match the equation of the line we were given. However, choice D's line has a y-intercept at (0,−5), which is exactly what we're looking for. Choice D is the only line with both the correct slope and y-intercept. Our final answer is D. Hurray, hurray! You've found your slopes, you know your lines! Congrats, congrats. The Take-Aways Once you have familiarized yourself with the basics of coordinate geometry, slopes should not be too far off field. Though the SAT will try to complicate problems as much as they are reasonably able, questions on lines and slopes are almost always easier than they appear. Keep your important formulas close to your heart and be vigilant with your negative signs, and you’ll do just fine when it comes to slopes and intercepts. What’s Next? Phew! You've learned all there is to learn about your slopes angles. Luckily for you (for a certain definition of "luck"), there is so much more to learn! Before you continue on, make sure you have a firm grasp of all the topics covered on SAT math so that you can see what to prioritize. If you're looking for a particular math topic, scope out our SAT math archive for individual topic guides like this one. We've got guides to solid geometry, probability, ratios, and more! If you don't know where to begin, make sure you take a practice test and see how your score ranks. This will give you a good sense of where your strengths and weaknesses are and how you should focus your study time. Looking to get a perfect score? Check out our guide to getting an 800, written by a perfect scorer! Trending Now How to Get Into Harvard and the Ivy League How to Get a Perfect 4.0 GPA How to Write an Amazing College Essay What Exactly Are Colleges Looking For? ACT vs. SAT: Which Test Should You Take? When should you take the SAT or ACT? Get Your Free eBook 5 Tips to Raise Your SAT Score 160+ Points 5 Tips to Raise Your ACT Score 4+ Points SAT Prep Find Your Target SAT Score Free Complete Official SAT Practice Tests How to Get a Perfect SAT Score, by an Expert Full Scorer Score 800 on SAT Math Score 800 on SAT Reading and Writing How to Improve Your Low SAT Score Score 600 on SAT Math Score 600 on SAT Reading and Writing Raise Your SAT Score by 160+ Points GUARANTEED. Start Your 5-Day Risk-Free Trial ACT Prep Find Your Target ACT Score Complete Official Free ACT Practice Tests How to Get a Perfect ACT Score, by a 36 Full Scorer Get a 36 on ACT English Get a 36 on ACT Math Get a 36 on ACT Reading Get a 36 on ACT Science How to Improve Your Low ACT Score Get a 24 on ACT English Get a 24 on ACT Math Get a 24 on ACT Reading Get a 24 on ACT Science Stay Informed Get the latest articles and test prep tips! Get Exclusive Tips for College Admissions Sign up to receive free guidance on everything from acing the SAT to writing a standout college admissions essay. SAT / ACT Blog Email Subscription × Thanks for Subscribing! To receive immediate updates on all our helpful guides and strategies, follow us on social media: Have friends who also need help with test prep? Share this article! About the Author Courtney Montgomery Courtney scored in the 99th percentile on the SAT in high school and went on to graduate from Stanford University with a degree in Cultural and Social Anthropology. She is passionate about bringing education and the tools to succeed to students from all backgrounds and walks of life, as she believes open education is one of the great societal equalizers. She has years of tutoring experience and writes creative works in her free time. Ask a Question Below Have any questions about this article or other topics? Ask below and we'll reply! WHO WE ARE SUCCESS STORIES ADVICE BLOG FOR SCHOOLS SAT ACT PSAT AP GRE GMAT TOEFL ADMISSIONS TERMS OF SERVICE PRIVACY POLICY © PrepScholar 2013-2025. All rights reserved SAT® is a registered trademark of the College Entrance Examination Board TM. The College Entrance Examination Board TM does not endorse, nor is it affiliated in any way with the owner or any content of this web site. ACT® is a registered trademark belonging to ACT Education Corporation ("ACT"). ACT is not involved with or affiliated with PrepScholar Inc., nor does ACT endorse or sponsor any of the products or services offered by PrepScholar Inc. GRE® and TOEFL® are registered trademarks of the ETS®. The ETS® does not endorse, nor is it affiliated in any way with, the owner or any content of this web site. GMAT® is a registered trademark of the Graduate Management Admissions Council®. The Graduate Management Admissions Council® does not endorse, nor is it affiliated in any way with, the owner or any content of this website.
15668
https://www.mathway.com/popular-problems/Calculus/593842
Enter a problem... Calculus Examples Popular Problems Calculus Find the Local Maxima and Minima y=e^x Step 1 Write as a function. Step 2 Differentiate using the Exponential Rule which states that is where =. Step 3 Differentiate using the Exponential Rule which states that is where =. Step 4 To find the local maximum and minimum values of the function, set the derivative equal to and solve. Step 5 Since there is no value of that makes the first derivative equal to , there are no local extrema. No Local Extrema Step 6 No Local Extrema Step 7 | | | | Please ensure that your password is at least 8 characters and contains each of the following: a number a letter a special character: @$#!%?&
15669
https://www.varsitytutors.com/practice/subjects/act-math/help/conversion-problems
Conversion Problems Help Questions ACT Math › Conversion Problems Questions 1 - 10 Tiffany wants to purchase an 18-foot length of rope to hang a tire swing. When she arrives at the store to buy the rope, all of the lengths are quoted in inches. How many inches should she purchase to have exactly 18 feet of rope? 184 201 216 Explanation When converting between units, it can be helpful to use dimensional analysis to set up your equation. You know that there are 12 inches in 1 foot, and that you need to convert 18 feet to a number of inches. If you then structure your math so that the units cancel via division, you can ensure that you're making the right "do I multiply or do I divide by 12" decision: Helps you determine that you need to multiply, because then the "feet" units cancel leaving you with the units you want, which is inches. This means you multiply 18 by 12, which gives you 216 inches. 2 Kendrick purchased 8 gallons of water for a camping trip, and plans to pour all of the water into insulated flasks that measure 1 quart each to keep the water cool. How many flasks will it take to contain all 8 gallons? (1 gallon = 4 quarts) 2 16 32 48 64 Explanation A great way to ensure you make the proper "do I multiply or divide?" decision on any conversion problem is to use dimensional analysis, which means to set up the units in your calculation so that the units you don't want in your answer cancel, and the unit(s) you do want in your answer remain. Here you're given gallons and asked to convert to quarts, so you want to set up the math so that gallons cancels. That means you'll use: This preserves the ratio of 4 quarts to every 1 gallon and allows the unit "gallons" to cancel via division so that you're left with quarts. That also means that you multiply 8 by 4 to get the answer, 32. 3 Rashid worked on his science project for a total of 450 minutes. How many hours did he spend working on his science project? 6.5 7 7.5 8 8.5 Explanation To convert between minutes and hours, you can use dimensional analysis to ensure that you are performing the right calculations. You know that there are 60 minutes for every 1 hour, and you start with minutes and need to get your answer in terms of hours. That means that you want to set up the equation so that the units "minutes" cancels and leaves you with just "hours": You can see here that "minutes" is in both numerator and denominator, allowing you to cancel those units. And it dictates that to get your answer you'll divide 450 by 60. That reduces as a fraction to 45 divided by 6, which comes out to 7.5. 4 To prepare for a bicycle race, Celeste wants to ride her bicycle for 20 miles. The app she is using to track her distance only cites distance in kilometers. Using the conversion that 1 kilometer = 0.62 miles, which of the following is closest to the exact number of kilometers she should ride? 15 21 28 32 39 Explanation When you're working with conversion problems, it is helpful to use dimensional analysis to make sure that you're making the proper "do I multiply or divide by this conversion" decision. Here you're given a number of miles (she wants to ride 20 miles) and you need to get to kilometers. So you can structure your math with miles in the denominator of the fraction, and that means that the units "miles" will cancel leaving you with just "kilometers": As you can see, this tells you to divide by 0.62, and when you've canceled the unit "miles" you're left with just "kilometers." Note, also, that the question asks for an estimate ("which of the following is closest") so you can safely divide by just 0.6, or 3/5, to get to 33.33, and the only answer that is close is 32. 5 Brionna uses a fitness app to track the number of steps she takes each day. For the month of February, she took a total of 320,000 steps. She estimates that 2,500 of her steps equates to one mile. Approximately how many miles did she walk in February? 100 128 144 156 172 Explanation When you're working with conversion problems, it is helpful to use dimensional analysis to ensure that you are properly applying the conversion (i.e. "should I multiply or divide?"). Here you're given a number of steps and you need to convert to miles using the conversion 2,500 steps = 1 mile. You can then set up your math so that the unit "steps" cancels and you're left with just "miles." That would mean: As you can see, steps will cancel due to division, meaning that you'll divide 320,000 by 2,500 and be left with the proper unit miles. To do this math by hand, you should also see that you can factor out 100 from both 320,000 and 2,500, leaving you with the problem 3200 divided by 25. This leads you to the correct answer, 128. 6 Caitlyn's goal for the upcoming track and field season is to extend her long jump personal best by 10 inches to break the school record. By approximately how many centimeters will she need to extend her personal best long jump? (1 inch = 2.54 centimeters) 4 10 15 25 54 Explanation Whenever you're working with unit conversions, it is a good idea to use dimensional analysis to structure your math to help you choose the right operation (multiply vs. divide). This means that you'll set up your equation to cancel the units that you don't want in your answer, and therefore you'll be left with the proper units. Here you're given inches and asked to convert to centimeters, so you'll set up the math with inches in the denominator so that the units cancel: This means that you'll multiply 10 by 2.54, and the inch units will cancel ensuring that you're properly converting to centimeters. 10 times 2.54 is 25.4, which rounds to 25. 7 Ajay plans to adopt a rescue dog from Europe, where the dogs' weights are measured in kilograms. He chooses a dog that weighs 25 kilograms; approximately how many pounds does that dog weigh? (1 kilogram = 2.2 pounds) 55 42 32 25 11 Explanation Whenever you're working with unit conversions, it is a good idea to use dimensional analysis to ensure that you are setting up your calculation - specifically the choice of whether you multiply or divide by the conversion ratio - properly. This means that you'll set up the math so that the units you're given cancel, leaving you with the units you need to arrive at in the end. Here you're given kilograms and want to get to pounds, so you'll multiply by a conversion with kilograms in the denominator so that kilograms cancel, leaving you with pounds: This means that you multiply 25 by 2.2 and that kilograms cancel and the only unit left is the one you want, pounds. 25 times 2.2 is 55, giving you the correct answer. 8 Peter is cooking a large batch of pancakes for a family breakfast. The recipe calls for 16 pints of buttermilk, but the store sells buttermilk in quarts. How many quarts does he need to buy to satisfy the recipe? (1 quart = 2 pints) 4 8 16 32 64 Explanation Whenever you're working with unit conversions, it is a good idea to use dimensional analysis to structure your math properly - specifically as it pertains to the question of whether you should multiply or divide by the conversion ratio provided. That means you'll set up the math so that the units you don't want cancel, leaving you with the units you do. Here you're given pints and asked to arrive at quarts, so you'll set up your math such that pints is in the denominator of the conversion ratio so that it cancels. That means that your math will look like: This means you'll divide 16 by 2 to get your answer, and since pints cancels via division, you know that you've converted properly to quarts. 16 divided by 2 is 8, making 8 the correct answer. 9 Aiden is purchasing a red carpet for a gala event. He wants carpet that is 18 feet long, but the vendor quotes all prices in inches. How many inches long does the carpet need to be? 48 144 216 Explanation Whenever you're working with unit conversions, it is a good idea to use dimensional analysis to ensure that you are structuring your calculation properly, as often the decision of whether to multiply or divide by the conversion ratio can be tricky. That means that you'll set up the math so that the units you don't want in your answer (here that's feet) will cancel, leaving you with the units you do want (here that's inches). Your math here would look like: Feet here will cancel via division, leaving you with inches. And this tells you to multiply 18 by 12 giving you an answer of 216. 10 Julie is a zookeeper responsible for feeding baby giraffes. Each giraffe should drink 12 quarts of milk per day, but Julie's milk containers measure in pints. How many pints should she feed each giraffe each day? (1 quart = 2 pints) 6 48 Explanation Whenever you're facing a unit conversion problem it is a good idea to use dimensional analysis to help you structure the math - whether you should multiply or divide by the provided conversion ratio - properly. That means that you'll set up the math such that the units you don't want in your answer cancel, leaving you with the units you do want. Here you're given quarts but asked to convert to pints, so you'll set up the math so that quarts are in the denominator and cancel, leaving you with pints: This means that quarts cancel leaving you with pints, and tells you that you have to multiply 12 by 2. The correct answer is 24. Return to subject
15670
https://www.geeksforgeeks.org/maths/secant-square-x-formula/
Secant Square x Formula - GeeksforGeeks Skip to content Tutorials Python Java DSA ML & Data Science Interview Corner Programming Languages Web Development CS Subjects DevOps Software and Tools School Learning Practice Coding Problems Courses DSA / Placements ML & Data Science Development Cloud / DevOps Programming Languages All Courses Tracks Languages Python C C++ Java Advanced Java SQL JavaScript Interview Preparation GfG 160 GfG 360 System Design Core Subjects Interview Questions Interview Puzzles Aptitude and Reasoning Data Science Python Data Analytics Complete Data Science Dev Skills Full-Stack Web Dev DevOps Software Testing CyberSecurity Tools Computer Fundamentals AI Tools MS Excel & Google Sheets MS Word & Google Docs Maths Maths For Computer Science Engineering Mathematics Switch to Dark Mode Sign In Number System and Arithmetic Algebra Set Theory Probability Statistics Geometry Calculus Logarithms Mensuration Matrices Trigonometry Mathematics Sign In ▲ Open In App Secant Square x Formula Last Updated : 07 Aug, 2025 Comments Improve Suggest changes Like Article Like Report Secant is one of the six fundamental identities in trigonometry, specifically a Pythagorean identity. There are various identities involving secant, one of which is the secant square x formula. This formula is expressed as: sec2x = 1 + tan2x This formula represents the relationship between the secant and tangent functions. In this article, we will discuss this formula in detail, including its proof as well as some solved examples related to it. Secant Trigonometric Ratio The ratio of the lengths of any two sides of a right triangle is called a trigonometric ratio. In trigonometry, these ratios link the ratio of sides of a right triangle to the angle. The secant ratio is expressed as the ratio of the hypotenuse (longest side) to the side corresponding to a given angle in a right triangle. It is the reciprocal of the cosine ratio and is denoted by the abbreviation sec. If θ is the angle that lies between the base and hypotenuse of a right-angled triangle then, sec θ = Hypotenuse/Base = 1/cos θ Here, hypotenuse is the longest side of right triangle and base is the side adjacent to the angle. Secant Square x Formula The secant square x ratio is denoted by the abbreviation sec 2 x. It's a trigonometric function that returns the square of the secant function value for an angle x. The period of the function sec x is 2π, but the period of sec 2 x is π. Its formula is equivalent to the sum of unity and the tangent square function. sec2x = 1 + tan2x Where, x is one of the angles of the right triangle, tan x is the tangent ratio for angle x. Derivation of Secant Square Formula The formula for secant square x is derived by using the identity of sum of squares of sine and cosine ratios. We know, sin 2 x + cos 2 x = 1. Dividing both sides by cos 2 x, we get (sin 2 x/cos 2 x) + (cos 2 x/cos 2 x) = 1/cos 2 x tan 2 x + 1 = sec 2 x sec 2 x = 1 + tan 2 x This derives the formula for secant square x ratio. Read More, Trigonometry Trigonometric Formulas Trigonometric Identities Trigonometric Identities Class 10 Sample Problems Secant Square Formula Problem 1. If tan x = 3/4, find the value of sec2x using the formula. Solution: We have, tan x = 3/4. Using the formula we get, sec 2 x = 1 + tan 2 x ⇒ sec 2 x = 1 + (3/4)2 ⇒ sec 2 x = 1 + 9/16 ⇒ sec 2 x = 25/16 Problem 2. If tan x = 12/5, find the value of sec2x using the formula. Solution: We have, tan x = 12/5. Using the formula we get, sec 2 x = 1 + tan 2 x ⇒ sec 2 x = 1 + (12/5)2 ⇒ sec 2 x = 1 + 144/25 ⇒ sec 2 x = 169/25 Problem 3. If sin x = 8/17, find the value of sec2x using the formula. Solution: We have, sin x = 8/17. Find the value of cos x using the formula sin 2 x + cos 2 x = 1. cos x = √(1 - (64/289)) ⇒ cos x = √(225/289) ⇒ cos x = 15/17 So, tan x = sin x/cos x = 8/15 Using the formula we get, sec 2 x = 1 + tan 2 x ⇒ sec 2 x = 1 + (8/15)2 ⇒ sec 2 x = 1 + 64/225 ⇒ sec 2 x = 289/225 Problem 4. If cot x = 8/15, find the value of sec2x using the formula. Solution: We have, cot x = 8/15. So, tan x = 1/cot x = 15/8 Using the formula we get, sec 2 x = 1 + tan 2 x ⇒ sec 2 x = 1 + (15/8)2 ⇒ sec 2 x = 1 + 225/64 ⇒ sec 2 x = 289/64 Problem 5. If cos x = 12/13, find the value of sec2x using the formula. Solution: We have, cos x = 12/13. Find the value of sin x using the formula sin 2 x + cos 2 x = 1. sin x = √(1 - (144/169)) ⇒ sin x = √(25/169) ⇒ sin x = 5/13 So, tan x = sin x/cos x = 5/12 Using the formula we get, sec 2 x = 1 + tan 2 x ⇒ sec 2 x = 1 + (5/12)2 ⇒ sec 2 x = 1 + 25/144 ⇒ sec 2 x = 169/144 What is the Secant Square Formula? The Secant Square Formula is a trigonometric identity that states: sec2x = 1 + tan2x How is the Secant Square Formula derived? The formula can be derived from the Pythagorean identity: sin2x + cos2x = 1 Where is the Secant Square Formula used? The Secant Square Formula is frequently used in calculus, particularly in integration and differentiation of trigonometric functions. Can the Secant Square Formula be applied to all angles? Yes, the Secant Square Formula is valid for all angles θ where cos⁡ θ ≠ 0. Comment More info J jatinxcx Follow Improve Article Tags : Mathematics School Learning Maths MAQ Trigonometry - MAQ Maths-Formulas +1 More Explore Maths 4 min read Basic Arithmetic What are Numbers? 15+ min readArithmetic Operations 9 min readFractions - Definition, Types and Examples 7 min readWhat are Decimals? 10 min readExponents 9 min readPercentage 4 min read Algebra Variable in Maths 5 min readPolynomials| Degree | Types | Properties and Examples 9 min readCoefficient 8 min readAlgebraic Identities 14 min readProperties of Algebraic Operations 3 min read Geometry Lines and Angles 9 min readGeometric Shapes in Maths 2 min readArea and Perimeter of Shapes | Formula and Examples 10 min readSurface Areas and Volumes 10 min readPoints, Lines and Planes 14 min readCoordinate Axes and Coordinate Planes in 3D space 6 min read Trigonometry & Vector Algebra Trigonometric Ratios 4 min readTrigonometric Equations | Definition, Examples & How to Solve 9 min readTrigonometric Identities 7 min readTrigonometric Functions 6 min readInverse Trigonometric Functions | Definition, Formula, Types and Examples 11 min readInverse Trigonometric Identities 9 min read Calculus Introduction to Differential Calculus 6 min readLimits in Calculus 12 min readContinuity of Functions 10 min readDifferentiation 2 min readDifferentiability of Functions 9 min readIntegration 3 min read Probability and Statistics Basic Concepts of Probability 7 min readBayes' Theorem 13 min readProbability Distribution - Function, Formula, Table 13 min readDescriptive Statistic 5 min readWhat is Inferential Statistics? 7 min readMeasures of Central Tendency in Statistics 11 min readSet Theory 3 min read Practice NCERT Solutions for Class 8 to 12 7 min readRD Sharma Class 8 Solutions for Maths: Chapter Wise PDF 5 min readRD Sharma Class 9 Solutions 10 min readRD Sharma Class 10 Solutions 9 min readRD Sharma Class 11 Solutions for Maths 13 min readRD Sharma Class 12 Solutions for Maths 13 min read Like Corporate & Communications Address: A-143, 7th Floor, Sovereign Corporate Tower, Sector- 136, Noida, Uttar Pradesh (201305) Registered Address: K 061, Tower K, Gulshan Vivante Apartment, Sector 137, Noida, Gautam Buddh Nagar, Uttar Pradesh, 201305 Company About Us Legal Privacy Policy Contact Us Advertise with us GFG Corporate Solution Campus Training Program Explore POTD Job-A-Thon Community Blogs Nation Skill Up Tutorials Programming Languages DSA Web Technology AI, ML & Data Science DevOps CS Core Subjects Interview Preparation GATE Software and Tools Courses IBM Certification DSA and Placements Web Development Programming Languages DevOps & Cloud GATE Trending Technologies Videos DSA Python Java C++ Web Development Data Science CS Subjects Preparation Corner Aptitude Puzzles GfG 160 DSA 360 System Design @GeeksforGeeks, Sanchhaya Education Private Limited, All rights reserved Improvement Suggest changes Suggest Changes Help us improve. Share your suggestions to enhance the article. Contribute your expertise and make a difference in the GeeksforGeeks portal. Create Improvement Enhance the article with your expertise. Contribute to the GeeksforGeeks community and help create better learning resources for all. Suggest Changes min 4 words, max Words Limit:1000 Thank You! Your suggestions are valuable to us. What kind of Experience do you want to share? Interview ExperiencesAdmission ExperiencesCareer JourneysWork ExperiencesCampus ExperiencesCompetitive Exam Experiences Login Modal | GeeksforGeeks Log in New user ?Register Now Continue with Google or Username or Email Password [x] Remember me Forgot Password Sign In By creating this account, you agree to ourPrivacy Policy&Cookie Policy. Create Account Already have an account ?Log in Continue with Google or Username or Email Password Institution / Organization Sign Up Please enter your email address or userHandle. Back to Login Reset Password
15671
https://www.nuclear-power.com/nuclear-power/reactor-physics/atomic-nuclear-physics/radioactive-decay/radioactive-equilibrium/bateman-equations/
Bateman Equations - Radioactive Decay | nuclear-power.com FacebookInstagramYoutubeTwitter Skip to contentKnowledge Base Nuclear Power Plant Nuclear Physics Reactor Physics Thermal Engineering Materials Radiation Protection Search Search Bateman Equations – Radioactive Decay In physics, the Bateman equations are a set of first-order differential equations which describe the time evolution of nuclide concentrations undergoing serial or linear decay chain. Ernest Rutherford formulated the model in 1905, and the analytical solution for the case of radioactive decay in a linear chain was provided by Harry Bateman in 1910. This model can also be used in nuclear depletion codes to solve nuclear transmutation and decay problems. For example, ORIGEN is a computer code system for calculating radioactive materials’ buildup, decay, and processing. ORIGEN uses a matrix exponential method to solve a large system of coupled, linear, first-order ordinary differential equations (similar to the Bateman equations) with constant coefficients. The Bateman equations for radioactive decay case of n – nuclide series in linear chain describing nuclide concentrations are as follows: Bateman Equations for Nuclear Transmutation Isotopic changes of 4% uranium-235 fuel as a function of fuel burnup. As was written, this model can also be used in nuclear depletion codes to solve nuclear transmutation and decayproblems. In the case of transmutation, the decay constants that govern Bateman equations for a decay case are substituted by transmutation constants. By the transmutation constant λ i,j, we understand the probability of the i th nuclide production per time unit from the j th nuclide destruction as a result of nuclear interaction with the whole spectrum of interacting particles or due to the natural nuclear decay. These equations are usually used for the exact evolution of isotopic changes in the nuclear fuel during fuel depletion. Fuel depletion is usually modeled mathematically as a set of differential equations known as evolution equations. Special Reference: Jerzy Cetnar, General solution of Bateman equations for nuclear transmutations. Annals of Nuclear Energy 33 (2006). January 2006. References: Radiation Protection: Knoll, Glenn F., Radiation Detection and Measurement 4th Edition, Wiley, 8/2010. ISBN-13: 978-0470131480. Stabin, Michael G., Radiation Protection and Dosimetry: An Introduction to Health Physics, Springer, 10/2010. ISBN-13: 978-1441923912. Martin, James E., Physics for Radiation Protection 3rd Edition, Wiley-VCH, 4/2013. ISBN-13: 978-3527411764. U.S.NRC, NUCLEAR REACTOR CONCEPTS U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993. Nuclear and Reactor Physics: J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983). J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317 W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467 G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965 Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414. See above: Radioactive Equilibrium Search for: Your browser does not support the video tag. Privacy Policy Our Website follows all legal requirements to protect your privacy. Visit our Privacy Policy page. The Cookies Statement is part of our Privacy Policy. Editorial note The information contained on this website is for general information purposes only.This website does not use any proprietary data.Visit our Editorial note. Copyright Notice It’s simple: 1) You may use almost everything for non-commercial and educational use. 2) You may not distribute or commercially exploit the content, especially on another website. See: Copyright Notice Contact us: If you want to get in touch with us, please do not hesitate to contact us via e-mail: info@nuclear-power.com © 2025 Nuclear Power • Built with GeneratePress
15672
https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:equilibrium/x2eef969c74e0d802:le-chatelier-s-principle/e/le-chatelier-s-principle
Le Châtelier's principle (practice) | Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. We're a nonprofit that relies on support from people like you. If everyone reading this gives $10 monthly, Khan Academy can continue to thrive for years. Please help keep Khan Academy free, for anyone, anywhere forever. Select gift frequency One time Recurring Monthly Yearly Select amount $10 $20 $30 $40 Other Give now By donating, you agree to our terms of service and privacy policy. Skip to lesson content AP®︎/College Chemistry Course: AP®︎/College Chemistry>Unit 7 Lesson 7: Le Châtelier's principle Le Chȃtelier’s principle: Changing concentration Le Chȃtelier’s principle: Changing volume Le Chȃtelier’s principle: Changing temperature Worked example: Calculating the equilibrium total pressure after a change in volume Worked example: Using Le Chȃtelier’s principle to predict shifts in equilibrium Le Châtelier's principle Science> AP®︎/College Chemistry> Equilibrium> Le Châtelier's principle © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Le Châtelier's principle AP.Chem: TRA‑8 (EU), TRA‑8.A (LO), TRA‑8.A.1 (EK), TRA‑8.A.2 (EK) Google Classroom Microsoft Teams Problem 2 SO A 2(g)+O A 2(g)⇌2 SO A 3(g)‍ After the system above reaches equilibrium, additional O A 2(g)‍ is injected into the reaction flask at a time t 1‍ while the temperature is held constant. Which of the following graphs best shows the rates of the forward and reverse reactions as a function of time? Choose 1 answer: Choose 1 answer: (Choice A) A (Choice B) B (Choice C) C (Choice D) D Related content Video 8 minutes 38 seconds 8:38 Le Chȃtelier’s principle: Changing concentration Video 11 minutes 9 seconds 11:09 Le Chȃtelier’s principle: Changing volume Video 11 minutes 16 seconds 11:16 Le Chȃtelier’s principle: Changing temperature Video 3 minutes 58 seconds 3:58 Worked example: Using Le Chȃtelier’s principle to predict shifts in equilibrium Report a problem Do 4 problems Skip Check Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Allow All Manage Consent Preferences Strictly Necessary Cookies Always Active Certain cookies and other technologies are essential in order to enable our Service to provide the features you have requested, such as making it possible for you to access our product and information related to your account. For example, each time you log into our Service, a Strictly Necessary Cookie authenticates that it is you logging in and allows you to use the Service without having to re-enter your password when you visit a new page or new unit during your browsing session. Functional Cookies [x] Functional Cookies These cookies provide you with a more tailored experience and allow you to make certain selections on our Service. For example, these cookies store information such as your preferred language and website preferences. Targeting Cookies [x] Targeting Cookies These cookies are used on a limited basis, only on pages directed to adults (teachers, donors, or parents). We use these cookies to inform our own digital marketing and help us connect with people who are interested in our Service and our mission. We do not use cookies to serve third party ads on our Service. Performance Cookies [x] Performance Cookies These cookies and other technologies allow us to understand how you interact with our Service (e.g., how often you use our Service, where you are accessing the Service from and the content that you’re interacting with). Analytic cookies enable us to support and improve how our Service operates. For example, we use Google Analytics cookies to help us measure traffic and usage trends for the Service, and to understand more about the demographics of our users. We also may use web beacons to gauge the effectiveness of certain communications and the effectiveness of our marketing campaigns via HTML emails. Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Reject All Confirm My Choices
15673
https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality?srsltid=AfmBOorOECEEl3SK3tqpcfma2ZqEaXTv08wOmWZlTZ8M8JEpMIEay95b
Art of Problem Solving Cauchy-Schwarz Inequality - 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 Cauchy-Schwarz Inequality Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Cauchy-Schwarz Inequality In algebra, the Cauchy-Schwarz Inequality, also known as the Cauchy–Bunyakovsky–Schwarz Inequality or informally as Cauchy-Schwarz, is an inequality with many ubiquitous formulations in abstract algebra, calculus, and contest mathematics. In high-school competitions, its applications are limited to elementary and linear algebra. Its elementary algebraic formulation is often referred to as Cauchy's Inequality and states that for any list of reals and , with equality if and only if there exists a constant such that for all , or if one list consists of only zeroes. Along with the AM-GM Inequality, Cauchy-Schwarz forms the foundation for inequality problems in intermediate and olympiad competitions. It is particularly crucial in proof-based contests. Its vector formulation states that for any vectors and in , where is the dot product of and and is the norm of , with equality if and only if there exists a scalar such that , or if one of the vectors is zero. This formulation comes in handy in linear algebra problems at intermediate and olympiad problems. The full Cauchy-Schwarz Inequality is written in terms of abstract vector spaces. Under this formulation, the elementary algebraic, linear algebraic, and calculus formulations are different cases of the general inequality. Contents 1 Proofs 2 Lemmas 2.1 Complex Form 2.2 A Useful Inequality 3 Real Vector Spaces 3.1 Proof 1 3.2 Proof 2 3.3 Proof 3 4 Complex Vector Spaces 4.1 Proof 5 Problems 5.1 Introductory 5.2 Intermediate 5.3 Olympiad 6 Other Resources 6.1 Books Proofs Here is a list of proofs of Cauchy-Schwarz. Consider the vectors and . If is the angle formed by and , then the left-hand side of the inequality is equal to the square of the dot product of and , or .The right hand side of the inequality is equal to . The inequality then follows from , with equality when one of is a multiple of the other, as desired. Lemmas Complex Form The inequality sometimes appears in the following form. Let and be complex numbers. Then This appears to be more powerful, but it follows from A Useful Inequality Also known as Sedrakyan's Inequality, Bergström's Inequality, Engel's Form or Titu's Lemma the following inequality is a direct result of Cauchy-Schwarz inequality: For any real numbers and where the following is true: Real Vector Spaces Let be a vector space, and let be an inner product. Then for any , with equality if and only if there exist constants not both zero such that . The following proofs assume the inner product to be real-valued and commutative, and so only apply to vector spaces over the real numbers. Proof 1 Consider the polynomial of This must always be greater than or equal to zero, so it must have a non-positive discriminant, i.e., must be less than or equal to , with equality when or when there exists some scalar such that , as desired. Proof 2 We consider Since this is always greater than or equal to zero, we have Now, if either or is equal to , then . Otherwise, we may normalize so that , and we have with equality when and may be scaled to each other, as desired. Proof 3 Consider for some scalar . Then: (by the Trivial Inequality) . Now, let . Then, we have: . Complex Vector Spaces For any two vectors in the complex vector space , the following holds: with equality holding only when are linearly dependent. Proof The following proof, a geometric argument that uses only the algebraic properties of the inner product, was discovered by Tarung Bhimnathwala in 2021. Define the unit vectors , as and . Put . In other words, is the complex argument of and lies on the unit circle. If any of the denominators are zero, the entire result follows trivially. Let and . Importantly, we have Since and , this calculation shows that and form an orthogonal basis of the linear subspace spanned by and . Thus we can think of and as lying on the unit sphere in this subspace, which is isomorphic to . Another thing to note is that The previous two calculations established that and are orthogonal, and that the sum of their squared norms is . Now we have Equality holds when either or , or equivalently when and . Lastly, multiplying each side by , we have Problems Introductory Consider the function , where is a positive integer. Show that . (Source) (APMO 1991 #3) Let , , , , , , , be positive real numbers such that . Show that Intermediate Let be a triangle such that where and denote its semiperimeter and inradius, respectively. Prove that triangle is similar to a triangle whose side lengths are all positive integers with no common divisor and determine those integers. (Source) Olympiad is a point inside a given triangle . are the feet of the perpendiculars from to the lines , respectively. Find all for which is least. (Source) Other Resources Wikipedia entry Books The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities by J. Michael Steele. Problem Solving Strategies by Arthur Engel contains significant material on inequalities. Retrieved from " Categories: Algebra Inequalities 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.
15674
https://chemistry.stackexchange.com/questions/29657/what-are-the-correct-resonance-structures-of-nitrous-oxide
What are the correct resonance structures of nitrous oxide? - Chemistry Stack Exchange Join Chemistry 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 Chemistry helpchat Chemistry 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 Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams 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 What are the correct resonance structures of nitrous oxide? Ask Question Asked 10 years, 5 months ago Modified7 years, 1 month ago Viewed 70k times This question shows research effort; it is useful and clear 19 Save this question. Show activity on this post. There was a question in one of my exam papers to draw the resonance structures for N X 2 O N X 2 O. These are the ones I drew, but they were not awarded marks: The ones in the marking scheme are as follows: Can anyone please explain why mine aren't correct? resonance Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Aug 31, 2017 at 17:43 orthocresol 72.7k 12 12 gold badges 259 259 silver badges 438 438 bronze badges asked May 1, 2015 at 6:42 lucy Glucy G 818 2 2 gold badges 11 11 silver badges 22 22 bronze badges 1 Related: antoine.frostburg.edu/chem/senese/101/inorganic/faq/…Martin - マーチン –Martin - マーチン♦ 2015-05-01 08:40:45 +00:00 Commented May 1, 2015 at 8:40 Add a comment| 3 Answers 3 Sorted by: Reset to default This answer is useful 22 Save this answer. Show activity on this post. Firstly, neither of the resonance structures that you drew for your test are possible because they both violate the octet rule. For the structure on the left, the leftmost N N is in control of only 6 electrons, and will not exist in this form. The structure on the right is not possible because central N N is participating in 5 bonds, which N N cannot do. The maximum number of bonds that N N can participate in is 3 (covalent bonding - like N H X 3 N H X 3) or 4 (coordinate bonding - like N H X 4 X+N H X 4 X+). As long as you satisfy the octet rule (or at least as best as you can - here are some exceptions) and you wish to check the stability of a molecule that you have drawn, simply use formula for formal charge. This is given by the equation: F C=V−(N B+B 2)F C=V−(N B+B 2) Wher F C F C is the formal charge, V V is the number of valence electrons that the atom under consideration usually has, N B N B is the number of non-bonded electrons, and B B is the number of electrons shared in a covalent bond on the atom. If a molecule is stable, the sum of the formal charges of each substituent atom should be 0. If you were to test this on each of the resonance structures that the test provides as answers, you would see that this holds true. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications edited Aug 31, 2017 at 17:44 orthocresol 72.7k 12 12 gold badges 259 259 silver badges 438 438 bronze badges answered May 1, 2015 at 7:22 ringoringo 24.2k 6 6 gold badges 90 90 silver badges 136 136 bronze badges 2 1 'If a molecule is stable, the sum of the formal charges of each substituent atom should be 0' — I'm not even going to try and list the number of ways in which that sentence is wrong. Even if we don't understand stable as a thermodynamic term, it still is wrong for molecular ions.Jan –Jan 2015-05-01 14:27:59 +00:00 Commented May 1, 2015 at 14:27 3 @Jan A molecular ion is not a molecule, it is a charged molecular entity and a molecule is by definition neutral. Hence the only flaw that is present in that sentence is, that it is redundant, as a molecule is also by definition always at least meta stable.Martin - マーチン –Martin - マーチン♦ 2015-05-07 14:06:15 +00:00 Commented May 7, 2015 at 14:06 Add a comment| This answer is useful 16 Save this answer. Show activity on this post. To append Ringo's good answer and to add some more insight into the bonding situation, I performed a calculation on the DF-BP86/def2-SVP level of theory. Since this is a linear molecule, there are symmetry restrictions. Its point group is C∞v C∞v, which means, that there are degenerated orbitals. As we will see, these correspond to π π bonds. Here is an image of the valence orbital scheme: The orbitals 1-3 are core orbitals (1s) at the respective elements. MO 4 corresponds to a O−N σ O−N σ bond. It is noteworthy, that this bond is delocalised and also describes part of the N−N σ N−N σ bond. Analogously this holds for MO 5, but the other way around. In MO 6 we can see the presence of the lone pair at oxygen. MOs 7 and 8 describe two delocalised π π bonds, bonding with respect to all atoms in the compound. MO 9 has mainly lone pair character and is located at the terminal nitrogen. Tho HOMOs are also π π orbitals, bonding with respect to N−N N−N and anti-bonding with respect to O−N O−N. This somewhat implies lone pair character for these orbitals, too. Please note, that there is no orbital with lone pair character at the central nitrogen. I further analysed the wave function with the natural bond orbital theory. This method localises the orbitals (and electron density) and compares it to the ideal electron density of a given Lewis structure. High percentages mean that the configuration agrees well with the Lewis concept. I included the structures you have suggested. You can see, that while configuration 1 is not an ideal Lewis structure, it is still agrees to a certain point with the Lewis concept. If someone would do a more rigorous approach with Valence Bond theory, I would expect this structure to have (at least a minor) contribution to the overall bonding situation. The basic flaw with this structure is the electron sextet at the terminal nitrogen, as Ringo already mentioned. The positive aspect of this structure is, that it does not need formal charges. I would not consider this structure wrong. The second structure, however, is not correct. This can only exist in an excited state. Nitrogen cannot have more than eight electrons binding around it. The third structure is the ideal structure, i.e. the one with the highest agreement. I would expect it to have the highest contribution to the overall bonding. Structure 3 accounts for delocalisation effects and is one of the major contributions to the overall resonance stabilisation of the molecule. Basically you can see the character of this in the MOs 7 and 8 above. The last structure (5) tries the same thing, but for my taste, this is a little bit far fetched. It still agrees somewhat with the Lewis-concept, but like configuration 1 I would only expect a smaller contribution. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered May 12, 2015 at 13:03 Martin - マーチン♦Martin - マーチン 45k 13 13 gold badges 163 163 silver badges 330 330 bronze badges Add a comment| This answer is useful 4 Save this answer. Show activity on this post. Nitrogen cannot expand its octet (it cannot accommodate more than eight valence electrons). In the resonance structures that you have drawn, both the nitrogen and oxygen on the ends are stable (having no charge) but nitrogen in the middle has five bonds hence violates the above rule. So, the maximum number of bonds for nitrogen in the middle is four. That is why the middle nitrogen atom (in the correct answer) has four bonds but with a charge. Also, an expanded octet is generally observed in elements which are located below the second period. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 18, 2018 at 15:58 Parth Chauhan 449 1 1 gold badge 5 5 silver badges 13 13 bronze badges answered May 1, 2015 at 7:33 Simon-Nail-ItSimon-Nail-It 4,731 19 19 gold badges 49 49 silver badges 75 75 bronze badges 7 "Expanded octet is used for elements which are located below period 2 cause they have d-orbitals to expand." Um, what?! How does Iodine expand octet then?M.A.R. –M.A.R. 2015-05-01 08:23:24 +00:00 Commented May 1, 2015 at 8:23 There always some expection Simon-Nail-It –Simon-Nail-It 2015-05-01 08:25:54 +00:00 Commented May 1, 2015 at 8:25 2 How does Bromine and Chlorine expand octet then? The point is, this statement is wrong! Wrong I say!M.A.R. –M.A.R. 2015-05-01 08:27:28 +00:00 Commented May 1, 2015 at 8:27 3 [...] Heck, even H has vacant d orbitals, but it takes a considerable amount of energy to make electrons reach there. So, all in all, this statement is misleading...if not totally based on a wrong premise.M.A.R. –M.A.R. 2015-05-01 14:37:47 +00:00 Commented May 1, 2015 at 14:37 1 @MARamezani: Yes, I can agree with that; the statement being incomplete. I hardly ever address elements based on the number of their period (or even group) but rather according to their relative positions in the periodic table … apparantly usage of that is different around the world.Jan –Jan 2015-05-01 15:44:01 +00:00 Commented May 1, 2015 at 15:44 |Show 2 more comments Protected question. To answer this question, you need to have at least 10 reputation on this site (not counting the association bonus). The reputation requirement helps protect this question from spam and non-answer activity. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions resonance See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked 0What is the hybridization of the oxygen in the nitrous oxide? 34Hypervalency and the octet rule 2Oxidation State of Nitrogen in N2O -4Structure of NO2 molecule Related 27How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine? 17How to determine the worst resonance structures out of a given set? 13Why is the original Pauling's theory of resonance that uses superposition of wavefunctions not used today? 10Difference between Resonance Effect and Mesomeric Effect 1how do electromeric / inductive effect theories compare with resonance 0How many resonance structures does the oxalate ion have? Hot Network Questions Why do universities push for high impact journal publications? What is a "non-reversible filter"? Does the curvature engine's wake really last forever? How different is Roman Latin? Identifying a movie where a man relives the same day Is direct sum of finite spectra cancellative? What is this chess h4 sac known as? Any knowledge on biodegradable lubes, greases and degreasers and how they perform long term? A time-travel short fiction where a graphologist falls in love with a girl for having read letters she has not yet written… to another man What were "milk bars" in 1920s Japan? If Israel is explicitly called God’s firstborn, how should Christians understand the place of the Church? What’s the usual way to apply for a Saudi business visa from the UAE? Is it possible that heinous sins result in a hellish life as a person, NOT always animal birth? What happens if you miss cruise ship deadline at private island? How do trees drop their leaves? The geologic realities of a massive well out at Sea Cannot build the font table of Miama via nfssfont.tex Storing a session token in localstorage Why include unadjusted estimates in a study when reporting adjusted estimates? Why are LDS temple garments secret? Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? How to locate a leak in an irrigation system? Where is the first repetition in the cumulative hierarchy up to elementary equivalence? An odd question Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Chemistry Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
15675
https://www.reddit.com/r/homeassistant/comments/1kglc2f/why_the_new_clock_card_show_leading_zero_for_12hr/
Why the new clock card show leading zero for 12hr time format? : r/homeassistant Skip to main contentWhy the new clock card show leading zero for 12hr time format? : r/homeassistant Open menu Open navigationGo to Reddit Home r/homeassistant A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to homeassistant r/homeassistant r/homeassistant Home Assistant is open source home automation that puts local control and privacy first. Powered by a worldwide community of tinkerers and DIY enthusiasts. Perfect to run on a Raspberry Pi or a local server. Available for free at home-assistant.io. 456K Members Online •5 mo. ago chimph Why the new clock card show leading zero for 12hr time format? I would have thought official integrations follow standard convention. Read more Share Related Answers Section Related Answers Energy monitoring setups using Home Assistant Home Assistant dashboards for beginners Home Assistant automations for pet care Home Assistant and MQTT integration tips Automating window blinds with Home Assistant 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 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
15676
https://www.youtube.com/watch?v=YZviwxCR4Jo
Trigonometric Identities & Sneaky Absolute Values Eddie Woo 1930000 subscribers 119 likes Description 8631 views Posted: 15 Jul 2015 More resources available at www.misterwootube.com 7 comments Transcript: let's have a look at the obvious part okay cot theta square root of one minus cos square theta what are the obvious things that jump out at you that we can simplify you can do cos on sine over there at the front what's happening underneath the square root good so this is this is one of the versions of the pythagorean identity right since we know that sine squared plus cos squared equals one right you can kick either of these guys over the other side in this case they've taken over cos squared right so 1 minus cos squared will just leave you with sine squared right so this is what you've got now you'll see oh i'm taking the square root of a square right so most of you will jump in and just say well clearly the square root of a square is just going to be whatever you are whatever you had there right but yes now here's the thing right and in the middle of a whole bunch of um you know other trig stuff maybe you didn't spot this but the square root of a square number right the square of a square number it's easy it's easy when you're dealing with positive things okay the square root of the square of five is five okay but the square root of the square of minus five is not minus five because you square it you lose the negative then you take the square root and the negative doesn't magically come back both of those are equal to five okay so what this is really saying is that and if you you think back to um when we introduced absolute value okay the square root of a square is in fact the absolute value of that squaring like i said it strips out the fact that it was negative and you lose that information okay so really what's going on here is this okay well then that that that leads to the question how do i do some cancelling business here okay how would i go about thinking about this let's think about forget about it being signed for a second let's just think about a simpler example if it was not trigonometric but if it was just something like this the absolute value of x over x okay now remember the absolute value of x is actually in disguise it's a piecemeal function right it's a piecemeal function what's it actually equal to that absolute value of x it's equal to plus x sometimes and it's equal to minus x other times so it'll be x over x sometimes and it'll be minus x over x other times when is it just equal to plus x but it's greater than zero right and by contrast when it's less than zero right and obviously you can put the um you can put the equality in whichever spot you like at zero okay so therefore this function would be equal to one in this domain and it would be equal to minus one and estimate so what does the graph of this look like we've done these piecemeal functions before right uh no this is a bit tricky okay i'll do the path that's obvious um you're gonna get that and that as for where the filled circle would go oh i know where the filled circle would go can't you see where the field circle would go there is no nowhere it's going to be open on both of them why because you can't have zero yeah good because i've been given it as a denominator it's on the it's on the denominator so i can't ever put it in okay so i've got i've got two hollow circles there's the graph okay now just take what you've seen of this and bring it now over here okay let's just write this in a bit of a clear way slightly you've got this absolute value over itself okay now because it's trig this now makes it a bit trickier okay because this is obvious when it's equal to x and when it's equal to minus x it's pretty clearly defined right now this is also clearly defined but we're not used to thinking in these terms are we when is sinus when is sine x positive when it's above the axis yeah between 0 and 180 okay 0 to 180 in there positive then it goes negative but because it's periodic of course it comes back doesn't it right this would be 360 up to 5 40. the other way right so this is getting towards general solution which we will look at later on okay but the whole point is the whole point is sometimes sign is positive namely in in these spots okay and other times it's negative yes yes it would be so i i just drew this is just sine theta okay so um i'm just trying to work out where it's positive where it's negative so i can get my cases and then i work about my boundary points later on so really what's happening here is you're going to get cos theta times 1 just like just like this okay when sine theta is positive positive right and that is what i've written the first case from not to 180 and then you get it from 360 to 450 and on a 540 and on and on and on yeah wouldn't it be multiplying by like an undefined thing then so wouldn't it not be equal to 0 or 180 uh yes yeah yeah yeah which is which is why i've put in this yeah it should be this not that i mean this this captures all of them really okay um or alternatively right it's equal to negative cos theta when sine theta is negative okay so just quickly before we leave off this what is this just for the sake of curiosity because why not what is this thing actually going to look like it's a piecemeal function right uh let's have a go we'll just do it for positive values okay so for not to 180 that's the first spot okay you're going to get cos theta what does cos theta look like from not to 180 yeah good you get it starts at the top and it comes down to the bottom but you've only gone halfway so it doesn't get to come back it doesn't complete the cycle yep okay now yeah yeah i've got some um yeah i think they're all hollow i'll double check that in a second okay now what happens when i cross over from 180 to 360. well i'm going to get minus cos theta right now this is what normal cos theta looks like right so minus cos theta will flip it upside down right so this guy will come up to here in fact i'm going to get the same thing again okay of course it's still periodic all my pieces are periodic it's just a different kind of periodic okay and in fact you're going to get the same shape because sine theta sine theta is equal to zero at zero and it's equal to zero at one eighty and it's equal to zero here and it's on the denominator no actually oh yeah it is it comes back to your first line right you've got a sine theta on the denominator in your first line so therefore there must be holy circles okay all right because think about cos theta what's the range of cos theta yeah that's right minus one to one okay so therefore no no stay with me stay with me if the range of this thing is minus one to one then when you square it what happens to its range uh something does happen it'll become zero to one that's a positive right so therefore this will always be between zero and one and all of those are okay does that make sense this thing will never ever ever be less than correct
15677
https://www.myphysicslab.com/explain/classify-diff-eq-en.html
Classifying Differential Equations When you study differential equations, it is kind of like botany. You learn to look at an equation and classify it into a certain group. The reason is that the techniques for solving differential equations are common to these various classification groups. And sometimes you can transform an equation of one type into an equivalent equation of another type, so that you can use easier solution techniques. Here then are some of the major classifications of differential equations: On this page we assume that x and y are functions of time, t : x = x(t) y = y(t) And the derivatives are with respect to t | | | --- | | d x | = x'(t) | | d t | Partial vs. Ordinary An ordinary differential equation (or ODE) has a discrete (finite) set of variables. For example in the simple pendulum, there are two variables: angle and angular velocity. A partial differential equation (or PDE) has an infinite set of variables which correspond to all the positions on a line or a surface or a region of space. For example in the string simulation we have a continuous set of variables along the string corresponding to the displacement of the string at each position. In practice we approximate the infinite set of variables with a finite set of variables spread across the string (or surface or volume) at each position. For an ODE, each variable has a distinct differential equation using "ordinary" derivatives. For a PDE, there is only one "partial" differential equation for each dimension. First Order, Second Order The order of a differential equation is equal to the highest derivative in the equation. The single-quote indicates differention. So x' is a first derivative, while x'' is a second derivative. x' = 1/x is first-order x'' = −x is second-order x'' + 2 x' + x = 0 is second-order Linear vs. Non-linear Linear just means that the variable in an equation appears only with a power of one. So x is linear but x2 is non-linear. Also any function like cos(x) is non-linear. In math and physics, linear generally means "simple" and non-linear means "complicated". The theory for solving linear equations is very well developed because linear equations are simple enough to be solveable. Non-linear equations can usually not be solved exactly and are the subject of much on-going research. Here is a brief description of how to recognize a linear equation. Recall that the equation for a line is y = m x + b where m, b are constants ( m is the slope, and b is the y -intercept). In a differential equation, when the variables and their derivatives are only multiplied by constants, then the equation is linear. The variables and their derivatives must always appear as a simple first power. Here are some examples. x'' + x = 0 is linear x'' + 2x' + x = 0 is linear x' + 1/x = 0 is non-linear because 1/x is not a first power x' + x2 = 0 is non-linear because x2 is not a first power x'' + sin(x) = 0 is non-linear because sin(x) is not a first power x x' = 1 is non-linear because x' is not multiplied by a constant Similar rules apply to multiple variable problems. x' + y' = 0 is linear x y' = 1 is non-linear because y' is not multiplied by a constant Note, however, that an exception is made for the time variable t (the variable that we are differentiating by). We can have any crazy non-linear function of t appear in the equation, but still have an equation that is linear in x. x'' + 2 x' + x = sin(t) is linear in x x' + t2x = 0 is linear in x sin(t) x' + cos(t) x = exp(t) is linear in x See the Wikipedia article on linear differential equations for more details. Homogeneous vs. Non-homogeneous This is another way of classifying differential equations. These fancy terms amount to the following: whether there is a term involving only time, t (shown on the right hand side in equations below). x'' + 2_x' + x = 0 is homogeneous x'' + 2_x' + x = sin(t) is non-homogeneous x' + t2x = 0 is homogeneous x' + t2x = t + t2 is non-homogeneous The non-homogeneous part of the equation is the term that involves only time. It usually corresponds to a forcing term in the physical model. For example, in a driven pendulum it would be the motor that is driving the pendulum. This web page was first published June 2001. © Erik Neumann, 2004-2021 revised Dec 19 2023 9:56 previous next
15678
https://www.kuniga.me/blog/2022/08/01/random-points-in-circumference.html
Random Points in Circumference kuniga.me>NP-Incompleteness> Random Points in Circumference Random Points in Circumference 01 Aug 2022 I recently wanted to generate points in the circumference of a unit circle. After some digging, I found a very elegant algorithm from John von Neumann and then a related method for generating normally distributed samples. In this post we’ll describe these methods starting with a somewhat related problem which provides some techniques utilized by the other methods. π π From Random Number Generator First, let’s explore a different problem: how can we estimate π π using only a random number generator? The idea is to define a point with independent random variables as coordinates (X,Y)(X,Y), uniformed sampled from [−1,1][−1,1]. Suppose we sampled values x x and y y. Then we compute the fraction of samples such that x 2+y 2≤1 x 2+y 2≤1. We claim that this ratio will be approximately π 4 π 4! The idea is that when x 2+y 2≤1 x 2+y 2≤1 they’re points inside the unit circle (centered in 0). Whereas (x,y)(x,y) are coordinates of a square of side 2. Since the samples are uniformly distributed, we’d expect that the proportion of points inside the circle over the total will be the ratio of the area of the circle, π r 2 π r 2 over that of the square, ℓ 2 ℓ 2. Since r=1 r=1 and ℓ=2 ℓ=2, we get π 4 π 4, which is approximately plaintext 0.78 . To gain better intuition, we can plot the points. We color the ones satisfying x 2+y 2≤1 x 2+y 2≤1 red, otherwise blue. Figure 1 shows an example with 10,000 samples. Figure 1: plot of 10,000 points with random coordinates each sampled from [-1, 1]. Points inside the unit circle are colored red (source: Observable) I learned about this trick many years ago from Brain Dump (in Portuguese). Random Points in Circumference Now let’s focus on our main problem. One naive idea to solve it using the method we just saw above is to generate sample points and discard all those not satisfying x 2+y 2=1 x 2+y 2=1. This would be incredibly inefficient and in theory the probability of a continuous random variable is actually 0. Another approach is to generate a random number Θ Θ in the interval [0,2 π[[0,2 π[. We can then compute x=cos Θ x=cos⁡Θ and y=sin Θ y=sin⁡Θ which will be in the circumference. It’s possible however to avoid computing sin sin and cos cos altogether, with a method proposed by von Neumann . First we generate a random point inside the unit circle. We just saw how to do this above: define random variables X X and Y Y, uniformly sampled from [−1,1][−1,1]. Discard sample if X 2+Y 2>1 X 2+Y 2>1. Recall that we can represent a point in cartesian coordinates (X,Y)(X,Y) with polar coordinates (R,Θ)(R,Θ), where X=R cos Θ X=R cos⁡Θ and Y=R sin Θ Y=R sin⁡Θ and R=√X 2+Y 2 R=X 2+Y 2. If X X and Y Y are random points in the unit circle, Θ Θ is uniformly distributed over the interval [0,2 π[[0,2 π[. Intuitively, there’s no reason to assume a given angle is more likely if we pick a point at random inside a circle. Notice this is not true if we sample inside a square! If we wish to generate coordinates (X′,Y′)(X′,Y′) on the circumference of the unit circle, we can compute X′=cos Θ X′=cos⁡Θ, which can be obtained as X′=cos Θ=X R=X√X 2+Y 2 X′=cos⁡Θ=X R=X X 2+Y 2 Similarly for Y′=sin Θ Y′=sin⁡Θ: Y′=sin Θ=Y R=Y√X 2+Y 2 Y′=sin⁡Θ=Y R=Y X 2+Y 2 Computing the square root is not ideal, so the trick is to use 2 Θ 2 Θ which is uniformly sampled between [0,4 π[[0,4 π[, but since trigonometric functions have period 2 π 2 π (e.g. sin(x)=sin(x+2 π)sin⁡(x)=sin⁡(x+2 π) and cos(x)=cos(x+2 π)cos⁡(x)=cos⁡(x+2 π)), effectively 2 Θ 2 Θ has the same distribution as Θ Θ! Assuming X′=cos(2 Θ)X′=cos⁡(2 Θ) and using the identity cos(2 Θ)=cos Θ 2−sin Θ 2 cos⁡(2 Θ)=cos⁡Θ 2−sin⁡Θ 2: X′=X 2−Y 2 R 2=X 2−Y 2 X 2+Y 2 X′=X 2−Y 2 R 2=X 2−Y 2 X 2+Y 2 Similarly Y′=sin(2 Θ)=2 sin(x)cos(x)Y′=sin⁡(2 Θ)=2 sin⁡(x)cos⁡(x): Y′=2 X Y R 2=2 X Y X 2+Y 2 Y′=2 X Y R 2=2 X Y X 2+Y 2 Which gets rid of the square root! We have to be careful with small values of X 2+Y 2 X 2+Y 2 in the denominator, but we can discard points when X 2+Y 2≤ϵ X 2+Y 2≤ϵ without affecting the distribution of Θ Θ. Experimentation We can generate multiple points using this method and plot them in the plane to obtain Figure 2. Figure 2: Plotting 400 random points in the circumference of a unit circle, point at the origin added for reference (source: Observable) Complexity The probability of a point falling inside the circle is π 4 π 4, so the number of tries until we find a valid point can be modeled as a geometric distribution with success rate p=π 4 p=π 4, with expected value given by 1 p=4 π=1.27 1 p=4 π=1.27, which is pretty efficient. Amazingly, it’s possible to use this technique to generate points with a normal distribution, as we’ll see next! Generating Normal Distribution If a random variable X X has a normal distribution (σ=1 σ=1, μ=0 μ=0), its cumulative probability distribution (CDF) can be given by: (1)P(X≤x)=F X(x)=√1 2 π∫x−∞e−u 2/2 d u(1)P(X≤x)=F X(x)=1 2 π∫−∞x e−u 2/2 d u Now, consider a random point (X,Y)(X,Y) inside the unit circle. We define (2)N 1=X√−2 ln S S(2)N 1=X−2 ln⁡S S(3)N 2=Y√−2 ln S S(3)N 2=Y−2 ln⁡S S Where S=X 2+Y 2 S=X 2+Y 2. We’ll show that N 1 N 1 and N 2 N 2 are normally distributed independent variables. As before, we represent (X,Y)(X,Y) in polar coordinates (R,Θ)(R,Θ), where X=R cos Θ X=R cos⁡Θ and Y=R sin Θ Y=R sin⁡Θ, R=√X 2+Y 2 R=X 2+Y 2 and thus S=R 2 S=R 2. We can write (2) as N 1=X R√−2 ln S=cos Θ√−2 ln S N 1=X R−2 ln⁡S=cos⁡Θ−2 ln⁡S and (3) as: N 2=Y R√−2 ln S=sin Θ√−2 ln S N 2=Y R−2 ln⁡S=sin⁡Θ−2 ln⁡S The pair (N 1,N 2)(N 1,N 2) forms a cartesian coordinate so they too can be written in polar form: N 1=R′cos Θ′N 1=R′cos⁡Θ′ and N 2=R′sin Θ′N 2=R′sin⁡Θ′, for R′>0 R′>0 and Θ′∈[0,2 π[Θ′∈[0,2 π[. Every cartesian coordinate can be uniquely represented in a polar form as α sin(ϕ)α sin⁡(ϕ) and α cos(ϕ)α cos⁡(ϕ), if α>0 α>0 and ϕ∈[0,2 π[ϕ∈[0,2 π[. So if we ignore samples with S=0 S=0 (making √−2 ln S>0−2 ln⁡S>0) and we can normalize Θ Θ to be ∈[0,2 π[∈[0,2 π[, this allows us to conclude that R′=√−2 ln S R′=−2 ln⁡S and Θ′=Θ Θ′=Θ. Let’s compute the CDF for R′R′, i.e. the probability that R′≤r R′≤r. Since R′=√−2 ln S R′=−2 ln⁡S, we need to find −2 ln S≤r 2−2 ln⁡S≤r 2 or ln S≥−r 2/2 ln⁡S≥−r 2/2. Applying exp()exp⁡() on both sides yields: S≥e−r 2/2 S≥e−r 2/2. We claim that S S is uniformly distributed in the range 0≤x≤1 0≤x≤1 (see Appendix for proof). Thus P(S≤e−r 2/2)=e−r 2/2 P(S≤e−r 2/2)=e−r 2/2 and since P(S≥e−r 2/2)=1−P(S≤e−r 2/2)P(S≥e−r 2/2)=1−P(S≤e−r 2/2) we have P(R′≤r)=1−e−r 2/2 P(R′≤r)=1−e−r 2/2. The probability R′R′ is in the interval [r,r+d r][r,r+d r] can be obtained by the derivative d P(R′≤r)d r d P(R′≤r)d r, which is r e−r 2/2 d r r e−r 2/2 d r. Θ′Θ′ is uniformly distributed in the range [0,2 π[[0,2 π[, so its CDF is F Θ′(θ)=θ 2 π F Θ′(θ)=θ 2 π and the probability Θ′Θ′ is in the interval [θ,θ+d θ][θ,θ+d θ] is d F Θ′(θ)d θ=(1/2 π)d θ d F Θ′(θ)d θ=(1/2 π)d θ. Because R′R′ and Θ′Θ′ are independent variables, we can compute their joint probability more easily, since: P(R′=r,Θ′=θ)=P(R′=r)P(Θ′=θ)=r e−r 2/2 d r(1/2 π)d θ P(R′=r,Θ′=θ)=P(R′=r)P(Θ′=θ)=r e−r 2/2 d r(1/2 π)d θ Thus, to compute the CDF over an interval: P(R′≤r′,Θ′≤θ′)=∫r′0 r e−r 2/2 d r∫θ′0(1/2 π)d θ=1 2 π∫r′0∫θ′0 r e−r 2/2 d r d θ P(R′≤r′,Θ′≤θ′)=∫0 r′r e−r 2/2 d r∫0 θ′(1/2 π)d θ=1 2 π∫0 r′∫0 θ′r e−r 2/2 d r d θ A further generalization is to combine both integrals into one but sum over a pair (r,θ)(r,θ) based on a predicate: 1 2 π∫r′0∫θ′0 r e−r 2/2 d r d θ=1 2 π∫(r,θ):r≤r′,θ≤θ′r e−r 2/2 d r d θ 1 2 π∫0 r′∫0 θ′r e−r 2/2 d r d θ=1 2 π∫(r,θ):r≤r′,θ≤θ′r e−r 2/2 d r d θ This predicate-based integral over a pair form is convenient for what we’ll do next. We now want to compute the joint probability P(N 1≤x 1,N 2≤x 2)P(N 1≤x 1,N 2≤x 2) which can be expressed in polar form: P(N 1≤x 1,N 2≤x 2)=P(R′cos Θ′≤x 1,R′cos Θ′≤x 2)P(N 1≤x 1,N 2≤x 2)=P(R′cos⁡Θ′≤x 1,R′cos⁡Θ′≤x 2) The predicate-based integral over (r,θ)(r,θ) is useful for this because while R′R′ and Θ′Θ′ are independent, the conditions R′cos Θ′≤x 1 R′cos⁡Θ′≤x 1 and R′cos Θ′≤x 2 R′cos⁡Θ′≤x 2 are binding them together, thus we can do: P(R′cos Θ′≤x 1,R′cos Θ′≤x 2)=1 2 π∫(r,θ):r cos Θ≤x 1,r sin Θ≤x 2 r e−r 2/2 d r d θ P(R′cos⁡Θ′≤x 1,R′cos⁡Θ′≤x 2)=1 2 π∫(r,θ):r cos⁡Θ≤x 1,r sin⁡Θ≤x 2 r e−r 2/2 d r d θ We can switch from polar coordinates to cartesian: =1 2 π∫(x,y):x≤x 1,y≤x 2 r e−(x 2+y 2)/2 d x d y=1 2 π∫(x,y):x≤x 1,y≤x 2 r e−(x 2+y 2)/2 d x d y In this form we can switch back to the interval-based integrals: =1 2 π∫x 1−∞∫x 2−∞r e−(x 2+y 2)/2 d x d y=1 2 π∫−∞x 1∫−∞x 2 r e−(x 2+y 2)/2 d x d y Which can be split into a product of integrals: =1 2 π(∫x 1−∞r e−x 2/2 d x)(∫x 2−∞r e−y 2/2 d y)=1 2 π(∫−∞x 1 r e−x 2/2 d x)(∫−∞x 2 r e−y 2/2 d y) Moving the 1/2 π 1/2 π in equal parts to each factor yields: P(N 1≤x 1,N 2≤x 2)=(√1 2 π∫x 1−∞r e−x 2/2 d x)(√1 2 π∫x 2−∞r e−y 2/2 d y)P(N 1≤x 1,N 2≤x 2)=(1 2 π∫−∞x 1 r e−x 2/2 d x)(1 2 π∫−∞x 2 r e−y 2/2 d y) Each factor now represents the CDF for a normal distribution as in (1)! And since the factors are separable their distributions are independent: P(N 1≤x 1,N 2≤x 2)=P(N 1≤x 1)P(N 2≤x 2)P(N 1≤x 1,N 2≤x 2)=P(N 1≤x 1)P(N 2≤x 2) So if we want to generate a normally distributed random value we can generate a sample (x,y)(x,y) in the unit circle and compute a number with either (2) or (3). This method was devided by Box, Marsaglia and Muller and is known as the Polar Method. Experimentation We can generate multiple points using (2) or (3) and build a histogram as in Figure 3. Note the bell-shaped curve which is what we’d expect from sampling from a normal distribution. Figure 3: Points sampled using (2) and plotted in a histogram. The bell shape is indicative of a normal distribution (source: Observable) We can also generate points (x,y)(x,y) with x x sampled from (2) and y y from (3) and plot them in a scatterplot to visualize their correlation as in Figure 4. Notice that the circle shape suggests the distributions used to generate the samples are not correlated. Figure 4: Samples generated with (2) plotted against samples generated with (3). The circular shape is indicative of lack of correlation between the distributions (2) andd (3). (source: Observable) Conclusion The process of investigating the question of generating points on the circumference let me to very interesting findings. I first found an article at MathWorld which provides the formula for the generator but not the theory behind it. It refers to a paper from von Neumman which luckily was available online but somewhat hard to understand. In another search attempt I ran into a Q&A (I forgot the link) where the OP mentioned Knuth’s TAOCP and looking it up led me to find the normally distribution sample via the polar method, which in turn made it easier to understand von Neumann’s method. Appendix Lemma 1.S S is uniformly distributed in the range 0≤r≤1 0≤r≤1. First we compute the CDF P(R≤r)P(R≤r) for r∈[0,1]r∈[0,1]. The odds we pick a point which falls with a circle of radius r r is the ratio of the area of such circle over the area of the unit circle, that is, (4)P(R≤r)=π r 2 π=r 2(4)P(R≤r)=π r 2 π=r 2 We now compute P(S≤r)P(S≤r) for r∈[0,1]r∈[0,1]. Since S=R 2 S=R 2 we have P(S≤r)=P(R 2≤r)=P(−√r≤R≤√r)P(S≤r)=P(R 2≤r)=P(−r≤R≤r). Since R R represents the radius, R≥0 R≥0 and thus P(R≤0)=0 P(R≤0)=0, so P(R≤−√r)=0 P(R≤−r)=0 and P(−√r≤R≤√r)=P(R≤√r)P(−r≤R≤r)=P(R≤r). From (4) we get P(S≤r)=P(R≤√r)=r P(S≤r)=P(R≤r)=r Which is the CDF for uniform distribution. QED. References Various Techniques Used in Connection With Random Digits, J. von Neumann. The Art of Computer Programming: Volume 2, 3.4.1, D. Knuth. (p. 122) Source code | RSS feed | Tags: geometry, probability, puzzle
15679
https://www.youtube.com/watch?v=JiXZBetJKgQ
Art of Problem Solving: Introduction to Infinite Geometric Series Art of Problem Solving 102000 subscribers 8 likes Description 744 views Posted: 9 Jan 2025 Art of Problem Solving's Richard Rusczyk introduces the sum of an infinite geometric series. This video is part of our AoPS Algebra curriculum. Take your math skills to the next level with our Algebra materials: 📚 AoPS Introduction to Algebra Textbook: 🖥️ AoPS Introduction to Algebra A Course (Textbook Chapters 1-13): 🖥️ AoPS Introduction to Algebra B Course (Textbook Chapters 10-22): 🔔 Subscribe to our channel for more engaging math videos and updates Transcript:
15680
https://www.youtube.com/watch?v=TYJ37Od_OFk
Related Rates – Two Cars Traveling in Different Directions John's Solution Set 2200 subscribers 26 likes Description 4029 views Posted: 29 Jun 2023 One car is traveling west and the other is traveling south in this example, and we want to know how fast the distance between them is increasing when t = 2hr. We use related rates and set up a right triangle. Next, we follow the steps to solve related rates problems: Read and re-read the problem to gain an intuitive understanding. Sketch a diagram, if possible. Identify rates that are given and the rate that is to be determined. Write an equation that relates the various quantities of the exercise. If necessary, use geometry to eliminate a variable. Use the Chain Rule to differentiate both sides with respect to t. Substitute known values and solve for the desired quantity. Check that the answer is reasonable. We use the Pythagorean Theorem to solve for the various unknown quantities and get a reasonable answer that the speed at which the distance between the two cars is increasing is 65 mph. ADA-compliant captions/subtitles. 2 comments Transcript: INSTRUCTOR: One standard related rates application that you see often is objects moving in such a way that makes a triangle. This is one such example two cars start moving from the same point. One travels south at 60 miles per hour, and the other travels west at 25 miles per hour. At what rate is the distance between the cars increasing two hours later? So step one is to deeply understand the exercise. And it really helps to draw a figure to help understand what's happening. So here's our starting point. We have one car moving south at 60 miles per hour, and we have another one moving west at 25 miles per hour. So let's call this east-wet distance x and the north-south distance y. And what we're interested in is the distance between them, which we can call z. So here, I have a pretty good idea of what's happening. These two cars are moving in 90-degree directions from each other. And the distance between them is increasing as they keep driving and driving, driving. Let's write down what's given and also, what our goal is. That's a good way to keep track of all this. So given, we have dx dt is a constant speed of 25 miles per hour. So I'll write that as miles per hour. And we have dy dt. That's the rate north-south, 60 miles per hour. And our goal is to find dz dt, the rate at which the distance between the two cars is changing, but when. That's always important for our goal at t equals 2 hours. Let's check in with our steps to see where we stand. OK, step one is done. We understand the problem. We have our diagram. We have our givens and our goal. Step two, we need to write an equation that relates the various quantities. Here, I think we're going to veer a little bit. It says, if necessary, use geometry to eliminate a variable. We can do this without eliminating a variable. You don't always have to eliminate a variable in step two. You can just jump to step three sometimes and be just fine. Use the chain rule to differentiate both sides. So let's do that in this case. Can you think of an equation that relates x, y, and z? You may have thought of the Pythagorean theorem, x squared plus y squared equals z squared. And we can jump straight to the derivative, 2x dx dt plus 2y dy dt. I'm using the chain rule because we're taking the derivative with respect to t instead of x, y, or z, so equals. And we have the chain rule one more time, 2z dz dt. And now we can solve all of this for dz dt. I'm just going to divide everything by 2z. That gives us 2x dx dt-- lots of writing for these-- plus 2y dyu dt, all over 2z. 3 minutes, 22 seconds And we want all of this evaluated at t equals 2 hours. Well, we know what dx dt and dy dt are. Those are given. We just need to find x, y, and z, which we can do. Oh, and look, the 2's cancel out there, as well, so that's nice, 2, 2, 2, save us a little bit of time. Well, we know the speed, so we can compute the distance after two hours. So let's see here. x equals 25 miles per hour. Miles per hour times two hours, that gives us 50 miles. Similarly for y, we have y equals 60 miles per hour times 2 hours again, which gives us 120 miles. So there we have x and y. We just need z. Well, we have this Pythagorean theorem relating x, y, and z. So we can solve for z. Let's take only the positive value, z equals x squared plus y squared, square root of that, square root of 50 squared plus 120 squared. That comes out nicely, it turns out, to 130 miles. So now we plug everything into this dz dt equation. So dz dt finally is-- OK, so we have x, so 50 times dx dt 25 plus y 120 dy dt 60, all over z, which we now know is 130. Computing all that out gives 65 miles per hour. The last thing we do is ask if 65 is a reasonable answer. And it is. You have two cars traveling, one at 25 miles per hour, one at 60 miles per hour. So 65 miles per hour is around what we'd expect for the distance between them. I'd be concerned if it was 1 mile per hour or 300 miles per hour. 5 minutes, 46 seconds
15681
https://chemistry.stackexchange.com/questions/26728/what-is-the-meaning-of-parentheses-in-a-condensed-molecular-formula
organic chemistry - What is the meaning of parentheses in a condensed molecular formula? - Chemistry Stack Exchange Join Chemistry 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 Chemistry helpchat Chemistry 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 Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams 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 What is the meaning of parentheses in a condensed molecular formula? Ask Question Asked 10 years, 7 months ago Modified8 years, 10 months ago Viewed 15k times This question shows research effort; it is useful and clear 5 Save this question. Show activity on this post. In the formula (C H X 3)X 3−C−C H(C H X 3)−C H X 2−C H X 3(C H X 3)X 3−C−C H(C H X 3)−C H X 2−C H X 3, what do the parenthesized portions represent? What is the molecular structure, and how do I name the compound? organic-chemistry nomenclature structural-formula Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited Nov 18, 2016 at 15:27 user7951 asked Feb 28, 2015 at 11:15 MarieMarie 53 1 1 gold badge 1 1 silver badge 4 4 bronze badges 0 Add a comment| 2 Answers 2 Sorted by: Reset to default This answer is useful 8 Save this answer. Show activity on this post. The parentheses just have a grouping function: (C H X 3)X 3−C−(C H X 3)X 3−C− represents a carbon atom to which 3 methyl groups are attached, −C H(C H X 3)−−C H(C H X 3)− is a carbon atom in the chain to which one hydrogen atom and a methyl group are attached. Draw the molecule, find the longest chain for the base name, determine the position of the substituents and you're done. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Feb 28, 2015 at 11:53 Klaus-Dieter WarzechaKlaus-Dieter Warzecha 44.5k 8 8 gold badges 106 106 silver badges 166 166 bronze badges 1 1 I think that the bad bad nasty (CH₃)₃−C− notation shouldn't be encouraged. I could live with (H₃C−)₃≡C−. (But the correct one in Loong's answer.)mykhal –mykhal 2018-10-16 09:43:21 +00:00 Commented Oct 16, 2018 at 9:43 Add a comment| This answer is useful 5 Save this answer. Show activity on this post. The Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), include recommendations for contracted labels (labels representing more than one non-hydrogen atom) and entire formulas (formulas may be considered as labels not connected to any bonds). Such labels and formulas rely on the fact that many elements have consistent and well-understood bonding patterns. They can be interpreted with a simple application of valence rules; i.e. they are interpreted to fill as many valences as possible. For example, when reading the contracted label for a carboxylic acid group (−C O O H−C O O H) from left to right, the carbon atom has three remaining valences. The next atom is oxygen, which has two available valences. Both valences are used to form a double bond with the carbon atom, which leaves the carbon atom with one remaining valence. The third atom is another oxygen, but only one of its valences can be used to create a single bond with the remaining valence on the carbon atom. That fills all of the valences for the carbon atom, but leaves one remaining valence on the oxygen atom. Finally, the fourth atom is a hydrogen atom. It fills the remaining valence on the oxygen atom. Explicit single bonds are generally not necessary within contracted labels and should be avoided. Therefore, the formula (C H X 3)X 3−C−C H(C H X 3)−C H X 2−C H X 3(C H X 3)X 3−C−C H(C H X 3)−C H X 2−C H X 3 that is given in the question should be written as (C H X 3)X 3 C C H(C H X 3)C H X 2 C H X 3(C H X 3)X 3 C C H(C H X 3)C H X 2 C H X 3. Note that the first explicit single bond in (C H X 3)X 3−C−C H(C H X 3)−C H X 2−C H X 3(C H X 3)X 3−C−C H(C H X 3)−C H X 2−C H X 3 is even wrong since a (C H X 3)X 3−C−(C H X 3)X 3−C− group is not possible. Parentheses are used to indicate branching. The corresponding IUPAC recommendations read as follows: Simple branching patterns may be implied by the basic valence rules described above, and do not require special notation. More complex branching may be clarified by placing parentheses around all elements within a branch. One valence for the first element within the parentheses is used for connecting the previous atom outside the parentheses; subsequent atoms within the parenthesized section are then bound to the first or subsequent atoms, even if an atom outside the parentheses has remaining open valences. For example, the −C H(C H X 3)−−C H(C H X 3)− segment of the formula that is given in the question includes parentheses that are placed around the atoms of a methyl branch. One valence of the carbon atom within the parentheses is used to form a bond with the previous carbon atom outside the parentheses. This leaves one remaining valence on the carbon atom outside the parentheses when reading from left to right. The three hydrogen atoms within the parentheses fill the remaining valences on the carbon atom within the parentheses. Therefore, the complete formula (C H X 3)X 3 C C H(C H X 3)C H X 2 C H X 3(C H X 3)X 3 C C H(C H X 3)C H X 2 C H X 3 corresponds to the structure of 2,2,3-trimethylpentane. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Nov 18, 2016 at 15:25 user7951 user7951 Add a comment| Your Answer Reminder: Answers generated by AI tools are not allowed due to Chemistry Stack Exchange's artificial intelligence policy Thanks for contributing an answer to Chemistry Stack Exchange! 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. Use MathJax to format equations. MathJax reference. 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 organic-chemistry nomenclature structural-formula See similar questions with these tags. Featured on Meta Spevacus has joined us as a Community Manager Introducing a new proactive anti-spam measure Linked 3Drawing the condensed structure of ethylcyclobutane -4Are CH₃COOH and HCOOCH₃ same? -1What can we deduce about a (complex, organic) molecule's structure from the order of elements in its formula? Related 4Why doesn't this structure represent a substituted pentane? 1How to correctly write the condensed formula for salicylic acid 1Naming CHCl(NH2)OH 3Drawing the condensed structure of ethylcyclobutane 2Are these alternate names for long chain hydrocarbons with complex substituents accepted? 6How to name aliphatic chain with carboxyl and amide terminal groups? 0What exactly does 'isomerism' mean? Hot Network Questions Transforming wavefunction from energy basis to annihilation operator basis for quantum harmonic oscillator What is this chess h4 sac known as? Can a cleric gain the intended benefit from the Extra Spell feat? How to start explorer with C: drive selected and shown in folder list? Clinical-tone story about Earth making people violent Storing a session token in localstorage Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? Making sense of perturbation theory in many-body physics What's the expectation around asking to be invited to invitation-only workshops? Determine which are P-cores/E-cores (Intel CPU) Calculating the node voltage Matthew 24:5 Many will come in my name! Proof of every Highly Abundant Number greater than 3 is Even What were "milk bars" in 1920s Japan? How many stars is possible to obtain in your savefile? Riffle a list of binary functions into list of arguments to produce a result I have a lot of PTO to take, which will make the deadline impossible An odd question How exactly are random assignments of cases to US Federal Judges implemented? Who ensures randomness? Are there laws regulating how it should be done? Is existence always locational? With with auto-generated local variables Bypassing C64's PETSCII to screen code mapping Can you formalize the definition of infinitely divisible in FOL? Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations? Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Chemistry Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
15682
https://courses.lumenlearning.com/coloradomesa-mathforliberalartscorequisite/chapter/union-intersection-and-complement/
Chapter 4: Logic and Set Theory Set Notation A set can be defined by describing its contents or by listing the elements of the set, enclosed in curly brackets. The symbol ∈ means “is an element of”. A set that contains no elements is called the empty set and is notated ∅. Examples Example 1: The set, A, of whole numbers between 1 and 4 (inclusive) can be written as: A = {1, 2, 3, 4}. To notate that 2 is element of the set, we’d write 2 ∈ A. Example 2: Suppose the universal set is U = {1, 2, 3, 4, 5, 6, 7, 8, 9}. Let A = {1, 2, 4, 6, 9} and B ={2, 4, 6, 8}. Construct the Venn diagram to visualize these sets. Find Ac. Find A ⋃ B. Find A ⋂ B. Show Solution Part 1: Part 2: Ac= {3, 5, 7, 8} Part 3: A ⋃ B ={1, 2, 4, 6, 8, 9} Part 4: A ⋂ B= {2, 4, 6} Try It Example Suppose that the universal set, U= {purple, red, green, blue, brown, yellow, orange}. Consider the sets: A = {red, green, blue} B = {red, yellow, orange} C = {red, orange, yellow, green, blue, purple} Find each of the following sets. Find A ⋃ B Find A ⋂ B Find Ac Find Ac⋂ C Show Solution The union contains all the elements in either set: A ⋃ B = {red, green, blue, yellow, orange}. Notice we only list red once. The intersection contains all the elements in both sets: A ⋂ B = {red} We need to find all of thet elements that are not in the set A but are in the universal set: Ac={yellow, orange, purple, brown} Here we’re looking for all the elements that are in the set Ac (found in part 3) and are also in C: Ac⋂ C = {orange, yellow, purple} Subsets Sometimes a collection might not contain all the elements of a set. For example, Chris owns three Beatles albums. While Chris’s collection is a set, we can also say it is a subset of the larger set of all Beatles albums. Subset The set Bis a subset of the set A if every element in Bis also an element of A. If B is a subset of A, we write B ⊆ A A proper subset is a subset that is not identical to the original set—it contains fewer elements. If B is a proper subset of A, we write B ⊂ A Example Consider these three sets: A = the set of all even numbers B = {2, 4, 6} C = {2, 3, 4, 6} Here B ⊆ A since every element of B is an even number and so is an element of A. More formally, we could say B ⊆ A since if x ∈ B, then x ∈ A. It is also true that B ⊆ C. Notice that C is not a subset of A, since C contains an element, 3, and 3 is not an element of A. See more on this example in the following video. Try It Example Suppose a set contains the plays “Much Ado About Nothing,” “MacBeth,” and “A Midsummer’s Night Dream.” What is a larger set this might be a subset of? Show Solution There are many possible answers here. One would be the set of plays by Shakespeare. This is also a subset of the set of all plays ever written. It is also a subset of all British literature. Example Consider the set [latex]A = {1, 3, 5}[/latex]. Then, [latex]A[/latex] is a subset of which of the following sets?[latex]X = {1, 3, 7, 5}[/latex][latex]Y = {1, 3 }[/latex][latex]Z = {1, m, n, 3, 5}[/latex] Show Solution [latex]X[/latex] and [latex]Z[/latex] Exercises Given the set: A = {a, b, c, d}. List all of the subsets of A Show Solution [latex]\emptyset,\ {a},\ {b},\ {c},\ {d},\ {a,b},\ {b},\ {c},\ {d},\ {a,b},\ {a,c},\ {a,d},\ {b,c},\ {b,d},\ {c,d},\ {a,b,c},\ {a,b,d},[/latex] [latex]{a,c,d},{b,c,d},\ {a,b,c,d}[/latex] You can see that there are 16 subsets. Candela Citations CC licensed content, Original Revision and Adaptation. Provided by: Lumen Learning. License: Public Domain: No Known Copyright Question ID 132343. Provided by: Lumen Learning. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL CC licensed content, Shared previously Math in Society. Authored by: Open Textbook Store, Transition Math Project, and the Open Course Library. Located at: License: CC BY-SA: Attribution-ShareAlike Sets: drawing a Venn diagram. Authored by: David Lippman. Located at: License: CC BY: Attribution Sets: drawing a Venn diagram. Authored by: David Lippman. Located at: License: CC BY: Attribution Question ID 6699. Authored by: Morales,Lawrence. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Question ID 125855. Authored by: Bohart, Jenifer. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Licenses and Attributions CC licensed content, Original Revision and Adaptation. Provided by: Lumen Learning. License: Public Domain: No Known Copyright Question ID 132343. Provided by: Lumen Learning. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL CC licensed content, Shared previously Math in Society. Authored by: Open Textbook Store, Transition Math Project, and the Open Course Library. Located at: License: CC BY-SA: Attribution-ShareAlike Sets: drawing a Venn diagram. Authored by: David Lippman. Located at: License: CC BY: Attribution Sets: drawing a Venn diagram. Authored by: David Lippman. Located at: License: CC BY: Attribution Question ID 6699. Authored by: Morales,Lawrence. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL Question ID 125855. Authored by: Bohart, Jenifer. License: CC BY: Attribution. License Terms: IMathAS Community License CC-BY + GPL
15683
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/The_Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/02%3A_Elements_Atoms_and_the_Periodic_Table/2.02%3A_Atomic_Theory
Skip to main content 2.2: Atomic Theory Last updated : Jun 9, 2025 Save as PDF 2.1: The Elements 2.3: The Structure of Atoms Page ID : 16143 Anonymous LibreTexts ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives Explain how all matter is composed of atoms. Describe the modern atomic theory . Recognize which elements exist as diatomic molecules . Take some aluminum foil. Cut it in half. Now you have two smaller pieces of aluminum foil. Cut one of the pieces in half again. Cut one of those smaller pieces in half again. Continue cutting, making smaller and smaller pieces of aluminum foil. It should be obvious that the pieces are still aluminum foil; they are just becoming smaller and smaller. But how far can you take this exercise, at least in theory ? Can you continue cutting the aluminum foil into halves forever, making smaller and smaller pieces? Or is there some limit, some absolute smallest piece of aluminum foil? (Thought experiments like this—and the conclusions based on them—were debated as far back as the fifth century BC.) The modern atomic theory , proposed about 1803 by the English chemist John Dalton (Figure 2.2.1), is a fundamental concept that states that all elements are composed of atoms. Previously, we defined an atom as the smallest part of an element that maintains the identity of that element. Individual atoms are extremely small; even the largest atom has an approximate diameter of only 5.4 × 10−10 m. With that size, it takes over 18 million of these atoms, lined up side by side, to equal the width of your little finger (about 1 cm). Most elements in their pure form exist as individual atoms. For example, a macroscopic chunk of iron (Fe) metal is composed, microscopically, of individual atoms. Some elements, however, exist as groups of atoms called molecules. Several important elements exist as two-atom combinations and are called diatomic molecules. In representing a diatomic molecule , we use the symbol of the element and include the subscript "2" to indicate that two atoms of that element are joined together. The elements that exist as diatomic molecules are hydrogen (H2), nitrogen (N2), fluorine (F2), oxygen (O2), iodine (I2), chlorine (Cl2) and bromine (Br2). A few other elements can exist as 3-atom molecules like ozone (O3) and 4-atom molecules like phosphorus (P4). The most common form of the element sulfur is composed of molecules that consist of eight atoms of sulfur; its molecular formula is S8 (Figure 2.2.2). It is important to note that a subscript following a symbol and a number in front of a symbol do not represent the same thing; for example, H2 and 2H represent distinctly different species. H2 is a molecular formula; it represents a diatomic molecule of hydrogen, consisting of two atoms of the element that are chemically bonded together. The expression 2H, on the other hand, indicates two separate hydrogen atoms that are not combined as a unit . The expression 2H2 represents two molecules of diatomic hydrogen (Figure 2.2.3). Example 2.2.1 Write the chemical formula of the following elements: oxygen carbon potassium Solution Memorizing the diatomic molecules is worthwhile for our future endeavors. A mnemonic device to help in the memorization of the diatomic elements is as follows: Have No Fear Of Ice Cold Beer. (hydrogen, nitrogen, fluorine, oxygen, iodine, chlorine and bromine). Oxygen is one of the seven diatomic molecular elements. Its chemical formula is O2. 2. Carbon is monatomic, hence its formula is C. 3. Potassium is monatomic hence its formula is K. Exercise 2.2.1 Write the chemical formula of the following elements: hydrogen nitrogen neon Answer a : Hydrogen is one of the seven diatomic molecular elements. Its chemical formula is H2. Answer b : Nitrogen is one of the seven diatomic molecular elements. Its chemical formula is N2. Answer c : Neon is a monatomic element, hence its formula is Ne. Looking Closer: Atomic Theory Dalton’s ideas are called the modern atomic theory because the concept of atoms is very old. The Greek philosophers Leucippus and Democritus originally introduced atomic concepts in the fifth century BC. (The word atom comes from the Greek word atomos, which means “indivisible” or “uncuttable.”) Dalton had something that the ancient Greek philosophers didn’t have, however; he had experimental evidence, such as the formulas of simple chemicals and the behavior of gases. In the 150 years or so before Dalton, natural philosophy had been maturing into modern science , and the scientific method was being used to study nature. So when Dalton announced a modern atomic theory , he was proposing a fundamental theory to describe many previous observations of the natural world; he was not just participating in a philosophical discussion. Concept Review Exercises What is the modern atomic theory ? 2. What are atoms? Answers The modern atomic theory states that all matter is composed of atoms. 2. Atoms are the smallest parts of an element that maintain the identity of that element. Key Takeaways Atoms are the ultimate building blocks of all matter . The modern atomic theory establishes the concepts of atoms and how they compose matter . The diatomic elements are hydrogen, nitrogen, fluorine, oxygen, iodine, chlorine and bromine. 2.1: The Elements 2.3: The Structure of Atoms
15684
https://medium.com/@yusufaksoyeng/finding-the-eulerian-cycle-with-hierholzers-algorithm-f60bb773db3c
Sitemap Open in app Sign in Sign in Finding the Eulerian Cycle with Hierholzer’s Algorithm Yusuf Aksoy 4 min readApr 26, 2022 In this blog post, I would like to explain the Eulerian Path/Cycle and how we can find the Eulerian Cycle with Hierholzer’s Algorithm by giving an example. Seven Bridges of Königsberg It all started with investigating the possibility of walking in Königsberg (now Kaliningrad) and crossing each bridge only once. ## Seven Bridges of Königsberg - Wikipedia The Seven Bridges of Königsberg is a historically notable problem in mathematics. Its negative resolution by Leonhard… en.wikipedia.org To reduce the problem to a graph, we can consider each land as a vertex and each bridge as an edge. Euler analyzed this problem and concluded. He realized that if one enters a vertex by the bridge, one leaves the same vertex by the bridge also. In conclusion, the number of the in-going edges to the vertex should be equal to out-going edges from the same vertex. We can represent the problem with a simple graph. As can be seen from the graph, one vertex has five edges and the other three. According to the observation of Euler, there is no solution to the problem, this was later proved by Carl Hierholzer. What are the Eulerian Path and Eulerian Cycle? According to Wikipedia, Eulerian Path (also called Eulerian Trail) is a path in a finite graph that visits every edge exactly once. The path may be started and ended at different graph vertices. As a precondition, all the non-zero degree vertices have to be connected as a single component for the Eulerian Path. There were puzzles in my childhood where you have to draw the shape on a paper without lifting your pencil and without going over what you have drawn before. Such puzzles must have the Euler Path to be solved. On the other hand, there is a concept named Eulerian Circuits (or Eulerian Cycle) that restricts Eulerian Path conditions further. It is still an Eulerian Path and it starts and ends at the same vertex. The degree of a vertex is the number of edges it has. In-degree edge means that an edge of a vertex is coming into the vertex and an out-degree edge is going out from the vertex. There are Euler Path conditions that graphs must have: For an undirected graph Every vertex should have an even degree or only two vertices should have odd degrees. For a directed graph Each vertex should have the same in-degree and out-degree except for two of them. One of these vertex will be the start vertex which has one more out-going edge than in-going edges. The other one will be the end vertex which has one more in-going edge than out-going edges. The following conditions determine whether a graph has an Eulerian Cycle or not on top of Eulerian Path conditions: For an undirected graph Each vertex should have an even degree. For a directed graph Every vertex should have equal in-degree and out-degree edges. Hierholzer’s Algorithm Now we know what Eulerian Cycle is, but how can we find it? In 1873, Hierholzer proposed an algorithm to find the Eulerian Cycle in linear time. The algorithm can be applied to both undirected and directed graphs. In Hierholzer Algorithm, we can follow the steps below to find the Eulerian Cycle in a graph: We can choose any arbitrary vertex as a starting point, We follow the outgoing edges of the vertex that we haven’t followed before and traverse the vertices (Initially, all the edges are unfollowed of course). We can follow whichever edge we want, the only rule is not to follow previously traversed ones. We should apply Step 2 until we are stuck. At some point, we will visit a vertex and there will be no edges to follow. Remember that Eulerian Cycle properties, every vertex should have even degrees or equal in-out degrees. If we are stuck the first time it means that we formed a cycle, and the vertex that we are stuck is starting vertex. we returned where we started. Push the vertex that we stuck to the top of the stack data structure which holds the Eulerian Cycle. Backtrack from this vertex to the previous one. If there are edges to follow, we have to return to Step 2. If there are no vertices left to traverse, now the stack holds the complete Eulerian Cycle, and we are done. Otherwise, we have to follow Step 5. Example A concrete example would make the algorithm clear. Let’s consider we are tourists and traveling to cities. The example starts from Istanbul and follows all the edges and returns to Istanbul again. It traverses the graph with a depth-first search and backtracks to the previously followed node if encounters a dead end. Note that the example may not be the best implementation, but it gives an idea of ​​how the algorithm works. The time complexity of this example is O(E), where E is the number of edges, since it traverses each edge once. This prints: Start --> Istanbul --> Madrid --> London --> Madrid --> Paris --> Madrid --> Istanbul --> Paris --> London --> Paris --> Berlin --> London --> Berlin --> Rome --> Berlin --> Paris --> Istanbul --> Berlin --> Istanbul --> Rome --> Istanbul --> End You can also try out the program in replit: ## EulerianCycleFinder A Nix (beta) repl by YusufAksoy2 replit.com References Thank you for reading. Graph Theory Euleriantrail DFS Backtracking Cpp ## Written by Yusuf Aksoy 33 followers ·55 following Software Engineer Responses (1) Write a response What are your thoughts? Jonathanchen Jul 3, 2024 ``` I believe your IsEulerian function doesn't take into account that in a directed graph, two nodes do not need to have equal in/out degrees. In a directed graph, the graph can still contain an eulerian circuit if there are two nodes where one node's… ``` More from Yusuf Aksoy Yusuf Aksoy ## Disjoint-Set (Union-Find) Data Structure with Optimizations In this blog post, I would like to introduce disjoint-set data structure, also known as union-find data structure, and implementations with… May 30, 2022 85 Yusuf Aksoy ## Function Objects (Functors) in C++ In this blog post, I would like to explain what are function objects and how they can be used to solve different problems by giving… Apr 7, 2022 144 1 See all from Yusuf Aksoy Recommended from Medium ThreadSafe Diaries ## He Was a Senior Developer, Until We Read His Pull Request When experience doesn’t translate to expertise, and how one code review changed everything Aug 3 5.3K 158 In Coding Beauty by Tari Ibaba ## This new IDE from Google is an absolute game changer This new IDE from Google is seriously revolutionary. Mar 11 6.2K 366 D4rsh ## Everything you need to know about Graph Theory The complete guide to graph data structures spanning fundamental principles, advanced algorithms, and effective implementation strategies Aug 2 480 4 Jordan Gibbs ## ChatGPT Is Poisoning Your Brain… Here‘s How to Stop It Before It’s Too Late. Apr 29 24K 1203 Rishabh Singh ## Classic Graph Algorithms (C++) Breadth First Search (BFS): Mar 3 4 In by Aditi Saxena ## 🛣️ Dijkstra’s Algorithm In Java: The Dev-Friendly Guide to Shortest Paths Learn how to implement Dijkstra’s Algorithm in Java using PriorityQueue, understand real-world use cases, and ace graph problems in coding… Jun 14 2 1 See more recommendations Text to speech
15685
https://dynref.engr.illinois.edu/rvs.html
Spherical coordinates Dynamics Home Reference Applications Spherical coordinates#rvs The spherical coordinate system extends polar coordinates into 3D by using an angle ϕ ϕ for the third coordinate. This gives coordinates (r,θ,ϕ)(r,θ,ϕ) consisting of: | coordinate | name | range | definition | --- --- | | r r | radius | 0≤r<∞0≤r<∞ | distance from the origin | | θ θ | azimuth | −π<θ≤π−π<θ≤π | angle from the x x-axis in the x x–y y plane | | ϕ ϕ | inclination | 0≤ϕ≤π 0≤ϕ≤π | angle from the positive z z axis | The diagram below shows the spherical coordinates of a point P P. By changing the display options, we can see that the basis vectors are tangent to the corresponding coordinate lines. Changing θ θ moves P P along the θ θ coordinate line in the direction ^e θ e^θ, and similarly for the other coordinates. reset Show:labels coordinates basis vectors Coordinate lines:r r θ θ ϕ ϕ Radius:r=r=4 Azimuth:θ=θ=45° Inclination:ϕ=ϕ=45° Spherical coordinates are defined with respect to a set of Cartesian coordinates, and can be converted to and from these coordinates using the atan2 function as follows. Conversion between spherical and Cartesian coordinates #rvs‑ec x=r cos θ sin ϕ r=√x 2+y 2+z 2 y=r sin θ sin ϕ θ=atan2(y,x)z=r cos ϕ ϕ=arccos(z/r)x=r cos⁡θ sin⁡ϕ r=x 2+y 2+z 2 y=r sin⁡θ sin⁡ϕ θ=atan2⁡(y,x)z=r cos⁡ϕ ϕ=arccos⁡(z/r) +Derivation#rvs‑ec‑d To find the conversion to Cartesian coordinates, we consider the projection of ⃗r r→ down onto the x x–y y plane, as shown. Then z=r cos ϕ z=r cos⁡ϕ and ℓ=r sin ϕ ℓ=r sin⁡ϕ, from which we obtain x=ℓ cos θ x=ℓ cos⁡θ and y=ℓ sin θ y=ℓ sin⁡θ. To convert from Cartesian coordinates, we use the same projection and read off the expressions for the spherical coordinates. This uses the atan2 function to handle the different quadrants for θ θ. The basis vectors are tangent to the coordinate lines and form an orthonormal basis ^e r,^e θ,^e ϕ e^r,e^θ,e^ϕ that depends on the current position P P as follows. Spherical basis vectors #rvs‑eb ^e r=cos θ sin ϕ^ı+sin θ sin ϕ^ȷ+cos ϕ^k^e θ=−sin θ^ı+cos θ^ȷ^e ϕ=cos θ cos ϕ^ı+sin θ cos ϕ^ȷ−sin ϕ^k^ı=cos θ sin ϕ^e r−sin θ^e θ+cos θ cos ϕ^e ϕ^ȷ=sin θ sin ϕ^e r+cos θ^e θ+sin θ cos ϕ^e ϕ^k=cos ϕ^e r−sin ϕ^e ϕ e^r=cos⁡θ sin⁡ϕ ı^+sin⁡θ sin⁡ϕ ȷ^+cos⁡ϕ k^e^θ=−sin⁡θ ı^+cos⁡θ ȷ^e^ϕ=cos⁡θ cos⁡ϕ ı^+sin⁡θ cos⁡ϕ ȷ^−sin⁡ϕ k^ı^=cos⁡θ sin⁡ϕ e^r−sin⁡θ e^θ+cos⁡θ cos⁡ϕ e^ϕ ȷ^=sin⁡θ sin⁡ϕ e^r+cos⁡θ e^θ+sin⁡θ cos⁡ϕ e^ϕ k^=cos⁡ϕ e^r−sin⁡ϕ e^ϕ +Derivation#rvs‑eb‑d We write the position vector ⃗r=r cos θ sin ϕ^ı+r sin θ sin ϕ^ȷ+r cos ϕ^k r→=r cos⁡θ sin⁡ϕ ı^+r sin⁡θ sin⁡ϕ ȷ^+r cos⁡ϕ k^ and then use the definition of coordinate basis vectors to find the non-normalized spherical basis vectors: ⃗e r=∂⃗r∂r=cos θ sin ϕ^ı+sin θ sin ϕ^ȷ+cos ϕ^k⃗e θ=∂⃗r∂θ=−r sin θ sin ϕ^ı+r cos θ sin ϕ^ȷ⃗e ϕ=∂⃗r∂ϕ=r cos θ cos ϕ^ı+r sin θ cos ϕ^ȷ−r sin ϕ^k e→r=∂r→∂r=cos⁡θ sin⁡ϕ ı^+sin⁡θ sin⁡ϕ ȷ^+cos⁡ϕ k^e→θ=∂r→∂θ=−r sin⁡θ sin⁡ϕ ı^+r cos⁡θ sin⁡ϕ ȷ^e→ϕ=∂r→∂ϕ=r cos⁡θ cos⁡ϕ ı^+r sin⁡θ cos⁡ϕ ȷ^−r sin⁡ϕ k^ To normalize these vectors we divide by their lengths, which we can compute to be ∥⃗e r∥=1‖e→r‖=1, ∥⃗e θ∥=r sin ϕ‖e→θ‖=r sin⁡ϕ, and ∥⃗e ϕ∥=r‖e→ϕ‖=r. To invert the basis change we first observe that we can take combinations of ^e r e^r and ^e ϕ e^ϕ to give: cos ϕ^e r−sin ϕ^e ϕ=^k sin ϕ^e r+cos ϕ^e ϕ=cos θ^ı+sin θ^ȷ cos⁡ϕ e^r−sin⁡ϕ e^ϕ=k^sin⁡ϕ e^r+cos⁡ϕ e^ϕ=cos⁡θ ı^+sin⁡θ ȷ^ Then: cos θ(sin ϕ^e r+cos ϕ^e ϕ)−sin θ^e θ=cos 2 θ^ı+sin 2 θ^ı=^ı sin θ(sin ϕ^e r+cos ϕ^e ϕ)+cos θ^e θ=sin 2 θ^ı+cos 2 θ^ȷ=^ȷ cos⁡θ(sin⁡ϕ e^r+cos⁡ϕ e^ϕ)−sin⁡θ e^θ=cos 2⁡θ ı^+sin 2⁡θ ı^=ı^sin⁡θ(sin⁡ϕ e^r+cos⁡ϕ e^ϕ)+cos⁡θ e^θ=sin 2⁡θ ı^+cos 2⁡θ ȷ^=ȷ^ Rearranging these gives the Cartesian basis vector expressions above. +Warning: (^e r,^e θ,^e ϕ)(e^r,e^θ,e^ϕ) is not right-handed#rvs‑wr Although it is common to write the spherical coordinates in the order (r,θ,ϕ)(r,θ,ϕ), this order gives a left-handed basis (^e r,^e θ,^e ϕ)(e^r,e^θ,e^ϕ), which we can see graphically from the fact that ^e r×^e θ=−^e ϕ e^r×e^θ=−e^ϕ. We can either work with this as a left-handed basis, or re-order the coordinates to give the right-handed basis (^e r,^e ϕ,^e θ)(e^r,e^ϕ,e^θ). If the spherical coordinates change with time then this causes the spherical basis vectors to rotate with the following angular velocity. Angular velocity of the spherical basis #rvs‑ew ⃗ω=˙ϕ^e θ+˙θ^k=˙θ cos ϕ^e r+˙ϕ^e θ−˙θ sin ϕ^e ϕ ω→=ϕ˙e^θ+θ˙k^=θ˙cos⁡ϕ e^r+ϕ˙e^θ−θ˙sin⁡ϕ e^ϕ +Derivation#rvs‑ew‑d Changing r r does not cause a rotation of the basis, while changing θ θ rotates about the vertical axis ^k k^ and changing ϕ ϕ rotates about ^e θ e^θ. Combining these angular velocities gives the expression for ⃗ω ω→. The rotation of the basis vectors caused by changing coordinates can be directly computed, giving the time derivatives below. Time derivatives of spherical basis vectors #rvs‑et ˙^e r=˙θ sin ϕ^e θ+˙ϕ^e ϕ˙^e θ=−˙θ sin ϕ^e r−˙θ cos ϕ^e ϕ˙^e ϕ=−˙ϕ^e r+˙θ cos ϕ^e θ e^˙r=θ˙sin⁡ϕ e^θ+ϕ˙e^ϕ e^˙θ=−θ˙sin⁡ϕ e^r−θ˙cos⁡ϕ e^ϕ e^˙ϕ=−ϕ˙e^r+θ˙cos⁡ϕ e^θ +Derivation#rvs‑et‑d We can either directly differentiate the basis vector expressions, or we can recall that ˙^e=⃗ω×^e e^˙=ω→×e^ for any basis vector ^e e^. This gives: ˙^e r=⃗ω×^e r=˙θ cos ϕ^e r×^e r+˙ϕ^e θ×^e r−˙θ sin ϕ^e ϕ×^e r˙^e θ=⃗ω×^e θ=˙θ cos ϕ^e r×^e θ+˙ϕ^e θ×^e θ−˙θ sin ϕ^e ϕ×^e θ˙^e ϕ=⃗ω×^e ϕ=˙θ cos ϕ^e r×^e ϕ+˙ϕ^e θ×^e ϕ−˙θ sin ϕ^e ϕ×^e ϕ e^˙r=ω→×e^r=θ˙cos⁡ϕ e^r×e^r+ϕ˙e^θ×e^r−θ˙sin⁡ϕ e^ϕ×e^r e^˙θ=ω→×e^θ=θ˙cos⁡ϕ e^r×e^θ+ϕ˙e^θ×e^θ−θ˙sin⁡ϕ e^ϕ×e^θ e^˙ϕ=ω→×e^ϕ=θ˙cos⁡ϕ e^r×e^ϕ+ϕ˙e^θ×e^ϕ−θ˙sin⁡ϕ e^ϕ×e^ϕ Now we evaluate the cross products graphically to obtain the final expressions. A point P P at a time-varying position (r,θ,ϕ)(r,θ,ϕ) has position vector ⃗r r→, velocity ⃗v=˙⃗r v→=r→˙, and acceleration ⃗a=¨⃗r a→=r→¨ given by the following expressions in spherical components. Position, velocity, and acceleration in spherical components #rvs‑ep ⃗r=r^e r⃗v=˙r^e r+r˙θ sin ϕ^e θ+r˙ϕ^e ϕ⃗a=(¨r−r˙θ 2 sin 2 ϕ−r˙ϕ 2)^e r+(r¨θ sin ϕ+2˙r˙θ sin ϕ+2 r˙θ˙ϕ cos ϕ)^e θ+(r¨ϕ+2˙r˙ϕ−r˙θ 2 sin ϕ cos ϕ)^e ϕ r→=r e^r v→=r˙e^r+r θ˙sin⁡ϕ e^θ+r ϕ˙e^ϕ a→=(r¨−r θ˙2 sin 2⁡ϕ−r ϕ˙2)e^r+(r θ¨sin⁡ϕ+2 r˙θ˙sin⁡ϕ+2 r θ˙ϕ˙cos⁡ϕ)e^θ+(r ϕ¨+2 r˙ϕ˙−r θ˙2 sin⁡ϕ cos⁡ϕ)e^ϕ +Derivation#rvs‑ep‑d Because ^e r e^r is a unit vector in the direction of the position vector ⃗r r→, we know that ⃗r=r^e r r→=r e^r. Then we can differentiate this expression to obtain: ⃗v=˙⃗r=d d t(r^e r)=˙r^e r+r˙^e r v→=r→˙=d d t(r e^r)=r˙e^r+r e^˙r and we substitute in the expression for ˙^e r e^˙rfrom above. Taking another derivative gives: ⃗a=˙⃗v=d d t(˙r^e r+r˙θ sin ϕ^e θ+r˙ϕ^e ϕ)=¨r^e r+˙r˙^e r+(˙r˙θ sin ϕ+r¨θ sin ϕ+r˙θ cos ϕ˙ϕ)^e θ+r˙θ sin ϕ˙^e θ+(˙r˙ϕ+r¨ϕ)^e ϕ+r˙ϕ˙^e ϕ a→=v→˙=d d t(r˙e^r+r θ˙sin⁡ϕ e^θ+r ϕ˙e^ϕ)=r¨e^r+r˙e^˙r+(r˙θ˙sin⁡ϕ+r θ¨sin⁡ϕ+r θ˙cos⁡ϕ ϕ˙)e^θ+r θ˙sin⁡ϕ e^˙θ+(r˙ϕ˙+r ϕ¨)e^ϕ+r ϕ˙e^˙ϕ and again we can substitute the basis vector derivatives. Notation note#rvs‑ic There are many different conventions for spherical coordinates notation, so it's important to check which variant is being used in any document. The convention used here is common in mathematics. In physics it is also common to use the same angles, but to reverse the symbol convention so that ϕ ϕ is the azimuth and θ θ is the inclination. Many different names for the coordinates are also used, with the inclination also being called the zenith angle, polar angle, or normal angle. Alternatively, the inclination may be replaced by the elevation measured up from the x x-y y plane. In geographic coordinates the azimuth is the longitude λ λ and the elevation is the latitude ϕ ϕ, while in celestial coordinates the azimuth is the right ascension α α and the elevation is the declination δ δ. The angles themselves may be given in radians or degrees, and are sometimes written in degrees (symbol °), arc-minutes (1/60 of a degree, symbol ′) and arc-seconds (1/60 of an arc-minute, symbol ″). Positive and negative angles are normally indicated by +/-, but sometimes use North/South (N/S) or East/West (E/W). Thus a longitude may be written as λ λ = 88°12′15″W = 88.2042°W = -88.2042° = -1.53945 rad. Related applications#rvs‑ia Shortest flight paths Copyright (C) 2012-2015 Matthew West
15686
https://mhchem.org/221/221PPT/pdf221/IsotopicAbundanceHandout.pdf
carbon 6 C 12.011 Isotopic Abundance Handout and Practice Problems Much of the information in this handout was adapted from the handouts "isotopic abundance - practice problems" (http:// maurermath.weebly.com/uploads/7/0/2/8/7028280/average_atomic_mass_and_percent_abundance.pdf) and "Calculate the isotopic abundances when given the average atomic weight and the isotopic weights" ( - my deep thanks to these authors for their wonderful content! The atomic mass for each element appearing on the periodic table represents the weighted average of masses for each individual isotope of an element. For example, the atomic mass of carbon is reported as 12.011 amu (atomic mass units). Carbon is composed primarily of two isotopes; carbon-12 and carbon-14. The atomic mass is calculated using both the relative abundance and the masses for each of these two isotopes. Using the equation below, the atomic mass for carbon can be calculated. atomic mass = (mass 1 × % 1) + (mass 2 × % 2 ) + . . . Carbon-12 accounts for 99.45% of all of the carbon atoms, while carbon-14 only accounts for the remaining 0.55%. Since the carbon-12 isotope is more abundant, its mass is weighted more in the calculation of carbon’s atomic mass. The calculation of the atomic mass is shown below. atomic mass = (12.000 × 0.9945) + (14.003 × 0.0055) atomic mass = (11.934) + (0.077) = 12.011 amu Directions: Use the equation for atomic mass to complete the following problems. 1. Argon has three naturally occurring isotopes: argon-36, argon-38, and argon-40. Based on argon’s reported atomic mass, which isotope exists as the most abundant in nature? Explain. 2. Copper exists as a mixture of two isotopes. Copper-63 is 69.17% abundant and it has a mass of 62.9296 amu. Copper-65 is 30.83% abundant and it has a mass of 64.9278 amu. Calculate the atomic mass (am) of copper. 3. Calculate the atomic mass (am) of silicon. The three silicon isotopes have atomic masses and relative abundances of 27.9769 amu (92.2297%), 28.9765 amu (4.6832%) and 29.9738 amu (3.0872%). 4. Gallium has two naturally occurring isotopes. The mass of gallium-69 is 68.9256 amu and it is 60.108% abundant. The mass of gallium-71 is 70.9247 amu and it is 39.892% abundant. Calculate the atomic mass (am) of gallium. 5. Bromine has two naturally occurring isotopes. Bromine-79 has a mass of 78.918 amu and is 50.69% abundant. Using the atomic mass reported on the periodic table (79.904), determine the mass of bromine-81, the other isotope of bromine. 6. Calculate the atomic mass of lead. The four lead isotopes have atomic masses and relative abundances of 203.973 amu (1.4000%), 205.974 amu (24.1010%), 206.976 amu (22.1000%) and 207.977 amu (52.3990%). 7. Antimony has two naturally occurring isotopes. The mass of antimony-121 is 120.904 amu and the mass of antimony-123 is 122.904 amu. Using the average mass from the periodic table, calculate the abundance of each isotope. Page IV-2a- / Isotopic Abundance Handout Study Guide 3 isotope % abundance mass (amu) carbon-12 99.45 12.000 carbon-14 0.55 14.003 Answers: 1. Argon has three naturally occurring isotopes: argon-36, argon-38, and argon-40. Based on argon’s reported atomic mass, which isotope exists as the most abundant in nature? Explain. argon has an atomic mass of 39.948 amu, which is closet to argon-40, suggesting that it is the most abundant isotope 2. Copper exists as a mixture of two isotopes. Copper-63 is 69.17% abundant and it has a mass of 62.9296 amu. Copper-65 is 30.83% abundant and it has a mass of 64.9278 amu. Calculate the atomic mass (am) of copper. am = (62.9296 x 0.6917) + (64.9278 x 0.3083) = 63.545645... amu = 63.55 amu 3. Calculate the atomic mass (am) of silicon. The three silicon isotopes have atomic masses and relative abundances of 27.9769 amu (92.2297%), 28.9765 amu (4.6832%) and 29.9738 amu (3.0872%). am = (27.9769 x 0.922297) + (28.9765 x 0.046832) + (29.9738 x 0.030872) = 28.0853895... amu = 28.085 amu 4. Gallium has two naturally occurring isotopes. The mass of gallium-69 is 68.9256 amu and it is 60.108% abundant. The mass of gallium-71 is 70.9247 amu and it is 39.892% abundant. Calculate the atomic mass (am) of gallium. am = (68.9256 x 0.60108) + (70.9247 x 0.39892) = 69.723080... amu = 69.723 amu 5. Bromine has two naturally occurring isotopes. Bromine-79 has a mass of 78.918 amu and is 50.69% abundant. Using the atomic mass reported on the periodic table (79.904), determine the mass of bromine-81, the other isotope of bromine. 79.904 = (78.918 x 0.5069) + (mass2 x 0.4931) → 39.9005 = mass20.4931 → mass2 = 80.91766... amu = 80.92 amu 6. Calculate the atomic mass of lead. The four lead isotopes have atomic masses and relative abundances of 203.973 amu (1.4000%), 205.974 amu (24.1010%), 206.976 amu (22.1000%) and 207.977 amu (52.3990%). atomic mass = (203.973 x 0.014000) + (205.974 x 0.241010) + (206.976 x 0.221000) + (207.977 x 0.523990) = 207.21169... amu = 207.22 amu 7. Antimony has two naturally occurring isotopes. The mass of antimony-121 is 120.904 amu and the mass of antimony-123 is 122.904 amu. Using the average mass from the periodic table, calculate the abundance of each isotope. 121.760 = 120.904(x) + 122.904(1-x) → 121.760 = 120.904x + 122.904 - 122.904x → -1.144 = -2.000x → x = 0.5720 121Sb = 57.20 %, and 123Sb = 100 - 57.20 = 42.80 % (more on these kinds of problems in the next section!) Page IV-2a- / Isotopic Abundance Handout Study Guide 4 Calculate the Isotopic Abundances When Given the Average Atomic Weight and the Isotopic Weights Let's start by calculating the average atomic weight for nitrogen. Nitrogen-14 has a mass of 14.003074 amu and an abundance of 99.63%, while nitrogen-15 has a mass of 15.000108 amu and an abundance of 0.3700%, so: (14.003074) (0.9963) + (15.000108) (0.003700) = 14.006763... = 14.007 amu The solution is laid out like this: (exact weight of isotope #1) (abundance of isotope #1) + (exact weight of isotope #2) (abundance of isotope #2) = average atomic weight of the element Note how the abundance values are given in decimal form, so make sure to divide the percentage values by 100%. Solving equations like this is possible because the sum of the percent abundances of the two isotopes add up to 100% (or, since we use decimal abundances in the calculation, 1.00). So, the trick is to express both abundances using only one unknown. This is how to do so using the above nitrogen example. Problem #1: Nitrogen is made up of two isotopes, N-14 and N-15. Given nitrogen's atomic weight of 14.007, what is the percent abundance of each isotope? Here's the solution: (14.003074) (x) + (15.000108) (1 − x) = 14.007 Notice that the abundance of N-14 is assigned 'x' and the N-15 is 'one minus x.' The two abundances always add up to one (or, if you prefer, 100%) Note: sometimes the exact weights of the isotopes are not provided in the problem. However, in this Internet era, it is easy to look up the values online. If you do not have access to the internet, notice that the mass numbers (14 and 15) are quite close to the exact values. The consequence of using 14 and 15 rather than the exact values is that we will get slightly approximate answers for the two abundances (and worse sig figs.) For example, I might have this: (14) (x) + (15) (1 − x) = 14.007 Solving gives: 14x + 15 − 15x = 14.007 After some more simple algebra, we get: x = 15 − 14.007 = 0.993 so 1 − x = 0.007 You may want to compare them to the more exact abundances which are in the equation set up at the top of this page. Page IV-2a- / Isotopic Abundance Handout Study Guide 5 Problem #2a: Copper is made up of two isotopes, Cu-63 (62.9296 amu) and Cu-65 (64.9278 amu). Given copper's atomic weight of 63.546, what is the percent abundance of each isotope? Solution: 1) Write the following equation: (62.9296) (x) + (64.9278) (1 − x) = 63.546 Once again, notice that 'x' and 'one minus x' add up to one. 2) Solve for x: 62.9296x + 64.9278 - 64.9278x = 63.546 -1.9982x = -1.3818 divide both sides by -1.9982 x = -1.3818 / -1.9982 = 0.691522... = 0.6915 (the decimal abundance for Cu-63, or 69.15 %) Cu-65 = 1 - 0.6915 = 0.3085, or 30.85% check: both isotopes should equal 100% (or be very close to 100%!), so: 69.15 + 30.85 = 100.00% good! Note that this calculation technique works only with two isotopes. If you have three or more, there are too many variables and not enough equations. Also note the four sig figs: this stems from 63.546 - 64.9278, the answer should stop at the thousandths position, even though we use all the values in the calculation until the end of the problem. Problem #2b: Chlorine is made up of two isotopes, Cl-35 (34.969 amu) and Cl-37 (36.966 amu). Given chlorine's atomic weight of 35.453 amu, what is the percent abundance of each isotope? Solution: 1) Write the following equation: (34.969) (x) + (36.966) (1 − x) = 35.453 2) Solve for x: 34.969x + 36.966 − 36.966x = 35.453 -1.997x = -1.513 x = -1.515 / -1.997 = 0.757636... = 0.7576 (the decimal abundance for Cl-35, or 75.76%) Cl-37 = 1 - 0.7576 = 0.2424, or 24.24% check: both isotopes should equal 100% (or be very close to 100%!), so: 75.76 + 24.24 = 100.00% good! Page IV-2a- / Isotopic Abundance Handout Study Guide 6 Problem #3: A sample of naturally occurring silicon consists Si-28 (amu = 27.9769), Si-29 (amu = 28.9765) and Si-30 (amu = 29.9738). If the atomic mass of silicon is 28.0855 and the natural abundance of Si-28 is 92.23%, what are the natural abundances of Si-29 and Si-30? Solution: 1) Set up a system of two equations in two unknowns: Let x = isotopic abundance of Si-28 (as a decimal) Let y = isotopic abundance of Si-30 (as a decimal) Therefore: (27.9769) (0.9223) + (28.9765) (x) + (29.9738) (y) = 28.0855, and 0.9223 + x + y = 1.000 (all abundances must equal 100%) 1) Rearrange the second equation to: y = 1.000 − 0.9223 − x = 0.0777 − x 2) Substitute into the first equation and solve: (27.9769) (0.9223) + (28.9765) (x) + (29.9738) (0.0777 − x) = 28.0855 This is one equation with one unknown (x), so time for some algebra! Note the 0.0777 will make the final x value 3 sig figs 25.80309 + 28.9765x + 2.32896 − 29.9738x = 28.0855 -0.9973x = -0.04655 x = 0.046676... = 0.0467 (the decimal abundance for Si-29, or 4.67%) Si-30 = 0.0777 - x = 0.0777 - 0.0467 = 0.0310, or 3.10% check: the three isotopes should equal 100% (or be very close to 100%!), so: 92.23 + 4.67 + 3.10 = 100.00% good! Page IV-2a- / Isotopic Abundance Handout Study Guide 7 Problem #4: Determine the percent abundance for Fe-57 and Fe-58, given the following data and the atomic mass of iron (55.845): Solution: 1) Assign the percent abundance of Fe-57 to the variable 'x' and Fe-58 to the variable 'y' 2) We need to get the percent abundance for Fe-58 in terms of x. Like this: 1 = 0.05845 + 0.91754 + x + y; solve for y y = 1 - (0.05845 + 0.91754 + x) = 1 - (0.97599 + x) and y = 1 - 0.97599 - x = 0.02401 - x We will use y = 0.02401 - x in the full equation below. 3) Set up (and solve) the following equation: (53.9396) (0.05845) + (55.9349) (0.91754) + (56.9354) (x) + [(57.9333) (0.02401 − x)] = 55.845 3.15277 + 51.32251 + 56.9354x + [1.39098 - 57.9333x] = 55.845 55.86626 - 0.9979x = 55.845 -0.9979x = -0.02126 x = -0.02126 / -0.9979 = 0.0213047... = 0.02130 (the decimal abundance for Fe-57, or or 2.130%) Fe-58 = y = 0.02401 - x = 0.02401 - 0.02130 = 0.00271, or 0.271% check: the four isotopes should equal 100% (or be very close to 100%!), so: 5.845 + 91.754 + 2.130 + 0.271 = 100.00% good! Isotope Atomic Weight Percent Abundance Fe-54 53.9396 5.845 Fe-56 55.9349 91.754 Fe-57 56.9354 ??? Fe-58 57.9333 ??? Page IV-2a- / Isotopic Abundance Handout Study Guide 8
15687
https://math.libretexts.org/Courses/University_of_California_Davis/UCD_Mat_21A%3A_Differential_Calculus/4%3A_Applications_of_Definite_Integrals/4.5%3A_Indeterminate_Forms_and_L'H%C3%B4pital's_Rule
Skip to main content 4.5: Indeterminate Forms and L'Hôpital's Rule Last updated : Apr 27, 2019 Save as PDF 4.4: Concavity and Curve Sketching 4.6: Applied Optimization Page ID : 8241 ( \newcommand{\kernel}{\mathrm{null}\,}) Learning Objectives Recognize when to apply L’Hôpital’s rule. Identify indeterminate forms produced by quotients, products, subtractions, and powers, and apply L’Hôpital’s rule in each case. Describe the relative growth rates of functions. In this section, we examine a powerful tool for evaluating limits. This tool, known as L’Hôpital’s rule, uses derivatives to calculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead of relying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and to determine its exact value. Applying L’Hôpital’s Rule L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider If and then However, what happens if and ? We call this one of the indeterminate forms, of type . This is considered an indeterminate form because we cannot determine the exact behavior of as without further analysis. We have seen examples of this earlier in the text. For example, consider and For the first of these examples, we can evaluate the limit by factoring the numerator and writing For we were able to show, using a geometric argument, that Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more importantly, it provides us with a way to evaluate many other limits that we could not calculate previously. The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions and such that and such that For near ,we can write and Therefore, Since is differentiable at , then is continuous at , and therefore . Similarly, . If we also assume that and are continuous at , then and . Using these ideas, we conclude that Note that the assumption that and are continuous at and can be loosened. We state L’Hôpital’s rule formally for the indeterminate form . Also note that the notation does not mean we are actually dividing zero by zero. Rather, we are using the notation to represent a quotient of limits, each of which is zero. L’Hôpital’s Rule (0/0 Case) Suppose and are differentiable functions over an open interval containing , except possibly at . If and then assuming the limit on the right exists or is or . This result also holds if we are considering one-sided limits, or if or Proof We provide a proof of this theorem in the special case when and are all continuous over an open interval containing . In that case, since and and are continuous at , it follows that . Therefore, Note that L’Hôpital’s rule states we can calculate the limit of a quotient by considering the limit of the quotient of the derivatives . It is important to realize that we are not calculating the derivative of the quotient . □ Example : Applying L’Hôpital’s Rule (0/0 Case) Evaluate each of the following limits by applying L’Hôpital’s rule. Solution a. Since the numerator and the denominator , we can apply L’Hôpital’s rule to evaluate this limit. We have b. As the numerator and the denominator Therefore, we can apply L’Hôpital’s rule. We obtain c. As , the numerator and the denominator . Therefore, we can apply L’Hôpital’s rule. We obtain d. As both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain Since the numerator and denominator of this new quotient both approach zero as , we apply L’Hôpital’s rule again. In doing so, we see that Therefore, we conclude that Exercise Evaluate Hint Answer We can also use L’Hôpital’s rule to evaluate limits of quotients in which and . Limits of this form are classified as indeterminate forms of type . Again, note that we are not actually dividing by . Since is not a real number, that is impossible; rather, is used to represent a quotient of limits, each of which is or . L’Hôpital’s Rule ( Case) Suppose and are differentiable functions over an open interval containing , except possibly at . Suppose (or ) and (or ). Then, assuming the limit on the right exists or is or . This result also holds if the limit is infinite, if or , or the limit is one-sided. Example : Applying L’Hôpital’s Rule () Case Evaluate each of the following limits by applying L’Hôpital’s rule. Solution a. Since and are first-degree polynomials with positive leading coefficients, and . Therefore, we apply L’Hôpital’s rule and obtain Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of x in the denominator. In doing so, we saw that L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit. b. Here, and . Therefore, we can apply L’Hôpital’s rule and obtain Now as . Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of to write Now and , so we apply L’Hôpital’s rule again. We find We conclude that Exercise Evaluate Hint Answer As mentioned, L’Hôpital’s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that to apply L’Hôpital’s rule to a quotient , it is essential that the limit of be of the form or . Consider the following example. Example : When L’Hôpital’s Rule Does Not Apply Consider Show that the limit cannot be evaluated by applying L’Hôpital’s rule. Solution Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply L’Hôpital’s rule. If we try to do so, we get and At which point we would conclude erroneously that However, since and we actually have We can conclude that Exercise Explain why we cannot apply L’Hôpital’s rule to evaluate . Evaluate by other means. Hint : Determine the limits of the numerator and denominator separately. Answer : Therefore, we cannot apply L’Hôpital’s rule. The limit of the quotient is Other Indeterminate Forms L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms and . However, we can also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions , and are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form or . Indeterminate Form of Type 0⋅∞ Suppose we want to evaluate , where and (or ) as . Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation to denote the form that arises in this situation. The expression is considered indeterminate because we cannot determine without further analysis the exact behavior of the product as . For example, let be a positive integer and consider and . As and . However, the limit as of varies, depending on . If , then . If , then . If , then . Here we consider another limit involving the indeterminate form and show how to rewrite the function as a quotient to use L’Hôpital’s rule. Example : Indeterminate Form of Type Evaluate Solution First, rewrite the function as a quotient to apply L’Hôpital’s rule. If we write we see that as and as . Therefore, we can apply L’Hôpital’s rule and obtain We conclude that Exercise Evaluate Hint : Write Answer Indeterminate Form of Type ∞-∞ Another type of indeterminate form is Consider the following example. Let be a positive integer and let and . As and . We are interested in . Depending on whether grows faster, grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since and , we write to denote the form of this limit. As with our other indeterminate forms, has no meaning on its own and we must do more analysis to determine the value of the limit. For example, suppose the exponent n in the function is , then On the other hand, if then However, if , then Therefore, the limit cannot be determined by considering only . Next we see how to rewrite an expression involving the indeterminate form as a fraction to apply L’Hôpital’s rule. Example : Indeterminate Form of Type Evaluate Solution By combining the fractions, we can write the function as a quotient. Since the least common denominator is we have . As , the numerator and the denominator Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have As , and . Since the denominator is positive as approaches zero from the right, we conclude that Therefore, Exercise Evaluate . Hint : Rewrite the difference of fractions as a single fraction. Answer : 0 Indeterminate Forms Involving Exponents Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions , and are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits. Now we examine how L’Hôpital’s rule can be used to evaluate limits involving these indeterminate forms. Since L’Hôpital’s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want to evaluate and we arrive at the indeterminate form . (The indeterminate forms and can be handled similarly.) We proceed as follows. Let Then, Therefore, Since we know that . Therefore, is of the indeterminate form , and we can use the techniques discussed earlier to rewrite the expression in a form so that we can apply L’Hôpital’s rule. Suppose , where may be or Then Since the natural logarithm function is continuous, we conclude that which gives us Example : Indeterminate Form of Type Evaluate Solution Let .Then, We need to evaluate . Applying L’Hôpital’s rule, we obtain Therefore, Since the natural logarithm function is continuous, we conclude that which leads to Hence, Exercise Evaluate Hint : Let and apply the natural logarithm to both sides of the equation. Answer Example : Indeterminate Form of Type Evaluate Solution Let Therefore, We now evaluate Since and , we have the indeterminate form . To apply L’Hôpital’s rule, we need to rewrite as a fraction. We could write or Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain Unfortunately, we not only have another expression involving the indeterminate form but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing and applying L’Hôpital’s rule, we obtain Using the fact that and , we can rewrite the expression on the right-hand side as We conclude that Therefore, and we have Hence, Exercise Evaluate . Hint : Let and take the natural logarithm of both sides of the equation. Answer : 1 Growth Rates of Functions Suppose the functions and both approach infinity as . Although the values of both functions become arbitrarily large as the values of become sufficiently large, sometimes one function is growing more quickly than the other. For example, and both approach infinity as . However, as Table shows, the values of are growing much faster than the values of . Table : Comparing the Growth Rates of and | | 10 | 100 | 1000 | 10,000 | | | 100 | 10,000 | 1,000,000 | 100,000,000 | | | 1000 | 1,000,000 | 1,000,000,000 | 1,000,000,000,000 | In fact, or, equivalently As a result, we say is growing more rapidly than as . On the other hand, for and , although the values of are always greater than the values of for , each value of is roughly three times the corresponding value of as , as shown in Table . In fact, Table : Comparing the Growth Rates of and | | 10 | 100 | 1000 | 10,000 | | | 100 | 10,000 | 1,000,000 | 100,000,000 | | | 341 | 30,401 | 3,004,001 | 300,040,001 | In this case, we say that and are growing at the same rate as More generally, suppose and are two functions that approach infinity as . We say grows more rapidly than as if On the other hand, if there exists a constant such that we say and grow at the same rate as . Next we see how to use L’Hôpital’s rule to compare the growth rates of power, exponential, and logarithmic functions. Example : Comparing the Growth Rates of , , and For each of the following pairs of functions, use L’Hôpital’s rule to evaluate and and Solution a. Since and , we can use L’Hôpital’s rule to evaluate . We obtain Since and , we can apply L’Hôpital’s rule again. Since we conclude that Therefore, grows more rapidly than as (See Figure and Table ) Table : Growth rates of a power function and an exponential function. | | 5 | 10 | 15 | 20 | | | 25 | 100 | 225 | 400 | | | 148 | 22,026 | 3,269,017 | 485,165,195 | b. Since and , we can use L’Hôpital’s rule to evaluate . We obtain Thus, grows more rapidly than as (see Figure and Table ). Table : Growth rates of a power function and a logarithmic function | | 10 | 100 | 1000 | 10,000 | | | 2.303 | 4.605 | 6.908 | 9.210 | | | 100 | 10,000 | 1,000,000 | 100,000,000 | Exercise Compare the growth rates of and . Hint : Apply L’Hôpital’s rule to . Answer : The function grows faster than . Using the same ideas as in Example . it is not difficult to show that grows more rapidly than for any . In Figure and Table , we compare with and as . Table : An exponential function grows at a faster rate than any power function | | 5 | 10 | 15 | 20 | | | 125 | 1000 | 3375 | 8000 | | | 625 | 10,000 | 50,625 | 160,000 | | | 148 | 22,026 | 3,269,017 | 485,165,195 | Similarly, it is not difficult to show that grows more rapidly than for any . In Figure and Table , we compare with and . Table : A logarithmic function grows at a slower rate than any root function | | 10 | 100 | 1000 | 10,000 | | | 2.303 | 4.605 | 6.908 | 9.210 | | | 2.154 | 4.642 | 10 | 21.544 | | | 3.162 | 10 | 31.623 | 100 | Key Concepts L’Hôpital’s rule can be used to evaluate the limit of a quotient when the indeterminate form or arises. L’Hôpital’s rule can also be applied to other indeterminate forms if they can be rewritten in terms of a limit involving a quotient that has the indeterminate form or The exponential function grows faster than any power function . The logarithmic function grows more slowly than any power function . Glossary indeterminate forms : When evaluating a limit, the forms , , and are considered indeterminate because further analysis is required to determine whether the limit exists and, if so, what its value is. L’Hôpital’s rule : If and are differentiable functions over an interval , except possibly at , and or and are infinite, then , assuming the limit on the right exists or is or . 4.4: Concavity and Curve Sketching 4.6: Applied Optimization
15688
https://www.youtube.com/watch?v=L3CHMlbGVrI
Find the shaded area in the equilateral triangle | A Very Nice Geometry Problem Math Booster 73500 subscribers 123 likes Description 4748 views Posted: 26 Sep 2025 Find the shaded area in the equilateral triangle | A Very Nice Geometry Problem 53 comments Transcript: इन दिस वीडियो वी हैव बीन गिवन दैट ABC इज एन इक्विलैटरल ट्रायंगल एंड BC इज 10 देन वी हैव टू फाइंड शेडेड एरिया सो वी हैव BC इज 10 एंड सपोज दिस पॉइंट इज S देन फ्रॉम पॉइंट A इफ वी मेक ए परपेंडिकुलर ऑन PQ एंड BC देन इट विल पास थ्रू पॉइंट s सपोज दिस पॉइंट इज m एंड दिस पॉइंट इज n देन an इट विल पास थ्रू पॉइंट एस डाइसिमेट्री एंड BN इट विल बी इक्वल टू CN दैट विल बी BC ओवर 2 एंड इट इज 10 ओवर 2 दैट विल बी फाइव सो BN इज 5 एंड CN इज फाइव एंड इन ट्रायंगल ABN इट इज A B एंड दिस एंगल इज इज 90° एंड BN इज 5 एंड वी हैव BC इज 10 सो AB इट विल बी आल्सो 10 एंड वी हैव ए सर्कल इंसाइड ABN n एंड दिस पॉइंट इज s एंड सपोज द रेडियस ऑफ दी सर्कल इज a एंड सपोज दिस पॉइंट इज e एंड दिस पॉइंट इज f देन nf इट विन बी a ns इट विन बी ए एंड दिस पॉइंट इज m एंड sm इट विल बी आल्सो ए एंड इन एनी सर्कल फ्रॉम एनी पॉइंट P इफ वी मेक टू टेंजेंट ऑन द सर्कल सपोज दिस पॉइंट इज A एंड दिस पॉइंट इज B देन PA इट विल बी इक्वल टू PB एंड BF इट विल बी 5 - A इट इज 5 - A सो BE इट विल बी आल्सो 5 - A एंड ae इट विल बी ab - be एंड इट इज 10 - 5 - a दैट विल बी 5 + ae इज 5 + a एंड फ्रॉम पॉइंट AE इट विल बी इक्वल टू AS सो AS इट विल बी ऑल सो 5 + Am इट विल बी 5 + a - a दैट विल बी 5 वी गेट am इज 5 एंड नाउ ए बी सी इट इज इक्विलैटरल ट्रायंगल दैट मींस दिस एंगल इट विल बी 60 डिग्री एंड दिस एंगल इज 60° दैट मींस दिस एंगल इट विल बी आल्सो 60° डिग्री एंड दिस एंगल इज 60° दैट मींस दिस एंगल इट विल बी आल्सो 60° एंड इन ट्रायंगल APQ इट इज A P Q दी थ्री एंगल्स आर इक्वल दैट मींस इफ AP इज X देन AQ इट विल बी आल्सो X एंड PQ इट विल बी आल्सो X एंड वी हैव AM इज 5 एंड PM इट विल बी x2 एंड qm इट विल बी x2 एंड a ps² इट विल बी am s² + pm s² एंड इट इज x² इट इज 5s स्क्वायर प्लस x 2² एंड x² इट इज 25 + x² ओवर 4 एंड x² - x² 4 इट विल बी 25 एंड इट इज 3x² ओवर 4 इज 25 सो x² इट विल बी 25 टाइम्स 4 ओवर 3 एंड x² इट इज 100 ओवर 3 दैट मींस x इट विल बी 10 ओवर 3 pq इट इज x दैट इज 10 ओवर 3 एंड नाउ शेडेड एरिया इट इज एरिया ऑफ़ APQ दैट विन बी 1 2 टाइम्स बेस इज़ PQ एंड हाइट इज AM एंड इट इज़ वन 1 2 टाइम्स pq इज 10 3 टाइम्स am इज 5 एंड 2 टाइम्स 5 इज 10 सो इट इज 25 ओवर 3 दैट विल बी 25 टाइम्स 3 ओवर 3 इट इज 25 टाइम्स 3 ओवर 3 [संगीत]
15689
https://www.doubtnut.com/qna/119405
A line passes through the origin and makes an angle of π4 with the line x−y+1=0. Then : More from this Exercise Step by step video & image solution for A line passes through the origin and makes an angle of pi/4 with the line x-y+1=0. Then : by Maths experts to help you in doubts & scoring excellent marks in Class 12 exams. Similar Questions The equation to the striaght lines passing through the origin and making an angle α with straight line y+x=0 are given by the equation of the line passing through (1,1) and makes an angle π4 which the line 2x−y+7=0 Knowledge Check The joint equation of pair of lines passing through the origin and making an angle of 45∘ with the line 3x+y=0 is The joint equation of pair of lines through the origin and making an angle of π6 with the line 3x+y−6=0 is The combined equation of the lines which pass through the origin and eahc of which makes an angle of 30∘ with the line 2x−y=0 is Equation of pair of lines passing through origin and making an angle tan−12 with the line 4x−3y+7=0 Equation of pair of lines passing through origin and making an angle π4 with the line 3x+2y+4=0 is (3x+2y)2−k(2x−3y)2=0 then k = (i) 1 (ii) −1 (iii) 3 (iv) 13 The equation of the lines passes through (2,3) and making an angle of 45∘with the line 2x−y+3=0 is The equation to the striaght lines passing through the origin and making an angle α with straight line y+x=0 are given by The equation of the lines passes through (2,3) and making an angle of 45∘with the line 2x−y+3=0 is Recommended Questions A line passes through the origin and makes an angle of pi/4 with the l... the equation of the line passing through (1,1) and makes an angle (pi)... A line passes through the origin and makes an angle of (pi)/(4) with t... The equation to the striaght lines passing through the origin and maki... The joint equation of pair of lines passing through the origin and mak... The joint equation of pair of lines through the origin and making an a... The combined equation of the lines which pass through the origin and e... Find the equation of the line passing through the point (4,5) and maki... The equation of the line passing through (1, 1) and makes an angle pi/... 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
15690
https://www.open.edu/openlearn/mod/oucontent/view.php?printable=1&id=4256
Using a scientific calculator Completion requirements View all sections of the document Printable page generated Sunday, 28 September 2025, 6:15 AM Use 'Print preview' to check the number of pages and printer settings. Print functionality varies between browsers. Unless otherwise stated, copyright © 2025 The Open University, all rights reserved. Printable page generated Sunday, 28 September 2025, 6:15 AM ◄ Using a scientific calculator Using a scientific calculator Introduction The course describes some of the main features of a scientific calculator and encourages you to use your calculator, both for everyday arithmetic and for more complicated calculations that use the function keys as well. Key sequences, which describe which keys to press, are included in all the activities, so you can try out the ideas straightaway. Due to the wide range of scientific calculators available, for the purposes of this course we will be concentrating on the Casio fx-83ES model. Other calculators may function differently to the methods described within this course. This calculator is used on the Open University courses Starting with maths (Y182) and Discovering mathematics (MU123), but would also be useful for many other courses requiring the use of a scientific calculator. This OpenLearn course is an adapted extract from the Open University course MU123 Discovering Mathematics. Learning outcomes After studying this course, you should be able to: understand the basic functions on your calculator understand which calculator functions are needed for a given problem understand what may go wrong when entering calculations and know how to fix them apply knowledge of calculator functions to a range of mathematical calculations. 1 Getting to know your calculator The first 11 sections describe how to use the calculator and how to perform different types of calculations. Section 12 contains a calculator reference guide that you can refer to as needed for some of the main key sequences. This course is not an exhaustive list of all the calculator’s features. If you are using a different calculator, you should use the corresponding features of your own calculator to do the activities in this guide. You may need to refer to your calculator manual to do this. You may be able to download the manual for your calculator from the manufacturer’s website. The first step in using your calculator effectively is to make sure that you are familiar with the layout of the keys on the keypad, and that you can understand the information on the display. Figure 1 shows the different parts of the Casio fx-83ES calculator. Figure 1 A typical scientific calculator Show description|Hide description The figure shows a photograph of a Casio scientific calculator f x - 83 E S (natural display) with annotations added to explain some of the features. The calculator screen at the top, annotated with ‘display’, has three rows. The top row of the screen contains the letter D (white text on black) in the centre, and the word ‘math’ with an upward pointing arrowhead towards the right end. The middle row of the screen has the mathematical text square root 12 multiplied by square root 6 multiplied by square root 3 all over 2 at the left-hand side. The bottom row has the mathematical text 3 square root 6 at the right-hand side. Beneath the display is a row of five blue buttons annotated, from left to right: ‘shift key’; ‘alpha key’; ‘cursor control button’ (this button is much larger and in the centre); ‘mode key’; ‘on key’. Below these are four rows of black buttons (twenty two buttons in total) with the annotation ‘function keys’. At the bottom left of the calculator an array of twelve buttons is annotated with ‘number keys’. To the right of these the red button labelled D E L (written as one word) is annotated with ‘delete key’ and the red button labelled A C (written as one word) is annotated with ‘all clear key’. Below these the multiply, divide, add, subtract keys are annotated with ‘basic operation keys’. Below these the button labelled Ans is annotated with ‘last answer key’ and the button labelled = is annotated with ‘equals key’. Figure 1 A typical scientific calculator If you are using a different model of calculator, make sure that you can identify similar functions on your model. The calculator is switched on using the key at the top right-hand corner of the keypad. Figure 2 shows the different elements of this calculator’s display. Throughout this section, calculator keys will be indicated using the symbol on the key enclosed in a box, for example . Figure 2 The calculator display Show description|Hide description The figure shows the display on a calculator screen with annotations added to explain some of the features. The top row of the screen contains the letter D (white text on black) in the centre, and the word ‘math’ with an upward pointing arrowhead towards the right end, and these are annotated with ‘display indicators’. The middle row of the screen has the mathematical text 1 add 2 at the left-hand side, annotated with ‘input’. The bottom row has the mathematical text 3 at the right-hand side, annotated with ‘output’. Figure 2 The calculator display The lower half of the keypad contains the number keys, keys for the basic operations of addition, subtraction, division and multiplication, and the key, which is pressed when you want the calculator to display the result of the calculation you have entered. The keys used to insert brackets into a calculation are in the centre of the row above the number keys. Many keys on the calculator have more than one use. The main function of a key is printed in white on the key itself. The second function of the key is printed in yellow above the key, and is accessed by pressing the button before pressing the key. When you press the button, the symbol ‘’ appears at the top left-hand corner of the calculator display to remind you that the button has been pressed. It disappears when you press another key. Some keys also have a third function, printed above the key in red. These functions allow numerical values stored in the calculator memories to be used within calculations and are accessed by pressing the button before the appropriate key. When the button is pressed, the symbol ‘’ is shown at the top of the calculator display. You will learn how to use the calculator memories later in section 4. The calculator manual describes this second function as the ‘alternate’ function of the key. Some calculator operations are accessed through a system of menus that are displayed on the calculator screen, as shown in Figure 3. The required menu option is selected by pressing the number key associated with the option, as given on the calculator screen. Figure 3 A typical calculator on-screen menu Show description|Hide description The figure shows the display on a calculator screen containing four lines of text. The top row of the screen contains a downward pointing arrowhead towards the right end. First line of text is 1 : M t h I O (written as one word) space 2 : Line I O (written as one word). Second line of text is 3 : Deg space 4 : Rad. Third line of text is 5 : Gra space 6 : Fix. Last line of text is 7 : Sci space 8 : Norm. Figure 3 A typical calculator on-screen menu When describing how to use various calculator functions, this guide gives the exact keys that you need to press using the symbols shown on the keys. This is known as a ‘key sequence’. If the key sequence accesses the second function of a key, or a function from a menu, the name of this function will be given in brackets at the appropriate point in the key sequence. Names in brackets are thus not keys that you press but simply describe the function that is accessed using the previous key sequence. For example, to turn off the calculator, press (OFF). In this notation, (OFF) is not a key that you press, but is the name of the second function of the key, which is accessed with the key. The calculator has many modes of operation that affect how mathematics is entered and displayed. These will be described later in this guide, but before progressing any further you should reset your calculator to the default course settings. Activity 1 Initialising your calculator To initialise your calculator to the default course settings, turn it on and then enter the following two key sequences: (CLR) (Setup) (Yes) (SETUP) (Norm) Note that in the first key sequence, ‘CLR’ (which is short for ‘clear’) is the second function of the key, and ‘Setup’ is the name of the on-screen menu option corresponding to the key. ‘Yes’ is the name of the on-screen menu option corresponding to the key. This key sequence clears all previous ‘setup’ settings on the calculator. In the second key sequence, ‘SETUP’ is the name of the second function of the key, and ‘Norm’ (short for ‘normal’) is the on-screen menu option corresponding to the key. Pressing the key selects to use ‘Normal 2’ mode, which will be described in more detail in section 5. Note the difference between ‘SETUP’, the second function of the key, and the menu option ‘Setup’. Your calculator will now be working in ‘Math’ mode, and the word Math will be shown near the right-hand side of the top of the calculator display, as shown in Figure 4 below. Math mode is the recommended way of using your calculator during this course as it allows mathematics to be entered and displayed in a similar way to how you would write it on paper. Figure 4 ‘Math’ mode Show description|Hide description The top row of the screen contains the letter D (white text on black) in the centre, and the word ‘math’ towards the right end. Figure 4 ‘Math’ mode 1.1 Basic calculations Basic calculations are entered into the calculator in exactly the same order as they are written on paper, as demonstrated in the following activity. The calculator displays the calculation that you enter. When you press , the answer is displayed at the bottom right of the screen. Activity 2 Sums, differences, products and quotients Use your calculator to work out the answers to the following calculations. Answer Note that the calculator uses the BIDMAS rules. These say that any expression within Brackets should be calculated first, then any Indices (often called powers), followed by Divisions and Multiplications and finally Additions and Subtractions. To calculate this correctly you need to remember to insert the brackets into the calculation using the keys. You may have noticed that in part (4) of the above activity, the calculation was too long to fit on the calculator display. In such circumstances, the symbols ‘’ or ‘’ appear at the left or right of the display to indicate that there is more information in that direction. This information can be seen by scrolling left or right using the and keys, which are found at the left and right sides of the large cursor control button (labelled with the word ‘REPLAY’) located under the calculator screen. If you type a very long calculation into your calculator,  then you may see the cursor  (which is usually shown as ‘ ’) change to ‘’. This means that you are allowed to type only at most 10 more characters. If you encounter this, you should break your calculation into smaller parts. The cursor is a flashing symbol indicating where the next item entered into the calculator will appear. 1.2 Fractions or decimals? In Activity 1 you set up your calculator to use Math mode. In this mode, when the result of a calculation is not a whole number, it will be displayed as a fraction, such as , wherever possible. To obtain the answer in decimal form, you need to press instead of , or you can toggle between the fractional and decimal outputs using the key. Remember, your calculator is in Math mode if the word Math is shown at the top of the calculator display. If your calculator is not in Math mode, repeat the steps of Activity 1. Activity 3 Fractions and decimals Use your calculator to find in both fractional and decimal forms. Answer In Math mode, calculating gives the result . Converting this to a decimal using gives 0.1814671815. 1.3 Powers There are several keys on the calculator that enable you to perform calculations involving powers. For small powers such as squares or cubes there are dedicated buttons, and , which are located in the function key area of the keypad. These are used in a similar manner to how you would write mathematics; for example, to enter you would press . The display also shows the maths in the same way as you would write it on paper. To calculate higher powers, for example , you need to use the more general power key . This is again used in a natural way. To enter , you use the key sequence . Note that after you press the key, a small box is shown on the calculator display containing the flashing cursor (‘ ’), which enables you to enter the power in the correct place. To move the cursor away from this box and back to the main line of the display once the power has been entered, press the right arrow key at the right-hand side of the large cursor control button. Figure 5 The general power key function Show description|Hide description The top row of the screen contains the letter D (white text on black) in the centre, and the word ‘math’ towards the right end. The second row of the screen has the mathematical text 2 with, immediately to the right, a raised small empty box. Figure 5 The general power key function Other models of calculator may have the button instead of and may not have specific and keys. Activity 4 Calculating powers Calculate each of the following using your calculator. Answer Here you need to use the general power key . Remember to use the right arrow key after inputting the power 8 to move out of the power position before entering the rest of the expression. If you obtained the answer 2243822356, you have calculated by mistake. 1.4 Making corrections If you make a mistake when entering a key sequence into the calculator, you can use the editing facilities to correct your error. The and keys on the large cursor control button enable you to move the cursor (shown on the display as ‘ ’) within a calculation on the calculator screen. Characters can then be inserted at the cursor location simply by pressing the appropriate buttons, and items to the left of the cursor can be deleted using the key. This can be done either before or after the key has been pressed. To re-evaluate an edited calculation, simply press at any time. In some circumstances, however, it may be easiest to abandon what you have typed and start again, by pressing the ‘all clear’ key! If a severe error is made when entering a calculation into the calculator, it may prevent the answer being calculated at all, as the calculation may not make mathematical sense. In such circumstances a ‘Syntax Error’ will be displayed as shown in Figure 6. The Syntax Error screen gives you two options: Press to abandon the calculation and clear the screen press either or to return to the erroneous calculation with the editing cursor placed at the point of the error, ready for a correction to be made Figure 6 Syntax Error Show description|Hide description The top row of the screen contains the letter D (white text on black) in the centre, and the word ‘math’ towards the right end. First line of text is Syntax ERROR (in uppercase letters). Second line of text is [ A C (written as one word in uppercase letters) ] : Cancel. Last line of text is [ left pointing arrowhead ] [ right pointing arrowhead ] : Goto. Figure 6 Syntax Error Other types of calculator error that you may encounter are: ‘Math Error’, when the calculation you entered makes mathematical sense but the result cannot be calculated, such as attempting to divide by zero, or when the result is too large for the calculator to handle. ‘Stack Error’, when your calculation is too complex to be handled in one go – in such circumstances, try to break the calculation into a number of simpler ones. Section 4 considers how you might do this. In these cases, the calculator will display a screen similar to that for the Syntax Error, allowing you to either abandon or correct your calculation. Activity 5 Making corrections Enter the following key sequence into your calculator in an (erroneous) attempt to calculate : What should the correct answer be, and why does this key sequence not give it? Use the calculator editing functions to correct the inputted key sequence. Answer The correct value of is 16. The key sequence given did not close the bracket before squaring the expression, so only the 3 was squared, giving . When correcting the expression, be careful to ensure that the cursor is immediately after the 3, and the same height as it – shown below – before inserting the missing bracket. Otherwise, the bracket may be inserted within the power. Figure 7 The cursor, placed after the 3 Show description|Hide description The top row of the screen contains the letter D (white text on black) in the centre, and the word ‘math’ towards the right end. The second row of the screen has the mathematical text ( 1 add 3 vertical cursor line superscript 2. Figure 7 The cursor, placed after the 3 2 Using your calculator for negative numbers There are two different mathematical uses for the minus sign (-): as the symbol for subtraction, as in to indicate a negative number such as . Corresponding to these there are two different minus sign keys on the calculator: , which is used for the operation of subtraction, as in , which is used for negative numbers, e.g.  . In fact, some calculators permit the key to be used for both purposes, but many other calculators require the equivalent of the key to be used for negative numbers. For this reason we shall use to input negative numbers throughout this guide. Note that if you attempt to use for subtraction, for example , you will generate a Syntax Error. Activity 6 Subtraction and negative numbers Calculate each of the following using your calculator. In each case, give your answer as a decimal not as a fraction. Answer Remember to press to obtain a decimal answer. You may have been surprised that the correct answer to part (5) is negative. According to the BIDMAS rules, the squaring is performed first, then the negative taken. If we wanted to calculate the square of , we write this mathematically as and would need to use the brackets when evaluating it on a calculator. 3 Using your calculator for fractions When your calculator is in Math mode, as recommended, fractions are entered using the button in the left-hand column of the function key area of the calculator keypad. This displays a fraction ‘template’ on the display – as shown in Figure 8 below – that contains boxes that need to be ‘filled in’. When the button is first pressed, the cursor is located in the top box ready for you to enter the numerator. To move to the bottom box to enter the denominator, use the cursor down key . If there are further parts of a calculation to be entered when the template has been completed, the right cursor key can be used to move out of the denominator in preparation for the input of the rest of the calculation. Figure 8 A fraction template Show description|Hide description The top row of the screen contains the letter D (white text on black) in the centre, and the word ‘math’ towards the right end. The second row of the screen contains the outline of a fraction consisting of a raised small empty box over a division line with another small empty box beneath the division line. Figure 8 A fraction template Remember, your calculator is in Math mode if the word Math is shown at the top of the calculator display. If your calculator is not in Math mode, repeat the steps of Activity 1. Mixed numbers such as can be entered similarly using the mixed number template obtained using the key sequence . This template provides three boxes to fill, one for the whole number part, and one each for the numerator and denominator of the fractional part. Any fractional answers to calculations will automatically be displayed in lowest terms. Activity 7 Fractions Use your calculator to: express in its simplest form calculate of 190. Answer of 190 can be written , which is equal to or 71.25. The decimal answer can be obtained by using . Remember to use the cursor right key to move the cursor out of the denominator of the fraction before entering the multiplication sign. If you obtained the answer , you calculated by mistake. You may have noticed that the results of both these exercises were displayed on the calculator as top-heavy fractions. This is the default behaviour of the calculator in Math mode. You can toggle between a top-heavy fraction and its mixed number equivalent using the key sequence . The default behaviour of the calculator can be changed as follows: To set the calculator to use mixed numbers by default, use the key sequence (SETUP)(ab/c). To set the calculator to use improper or top-heavy fractions by default, use the key sequence (SETUP) (d/c). Here, the key is used to access part of the on-screen menu that is not initially visible. Activity 8 Mixed numbers Use your calculator to: express as a mixed number in its simplest form express as a top-heavy fraction. Answer Remember to use to toggle between the top-heavy fraction and mixed number answers. Remember to use the template obtained using and to use the cursor arrow keys to move between the boxes when inputting the mixed number. 4 Doing longer calculations using your calculator The volume of wood (in cubic metres) contained in a log of length metres with a distance around its middle of metres is given by the formula For a log of length 1.5 m with a distance around the middle of 92 cm, this becomes In this section we consider several different approaches that can be used to evaluate this and other more complex expressions using different functions on your calculator. While the first method – considered in Activity 9 – is probably the most straightforward for this relatively simple expression, it is useful to see how you might use other calculator functions when you are faced with more complicated expressions to evaluate. The expression for the volume of wood requires the value of . You could enter an approximate value for by hand, but this is time-consuming and may be prone to error. The calculator has an approximation for built into it, which is obtained using the key sequence . The key is located on the bottom row of the keypad. Activity 9 Using the fraction key The most obvious way of calculating is to enter it as a fraction on your calculator. What key sequence is needed, and what is the final answer to 3 significant figures? Answer (to 3 significant figures). The key sequence used was . Note that it is not strictly necessary to include the multiplication between the 4 and the in the denominator since if the sign is omitted, it will be assumed by the calculator. Another way to carry out the calculation in Activity 9 is to use the key. Activity 10 Using the key You will not obtain the correct answer to the calculation if you type into your calculator and press . Can you explain why? Insert a pair of brackets into the expression with the ÷ sign so that it will give the correct answer. Then type this new expression into the calculator and check that you obtain the same answer as in the activity above. Answer Typing into the calculator and pressing will not give the correct answer because the calculator will follow the BIDMAS rules and divide by 4 and then multiply by , instead of dividing by . To obtain the correct result, you have to type . (Alternatively, you can type .) (Note that on some later models of the calculator, the correct answer is obtained without adding the brackets to the denominator; however it is good practice to add the brackets to ensure the correct calculation is carried out). 4.1 Reusing a previous result An alternative approach to our calculation is to calculate the denominator of the fraction first, and then divide the numerator by this. You could write down the answer to the first part of the calculation on paper, and enter it into the calculator again. However, it is possible that you may make an error either in writing down the number or in typing it into the calculator. A better method is to use the fact that the calculator retains the last calculated answer, which can then be inserted in the subsequent calculation using the key located at the bottom of the keypad. Note that the key only remembers the result of your last calculation. Activity 11 Bottom first! Use your calculator to calculate the value of the denominator of , then complete the calculation by finding the value of to 3 significant figures. Answer The value of the denominator is 12.566…, and the final answer is 0.101 to 3 significant figures. 4.2 Using the calculator memory A variation on the above method is to break the calculation into two parts, and use the memory functions of the calculator to store the result of the first part. The calculator memory is particularly useful when you want to calculate the values of several expressions that have a common part. This common part need be entered only once and its value reused several times subsequently. For example, rewriting the formula for the volume of wood contained in a log as we can see that no matter what the values of and , the formula always requires the value of . If we wished to calculate the volume of wood contained in several different logs, it might be efficient to calculate the value of once, store it in memory and reuse this value in the subsequent calculations. The calculator has several different memories. We shall first consider the ‘M’ memory, which is accessed using the key (and its associated functions) at the bottom right-hand corner of the function key area. Before using the calculator memory, it is good practice to always clear any previous data stored in the calculator using the key sequence (CLR) (Memory) (Yes) . Note that this clears all the calculator memories. To store the result of an expression just calculated (i.e. an answer displayed in the output area of the calculator screen) in the ‘M’ calculator memory, use the key sequence (STO) (M). Here we are using the second function of the (or recall) button, which is called ‘STO’ (or store). After selecting the store function, we need to tell the calculator which memory the value is to be stored in. These memories are labelled in red on some of the calculator keys, and the ‘M’ memory is obtained by pressing the key. We could read the key sequence as ‘store the current result into the M memory’. Once (or (STO)) has been pressed, the display indicator RCL (or STO) is shown on the display to indicate that the calculator is waiting to know which memory to recall (store) the value from (in). To display the current contents of the ‘M’ memory, press (M). The value stored in memory can also be used as part of a subsequent calculation by inserting the ‘letter’ M into the appropriate point of the expression using (M). For example, to find the square of the value currently stored in the ‘M’ memory, , we can use the key sequence (M) . When there is a value stored in the ‘M’ memory, the display indicator M is shown at the top of the display. Activity 12 Using memory Store the value of in the ‘M’ memory of the calculator and then use this stored value to evaluate to 3 significant figures. Answer The value of (which equals 0.0796 to 3 significant figures) can be calculated using and stored in the ‘M’ memory using the key sequence (STO) (M). This value can then be used to find the final result using (M) , which gives 0.101 to 3 significant figures, as expected. 4.3 Other ‘M’ memory operations The value stored in the ‘M’ memory can also be changed by adding or subtracting the result of a further calculation: To add the result of the latest calculation to the value currently in the memory, press . To subtract the value of the latest calculation from the value currently in the memory, use the key sequence (M-). Expressions can also be stored in, added to or subtracted from the memory at the same time as they are evaluated by replacing the at the end of a calculation with one of the memory access sequences. For example, to calculate and store the result straight into the memory, use the key sequence (STO) (M). To clear the ‘M’ memory alone, simply store the value 0 in it using the key sequence (STO) (M). 4.4 Other memories The calculator also has 6 other memories, labelled ‘A’, ‘B’, ‘C’, ‘D’, ‘X’ and ‘Y’, which are accessed using several of the keys in the lower half of the function key area of the calculator. Each memory name is printed in red above the key used to access it. These memories can be used in exactly the same way as the ‘M’ memory, except that there are no equivalents to the ‘add to memory’ () and ‘subtract from memory’ ((M-)) functions, and no display indicators. 5 Scientific notation on your calculator Displaying numbers in scientific notation on your calculator If the result of a calculation is a number greater than or equal to (i.e. ), the calculator will automatically display the result using scientific notation. For example, calculating gives the answer , which is displayed on the calculator screen as it is written here. Small numbers are also automatically displayed using scientific notation. However, how small the number needs to be for this to happen depends on the mode the calculator is working in: ‘Norm 1’ mode uses scientific notation for any number less than but greater than . ‘Norm 2’ mode uses scientific notation for any number less than but greater than . ‘Norm’ is short for ‘normal’. In Activity 1 you will have already set your calculator to use Norm 2 mode, and we suggest that for the moment you continue to use this. To change the mode, use the key sequence (SETUP) (Norm) followed by (for Norm 1) or (for Norm 2). You can also set the calculator to always display results using scientific notation with a set number of significant figures using the key sequence (SETUP) (Sci) followed by the number of significant figures required, for example . When your calculator is set in this fashion, the display indicator SCI is displayed at the top of the screen. To cancel such a setting, use one of the key sequences given above to return to a ‘Norm’ mode. 5.1 Inputting numbers in scientific notation to your calculator Numbers expressed in scientific notation can be input directly to the calculator by using the key on the bottom row of keys. For example, can be entered using the key sequence . Activity 13 Calculating with scientific notation Use the scientific notation functions of your calculator to calculate each of the following, giving your answer in both scientific and ordinary forms. Answer or 85570 or 0.129258 or 2343750 6 Powers and surds on your calculator Inputting fractional and negative powers In Activity 4 you saw how to use the key to input powers on the calculator. The key can be used with other functions, such as the fraction template , to calculate fractional and negative powers. Activity 14 Calculating more powers Calculate each of the following using your calculator, giving your answer correct to 3 significant figures. Answer (to 3 significant figures) using the key sequence . (to 3 significant figures) using the key sequence . (to 3 significant figures) using the key sequence . 6.1 Using roots on your calculator Just as there are keys on your calculator for entering powers, roots can also be entered directly. Square roots can be calculated using the key. For example, can be entered using . Cube roots are entered using the second function of this key. For higher roots, such as fourth or fifth roots you need to use the more general template, which is the second function of the key. This template is filled in by using the number and arrow keys (and ) in a way similar to that used when the fraction template is completed. Activity 15 Calculating roots Calculate each of the following using your calculator, giving your answer correct to 3 significant figures. Answer (to 3 significant figures) using the key sequence . (to 3 significant figures) using the key sequence . (to 3 significant figures) using the key sequence . or 18.4 (to 3 significant figures) using the key sequence (and using to find the decimal result). You will notice from the result of Activity 15, part 4 that the calculator sometimes presents answers using surds. This is true only if the calculator is in the recommended Math mode. To find the decimal equivalent of an answer like this, you can use the or keys that you used earlier to find the decimal forms of fractional answers. 6.2 Inserting a missing root Sometimes when entering into your calculator an expression involving roots, you may accidentally forget to press the appropriate function key. However, moving the cursor to the correct point and pressing the missing key, as in section 1, will not work as this simply inserts an empty template. If you wish to edit an expression to insert a missing root, first move the cursor to the correct place – that is, to the left of the number. Then activate the ‘Insert’ function by pressing (INS), and finally press the appropriate root key. 7 Trigonometric ratios on your calculator There are various different units in which an angle can be measured, degrees being one of the possibilities. Before using your calculator to find the values of the trigonometric ratios of angles measured in degrees, you need to ensure that it is set to use the correct units. Always check that your calculator is using the correct system of angle measurement before using trigonometric ratios. Your calculator is set to use degrees if the display indicator is shown at the top of the screen. If you see the indicator or , then your calculator is set to use different units for measuring angles. Figure 9 The degrees setting Show description|Hide description The top row of the screen contains the letter D (white text on black) in the centre, and the word ‘math’ towards the right end. Figure 9 The degrees setting To set your calculator to work in degrees, use the key sequence (SETUP) (Deg). To calculate the sine, cosine or tangent of an angle, press the , or key and then type in the size of the angle. Note that the , and keys automatically open a bracket for you. If you are simply calculating the sine, cosine or tangent of an angle, just press after entering the angle – there is no need to close the bracket. If you are using these ratios as part of a larger calculation, then you will need to remember to close the bracket yourself (by pressing ) before entering the remainder of the calculation. Some older models of calculator require the angle to be input first, followed by the , or button. Activity 16 Trigonometric ratios on your calculator Calculate the value of each of the following using your calculator, giving your answers correct to 3 significant figures. Answer (to 3 significant figures). If you got 0.657, then your calculator is currently set to calculate in radians, so reset it to work in degrees using (SETUP) (Deg). (to 3 significant figures). or or 0.707 (to 3 significant figures). (to 3 significant figures). (to 3 significant figures). Remember to close the bracket that pressing opens before entering the ‘+5’. If you obtained the answer 0.845 (to 3 significant figures), you probably calculated instead. Note that when entering this expression in your calculator, it is possible to omit explicitly entering the multiplication between the 2 and since the calculator will assume it. 6. . Note that means first find the sine of , then square the answer. The key sequence to enter into the calculator is thus . The first is necessary to close the bracket automatically opened when pressing , and the second closes the bracket opened at the start of the sequence. Since the calculator evaluates the sine as soon as it encounters the first closing bracket, it is possible to enter this expression using the alternative sequence , but this is not recommended as the former is more clear. It is a property of trigonometric ratios that for any angle , . You will notice from the answer to part (3) that the calculator displays the ratios of some angles as fractions, involving surds where needed, and not in decimal form. The decimal form can be found using or . 8 Finding angles from trigonometric ratios Inverse trigonometric values can be found using the second functions , and of the , and keys. These functions are used in a similar manner to , and . Activity 17 Finding angles from trigonometric ratios Calculate the value of each of the following expressions using your calculator, where possible, giving your answers correct to 1 decimal place. Answer . If you obtained the answer or 0.5236 (to 4 decimal places), then your calculator is set to work in radians. Make sure that you are working in degrees. 2. (to 1 decimal place). 3. (to 1 decimal place). 4. The calculation is not possible, and your calculator will give a ‘Math Error’. There is no angle whose sine is 2, because the hypotenuse is always the largest side of a right-angled triangle, and hence the maximum value the sine of an angle can be is 1. In part (1) of the activity above, you used your calculator to find an angle whose sine is 0.5. This is not the only angle whose sine is 0.5, but you can use this angle to find the other angles. Similar remarks apply to parts (2) and (3). 9 Radians on your calculator Your calculator can be set to calculate trigonometric functions using the radian measure for angles, instead of degrees, by using the key sequence (SETUP) (Rad). When in this mode, the display indicator is shown. Activity 18 Radians on your calculator In this activity, the angles are measured in radians. Find the values of the following expressions, giving your answers correct to 3 significant figures. Answer Remember to set your calculator to work in radians before entering these calculations. (to 3 significant figures). . Remember: can be input to the calculator using . 3. or 0.785 (to 3 significant figures). Notice from the final example in this activity that where an answer is a simple (possibly fractional) multiple of , the answer is displayed in terms of rather than as a decimal number. 10 Logarithms on your calculator Logarithms to base 10 of numbers can be found using the key. For example, can be calculated using the key sequence . Note that as with the trigonometric functions, the key automatically opens a bracket that must be closed if you are using the calculated logarithm as part of a longer calculation. The second function of the key , accessed using , can be used as an alternative to when calculating powers of 10. Activity 19 Calculating logarithms Use your calculator to find the values of the following. , , (1 billion) , Answer All the calculations are logarithms to base 10, so we use the key. , , (1 billion) , . 11 Natural logarithms and powers of e on your calculator Natural logarithms, for example , can be evaluated on your calculator using the key. The second function of this key () permits the calculation of powers of . Note that an approximate value for itself can be obtained using the key sequence (). Remember that means the same as ln. Some calculators have a key instead of . Activity 20 Calculating natural logarithms and powers of e Calculate the value of each of the following using your calculator, giving each answer to 3 significant figures. Answer (to 3 significant figures). (to 3 significant figures). (to 3 significant figures). Notice that this is about three times the answer to part (2). In fact, the exact value of is precisely three times the exact value of . This is because , and for any base of logarithm, . This is an important result for logarithms. In this case we have . 4. (to 3 significant figures). To calculate this answer correctly, you need to remember to close the bracket after the 3 on your calculator display. 5. (to 3 significant figures). 12 Calculator reference guide Calculator modes General modes The calculator can operate in several different modes: COMP, which is used for general calculations STAT, which is used for statistical calculations TABLE, which is used for generating tables of numbers. COMP is short for ‘computation’, and STAT is short for ‘statistics’. Comp mode is selected by using the key sequence (COMP). Mathematics modes There are two different ways in which mathematics can be input to and displayed on the calculator: Math mode, in which fractions are entered and displayed in their proper mathematical form – for example, – and some irrational numbers are displayed as surds (such as ) or multiples of (such as ). Math mode is selected by using the key sequence (SETUP) (MthIO). Linear mode, in which fractions such as are entered and displayed using a linear notation – for example, 2 3 – and all irrational numbers (such as and ) are displayed as decimal approximations. Linear mode is selected by using the key sequence (SETUP) (LineIO). Put simply, an irrational number is one that cannot be expressed as a simple fraction. You know that you are in Math mode if the word Math is shown near the right-hand side of the top of the calculator display. If this is not shown, you are using Linear mode. In Math mode, you can force an answer to be displayed as a decimal using , or you can toggle between the mathematical and decimal outputs using . Fraction display modes It is possible to set the calculator so that answers that are top-heavy fractions (such as ) are always displayed as mixed numbers (such as ). To set the calculator to display as mixed numbers, use the key sequence (SETUP) (ab/c). To set the calculator to display as top-heavy fractions, use the key sequence (SETUP) (d/c). Decimal display modes The calculator can be set to display decimal numbers in various different ways: Normal 1 (Norm 1) mode uses scientific notation for any number less than 0.01 but greater than −0.01, and otherwise displays answers to however decimal places they have, subject to the maximum that will fit on the display. This mode is entered using (SETUP) (Norm). Normal 2 (Norm 2) mode uses scientific notation for any number less than 0.000000001 but greater than −0.000000001, but otherwise displays answers to however many decimal places they have, subject to the maximum that will fit on the display. This mode is entered using (SETUP) (Norm). Scientific notation mode (Sci) displays all decimal numbers in scientific notation using a specified number of significant figures. This mode is entered using (SETUP) (Sci) followed by the number of significant figures required, for example . When your calculator is set in this mode, the display indicator SCI is shown. Fixed decimal place mode (Fix) displays all decimal numbers to a given number of decimal places (unless scientific notation is needed to fit the result on the display). This mode is entered using (SETUP) (Fix) followed by the number of decimal places required, for example . When your calculator is set in this mode, the display indicator FIX is shown. 12.1 Display indicators These are symbols shown on the calculator display to indicate its current state of operation. | Symbol on display | Meaning | --- | | The key has been pressed. | | The key has been pressed. | | M | A value is stored in the ‘M’ memory. | | STO | The STORE key ((STO)) has been pressed. | | RCL | The RECALL () key has been pressed. | | The calculator is set to measure angles in degrees. | | The calculator is set to measure angles in radians. | | The calculator is set to measure angles in gradians. | | FIX | The calculator is set to display answers to a fixed number of decimal places. | | SCI | The calculator is set to display answers in scientific notation with a fixed number of significant figures. | | | The calculator is set to use Math mode for input and display. | | | The calculator is in statistics mode. | | There is more information available than can be displayed, and this can be accessed using the up/down cursor keys. | | The line displayed is longer than can fit on the display. Other parts of the line can be displayed by scrolling using the left/right cursor keys: and . | 12.2 Common operations | To | Key sequence | --- | | Turn on | | Turn off | Note that the calculator will automatically turn off if not used for about 6 minutes. | | Adjust the display contrast | (SETUP) (◄CONT►) then use and to adjust, and press when finished. | | Restore the factory settings | (CLR) (Setup) (Yes) | | Clear the contents of all memories | (CLR) (Memory) (Yes) | | Restore the default settings and clear all memories | (CLR) (All) (Yes) | | Cancel a calculation, or exit menus | | Obtain an answer | | Obtain a decimal answer in Math mode | | Toggle between exact and decimal answers in Math mode | | Toggle between displaying fractions as improper fractions or mixed numbers in Math mode | | Use the result of the previous calculation… | | … e.g. to find 42 minus the previous result | | Store a value in ‘M’ memory | (STO) (M) | | Clear a value stored in the M memory | (STO) (M) | | Add a value to that in the ‘M’ memory | | Subtract a value from that in the ‘M’ memory | | | Display the value stored in the ‘M’ memory | | | Use the value in the ‘M’ memory in a calculation… | | … e.g. to find 42 minus the value in the ‘M’ memory | (M) | | Set angles to be measured in degrees | (SETUP) (Deg) | | Set angles to be measured in radians | (SETUP) (Rad) | | Set answers to be displayed to 3 decimal places (To use a different number of decimal places, replace with that number) | (SETUP) (Fix) | | Set answers to be displayed in scientific notation with 3 significant figures (To use a different number of significant figures, replace with that number) | (SETUP) (Sci) | | Cancel any display rounding settings | (SETUP) (Norm) | The A, B, C, D, X, Y memories are used similarly to the M memory, except that these have no ‘add to’ and ‘subtract from’ functions. 12.3 Entering mathematics | To enter | Key sequence | --- | | | | or | | or | | In Math mode: In Linear mode: | | In Math mode: In Linear mode: | | | and are mathematical constants. When using trigonometric functions, ensure that your calculator is set to use the correct units: degrees or radians. and are calculated in a similar way to . Conclusion This free course provided an introduction to studying Mathematics and Statistics. It took you through a series of exercises designed to develop your approach to study and learning at a distance and helped to improve your confidence as an independent learner. Keep on learning Study another free course There are more than 800 courses on OpenLearn for you to choose from on a range of subjects. Find out more about all our free courses. Take your studies further Find out more about studying with The Open University by visiting our online prospectus. If you are new to university study, you may be interested in our Access Courses or Certificates. What’s new from OpenLearn? Sign up to our newsletter or view a sample. For reference, full URLs to pages listed above: OpenLearn – www.open.edu/ openlearn/ free-courses Visiting our online prospectus – www.open.ac.uk/ courses Access Courses – www.open.ac.uk/ courses/ do-it/ access Certificates – www.open.ac.uk/ courses/ certificates-he Newsletter ­– www.open.edu/ openlearn/ about-openlearn/ subscribe-the-openlearn-newsletter Acknowledgements Grateful acknowledgement is made to the following: Course image: kaboompics.com in Pexels made available under Creative Commons Public Domain 1.0 Licence. Figure 1: © Casio Electronics Co Ltd. Don't miss out: If reading this text has inspired you to learn more, you may be interested in joining the millions of people who discover our free learning resources and qualifications by visiting The Open University - www.open.edu/ openlearn/ free-courses ◄ Using a scientific calculator Copyright © 2016 The Open University
15691
https://pmc.ncbi.nlm.nih.gov/articles/PMC11266805/
Obstructive Shock, from Diagnosis to Treatment - 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 Rev Cardiovasc Med . 2022 Jun 29;23(7):248. doi: 10.31083/j.rcm2307248 Search in PMC Search in PubMed View in NLM Catalog Add to search Obstructive Shock, from Diagnosis to Treatment Viviane Zotzmann Viviane Zotzmann 1 Interdisciplinary Medical Intensive Care (IMIT), Medical Center - University of Freiburg, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany Find articles by Viviane Zotzmann 1, Felix A Rottmann Felix A Rottmann 1 Interdisciplinary Medical Intensive Care (IMIT), Medical Center - University of Freiburg, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany 2 Department of Medicine IV, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany Find articles by Felix A Rottmann 1,2, Katharina Müller-Pelzer Katharina Müller-Pelzer 3 Department of Diagnostic and Interventional Radiology, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany Find articles by Katharina Müller-Pelzer 3, Christoph Bode Christoph Bode 1 Interdisciplinary Medical Intensive Care (IMIT), Medical Center - University of Freiburg, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany 4 Department of Cardiology and Angiology I, Heart Center Freiburg University, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany Find articles by Christoph Bode 1,4, Tobias Wengenmayer Tobias Wengenmayer 1 Interdisciplinary Medical Intensive Care (IMIT), Medical Center - University of Freiburg, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany Find articles by Tobias Wengenmayer 1, Dawid L Staudacher Dawid L Staudacher 1 Interdisciplinary Medical Intensive Care (IMIT), Medical Center - University of Freiburg, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany Find articles by Dawid L Staudacher 1, Editors: Karim Bendjelid, Raphael Giraud Author information Article notes Copyright and License information 1 Interdisciplinary Medical Intensive Care (IMIT), Medical Center - University of Freiburg, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany 2 Department of Medicine IV, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany 3 Department of Diagnostic and Interventional Radiology, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany 4 Department of Cardiology and Angiology I, Heart Center Freiburg University, Faculty of Medicine, University of Freiburg, 79106 Freiburg, Germany Correspondence: dawid.staudacher@uniklinik-freiburg.de (Dawid L. Staudacher) Roles Karim Bendjelid: Academic Editor Raphael Giraud: Academic Editor Received 2022 Feb 25; Revised 2022 Apr 30; Accepted 2022 May 26; Collection date 2022 Jul. Copyright: © 2022 The Author(s). Published by IMR Press. This is an open access article under the CC BY 4.0 license. PMC Copyright notice PMCID: PMC11266805 PMID: 39076909 Abstract Shock is a life threatening pathological condition characterized by inadequate tissue oxygen supply. Four different subgroups of shock have been proposed according to the mechanism causing the shock. Of these, obstructive shock is characterized by reduction in cardiac output due to noncardiac diseases. The most recognized causes include pulmonary embolism, tension pneumothorax, pericardial tamponade and aortic dissection. Since obstructive shock typically cannot be stabilized unless cause for shock is resolved, diagnosis of the underlying disease is eminent. In this review, we therefore focus on diagnosis of obstructive shock and suggest a structured approach in three steps including clinical examination, ultrasound examination using the rapid ultrasound in shock (RUSH) protocol and radiological imaging if needed. Keywords: obstructive shock, circulatory shock, review 1. Introduction 1.1 Definition of Obstructive Shock In general, shock is a circulatory failure that results in inadequate cellular oxygen utilization . According to the mechanism causing the shock, four different types have been defined, including cardiogenic, hypovolemic, distributive and obstructive shock based on a classification by Cox and Hinshaw from 1972. The classification of shock has hardly changed over the past few decades [3, 4]. Common to all types of shock is a mismatch of oxygen supply and consumption, which ultimately leads to poor perfusion and multiple organ failure . Three components affect cardiac output: the blood volume, the cardiac output and vascular resistance . A significant disturbance in any of these three factors may result in a critical undersupply of the whole body, a state which is defined as shock . While cardiogenic and obstructive shock both result in shock due to undersupply with blood , it is important to distinguish between those two entities since cardiogenic shock is caused by primary cardiac dysfunction while in obstructive shock is caused by extra cardiac diseases (like cardiac tamponade). The etiology of shock is of immense importance since treatment and underlying diseases differ. The most frequent causes of obstructive shock are given in Table 1 and include pulmonary embolism, pneumothorax and cardiac tamponade. Table 1. Causes of obstructive shock. Cause Effect Pathology Pathology Intravasal/intraluminal Extravasal/extraluminal Disorders of diastolic filling RV-Preload Tension pneumothorax Cardiac tamponade Caval compression syndrome Ventilation with high PEEP Obstruction in the pulmonary circulation RV-Afterload Pulmonary embolism Pulmonary compression syndrome by mediastinal mass LV-Preload Intracardiac mass Obstruction in the aortic circulation LV-Afterload Leriche Syndrome Aortic dissection Open in a new tab Abbreviations: RV, right ventricle; LV, left ventricle; PEEP, positive end-expiratory pressure. 1.2 Epidemiology There are no reliable data on the frequency of obstructive shock. The incidences of the most common causes of obstructive shock are given below and can serve as a surrogate for obstructive shock incidence. A population-based incidence for aortic dissection suggests an incidence of 2.3–16.3/100,000 inhabitants per year . Venous thromboembolism, which includes thrombosis of deep leg and pelvic veins and pulmonary embolism is much more frequent and varies between 100–200/100,000 inhabitants per year. Of these, about one third present with pulmonary embolism . The overall person consulting rate for pneumothorax (primary and secondary combined) in the Great Britain was 24.0/100,000 per year for men and 9.8/100,000 for women . A pericardial effusion is a frequent finding in patients with pulmonary hypertension, AIDS or malignoma. In a study of patients presenting in the emergency department with unexplained dyspnea, 13.6% had a pericardial effusion . However, the frequency of obstructive shock cannot be derived from the incidence of these underlying diseases alone. Therefore, more data is required on this matter in order to understand the incidence of obstructive shock. 1.3 Pathophysiology Etiologically, obstructive shock is caused by an impaired diastolic filling and thus a reduced cardiac RV- or LV-preload (venous return). A reduced preload is caused by tension pneumothorax, V. cava compression syndrome, mediastinal tumors, pericardial effusion or ventilation with a very high PEEP level. On the other hand, diseases which lead to an increased afterload and thus cardiac output may lead to obstructive shock. Causes associated with an increase in afterload are, for example, an aortic dissection, pulmonary embolism or Leriche syndrome. A pulmonary embolism or mediastinal space-occupying mass increases right-ventricular afterload, while decreasing left ventricular preload. The same mechanisms may occur in case of obstructive intracardial mass as displayed in Fig. 1, (Ref. ). Obstruction of the aortic blood flow however increases left ventricular afterload (e.g., Leriche syndrome [aortoiliac occlusive disease], aortic dissection) . Fig. 1. Open in a new tab Transversal (A,B) and coronal (C) reconstructions of a computed tomography (CT) angiography shwoeing a rare cause for obstructive shock. (A) Tumor mass (TU) infiltrating the right ventricular wall and left ventricular ouitflow tract. (B) The tumor (TU) mass compresses the right artrium (RA) and right ventricle (RV). The tumor has no direct contact to the left ventricle (LV). (C) Demonstrates tumor grophs along the right heart in coronal reconstruction. Adapted from . 2. Prognoses The prognosis of obstructive shock strictly depends on the appropriate treatment of the underlying cause and the extent of the tissue damage which had occurred by the time the shock was treated. Since no resilient data on obstructive shock mortality is available, mortality rates of the three most common causes of obstructive shock are discussed below. 2.1 Tension Pneumothorax Prognosis depends on the extend of the tension component (valve mechanism) and how quickly therapy is initiated. Without treatment, a fulminant tension pneumothorax is fatal due to a cardiac arrest. If a tension pneumothorax is treated appropriately however, prognosis of the pneumothorax itself is very good. Based on the cause-of-death statistics, disease-specific mortality for spontaneous pneumothorax is 0.094/100,000 population annually, with a lethality of 0.7% , and might be even higher in secondary pneumothorax. Mortality rates in Great Britain between 1950–1997 were 1.26/million per year for men and 0.62/million per year for women . It is important however to consider that most of these patients were not in obstructive shock. 2.2 Pulmonary Embolism According to an epidemiological study from 6 European countries, more than 317,000 deaths were associated with pulmonary embolism in 2004 . Of these, 34% presented with fatal pulmonary embolism; 59% of the deaths were caused by pulmonary embolism that had not been diagnosed during lifetime. The average mortality rate from pulmonary embolism is given as 11% (40,000 deaths/year in Germany), but is significantly higher in patients in shock (40–50%) or in patients requiring resuscitation (70–80%) . 2.3 Aortic Dissection Due to the high preclinical mortality and missing forensic and clinical autopsy data, the number of unreported cases can safely be assumed to be very high . 26% of the patients who receive surgical treatment due to a type A dissection die within the clinic. The mortality of the conservatively treated patients was reported 58% . 3. Diagnosis 3.1 Anamnesis The diagnostic algorithm in case of shock is guided by patient history and is a crucial part of the diagnostic workup. Typical patient histories for pulmonary embolism include previous arterial embolisms, thrombosis, malignant disease, immobilization or surgery in the preceding four weeks . Cardiac tamponade is suspected primarily in trauma patients as possible sequel of rupture of the heart and those who have a history of previous, slowly accumulating cardiac effusion related to preconditions such as viral infections, tuberculosis, uremia or neoplasia or in patients after heart surgery . Tension pneumothorax (and hemothorax) must be suspected especially in trauma patients while spontaneous pneumothorax rarely causes relevant intrathoracic tension . In aortic dissection, sudden onset of chest, back or abdominal pain is the most common symptom and often combined with a neurological deficit as a consequence of reduced cerebral perfusion . Ventilation with high PEEP should also be considered as possible reason for vascular obstruction . An obstructive shock by mediastinal mass can only be suspected in patients with known intrathoracic tumor such as lymphomas. Collapse of large vessels or airways by a mediastinal mass is a known problem under general anesthesia and should be prepared for accordingly . 3.2 Clinical Examination 3.2.1 Physical Examination Clinical findings in shock include shivering, paleness of skin, arterial hypotension, tachycardia, centralization, dyspnea, hypoxemia, impaired mental status including syncope and reduced urine output and can all rapidly lead up to circulatory death [16, 23, 25]. Thus clinical findings specific for different types of obstructive shock are crucial to direct rapid treatment . If an upper inflow congestion (bulging neck veins, cyanotic-livid complexion) is present, obstructive shock should be suspected (pneumothorax, cardiac tamponade). Clinical examination can also guide towards Leriche syndrome (lack of foot, popliteal and femoral pulses, cold legs on both sides, pale lower extremities). A juxtaposition of focal neurological symptoms and shock can hint towards aortic dissection. 3.2.2 Auscultation Decreased or missing respiratory sounds on one hemithorax are typical signs of pneumo- and hemothorax . In auscultation of the heart Pulsus paradoxus as a possible correlate of obstructive shock was described as early as 1873 in the context of constrictive pericarditis . Pulsus paradoxus describes a pathological increase in the physiological decrease of systolic blood pressure during inspiration and includes the extreme of no palpable pulse during inspiration while the formal definition only requires a decrease of systolic blood pressure by 10 mmHg during inspiration . Even though first described in constrictive pericarditis, Pulsus paradoxus can also be found in pulmonary embolism as well as cardiac tamponade and other causes for increased intrathoracic pressure such as severe asthma . Physiologically inspiration reduces intrathoracic pressure, increases blood pooling in the lung and consequently reduces left ventricular filling and cardiac output [23, 29]—the mechanisms that decrease systolic blood pressure during inspiration even further in events such as cardiac tamponade and pulmonary embolism are still debated . 3.2.3 Inspection and Palpation Subcutaneous emphysema or asymmetrical breathing patterns are signs of tension pneumothorax. Regarding obstructive shock, hemoptysis is commonly seen in pulmonary embolism . Classical clinical finding in aortic dissection is a pulse deficit but can only be found in a minority of cases . In aortic dissection as well as Leriche’s syndrome, skin mottling or acral gangrene exposes inadequately perfused tissue . In aortic dissection obvious ischemia of the lower limbs is rare . 3.2.3 Electrocardiogram (ECG) Common ECG-findings in pulmonary embolism include sinus tachycardia, atrial fibrillation, (in-) complete right bundle branch block, a S1Q3T3 pattern and T-wave inversion in V1-4 but are all reflective of right ventricular strain and consequently could be found in other types of obstructive shock . Electrical alternans as correlate of the intrapericardial fluid collection can be found in cardiac tamponade . Low-voltage as a possible ECG-correlate of tamponade occurs in pleural effusion, emphysema, obesity and anasarca as well. Aortic dissection can present with signs of myocardial ischemia . 3.3 Imaging 3.3.1 Sonography It has been shown decades ago that transthoracic echo (TTE) is sensitive and specific for cardiac tamponade and can be used to measure the effusion volume [37, 38]. Flow over the mitral- and trikuspidal-valve are dependent on in- and expiration and differences are increased in pericardial effusion . Today, hemodynamic relevant tamponades are characterized by complete collapse of the right atrium (1/3 of cardiac cycle), circumferential pericardial 2 cm (in diastole), dilated V. cava inferior (2.5 cm, 50% inspiratory collapse), right ventricular collapse, left atrial collapse, increased mitral- and tricuspidal valve flow variations . In suspected high risk pulmonary embolism which is characterized by hemodynamic instability computed tomography pulmonary angiogram (CTPA) or bedside echocardiography remain gold standard for diagnosis [41, 42]. Right ventricular failure due to pressure overload can be shown in echocardiography but cannot differentiate different types of right ventricular afterload increase. Classical echocardiographic criteria are right ventricular dilation and increased tricuspid regurgitation (velocity 2.7 ms/s) . A reduced tricuspid annular plane systolic excursion (TAPSE) can occur in pulmonary embolism as well . The most specific if not sensitive echocardiographic signs for pulmonary embolism are the “60/60 sign”, McConnell sign and right heart thrombi . On the other hand, missing signs of right ventricular overload can exclude pulmonary embolism as the cause for hemodynamic instability . If performed by a skilled operator bedside sonography can be used to accurately diagnose pneumothorax by demonstrating missing lung sliding and showing a lung point [48, 49]. If the patient is stable upright chest x-ray remains a much less operator dependent alternative [50, 51]. For visual examples see Fig. 2. Fig. 2. Open in a new tab Ultrasound in obstructive shock. (A) TTE shows a wormously configured thrombus in the right atrium with clearly dilated right ventricles and completely emptied left ventricle. RA, right atrium; RV, right ventricle; LA, left atrium. (B) Lung sonography (LUS) shows A-lines and a missing lung sliding (white arrow), while lung on the right side is normal. (C) TTE shows a large pericardial effusion (white arrows) with swinging heart. RA, right atrium; RV, right ventricle; LA, left atrium; LV, left ventricle. We recommend using a structured approach for the ultrasound examination, which is usefull in any shock patient and might be lifesaving in case of obstructive shock. Point of care ultrasound (POCUS) allows for rapid, real-time evaluation of cardiovascular and respiratory pathology . The RUSH Protocol (Rapid Ultrasound in Shock) was developed by Perera et al. , see Table 2 (Ref. ). Table 2. Rapid Ultrasound in Shock (RUSH) protocol: ultrasonographic findings seen with classic shock states, modified from . RUSH Evaluation Obstructive shock Hypovolemic shock Cardiogenic shock Distributive shock Pump Hypercontractile heart Hypercontractile heart Hypocontractile heart Hypercontractile heart (early sepsis) Pericardial effusion Small chamber size Dilated heart Hypocontractile heart (late sepsis) Cardiac tamponade RV strain Cardiac thrombus Tank Distended IVC Flat IVC Distended IVC Normal or small IVC (early sepsis) Distended jugular veins Flat jugular veins Distended jugular veins Peritoneal fluid (sepsis source?) Absent lung-sliding (pneumothorax)Peritoneal fluid (fluid loss)B-Lines (pulmonary edema)Pleural fluid (sepsis source) Stratosphere- Phenomenon (pneumothorax)Pleural or peritoneal fluid (ascites) Distended IVC Pipes DVT (pulmonary embolism)(Abdominal aneurysma)Normal Normal Aortic dissection(Aortic dissection)Possible vasoconstricted arteries Complete or incomplete occlusion of the aorta distal to the renal arteries (Leriche) Open in a new tab Abbreviations: RV, right ventricle; IVC, inferior vena cava; DVT, deep vein thrombosis. 3.3.2 Radiology If the patient is stable enough for radiological imaging and the diagnosis could not be clarified by ultrasound, CT might help identifying the cause for shock. CT-angiography has been proven superior in sensitivity compared to TTE/TEE for diagnosis of aortic dissection or pulmonary embolism. TTE shows complications such as aortic valve regurgitation or hemorrhagic cardiac tamponade but if it shows the typical intimal flap it can be faster than CT regarding diagnosis . Aortic dissection has been grouped into Stanford A and B ever since 1970 to derive treatment necessity straight from classification, which can be derived from CT imaging—but causal therapy in shock remains limited to surgery [40, 57]. 3.3.3 Blood Tests In patients presenting with any form of shock, several laboratory tests should be considered to detect the cause for shock or complications of shock. Importantly, no lab test is specific for obstructive shock and therefore should not delay diagnosis and therapy. D-dimers is not primarily recommended in shock suspected to be caused by pulmonary embolism or aortic dissection as CT-angiography is the faster diagnostic tool in both cases. Upcoming diagnostic tests for aortic dissection are related to vascular damage include calponin , plasma matrix metalloproteinase 8 and tenascin-C and may enter clinical routine diagnostic in the near future. There is no diagnostic laboratory test specific for cardiac tamponade or pneumothorax. Blood tests to consider might include blood gas analysis (to detect hypoxemia and acidosis), Lactate (reduced tissue perfusion ), glucose (hypoglycaemia), blood cell count (anemia), procalcitonin (inflammation), coagulation and platelet count (bleeding tendency), d-dimer (pulmonary embolism or aortic dissection ), troponin, pro-brain natriuretic peptide (myocardial infarction or peulmonary embolism [64, 65, 66]), creatinine, aspartate transaminase and aspartat-aminotransferase (end-organ damage). 4. Diagnostic Algorithm A pragmatic approach in obstructive shock is reasonable in order to adequately identify and quickly address the causes of shock. This diagnostic algorithm includes three main steps: first clinical examination, second ultrasound studies following the RUSH protocol and third radiological imaging, see Table 3 and Fig. 3. Table 3. Diagnostic findings in obstructive shock. Diagnostic modality Pericardial tamponade Pneumothorax Pulmonary embolism Clinical investigation Pulsus paradoxus Silent lung on one side Signs for deep vein thrombosis Pulsus paradoxus Venous congestion (jugular) Venous congestion (jugular) Pain TTE Hypercontractile heart No window in cardiac echo Hypocontractile RV Pericardial effusion Hypercontractile heart RV volume elevated Cardiac tamponade RV strain Cardiac thrombus Sonography Distended IVC Absent lung-sliding Distended IVC Distended jugular veins Stratosphere-phenomenon Distended jugular veins Radiological imaging Not required Chest X-ray/CT confirms diagnosis CT confirms diagnosis POCT/Lab Lactate Hypoxemia Hypoxemia Liver failure D-dimer rule out CVP elevated CVP elevated Open in a new tab Abbreviations: TTE, transthoracic echocardiogram; POCT, point of care testing; CVP, central venous pressure; IVC, inferior vena cava; CT, computed tomography; RV, right ventricle. Fig. 3. Open in a new tab Diagnostic algorithm in obstructive shock. 5. Treatment 5.1 General in Shock Shock is a life threatening disease and requires immediate treatment. Since the hallmark of shock is undersupply of the whole body, treatment has to support multiple organ systems to different degrees. When shock is presumed, resuscitation should be started while investigating the cause . In obstructive shock, a causal therapy (e.g., paracentesis or lysis) often is the only lifesaving intervention and must not be delayed. 5.2 Fluid Therapy It is important to note that fluid therapy by itself will not be able to reverse an obstructive shock. In tension pneumothorax, venous return (preload) is limited by an increase in intrapleural and secondarily intrathoracic pressure. When intrathoracic pressure exceeds intraventricular pressure, cardiac output runs dry [67, 68] and cannot be restored by fluid therapy. Especially in obstructive shock with right heart dysfunction, volume expansion can aggravate right ventricular failure . Even though obstructive shock is developed in euvolemic patients with normal myocardial function, most patients are to a certain degree volume responsive . Opposing the fluid resuscitation strategies in septic shock, fluid therapy in obstructive shock therefore can only buy time for a causal therapy. Choosing the right clinical surrogate for evaluating volume responsiveness however is hard and the gold standard of measuring cardiac output is dependent on the availability of advanced hemodynamic monitoring or experience in cardiac ultrasound. Passive leg raise maneuver should be used in order to test fluid responsiveness in obstructive shock. This intervention can be performed rather safely even in patients in shock . Performing a passive leg raise maneuver correctly however might be challenging in unstable patients since effects on hemodynamics are transient . In case passive leg raise maneuver is not feasible, a mini-fluid challenge of 100–200 mL crystalloid fluid over 10 minutes can be used in order to test volume responsiveness. Larger volumes should be avoided since fluid overload negatively impacts the outcome . 5.3 Vasopressors Like in most critically ill patients, vasopressors are frequently used in patients with shock. While vasoplegia is a key feature in septic and anaphylactic shock, it is not in obstructive shock , which is more similar to cardiogenic shock presenting with centralization and elevated total vascular resistance. Here, elevated afterload can decrease cardiac output and aggravate the shock . Still, maintaining a mean arterial pressure of 65 mmHg seems reasonable in shock which might necessitate some vasopressor therapy. There are concerns with increased pulmonary resistance in obstructive shock like pneumothorax which might aggravate shock . Some data suggests that vasopressin might not affect pulmonary resistance [74, 75] therefore offering a superior vasopressor in obstructive shock. There are however also studies suggesting a similar inert effect of norepinephrine on pulmonal vascular resistance and the Guideline of the European Society of Cardiology suggests norepinephrine as first line vasopressor in obstructive shock caused by pulmonary embolism . 5.4 Inotropes In obstructive shock, low cardiac output is not caused by myocardial dysfunction. Therefore, inotropes are often not required for stabilizing patients in obstructive shock and can be detrimental since inotropes can cause vasodilatation and arrhythmias . Inotropes like dobutamine or levosimendan might be however reasonable in case of chronically reduced cardiac function or in patients with increased pulmonary afterload as seen in pulmonary embolism [78, 79] or pulmonary hypertension [79, 80]. Also, inotropes in conjunction with fluid therapy might be able to increase cardiac output in obstructive shock . 5.5 Mechanical Ventilation Patients in severe shock are frequently intubated . Mechanical ventilation by itself increases afterload of the right ventricle , similar in nature to tension pneumothorax . This fact highlights the potential detrimental effects of high pressure ventilation in obstructive shock. In case of mechanical ventilation, a low-volume (following recommendations for ARDS) and low PEEP (positive end expiratory pressure) ventilation should be aimed for in patients with shock due to high right ventricular afterload . Since hypercapnia can further increase pulmonary resistance, it should be avoided . 5.6 Extracorporeal Membrane Oxygenation (ECMO) Venoarterial extracorporeal membrane oxygenation can be a lifesaving therapy in obstructive shock caused by pericardial tamponade [85, 86]. It is important to remember that ECMO might also cause pericardial effusion . There are several reports on decreased venoarterial ECMO flow in case of development of obstructive shock in neonates as well as in adults . On the other hand, it has been suggested that RV dysfunction can be improved by venovenous ECMO in case of pulmonary embolism by reduction on hypoxia and hypercapnia induced pulmonary vasoconstriction . Similarly, venovenous ECMO unloads the right ventricle in patients with pneumonia by reducing mean pulmonary artery pressure . On the other hand, venovenous ECMO is contraindicated in severe right heart failure since oxygenated blood will not be pumped into the systemic circulation . Therefore, ECMO configuration and cannula placement has to be carefully evaluated in obstructive shock. 6. Specific 6.1 Pulmonary Artery Embolism In pulmonary embolism, treatment is guided by short term (in-hospital or 30 day) mortality as proposed by the European Society of Cardiology . Importantly, all patients with hemodynamic instability are considered high risk and need urgent reperfusion therapy either by medical thrombolysis (systemic or catheter guided) or surgical embolectomy [93, 94, 95, 96]. Since best reperfusion strategy depends on patient specific factors including severity of disease and bleeding risk an interdisciplinary team approach is advocated in order to optimize outcomes and might be best organized as a pulmonary embolism response team (PERT) . Independent of the reperfusion strategy, the underlying disease causing PE (carcinoma, blood diseases, etc.) should be investigated and long term anticoagulation should be evaluated. 6.2 Pericardial Tamponade In patients with acute pericardial tamponade, cardiac index improves and right atrial pressure decreases after pericardiocentesis . The European Society of Cardiology suggests percutaneous pericardiocentesis in acute tamponade can be safely performed using echocardiographic or fluoroscopic guidance. In post-cardiotomy patients (or those with recurrent effusion), surgical pericardiocentesis is frequently indicated. Caution might be reasonable in patients with very high bleeding risk or in those with severe pulmonary hypertension, since pericardiocentesis might lead to dilatation of the right ventricle with poor prognosis . A recent review however found the pericardial decompression syndrome (PDS) to be a rare complication in pericardiocentesis . 6.3 Tension Pneumothorax Unstable patients with tension pneumothorax require immediate intervention. A needle decompression is a fast and feasible treatment, provided the operator is trained and uses the right equipment [101, 102]. Chest tube thoracostomy is the treatment of choice in the hospital [103, 104]. Noteworthy, spontaneous pneumothorax rarely leads to a tension pneumothorax due to the lack of a precipitating cause (mechanical ventilation, lung disease or trauma) . Further diagnostics are frequently indicated in patients with tension pneumothorax in order to identify the underlying disease. 7. Conclusions Obstructive shock is a life threatening pathological condition. Since the patient typically cannot be stabilized unless the underlying cause for shock is resolved, exact diagnosis is eminent. We suggest a structured approach in three steps including clinical examination, ultrasound examination using the RUSH protocol and radiological imaging if needed. Acknowledgment Not applicable. Footnotes Publisher’s Note: IMR Press stays neutral with regard to jurisdictional claims in published maps and institutional affiliations. Disclosure of Relationships and Activities The authors declare no relevant activities. Author Contributions Conception and design—VZ, FAR, KMP, CB, TW, DLS; drafting the manuscript—VZ, FAR, DLS; revision—VZ, FAR, KMP, CB, TW, DLS; final approval—VZ, FAR, KMP, CB, TW, DLS. Ethics Approval and Consent to Participate Not applicable. Funding We acknowledge support by the Open Access Publication Fund of the University of Freiburg. Conflict of Interest The authors declare no conflict of interest. References .Vincent JL, De Backer D. Circulatory Shock. New England Journal of Medicine . 2013;369:1726–1734. doi: 10.1056/NEJMra1208943. [DOI] [PubMed] [Google Scholar] .Hinshaw LB, Cox BG. The Fundamental mechanisms of shock; proceedings of a symposium held in Oklahoma City, Oklahoma, October 1–2, 1971 . Plenum Press; New York: 1972. [Google Scholar] .Kumar A, Parrillo JE. Shock: Classification, Pathophysiology, and Approach to Management. Critical Care Medicine . 2008:371–420. [Google Scholar] .Adams HA, Baumann G, Gänsslen A, Janssens U, Knoefel W, Koch T, et al. Die Definitionen der Schockformen. Intensivmedizin Und Notfallmedizin . 2001;38:541–553. doi: 10.1055/s-2001-18174. [DOI] [PubMed] [Google Scholar] .Astiz ME. Pathophysiology and Classification of Shock States . Elsevier; Philadelphia, PA: 2005. p. 677. [Google Scholar] .Guyton AC. The relationship of cardiac output and arterial pressure control. Circulation . 1981;64:1079–1088. doi: 10.1161/01.cir.64.6.1079. [DOI] [PubMed] [Google Scholar] .Adams HA, Baumann G, Cascorbi I, Ebener C, Emmel M, Geiger S, et al. Empfehlungen zur Diagnostik und Therapie der Schockformen der IAG Schock der DIVI. Intensivmedizin + Notfallmedizin . 2005;42:196–210. [Google Scholar] .Wundram M, Falk V, Eulert-Grehn J, Herbst H, Thurau J, Leidel BA, et al. Incidence of acute type a aortic dissection in emergency departments. Scientific Reports . 2020;10:7434. doi: 10.1038/s41598-020-64299-4. [DOI] [PMC free article] [PubMed] [Google Scholar] .Hagan PG, Nienaber CA, Isselbacher EM, Bruckman D, Karavite DJ, Russman PL, et al. The International Registry of Acute Aortic Dissection (IRAD): new insights into an old disease. The Journal of the American Medical Association . 2000;283:897–903. doi: 10.1001/jama.283.7.897. [DOI] [PubMed] [Google Scholar] .Gupta D, Hansell A, Nichols T, Duong T, Ayres JG, Strachan D. Epidemiology of pneumothorax in England. Thorax . 2000;55:666–671. doi: 10.1136/thorax.55.8.666. [DOI] [PMC free article] [PubMed] [Google Scholar] .Blaivas M. Incidence of Pericardial Effusion in Patients Presenting to the Emergency Department with Unexplained Dyspnea. Academic Emergency Medicine . 2001;8:1143–1146. doi: 10.1111/j.1553-2712.2001.tb01130.x. [DOI] [PubMed] [Google Scholar] .Zotzmann V, Wengenmayer T, Lang CN, Staudacher DL, Mueller-Peltzer K, Bamberg F, et al. Case Report: Refusal of an Veno-Arterial Extracorporeal Membrane Oxygenation Due to Malignant Disease? - An Extremely Rare Form of Cardiac Involvement in Acute Myeloid Leukemia. Frontiers in Medicine . 2021;8:584507. doi: 10.3389/fmed.2021.584507. [DOI] [PMC free article] [PubMed] [Google Scholar] .Boyd JL, Stanford GG, Chernow B. The Pathophysiology of Shock. In: Tinker J, Zapol WM, editors. Care of the Critically Ill Patient . Springer London; London: 1992. pp. 243–257. [Google Scholar] .Schnell J, Koryllos A, Lopez-Pastorini A, Lefering R, Stoelben E. Spontaneous Pneumothorax: Epidemiology and Treatment in Germany Between 2011 and 2015. Deutsches Ärzteblatt International . 2017;114:739–744. doi: 10.3238/arztebl.2017.0739. [DOI] [PMC free article] [PubMed] [Google Scholar] .Agnelli G, Anderson F, Arcelus J, Bergqvist D, Brecht J, Greer I, et al. Venous thromboembolism (VTE) in Europe. The number of VTE events and associated morbidity and mortality. Thrombosis and Haemostasis . 2007;98:756–764. doi: 10.1160/TH07-03-0212. [DOI] [PubMed] [Google Scholar] .Standl T, Annecke T, Cascorbi I, R. Heller A, Sabashnikov A, Teske W. The Nomenclature, Definition and Distinction of Types of Shock. Deutsches Ärzteblatt International . 2018;115:757–768. doi: 10.3238/arztebl.2018.0757. [DOI] [PMC free article] [PubMed] [Google Scholar] .van Belle A, Büller HR, Huisman MV, Huisman PM, Kaasjager K, Kamphuisen PW, et al. Effectiveness of Managing Suspected Pulmonary Embolism Using an Algorithm Combining Clinical Probability, D-Dimer Testing, and Computed Tomography. The Journal of the American Medical Association . 2006;295:172–179. doi: 10.1001/jama.295.2.172. [DOI] [PubMed] [Google Scholar] .Spodick DH. Acute Cardiac Tamponade. 2009. [(Accessed: 6 February 2022)]. Available at: .Permanyer-Miralda G. Acute pericardial disease: approach to the aetiologic diagnosis. Heart . 2004;90:252–254. doi: 10.1136/hrt.2003.024802. [DOI] [PMC free article] [PubMed] [Google Scholar] .Chuttani K, Tischler MD, Pandian NG, Lee RT, Mohanty PK. Diagnosis of cardiac tamponade after cardiac surgery: Relative value of clinical, echocardiographic, and hemodynamic signs. American Heart Journal . 1994;127:913–918. doi: 10.1016/0002-8703(94)90561-4. [DOI] [PubMed] [Google Scholar] .Kulshrestha P, Munshi I, Wait R. Profile of Chest Trauma in a Level I Trauma Center. The Journal of Trauma . 2004;57:576–581. doi: 10.1097/01.ta.0000091107.00699.c7. [DOI] [PubMed] [Google Scholar] .Noppen M, De Keukeleire T. Pneumothorax. Respiration . 2008;76:121–127. doi: 10.1159/000135932. [DOI] [PubMed] [Google Scholar] .Pich H, Heller AR. Obstruktiver Schock. Der Anaesthesist . 2015;64:403–419. doi: 10.1007/s00101-015-0031-9. [DOI] [PubMed] [Google Scholar] .Blank RS, de Souza DG. Anesthetic management of patients with an anterior mediastinal mass: Continuing Professional Development. Canadian Journal of Anesthesia/Journal Canadien D’AnesthéSie . 2011;58:853–867. doi: 10.1007/s12630-011-9539-x. [DOI] [PubMed] [Google Scholar] .Soar J, Böttiger BW, Carli P, Couper K, Deakin CD, Djärv T, et al. European Resuscitation Council Guidelines 2021: Adult advanced life support. Resuscitation . 2021;161:115–151. doi: 10.1016/j.resuscitation.2021.02.010. [DOI] [PubMed] [Google Scholar] .Chen SC, Chang KJ, Hsu CY. Accuracy of auscultation in the detection of haemopneumothorax. The European Journal of Surgery . 1998;164:643–645. doi: 10.1080/110241598750005516. [DOI] [PubMed] [Google Scholar] .Kussmaul A. Ueber schwielige Mediastino-Pericarditis und den paradoxen Puls. Berliner Klinische Wochenschrift . 1873;10:433–435, 445. [Google Scholar] .Bilchick KC, Wise RA. Paradoxical physical findings described by Kussmaul: pulsus paradoxus and Kussmaul’s sign. The Lancet . 2002;359:1940–1942. doi: 10.1016/S0140-6736(02)08763-9. [DOI] [PubMed] [Google Scholar] .Alpert JS. The effect of Right Ventricular Dysfunction on Left Ventricular Form and Function. Chest . 2001;119:1632–1633. doi: 10.1378/chest.119.6.1632. [DOI] [PubMed] [Google Scholar] .Melhorn J, Davies HE. The Management of Subcutaneous Emphysema in Pneumothorax: a Literature Review. Current Pulmonology Reports . 2021;10:92–97. [Google Scholar] .Pollack CV, Schreiber D, Goldhaber SZ, Slattery D, Fanikos J, O’Neil BJ, et al. Clinical Characteristics, Management, and Outcomes of Patients Diagnosed with Acute Pulmonary Embolism in the Emergency Department: Initial Report of EMPEROR (Multicenter Emergency Medicine Pulmonary Embolism in the Real World Registry) Journal of the American College of Cardiology . 2011;57:700–706. doi: 10.1016/j.jacc.2010.05.071. [DOI] [PubMed] [Google Scholar] .Pulido-Pérez A, Bergón-Sendín M, Suárez-Fernández R, Muñoz-Martín P, Bouza E. Skin and sepsis: contribution of dermatology to a rapid diagnosis. Infection . 2021;49:617–629. doi: 10.1007/s15010-021-01608-7. [DOI] [PubMed] [Google Scholar] .Erbel R, Aboyans V, Boileau C, Bossone E, Bartolomeo RD, Eggebrecht H, et al. 2014 ESC Guidelines on the diagnosis and treatment of aortic diseases: Document covering acute and chronic aortic diseases of the thoracic and abdominal aorta of the adultThe Task Force for the Diagnosis and Treatment of Aortic Diseases of the European Society of Cardiology (ESC) European Heart Journal . 2014;35:2873–2926. doi: 10.1093/eurheartj/ehu281. [DOI] [PubMed] [Google Scholar] .Shopp JD, Stewart LK, Emmett TW, Kline JA. Findings from 12-lead Electrocardiography that Predict Circulatory Shock from Pulmonary Embolism: Systematic Review and Meta-analysis. Academic Emergency Medicine . 2015;22:1127–1137. doi: 10.1111/acem.12769. [DOI] [PMC free article] [PubMed] [Google Scholar] .Jehangir W, Osman M. Electrical Alternans with Pericardial Tamponade. New England Journal of Medicine . 2015;373:e10. doi: 10.1056/NEJMicm1408805. [DOI] [PubMed] [Google Scholar] .Madias JE, Bazaz R, Agarwal H, Win M, Medepalli L. Anasarca-mediated attenuation of the amplitude of electrocardiogram complexes: a description of a heretofore unrecognized phenomenon. Journal of the American College of Cardiology . 2001;38:756–764. doi: 10.1016/s0735-1097(01)01429-2. [DOI] [PubMed] [Google Scholar] .Horowitz MS, Schultz CS, Stinson EB, Harrison DC, Popp RL. Sensitivity and Specificity of Echocardiographic Diagnosis of Pericardial Effusion. Circulation . 1974;50:239–247. doi: 10.1161/01.cir.50.2.239. [DOI] [PubMed] [Google Scholar] .D’Cruz IA, Hoffman PK. A new cross sectional echocardiographic method for estimating the volume of large pericardial effusions. British Heart Journal . 1991;66:448–451. doi: 10.1136/hrt.66.6.448. [DOI] [PMC free article] [PubMed] [Google Scholar] .Appleton CP, Hatle LK, Popp RL. Cardiac tamponade and pericardial effusion: Respiratory variation in transvalvular flow velocities studied by Doppler echocardiography. Journal of the American College of Cardiology . 1988;11:1020–1030. doi: 10.1016/s0735-1097(98)90060-2. [DOI] [PubMed] [Google Scholar] .Ristić AD, Imazio M, Adler Y, Anastasakis A, Badano LP, Brucato A, et al. Triage strategy for urgent management of cardiac tamponade: a position statement of the European Society of Cardiology Working Group on Myocardial and Pericardial Diseases. European Heart Journal . 2014;35:2279–2284. doi: 10.1093/eurheartj/ehu217. [DOI] [PubMed] [Google Scholar] .Kucher N, Luder CM, Dörnhöfer T, Windecker S, Meier B, Hess OM. Novel management strategy for patients with suspected pulmonary embolism. European Heart Journal . 2003;24:366–376. doi: 10.1016/s0195-668x(02)00476-1. [DOI] [PubMed] [Google Scholar] .Konstantinides SV, Meyer G, Becattini C, Bueno H, Geersing G, Harjola V, et al. 2019 ESC Guidelines for the diagnosis and management of acute pulmonary embolism developed in collaboration with the European Respiratory Society (ERS) European Heart Journal . 2020;41:543–603. doi: 10.1093/eurheartj/ehz405. [DOI] [PubMed] [Google Scholar] .Bova C, Greco F, Misuraca G, Serafini O, Crocco F, Greco A, et al. Diagnostic utility of echocardiography in patients with suspected pulmonary embolism. The American Journal of Emergency Medicine . 2003;21:180–183. doi: 10.1016/s0735-6757(02)42257-7. [DOI] [PubMed] [Google Scholar] .Pruszczyk P, Goliszek S, Lichodziejewska B, Kostrubiec M, Ciurzyński M, Kurnicka K, et al. Prognostic Value of Echocardiography in Normotensive Patients with Acute Pulmonary Embolism. JACC: Cardiovascular Imaging . 2014;7:553–560. doi: 10.1016/j.jcmg.2013.11.004. [DOI] [PubMed] [Google Scholar] .Kurzyna M, Torbicki A, Pruszczyk P, Burakowska B, Fijałkowska A, Kober J, et al. Disturbed right ventricular ejection pattern as a new Doppler echocardiographic sign of acute pulmonary embolism. The American Journal of Cardiology . 2002;90:507–511. doi: 10.1016/s0002-9149(02)02523-7. [DOI] [PubMed] [Google Scholar] .Barrios D, Rosa-Salazar V, Jiménez D, Morillo R, Muriel A, del Toro J, et al. Right heart thrombi in pulmonary embolism. European Respiratory Journal . 2016;48:1377–1385. doi: 10.1183/13993003.01044-2016. [DOI] [PubMed] [Google Scholar] .Dresden S, Mitchell P, Rahimi L, Leo M, Rubin-Smith J, Bibi S, et al. Right Ventricular Dilatation on Bedside Echocardiography Performed by Emergency Physicians Aids in the Diagnosis of Pulmonary Embolism. Annals of Emergency Medicine . 2014;63:16–24. doi: 10.1016/j.annemergmed.2013.08.016. [DOI] [PubMed] [Google Scholar] .Zhang M, Liu ZH, Yang JX, Gan JX, Xu SW, You XD, et al. Rapid detection of pneumothorax by ultrasonography in patients with multiple trauma. Critical Care . 2006;10:R112. doi: 10.1186/cc5004. [DOI] [PMC free article] [PubMed] [Google Scholar] .Shostak E, Brylka D, Krepp J, Pua B, Sanders A. Bedside Sonography for Detection of Postprocedure Pneumothorax. Journal of Ultrasound in Medicine . 2013;32:1003–1009. doi: 10.7863/ultra.32.6.1003. [DOI] [PMC free article] [PubMed] [Google Scholar] .Seow A, Kazerooni EA, Pernicano PG, Neary M. Comparison of upright inspiratory and expiratory chest radiographs for detecting pneumothoraces. American Journal of Roentgenology . 1996;166:313–316. doi: 10.2214/ajr.166.2.8553937. [DOI] [PubMed] [Google Scholar] .Thomsen L, Natho O, Feigen U, Schulz U, Kivelitz D. Value of Digital Radiography in Expiration in Detection of Pneumothorax. Rofo . 2014;186:267–273. doi: 10.1055/s-0033-1350566. [DOI] [PubMed] [Google Scholar] .Kirkpatrick JN, Grimm R, Johri AM, Kimura BJ, Kort S, Labovitz AJ, et al. Recommendations for Echocardiography Laboratories Participating in Cardiac Point of Care Cardiac Ultrasound (POCUS) and Critical Care Echocardiography Training: Report from the American Society of Echocardiography. Journal of the American Society of Echocardiography . 2020;33:409–422. doi: 10.1016/j.echo.2020.01.008. e4. [DOI] [PubMed] [Google Scholar] .Perera P, Mailhot T, Riley D, Mandavia D. The RUSH Exam: Rapid Ultrasound in SHock in the Evaluation of the Critically lll. Emergency Medicine Clinics of North America . 2010;28:29–56, vii. doi: 10.1016/j.emc.2009.09.010. [DOI] [PubMed] [Google Scholar] .Sommer T, Fehske W, Holzknecht N, Smekal AV, Keller E, Lutterbey G, et al. Aortic dissection: a comparative study of diagnosis with spiral CT, multiplanar transesophageal echocardiography, and MR imaging. Radiology . 1996;199:347–352. doi: 10.1148/radiology.199.2.8668776. [DOI] [PubMed] [Google Scholar] .Gilon D, Mehta RH, Oh JK, Januzzi JL, Bossone E, Cooper JV, et al. Characteristics and in-Hospital Outcomes of Patients with Cardiac Tamponade Complicating Type a Acute Aortic Dissection. The American Journal of Cardiology . 2009;103:1029–1031. doi: 10.1016/j.amjcard.2008.12.013. [DOI] [PubMed] [Google Scholar] .Erbel R, Engberding R, Daniel W, Roelandt J, Visser C, Rennollet H. Echocardiography in diagnosis of aortic dissection. The Lancet . 1989:457–461. doi: 10.1016/s0140-6736(89)91364-0. [DOI] [PubMed] [Google Scholar] .Daily PO, Trueblood HW, Stinson EB, Wuerflein RD, Shumway NE. Management of Acute Aortic Dissections. The Annals of Thoracic Surgery . 1970;10:237–247. doi: 10.1016/s0003-4975(10)65594-4. [DOI] [PubMed] [Google Scholar] .Suzuki T, Distante A, Zizza A, Trimarchi S, Villani M, Salerno Uriarte JA, et al. Preliminary experience with the smooth muscle troponin-like protein, calponin, as a novel biomarker for diagnosing acute aortic dissection. European Heart Journal . 2008;29:1439–1445. doi: 10.1093/eurheartj/ehn162. [DOI] [PubMed] [Google Scholar] .Giachino F, Loiacono M, Lucchiari M, Manzo M, Battista S, Saglio E, et al. Rule out of acute aortic dissection with plasma matrix metalloproteinase 8 in the emergency department. Critical Care . 2013;17:R33. doi: 10.1186/cc12536. [DOI] [PMC free article] [PubMed] [Google Scholar] .Nozato T, Sato A, Hirose S, Hikita H, Takahashi A, Endo H, et al. Preliminary study of serum tenascin-C levels as a diagnostic or prognostic biomarker of type B acute aortic dissection. International Journal of Cardiology . 2013;168:4267–4269. doi: 10.1016/j.ijcard.2013.04.211. [DOI] [PubMed] [Google Scholar] .Kraut JA, Madias NE. Lactic Acidosis. New England Journal of Medicine . 2014;371:2309–2319. doi: 10.1056/NEJMra1309483. [DOI] [PubMed] [Google Scholar] .Kearon C, de Wit K, Parpia S, Schulman S, Afilalo M, Hirsch A, et al. Diagnosis of Pulmonary Embolism with d-Dimer Adjusted to Clinical Probability. New England Journal of Medicine . 2019;381:2125–2134. doi: 10.1056/NEJMoa1909159. [DOI] [PubMed] [Google Scholar] .Rogers AM, Hermann LK, Booher AM, Nienaber CA, Williams DM, Kazerooni EA, et al. Sensitivity of the Aortic Dissection Detection Risk Score, a Novel Guideline-Based Tool for Identification of Acute Aortic Dissection at Initial Presentation. Circulation . 2011;123:2213–2218. doi: 10.1161/CIRCULATIONAHA.110.988568. [DOI] [PubMed] [Google Scholar] .Sato Y, Fujiwara H, Takatsu Y. Cardiac troponin and heart failure in the era of high-sensitivity assays. Journal of Cardiology . 2012;60:160–167. doi: 10.1016/j.jjcc.2012.06.007. [DOI] [PubMed] [Google Scholar] .Mueller C. Biomarkers and acute coronary syndromes: an update. European Heart Journal . 2014;35:552–556. doi: 10.1093/eurheartj/eht530. [DOI] [PubMed] [Google Scholar] .Shah ASV, Anand A, Strachan FE, Ferry AV, Lee KK, Chapman AR, et al. High-sensitivity troponin in the evaluation of patients with suspected acute coronary syndrome: a stepped-wedge, cluster-randomised controlled trial. The Lancet . 2018;392:919–928. doi: 10.1016/S0140-6736(18)31923-8. [DOI] [PMC free article] [PubMed] [Google Scholar] .Funk DJ, Jacobsohn E, Kumar A. Role of the Venous Return in Critical Illness and Shock. Critical Care Medicine . 2013;41:573–579. doi: 10.1097/CCM.0b013e31827bfc25. [DOI] [PubMed] [Google Scholar] .Barton ED, Rhee P, Hutton KC, Rosen P. The pathophysiology of tension pneumothorax in ventilated swine. The Journal of Emergency Medicine . 1997;15:147–153. doi: 10.1016/s0736-4679(96)00312-5. [DOI] [PubMed] [Google Scholar] .Ghignone M, Girling L, Prewitt RM. Volume Expansion versus Norepinephrine in Treatment of a Low Cardiac Output Complicating an Acute Increase in Right Ventricular Afterload in Dogs. Anesthesiology . 1984;60:132–135. doi: 10.1097/00000542-198402000-00009. [DOI] [PubMed] [Google Scholar] .Toppen W, Aquije Montoya E, Ong S, Markovic D, Kao Y, Xu X, et al. Passive Leg Raise: Feasibility and Safety of the Maneuver in Patients with Undifferentiated Shock. Journal of Intensive Care Medicine . 2020;35:1123–1128. doi: 10.1177/0885066618820492. [DOI] [PMC free article] [PubMed] [Google Scholar] .Monnet X, Teboul J. Passive leg raising: five rules, not a drop of fluid. Critical Care . 2015;19:18. doi: 10.1186/s13054-014-0708-5. [DOI] [PMC free article] [PubMed] [Google Scholar] .Malbrain MLNG, Marik PE, Witters I, Cordemans C, Kirkpatrick AW, Roberts DJ, et al. Fluid overload, de-resuscitation, and outcomes in critically ill or injured patients: a systematic review with suggestions for clinical practice. Anaesthesiology Intensive Therapy . 2014;46:361–380. doi: 10.5603/AIT.2014.0060. [DOI] [PubMed] [Google Scholar] .Alhazzani W, Møller MH, Arabi YM, Loeb M, Gong MN, Fan E, et al. Surviving Sepsis Campaign: guidelines on the management of critically ill adults with Coronavirus Disease 2019 (COVID-19) Intensive Care Medicine . 2020;46:854–887. doi: 10.1007/s00134-020-06022-5. [DOI] [PMC free article] [PubMed] [Google Scholar] .Wallace AW, Tunin CM, Shoukas AA. Effects of vasopressin on pulmonary and systemic vascular mechanics. American Journal of Physiology-Heart and Circulatory Physiology . 1989;257:H1228–H1234. doi: 10.1152/ajpheart.1989.257.4.H1228. [DOI] [PubMed] [Google Scholar] .Mizota T, Fujiwara K, Hamada M, Matsukawa S, Segawa H. Effect of arginine vasopressin on systemic and pulmonary arterial pressure in a patient with pulmonary hypertension secondary to pulmonary emphysema: a case report. JA Clinical Reports . 2017;3:1. doi: 10.1186/s40981-016-0072-3. [DOI] [PMC free article] [PubMed] [Google Scholar] .Robie NW, Nutter DO, Moody C, Mcnay JL. In Vivo Analysis of Adrenergic Receptor Activity of Dobutamine. Circulation Research . 1974;34:663–671. doi: 10.1161/01.res.34.5.663. [DOI] [PubMed] [Google Scholar] .Tarján J, Nagy L, Liziczai I, Junger E. Arrhythmic effects of intermittent dobutamine therapy in chronic heart disease failure. The Working Group of Cardiology of the Academic Committee of Veszprém, Hungary. American Journal of Therapeutics . 1998;5:405–411. doi: 10.1097/00045391-199811000-00009. [DOI] [PubMed] [Google Scholar] .Manier G, Castaing Y. Influence of Cardiac Output on Oxygen Exchange in Acute Pulmonary Embolism. American Review of Respiratory Disease . 1992;145:130–136. doi: 10.1164/ajrccm/145.1.130. [DOI] [PubMed] [Google Scholar] .Kerbaul F, Gariboldi V, Giorgi R, Mekkaoui C, Guieu R, Fesler P, et al. Effects of levosimendan on acute pulmonary embolism-induced right ventricular failure. Critical Care Medicine . 2007;35:1948–1954. doi: 10.1097/01.CCM.0000275266.33910.8D. [DOI] [PubMed] [Google Scholar] .Vizza CD, Della Rocca G, Roma DA, Iacoboni C, Pierconti F, Venuta F, et al. Acute hemodynamic effects of inhaled nitric oxide, dobutamine and a combination of the two in patients with mild to moderate secondary pulmonary hypertension. Critical Care . 2001;5:355–361. doi: 10.1186/cc1069. [DOI] [PMC free article] [PubMed] [Google Scholar] .Delbove A, Darreau C, Hamel JF, Asfar P, Lerolle N. Impact of endotracheal intubation on septic shock outcome: a post hoc analysis of the SEPSISPAM trial. Journal of Critical Care . 2015;30:1174–1178. doi: 10.1016/j.jcrc.2015.08.018. [DOI] [PubMed] [Google Scholar] .Jardin F, Delorme G, Hardy A, Auvert B, Beauchet A, Bourdarias J. Reevaluation of Hemodynamic Consequences of Positive Pressure Ventilation: emphasis on cyclic right ventricular afterloading by mechanical lung inflation. Anesthesiology . 1990;72:966–970. doi: 10.1097/00000542-199006000-00003. [DOI] [PubMed] [Google Scholar] .Tsapenko M. Arterial pulmonary hypertension in noncardiac intensive care unit. Vascular Health and Risk Management . 2008;4:1043–1060. doi: 10.2147/vhrm.s3998. [DOI] [PMC free article] [PubMed] [Google Scholar] .Balanos GM, Talbot NP, Dorrington KL, Robbins PA. Human pulmonary vascular response to 4 h of hypercapnia and hypocapnia measured using Doppler echocardiography. Journal of Applied Physiology . 2003;94:1543–1551. doi: 10.1152/japplphysiol.00890.2002. [DOI] [PubMed] [Google Scholar] .Kondo T, Morimoto R, Yokoi T, Yamaguchi S, Kuwayama T, Haga T, et al. Hemodynamics of cardiac tamponade during extracorporeal membrane oxygenation support in a patient with fulminant myocarditis. Journal of Cardiology Cases . 2019;19:22–24. doi: 10.1016/j.jccase.2018.08.009. [DOI] [PMC free article] [PubMed] [Google Scholar] .Morcos M, Vincent L, Harari R, Badulak J, Chen M. Cardiac tamponade in venoarterial extracorporeal membrane oxygenation. Echocardiography . 2021;38:1465–1470. doi: 10.1111/echo.15145. [DOI] [PubMed] [Google Scholar] .AlAnsari S, Mahalwar G, Furqan M, Warner D, Soe W, Klein A, et al. Incidence, Characteristics and Outcomes of Cardiac Tamponade in Patients on Extracorporeal Membrane Oxygenation for the Treatment of Refractory Cardiogenic Shock. Journal of Cardiac Failure . 2020;26:S148–S149. [Google Scholar] .Zwischenberger JB, Cilley RE, Hirschl RB, Heiss KF, Conti VR, Bartlett RH. Life-threatening intrathoracic complications during treatment with extracorporeal membrane oxygenation. Journal of Pediatric Surgery . 1988;23:599–604. doi: 10.1016/s0022-3468(88)80626-2. [DOI] [PubMed] [Google Scholar] .Holoyda K. Tension Pneumothorax on Extracorporeal Membrane Oxygenation Leading to Significant Pneumoperitoneum. The Open Cardiovascular and Thoracic Surgery Journal . 2012;5:31–34. [Google Scholar] .Giraud R, Banfi C, Siegenthaler N, Bendjelid K. Massive pulmonary embolism leading to cardiac arrest: one pathology, two different ECMO modes to assist patients. Journal of Clinical Monitoring and Computing . 2016;30:933–937. doi: 10.1007/s10877-015-9796-2. [DOI] [PubMed] [Google Scholar] .Miranda DR, van Thiel R, Brodie D, Bakker J. Right Ventricular Unloading after Initiation of Venovenous Extracorporeal Membrane Oxygenation. American Journal of Respiratory and Critical Care Medicine . 2015;191:346–348. doi: 10.1164/rccm.201408-1404LE. [DOI] [PubMed] [Google Scholar] .Harnisch LO, Moerer O. Contraindications to the Initiation of Veno-Venous ECMO for Severe Acute Respiratory Failure in Adults: A Systematic Review and Practical Approach Based on the Current Literature. Membranes . 2021;11:584. doi: 10.3390/membranes11080584. [DOI] [PMC free article] [PubMed] [Google Scholar] .Becattini C, Agnelli G, Salvi A, Grifoni S, Pancaldi LG, Enea I, et al. Bolus tenecteplase for right ventricle dysfunction in hemodynamically stable patients with pulmonary embolism. Thrombosis Research . 2010;125:e82–e86. doi: 10.1016/j.thromres.2009.09.017. [DOI] [PubMed] [Google Scholar] .Goldhaber SZ, Come PC, Lee RT, Braunwald E, Parker JA, Haire WD, et al. Alteplase versus heparin in acute pulmonary embolism: randomised trial assessing right-ventricular function and pulmonary perfusion. The Lancet . 1993;341:507–511. doi: 10.1016/0140-6736(93)90274-k. [DOI] [PubMed] [Google Scholar] .Lin PH, Annambhotla S, Bechara CF, Athamneh H, Weakley SM, Kobayashi K, et al. Comparison of Percutaneous Ultrasound-Accelerated Thrombolysis Versus Catheter-Directed Thrombolysis in Patients with Acute Massive Pulmonary Embolism. Vascular . 2009;17:137–147. doi: 10.2310/6670.2009.00063. [DOI] [PubMed] [Google Scholar] .Lee T, Itagaki S, Chiang YP, Egorova NN, Adams DH, Chikwe J. Survival and recurrence after acute pulmonary embolism treated with pulmonary embolectomy or thrombolysis in New York State, 1999 to 2013. The Journal of Thoracic and Cardiovascular Surgery . 2018;155:1084–1090. doi: 10.1016/j.jtcvs.2017.07.074. e12. [DOI] [PubMed] [Google Scholar] .Monteleone PP, Rosenfield K, Rosovsky RP. Multidisciplinary pulmonary embolism response teams and systems. Cardiovascular Diagnosis and Therapy . 2016;6:662–667. doi: 10.21037/cdt.2016.11.05. [DOI] [PMC free article] [PubMed] [Google Scholar] .Reddy PS, Curtiss EI, O’Toole JD, Shaver JA. Cardiac tamponade: hemodynamic observations in man. Circulation . 1978;58:265–272. doi: 10.1161/01.cir.58.2.265. [DOI] [PubMed] [Google Scholar] .Hemnes AR, Gaine SP, Wiener CM. Poor Outcomes Associated with Drainage of Pericardial Effusions in Patients with Pulmonary Arterial Hypertension. Southern Medical Journal . 2008;101:490–494. doi: 10.1097/SMJ.0b013e31816c0169. [DOI] [PubMed] [Google Scholar] .Pradhan R, Okabe T, Yoshida K, Angouras DC, DeCaro MV, Marhefka GD. Patient characteristics and predictors of mortality associated with pericardial decompression syndrome: a comprehensive analysis of published cases. European Heart Journal: Acute Cardiovascular Care . 2015;4:113–120. doi: 10.1177/2048872614547975. [DOI] [PubMed] [Google Scholar] .Netto FACS, Shulman H, Rizoli SB, Tremblay LN, Brenneman F, Tien H. Are needle decompressions for tension pneumothoraces being performed appropriately for appropriate indications. The American Journal of Emergency Medicine . 2008;26:597–602. doi: 10.1016/j.ajem.2007.08.016. [DOI] [PubMed] [Google Scholar] .Cantwell K, Burgess S, Patrick I, Niggemeyer L, Fitzgerald M, Cameron P, et al. Improvement in the prehospital recognition of tension pneumothorax: the effect of a change to paramedic guidelines and education. Injury . 2014;45:71–76. doi: 10.1016/j.injury.2013.06.010. [DOI] [PubMed] [Google Scholar] .Mattox KL, Allen MK. Systematic approach to pneumothorax, haemothorax, pneumomediastinum and subcutaneous emphysema. Injury . 1986;17:309–312. doi: 10.1016/0020-1383(86)90152-x. [DOI] [PubMed] [Google Scholar] .MacDuff A, Arnold A, Harvey J. Management of spontaneous pneumothorax: British Thoracic Society pleural disease guideline 2010. Thorax . 2010;65:ii18–ii31. doi: 10.1136/thx.2010.136986. [DOI] [PubMed] [Google Scholar] Articles from Reviews in Cardiovascular Medicine are provided here courtesy of IMR Press ACTIONS View on publisher site PDF (950.9 KB) Cite Collections Permalink PERMALINK Copy RESOURCES Similar articles Cited by other articles Links to NCBI Databases On this page Abstract 1. Introduction 2. Prognoses 3. Diagnosis 4. Diagnostic Algorithm 5. Treatment 6. Specific 7. Conclusions Acknowledgment Footnotes Disclosure of Relationships and Activities Author Contributions Ethics Approval and Consent to Participate Funding Conflict of Interest References Cite Copy Download .nbib.nbib Format: Add to Collections Create a new collection Add to an existing collection Name your collection Choose a collection Unable to load your collection due to an error Please try again Add Cancel Follow NCBI NCBI on X (formerly known as Twitter)NCBI on FacebookNCBI on LinkedInNCBI on GitHubNCBI RSS feed Connect with NLM NLM on X (formerly known as Twitter)NLM on FacebookNLM on YouTube National Library of Medicine 8600 Rockville Pike Bethesda, MD 20894 Web Policies FOIA HHS Vulnerability Disclosure Help Accessibility Careers NLM NIH HHS USA.gov Back to Top
15692
https://oercommons.org/groups/openstax-elementary-algebra/1443/
OpenStax Elementary Algebra Group Resources | OER Commons Donate to ISKME Discover Resources Collections Providers Hubs Sign in to see your Hubs Login Featured HubsAI & OER Community Hub Open Textbooks #GoOpen Professional Learning Climate Education California Community Colleges Hub See all Hubs Groups Sign in to see your Groups Login Featured GroupsAdult Education Open Community of Resources OpenStax Biology 2e PA STEM Toolkit Pathways Project | Language Teaching Repository @ Boise State Student Advocacy See all Groups Learn More About Help Center About Hubs Services OER 101 Add OER Open Author Create a standalone learning module, lesson, assignment, assessment or activity Create Resource Submit from Web Submit OER from the web for review by our librarians Add Link Learn more about creating OER Add OER Add Link Create Resource About creating OER Search Advanced Search Notifications Sign In/Register Sign In/Register Discover Resources Collections Providers Hubs Sign in to see your Hubs Login Featured HubsAI & OER Community Hub Open Textbooks #GoOpen Professional Learning Climate Education California Community Colleges Hub See all Hubs Groups Sign in to see your Groups Login Featured GroupsAdult Education Open Community of Resources OpenStax Biology 2e PA STEM Toolkit Pathways Project | Language Teaching Repository @ Boise State Student Advocacy See all Groups Learn More About Help Center About Hubs Services OER 101 Add OER Open Author Create a standalone learning module, lesson, assignment, assessment or activity Create Resource Submit from Web Submit OER from the web for review by our librarians Add Link Learn more about creating OER Add OER Add Link Create Resource About creating OER OpenStax Elementary Algebra Access, develop and share resources created by the OpenStax community that align with OpenStax Elementary Algebra. 68 members |9 affiliated resources Join this group Group Resources Discussions Members Information Recent Activity All Group Resources All resources in OpenStax Elementary Algebra(9) Search Per page How to Use the OpenStax Community Hub (View Complete Item Description) Tips and guidelines to help you create resources that align with OpenStax titles Material Type:Lesson Author:Sonya Bennett-Brandt Algebra Tiles (View Complete Item Description) Use tiles to represent variables and constants, learn how to represent and solve algebra problems. Solve equations, substitute in variable expressions, and expand and factor. Flip tiles, remove zero pairs, copy and arrange, and make your way toward a better understanding of algebra. Material Type:Diagram/Illustration,Interactive,Simulation Author:Illuminations National Council of Teachers of Mathematics OpenStax Algebra & Trigonometry Recorded Lectures and Notes for Precalculus (View Complete Item Description) This resource is a collection of full hour-long lectures and accompanying notes covering most of chapters 6-10 of the OpenStax Algebra and Trigonometry book. The notes are provided in both .DOCX and .PDF format to accommodate downloading ease. Students should be encouraged to print notes and fill them out as they watch through the lectures. Material Type:Full Course Author:Libby Gore Polynomials Investigation (View Complete Item Description) The purpose of this lesson is for students to discover the connection between the algebraic and the graphical structure of polynomial functions. This lesson leads to students being able to sketch a graph by identifying the end behavior, intercepts, and multiplicities from a given polynomial equation. It also leads to students being able to write a possible equation by determining the sign of the leading coefficient, minimum possible degree, x-intercepts and y-intercept from a given polynomial graph. Material Type:Activity/Lab,Lesson,Lesson Plan Author:Victoria Olinde Remix Adding & Subtracting Polynomials (View Complete Item Description) Understanding adding and subtracting polynomials.StandardsCC.2.2.HS.D.3Extend the knowledge of arithmetic operations and apply to polynomials. Material Type:Homework/Assignment Author:Bailey Hauswirth Linear Functions & Their Inverses (View Complete Item Description) Desmos activities are interactive, math based activities that can be altered to fit your needs. This particular Desmos activity works through inverses of linear equations for Secondary Math III students. Material Type:Activity/Lab Author:Melissa Hesterman Illustrative Mathematics Algebra 1 - Teachers (View Complete Item Description) Per April Material Type:Activity/Lab,Assessment,Full Course,Homework/Assignment,Teaching/Learning Strategy,Textbook Author:Illustrative Mathematics and Kendall Hunt Illustrative Mathematics - Algebra 1 Supports Students (View Complete Item Description) Per April Material Type:Activity/Lab,Assessment,Full Course,Homework/Assignment,Textbook Author:Illustrative Mathematics and Kendall Hunt Illustrative Mathematics - Algebra 1 Students (View Complete Item Description) Per April Material Type:Activity/Lab,Assessment,Full Course,Homework/Assignment,Textbook Author:Illustrative Mathematics and Kendall Hunt Discover Resources Collections Providers Community All Hubs All Groups Create Open Author Submit a Resource Our Services About Hubs About OER Commons OER 101 Help Center My Account My Items My Groups My Hubs Subscribe to OER Newsletter Subscribe Connect with OER CommonsFacebook, Opens in new windowTwitter, Opens in new window Donate to ISKME Powered By Privacy PolicyTerms of ServiceCollection PolicyDMCA © 2007 - 2025, OER Commons A project created by ISKME. Except where otherwise noted, content on this site is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 License. ` × Sign in / Register Your email or username: Did you mean ? Password: Forgot password?- [x] Show password or Sign In through your Institution Create an account Register or Sign In through your Institution
15693
https://www.sciencedirect.com/science/article/pii/S0033838922020668
THE RADIOLOGIC SIGNS OF DISEASE IN THE LESSER PERITONEAL SAC - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Patient Access Other access options Search ScienceDirect Article preview Abstract References (15) Radiologic Clinics of North America Volume 2, Issue 1, April 1964, Pages 107-120 THE RADIOLOGIC SIGNS OF DISEASE IN THE LESSER PERITONEAL SAC Author links open overlay panel HARRY Z.MELLINS M.D. Show more Add to Mendeley Share Cite rights and content SUMMARY The contour of the lesser peritoneal sac is readily identified on supine and erect plain films of the abdomen if the cavity contains moderate or large amounts of gas. Small amounts of gas or loculated collections cannot be identified as being in the lesser peritoneal sac without lateral views of the abdomen after opacification of the stomach. Almost always, midline subdiaphragmatic gas collections will be shown to be intraperitoneal. Fluid collections in the lesser sac displace the stomach anteriorly, upward and to the right. Typically there is a large left-sided soft tissue density with an oblique lower border that runs laterally and downward. Abscesses originating in the pancreas may fill the lesser sac with multiple small gas collections or, if coalescent, with long retrogastric air-fluid levels. The distal ileum or right side of the colon may herniate through the foramen of Winslow into the lesser peritoneal sac. The small intestine proximally shows the signs of mechanical obstruction. The herniated segment usually undergoes obstruction of the closed-loop type. Often a characteristic extrinsic obstruction of the distal limb of the hepatic flexure is produced when the herniating loop of small intestine crosses the colon on its way to the epiploic foramen. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Recommended articles References (15) J.E. Bannen Investigation of free gas in the peritoneal cavity >Brit. J. Radiol. (1945) A. Bittdorf Über Abszesse im Saccus omentalis nach Pankreasnekrose >Mitt. a.d. Grenzgeb. d. Med. u. Chir. (1913) J. Buckstein The Digestive Tract in Roentgenology (1953) C.V. Cimmino Lesser sac hernia via the foramen of Winslow >Radiology (1953) C.V. Cimmino et al. Gas in the lesser sac in perforated peptic ulcer >Am. J. Roentgenol. (1952) M.S. Hollenberg Radiographic diagnosis of hernia into lesser peritoneal sac through the foramen of Winslow >Surgery (1945) H. Kopf Traumatische Bursa-omentalis- “Zyste.” Zentralbl >Chir. (Leipzig) (1953) There are more references available in the full text version of this article. Cited by (0) From the Departments of Radiology of the State University of New York Downstate Medical Center and the Kings County Hospital Center, Brooklyn, New York View full text Copyright © 1964 Elsevier B.V. Company. Published by Elsevier Inc. Published by Elsevier Inc. All rights reserved. Recommended articles Sugar-based systems Engineering of Biomaterials for Drug Delivery Systems, 2018, pp. 273-297 Rajesh A.Shenoi ### Disruption of intracellular signaling An Introduction to Interdisciplinary Toxicology, 2020, pp. 81-96 Anumantha Kanthasamy, …, Adhithiya Charli ### Caenorhabitidis elegans as an animal model in toxicological studies An Introduction to Interdisciplinary Toxicology, 2020, pp. 533-544 Marina Lopes Machado, …, Félix Alexandre Antunes Soares ### Fabrication of microfluidic chips Multidisciplinary Microfluidic and Nanofluidic Lab-on-a-chip, 2022, pp. 3-35 Hui Chen, …, Zhejun Yang ### Skin Aging Cosmetic Science and Technology, 2017, pp. 711-728 S.Inoue ### Scouting to Meet Unmet Needs Cosmetic Science and Technology, 2017, pp. 77-86 C.M.Rocafort Show 3 more articles About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie Settings All content on this site: Copyright © 2025 Elsevier B.V., 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. Cookie Preference Center We use cookies which are necessary to make our site work. We may also use additional cookies to analyse, improve and personalise our content and your digital experience. For more information, see our Cookie Policy and the list of Google Ad-Tech Vendors. You may choose not to allow some types of cookies. However, blocking some types may impact your experience of our site and the services we are able to offer. See the different category headings below to find out more or change your settings. Allow all Manage Consent Preferences Strictly Necessary Cookies Always active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. Cookie Details List‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. Cookie Details List‎ Targeting Cookies [x] Targeting Cookies These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. If you do not allow these cookies, you will experience less targeted advertising. Cookie Details List‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Confirm my choices Your Privacy [dialog closed]
15694
https://www.hmhco.com/blog/folding-shapes-activities?srsltid=AfmBOorCHEqtXDTOPgsrZKA1VluaFi282XA0Ke7D4jQ-lxtLrqTGi0KZ
Folding Shapes Activities | HMH Curriculum Literacy Core Curriculum Into Literature, 6-12 Into Reading, K-6 See all Literacy Performance Suite Intervention English 3D, K-12 Read 180, 3-12 See all Reading Intervention Assessment Supplemental A Chance in the World SEL, 8-12 Amira Learning, PreK–8 Classcraft, K-8 JillE Literacy, K-3 Waggle, K-8 Writable, 3-12 See all Reading Supplemental Personalized Path Math Core Curriculum Arriba las Matematicas, K-8 Go Math!, K-6 Into Algebra 1, Geometry, Algebra 2, 8-12 Into Math, K-8 Math in Focus, K-8 See all Math Performance Suite Supplemental Classcraft, K-8 Waggle, K-8 See all Supplemental Math Personalized Path Intervention Math 180, 3-12 See all Math Intervention See all Assessment Science Core Curriculum Into Science, K-5 Into Science, 6-8 Science Dimensions, K-12 See all Science Readers ScienceSaurus, K-8 Social Studies Core Curriculum HMH Social Studies, 6-12 See all Social Studies Supplemental Writable Professional Development For Teachers Coachly Teacher's Corner Live Online Courses Program-Aligned Courses See all Professional Development For Leaders The Center for Model Schools More AI Tools on HMH Ed Assessment Early Learning English Language Development Homeschool Intervention Literacy Mathematics Professional Development Science School Improvement Social Studies Special Education Summer School See all Solutions Resources Browse Resources AI Tools Classroom Activities Customer Resource Hub Customer Success Stories Digital Samples Events Grants & Funding Family & Caregiver Resources International Product Updates Research Library Shaped, HMH Blog Webinars Customer Support Contact Sales Customer Service & Technical Support Portal Platform Login Company Learn about us About Corporate Responsibility Leadership News Announcements Our Culture Our Legacy Join us Careers Educator Input Panel Suppliers Divisions Center for Model Schools Heinemann NWEA Shop Support Platform Login From setup to support, HMH will ensure your back-to-school season runs smoothly. Let’s go. Curriculum Literacy Math Science Social Studies Professional Development More Core Curriculum Into Literature, 6-12 Into Reading, K-6 See all Literacy Performance Suite Intervention English 3D, K-12 Read 180, 3-12 See all Reading Intervention Assessment Supplemental A Chance in the World SEL, 8-12 Amira Learning, PreK–8 Classcraft, K-8 JillE Literacy, K-3 Waggle, K-8 Writable, 3-12 See all Reading Supplemental Personalized Path HMH Personalized Path Discover a solution that provides K–8 students in Tiers 1, 2, and 3 with the adaptive practice and personalized intervention they need to excel. AI in the Classroom: A Quick Guide to Harness Benefits and Minimize Risks Download our quick guide of best practices for integrating AI in schools in a way that targets educational goals while minimizing the risks. Core Curriculum Arriba las Matematicas, K-8 Go Math!, K-6 Into Algebra 1, Geometry, Algebra 2, 8-12 Into Math, K-8 Math in Focus, K-8 See all Math Performance Suite Supplemental Classcraft, K-8 Waggle, K-8 See all Supplemental Math Personalized Path Intervention Math 180, 3-12 See all Math Intervention See all Assessment HMH Personalized Path Discover a solution that provides K–8 students in Tiers 1, 2, and 3 with the adaptive practice and personalized intervention they need to excel. Optimizing the Math Classroom: 6 Best Practices Our compilation of math best practices highlights six ways to optimize classroom instruction and make math something all learners can enjoy. Core Curriculum Into Science, K-5 Into Science, 6-8 Science Dimensions, K-12 See all Science Readers ScienceSaurus, K-8 2025-2026 Classroom Activities and Holidays Calendar Download our free activities calendar for fun lessons to do with your class, plus engaging ways to commemorate and celebrate events throughout the year. Core Curriculum HMH Social Studies, 6-12 See all Social Studies Supplemental Writable 2025-2026 Classroom Activities and Holidays Calendar Download our free activities calendar for fun lessons to do with your class, plus engaging ways to commemorate and celebrate events throughout the year. For Teachers Coachly Teacher's Corner Live Online Courses Program-Aligned Courses See all Professional Development For Leaders The Center for Model Schools Building Your School Culture: An Administrator's Guide Get our free administrator’s guide to building a positive school culture, filled with practical advice, real-world examples, and resources for further exploration. AI Tools on HMH Ed Assessment Early Learning English Language Development Homeschool Intervention Literacy Mathematics Professional Development Science School Improvement Social Studies Special Education Summer School See all Solutions Reading Intervention Handbook for Educators Download the Reading Intervention Handbook, filled with expert advice and research-backed strategies that will set students up for success. Math Intervention Teacher's Handbook Download the Math Intervention Handbook, filled with expert advice and research-backed strategies that will set students up for success. Resources Browse Resources AI Tools Classroom Activities Customer Resource Hub Customer Success Stories Digital Samples Events Grants & Funding Family & Caregiver Resources International Product Updates Research Library Shaped, HMH Blog Webinars Customer Support Contact Sales Customer Service & Technical Support Portal Platform Login 2025-2026 Classroom Activities and Holidays Calendar Download our free activities calendar for fun lessons to do with your class, plus engaging ways to commemorate and celebrate events throughout the year. Building Your School Culture: An Administrator's Guide Get our free administrator’s guide to building a positive school culture, filled with practical advice, real-world examples, and resources for further exploration. Company Learn about us About Corporate Responsibility Leadership News Announcements Our Culture Our Legacy Join us Careers Educator Input Panel Suppliers Divisions Center for Model Schools Heinemann NWEA HMH Careers Exploring a career at HMH? Do work that matters. Learn about our culture, benefits, and available job opportunities. Accessibility Explore HMH’s approach to designing affirming and accessible curriculum materials and learning tools for students and teachers. Shop Support 0 Log in Back to Shaped Math Folding Shape Activities to Practice Geometry Mikayla Capestrani Alyssa Fuller January 10, 2023 10 Min Read What Is Symmetry? In first and second grade we introduce the concept of geometry by teaching students the main attributes of shapes. We begin by teaching the common shapes, like circles, squares, triangles, and rectangles. A lesson may include identifying how many sides and angles a shape has, the length of its sides, and similarities and differences between shapes. An important attribute to identify is whether or not the shape is symmetrical. One way to teach the concept of symmetry to young students is by showing them that when folding paper shapes, each “equal side” of the shape should be the same, or mirror itself. When a shape has at least one line of symmetry, it is considered a symmetrical shape. Key Standard:Identify a shape that is symmetrical or asymmetrical Folding Common Paper Shapes Activity For this folding shapes activity, students will need cut-out paper shapes and a highlighter. We recommend having a variety of common shapes pre-cut and ready for your students to fold. The pre-cut shapes will give the students a better opportunity to identify symmetry. Feel free to use our free cut-outs of common shapes. To start this activity, you will want to teach your students the key vocabulary term: line(s) of symmetry. A way to learn what a line of symmetry is by folding paper shapes and then examining the result. Have you and your students grab the paper square and ask them to identify which shape it is. Then ask if they think a square can be folded in half so that one half of the fold fits exactly on top of the other half of the fold. Demonstrate, and walk students through how to fold a shape in half, explaining that when folding shapes in half, the edges and corners need to be lined up and touching. After that first fold, have students open their shape back up and explain to the students that the folded line is known as a line of symmetry. Prompt students to look for different ways they could fold the same square to create another line of symmetry. Guide students in discovering that a square has four different lines of symmetry. Ways to support students: Have students use a highlighter to highlight the folded line each time they identify it as a line of symmetry. When students independently explore their paper shapes, they may create a lot of folded lines that are not lines of symmetry. By having the students mark them, they will easily be able to look back at that shape and count how many lines of symmetry they identified. To differentiate instruction for some students, provide a “dot” at the center of the shape to help them identify where to fold to find half. The students will be able to reference whether the dot was located on the folded line. The circle is a special shape that has an infinite number of lines of symmetry. Before your students begin folding their circles, ask them to guess how many lines of symmetry a circle has. Record some of the answers you receive on the board so you can compare what they guessed and what they discovered after they folded their shapes. Allow your students the time to explore their circle and see how many lines of symmetry they can create. After everyone has completed folding their circle, look back and have a discussion about their guesses and discoveries. Point out no matter which way you fold a circle in half it will always be considered a line of symmetry. For students who are ready, this can lead to a discussion on what infinity means. Mathematically, there are infinitely many lines of symmetry. But in reality, it is impossible to fold a circle an infinite number of times. Can students think of other examples of infinity? For example, how large numbers can get or how many points are on a line. Progressing Through Common Shapes One progression through the shapes could be to start with the easiest shapes to identify lines of symmetry for, and to end with the more challenging shapes. An example of this progression could be square, diamond, rectangle, trapezoid, circle, and ending with different types of triangles (equilateral, right, and isosceles). Through the progression, students may recognize that shapes with equal side lengths will often have more lines of symmetry than shapes with diverse side lengths. For instance, a square has four sides of equal length and four lines of symmetry. The rectangle is a great example to introduce students to shapes that can be folded in half and have folds that are not lines of symmetry as well as folds that are. The rectangle has exactly two lines of symmetry. Ask all the students to grab their paper rectangle and fold it vertically. Remind students that they need to fold their shapes exactly in half, meaning that the edges and corners of the paper need to be lined up and touching. Shapes can have more than one line of symmetry. After students have folded the rectangle once, have them independently explore the shape by unfolding it then trying to fold it in different directions to identify other lines of symmetry. Students can highlight their lines of symmetry with a highlighter to not confuse them from other folding lines that might not have worked out to be symmetrical. Take a moment with your students to discuss what led them to create folded lines that weren’t lines of symmetry. Avoid telling students they were right or wrong, instead seeking to elicit their thinking about lines and shapes and compare their thinking with their peers. Lastly, ask the students to unfold their shape and count how many lines of symmetry they found. Take a moment to wrap up this lesson by reviewing some key points: What is a line of symmetry, and how do we find it? What are some shapes that always have lines of symmetry when folded in half? What are some shapes that don’t have lines of symmetry when folded in half? Why is a circle special? Folding Uncommon Paper Shapes Activity In this activity students will work with symmetrical and non-symmetrical shapes. Provide students with pre-cut uncommon shapes and a highlighter. Cut-outs for some uncommon shapes are available below. Similar to the Folding Common Paper Shapes Activity, you will work through different shapes and identify different lines of symmetry. To start this activity, you will want to review what symmetry is and how to find out if a shape has lines of symmetry. If the folded shapes have even one line of symmetry, that makes the shape symmetrical. Have students grab the shape that resembles a plus sign, then allow them time to fold it and discover if the shape has any lines of symmetry. The next progression of this activity is to introduce the students to an uncommon shape that has only one line of symmetry. Have students grab the paper shape that resembles an arrow and give them time to explore this shape to see if they can identify any lines of symmetry. The last part of this activity is better for students who are ready for more advanced geometry ideas. We find that students enjoy the opportunity to create their own shapes. Have your students cut out any shape they want from a piece of paper. Encourage students to make shapes that are funny and unique. This is a great time for them to be creative. If it helps, you can demonstrate how to make random cuts that will result in an odd-looking shape. Although a goal of this part of the activity is to encourage student exploration, if you think there won’t be enough time, you can provide them with a pre-cut uncommon shape. Once all students have created their folded shape, ask them if they can find any lines of symmetry. The goal of this exercise is for the students to create shapes that won’t contain any lines of symmetry. If any students create a shape that is actually symmetrical, explore how that happened and discuss the differences between shapes with and without symmetry. Take a moment to wrap up this lesson by reviewing some key points: Can uncommon shapes be symmetrical? What is the folded line called when we can fold a shape perfectly in half? Are there shapes that are not symmetrical and how would you be able to tell? Extension Activity For an extension activity or an exit ticket, assemble a grid of 10–15 different shapes taken from the real world, for example stop signs, leaves and flowers, and outlines for tools or toys. Students can identify lines of symmetry on each shape. Students can then highlight as many lines of symmetry that they can find in the symmetrical shapes, while leaving the asymmetrical shapes blank. Additional Symmetry Activities An engaging way to have students continue to identify lines of symmetry is to have them finish drawing the other half of a pre-drawn image. This activity will familiarize students with finding the halfway point in a figure as well as getting them to realize that symmetrical shapes should look the exact same on each side of the line of symmetry. Find a few images that you believe your students would enjoy completing. The easiest way to create this activity is to print the full image, then fold it in half and make copies of one of the folded sides. Keep in mind the complexity of the image; we suggest using images that aren’t super detailed. Once the activity is over, don‘t forget to point out symmetry in the world around you throughout the year! One of the joys of mathematics is learning about an idea and then seeing all the ways that it can describe phenomena, objects, and ideas. Students might start to see their classrooms, living environments, and natural world just a little bit differently now. The views expressed in this article are those of the authors and do not necessarily represent those of HMH. Grow student confidence in mathematics withHMH Into Math, our core math solution for Grades K–8. Math Grades 1-2 Activities & Lessons Related Reading 7 Classroom Math Center Ideas and Activities Ana Berry Instructional Designer, HMH September 15, 2025 35 Fun Halloween Writing Prompts & Ideas Carey Blankenship-Kramer Shaped Staff September 15, 2025 Classroom Rules and Procedures for Elementary Students Beckah Sipes Fifth-Grade Teacher September 15, 2025 FREE RESOURCE Math Best Practices Guide Get our FREE guide "Optimizing the Math Classroom: 6 Best Practices." Access Now Math Grades 1-2 Activities & Lessons Curriculum Connected Teaching & Learning Literacy Math Science Social Studies Intervention Supplemental Professional Development See All Solutions Resources Contact Sales Customer Support Digital Samples International Platform Login Research Library Shaped - HMH Blog Company About Accessibility Careers Leadership News Announcements Our Legacy Policy Framework Divisions Center for Model Schools (formerly ICLE) Heinemann NWEA SUBSCRIBE NOW Stay up-to-date with the latest HMH news and solutions. Sign Up Terms of Use Terms of Purchase Privacy Policy Trademark Credits Permissions Request © 2025 HMH Education Company. All rights reserved. Get our FREE guide "Optimizing the Math Classroom: 6 Best Practices." Download Privacy Preference Center Your Privacy Your Privacy When you visit any web site, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalised web experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and change our default settings. However, blocking some types of cookies may impact your experience of the site and the services we are able to offer. More information Strictly Necessary Cookies #### Strictly Necessary Cookies Always Active - [x] Strictly Necessary Cookies These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. ### Sale of Personal Data Sale of Personal Data [x] Sale of Personal Data Under the California Consumer Privacy Act, you have the right to opt-out of the sale of your personal information to third parties. These cookies collect information for analytics and to personalize your experience with targeted ads. You may exercise your right to opt out of the sale of personal information by using this toggle switch. If you opt out we will not be able to offer you personalised ads and will not hand over your personal information to any third parties. Additionally, you may contact our legal department for further clarification about your rights as a California consumer by using this Exercise My Rights link. If you have enabled privacy controls on your browser (such as a plugin), we have to take that as a valid request to opt-out. Therefore we would not be able to track your activity through the web. This may affect our ability to personalize ads according to your preferences. Performance Cookies [x] Targeting Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Targeting Cookies [x] Targeting Cookies These cookies may be set through our site by our advertising partners. They may be used by those companies to build a profile of your interests and show you relevant adverts on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. If you do not allow these cookies, you will experience less targeted advertising. Advertising Cookies Consent Leg.Interest [x] Select All Vendors [x] Select All Vendors [x] Select All Hosts Select All Clear Filters [x] Information storage and access Apply Save Settings Allow All
15695
https://math.stackexchange.com/questions/4360564/algebraic-proof-that-sum-limits-k-0n-n-choose-k-cdot-frac-1kk
combinatorics - Algebraic proof that $\sum\limits_{k=0}^{n} {n \choose k}\cdot \frac{(-1)^k}{(k+1)(k+2)} = \frac{1}{n+2}$ - Mathematics Stack Exchange Join Mathematics 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 Mathematics helpchat Mathematics 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 Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams 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 Algebraic proof that ∑k=0 n(n k)⋅(−1)k(k+1)(k+2)=1 n+2∑k=0 n(n k)⋅(−1)k(k+1)(k+2)=1 n+2 Ask Question Asked 3 years, 8 months ago Modified1 year, 4 months ago Viewed 418 times This question shows research effort; it is useful and clear 9 Save this question. Show activity on this post. I tried evaluating the integral ∫[0,1]n min(x 1,x 2,…,x n)|d n x|∫[0,1]n min(x 1,x 2,…,x n)|d n x|. In my first attempt, I used a recursive approach and managed to defined the integral as being I k n=∫[0,1]n min(x 1,x 2,…,x n)k|d n x|I n k=∫[0,1]n min(x 1,x 2,…,x n)k|d n x|, where I could evaluate I k n=I k n−1−k k+1⋅I k+1 n−1 I n k=I n−1 k−k k+1⋅I n−1 k+1, and I k 1=1 k+1 I 1 k=1 k+1. Using this recursive approach, I managed to extract the I 1 n=∑k=0 n−1(n−1 k)⋅(−1)k(k+1)(k+2)I n 1=∑k=0 n−1(n−1 k)⋅(−1)k(k+1)(k+2) I plugged numbers into this sum for multiple values of n n, and saw that I constantly get 1 n+1 1 n+1. During my second attempt, I managed to eventually find the integral by splitting the area into n!n! areas in which there exists some order for each element x 0 x 0 such that x i 1 0≤x i 2 0≤…≤x i n 0 x 0 i 1≤x 0 i 2≤…≤x 0 i n, and proved that the integral over each of these area is 1(n+1)!1(n+1)! which gave me showed me more definitively that the integral is equal to 1 n+1 1 n+1. That said, after trying for a while, I couldn't come up with any combinatorial/algebraic proof that the sum I found is indeed 1 n+1 1 n+1. I tried evaluating it as a telescoping sum, giving me the expression ∑n 2 k=0(n 2 k)⋅(4 k+3−n)(2 k+3)(2 k+2)(2 k+1)∑k=0 n 2(n 2 k)⋅(4 k+3−n)(2 k+3)(2 k+2)(2 k+1), but couldn't expand this sum to anything useful either. I haven't worked much with sums of this form and was wondering whether I'm missing something that can help me show this without the integral. combinatorics summation binomial-coefficients Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited May 2, 2024 at 13:25 Martin Sleziak 56.3k 20 20 gold badges 211 211 silver badges 391 391 bronze badges asked Jan 19, 2022 at 8:55 jua jajua ja 93 3 3 bronze badges 1 You can show that (1+x)n+2−1−(n+2)x(n+1)(n+2)=∑n k=0(n k)x k+2(k+1)(k+2)(1+x)n+2−1−(n+2)x(n+1)(n+2)=∑k=0 n(n k)x k+2(k+1)(k+2) by using the rules of differentiation of polynomials and the binomial theorem (ie. no analysis).reuns –reuns 2022-01-19 09:10:36 +00:00 Commented Jan 19, 2022 at 9:10 Add a comment| 7 Answers 7 Sorted by: Reset to default This answer is useful 10 Save this answer. Show activity on this post. Since I'm currently studying Discrete Mathematics, I present to you a proof using tools from this domain. More precisely, we could make use of the following result: Difference formula: If f(x)f(x) is defined for all integers x x, then Δ n f(x)=∑k=0 n(n k)(−1)n−k f(x+k).Δ n f(x)=∑k=0 n(n k)(−1)n−k f(x+k). Here, Δ f(x)=f(x+1)−f(x)Δ f(x)=f(x+1)−f(x), and Δ n Δ n is the application of the Δ Δ-operator n n times. Choosing f(x)=1(x+1)(x+2)=x−2–––f(x)=1(x+1)(x+2)=x−2 _ (x x to the −2−2 falling) and x=0 x=0, the RHS equals (−1)n∑k=0 n(n k)(−1)k 1(k+1)(k+2).(−1)n∑k=0 n(n k)(−1)k 1(k+1)(k+2). Thus, your sum equals the LHS multiplied by (−1)n(−1)n, that is (−1)n Δ n f(0)=(−1)n(−1)n(n+1)!(n+2)!=1 n+2.(−1)n Δ n f(0)=(−1)n(−1)n(n+1)!(n+2)!=1 n+2. I made use of the fact that Δ x m––=m x m−1–––––Δ x m =m x m−1 . Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jan 22, 2022 at 17:20 answered Jan 19, 2022 at 9:19 ZuyZuy 5,121 1 1 gold badge 9 9 silver badges 23 23 bronze badges 0 Add a comment| This answer is useful 9 Save this answer. Show activity on this post. ∑k=0 n(n k)(−1)k(k+1)(k+2)=∑k=0 n(n k)∫0−1 x k(1+x)d x=∫0−1(1+x)∑k=0 n(n k)x k d x=∫0−1(1+x)(1+x)n d x=[(1+x)n+2 n+2]0−1=1 n+2∑k=0 n(n k)(−1)k(k+1)(k+2)=∑k=0 n(n k)∫−1 0 x k(1+x)d x=∫−1 0(1+x)∑k=0 n(n k)x k d x=∫−1 0(1+x)(1+x)n d x=[(1+x)n+2 n+2]−1 0=1 n+2 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 19, 2022 at 9:04 CHAMSICHAMSI 9,232 9 9 silver badges 28 28 bronze badges Add a comment| This answer is useful 4 Save this answer. Show activity on this post. Proof that ∑k=0 n(n k)⋅(−1)k(k+1)(k+2)=1 n+2.∑k=0 n(n k)⋅(−1)k(k+1)(k+2)=1 n+2. Begin with the binomial theorem ∑k=0 n(n k)x k=(1+x)n.∑k=0 n(n k)x k=(1+x)n. Integrate ∫y 0∫0 y: ∑k=0 n(n k)y k+1 k+1=(1+y)n+1−1 n+1.∑k=0 n(n k)y k+1 k+1=(1+y)n+1−1 n+1. Integrate ∫z 0∫0 z: ∑k=0 n(n k)z k+2(k+1)(k+2)=(1+z)n+2−1−(n+2)z(n+1)(n+2).∑k=0 n(n k)z k+2(k+1)(k+2)=(1+z)n+2−1−(n+2)z(n+1)(n+2). Substitute z=−1 z=−1: ∑k=0 n(n k)(−1)k(k+1)(k+2)=n+1(n+1)(n+2)=1 n+2.∑k=0 n(n k)(−1)k(k+1)(k+2)=n+1(n+1)(n+2)=1 n+2. Perhaps the same proof shows ∑k=0∞(α k)⋅(−1)k(k+1)(k+2)=1 α+2∑k=0∞(α k)⋅(−1)k(k+1)(k+2)=1 α+2 for real (or complex) α≠−2 α≠−2. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 19, 2022 at 9:29 GEdgarGEdgar 118k 9 9 gold badges 128 128 silver badges 274 274 bronze badges 1 1 Your last sentence inspired me. It works like a charm ! Cheers & thanks :-)Claude Leibovici –Claude Leibovici 2022-01-19 10:01:17 +00:00 Commented Jan 19, 2022 at 10:01 Add a comment| This answer is useful 2 Save this answer. Show activity on this post. After @GEdgar's answer, inspired by the approach, a generalization is ∑k=0∞(a k)(−1)k(k+b)(k+c)=Γ(a+1)c−b(Γ(b)Γ(a+b+1)−Γ(c)Γ(a+c+1))∑k=0∞(a k)(−1)k(k+b)(k+c)=Γ(a+1)c−b(Γ(b)Γ(a+b+1)−Γ(c)Γ(a+c+1)) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 19, 2022 at 9:59 Claude LeiboviciClaude Leibovici 294k 55 55 gold badges 130 130 silver badges 316 316 bronze badges Add a comment| This answer is useful 2 Save this answer. Show activity on this post. Multiplying both sides by (n+1)(n+2)(n+1)(n+2) yields the equivalent identity ∑k=0 n(−1)k(n+2 k+2)=n+1,∑k=0 n(−1)k(n+2 k+2)=n+1, which follows directly from the well-known identity ∑j=0 m(−1)j(m j)=0∑j=0 m(−1)j(m j)=0 by taking m=n+2 m=n+2 and stripping off the first two terms. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 19, 2022 at 15:12 RobPrattRobPratt 51.8k 4 4 gold badges 32 32 silver badges 69 69 bronze badges Add a comment| This answer is useful 1 Save this answer. Show activity on this post. This answer comes a bit late but I thought it is worth mentioning it because it shows a simple algebraic proof of the formula in question. First note the following two facts: (1): 1(k+1)(k+2)(n k)=1(n+1)(n+2)(n+2 k+2)1(k+1)(k+2)(n k)=1(n+1)(n+2)(n+2 k+2), Indeed 1(k+1)(k+2)(n k)===n!k!(k+1)(k+2)(n−k)!1(n+1)(n+2)(n+2)!(k+2)!((n+2)−(k+2))!1(n+1)(n+2)(n+2 k+2)1(k+1)(k+2)(n k)=n!k!(k+1)(k+2)(n−k)!=1(n+1)(n+2)(n+2)!(k+2)!((n+2)−(k+2))!=1(n+1)(n+2)(n+2 k+2) (2): According to the binomial formula, we have 0=(1−1)n+2=∑k=0 n+2(−1)k(n+2 k)0=(1−1)n+2=∑k=0 n+2(−1)k(n+2 k) Now, we have ∑k=0 n(n k)⋅(−1)k(k+1)(k+2)=(1)=(−1)k+2=(−1)k=(2)=1(n+1)(n+2)∑k=0 n(−1)k⋅(n+2 k+2)1(n+1)(n+2)∑k=2 n+2(−1)k⋅(n+2 k)1(n+1)(n+2)(0−(n+2 0)+(n+2 1))n+1(n+1)(n+2)=1 n+2∑k=0 n(n k)⋅(−1)k(k+1)(k+2)=(1)1(n+1)(n+2)∑k=0 n(−1)k⋅(n+2 k+2)=(−1)k+2=(−1)k 1(n+1)(n+2)∑k=2 n+2(−1)k⋅(n+2 k)=(2)1(n+1)(n+2)(0−(n+2 0)+(n+2 1))=n+1(n+1)(n+2)=1 n+2 Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jan 19, 2022 at 14:04 trancelocationtrancelocation 33.5k 1 1 gold badge 22 22 silver badges 48 48 bronze badges Add a comment| This answer is useful 1 Save this answer. Show activity on this post. A variation based upon telescoping. We obtain ∑k=0 n(n k)(−1)k(k+1)(k+2)=∑k=0 n(n k)(−1)k(1 k+1−1 k+2)=∑k=0 n(n k)(−1)k 1 k+1−∑k=1 n+1(n k−1)(−1)k−1 1 k+1=∑k=0 n+1(n+1 k)(−1)k 1 k+1=1 n+2∑k=0 n+1(n+2 k+1)(−1)k=−1 n+2∑k=1 n+2(n+2 k)(−1)k=−1 n+2(1−1)n+2+1 n+2=1 n+2(1)(2)(3)(4)(5)∑k=0 n(n k)(−1)k(k+1)(k+2)=∑k=0 n(n k)(−1)k(1 k+1−1 k+2)(1)=∑k=0 n(n k)(−1)k 1 k+1−∑k=1 n+1(n k−1)(−1)k−1 1 k+1(2)=∑k=0 n+1(n+1 k)(−1)k 1 k+1(3)=1 n+2∑k=0 n+1(n+2 k+1)(−1)k(4)=−1 n+2∑k=1 n+2(n+2 k)(−1)k(5)=−1 n+2(1−1)n+2+1 n+2=1 n+2 and the claim follows. Comment: In (1) we shift the index of the right-hand sum to start with k=1 k=1. In (2) we use (p q)+(p q−1)=(p+1 q)(p q)+(p q−1)=(p+1 q). In (3) we use p+1 q+1(p q)=(p+1 q+1)p+1 q+1(p q)=(p+1 q+1). In (4) we shift the index to start with k=1 k=1. In (5) we apply the binomial theorem. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Jan 21, 2022 at 13:52 answered Jan 20, 2022 at 23:12 Markus ScheuerMarkus Scheuer 113k 7 7 gold badges 106 106 silver badges 251 251 bronze badges 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 combinatorics summation binomial-coefficients 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 Report this ad Related 5Algebraic Proof that ∑k=0 m(r k)(m+n−r m−k)=(m+n m)∑k=0 m(r k)(m+n−r m−k)=(m+n m) 8Calculate the binomial sum I n=∑n i=0(−1)i(2 n+1−i i)I n=∑i=0 n(−1)i(2 n+1−i i) 3Combinatorial proof of ∑i=0 m 2 n−i(n i)(m i)=∑i=0 m(n+m−i m)(n i)∑i=0 m 2 n−i(n i)(m i)=∑i=0 m(n+m−i m)(n i) 0Evaluate ∑n=0∞∑r=0 n(1(n−r)!a n−r)(1 r!b r)∑n=0∞∑r=0 n(1(n−r)!a n−r)(1 r!b r) 0Is there known generating function having this form: ∑m−(i 1+…+i k)i 1!…i k!x i 1 1…x i k k∑m−(i 1+…+i k)i 1!…i k!x 1 i 1…x k i k? 3Evaluating the expression: ∑1 n(−1)k−1(n k)k 2∑1 n(−1)k−1(n k)k 2 Hot Network Questions Should I let a player go because of their inability to handle setbacks? Is it safe to route top layer traces under header pins, SMD IC? Riffle a list of binary functions into list of arguments to produce a result How do you create a no-attack area? Does the curvature engine's wake really last forever? Are there any legal strategies to modify the nature and rules of the UN Security Council in the face of one, or more SC vetos? Making sense of perturbation theory in many-body physics Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward" I have a lot of PTO to take, which will make the deadline impossible Languages in the former Yugoslavia Where is the first repetition in the cumulative hierarchy up to elementary equivalence? Passengers on a flight vote on the destination, "It's democracy!" How to rsync a large file by comparing earlier versions on the sending end? Overfilled my oil Is it possible that heinous sins result in a hellish life as a person, NOT always animal birth? How to home-make rubber feet stoppers for table legs? On being a Maître de conférence (France): Importance of Postdoc Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations? How to convert this extremely large group in GAP into a permutation group. Countable and uncountable "flavour": chocolate-flavoured protein is protein with chocolate flavour or protein has chocolate flavour Interpret G-code Suggestions for plotting function of two variables and a parameter with a constraint in the form of an equation What is the feature between the Attendant Call and Ground Call push buttons on a B737 overhead panel? For every second-order formula, is there a first-order formula equivalent to it by reification? Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.29.34589 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
15696
https://siers.weebly.com/uploads/5/4/3/0/5430561/7-4_solving_logarithmic_equations_and_inequalities.pdf
Solve each equation. 1. SOLUTION: 2. SOLUTION: 3. MULTIPLE CHOICE Solve log5 (x2 − 10) = log5 3x. A 10 B 2 C 5 D 2, 5 SOLUTION: Substitute each value into the original equation. The domain of a logarithmic function cannot be 0 so 3. MULTIPLE CHOICE Solve log5 (x2 − 10) = log5 3x. A 10 B 2 C 5 D 2, 5 SOLUTION: Substitute each value into the original equation. The domain of a logarithmic function cannot be 0, so log5 (–6) is undefined and –2 is an extraneous solution. C is the correct option. Solve each inequality. 4. log5 x > 3 SOLUTION: Thus, solution set is {x | x > 125}. 5. log8 x ≤ −2 SOLUTION: Thus, solution set is . eSolutions Manual - Powered by Cognero Page 1 7-4 Solving Logarithmic Equations and Inequalities Thus, solution set is {x | x > 125}. 5. log8 x ≤ −2 SOLUTION: Thus, solution set is . 6. log4 (2x + 5) ≤ log4 (4x − 3) SOLUTION: Thus, solution set is {x | x ≥ 4}. 7. log8 (2x) > log8 (6x − 8) SOLUTION: Exclude all values of x for which So, Thus, solution set is . CCSS STRUCTURE Solve each equation. 8. SOLUTION: So, Thus, solution set is . CCSS STRUCTURE Solve each equation. 8. SOLUTION: 9. SOLUTION: 10. SOLUTION: 11. SOLUTION: eSolutions Manual - Powered by Cognero Page 2 7-4 Solving Logarithmic Equations and Inequalities 11. SOLUTION: 12. SOLUTION: 13. SOLUTION: 14. log3 (3x + 8) = log3 (x2 + x) SOLUTION: S b tit t h l i t th i i l ti 14. log3 (3x + 8) = log3 (x2 + x) SOLUTION: Substitute each value into the original equation. Thus, x = –2 or 4. 15. log12 (x2 − 7) = log12 (x + 5) SOLUTION: Substitute each value into the original equation. Thus, x = –3 or 4. 16. log6 (x2 − 6x) = log6 (−8) SOLUTION: eSolutions Manual - Powered by Cognero Page 3 7-4 Solving Logarithmic Equations and Inequalities Thus, x = –3 or 4. 16. log6 (x2 − 6x) = log6 (−8) SOLUTION: Substitute each value into the original equation. log6 (–8) is undefined, so 4 and 2 are extraneous solutions. Thus, no solution. 17. log9 (x2 − 4x) = log9 (3x − 10) SOLUTION: Substitute each value into the original equation. log9 (–4) is undefined and 2 is extraneous solution. Thus, x = 5. 18. log4 (2x2 + 1) = log4 (10x − 7) SOLUTION: log9 (–4) is undefined and 2 is extraneous solution. Thus, x = 5. 18. log4 (2x2 + 1) = log4 (10x − 7) SOLUTION: Substitute each value into the original equation. Thus, x = 1 or 4. 19. log7 (x2 − 4) = log7 (− x + 2) SOLUTION: Substitute each value into the original equation. Since you can not have a log of 0, x = 3 is the solution. SCIENCE The equation for wind speed w, in miles per hour, near the center of a tornado is w = 93 log10 d + 65, where d is the distance in miles that the tornado travels. 20. Write this equation in exponential form. SOLUTION: eSolutions Manual - Powered by Cognero Page 4 7-4 Solving Logarithmic Equations and Inequalities Since you can not have a log of 0, x = 3 is the solution. SCIENCE The equation for wind speed w, in miles per hour, near the center of a tornado is w = 93 log10 d + 65, where d is the distance in miles that the tornado travels. 20. Write this equation in exponential form. SOLUTION: 21. In May of 1999, a tornado devastated Oklahoma City with the fastest wind speed ever recorded. If the tornado traveled 525 miles, estimate the wind speed near the center of the tornado. SOLUTION: Substitute 525 for d in the equation and simplify. Solve each inequality. 22. log6 x < −3 SOLUTION: The solution set is . 23. log4 x ≥ 4 SOLUTION: The solution set is {x | x ≥ 256}. 24. log3 x ≥ −4 SOLUTION: The solution set is {x | x ≥ 256}. 24. log3 x ≥ −4 SOLUTION: The solution set is . 25. log2 x ≤ −2 SOLUTION: The solution set is . 26. log5 x > 2 SOLUTION: The solution set is . 27. log7 x < −1 SOLUTION: The solution set is . 28. log2 (4x − 6) > log2 (2x + 8) SOLUTION: The solution set is . eSolutions Manual - Powered by Cognero Page 5 7-4 Solving Logarithmic Equations and Inequalities The solution set is . 28. log2 (4x − 6) > log2 (2x + 8) SOLUTION: The solution set is . 29. log7 (x + 2) ≥ log7 (6x − 3) SOLUTION: Exclude all values of x for which So, The solution set is . 30. log3 (7x – 6) < log3 (4x + 9) SOLUTION: Exclude all values of x for which So, The solution set is . 31. log5 (12x + 5) ≤ log5 (8x + 9) SOLUTION: So, The solution set is . 31. log5 (12x + 5) ≤ log5 (8x + 9) SOLUTION: Exclude all values of x for which So, The solution set is . 32. log11 (3x − 24) ≥ log11 (−5x − 8) SOLUTION: The solution set is {x | x ≥ 2}. 33. log9 (9x + 4) ≤ log9 (11x − 12) SOLUTION: The solution set is {x | x ≥ 8}. 34. CCSS MODELING The magnitude of an earthquake is measured on a logarithmic scale called the Richter scale. The magnitude M is given by M = log10 x, where x represents the amplitude of the seismic wave causing ground motion. a. How many times as great is the amplitude caused by an earthquake with a Richter scale rating of 8 as an aftershock with a Richter scale rating of 5? b. In 1906, San Francisco was almost completely destroyed by a 7.8 magnitude earthquake. In 1911, an earthquake estimated at magnitude 8.1 occurred along the New Madrid fault in the Mississippi River Valley. How many times greater was the New M d id h k h h S F i eSolutions Manual - Powered by Cognero Page 6 7-4 Solving Logarithmic Equations and Inequalities The solution set is {x | x ≥ 8}. 34. CCSS MODELING The magnitude of an earthquake is measured on a logarithmic scale called the Richter scale. The magnitude M is given by M = log10 x, where x represents the amplitude of the seismic wave causing ground motion. a. How many times as great is the amplitude caused by an earthquake with a Richter scale rating of 8 as an aftershock with a Richter scale rating of 5? b. In 1906, San Francisco was almost completely destroyed by a 7.8 magnitude earthquake. In 1911, an earthquake estimated at magnitude 8.1 occurred along the New Madrid fault in the Mississippi River Valley. How many times greater was the New Madrid earthquake than the San Francisco earthquake? SOLUTION: a. The amplitude of the seismic wave with a Richter scale rating of 8 and 5 are 108 and 105 respectively. Divide 108 by 105. The scale rating of 8 is 103 or 1000 times greater than the scale rating of 5. b. The amplitudes of San Francisco earthquake and New Madrid earthquake were 107.8 and 108.1 respectively. Divide 108.1 by 107.8. The New Madrid earthquake was 100.3 or about 2 times greater than the San Francisco earthquake. 35. MUSIC The first key on a piano keyboard corresponds to a pitch with a frequency of 27.5 cycles per second. With every successive key, going up the black and white keys, the pitch multiplies by a constant. The formula for the frequency of the pitch sounded when the nth note up the keyboard is played is given by a A note has a frequency of 220 cycles per second The New Madrid earthquake was 100.3 or about 2 times greater than the San Francisco earthquake. 35. MUSIC The first key on a piano keyboard corresponds to a pitch with a frequency of 27.5 cycles per second. With every successive key, going up the black and white keys, the pitch multiplies by a constant. The formula for the frequency of the pitch sounded when the nth note up the keyboard is played is given by a. A note has a frequency of 220 cycles per second. How many notes up the piano keyboard is this? b. Another pitch on the keyboard has a frequency of 880 cycles per second. After how many notes up the keyboard will this be found? SOLUTION: a. Substitute 220 for f in the formula and solve for n. b. Substitute 880 for f in the formula and solve for n. 36. MULTIPLE REPRESENTATIONS In this problem, you will explore the graphs shown: y = log4 x and a. ANALYTICAL How do the shapes of the graphs compare? How do the asymptotes and the x-intercepts of the graphs compare? b. VERBAL Describe the relationship between the graphs. c. GRAPHICAL Use what you know about transformations of graphs to compare and contrast eSolutions Manual - Powered by Cognero Page 7 7-4 Solving Logarithmic Equations and Inequalities a. ANALYTICAL How do the shapes of the graphs compare? How do the asymptotes and the x-intercepts of the graphs compare? b. VERBAL Describe the relationship between the graphs. c. GRAPHICAL Use what you know about transformations of graphs to compare and contrast the graph of each function and the graph of y = log4 x. 1. y = log4 x + 2 2. y = log4 (x + 2) 3. y = 3 log4 x d. ANALYTICAL Describe the relationship between y = log4 x and y = −1(log4 x). What are a reasonable domain and range for each function? e. ANALYTICAL Write an equation for a function for which the graph is the graph of y = log3 x translated 4 units left and 1 unit up. SOLUTION: a. The shapes of the graphs are the same. The asymptote for each graph is the y-axis and the x-intercept for each graph is 1. b. The graphs are reflections of each other over the x-axis. c. 1. The second graph is the same as the first, except it is shifted horizontally to the left 2 units. [−2, 8] scl: 1 by [−5, 5] scl: 1 2. The second graph is the same as the first, except it is shifted vertically up 2 units. [−2, 8] scl: 1 by [−5, 5] scl: 1 2. The second graph is the same as the first, except it is shifted vertically up 2 units. [−4, 8] scl: 1 by [−5, 5] scl: 1 3. Each point on the second graph has a y-coordinate 3 times that of the corresponding point on the first graph. [−2, 8] scl: 1 by [−5, 5] scl: 1 d. The graphs are reflections of each other over the x-axis. D = {x | x > 0}; R = {all real numbers} e. where h is the horizontal shift and k is the vertical shift. Since there is a horizontal shift of 4 and vertical shift of 1, h = 4 and k = 1. y = log3 (x + 4) + 1 37. SOUND The relationship between the intensity of sound I and the number of decibels β is , where I is the intensity of sound in watts per square meter. a. Find the number of decibels of a sound with an intensity of 1 watt per square meter. b. Find the number of decibels of sound with an intensity of 10−2 watts per square meter. c. The intensity of the sound of 1 watt per square meter is 100 times as much as the intensity of 10−2 watts per square meter. Why are the decibels of sound not 100 times as great? SOLUTION: a. Substitute 1 for I in the given equation and solve for β. eSolutions Manual - Powered by Cognero Page 8 7-4 Solving Logarithmic Equations and Inequalities e. where h is the horizontal shift and k is the vertical shift. Since there is a horizontal shift of 4 and vertical shift of 1, h = 4 and k = 1. y = log3 (x + 4) + 1 37. SOUND The relationship between the intensity of sound I and the number of decibels β is , where I is the intensity of sound in watts per square meter. a. Find the number of decibels of a sound with an intensity of 1 watt per square meter. b. Find the number of decibels of sound with an intensity of 10−2 watts per square meter. c. The intensity of the sound of 1 watt per square meter is 100 times as much as the intensity of 10−2 watts per square meter. Why are the decibels of sound not 100 times as great? SOLUTION: a. Substitute 1 for I in the given equation and solve for β. b. Substitute 10−2 for I in the given equation and solve for β. c. Sample answer: The power of the logarithm only changes by 2. The power is the answer to the logarithm. That 2 is multiplied by the 10 before the logarithm. So we expect the decibels to change by 20. 38. CCSS CRITIQUE Ryan and Heather are solving log3 x ≥ −3. Is either of them correct? Explain your reasoning. c. Sample answer: The power of the logarithm only changes by 2. The power is the answer to the logarithm. That 2 is multiplied by the 10 before the logarithm. So we expect the decibels to change by 20. 38. CCSS CRITIQUE Ryan and Heather are solving log3 x ≥ −3. Is either of them correct? Explain your reasoning. SOLUTION: Sample answer: Ryan; Heather did not need to switch the inequality symbol when raising to a negative power. 39. CHALLENGE Find log3 27 + log9 27 + log27 27 + log81 27 + log243 27. SOLUTION: 40. REASONING The Property of Inequality for Logarithmic Functions states that when b > 1, logb x > logb y if and only if x > y. What is the case for when 0 < b < 1? Explain your reasoning. SOLUTION: Sample answer: When 0 < b < 1, logb x > logb y if and only if x < y. The inequality symbol is switched because a fraction that is less than 1 becomes smaller when it is taken to a greater power. 41. WRITING IN MATH Explain how the domain and range of logarithmic functions are related to the domain and range of exponential functions. SOLUTION: The logarithmic function of the form y = logb x is the inverse of the exponential function of the form y = eSolutions Manual - Powered by Cognero Page 9 7-4 Solving Logarithmic Equations and Inequalities Sample answer: When 0 < b < 1, logb x > logb y if and only if x < y. The inequality symbol is switched because a fraction that is less than 1 becomes smaller when it is taken to a greater power. 41. WRITING IN MATH Explain how the domain and range of logarithmic functions are related to the domain and range of exponential functions. SOLUTION: The logarithmic function of the form y = logb x is the inverse of the exponential function of the form y = bx. The domain of one of the two inverse functions is the range of the other. The range of one of the two inverse functions is the domain of the other. 42. OPEN ENDED Give an example of a logarithmic equation that has no solution. SOLUTION: Sample answer: log3 (x + 4) = log3 (2x + 12) 43. REASONING Choose the appropriate term. Explain your reasoning. All logarithmic equations are of the form y = logb x. a. If the base of a logarithmic equation is greater than 1 and the value of x is between 0 and 1, then the value for y is (less than, greater than, equal to) 0. b. If the base of a logarithmic equation is between 0 and 1 and the value of x is greater than 1, then the value of y is (less than, greater than, equal to) 0. c. There is/are (no, one, infinitely many) solution(s) for b in the equation y = logb 0. d. There is/are (no, one, infinitely many) solution (s) for b in the equation y = logb 1. SOLUTION: a. less than b. less than c. no d. infinitely many 44. WRITING IN MATH Explain why any logarithmic function of the form y = logb x has an x-intercept of (1, 0) and no y-intercept. SOLUTION: The y-intercept of the exponential function y = bx is (0, 1). When the x and y coordinates are switched, the y-intercept is transformed to the x-intercept of (1, 0). There was no x-intercept (1, 0) in the exponential function of the form y = bx. So when the x and y-coordinates are switched there would be no point on the inverse of (0, 1), and there is no y-intercept. 45. Find x if A 3 4 a. less than b. less than c. no d. infinitely many 44. WRITING IN MATH Explain why any logarithmic function of the form y = logb x has an x-intercept of (1, 0) and no y-intercept. SOLUTION: The y-intercept of the exponential function y = bx is (0, 1). When the x and y coordinates are switched, the y-intercept is transformed to the x-intercept of (1, 0). There was no x-intercept (1, 0) in the exponential function of the form y = bx. So when the x and y-coordinates are switched there would be no point on the inverse of (0, 1), and there is no y-intercept. 45. Find x if A 3.4 B 9.4 C 11.2 D 44.8 SOLUTION: C is the correct choice. 46. The monthly precipitation in Houston for part of a year is shown. Find the median precipitation. F 3.60 in. G 4.22 in. H 3.83 in. J 4.25 in. SOLUTION: Arrange the data in ascending order. 3.18, 3.60, 3.83, 5.15, 5.35 The median is the middle value. So, 3.83 is the median precipitation. H is the correct choice. 47. Clara received a 10% raise each year for 3 eSolutions Manual - Powered by Cognero Page 10 7-4 Solving Logarithmic Equations and Inequalities C is the correct choice. 46. The monthly precipitation in Houston for part of a year is shown. Find the median precipitation. F 3.60 in. G 4.22 in. H 3.83 in. J 4.25 in. SOLUTION: Arrange the data in ascending order. 3.18, 3.60, 3.83, 5.15, 5.35 The median is the middle value. So, 3.83 is the median precipitation. H is the correct choice. 47. Clara received a 10% raise each year for 3 consecutive years. What was her salary after the three raises if her starting salary was $12,000 per year? A $14,520 B $15,972 C $16,248 D $16,410 SOLUTION: Use the compound interest formula. Substitute $12,000 for P, 0.10 for r, 1 for n and 3 for t and simplify. B is the correct choice. 48. SAT/ACT A vendor has 14 helium balloons for sale: 9 are yellow, 3 are red, and 2 are green. A balloon is selected at random and sold. If the balloon sold is yellow, what is the probability that the next balloon, selected at random, is also yellow? B is the correct choice. 48. SAT/ACT A vendor has 14 helium balloons for sale: 9 are yellow, 3 are red, and 2 are green. A balloon is selected at random and sold. If the balloon sold is yellow, what is the probability that the next balloon, selected at random, is also yellow? F G H J K SOLUTION: The probability of selecting an yellow balloon next is: So, the correct answer choice is J. Evaluate each expression. 49. log4 256 SOLUTION: 50. SOLUTION: 51. log6 216 SOLUTION: 52. log3 27 SOLUTION: 53. SOLUTION eSolutions Manual - Powered by Cognero Page 11 7-4 Solving Logarithmic Equations and Inequalities 51. log6 216 SOLUTION: 52. log3 27 SOLUTION: 53. SOLUTION: 54. log7 2401 SOLUTION: Solve each equation or inequality. Check your solution. 55. 52x + 3 ≤ 125 SOLUTION: 56. 33x − 2 > 81 SOLUTION: 57. 44a + 6 ≤ 16a SOLUTION: 58. SOLUTION: 57. 44a + 6 ≤ 16a SOLUTION: 58. SOLUTION: 59. SOLUTION: 60. SOLUTION: 61. SHIPPING The height of a shipping cylinder is 4 feet more than the radius. If the volume of the cylinder is 5π cubic feet, how tall is it? Use the formula V = πr2h. SOLUTION: Substitute 5π for V and r + 4 for h in the formula and simplify. eSolutions Manual - Powered by Cognero Page 12 7-4 Solving Logarithmic Equations and Inequalities 61. SHIPPING The height of a shipping cylinder is 4 feet more than the radius. If the volume of the cylinder is 5π cubic feet, how tall is it? Use the formula V = πr2h. SOLUTION: Substitute 5π for V and r + 4 for h in the formula and simplify. The equation has one real root r = 1. Thus, the height of the shipping cylinder is 1 + 4 = 5 ft. 62. NUMBER THEORY Two complex conjugate numbers have a sum of 12 and a product of 40. Find the two numbers SOLUTION: The equations that represent the situation are: Solve equation (1). Solve equation (2). Thus, the two numbers are 6 + 2i and 6 – 2i. Simplify. Assume that no variable equals zero. 63. x5 · x3 SOLUTION: 64. a2 · a6 SOLUTION: Thus, the two numbers are 6 + 2i and 6 – 2i. Simplify. Assume that no variable equals zero. 63. x5 · x3 SOLUTION: 64. a2 · a6 SOLUTION: 65. (2p 2n)3 SOLUTION: 66. (3b3c2)2 SOLUTION: 67. SOLUTION: 68. SOLUTION: eSolutions Manual - Powered by Cognero Page 13 7-4 Solving Logarithmic Equations and Inequalities
15697
https://www.fourmilab.ch/rpkp/experiments/analysis/zCalc.html
z score Calculator The probability of a result x in an experiment consisting of a large number of equally probable independent trials n is approximated by the normal probability density function: where μ, the mean value, is n/2 and σ, the standard deviation, is a measure of the breadth of the curve which, for experiments with two equally probable outcomes of each trial is: Since the mean value and standard deviation depend upon the number of trials in the experiment, comparison between experiments with differing numbers of trials is facilitated by standardising the result: transforming it to a distribution with mean value zero and standard deviation of 1. A normally distributed experimental result x is thus standardised by subtracting the mean and dividing by the standard deviation of the experiment: This z-value or z score expresses the divergence of the experimental result x from the most probable result μ as a number of standard deviations σ The larger the value of z, the less probable the experimental result is due to chance. The probability can be calculated from the cumulative standard normal distribution: Which gives the probability P that an experimental result with a z value less than or equal to that observed is due to chance. Subtracting P from one: gives Q, the probability that the observed z score is due to chance. Unfortunately, there is no closed form solution for P, so it must be evaluated numerically. This page allows you to calculate the probability of chance occurrence of a given z, or to calculate z given a probability Q. All calculations are performed with six decimal places of accuracy; the maximum z accepted is 6. Note that the probability calculated from the z is an approximation which is valid only for large values of n, and is therefore only meaningful when calculated for experiments with a large number of individual trials. In order to use this page, your browser must support JavaScript. The text field below indicates whether JavaScript is available; if not, consider switching to a browser which implements it. Calculate probability Q from z One of the most common probability calculations is determining, given the measured z value from an experiment or set of experiments, the probability of the result being due to chance. Enter the z value in the box below, press the Return key or the Calculate button, and the probability will appear in the Q box. Calculate z from probability Q To determine the z score indicating a probability Q of non-chance occurrence for an experiment, enter Q in the box below and press the Return key or the Calculate button. Mathematical Details Chi-square Calculator Table of Contents Run Experiments RetroPsychoKinesis Project Home by John Walker
15698
https://math.stackexchange.com/questions/1304787/inverse-function-theorem-for-analytic-functions-whose-derivative-might-vanish
real analysis - inverse function theorem for analytic functions whose derivative might vanish - Mathematics Stack Exchange Join Mathematics 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 Mathematics helpchat Mathematics 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 Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams 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 inverse function theorem for analytic functions whose derivative might vanish Ask Question Asked 10 years, 4 months ago Modified10 years, 4 months ago Viewed 252 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Suppose x(t),y(t)x(t),y(t) are monotone increasing functions, and f f and g g are real-analytic functions that are not identically zero. If f(x(t))=g(y(t))f(x(t))=g(y(t)) for all t t, does it follow that x x is an analytic function of y y? If f′≠0 f′≠0 I could use the inverse function theorem to patch together an analtyic x=f−1(g(y))x=f−1(g(y)); can anything go wrong if f′f′ vanishes at isolated points? real-analysis Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications edited May 30, 2015 at 5:58 user7530user7530 asked May 30, 2015 at 2:08 user7530user7530 50.7k 11 11 gold badges 92 92 silver badges 159 159 bronze badges 3 maybe I'm misreading something, but, take the bump function (smooth but, not analytic at zero) and add t/2 t/2. I think that makes an increasing function. Take x(t)=y(t)x(t)=y(t) the modified bump and let f=g f=g then certainly f(x(t))=g(y(t))f(x(t))=g(y(t)) for all t t and it does not follow that x x or y y is analytic. Surely I misinterpret the spirit of the question.James S. Cook –James S. Cook 2015-05-30 04:43:39 +00:00 Commented May 30, 2015 at 4:43 @JamesS.Cook I've clarified the question a bit... I'm asking if x x is an analytic function of y y; in your example if I understand correctly x(y)=y x(y)=y which is certainly analytic.user7530 –user7530 2015-05-30 05:55:27 +00:00 Commented May 30, 2015 at 5:55 I see, and I'm tempted to follow your solution. Certainly we know that for such t t which give f′(x(t))=0 f′(x(t))=0 then g′(y(t))=0 g′(y(t))=0 as well (since x′(t),y′(t)x′(t),y′(t) are nonzero and the chain rule tells us f′(x(t))x′(t)=g′(y(t))y′(t)f′(x(t))x′(t)=g′(y(t))y′(t)). Perhaps that is useful?James S. Cook –James S. Cook 2015-05-30 06:33:23 +00:00 Commented May 30, 2015 at 6:33 Add a comment| 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Yes, things can go wrong if f′f′ vanishes. For example: x(t)=t x(t)=t, y(t)=t 3 y(t)=t 3, f(x)=x 3 f(x)=x 3, g(x)=x g(x)=x. Then f(x(t))=g(y(t))=t 3 f(x(t))=g(y(t))=t 3, but x x cannot be written as a differentiable function of y y. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered May 31, 2015 at 15:54 user147263 user147263 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 real-analysis 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 Report this ad Related 2Application of the Inverse Function Theorem 25The Identity Theorem for real analytic functions 2Identity theorem for forms (on analytic manifolds) 5A question about real-analytic functions vanishing on an open set 0Is monotonicity a necessary condition for the inverse function theorem? 6Is there a real-analytic monotone function f:(0,∞)→R f:(0,∞)→R which vanishes at infinity, but whose derivative admits no limit? Hot Network Questions My dissertation is wrong, but I already defended. How to remedy? Childhood book with a girl obsessed with homonyms who adopts a stray dog but gives it back to its owners Are there any world leaders who are/were good at chess? Cannot build the font table of Miama via nfssfont.tex Making sense of perturbation theory in many-body physics Who is the target audience of Netanyahu's speech at the United Nations? What happens if you miss cruise ship deadline at private island? Should I let a player go because of their inability to handle setbacks? Does the Mishna or Gemara ever explicitly mention the second day of Shavuot? How can the problem of a warlock with two spell slots be solved? Do we declare the codomain of a function from the beginning, or do we determine it after defining the domain and operations? Can peaty/boggy/wet/soggy/marshy ground be solid enough to support several tonnes of foot traffic per minute but NOT support a road? With with auto-generated local variables On being a Maître de conférence (France): Importance of Postdoc How to home-make rubber feet stoppers for table legs? How to start explorer with C: drive selected and shown in folder list? Overfilled my oil Why multiply energies when calculating the formation energy of butadiene's π-electron system? Explain answers to Scientific American crossword clues "Éclair filling" and "Sneaky Coward" Is direct sum of finite spectra cancellative? Xubuntu 24.04 - Libreoffice Is it safe to route top layer traces under header pins, SMD IC? What were "milk bars" in 1920s Japan? Origin of Australian slang exclamation "struth" meaning greatly surprised more hot questions Question feed Subscribe to RSS Question feed To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Why are you flagging this comment? It contains harassment, bigotry or abuse. This comment attacks a person or group. Learn more in our Code of Conduct. It's unfriendly or unkind. This comment is rude or condescending. Learn more in our Code of Conduct. Not needed. This comment is not relevant to the post. Enter at least 6 characters Something else. A problem not listed above. Try to be as specific as possible. Enter at least 6 characters Flag comment Cancel You have 0 flags left today Mathematics Tour Help Chat Contact Feedback Company Stack Overflow Teams Advertising Talent About Press Legal Privacy Policy Terms of Service Your Privacy Choices Cookie Policy Stack Exchange Network Technology Culture & recreation Life & arts Science Professional Business API Data Blog Facebook Twitter LinkedIn Instagram Site design / logo © 2025 Stack Exchange Inc; user contributions licensed under CC BY-SA. rev 2025.9.26.34547 By clicking “Accept all cookies”, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Accept all cookies Necessary cookies only Customize settings Cookie Consent Preference Center When you visit any of our websites, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences, or your device and is mostly used to make the site work as you expect it to. The information does not usually directly identify you, but it can give you a more personalized experience. Because we respect your right to privacy, you can choose not to allow some types of cookies. Click on the different category headings to find out more and manage your preferences. Please note, blocking some types of cookies may impact your experience of the site and the services we are able to offer. Cookie Policy Accept all cookies Manage Consent Preferences Strictly Necessary Cookies Always Active These cookies are necessary for the website to function and cannot be switched off in our systems. They are usually only set in response to actions made by you which amount to a request for services, such as setting your privacy preferences, logging in or filling in forms. You can set your browser to block or alert you about these cookies, but some parts of the site will not then work. These cookies do not store any personally identifiable information. Cookies Details‎ Performance Cookies [x] Performance Cookies These cookies allow us to count visits and traffic sources so we can measure and improve the performance of our site. They help us to know which pages are the most and least popular and see how visitors move around the site. All information these cookies collect is aggregated and therefore anonymous. If you do not allow these cookies we will not know when you have visited our site, and will not be able to monitor its performance. Cookies Details‎ Functional Cookies [x] Functional Cookies These cookies enable the website to provide enhanced functionality and personalisation. They may be set by us or by third party providers whose services we have added to our pages. If you do not allow these cookies then some or all of these services may not function properly. Cookies Details‎ Targeting Cookies [x] Targeting Cookies These cookies are used to make advertising messages more relevant to you and may be set through our site by us or by our advertising partners. They may be used to build a profile of your interests and show you relevant advertising on our site or on other sites. They do not store directly personal information, but are based on uniquely identifying your browser and internet device. Cookies Details‎ Cookie List Clear [x] checkbox label label Apply Cancel Consent Leg.Interest [x] checkbox label label [x] checkbox label label [x] checkbox label label Necessary cookies only Confirm my choices
15699
https://www.irishmathsoc.org/bull89/wef/Classroom/Holland/Holland-wef.pdf
Irish Math. Soc. Bulletin Number 89, Summer 2022, 57–64 ISSN 0791-5578 Abel’s limit theorem, its converse, and multiplication formulae for Γ(x) FINBARR HOLLAND Abstract. Abel’s well-known limit theorem for power series, and its corrected con-verse due to J. E. Littlewood, form the basis for a general identity that is prre-sented here, which is shown to be equivalent to Gauss’s multiplication theorem for the Gamma function. 1. Introduction An incomplete solution of a problem of mine, numbered Problem 86.3 in , (that was presented in ) prompted this note about Abel’s limit theorem on power series, and two of its partial converses due, respectively, to Tauber and Littlewood. These are landmark results in the development of Real and Complex Analysis. For instance, Abel’s theorem initiated the study of the boundary behaviour of analytic functions on the unit disc, and, in conjunction with Ces´ aro’s consistency theorem on the convergence of arithmetic means of a convergent sequence, paved the way for summing series by different methods dealt with in , while the theorems of Tauber and Littlewood gave rise to the beautiful sub-topic of Wiener’s Tauberian Analysis, also exposed in . Students of Analysis who are desirous of learning “the tricks of the trade” would do well to study proofs of Abel’s theorem and Tauber’s, and at least acquaint themselves with the more profound result of Littlewood. All three theorems are simply expounded in . In this note, we’ll state and provide standard proofs of the theorems of Abel and Tauber, and state, but not prove, Littlewood’s deeper result; instead, we’ll illustrate its utility by means of a simple example. These theorems will be discussed in Sections 2, 3 and 4, respectively. As an illustration of the underlying ideas we’ll derive a general theorem in Section 5, which is motivated by the aforementioned journal problem, and show that a special case of it is equivalent to Gauss’s multiplication formula for the Gamma function (see the example in Section 9.56 of ). 2. Abel’s limit theorem Throughout the note, f stands for a generic power series P∞ n=0 anxn whose radius of convergence is 1, though the coefficients will differ from time to time. According to Abel: if the series P∞ n=0 an is convergent, then lim x→1−f(x) = ∞ X n=0 an. We sketch the standard proof of this. 2010 Mathematics Subject Classification. 30B30, 40E05, 33B15. Key words and phrases. power series, sequence, coefficients, convergent series, entire function, roots of unity. Received on 5-4-2022; revised 6-5-2022. DOI:10.33232/BIMS.0089.57.64. c ⃝2022 Irish Mathematical Society 57 58 HOLLAND Proof. Let sn = n X k=0 ak, n = 0, 1, 2 . . . , and x ∈[0, 1). As a first step we express f(x) as a convex combination of the sequence s0, s1, s2, . . .. This is easy to do since by pointwise multiplication of two absolutely convergent power series (1 + x + x2 + · · · )(a0 + a1x + a2x2 + · · · ) = a0 + (a0 + a1)x + (a0 + a1 + a2)x2 + · · · = s0 + s1x + s2x2 + · · · so that f(x) = (1 −x)  ∞ X n=0 xn ∞ X n=0 anxn = (1 −x) ∞ X n=0 snxn. Accordingly, if s = limn→∞sn, and 0 ≤x < 1, f(x) −s = (1 −x) ∞ X n=0 snxn −s(1 −x) ∞ X n=0 xn = (1 −x) ∞ X n=0 (sn −s)xn, from which it follows that |f(x) −s| ≤(1 −x) ∞ X n=0 |sn −s|xn ≤sup{|sn −s| : n = 0, 1, 2, . . .}. Hence sup{|f(x) −s| : 0 ≤x < 1} ≤sup{|sn −s| : n = 0, 1, 2, . . .}, a step in the right direction, but not the final one! To obtain the desired result, we refine the argument just given by splitting the sum (1−x) P∞ n=0 |sn−s|xn in two appropriately. To achieve this, let ǫ > 0, and choose an integer n0 so that |sn−s| < ǫ, ∀n > n0, whence for any x ∈(0, 1), (1 −x) ∞ X n=n0+1 |sn −s|xn ≤ǫ(1 −x) ∞ X n=n0+1 xn ≤ǫ. Consequently, |f(x) −s| ≤(1 −x) n0 X n=0 |sn −s|xn + ǫ, and so, on letting x tend to 1 from the left, lim sup x→1−|f(x) −s| ≤ǫ. Since ǫ is an arbitrary positive number, this means that limx→1−f(x) = s = P∞ n=0 an, as claimed. □ 3. Tauber’s converse As the example 1 1 + x = ∞ X n=0 (−1)nxn, |x| < 1, shows, the direct converse of Abel’s theorem is false. Tauber proved a conditional converse according to which, if limx→1−f(x) = s, and limn→∞nan = 0, then P∞ n=0 an is convergent and its sum is s. Abel’s limit theorem 59 Proof. To see this, note that f(x) −sn = ∞ X k=0 akxk − n X k=0 ak = n X k=1 ak(xk −1) + ∞ X k=n+1 akxk, for any x ∈(0, 1), and any positive integer n. Now ∞ X k=n+1 akxk = ∞ X k=n+1 (kak)1 kxk ≤ 1 n + 1 ∞ X k=n+1 k|ak|xk ≤ 1 (n + 1)(1 −x) max{k|ak| : k ≥n + 1}, and n X k=1 ak(xk −1) ≤ n X k=1 |ak|(1 −xk) ≤(1 −x) n X k=1 k|ak|. Combining these estimates we have that |f(x) −sn| ≤(1 −x) n X k=1 k|ak| + 1 (n + 1)(1 −x) max{k|ak| : k ≥n}. Bearing in mind that x and n are at our disposal, to be chosen as we see fit, it’s now convenient to set x ≡xn = 1 − 1 n+1. With this choice we have |f(xn) −sn| ≤ 1 n + 1 n X k=1 k|ak| + max{k|ak| : k ≥n}, an expression that tends to zero as n →∞, its first term by Ces´ aro’s theorem, and its second by hypothesis. Therefore ∞ X n=0 an = lim n→∞sn = lim n→∞f(xn) = lim x→1−f(x) = s, as we wanted to show. □ 4. Littlewood’s converse Tauber’s result was considerably strengthened by Littlewood who proved that if limx→1−f(x) = s, and the sequence nan is merely bounded, then the series P∞ n=0 an is convergent and its sum is s. We won’t give the proof of this, but instead provide a simple example to illustrate its utility. Example 4.1. Suppose 0 < θ < 2π. Then ∞ X n=1 cos nθ n = −ln  2 sin θ 2  . 60 HOLLAND Proof. Fix θ ∈(0, 2π), and consider the power series expansion about the origin of f(x) = ln(1 −2 cos θx + x2), namely, if |x| < 1, then f(x) = ln (1 −eiθx)(1 −e−iθx) = − ∞ X n=1 einθ n xn − ∞ X n=1 e−inθ n xn = −2 ∞ X n=1 cos nθ n xn. Clearly, f(x) →ln(2 −2 cos θ) as x →1−, and the coefficients of the last displayed power series satisfy Littlewood’s condition. Hence, when x = 1 the displayed series is convergent and its sum is ln 4 sin2 θ 2  = 2 ln 2 sin θ 2  , which yields the result. □ 5. Gauss’s multiplication formula for Γ(x) As a precursor to this, we first establish the next result which relies on the theorems just described of both Abel and Littlewood. Theorem 5.1. Let f(x) = P∞ n=0 anxn. Suppose the sequence an satisfies Littlewood’s condition, and m is a positive integer. Then f(x) −f(xm) converges to s as x →1−iff the series ∞ X n=0 h m−1 X r=0 anm+r  −an i is convergent and its sum is s. Proof. For |x| < 1, f(x) −f(xm) = ∞ X n=0 anxn − ∞ X n=0 anxnm = ∞ X n=0 m−1 X r=0 anm+rxnm+r − ∞ X n=0 anxnm = ∞ X n=0 xmn m−1 X r=0 amn+rxr −an  = ∞ X n=0 xmn (amn −an) + m−1 X r=1 amn+rxr = ∞ X n=0 cnxn, where, for n = 0, 1, . . ., cnm+r =  amn −an, if r = 0, anm+r, if r = 1, . . . , m −1. Suppose now that f(x) −f(xm) converges to s as x →1−. Then the series P∞ n=0 cnxn satisfies the hypotheses of Littlewood’s theorem, and so s = P∞ n=0 cn. In other words, if Cn denotes the nth partial sum of this series, Cn →s, whence, in particular, s = limn→∞Cmn, i.e. s = lim n→∞ n X k=0 m−1 X r=0 ckm+r = ∞ X n=0  m−1 X r=0 anm+r −an  , Abel’s limit theorem 61 as desired. Conversely, suppose the last displayed series is convergent. Let bn = m−1 X r=0 anm+r −an, n = 0, 1, . . . , and F(x) = ∞ X n=0 bnxn. Then from above, f(x) −f(xm) = ∞ X n=0 xmn m−1 X r=0 amn+rxr −an  = ∞ X n=0 bnxmn + ∞ X n=0 xmn m−1 X r=1 amn+r(xr −1) = F(xm) + m−1 X r=1 (xr −1) ∞ X n=0 amn+rxmn = F(xm) + m−1 X r=1 (xr −1)hr(x), where, for r = 1, .., m −1, hr(x) = ∞ X n=0 amn+rxmn = O(1) log 1 1 −x, (x →1−). As a result, m−1 X r=1 (xr −1)hr(x) = O(1)(1 −x) log 1 1 −x = o(1), (x →1−). Thus f(x) −f(xm) = F(xm) + o(1) (x →1−). By Abel, limx→1−F(xm) = s, and so f(x) −f(xm) converges to s as x →1−. This completes the proof. □ The folowing example is a direct consequence of this theorem. Example 5.2. Let m be any positive integer. Then, for all a > 0, ∞ X n=0 h m−1 X r=0 1 nm + r + a  − 1 n + a i = ln m. (1) Proof. Let an = 1/(n + a), n = 0, 1, 2, . . ., and f(x) = P∞ n=0 anxn. Since the series in (1) is plainly convergent, by the theorem its sum is equal to the limit of f(x) −f(xm) as x →1−. To calculate this, notice first that if |x| < 1, then f(x) = ∞ X n=1 1 nxn + 1 a + ∞ X n=1  1 n + a −1 n  xn = ln 1 1 −x + g(x), say, and then that f(x)−f(xm) = ln(1+x+· · ·+xm−1)+g(x)−g(xm) which converges to ln m as x →1−, since, by Abel, limx→1−g(x) exists. □ 62 HOLLAND The special case of this example, with m = 3 and a = 1, leads to the conclusion that ∞ X n=0 9n + 5 9n3 + 18n2 + 11n + 2 = 3 ln 3, a proof of which was sought in . What’s noteworthy about (1), and surprising perhaps, is that, for each fixed integer m > 1, the series is convergent and its sum function is independent of a! What’s the explanation for that? The reason is because—as we shall proceed to demonstrate—it’s equivalent to Gauss’s multiplication theorem for the Gamma function, Γ(x), according to which if m is a positive integer, then mmx−1 2 m−1 Y r=0 Γ(x + r m) = (2π) m−1 2 Γ(mx), ∀x > 0. (2) This is an extension of the more familiar duplication formula due to Legendre: 22x−1Γ(x)Γ(x + 1 2) = √πΓ(2x). To explain the connection between (1) and (2), recall that the reciprocal of Γ(z) is an entire function of the complex variable z, with simple zeros at the integers 0, −1, −2, . . ., that admits of the canonical factorization 1 Γ(z) = zeγz ∞ Y n=1 (1 + z n)e−z/n, where γ is Euler’s constant limn→∞  Pn k=1 1 k −ln n  . Hence, denoting by ψ the deriv-ative of ln Γ, −ψ(x) = −Γ′(x) Γ(x) = 1 x + γ + ∞ X n=1  1 n + x −1 n  . Therefore, if m is a positive integer, and x > 0, then −mΓ′(mx) Γ(mx) + m−1 X r=0 Γ′(x + r m) Γ(x + r m) = 1 x + mγ + m ∞ X n=1  1 n + mx −1 n  − m−1 X r=0 h 1 x + r m + γ + ∞ X n=1  1 n + x + r m −1 n i = − m−1 X r=1 1 x + r m + m ∞ X n=1  1 n + mx −1 n  − m−1 X r=0 ∞ X n=1  1 n + x + r m −1 n  = − m−1 X r=1 1 x + r m − ∞ X n=1 h m−1 X r=0 1 n + x + r m − m n + mx i = −m  m−1 X r=1 1 mx + r + ∞ X n=1 h m−1 X r=0 1 mn + r + mx  − 1 n + mx i = −m ∞ X n=0 h m−1 X r=0 1 mn + r + mx  − 1 n + mx i = −m ln m, Abel’s limit theorem 63 by (1), with a = mx. In other words, for x > 0, assuming (1) holds, d dx  m−1 X r=0 ln Γ(x + r m) −ln Γ(mx)  = −m ln m. Thus, for some constant C(m), Qm−1 r=0 Γ(x + r m) Γ(mx) = m−mxC(m), ∀x > 0. But, from the product formula for 1/Γ(x), it’s clear that lim x→0+ 1 xΓ(x) = 1, whence lim x→0+ Γ(x) Γ(mx) = m. Hence C(m) = m m−1 Y r=1 Γ( r m). It remains to compute the product p(m) = Qm−1 k=1 Γ( k m). To do this, we adapt Gauss’s ploy (which legend says he used in kindergarten one day to add the first 100 natural numbers) and determine the geometric mean of p(m) and the product of its factors in reverse order, namely, Qm−1 k=1 Γ( m−k m ), also p(m), of course. So, we consider p(m)2 = m−1 Y k=1 Γ( k m)Γ(1 −k m) = m−1 Y k=1  π sin kπ m  = πm−1 m−1 Y k=1 1 sin kπ m , by the reflection property of the Gamma function: Γ(z)Γ(1 −z) = π sin πz . To compute the product of the numbers sin kπ n , k = 1, 2, . . . , n −1, note that 4m−1 m−1 Y k=1 sin kπ m 2 = m−1 Y k=1  4 sin2 kπ m  = m−1 Y k=1 1 −e 2ikπ m 2. But the m numbers e 2ikπ m , k = 0, 1, . . . , m −1, are precisely the mth roots of unity, and so zm −1 = (z −1) m−1 Y k=1 (z −e 2ikπ m ). Hence, m = m−1 Y k=1 (1 −e 2ikπ m ), m2 = m−1 Y k=1 1 −e 2ikπ m 2. 64 HOLLAND Consequently, 22(m−1) m−1 Y k=1 sin kπ m 2 = m2, from which it follows that m−1 Y k=1 sin kπ m = m 2m−1 , since sin kπ m > 0 for k = 1, 2, . . . , m −1. Hence p(m)2 = πm−12m−1 m , p(m) = (2π) m−1 2 √m , and so C(m) = √m(2π) m−1 2 , whence we obtain Gauss’s formula: m−1 Y r=0 Γ(x + r m) = (2π) m−1 2 m 1 2 −mxΓ(mx). Thus, with a = mx, the identity (1) implies (2). Since we can easily reverse the steps just made from (1) to (2), it should be clear that (1) is a consequence of (2). To sum up: if m is any positive integer, Gauss’s multiplication statement for the Gamma function that mmx m−1 Y r=0 Γ(x + r m) = m−1 Y r=1 Γ( r m)Γ(mx) = √m(2π) m−1 2 Γ(mx), ∀x > 0, is equivalent to the statement that ∞ X n=0  m−1 X r=0 1 mn + r + x − 1 n + x  = ln m, ∀x > 0. References E. T. Copson: Theory of Functions of a Complex Variable, Oxford University Press, 1935. G. H. Hardy: Divergent Series, Oxford University Press, 1949. J. E. Littlewood: The Converse of Abel’s theorem on power series, Proc. London. Math. Soc. (2),9 (1911),434–48. I. Short, Problems, Irish Math. Soc. Bulletin, Number 86, Winter 2020, 114–116. I. Short, Problems, Irish Math. Soc. Bulletin, Number 88, Winter 2021, 85–87. E. C. Titchmarsh: The Theory of Functions, Oxford University Press, 1968. Finbarr Holland received the degree of BSc in 1961 at University College Cork, where he is Professor Emeritus of Mathematics. The following year the NUI awarded him an MSc degree and a Travelling Studentship, which enabled him to study under the direction of Lionel Cooper for a PhD at University College, Cardiff, which he obtained from the National University of Wales in 1964. He contributes regularly to the problem pages of several journals. School of Mathematical Sciences, University College Cork E-mail address: f.holland@ucc.ie