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arctan(x) + arctan(y) = arctan((\frac{x + y}{1 - xy}))
We will learn how to prove the
property of the inverse trigonometric function arctan(x) + arctan(y) = arctan((\frac{x
+ y}{1 - xy})), (i.e., tan(^{-1}) x
+ tan(^{-1}) y
= tan(^{-1}) ((\frac{x
+ y}{1 - xy}))if
x > 0, y > 0 and xy < 1.
1. Prove that arctan(x) + arctan(y) = arctan((\frac{x + y}{1 - xy})), if x > 0, y > 0 and xy < 1.
Proof:
Let, tan(^{-1}) x = α and tan(^{-1}) y = β
From tan(^{-1}) x = α we get,
x = tan α
and from tan(^{-1}) y = β we get,
y = tan β
Now, tan (α + β) = ((\frac{tan
α + tan β}{1 - tan α tan β}))
tan (α + β) = (\frac{x +
y}{1 - xy})
⇒ α + β = tan(^{-1}) ((\frac{x
+ y}{1 - xy}))
⇒ tan(^{-1}) x
+ tan(^{-1}) y
= tan(^{-1}) ((\frac{x
+ y}{1 - xy}))
Therefore, tan(^{-1}) x + tan(^{-1}) y
= tan(^{-1}) ((\frac{x
+ y}{1 - xy})), if x > 0, y > 0 and xy < 1.
2. Prove that arctan(x)
+ arctan(y) = π + arctan((\frac{x + y}{1 - xy})),
if x > 0, y > 0 and xy > 1. And
arctan(x) + arctan(y) = arctan((\frac{x + y}{1 - xy}))
- π,
if x < 0, y < 0 and xy > 1.
Proof: If x > 0, y >
0 such that xy > 1, then (\frac{x + y}{1 - xy}) is positive and therefore,
(\frac{x + y}{1 - xy}) is positive angle between 0° and 90°.
Similarly, if x
< 0, y < 0 such that xy > 1, then (\frac{x + y}{1 - xy})
is
positive and therefore, tan(^{-1}) ((\frac{x
+ y}{1 - xy})) is a negative angle while tan(^{-1}) x + tan(^{-1}) y
is a positive angle while tan(^{-1})
x
+ tan(^{-1}) y
is a non-negative angle. Therefore, tan(^{-1}) x + tan(^{-1}) y
= π
+ tan(^{-1}) ((\frac{x
+ y}{1 - xy})), if x > 0, y > 0 and xy > 1 and
arctan(x) + arctan(y) = arctan((\frac{x + y}{1 - xy}))
- π,
if x < 0, y < 0 and xy > 1.
Solved examples on property of inverse
circular function tan(^{-1}) x
+ tan(^{-1}) y
= tan(^{-1}) ((\frac{x
+ y}{1 - xy}))
1. Prove that 4 (2 tan(^{-1}) (\frac{1}{3}) +
tan(^{-1}) (\frac{1}{7}))
= π
Solution:
2 tan(^{-1}) (\frac{1}{3})
= tan(^{-1}) (\frac{1}{3})
+ tan(^{-1}) (\frac{1}{3})
= tan(^{-1}) ((\frac{\frac{1}{3}
+ \frac{1}{3}}{1 - \frac{1}{3} • \frac{1}{3}}))
= tan(^{-1}) (\frac{3}{4})
Now L. H. S. = 4 (2 tan(^{-1}) (\frac{1}{3})
+ tan(^{-1}) (\frac{1}{7}))
= 4 (tan(^{-1}) (\frac{3}{4})
+ tan(^{-1}) (\frac{1}{7}))
= 4 tan(^{-1}) ((\frac{\frac{3}{4}
+ \frac{1}{7}}{1 - \frac{3}{4} • \frac{1}{7}}))
= 4 tan(^{-1}) ((\frac{25}{28})
x (\frac{28}{25}))
= 4 tan(^{-1}) 1
= 4 · (\frac{π}{4})
= π = R.H.S. Proved.
2.Prove
that, tan(^{-1}) (\frac{1}{4})
+ tan(^{-1}) (\frac{2}{9})
+ tan(^{-1}) (\frac{1}{5})
+ tan(^{-1}) (\frac{1}{8})
= π/4.
Solution:
L.
H. S. = tan(^{-1}) (\frac{1}{4})
+ tan(^{-1}) (\frac{2}{9})
+ tan(^{-1}) (\frac{1}{5})
+ tan(^{-1}) (\frac{1}{8})
= tan(^{-1}) (\frac{\frac{1}{4} +
\frac{2}{9}}{1 - \frac{1}{4} • \frac{2}{9}}) + tan(^{-1}) (\frac{\frac{1}{5} +
\frac{1}{8}}{1 - \frac{1}{5} • \frac{1}{8}})
= tan(^{-1}) ((\frac{17}{36})
x (\frac{36}{34})) + tan(^{-1})
((\frac{13}{40})
x (\frac{40}{39}))
= tan(^{-1}) (\frac{1}{2})
+ tan(^{-1}) (\frac{1}{3})
= tan(^{-1}) (\frac{\frac{1}{2} +
\frac{1}{3}}{1 - \frac{1}{2} • \frac{1}{3}})
= tan(^{-1}) 1
= (\frac{π}{4}) = R. H. S. Proved.
⇒ α + β = tan(^{-1}) ((\frac{x
+ y}{1 - xy}))
⇒ tan(^{-1}) x
+ tan(^{-1}) y
= tan(^{-1}) ((\frac{x
+ y}{1 - xy}))
Therefore, tan(^{-1}) x + tan(^{-1}) y
= tan(^{-1}) ((\frac{x
+ y}{1 - xy})), if x > 0, y > 0 and xy < 1.
2. Prove that arctan(x)
+ arctan(y) = π + arctan((\frac{x + y}{1 - xy})),
if x > 0, y > 0 and xy > 1. And
arctan(x) + arctan(y) = arctan((\frac{x + y}{1 - xy}))
- π,
if x < 0, y < 0 and xy > 1.
Proof: If x > 0, y >
0 such that xy > 1, then (\frac{x + y}{1 - xy}) is positive and therefore,
(\frac{x + y}{1 - xy}) is positive angle between 0° and 90°.
Similarly, if x
< 0, y < 0 such that xy > 1, then (\frac{x + y}{1 - xy})
is
positive and therefore, tan(^{-1}) ((\frac{x
+ y}{1 - xy})) is a negative angle while tan(^{-1}) x + tan(^{-1}) y
is a positive angle while tan(^{-1})
x
+ tan(^{-1}) y
is a non-negative angle. Therefore, tan(^{-1}) x + tan(^{-1}) y
= π
+ tan(^{-1}) ((\frac{x
+ y}{1 - xy})), if x > 0, y > 0 and xy > 1 and
arctan(x) + arctan(y) = arctan((\frac{x + y}{1 - xy}))
- π,
if x < 0, y < 0 and xy > 1.
Solved examples on property of inverse
circular function tan(^{-1}) x
+ tan(^{-1}) y
= tan(^{-1}) ((\frac{x
+ y}{1 - xy}))
1. Prove that 4 (2 tan(^{-1}) (\frac{1}{3}) +
tan(^{-1}) (\frac{1}{7}))
= π
Solution:
2 tan(^{-1}) (\frac{1}{3})
= tan(^{-1}) (\frac{1}{3})
+ tan(^{-1}) (\frac{1}{3})
= tan(^{-1}) ((\frac{\frac{1}{3}
+ \frac{1}{3}}{1 - \frac{1}{3} • \frac{1}{3}}))
= tan(^{-1}) (\frac{3}{4})
Now L. H. S. = 4 (2 tan(^{-1}) (\frac{1}{3})
+ tan(^{-1}) (\frac{1}{7}))
= 4 (tan(^{-1}) (\frac{3}{4})
+ tan(^{-1}) (\frac{1}{7}))
= 4 tan(^{-1}) ((\frac{\frac{3}{4}
+ \frac{1}{7}}{1 - \frac{3}{4} • \frac{1}{7}}))
= 4 tan(^{-1}) ((\frac{25}{28})
x (\frac{28}{25}))
= 4 tan(^{-1}) 1
= 4 · (\frac{π}{4})
= π = R.H.S. Proved.
2.Prove
that, tan(^{-1}) (\frac{1}{4})
+ tan(^{-1}) (\frac{2}{9})
+ tan(^{-1}) (\frac{1}{5})
+ tan(^{-1}) (\frac{1}{8})
= π/4.
Solution:
L.
H. S. = tan(^{-1}) (\frac{1}{4})
+ tan(^{-1}) (\frac{2}{9})
+ tan(^{-1}) (\frac{1}{5})
+ tan(^{-1}) (\frac{1}{8})
= tan(^{-1}) (\frac{\frac{1}{4} +
\frac{2}{9}}{1 - \frac{1}{4} • \frac{2}{9}}) + tan(^{-1}) (\frac{\frac{1}{5} +
\frac{1}{8}}{1 - \frac{1}{5} • \frac{1}{8}})
= tan(^{-1}) ((\frac{17}{36})
x (\frac{36}{34})) + tan(^{-1})
((\frac{13}{40})
x (\frac{40}{39}))
= tan(^{-1}) (\frac{1}{2})
+ tan(^{-1}) (\frac{1}{3})
= tan(^{-1}) (\frac{\frac{1}{2} +
\frac{1}{3}}{1 - \frac{1}{2} • \frac{1}{3}})
= tan(^{-1}) 1
= (\frac{π}{4}) = R. H. S. Proved.
● Inverse Trigonometric Functions
11 and 12 Grade Math
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8901 | https://www.publisso.de/en/advice/publishing-advice-faqs/peer-review | jump to pagenavigation
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Peer Review: Why is it important?
Peer Review: Why is it important?
Scientific findings and discoveries can have far-reaching implications for individuals and society. This is one reason why they undergo a process of quality control known as 'peer review' before they are published.
Peer review involves subjecting the author's scholarly work and research to the scrutiny of other experts in the same field to check its validity and evaluate its suitability for publication. A peer review helps the publisher decide whether a work should be accepted.
How the peer review process works
When a scholarly work is submitted to a scientific journal, it first undergoes a preliminary check known as a desk review. The editors decide if the manuscript should be sent for peer review or be immediately rejected. At this stage, the focus is on assessing to what extent an article aligns with the journal’s scope and whether the manuscript is broadly ready for publication. The next step is to select experts from the same field who are qualified and able to review the work impartially. Ideally, the work is evaluated by multiple experts.
The primary goals of peer review are to determine whether a scholarly work falls within the journal’s scope, to check whether the research topic has been clearly formulated, and to decide whether a suitable approach has been taken to address the scientific issues involved. In addition, reviewers check whether the methodology reflects current best practices, whether the results and their analysis are presented transparently and credibly, and whether sufficient information is provided to enable replication. They also assess the originality and novelty of the research findings. When studies involve patients or animals, peer review further considers the ethical aspects of the work. Finally, the reviewer will also rate the readability of the work, assessing how logically the argument has been constructed and whether the conclusions are well-founded. In addition, the author of the work will generally receive useful advice on how to improve their work.
Reviewers normally prepare their evaluations using a questionnaire or checklist, and then send their reports back to the editors. This forms the basis for deciding whether the work should be accepted, considered acceptable with revisions, or rejected. Submissions with serious failings will be rejected, though they can be re-submitted once they have been thoroughly revised. In principle, editors have the final say in how many reviewer reports they obtain and to what extent they follow the recommendations therein. They are not strictly bound by reviewer opinions and can ultimately decide independently whether to publish an article or not.
If a work is rejected, this does not necessarily mean it is of poor quality. A paper may also be rejected because it doesn’t fall within the journal’s area of specialisation or because it doesn’t meet the high standards of novelty and originality required by the journal in question.
Another possible reason for rejection is that an innovative approach may not be recognised as such by the reviewers, or the innovation may not have been explained in sufficient detail. Some prestigious journals have rejection rates of over 90 percent.
Across scientific journals as a whole, roughly half of all submitted manuscripts are rejected. The figures published on journal websites should be seen as indicative rather than definitive. In most cases, authors subsequently submit their rejected work to a different journal. Whether resubmission to the same journal is possible depends on the journal’s specific policy, or should be clarified directly with the editor. That said, if a manuscript is clearly outside the scope of a journal, resubmitting it is unlikely to be worthwhile.
If reviewers suggest changes or additions to the manuscript, the authors are expected to address these within a given timeframe. In a formal response – often called a rebuttal letter – authors explain how they have dealt with reviewers’ suggestions. It isn’t compulsory to accept every suggestion, but it is best practice to justify why a point was not implemented and to indicate whether it will be incorporated in future work.
If authors themselves have identified new changes they wish to make in the meantime, these can also be incorporated. However, if the changes are so substantial that they alter the central message of the article, this must be explicitly reported to the editors so they can decide whether a new round of peer review is needed. There is no problem with simply correcting typos or stylistic slips.
Minor corrections can still be made just before the proofs are finalised, but major changes at a late stage of the process may delay publication. Hence the importance of careful revision after the peer-review stage.
At any stage of the review process, authors can contact the editors if anything is unclear. However, queries about how the process is progressing should be made sparingly, as answering them ties up editorial resources. The manuscript submission system usually allows authors to track the stage their paper has reached.
Different types of peer review
The term peer review actually encompasses a number of different approaches, the most common of which are the following:
single-blind peer review: the name of the reviewer is hidden from the author;
double-blind peer review: both the reviewer and the author remain anonymous to each other.
There are also considerable differences in the level of detail with which papers are evaluated. For example, some journals make additional use of anti-plagiarism software, organise separate reviews of the author's methods and statistics, or examine the submitted illustrations to detect whether they have been manipulated.
There is an increasing number of journals that focus on scientific software or research datasets. Peer review is also used by conference organisers to select which contributions to include in the conference, and funding bodies even use peer review methods to assess the eligibility of research proposals for funding. In all of these cases, peer-review guidelines, questionnaires and checklists are adapted accordingly.
Criticisms of peer review
All the methods mentioned above have their advantages and disadvantages. For example, critics of the double-blind method argue that reviewers can guess who the authors are by looking at the references they cite. They suggest that this could undermine their neutral stance. The evaluation of an academic paper is also affected by the reviewer's scientific beliefs and by the care and effort they choose to invest in the process.
Peer review has recently come in for major criticism following cases where reviewers failed to spot serious errors in the author's methodology. The reasons for 'failures' in the peer review process include peer reviewers' heavy workloads as more papers are published and poor selection of reviewers by editorial boards.
Another objection that is frequently raised is that peer review is not transparent enough, not just because the reviews are inevitably subjective (especially if reviewers are unable to separate themselves adequately from their particular schools of thought), but also because reviewers may not appreciate the value of a new idea or may withhold – or simply not be asked to provide – relevant information on conflicts of interest.A further key criticism of peer review is that the process may stretch over a long period of time, generally weeks or months, but occasionally even years.
Some people also suspect that journals which claim to have implemented peer review actually carry out very superficial assessments, or none at all. For further information see the FAQ on "Predatory Publishing".
It is generally accepted that peer review cannot completely eliminate cases of fraud and the publication of low-quality papers. Nevertheless, peer review continues to be favoured despite all the criticism because it has ultimately proved its worth and shown that in most cases it can help improve the quality of publications – especially if authors are able to view the report and work through the comments. In the end, of course, responsibility lies with the authors who are required to demonstrate rigour, probity and scientific reproducibility as part of the scientific process. The peer review concept is also constantly being adapted to counter criticism such as the points mentioned above.
Alternatives to the standard peer review process
The variants described above are all forms of pre-publication peer review, in which an article is peer-reviewed prior to its formal publication. But efforts are also being made to experiment with different alternatives, with the hope that these may address some of the criticisms mentioned above:
Open peer review (or crowd-sourced peer review) is a collective term for various methods that seek to open up the peer-review process, for example by revealing the identity of authors and peer reviewers or even providing access to the reviews themselves.
Post-publication peer review is a generic term for all forms of feedback and commentary on an article that take place after publication. It can therefore be combined with open peer review. The basic idea is to publish a manuscript after nothing more than a rough, preliminary review – or no review at all – and to leave the corresponding evaluation and assessment to the scientific community. The F1000 platform is a well-known example of this approach. In addition, post-publication peer review generally allows people to publish comments and feedback on articles that have already been published. These comments are submitted within a formalised process, either on the publishing platform itself or on external platforms such as PubPeer.
Other alternatives include redirecting rejected manuscripts to another, potentially more suitable journal together with the review reports, thereby shortening the subsequent peer-review process. This is typically referred to as cascading peer review when the manuscript is submitted to another journal from the same publisher, or portable peer review when the manuscript and its peer-review reports are also submitted to a different publisher. The goal is to avoid needlessly duplicated review effort. A similar approach is taken by the Review Commons platform, an initiative that works in affiliation with a consortium of life science journals. Review Commons makes manuscripts available on a preprint server before submission to a journal, where they can be peer-reviewed independently of any particular journal. The authors then use the peer-review reports as a basis for deciding which of the participating journals is the best choice for their manuscript. They then submit this to the journal together with the peer-review reports.
All of these alternatives seek to address specific problems of the peer-review process, yet some of them may raise new challenges of their own.
Informal methods of obtaining feedback
Another option is to seek informal feedback on a manuscript, a process that is typically initiated by the author themselves. One possibility is to simply send the manuscript to colleagues and ask for their feedback; another is to upload the manuscript to a preprint server and share the link on a blog, via email, or on social media channels such as Twitter. The advantage of this approach is that the author can quickly get feedback on their manuscript before they submit it. The downside is that the author cannot be sure that their manuscript has been thoroughly reviewed, and the advice and comments they do get may offer little help in improving the manuscript. What is more, this approach is not recognised as a valid form of peer review in the strictest sense of the word.
Rapid peer review or fast-track peer review
The traditional peer review process generally takes weeks or months and often also depends on how quickly enough reviewers can be found and how promptly they write their reports. Some life-science journals have now introduced a fast-track peer-review process for submissions on specific topics, for example those relating to public health. This “accelerated” process draws on a pool of reviewers who have agreed to provide quicker reviews while maintaining the same degree of thoroughness. This fast-track process provides the fastest possible access to scientific findings in the area concerned without compromising on quality assurance. To find out whether a journal offers this service, please consult its website.
How to become a peer reviewer
In order to evaluate other researchers’ manuscripts, you need a certain amount of expertise in the subject area, otherwise the invitation to review should be declined. Editors mostly become aware of potential reviewers through publications or lectures. In addition, major publishers also maintain reviewer databases that contain each person’s specialist subject areas and contact details. Some journals also allow authors to suggest reviewers when they submit their manuscript. This makes the editor’s work easier, and they are not obliged to accept the suggestion.
Carrying out reviews
As already mentioned, the purpose of reviewing is to assess the manuscript for readability, coherence and whether the research has been conducted according to accepted standards. Reviewing also involves checking whether the cited literature is sufficient and relevant as supporting evidence. If the journal requires access to underlying datasets as part of its policy, these too should be included in the review.
Journals often provide a questionnaire, checklist or some other kind of template to structure the review and ensure all relevant aspects are covered. Special questionnaires exist for particular study types, such as AMSTAR‑2 for appraising systematic reviews or the PRESS checklist for reviewing electronic search strategies. Reviewers may also upload additional files with supplementary comments. Some questionnaires also ask reviewers to self‑assess their level of expertise in the subject area.
The peer-review process is not designed to detect scientific misconduct, although if reviewers suspect such misconduct they should inform the editors.Reviewers are expected to remain objective and reflect this in the tone of their reports.
In blind review processes, anonymity is essential and must be ensured by the editors. In double‑blind settings, attempts to uncover the author’s identity through additional searching should be avoided. In single‑blind review, reviewers should not contact authors directly if they have a question; all communication should go through the editors. Likewise, reviewers should not reveal their identity publicly through, for example, social-media activity, and should not share their reviews without permission. Conversely, if authors learn a reviewer’s identity, they should not contact them directly to clarify issues, to suggest collaborations or for any other reasons.
Review deadlines are usually specified in advance. If the reviewer is unable to meet the deadline, they should let the editors know to help them plan accordingly.
Reviews should be conducted personally by the invited reviewer. It is not unusual for senior reviewers to delegate this work to more junior colleagues as a learning opportunity, but this is only beneficial to both sides if proper instructions and supervision are provided, and the review should never be submitted without being checked beforehand. Some journals even require a formal declaration confirming that the invited reviewer authored the report, and request further information if this was not the case.
For early‑career reviewers, many publishers provide guidance on how to write reviews. These resources are often available on the publisher’s website, and additional guidance materials are linked in the navigation panel on the right‑hand side.
Larger publishers typically channel the whole review process through their manuscript submission systems, which also host the relevant forms and checklists.
When revised versions of manuscripts are submitted, reviewers are often asked to assess the changes again, and a willingness to do so is usually confirmed when the initial review is submitted.
How many peer reviews should I write?
There is no standard answer to this question because it largely depends on each individual’s interests, resources and available capacity. As a rule of thumb, you should aim to write as many peer reviews as you have received for your own work. Declining a review for personal or ethical reasons is acceptable, but it is worth weighing carefully whether your input might nonetheless be helpful for the progress of the field.
Acknowledging peer review as a service
As a rule, peer reviewers do not get paid for their work, because peer review is an integral part of the scientific process. Some publishers “reward” their reviewers by granting them free access to the publisher’s archive for a limited period of time or by offering vouchers that reduce publication fees if the reviewer later submits their own manuscript to the journal. A few journals and publishers are even experimenting with paying reviewers directly for their services.
Peer reviewers can also increase the visibility of the services they provide by making use of platforms such as Publons, which give researchers an opportunity to present their contributions to the scientific quality-assurance process. Publons allows researchers to indicate how many manuscripts they have peer-reviewed for each journal. However, the peer-review reports themselves can only be viewed if the journal has installed an open peer-review process and if the authors and reviewers consent to their publication. The platform can also be used to search for qualified reviewers.
Researchers can also refer to their peer-review activities on their CV.
Peer reviews and good research practice
Because peer review is part of the self‑governance of science, it is subject to the general rules of good research practice. This means candidates are expected to disclose any lack of expertise or possible conflicts of interest to the editorial board. It also means that any manipulation of the review process – for instance, through paper mills or biased or discriminatory reviews – counts as scientific misconduct. Hostile or unconstructive comments in reviews are also problematic and should be reported to editors so they can either speak with the reviewers or remove them from the reviewer pool. Authors, likewise, should avoid making such comments in their rebuttal letters.
Manipulating the review process also includes deliberately delaying submission of a review until one’s own similar results have been published, passing off findings from still‑unpublished manuscripts as one’s own, or otherwise exploiting privileged access to those manuscripts for personal advantage.It is also considered unethical to accept a review invitation simply to collect publisher incentives, or to suggest citations purely to inflate one’s own h‑index or a journal’s impact factor. Editors, too, should avoid such citation pushing.
Peer review and generative AI
Peer review is designed to be confidential. With common AI tools, it is impossible to guarantee that reviews or manuscripts fed into them will not enter the pool of AI-training documents – potentially making them indirectly public. Consequently, most journals explicitly forbid the use of generative AI for peer reviews. Moreover, generative AI is often unable to reliably assess all the aspects considered during peer review, so its use should, in any case, be limited to language polishing or summarising one’s own notes.
Details of the extent to which publishers or individual journals permit the use of AI tools can be found in each journal’s peer-review policy or guidelines.
See also
Journal quality and standing: which aspects are relevant to open access? Faster dissemination of research findings – key facts about preprints An introduction to good research practice, research misconduct and academic integrity
Disclaimer
Important note: The information and links provided here do not represent any form of binding legal advice. They are solely intended to provide an initial basis to help get you on the right track. ZB MED – Information Centre for Life Sciences has carefully checked the information included in the list of FAQs. However, we are unable to accept any liability whatsoever for any errors it may contain. Unless indicated otherwise, any statements concerning individual statutory norms or regulations refer to German law (FAQ updated 08/2025).
Contact
Dr. Jasmin Schmitz Head of Publication Advisory Services
Phone: +49 (0)221 999 892 665 Send mail
further information about Dr. Jasmin Schmitz
References
Schmitz, J. (2014). Peer Review. In Technische Informationsbibliothek (TIB) (Hrsg.), CoScience – gemeinsam forschen und publizieren mit dem Netz. Hannover, Technische Informationsbibliothek. (German only) Baxter, L. (2025). How to Peer Review a Systematic Review: A Peer‐Reviewer’s Guide to Reviewing Reviews. The Journal of Clinical Pharmacology. Wilson, J. (2017). PEER REVIEW. The Nuts and Bolts: A Guide for Early Career Researchers. London, Sense about Science. (3. ed.) Anderson, K. (2014). Your Question of the Day – What is „Peer Review“? The Scholarly Kitchen, 24. Juli 2014. (accessed 05/12/2022) Harnad, S. (2014). Crowd-Sourced Peer Review: Substitute or supplement fort he current outdated system? LSE – The Impact of Science Blog, 21. August 2014. (accessed 05/12/2022) Smith, K. L. (2014) Just the Tip of the Iceberg | Peer to Peer Review. Library Journal, 13. März 2014. (accessed 19/12/2022) Wagner, E. (2006). Ethics: What is it for? Nature.com Blogs, 14. Juni 2006. (accessed 05/12/2022) Table 1: Most common types of research misconduct. In Mousavi, T. & Abdollahi, M. (2020). A review of the current concerns about misconduct in medical sciences publications and consequences. DARU Journal of Pharmaceutical Sciences, 28, 359–369. Ross-Hellauer, T. (2017). What is open peer review? A systematic review. F1000Research, 6(588). Else, H. (2025). Publishers trial paying peer reviewers — what did they find? Nature. Gruda, D. (2025). Three AI-powered steps to faster, smarter peer review. Nature.
Related links
Publons F1000Research PubPeer Review Commons
PRESS – Peer Review of Electronic Search Strategies: 2015 Guideline Explanation and Elaboration (PRESS E&E) from January 2016, CADTH
Further Information
Schmitz, J. (2016). Picturing Peer Review [poster presentation]. EA Annual Conference, 27 January 2016, Mainz.
Schmitz, J. & Schroeder, C. (2022). Gefälschte Ergebnisse in Fachjournalen [radio report]. Deutschlandfunk Kultur, 24 March 2022. (German only)
Proctor, D. M., Abraham, R., & Righi, S. E. (2025). From novice to expert: preparing your peer review. mBio, 16(7).
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8902 | https://math.old.naboj.org/problems_new.php?competition=74 | Competition problems 2017 – Náboj
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Competition problems 2017
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The windows on an old tram look like shown in the picture. All curves forming the round corners are arcs of a quarter circle with radius 10 cm. A portion of the sliding window is opened 10 cm as you can see in the second picture. The height of the open section is 13 cm. What is the area of the opening in cm 2?
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2 J
A rectangle is subdivided into nine smaller rectangles as shown in the picture. The number written inside a small rectangle denotes its perimeter. Find the perimeter of the large rectangle.
Show result
42
Show solution Loading…
3 J
Peter deleted one digit from a four-digit prime number and obtained 630. What was the prime number?
Show result
6301
Show solution Loading…
4 J
Star architect Pegi wants to build a very modern pentagonal mansion on a rectangular lot of land having side lengths 35 m and 25 m. The floor area of the mansion fits into the lot as can be seen in the picture (the dots on the boundary mark 5 m distance). What fraction of the area of the lot does the floor area of the mansion cover?
Show result
41 70
Show solution Loading…
5 J
A square grid of 16 dots, as can be seen in the picture, contains the corners of nine 1×1-squares, four 2×2-squares, and one 3×3-square, for a total of 14 squares whose sides are parallel to the sides of the grid. What is the smallest possible number of dots you can remove so that, after removing those dots, each of the 14 squares is missing at least one corner?
Show result
4
Show solution Loading…
6 J
Find the digit at the units place of the sum of squares
1 2+2 2+3 2+⋯+2017 2.
Show result
5
Show solution Loading…
7 J
Express the quotient
0.2˙0.2˙4˙
as a fraction a b in lowest terms with positive integers a and b.
Note: The dot denotes periodic decimal expansion, for example 0.1˙2 3˙=0.123123….
Show result
11 12
Show solution Loading…
8 J
Passau has a railway station in the form of a triangle. Anna, Boris, and Cathy observe the railway traffic in Linz, Regensburg, and Waldkirchen, respectively, on the rails coming from Passau. Anna counts 190, Boris 208 and Cathy 72 incoming and outgoing trains in total. How many trains went from Linz to Regensburg or vice versa if no train starts, ends or reverses its direction in Passau?
Show result
163
Show solution Loading…
9 J
Find all positive integers x<10,000 such that x is a fourth power of some even integer and one can permute the digits of x in order to obtain a fourth power of some odd integer. The result of permuting may not begin with zero.
Show result
256
Show solution Loading…
10 J
In parallelogram A B C D a line through point C meets side A B in point E such that E B=1 5 A E. The line segment C E intersects the diagonal B D in point F. Find the ratio B F:B D.
Show result
1:7
Show solution Loading…
11 J / 11 S
A big house consists of 100 numbered flats. In every flat, there lives one person or there live two or three persons. The total number of people living in the flats No. 1 to No. 52 is 56 and the total number of people living in the flats No. 51 to No. 100 is 150. How many people live in this house?
Show result
200
Show solution Loading…
12 J / 12 S
In the first stage, Nicholas wrote the number 3 with a red pencil and 2 with a green pencil on a sheet of paper. In the following stages, he used the red pencil for writing the sum of the two numbers from the previous stage and the green pencil for their (positive) difference. What number did he write in red in the 2017th stage?
Show result
3⋅2 1008
Show solution Loading…
13 J / 13 S
Little Red Riding Hood finds herself at the entrance of the “Rectangular Forest”. Starting at point A, she has to reach point B as fast as possible. One possibility is to walk along the edges of the woods which will be 140 m in total. Of course, she knows that according to the triangle inequality a direct path would be shorter. Unfortunately, there is only a path which has the form of a zigzag with two right-angled turns as can be seen in the picture. If she knew that this way was shorter than 140 m, she would dare to take the shortcut. Find the length (in meters) of the zigzag through the forest!
Show result
124
Show solution Loading…
14 J / 14 S
Eight-digit palindromes are numbers of the form a b c d d c b a⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ where a, b, c, and d are not necessarily distinct digits. How many eight-digit palindromes have the property that we can delete some digits in such a way that the resulting number would be 2017?
Show result
8
Show solution Loading…
15 J / 15 S
For a positive integer n the sum of its digits is denoted by S(n), and the product of its digits by P(n). How many positive integers n are there having the property n=S(n)+P(n)?
Show result
9
Show solution Loading…
16 J / 16 S
A factory owner hires 100 employees. Each work team leader earns £ 5,000 per month, each labourer £ 1,000 per month and each part-time worker earns £ 50 per month. In total, the factory owner pays £ 100,000 monthly to his employees and there is at least one employee of each type. How many work team leaders are there in his factory?
Show result
19
Show solution Loading…
17 J / 17 S
The mosaic below is composed of regular polygons. The hexagon and the dark grey triangle are inscribed in the same circle. If the area of a striped triangle is 17, determine the area of the dark grey triangle.
Show result
51
Show solution Loading…
18 J / 18 S
In the course of renovating the train station of Passau, special paving for visually handicapped people is being built in. The shape of the paving is shown in the picture below. Unfortunately, only tiles of the size 1×2 are available. In how many ways can the paving be filled with tiles? The tiles are indistinguishable and two pavings are considered to be different if at some spot tiles are in a different position.
Show result
15
Show solution Loading…
19 J / 19 S
Michael picked a positive integer n. Then he chose a (positive) divisor of n, multiplied it by 4, and subtracted this result from n, getting 2017. Find all numbers that Michael could have picked.
Show result
2021, 10085
Show solution Loading…
20 J / 20 S
Grisha and Vanechka are very good friends, so whenever they sit next to each other they begin chatting. Five students (including Grisha and Vanechka) want to have a constructive discussion, so they want to take five chairs around a round table in such a way that Grisha and Vanechka are not neighbours. In how many ways can they do it? Arrangements that differ by rotations are different.
Show result
60
Show solution Loading…
21 J / 21 S
In the figure, A B is a diameter of a circle with center M. The two points D and C are on the circle in such a way that A C⊥D M and ∠M A C=56∘. Find the size of the acute angle between the lines A C and B D in degrees.
Show result
73∘
Show solution Loading…
22 J / 22 S
An object is built from unit squares by successively composing its copies as indicated in the figure below. What is the length of the thick boundary of the object in the stage 6?
Show result
488
Show solution Loading…
23 J / 23 S
All the 2017 seats around a very large round table are occupied by superheroes and villains. The superheroes always tell the truth, whereas the villains always lie. Each person sitting at the table reported that he or she was sitting between a superhero and a villain. For unknown reasons, exactly one superhero made a mistake. How many superheroes are there?
Show result
1345
Show solution Loading…
24 J / 24 S
Find all positive real numbers x such that
x 2017 x=(2017 x)x.
Show result
2017⎯⎯⎯⎯⎯⎯⎯⎯√2016
Show solution Loading…
25 J / 25 S
Leo wants to colour the edges of a regular dodecahedron in a special way: He chooses a felt pen of one colour, starts at one vertex of the dodecahedron and moves along connected edges without lifting his pen and without colouring any edge twice until he wants to or is forced to stop. Then he takes another felt pen and begins colouring some connected uncoloured edges. He continues this procedure using one colour at a time until every edge of the dodecahedron is coloured exactly once. What is the minimum number of colours which can be used?
Note: Regular dodecahedron is a regular solid with twelve pentagonal faces as shown in the picture:
Show result
10
Show solution Loading…
26 J / 26 S
Landscape gardener Joe designed a new gravel path around a lake. The triangle A B C marks the middle of the path, its side lengths are a=80 m, b=100 m and c=120 m. The borders of the path have distance 1 m from this triangle, as can be seen in the following picture. How many m 3 of fine loose gravel must Joe order if the covering with gravel should be of height 4 cm in average?
Show result
24
Show solution Loading…
27 J / 27 S
Find all four-digit squares of integers such that the first two digits are equal and the last two digits are also equal.
Show result
7744
Show solution Loading…
28 J / 28 S
At a fun fair there is a lottery drawing with the following rules: a participant has to choose one of four indistinguishable boxes and afterwards he draws one ball out of the chosen box. If this ball is white, the participant is a winner, if it is black, he loses. For example, if the distribution of the balls in the four boxes is
(6,6),(5,3),(4,0),(3,5),
where each pair (w,b) represents a box with w white and b black balls inside, then a participant wins with a probability of 5 8. Every 1000 th participant receives a super-joker: he or she is allowed to redistribute all the balls among the boxes arbitrarily, putting at least one ball in each box. Then the boxes are mixed again and the participant can choose a box and draw one ball. Johanna is a lucky girl and has won the super-joker. What is the largest probability for a win she can achieve by using an appropriate distribution of the balls given in the example?
Show result
51 58
Show solution Loading…
29 J / 29 S
A bus, a truck, and a motorcycle move at constant speeds and pass a stationary observer in this order at equal time intervals. They pass another observer farther down the road at the same equal time intervals, but in different order: this time the order is bus, motorcycle, truck. Find the speed of the bus in km/h, if the speed of the truck is 60 km/h, and the speed of the motorcycle is 120 km/h.
Show result
80
Show solution Loading…
30 J / 30 S
Find all ways to fill all gaps in the statement below with positive integers in order to make the statement true: “In this statement, 00% of digits is greater than 4, 00% of digits is less than 5, and 00% of digits is equal to either 4 or 5”.
Show result
50, 50, 60
Show solution Loading…
31 J / 31 S
Paul’s cattle grazes on a triangular meadow A B C. Since his spotted cows and those without spots did not go well together, Paul built a twenty-meter long fence perpendicular to the side A C, starting at point P on this side and ending in B, dividing the meadow into two right-angled triangles. However, this was soon met with protests from the spotted cows (grazing in the part with point A), which pointed out that A P:P C=2:7 and demanded a fairer division. Consequently, Paul replaced the fence with a new one, parallel to the old one and with ends on the borders of the meadow, but dividing the meadow into two parts of equal area. What was the length of the new fence in meters?
Show result
30 2 7⎯⎯⎯√
Show solution Loading…
32 J / 32 S
There were five (not necessarily distinct) real numbers written on a blackboard. For every pair of these numbers, Wendy calculated their sum and wrote the ten results
1,2,3,5,5,6,7,8,9,10
on the blackboard, erasing the initial numbers. Determine all possible values of the product of the erased numbers.
Show result
−144
Show solution Loading…
33 J / 33 S
Write 333 as the sum of squares of (arbitrarily many) distinct positive odd integers.
Show result
3 2+5 2+7 2+9 2+13 2
Show solution Loading…
34 J / 34 S
Ellen picked three real numbers a, b, c and defined the operation ⊙ as
x⊙y=a x+b y+c x y.
As an exercise, she computed that 1⊙2=3 and 2⊙3=4. After further investigation, she noticed that there was a non-zero real number u such that z⊙u=z for every real z. What was the value of u?
Show result
4
Show solution Loading…
35 J / 35 S
Three circles of radius r and one circle of radius 1 touch each other and a straight line as shown in the picture. Find r.
Show result
1+5√2
Show solution Loading…
36 J / 36 S
On an old concrete wall, someone had sprayed five (not necessarily distinct) real numbers, the sum of which was 20. For each pair of these numbers, Harry computed their sum and rounded it down (i.e. took the greatest non-exceeding nearest integer), thus obtaining ten integers. Finally, he added together all these new numbers. What was the smallest possible value of the sum that Harry could get?
Show result
72
Show solution Loading…
37 J / 37 S
For a composite positive integer n, let ξ(n) denote the sum of the least three divisors of n, and ϑ(n) the sum of the greatest two divisors of n. Find all composite numbers n such that ϑ(n)=(ξ(n))4.
Note: By a divisor we mean a positive, not necessarily proper divisor.
Show result
864
Show solution Loading…
38 J / 38 S
On a 3×3 board, a mouse sits in the bottom left square. Its goal is to get to a piece of cheese in the top right square using only moves between adjacent squares. In how many ways can we fill some (possibly none) of the free squares with obstacles so that the mouse is still able to reach the cheese?
Show result
51
Show solution Loading…
39 J / 39 S
Among all the pairs (x,y) of real numbers satisfying
x 2 y 2+6 x 2 y+10 x 2+y 2+6 y=42,
let (x 0,y 0) be the one with minimal x 0. Find y 0.
Show result
−3
Show solution Loading…
40 J / 40 S
Let △A B C be a right-angled triangle with right angle at C. Let D and E be points on A B with D between A and E such that C D and C E trisect ∠A C B. If D E:B E=8:15, find A C:B C.
Show result
4 3√11
Show solution Loading…
41 J / 41 S
Mike plays the following game: His task is to find an integer between 1 and N (inclusive). In each turn, he picks an integer from this interval; if it is the correct one, the game ends, otherwise he is told if his choice was too large or too small. However, if the picked number is too large, he has to pay £ 1, and if it is too small, he pays £ 2 (and he pays nothing if his choice is correct). What is the largest integer N such that Mike can always finish the game provided that he can spend at most £ 10?
Show result
232
Show solution Loading…
42 J / 42 S
Positive integers a, b, c satisfy a≥b≥c and
a+b+c+2 a b+2 b c+2 c a+4 a b c=2017.
Find all possible values of a.
Show result
134
Show solution Loading…
43 J / 43 S
An alien spacecraft has the shape of a perfect ball of radius R, supported by three parallel vertical straight legs of length 1 af (alien fathom) and negligible width. The lower ends of the legs form an equilateral triangle of side length 9 af and when the spacecraft rests on a flat surface, the lowest point of the ball precisely touches the surface. What is the radius R (in af)?
Show result
14
Show solution Loading…
44 J / 44 S
Four brothers, Allan, Bert, Charlie, and Daniel, had collected a large amount of hazelnuts in a forest. At night, Allan woke up with the irrestible urge to eat, so he decided to eat some of them. After counting the hazelnuts, he found out that if one hazelnut was removed, the rest could have been divided into four equal parts, so he discarded the extra hazelnut and ate his share (one fourth of the rest); then he returned to bed. Later that night, Bert woke up because of hunger, too, and found out that there was again an extra nut, which he removed and ate one fourth of the remaining nuts. Until morning, exactly the same happened to Charile and Daniel. When all brothers met in the morning, they found out that the number of the remaining nuts was still such that it was divisible by four after removing one nut. What was the minimal number of hazelnuts the brothers could have collected?
Show result
1021
Show solution Loading…
45 J / 45 S
We call a positive integer convenient if all its prime divisors are among 2, 3, and 7. How many convenient numbers are among 1000,1001,…,2000?
Show result
19
Show solution Loading…
46 J / 46 S
Emperor Decimus has banned using the digit 0 (introduced by his predecessor Nullus) and ordered the digit D, representing the quantity 10, to be used instead, thereby founding the Decimus notation. Each positive integer still has a unique representation, e.g.
3 D D 6=3⋅1000+10⋅100+10⋅10+6⋅1=4106.
To smoothen the transfer to the new system, a list of all integers from 1 to D D D (inclusive) has been written down. How many occurrences of the new digit D are there in the list? Multiple occurrences in a single number are counted with multiplicity, so e.g. D D is counted as two D’s.
Show result
321
Show solution Loading…
47 J / 47 S
Square A B C D has inscribed circle ω, which touches the square at points W, X, Y, Z lying on sides A B, B C, C D, D A respectively. Let E be an inner point of the (shorter) arc of ω between W and X and F the intersection of B C and E Y. Provided that E F=5 and E Y=7, determine the area of triangle F Y C.
Show result
21
Show solution Loading…
48 J / 48 S
In the cryptogram
W E⋅L I K E=N A B O J
different letters stand for different digits. Furthermore, we are given S(W E)=11, S(L I K E)=23, and S(N A B O J)=19, where S(n) denotes the sum of digits of a positive integer n. None of the three numbers may start with zero. Find the five-digit number N A B O J.
Show result
60,724
Show solution Loading…
49 J / 49 S
Find all positive integers n with the property that the sum of all non-trivial divisors is 63.
Note: A non-trivial divisor d of n fulfils 1<d<n.
Show result
56, 76, 122
Show solution Loading…
50 J / 50 S
Bob has a cool car with square-shaped rear wheels (the front wheels are the standard round ones). Such a car would usually be rather unpleasant to drive, but Bob has installed very good shock absorbers for the rear wheels so that the car stays in a fixed position parallel to the surface while driving on a flat road. The side length of the rear wheel is 40 cm and its axle is fixed with respect to the car in the horizontal direction. What is the radius of the front wheel (in cm) provided that when the car is moving with constant forward speed, the axle of the rear wheel is exactly half of the time lower and half of the time higher from the surface than the axle of the front wheel?
Show result
10 7⎯⎯√
Show solution Loading…
51 J / 51 S
The City of the Future has the shape of a regular 2017-gon. In vertices of the city there are 2017 metro stations, labelled 1,2,…,2017 counterclockwise on the city map. There are two metro lines: side and diagonal. The side line provides direct connection from station a to b (but not in the opposite direction) if and only if a−b+1 is divisible by 2017 and one such ride lasts 1 minute. The diagonal line provides direct connection from station a to b if and only if 2 b−2 a+1 is divisible by 2017 and one ride lasts 15 minutes. Ferdinand is an avid metro traveller and starting from station 1 he wants to get to the station n with the following property: the shortest possible time needed for getting to this station is the longest among all the stations. Find values n of all possible destinations of Ferdinand.
Show result
1984, 1985
Show solution Loading…
52 J / 52 S
Let f(n) be the number of positive integers that have exactly n digits and whose digits have a sum of 5. Determine, how many of the 2017 integers f(1),f(2),…,f(2017) have the units digit equal to 1.
Show result
202
Show solution Loading…
53 J / 53 S
The picture shows two (indirectly) similar hexangular figures having some sides of length a, b, c, d and A, B, C, D, respectively. If these two figures are put together as in the picture, we get a new hexangular figure being similar to each of the two smaller ones (directly similar to the right-hand one). Find the ratio A:a.
Show result
1+5√2⎯⎯⎯⎯⎯⎯⎯⎯⎯√
Show solution Loading…
54 J / 54 S
Determine all pairs of positive integers (a,b) such that all the roots of both the equations
x 2−a x+a+b−3 x 2−b x+a+b−3=0,=0
are also positive integers.
Show result
(2,2), (6,6), (7,8), (8,7)
Show solution Loading…
55 J / 55 S
Triangle A B C with A B=3, B C=7, and A C=5 is inscribed in circle ω. The bisector of angle ∠B A C meets side B C at D and circle ω at a second point E. Let γ be the circle with diameter D E. Circles ω and γ meet at E and a second point F. Determine the length of A F.
Show result
30 19√
Show solution Loading…
56 J / 56 S
Determine the number of ordered triples (x,y,z), where x, y, z are non-negative integers smaller than 2017 such that
(x+y+z)2−704 x y z
is divisible by 2017.
Show result
2017 2+1=4,068,290
Show solution Loading…
57 J / 57 S
Jane and Thomas play a game with a fair dice, which has two faces painted red, two green, and two blue. They alternately roll the dice until one of them has seen all three colors during his/her turns, this player being declared the winner. With what probability does Jane win the game, provided that she is the first one to roll the dice?
Show result
81/140
Show solution Loading…
58 J / 58 S
Given the equation
k(k+1)(k+3)(k+6)=n(n+1),
find the largest integer n for which there exists an integer solution (k,n).
Show result
104
Show solution Loading…
59 J / 59 S
The Quarter-Pizzeria delivers pizza in special pentagonal boxes, which are suitable both for a quarter of a big pizza and for three quarters of a small pizza (as shown in the picture). What is the radius of a small pizza (in cm) if the radius of a big pizza is 30 cm?
Show result
5(1+7⎯⎯√−2 7⎯⎯√−4⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√)
Show solution Loading…
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8903 | https://www.insideprecisionmedicine.com/topics/oncology/study-finds-a-link-between-lung-cancer-and-air-pollution/ | Study Finds a Link Between Lung Cancer and Air Pollution | Inside Precision Medicine
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Inside Precision MedicineOncologyStudy Finds a Link Between Lung Cancer and Air Pollution
Study Finds a Link Between Lung Cancer and Air Pollution
February 5, 2025
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Credit: Mohammed Haneefa Nizamudeen/Getty Images
Credit: Mohammed Haneefa Nizamudeen/Getty Images
Lung cancer in people who have never smoked is now estimated to be the fifth highest cause of cancer deaths worldwide. Lu ng cancer is still the leading cause of cancer incidence and mortality worldwide, with about 2.5 million people diagnosed with the disease in 2022 alone. But adenocarcinoma is now the dominant subtype particularly in women, a new study finds.
While lung cancer incidence rates among men have generally decreased in most countries during the past 30–40 years, rates among women have continued to rise.
The explanation for this may be air pollution.
This data comes from work done by scientists from the International Agency for Research on Cancer (IARC). They analyzed global variations in lung cancer incidence in 2022 and over time according to histological subtype.
Their studywas published this week in The Lancet Respiratory Medicine. The study focused on four main lung cancer subtypes: adenocarcinoma, squamous cell carcinoma, small-cell carcinoma, and large-cell carcinoma.
Drawing mainly on national Global Cancer Observatory (GLOBOCAN) incidence estimates for 2022 and recorded data included in successive volumes of the Cancer Incidence in Five Continents series (Volumes VII–XII), the study shows that lung adenocarcinoma has emerged as the predominant subtype in recent years, with increasing risks observed among younger generations, particularly females, in most countries assessed.
“This population-based study seeks to better understand variations in lung cancer incidence by place and time according to its constituent subtypes. We examine changes in risk in different countries across successive generations and assess the potential burden of lung adenocarcinoma linked to ambient PM pollution,” said Freddie Bray, head of the Cancer Surveillance Branch at IARC and lead author of the article.
“The results provide important insights as to how both the disease and the underlying risk factors are evolving, offering clues as to how we can optimally prevent lung cancer worldwide.”
In 2022, there were an estimated 2,480,675 new cases of lung cancer worldwide. Of the estimated 1,572,045 new cases among men, 45.6% were adenocarcinoma.
Among women, there were an estimated 908,630 new cases of lung cancer worldwide of which 59.7% were adenocarcinoma.
Geographical variations in 2022 in men, the highest age-standardized incidence rates (ASRs) were in East Asia for adenocarcinoma (27.12 per 100,000 people), in eastern Europe for SCC (21.70 per 100,000 people) and small-cell carcinoma (9.85 per 100,000 people), and in North Africa for large-cell carcinoma (4.34 per 100,000 people).
Globally, an estimated 114,486 adenocarcinoma cases among men and 80,378 among women were attributable to ambient PM pollution, corresponding to ASRs of 2.35 per 100,000 men and 1.46 per 100,000 women.
“Changes in smoking patterns and exposure to air pollution are among the main determinants of the changing risk profile of lung cancer incidence by subtype that we see today,” said Bray. “The diverging trends by sex in recent generations offer insights to cancer prevention specialists and policy-makers seeking to develop and implement tobacco and air pollution control strategies tailored to high-risk populations.”
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8904 | https://en.wikipedia.org/wiki/Neurilemma | Neurilemma - Wikipedia
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Neurilemma
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From Wikipedia, the free encyclopedia
Layer present on Schwann cells of PNS neurons
| Neurilemma |
| Diagram of longitudinal sections of medullated nerve fibers. |
| Cross section of an axon. 1. Axon 2. Nucleus of Schwann cell 3. Schwann cell 4. Myelin sheath 5. Neurilemma |
| Details |
| System | Peripheral nervous system |
| Location | Schwann cell |
| Identifiers |
| MeSH | D009441 |
| TH | H2.00.06.1.00002 |
| Anatomical terms of microanatomy [edit on Wikidata] |
Neurilemma (also known as neurolemma, sheath of Schwann, or Schwann's sheath) is the outermost nucleatedcytoplasmic layer of Schwann cells (also called neurilemmocytes) that surrounds the axon of the neuron. It forms the outermost layer of the nerve fiber in the peripheral nervous system.
The neurilemma is underlain by the myelin sheath (also known as the medullary sheath). In the central nervous system, axons are myelinated by oligodendrocytes, thus lack neurilemma. The myelin sheaths of oligodendrocytes do not have neurilemma because excess cytoplasm is directed centrally toward the oligodendrocyte cell body.
Neurilemma serves a protective function for peripheral nerve fibers. Damaged nerve fibers may regenerate if the cell body is not damaged and the neurilemma remains intact. The neurilemma forms a regeneration tube through which the growing axon re-establishes its original connection.
Neurilemoma is a tumor of the neurilemma.
References
[edit]
^ Jump up to: abAlbert, Daniel (2012). Dorland's illustrated medical dictionary (32nd ed.). Philadelphia, PA: Saunders/Elsevier. pp.1262–1263. ISBN9781416062578.
^Elaine N. Marieb; Katja Hoehn (2007). Human Anatomy & Physiology (7th Ed.). Pearson. pp.394–5. ISBN978-0-8053-5909-1.
External links
[edit]
Histology at dmacc.edu
| hide v t e Nervous tissue |
| CNS | | Tissue Types | Grey matter White matter Projection fibers Association fiber Commissural fiber Lemniscus Nerve tract Decussation Neuropil Meninges | | Cell Types | | Neuronal | Pyramidal Purkinje Granule Von Economo Medium spiny Interneuron | | Glial | Astrocyte Ependymal cells Tanycyte Oligodendrocyte progenitor cell Oligodendrocyte Microglia | | |
| PNS | | General | Dorsal Root Ganglion Ramus Ventral Root Ramus Ramus communicans Gray White Autonomic ganglion (Preganglionic nerve fibers Postganglionic nerve fibers) Nerve fascicle Funiculus | | Connective tissues | Epineurium Perineurium Endoneurium | | Neuroglia | Myelination: Schwann cell Neurilemma Myelin incisure Node of Ranvier Internodal segment Satellite glial cell | |
| Neurons/ nerve fibers | | Parts | | Soma | Axon hillock | | Axon | Telodendron Axon terminals Axoplasm Axolemma Neurofibril/neurofilament | | Dendrite | Nissl body Dendritic spine Apical dendrite/Basal dendrite | | | Types | Bipolar Unipolar Pseudounipolar Multipolar Interneuron Renshaw | | Afferent nerve fiber/ Sensory neuron | GSA GVA SSA SVA fibers Ia or Aα Ib or Golgi or Aα II or Aβ and Aγ III or Aδ or fast pain IV or C or slow pain | | Efferent nerve fiber/ Motor neuron | GSE GVE SVE Upper motor neuron Lower motor neuron α motorneuron β motorneuron γ motorneuron | |
| Termination | | Synapse | Electrical synapse/Gap junction Chemical synapse Synaptic vesicle Active zone Postsynaptic density Autapse Ribbon synapse Neuromuscular junction | | Sensory receptors | Meissner's corpuscle Merkel nerve ending Pacinian corpuscle Ruffini ending Muscle spindle Free nerve ending Nociceptor Olfactory receptor neuron Photoreceptor cell Hair cell Taste receptor | |
| Authority control databases: National | France BnF data |
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8906 | https://www.studocu.com/en-us/document/brown-university/inorganic-chemistry/miessler-fischer-tarr5e-sm-ch-04-cm-final/83185400 | Miessler-Fischer-Tarr5e SM Ch 04 CM Final Exam Notes on Symmetry and Group Theory - Studocu
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Miessler-Fischer-Tarr5e SM Ch 04 CM Final Exam Notes on Symmetry and Group Theory
Solutions to MFT Chapter 4 symmetry questions
Original title:
Miessler-Fischer-Tarr5e SM Ch 04 CM Final
Course
Inorganic Chemistry (CHEM 0500)
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Brown University
Academic year:2023/2024
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C HAPTER 4: SYMMETRY AND G ROUP THEORY
4 a. Ethane in the staggered conformation has 2 C 3 axes (the C–C line), 3 perpendicular C 2 axes bisecting the C–C line, in the plane of the two C’s and the H’s on opposite sides of the two C’s. No h, 3d, i, S 6. D 3 d.
b. Ethane in eclipsed conformation has twoC 3 axes (the C–C line), three perpendicular C 2 axes bisecting the C–C line, in the plane of the two C’s and the H’s on the same side of the two C’s. Mirror planes include h and 3d. D 3 h.
c in the staggered conformation has only one mirror plane, through both C’s, the Cl, and the opposite H on the other C. Cs.
d. 1,2-dichloroethane in the trans conformation has a C 2 axis perpendicular to the C–C bond and perpendicular to the plane of both Cl’s and both C’s, a h plane through both Cl’s and both C’s, and an inversion center. C 2 h.
4 a. Ethylene is a planar molecule, with C 2 axes through the C’s and perpendicular to the C–C bond both in the plane of the molecule and perpendicular to it. It also has a h plane and two d planes (arbitrarily assigned). D 2 h.
b. Chloroethylene is also a planar molecule, with the only symmetry element the mirror plane of the molecule. Cs.
c. 1,1-dichloroethylene has a C 2 axis coincident with the C–C bond, and two mirror planes, one the plane of the molecule and one perpendicular to the plane of the molecule through both C’s. C 2 v. cis-1,2-dichloroethylene has a C 2 axis perpendicular to the C–C bond, and in the plane of the molecule, two mirror planes (one the plane of the molecule and one perpendicular to the plane of the molecule and perpendicular to the C–C bond). C 2 v. trans-1,2-dichloroethylene has a C 2 axis perpendicular to the C–C bond and perpendicular to the plane of the molecule, a mirror plane in the plane of the molecule, and an inversion center. C 2 h.
4 a. Acetylene has a C axis through all four atoms, an infinite number of perpendicular C 2 axes, a h plane, and an infinite number of d planes through all four atoms. Dh.
b. Fluoroacetylene has only theC axis through all four atoms and an infinite number of mirror planes, also through all four atoms. Cv.
c has a C 3 axis through the carbons and three v planes, each including one hydrogen and all three C’s. C 3 v.
d-Chloropropene (assuming a rigid molecule, no rotation around the C–C bond) has no rotation axes and only one mirror plane through Cl and all three C atoms. Cs.
e (again assuming no internal rotation) has a C 2 axis down the long axis of the molecule and two mirror planes, one the plane of the benzene ring and the other perpendicular to it. C 2 v
4 a. Napthalene has three perpendicular C 2 axes, and a horizontal mirror plane (regardless of which C 2 is taken as the principal axis), making it a D 2 h molecule.
b. 1,8-dichloronaphthalene has only oneC 2 axis, the C–C bond joining the two rings, and two mirror planes, making it a C 2 v molecule.
c,5-dichloronaphthalene has one C 2 axis perpendicular to the plane of the molecule, a horizontal mirror plane, and an inversion center; overall, C 2 h.
d,2-dichloronaphthalene has only the mirror plane of the molecule, and is a C s molecule.
4 a. 1,1-dichloroferrocene has a C 2 axis parallel to the rings, perpendicular to the Cl–Fe–Cl h mirror plane. It also has an inversion center. C 2 h.
b. Dibenzenechromium has collinear 6 , C 3 , and C C 2 axes perpendicular to the rings, six perpendicular C 2 axes and a h plane, making it a D 6 h molecule. It also has three v and three d planes, S 3 and S 6 axes, and an inversion center.
c has a mirror plane through the Cr and the biphenyl bridge bond and no other symmetry elements, so it is a C s molecule.
d 3 O+ has the same symmetry as NH 3 : a C 3 axis, and three v planes for a C 3 v molecule.
e 2 F 2 has a C 2 axis perpendicular to the O–O bond and perpendicular to a line connecting the fluorines. With no other symmetry elements, it is a C 2 molecule.
f has a C 2 axis collinear with the C=O bond, a mirror plane including all the atoms, and another perpendicular to the first and including the C and O atoms. C 2 v.
g 8 has C 4 and C 2 axes perpendicular to the average plane of the ring, four C 2 axes through opposite bonds, and four mirror planes perpendicular to the ring, each including two S atoms. D 4 d.
h has a C 3 axis perpendicular to the plane of the ring, three perpendicular C 2 axes, and a horizontal mirror plane. D 3 h.
i(oxalato)chromate(III) has a C 3 axis and three perpendicular C 2 axes, each splitting a C–C bond and passing through the Cr. D 3.
j tennis ball has three perpendicular C 2 axes (one through the narrow portions of each segment, the others through the seams) and two mirror planes including the first rotation axis. D 2 d.
4 a. Cyclohexane in the chair conformation has a C 3 axis perpendicular to the average plane of the ring, three perpendicular C 2 axes between the carbons, and three v planes, each including the C 3 axis and one of the C 2 axes. D 3 d.
b. Tetrachloroallene has three perpendicularC 2 axes, one collinear with the double bonds and the other two at 45° to the Cl—C—Cl planes. It also has two v planes, one defined by each of the Cl–C–Cl groups. Overall, D 2 d. (Note that the ends of tetrachlorallene are staggered.)
j metal washer has a C axis, an infinite number of perpendicular C 2 axes, an infinite number of v mirror planes, and a horizontal mirror plane. Dh.
4 a. D 2 h f. C 3
b 4 h (note the four knobs) g. C 2 h
c h. C 8 v
d 2 v i. Dh
e. C 6 v j. C 3 v
4 a. D 3 h f. Cs (note holes)
b 4 h g. C 1
c s h. C 3 v
d 3 i. Dh
e 2 v j. C 1
4 Hands (of identical twins): C 2 Baseball: D 2 d Atomium: C 3 v
Eiffel Tower: C 4 v Dominoes: 6 × 6: C 2 v 3 × 3: C 2 5 × 4: Cs
Bicycle wheel: The wheel shown has 32 spokes. The point group assignment depends on how the pairs of spokes (attached to both the front and back of the hub) connect with the rim. If the pairs alternate with respect to their side of attachment, the point group is D 8 d. Other arrangements are possible, and different ways in which the spokes cross can affect the point group assignment; observing an actual bicycle wheel is recommended. (If the crooked valve is included, there is no symmetry, and the point group is a much less interesting C 1 .)
4 a. Problem 3 : a. VOCl 3 : C 3 v b. PCl 3 : C 3 v c. SOF 4 : C 2 v
d. SO 3 : D 3 h e. ICl 3 : C 2 v f. SF 6 : Oh
g. IF 7 : D 5 h h. XeO 2 F 4 : D 4 h i. CF 2 Cl 2 : C 2 v
j. P 4 O 6 : Td
Incorrectly cited as problem 4 in first printing of text.
b. Problem 3 : a. PH 3 : C 3 v b. H 2 Se: C 2 v c. SeF 4 : C 2 v
d. PF 5 : D 3 h e. IF 5 : C 4 v f. XeO 3 : C 3 v
g. BF 2 Cl: C 2 v h. SnCl 2 : C 2 v i. KrF 2 : Dh
j. IO 2 F 5 2– : D 5 h
4 a. Figure 3: a. CO 2 : Dh b. 3 : D 3 h SO c. 4 : T d CH
d. PCl 5 : D 3 h e. 6 : Oh SF 7 : Df. 5 h IF
g. TaF 83– : D 4 d
b. Figure 3:a. CO 2 : Dh b. COF 2 : C 2 v c. 2 NO
–: C 2 v
d. SO 3 : D 3 h SNFe. 3 : C 3 v f. SO 2 Cl 2 : C 2 v
g. XeO 3 : C 3 v h– : TSOd 4 i. : C 2 vSOF
j. ClO 2 F 3 : C 2 v k. XeO 3 F 2 : D 3 h l. IOF 5 : C 4 v
4 a. p x has Cv symmetry. (Ignoring the difference in sign between the two lobes, the point
group would be Dh.)
b xy has D 2 h symmetry. (Ignoring the signs, the point group would be D 4 h.)
c x 2 –y 2 has D 2 h symmetry. (Ignoring the signs, the point group would be D 4 h.)
d z 2 has Dh symmetry.
e xyz has T d symmetry.
4 a. The superimposed octahedron and cube show the matching symmetry elements. The descriptions below are for the elements of a cube; each element also applies to the octahedron.
E Every object has an identity operation.
8 C 3 Diagonals through opposite corners of the cube are C 3 axes.
6 C 2 Lines bisecting opposite edges are C 2 axes.
6 C 4 Lines through the centers of opposite faces are C 4 axes. Although there are only three such lines, there are six axes, counting the C 43 operations.
3 C 2 (=C 42 ) The lines through the centers of opposite faces are C 4 axes as well as C 2 axes.
Incorrectly cited as problem 3 in first printing of text.
C 3
C 2 , C 4
4 a. on-deck circle Dh e. home plate C 2 v
b. batter’s boxD 2 h f. baseball D 2 d (see Figure 4)
c C s g. pitcher C 1
d. batCv
4 a. D 2 h d 4 h g 2 h
b 2 v Ce. 5 h Dh. 4 h
c 2 v Cf. 2 v Ci. 2
4 SNF 3
S
N
F F
F
N F 1
F 2
F 3
x
y
(top view)
Symmetry Operations:
N F 1
F 2
F 3
N
F 1
F 2
F 3 N F 1
F 2
F 3
after E after C 3 after v (xz)
Matrix Representations (reducible):
E:
1 0 0
0 1 0
0 0 1
C 3 :
cos
2 3
sin 2
3
0
sin 2
3
cos 2
3
0
0 0 1
– 1
2
–
3
2
0
3
2
– 1
2
0
0 0 1
v(xz):
1 0 0
0 1 0
0 0 1
Characters of Matrix Representations:
3 0 1
(continued on next page)
Block Diagonalized Matrices:
Irreducible Representations:
E 2 C 3 3 v Coordinates Used 2 –1 0 (x, y) 1 1 1 z
Character Table:
C 3 v E 2 C 3 3 v Matching Functions A 1 1 1 1 z x 2 + y 2 , z 2 A 2 1 1 –1 R z E 2 –1 0 (x, y), (R x, R y) (x 2 – y 2 , xy)(xz, yz)
4 a. C 2 h molecules have E, C 2 , i, and h operations.
b. : E C 2 : i: h: 1 0 0 0 1 0 0 0 1
1 0 0 0 1 0 0 0 1
1 0 0 0 1 0 0 0 1
1 0 0 0 1 0 0 0 1
c. These matrices can be block diagonalized into three 1 × 1 matrices, with the representations shown in the table.
(E) (C 2 ) (i) (h) B u 1 –1 –1 1 from the x and y coefficients A u 1 1 –1 –1 from the z coefficients
The total is = 2B u +Au.
d B u and Au: 1 × 1 + (–1) × 1 + (–1) × (–1) + 1 × (–1) = 0, proving they are orthogonal.
C C
Cl
H
H
Cl
d. 1 = 2A 1 + B 1 + B 2 + E:
4 C 3 v
Oh
A 1 : 1/8[1 × 6 + 2 × 1 × 0 + 1 × 2 + 2 × 1 × 2 + 2 × 1 × 2] =
A 2 : 1/8[1 × 6 + 2 × 1 × 0 + 1 × 2 + 2 × (–1) × 2 + 2 × (–1) × 2] =
B 1 : 1/8[1 × 6 + 2 × (–1) × 0 + 1 × 2 + 2 × 1 × 2 + 2 × (–1) × 2] =
B 2 : 1/8[1 × 6 + 2 × (–1) × 0 + 1 × 2 + 2 × (–1) × 2 + 2 × 1 × 2] =
E: 1/8[2 × 6 + 2 × 0 × 0 + (–2) × 2 + 2 × 0 × 2 + 2 × 0 × 2] =
A 1 : 1/8[1 × 6 + 2 × 1 × 4 + 1 × 6 + 2 × 1 × 2 + 2 × 1 × 0] = 2 = 3 A 1 + 2A 2 + B 1 :
A 2 : 1/8[1 × 6 + 2 × 1 × 4 + 1 × 6 + 2 × (–1) × 2 + 2 × (–1) × 0] =
B 1 : 1/8[1 × 6 + 2 × (–1) × 4 + 1 × 6 + 2 × 1 × 2 + 2 × (–1) × 0] =
B 2 : 1/8[1 × 6 + 2 × (–1) × 4 + 1 × 6 + 2 × (–1) × 2 + 2 × 1 × 0] =
E: 1/8[2 × 6 + 2 × 0 × 4 + (–2) × 6 + 2 × 0 × 2 + 2 × 0 × 0] =
A 1 : 1/6[1 × 6 + 2 × 1 × 3 + 3 × 1 × 2] = 1 = 3A 1 + A 2 + E:
A 2 : 1/6[1 × 6 + 2 × 1 × 3 + 3 × (–1) × 2] =
E: 1/6[2 × 6 + 2 × (–1) × 3 + 3 × 0 × 2] =
A 1 : 1/6[1 × 5 + 2 × 1 × (–1) + 3 × 1 × (–1)] = 2 = A 2 + E:
A 2 : 1/6[1 × 5 + 2 × 1 × (–1) + 3 × (–1) × (–1)] =
E: 1/6[2 × 5 + 2 × (–1) × (–1) + 3 × 0 × (–1)] =
A 1 g: 1/48[6 + 0 + 0 + 12 + 6 + 0 + 0 + 0 + 12 + 12] = 3 = A 1 g + Eg + T 1 u:
A 2 g: 1/48[6 + 0 + 0 – 12 + 6 + 0 + 0 + 0 + 12 – 12] =
E g: 1/48[12 + 0 + 0 + 0 + 12 + 0 + 0 + 0 + 24 + 0] =
T 1 g: 1/48[18 + 0 + 0 + 12 – 6 + 0 + 0 + 0 – 12 – 12] =
T 2 g: 1/48[18 + 0 + 0 – 12 – 6 + 0 + 0 + 0 – 12 + 12] =
A 1 u: 1/48[6 + 0 + 0 + 12 + 6 + 0 + 0 + 0 – 12 – 12] =
A 2 u: 1/48[6 + 0 + 0 – 12 + 6 + 0 + 0 + 0 – 12 + 12] =
E u: 1/48[12 + 0 + 0 + 0 + 12 + 0 + 0 + 0 – 24 + 0] =
T 1 u: 1/48[18 + 0 + 0 + 12 – 6 + 0 + 0 + 0 + 12 + 12] =
T 2 u: 1/48[18 + 0 + 0 – 12 – 6 + 0 + 0 + 0 + 12 – 12] =
4 The d xy characters match the characters The d x2-y 2 characters match the characters of the B 2 g representation: of the B 1 g representation:
E
C 4
C 2
C 2
C 2
i
S 4
h
v
d
1 1 1 1 1 1
dxy
x
y
d xy
1 1 1 1 1 1
E
C 4
C 2
C 2
C 2
i
S 4
h
v
d
d x 2 –y 2
x
y
d x 2 –y 2
4 Chiral: 4: O 2 F 2 , [Cr(C 2 O 4 ) 3 ]3– 4: none 4: screw, Wilkins Micawber 4: recycle symbol 4: set of three wind turbine blades, Flying Mercury sculpture, coiled spring
4 a. Point group: C 4 v
C 4 v E 2 C 4 C 2 2 v 2 d 18 2 –2 4 A 1 1 1 1 1 1 z A 2 1 1 1 –1 –1 R z B 1 1 –1 1 1 – B 2 1 –1 1 –1 1 E 2 0 –2 0 0 (x, y), (R x, R y)
b. = 4 A 1 + A 2 + 2B 1 + B 2 + 5E
c. Translation: A 1 + E (match x, y, and z) Rotation: A 2 + E (match R x, R y, and R z) Vibration: all that remain: 3 A 1 + 2B 1 + B 2 + 3E
d. The character for each symmetry operation for the Xe–O stretch is +1. This corresponds to the A 1 irreducible representation, which matches the function z and is therefore IR-active.
Xe F F
F F
O
Xe
O
F F
F F
Omitting the operations that have zeroes in :
n(A 2 u) = 1/16[4 × 1 + 2 × 2 × (–1) + 4 × (–1) + 2 × 2 × 1] = 0 n(E u) = 1/16[4 × 2 + 2 × 2 × 0 + 4 × 2 + 2 × 2 × 0] = 1 (IR active)
Note: In checking for IR-active bands, it is only necessary to check the irreducible representations having the same symmetry as x, y, or z, or a combination of them.
c(CO) 5 has D 3 h symmetry.
The vectors for C–O stretching have the following representation :
D 3 h E 2 C 3 3 C 2 h 2 S 3 3 v 5 2 1 3 0 3 E 2 –1 0 2 –1 0 (x, y) A 2 1 1 –1 –1 –1 1 z
n(E) = 1/12 [ (5 × 2) + (2 × 2 × –1) + (3 × 2)] = 1 n(A 2 ) = 1/12 [(5 × 1) + (2 × 2 × 1) + (3 × 1 × –1) + (3 × –1) + (3 × 3 × 1)] = 1
There are two bands, one matching Eand one matching A 2 . These are the only irreducible representations that match the coordinates x, y, and z.
4 In 4, the symmetries of the CO stretching vibrations of cis-Fe(CO) 4 Cl 2 (C 2 v symmetry) are determined as 2 A 1 + B 1 + B 2. Each of these representations matches Raman-active functions: A 1 (x 2 , y 2 , z 2 ) ; A 2 (xy), B 1 (xz); and B 2 (yz), so all are Raman-active.
In 4, the symmetries of the CO stretching vibrations of trans-Fe(CO) 4 Cl 2 (D 4 h symmetry) are A 1 g + B 1 g + E u. Only A 1 g (x 2 + y 2 , z 2 ) and B 1 g (x 2 – y 2 ) match Raman active functions; this complex exhibits two Raman-active CO stretching vibrations.
In 4, the symmetries of the CO stretching vibrations of Fe(CO) 5 (D 3 h symmetry) are 2 A 1 + E + A 2 . Only A 1 (x 2 + y 2 , z 2 ) and E(x 2 – y 2 , xy) match Raman-active functions; this complex exhibits four Raman-active CO stretching vibrations.
4 a. The point group is C 2 h.
b. Using the Si–I bond vectors as a basis generates the representation:
C 2 h E C 2 i h 4 0 0 0 A g 1 1 1 1 x 2 + y 2 , z 2 B g 1 –1 1 –1 xz, yz A u 1 1 –1 –1 z B u 1 –1 –1 1 x, y
= A g + B g + A u + B u
The A u and Bu vibrations are infrared active.
c A g and Bg vibrations are Raman active.
Fe
C
C
C
C CO
O O
O
O
4 trans isomer (D 4 h): cis isomer (C 2 v):
The simplest approach is to consider if the number of infrared-active I–O stretches is different for these structures. (Alternatively, one could also determine the number of IR-active I–F stretches, a slightly more complicated task.)
trans:
D 4 h E 2 C 4 C 2 C 2 C 2 i 2 S 4 h 2 v 2 d 2 2 2 0 0 0 0 0 2 2 A 1 g 1 1 1 1 1 1 1 1 1 1 A 2 u 1 1 1 –1 –1 –1 –1 –1 1 1 z
There is only a single IR-active I–O stretch (the antisymmetric stretch), A 2 u.
cis: C 2 v E C 2 (xz) (yz) 2 0 2 0 A 1 1 1 1 1 z B 1 1 –1 1 –1 x
There are two IR-active I–O stretches, the A 1 and B 1 (symmetric and antisymmetric). Infrared spectra should therefore be able to distinguish between these isomers. (Reversing the x and y axes would give A 1 + B 2. Because B 2 matches y, it would also represent an IR-active vibration.)
Because these isomers would give different numbers of IR-active absorptions, infrared spectra should be able to distinguish between them. The reference provides detailed IR data.
4 a. One way to deduce the number of Raman-active vibrations of AsP 3 is to first determine the symmetries of all the degrees of freedom. This complex exhibits C 3 symmetry, with the C 3 axis emerging from the As atom. The (E) is 12; the x, y, and z axes of the four atoms do not shift when the identity operation is carried out. Only the As atom contributes to the character of the C 3 transformation matrix; the P atoms shift during rotation about the C 3 axis. The general transformation matrix for rotation about the z axis (Section 4.3) affords 0 as (C 3 ) for The As atom and one P atom do not shift when a v reflection is carried out, and ( v) = 2 (see the v(xz) transformation matrix in Section 4 for the nitrogen atom of NH 3 as a model of how the two unshifted atoms of AsP 3 will contribute to the character of the v transformation matrix).
C 3 v E C 3 v 12 0 2 A 1 1 1 1 z x 2 + y 2 , z 2 A 2 1 1 –1 R x E 2 –1 0 (x, y), (R x , R y) (x 2 – y 2 , xy), (xz, yz)
P P
As
P
As in part a, an alternative approach is to select the set of six bonds as the basis for a representation focused specifically on stretching vibrations. This approach generates the following representation:
This representation reduces to 3 A 1 + A 2 + B 1 + B 2 , the same result as obtained by first considering all degrees of freedom, then subtracting the translational and rotational modes.
c. The issue here is whether or not P 4 (Td) exhibits 4 Raman-active vibrations as does AsP 3. We will first determine the symmetries of all the degrees of freedom. The (E) of the transformation matrix is 12. Only one P 4 atom is fixed upon rotation about each C 3 axis; (C 3 ) = 0. All four atoms are shifted upon application of the C 2 and S 4 axes; (C 2 ) = (S 4 ) = 0. Two atoms do not shift upon reflection through each of the d planes. The contribution to the character of the transformation matrix for each of these unshifted atoms is 1, and (d) = 2.
Reduction of the reducible representation affords A 1 + E + T 1 + 2 T 2. Since the symmetries of the translational modes and rotational modes are T 2 and T 1 , respectively, the symmetries of the vibrational modes are A 1 + E + T 2 , all of these modes are Raman- active so three Raman absorptions are expected for P 4 , and P 4 could potentially be distinguished from AsP 3 solely on the basis of the number of Raman absorptions.
The alternative approach, using the set of six P–P bonds as the basis for a representation focused specifically on bond stretches, generates the following representation:
This representation reduces to A 1 + E + T 2 , the same result as obtained by first considering all degrees of freedom, then subtracting the translational and rotational modes.
C 2 v E C 2 v(xz) v(yz) 6 2 2 2
T d E 8 C 3 3 C 2 6 S 4 6 d 12 0 0 0 2 A 1 1 1 1 1 1 x 2 + y 2 + z 2 A 2 1 1 1 –1 – E 2 –1 2 0 0 (2z 2 – x 2 – y 2 , x 2 – y 2 ) T 1 3 0 –1 1 –1 (R x , R y , Rz ) T 2 3 0 –1 –1 1 (x, y, z) (xy, xz, yz)
T d E 8 C 3 3 C 2 6 S 4 6 d 6 0 2 0 2
P P
P
P
4 The possible isomers are as follows, with the triphenylphosphine ligand in either the axial (A) or equatorial (B) sites.
OC Fe CO
CO
Ph 3 P
CO
OC Fe PPh 3
CO
CO
CO
(A) C3v (B) C2v Note that the triphenylphosphine ligand is approximated as a simple L ligand for the sake of the point group determination. Rotation about the Fe–P bond in solution is expected to render the arrangement of the phenyl rings unimportant in approximating the symmetry of these isomers in solution. The impact of the phenyl rings would likely be manifest in the IR ν(CO) spectra of these isomers in the solid-state.
For A, we consider each CO bond as a vector to deduce the expected number of carbonyl stretching modes. The irreducible representation is as follows:
C 3 v E C 3 v 4 1 2 A 1 1 1 1 z x 2 + y 2 , z 2 A 2 1 1 –1 R x E 2 –1 0 (x, y), (R x, R y) (x 2 – y 2 , xy), (xz, yz)
Reduction of the reducible representation affords 2 A 1 + E. These stretching modes are IR-active and three ν(CO) absorptions are expected for A.
For B, a similar analysis affords the following irreducible representation:
Reduction of the reducible representation affords 2 A 1 + B 1 + B 2. These stretching modes are IR- active, and four ν(CO) absorptions are expected for A.
The reported ν(CO) IR spectrum is consistent with formation of isomer A, with the triphenylphosphine ligand in the axial site.
C 2 v E C 2 v(xz) v(yz) 4 0 2 2 A 1 1 1 1 1 z x 2 , y 2 , z 2 A 2 1 1 –1 –1 R z xy B 1 1 –1 1 –1 x, R y xz B 2 1 –1 –1 1 y, R x yz
For the D 4 h fragment, the reducible representation for the set of vectors associated with the CO bonds is:
Reduction of the reducible representation affords A 1 g + B 1 g + Eu. Only the E modes are IR-active, and a complex with a square planar titanium tetracarbonyl fragment is expected to exhibit a single IR ν(CO) stretching absorption. This D 4 h possibility can therefore be ruled out on the basis of the spectrum.
4 A reasonable product is the C 4 v molecule Mo(CO) 5 (P(OPh) 3 ), with CO replacing a triphenylphosphite ligand.
The reducible representation for the five vectors associated with the CO bonds in this molecule is:
Reduction of this representation affords 2 A 1 + B 1 + E. The A 1 and E modes are IR-active, and three IR ν(CO) stretching absorptions are expected. The reported IR spectrum features three strong ν(CO) absorptions, and one “very weak” absorption attributed to the forbidden B 1 mode in D. J. Darensbourg, T. L. Brown, Inorg. Chem., 1968 , 7 , 959.
4 hasI C 2 symmetry, with a C 2 axis running right to left, perpendicular to the Cl, N, Cl, N and Cl, P, Cl, P faces. II also has C 2 symmetry, with the same C 2 axis as I. (Lower left corner occupied by Cl, not C.) III has only an inversion center and Ci symmetry.
4 An example for each of the five possible point groups:
T d: C
F
F F
F
C 3 v: C
H
F F
F
C 2 v: C
H
F H
F
C s: C
F
Br H
Br
C 1 :
C
F
Cl H
Br
4 a. The S–C–C portion is linear, so the molecule has a C 3 axis along the line of these three atoms, three v planes through these atoms and an F atom on each end, but no other symmetry elements. C 3 v
b. The molecule has only an inversion center, so it isCi. The inversion center is equivalent to an S 2 axis perpendicular to the average plane of the ring.
c 2 Cl 6 Br 4 is Ci.
d complex has a C 3 axis, splitting the three N atoms and the three P atoms (almost as drawn), but no other symmetry elements. C 3
D 4 h E 2 C 4 C 2 2 C 2 2 C 2 i 2 S 4 h 2 v 2 d 4 0 0 2 0 0 0 4 2 0
C 4 v E 2 C 4 C 2 2 v 2 d 5 1 1 3 1
Mo
P
C
C C
C C
(OPh) 3
O O
O
O
O
e. The most likely isomer has the less electronegative Cl atoms in equatorial positions. Point group: C 2 v
4 The structures on the top row are D 2 d (left) and Cs (right). Those on the bottom row are C 2 h (left) and C 4 v (right).
4 a. C 3 v
b 5 h
c. Square structure: D 2 d (bottom ligand on lower left Re should be CO instead of L); Corner structure: Cs
d 3 d
4 Web of Science and SciFinder Scholar should be helpful, but simply searching for these symmetries using a general Internet search should provide examples of these point groups. Some examples:
a. S 6 : Mo 2 (SC 6 H 2 Me 3 ) 6 (M. H. Chisholm, J. F. Corning, and J. C. Huffman, J. Am. Chem. Soc., 1983 , 105 , 5924) Mo 2 (NMe 2 ) 6 (M. H. Chisholm, R. A. Cotton, B. A. Grenz, W. W. Reichert, L. W. Shive, and B. R. Stults, J. Am. Chem. Soc., 1976 , 98 , 4469) [NaFe 6 (OMe) 12 (dbm) 6 ]+ (dbm = dibenzoylmethane, C 6 H 5 COCCOC 6 H 5 ) (F. L. Abbati, A. Cornia, A. C. Fabretti, A. Caneschi, and D. Garreschi, Inorg. Chem., 1998 , 37 , 1430)
b. T Pt(CF 3 ) 4 , C 44
c. I h C 20 , C 80
d. T h [Co(NO 2 ) 6 ]3– , Mo(NMe 2 ) 6
In addition to examples that can be found using Web of Science, SciFinder, and other Internet search tools, numerous examples of these and other point groups can be found in I. Hargittai and M. Hargittai, Symmetry Through the Eyes of a Chemist, as listed in the General References section.
P
F
F
ClCl F
Miessler-Fischer-Tarr5e SM Ch 04 CM Final Exam Notes on Symmetry and Group Theory
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Chapter 4 Symmetry and Gr oup Theory 33
Copyright © 2014 Pearson Education, Inc.
C HAPTER 4:S YMMETRY AND G ROUP T HEORY
4.1 a.Ethane in the staggered conformation has 2 C 3 axes (the C–C line), 3 perpendicular C 2
axes bisecting the C–C line, in the plane of th e two C’s and the H’s on opposite sides of
the two C’s. No h, 3d, i, S 6. D 3 d.
b.Ethane in eclipsed conformation has two C 3 axes (the C–C line), three perpendicular C 2
axes bisecting the C–C line, in the plane of th e two C’s and the H’s on the same side of
the two C’s. Mirror planes include h and 3d. D 3 h.
c.Chloroethane i n the staggered conformation has only one mirror plane, throu gh both C’s,
the Cl, and the opposite H on the other C. C s.
d.1,2-dichloroethane in the trans conformation has a C 2 axis perpendicular to the
C–C bond and perpendi cular to the plane of both Cl’s and b oth C’s, a h plane through
both Cl’s and both C’s, and an inversion center. C 2 h.
4.2 a.Ethylene is a planar molecule, with C 2 axes through the C’s and perpendicular to the C–C
bond both in the plane of the molecule and p erpendicular to it. It also has a h plane and
two d planes (arbitrarily assigned). D 2 h.
b.Chloroethylene is also a planar molecule, with the only symmetry element the mirror
plane of the molecule. C s.
c.1,1-dichloroethylene has a C 2 axis coincident with the C–C bond, and two mirror planes,
one the plane of the molecule and one perpendicular to the plane of t he molecule through
both C’s. C 2 v.
cis-1,2-dichloroethylene has a C 2 axis perpendicu lar to the C–C bond, and in the plane of
the molecule, two mirror planes (one the plane of the molecule and one perpendicular to
the plane of the molecule and p erpendicular to the C–C bond). C 2 v.
trans-1,2-dichloroethylene has a C 2 axis perpendicular to the C–C bond and
perpendicular to the plane of the molecule, a mirror plane in the plane of the molecule,
and an inversion center. C 2 h.
4.3 a.Acetylene has a C axis through all four atoms, an infinite num ber of perpendicular
C 2 axes, a h plane, and an infinite number of d planes through all four atoms. Dh.
b.Fluoroacetylene has only the C axis through all four atoms and an in finite number of
mirror planes, also through all four atoms. Cv.
c. Methylacetylene has a C 3 axis through the carbons and three v planes, each
including one hydrogen and all three C’s. C 3 v.
d. 3-Chloropropene (assumin g a rigid molecule, no rotation around the C–C bond) has
no rotation axes and only one mirror plane through Cl and al l three C atoms. C s.
e. Phenylacetylene (again assu ming no internal rotation) has a C 2 axis down the long ax is
of the molecule and two mirror planes, one the plane of the benzene ring and the other
perpendicular to it. C 2 v
34 Chapter 4 Symm etry and Group Theory
Copyright © 2014 Pearson Education, Inc.
4.4 a.Napthalene has three perpendicular C 2 axes, and a horizontal mirror plane (regar dless
of which C 2 is taken as the principal axis), m aking it a D 2 h molecule.
b.1,8-dichloronaphthalene has only one C 2 axis, the C–C bond joinin g the two rings, and
two mirror planes, making it a C 2 v molecule.
c. 1,5-dichloronaphthalene has on e C 2 axis perpendicular to the plane of the molecule, a
horizontal mirror plane, and an inversion center; overall, C 2 h.
d. 1,2-dichloronaphthalene has only th e mirror plane of the molecule, and is a C s molecule.
4.5 a. 1,1-dichloroferrocene has a C 2 axis paralle l to the rings, perpendicular to the Cl–Fe–Cl
h mirror plane. It also has an inversion center. C 2 h.
b.Dibenzenechromium has collinear C 6, C 3, and C 2 axes perpendicular to the rings, six
perpendicular C 2 axes and a h plane, making it a D 6 h molecule. It also has three v
and three d planes, S 3 and S 6 axes, and an inversion center.
c. Benzenebiphenylchromium has a mirror plane through the Cr and the biphenyl bridge
bond and no other symmetry elements, so it is a C s molecule.
d. H 3 O+ has the same symmetry as NH 3: a C 3 axis, and three v planes for a C 3 v molecule.
e. O 2 F 2 has a C 2 axis perpendicular to the O–O bond an d perpendicul ar to a line connecting
the fluorines. With no other symmetry elements, it is a C 2 m olecule.
f. Formaldehyde has a C 2 axis collinear with the C=O bond, a mirror plane inclu ding all the
atoms, and another perpen dicular to th e first and including the C and O atoms. C 2 v.
g. S 8 has C 4 and C 2 axes perpendicular to the aver age plane of the ring, four C 2 axes through
opposite bonds, and four mirror plane s perp endicular to the ring, each including two S
atoms. D 4 d.
h. Borazine has a C 3 axis perpendicular to the plane of the ring, three perpendicular C 2 axes,
and a horizontal mirror plane. D 3 h.
i. Tris(oxalato)chromate(III) has a C 3 axis and three perpendicular C 2 axes, each splitting
a C–C bond and passing through the Cr. D 3.
j. A tennis ball has three perpendicular C 2 axes (one through the narrow portions of each
segment, the others through the seams) and two mirror planes including the fir st rotation
axis. D 2 d.
4.6 a.Cyclohexane in the chair conformation has a C 3 axis perpendicular to the average plane
of the ring, three perpendicular C 2 axes between the carbons, and three v planes, each
including the C 3 axis and one of the C 2 axes. D 3 d.
b.Tetrachloroallene has three perpendicular C 2 axes, one collinear with the double bonds
and the other two at 45° to the Cl—C—Cl pl anes. It also has two v planes, one de fined
by each of the Cl–C–Cl groups. Overall, D 2 d. (Note that the ends of tetrachlorallene are
staggered.)
Chapter 4 Symmetry and Gr oup Theory 35
Copyright © 2014 Pearson Education, Inc.
c. The sulfate ion is tetrahedral. T d.
d. Most snowflakes have hexagonal sy mmetry (Figure 4.2), and have coll inear C 6, C 3, and
C 2 axes, six perpendicular C 2 axes, and a horizontal mirror plane. Overall, D 6 h. (For high
quality images of snowflakes, inc luding some that have different shapes, see K. G.
Libbrecht, Snowflakes, Voyageur Press, Minneapolis, MN, 2008.)
e. Diborane has three perpendicular C 2 axes and three perpendicular mirror planes. D 2 h.
f. 1,3,5-tribromobenzene has a C 3 axis perpendicular to the plane of the ring, three
perpendicular C 2 axes, and a horizontal mirror plane. D 3 h.
1,2,3-tribromobenzene has a C 2 axis through the middle Br and two pe rpendicular
mirror planes that include this axis. C 2 v
1,2,4-tribromobenzene has only the plane of the ring as a mirror plane. C s.
g. A tetrahedron inscribed in a cube has T d symmetry (see Figure 4.6).
h. The left and right ends of B 3 H 8 are staggered with respect to each other. There is a
C 2 axis through the borons. In addition, there are two planes of symmetry, each
containing four H atoms, and two C 2 axes between these planes and perpendicular to
the original C 2. The point group is D 2 d.
i. A mountain swallowtail butterfly has only a mirro r that cuts through the head, thorax, and
abdomen. C s
j. The Golden Gate Bridge has a C 2 axis and two perpendicular mirror planes that include
this axis. C 2 v
4.7 a.A sheet of typing paper has three perpendicular C 2 axes and three perpendicular mirror
planes. D 2 h.
b. An Erlenmeyer flask has an infinite-fold rotation axis and an infinite num ber of v
planes, Cv.
c. A screw has no symmetry operations other than the identity, for a C 1 classification.
d. The number 96 (with the correct type font) has a C 2 axis perpendicular to the plane
of the paper, making it C 2 h.
e. Your choice—the list is too long to attempt to answer it here.
f. A pair of eyeglasses has only a vertical mirror plane. C s.
g. A five-pointed star has a C 5 axis, five perpendicular C 2 axes, one horizontal and f ive
vertical mirror planes. D 5 h.
h. A fork has only a mirror plane. C s.
i. Wilkins Micawber has no symmetry ope ration other than the i dentity. C 1.
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36 Chapter 4 Symm etry and Group Theory
Copyright © 2014 Pearson Education, Inc.
j. A metal washer has a Caxis, an infinite number of perpendicular C 2 axes, an infinite
number of v mirror planes, and a horizontal mirror plane. Dh.
4.8 a.D 2 h f. C 3
b. D 4 h(note the four knobs) g.C 2 h
c.C s h.C 8 v
d.C 2 v i.Dh
e.C 6 v j.C 3 v
4.9 a.D 3 h f. C s (note holes)
b.D 4 h g.C 1
c. C s h.C 3 v
d.C 3 i.Dh
e.C 2 v j.C 1
4.10 Hands (of identical twins): C 2 Baseball: D 2 d Atomium: C 3 v
Eiffel Tower: C 4 v Dominoes: 6× 6: C 2 v 3× 3: C 2 5× 4: C s
Bicycle wheel: The wheel shown has 32 spokes. The point group assignment depends on how
the pairs of spokes (attached to both the front and back of the hub) connect with the rim. If the
pairs alternate with respect to their side of attachment, the point group is D 8 d. Other arrangements
are possible, and different ways in which the spokes cross can affect the point group assignment;
observing an actual bicycle wheel is recommended. (If the crooked valve is included, there is no
symmetry, and the point group is a much less interesting C 1.)
4.11 a. Problem 3.41: a. VOCl 3: C 3 v b. PCl 3: C 3 v c. SOF 4: C 2 v
d. SO 3: D 3 h e. ICl 3: C 2 v f. SF 6: O h
g. IF 7: D 5 h h. XeO 2 F 4: D 4 h i. CF 2 Cl 2: C 2 v
j. P 4 O 6: T d
Incorrectly cited a s problem 4.30 in first printing of text.
Chapter 4 Symmetry and Gr oup Theory 37
Copyright © 2014 Pearson Education, Inc.
b. Problem 3.42: a. PH 3: C 3 v b. H 2 Se: C 2 v c. SeF 4: C 2 v
d. PF 5: D 3 h e. IF 5: C 4 v f. XeO 3: C 3 v
g. BF 2 Cl: C 2 v h. SnCl 2: C 2 v i. KrF 2: Dh
j. IO 2 F 5
2–: D 5 h
4.12 a. Figure 3.8: a. CO 2: Dh b. SO 3: D 3 h c. CH 4: T d
d.PCl 5: D 3 h e. SF 6: O h f. IF 7:D 5 h
g.TaF 8
3–: D 4 d
b. Figure 3.15: a. CO 2: Dh b. COF 2: C 2 v c. NO 2
–: C 2 v
d.SO 3: D 3 h e. SNF 3: C 3 v f. SO 2 Cl 2: C 2 v
g.XeO 3: C 3 v h. SO 4
2–: T d i. SOF 4: C 2 v
j.ClO 2 F 3: C 2 v k. XeO 3 F 2: D 3 h l. IOF 5: C 4 v
4.13 a. p x has Cv symmetry. (Ignoring the difference in sign betw een the two lobes, the point
group would be Dh.)
b.d xy has D 2 h symmetry. (Ignoring the signs, t he point group would be D 4 h.)
c. d x 2–y 2 has D 2 h symmetry. (Ignoring the signs, the point group would be D 4 h.)
d. d z 2 has Dh symmetry.
e. f xyz has T d symmetry.
4.14 a.The superimposed octahedron and cube
show the matching symmetry elements.
The descriptions below are for the
elements of a cube; each element also
applies to the octahedron.
E Every object has an identity
operation.
8 C 3 Diagonals through opposite
corners of the cube are C 3 axes.
6 C 2 Lines bisecting opposite edges are C 2 axes.
6 C 4 Lines through the centers of opposite faces are C 4 axes. Although there are only
three such lines, there are six axes, counting the C 4
3 operations.
3 C 2 (=C 4
2) The lines through the centers of opposite faces are C 4 axes as well as
C 2 axes.
Incorrectly cited as problem 3.41 in first printing of text.
C
3
C
2
,
C
4
38 Chapter 4 Symm etry and Group Theory
Copyright © 2014 Pearson Education, Inc.
i The center of the cube is the inversion center.
6 S 4 The C 4 axes are also S 4 axes.
8 S 6 The C 3 axes are also S 6 axes.
3h These mirror planes are parallel to the faces of the cube.
6d These mirror planes are through two opposite edges.
b.O h
c. O
4.15 a.There are three possible orientations of the two blue faces.
If the blue faces are opposite each other, a C 3 axis connects the centers of the
blue faces. This axis has 3 perpendicular C 2 axes, and contains three vertical
mirror places (D 3 d).
If the blue faces share one vertex of the octahedron, a C 2 axis includes this
vertex, and this axis includes two vertical mirror planes (C 2 v).
The third possibility is for the blue faces to share an edge of the octahedron. In
this case, a C 2 axis bisects this shared edge, and includes two vertical mirror
planes (C 2 v).
b. There are three unique orientations of the three blue faces.
If one blue face is arranged to form edges with each of the two remaining blue
faces, the only symmetry operations ar e identity and a single mirror plane (C s).
If the three blue faces are arranged such that a single blue face shares an edge
with one blue face, but only a vertex w ith the other blue face, the only symmetry
operation is a mirror plane that pass es through the center of the blue faces, an d
the point group is C s.
If the three blue faces each share an edge with the sa me yellow face, a C 3 axis
emerges from the center of this yellow face, and this axis in cludes three vertical
mirror planes (C 3 v).
c. If there are four different colors, and each pair of opposite faces has the identical color,
the only symmetry operations are identity and inversion (C i).
4.16 Four point groups are represented by the symbols of the chemical elements. Most
symbols have a single mirror in the plane of the symbol (C s), for example, Cs! Two
symbols have D 2 h symmetry (H, I), and two more (, S) have C 2 h. Seven exhibit C 2 v
symmetry (B, C, K, V, Y, W, U). In some case s, the choice of font may affect the point
group. For example, the sy mbol for nitrogen may have C 2 h in a sans serif font () but
otherwise C s (N). The symbol of oxygen has D∞h symmetry if shown as a circle but D 2 h
if oval.
Chapter 4 Symmetry and Gr oup Theory 39
Copyright © 2014 Pearson Education, Inc.
4.17 a. on-deck circle Dh e. home plate C 2 v
b. batter’s box D 2 h f. baseball D 2 d (see Figure 4.1)
c. cap C s g. pitcher C 1
d. bat Cv
4.18 a. D 2 h d. D 4 h g. D 2 h
b. C 2 v e. C 5 h h. D 4 h
c. C 2 v f. C 2 v i. C 2
4.19 SNF 3
S
N
F
F
F
N F
1
F
2
F
3
x
y
(t op v iew)
Symmetry Operations:
N F
1
F
2
F
3
N
F
1
F
2
F
3
N F
1
F
2
F
3
after E after C 3 after v(xz)
Matrix Representations (reducible):
E:
1 0 0
0 1 0
0 0 1
C 3:
cos
2
3
sin 2
3 0
sin 2
3 cos 2
3 0
0 0 1
–1
2–3
2 0
3
2–1
2 0
0 0 1
v(xz):
1 0 0
01 0
0 0 1
Characters of Matrix Representations:
3 0 1
(continued o n next page)
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8907 | https://peacemaker.un.org/sites/default/files/document/files/2022/07/sgguidancenote-approachrol2008.pdf | UNITED NATIONS NATIONS UNIES APRIL 2008 GUIDANCE NOTE OF THE SECRETARY-GENERAL UN Approach to Rule of Law Assistance U N I T E D N A T I O N S N A T I O N S U N I E S GUIDANCE NOTE OF THE SECRETARY-GENERAL UN Approach to Rule of Law Assistance 1 SUMMARY This note provides the guiding principles and framework for UN rule of law activities at the national level that apply in all circumstances, including in crisis, post-crisis, conflict-prevention, conflict, post-conflict and development contexts. In order to ensure a comprehensive and coherent UN approach, the note is derived from United Nations norms, standards and guidance, and the framework outlines the fundamental constituent elements of rule of law efforts. For the United Nations system, the rule of law is a principle of governance in which all persons, institutions and entities, public and private, including the State itself, are accountable to laws that are publicly promulgated, equally enforced and independently adjudicated, and which are consistent with international human rights norms and standards. It requires as well measures to ensure adherence to the principles of supremacy of the law, equality before the law, accountability to the law, fairness in the application of the law, separation of powers, participation in decision-making, legal certainty, avoidance of arbitrariness, and procedural and legal transparency. Justice is an ideal of accountability and fairness in the protection and vindication of rights and the prevention and punishment of wrongs. Its administration involves both formal judicial and informal/customary/traditional mechanisms. Within this paradigm, a range of terms are used to describe various approaches and activities that strengthen the rule of law, such as security sector reform, administration of justice, protection, combating impunity. A. Guiding Principles 1. Base assistance on international norms and standards 2. Take account of the political context 3. Base assistance on the unique country context 4. Advance human rights and gender justice 5. Ensure national ownership 6. Support national reform constituencies 7. Ensure a coherent and comprehensive strategic approach 8. Engage in effective coordination and partnerships B. Framework for Strengthening the Rule of Law 1. A constitution or equivalent 2. A legal framework and the implementation thereof 3. An electoral system 4. Institutions of justice, governance, security and human rights 5. Transitional justice processes and mechanisms 6. A public and civil society that contributes to strengthening the rule of law and holds public officials and institutions accountable 2 INTRODUCTION As set forth in the Charter of the United Nations, the three pillars of the Organization’s mission are to maintain international peace and security; to achieve economic and social progress, and development; and to encourage respect for human rights and fundamental freedoms. The promotion of the rule of law is fundamental to this mission. This note provides the guiding principles and framework for UN rule of law activities at the national level that apply in all circumstances, including in crisis, post-crisis, conflict-prevention, conflict, post-conflict and development contexts. UN rule of law assistance is based normatively on the Charter, international law, and the host of UN treaties, declarations, guidelines and bodies of principles developed in furtherance of national societies and an international order based on the rule of law. All human rights, the rule of law and democracy are interlinked and mutually reinforcing and they belong to the universal and indivisible core values and principles of the United Nations. Strengthening the rule of law at the international level involves respect for the Charter and international law, including the use of force, respect for the sovereign equality of States, and recognition of the responsibility of States to protect their populations from genocide, crimes against humanity, ethnic cleansing and war crimes. It requires participation in, and effective national implementation of, the main bodies of international law at both the national and international levels. At the national level, the work of the UN on rule of law is based operationally on technical assistance and capacity-building carried out for the benefit of Member States, at their request and/or as mandated by the Security Council, and in accordance with their national policies, priorities and plans. This enables the UN to respond to the needs of countries in a flexible manner, eschewing one-size-fits-all formulas and the importation of foreign models, and instead, to base our support on national assessments, local needs and aspirations, and broad participation. A. GUIDING PRINCIPLES 1. Base assistance on international norms and standards The normative foundation for UN rule of law work is the Charter, together with four of the main pillars of the international legal system: international human rights law, international humanitarian law, international criminal law and international refugee law. The countless UN treaties, declarations, guidelines and bodies of principles represent universally applicable standards. As such, they incorporate a legitimacy that cannot be said to attach to exported national models reflecting the values or experience of donors and assistance providers. These standards form the normative parameters for UN engagement, for example: the UN will neither establish nor directly participate in any tribunal that allows for capital punishment nor endorse peace agreements that allow for amnesties for genocide, war crimes, crimes against humanity or gross violations of human rights. All UN approaches to rule of law issues should take their guidance from, and be developed in conformity with, the applicable international standards. 2. Take account of the political context The international community has sometimes underestimated the extent of political will necessary to support effective rule of law development and invested inadequately in political dialogue on rule of law promotion. Rule of law activities take place in neither an economic nor a political 3 void, and require changes in the legal framework and the institutional structures of governance and their functioning. Rule of law development, like all national reforms, generates winners and losers. They are therefore political questions as well as technical ones. Rule of law assistance has often overemphasized technical dimensions and paid less attention to political and strategic considerations. Until national stakeholders see the utility in supporting rule of law development, technical assistance will have little impact. Senior UN representatives in the field need to understand the political nature of strengthening the rule of law, and dedicate attention to supporting both the political and institutional aspects of rule of law development. In cooperation with Headquarters and in partnership with the national political leadership and other stakeholders, UN leadership at the field level is responsible for fostering political space for reform and insulating the rule of law from inappropriate political influence or abuse. 3. Base assistance on the unique country context Effective and sustainable approaches to strengthening the rule of law must begin with a thorough analysis of national needs and capacities, and the mobilization of national expertise to the greatest extent possible. The UN must consider carefully the particular rule of law and justice needs in the specific context of each country, including the condition and nature of the country’s rule of law system (both formal and informal/customary/traditional) and the culture, traditions and institutions that underlie that system, including the role of gender in society, the position of minority groups, and the situation of children. In addition, rule of law assistance must be consistent with the international legal obligations applicable to the country. The UN must assess carefully the many factors giving rise to the need for rule of law assistance, such as the nature and causes of any recent conflict or any history of human rights abuse, and construct its assistance accordingly. The UN needs to promote this approach among all stakeholders in strengthening the rule of law. 4. Advance human rights and gender justice The UN faces many challenges in developing effective responses to sensitive political, cultural and operational contexts in its support for rule of law development. In providing assistance, the UN must not overlook the entitlements that have been established under international law for women, children, minorities, refugees and displaced persons, and other groups that may be subjected to marginalization and discrimination in the country. Gender-based discrimination permeates all cultures, and dedicated attention to gender equality issues is needed in all dimensions of rule of law work. The UN maintains a responsibility to help establish the rule of law for all on a basis of equality, and with due attention to the rights and specific vulnerabilities of children. Solutions to rule of law challenges that serve to advance the rights of dominant social groups while leaving others behind must be avoided. 5. Ensure national ownership No rule of law programme can be successful in the long term if imposed from the outside. Process leadership and decision-making must be in the hands of national stakeholders. Rule of law development requires the full and meaningful participation and support of national stakeholders, inter alia, government officials, justice and other rule of law officials, national legal professionals, traditional leaders, women, children, minorities, refugees and displaced persons, other marginalized groups and civil society. Experience indicates that the rule of law is strengthened if reform efforts are focused on assisting the State to apply its international legal obligations, and are credible and adhere to the principles of inclusion, participation and 4 transparency, facilitating increased legitimacy and national ownership. Meaningful ownership requires the legal empowerment of all segments of society. 6. Support national reform constituencies UN programmes must identify, support and empower national reform constituencies. The UN must facilitate the processes through which various national stakeholders debate and outline the elements of their country’s plan to strengthen the rule of law and secure sustainable justice. The aim is to help national stakeholders to develop their own vision, agenda, approaches to reform and programmes. Public consultation and public understanding of and support for reform are essential. The UN must encourage outreach to all groups in society, and support public awareness and education campaigns and public consultation initiatives. Civil society organizations, women’s groups, national legal associations, human rights groups and advocates of victims and of prisoners, as well as those who might otherwise be excluded (e.g., non-criminal former regime members and ex-combatants) must all be given a voice in these processes. Children and adolescents must also be enabled to participate and their role as constructive agents of change be acknowledged. 7. Ensure a coherent and comprehensive strategic approach Rule of law promotion is more than the provision of technical legal expertise. UN rule of law assistance is most effective if it draws on a wide-range of expertise and perspectives, inter alia, in the political, legal, human rights, development and social science fields within and outside the UN system. A comprehensive approach that supports all aspects of effective and efficient justice systems, and their management and oversight is likewise necessary. The UN must develop a holistic and strategic approach, within the context of existing planning processes, that involves: 1) conducting joint and thorough assessments with the full and meaningful participation of national stakeholders to determine rule of law needs and challenges; 2) supporting the development of a comprehensive rule of law strategy based on the results of the assessment; 3) developing a joint UN rule of law programme guided by the strategy; and 4) assigning accountabilities and implementation responsibilities. 8. Engage in effective coordination and partnerships Strengthening the rule of law encompasses a multitude of activities carried out by many entities across the UN system and the wider international community. Past engagement has sometimes been piece-meal, incongruous and donor-driven, resulting in uneven and contradictory development of rule of law institutions and short-term, superficial gains at the cost of longer-term, sustainable reform. Successful rule of law assistance requires the support and active engagement of all stakeholders working through a comprehensive strategy in a coordinated fashion. All UN entities involved in rule of law assistance must operate according to shared approaches and strategies, including effective coordination, and must recognize that the success of UN engagement is linked to the extent of its partnerships and support for national ownership. B. FRAMEWORK FOR STRENGTHENING THE RULE OF LAW 1. A Constitution or equivalent, which, as the highest law of the land, inter alia: • Incorporates internationally recognized human rights and fundamental freedoms as set out in international treaties, provides for their applicability in domestic law, and establishes effective and justiciable remedies at law for violations; 5 • Provides for non-discrimination on the basis of race, color, gender, language, religion, political or other opinion, national or social origin, property, birth or other status, and which protects national minorities; • Provides for the equality of men and women; • Defines and limits the powers of government and its various branches, vis-à-vis each other, and the people; • Limits emergency powers and derogations of human rights and freedoms under states of emergency to those permissible under international standards; • Empowers an independent and impartial judiciary. 2. A legal framework, and the implementation thereof, consistent with international norms and standards, which protects human rights and provides for effective redress, including: • Fair immigration, nationality and asylum laws; • Penal laws, including for transnational crimes, and criminal procedure laws that ensure the effective and fair administration of justice for perpetrators, including juveniles in conflict with the law as well as victims and witnesses, consistent with, among others, the Basic Principles of Justice for Victims of Crime and Abuse of Power; • Prison laws and regulations that are consistent with, among others, the Standard Minimum Rules for the Treatment of Prisoners; • Laws for the protection of minorities, children, displaced and returning populations, and other marginalized or vulnerable groups that take into account their special status and international standards for their protection, and that outlaw and address the effects of discrimination; • Laws that establish legal protection for the rights of women on an equal basis with men, and that ensure through competent national tribunals and other public institutions the effective protection of women against any act of discrimination; • Laws protecting free association and assembly, and guarantees that press, libel, broadcasting and other laws respect free expression, opinion and information; • Security legislation that protects non-derogable human rights, and ensures civilian control and oversight; • Laws on the judiciary, legal practice and prosecution that reflect, among others, the standards embodied in the Basic Principles on the Independence of the Judiciary, Basic Principles on the Role of Lawyers, and Guidelines on the Role of Prosecutors; • Laws, guidelines and directives that govern the conduct of police and other security forces consistent with, among others, the Code of Conduct for Law Enforcement Officials and Basic Principles on the Use of Force and Firearms by Law Enforcement Officials; • Fair procedures for the settlement of civil entitlements and disputes under the law and fair administration of laws, regulations, procedures and institutions. 3. An electoral system, which, inter alia: • Assures, through periodic and genuine elections, that the will of the people shall be the basis of the authority of government; 6 • Assures the right of everyone to take part in the government of his or her country, either directly or through freely chosen representatives, including through the application of temporary special measures; • Assures equal access to public service, including elective public service; • Guarantees universal and equal suffrage, and secrecy of the ballot; • Provides for non-discrimination in the area of political rights, and secures an electoral atmosphere that is free of intimidation and respectful of certain prerequisite rights, such as freedom of opinion, expression, information, assembly and association; • Provides for objective, unbiased and independent electoral administration, and independent review of alleged irregularities; • Provides for the transfer of power to victorious parties and candidates under the law. 4. Institutions of justice, governance, security and human rights that are well-structured and financed, trained and equipped to make, promulgate, enforce and adjudicate the law in a manner that ensures the equal enjoyment of all human rights for all, including: • A legislative institution or mechanism for the formulation and public promulgation of laws in a procedurally transparent manner; • Effective oversight institutions or mechanisms (e.g., anti-corruption bodies, parliamentary committees, national human rights institutions, independent commissions on human rights and ombudsman offices consistent with the Paris Principles); • A judiciary, which is independent, impartial and adequately empowered to adjudicate the law with integrity and ensure its equal application to all within its jurisdiction; • State institutional capacities to make policy for and manage the effective administration of justice, the provision of security, crime prevention, and to investigate and prosecute violations of the law; • Police and other law enforcement agencies that protect individuals and communities, enforce the law without discrimination and take appropriate action against alleged violations of the law, including appropriate oversight mechanisms; • Corrections services that provide for a safe, secure and humane prison and rehabilitation system, including alternatives to deprivation of liberty and diversion measures; • An accessible capacity to provide legal and paralegal assistance to those unable to afford it, and adequate and effective defense for those alleged to have violated the law; • A social service capacity to assist victims and witnesses of crime and abuse of power, including children, to participate effectively in the administration of justice in a manner that ensures redress for harm suffered; • A system to effectively adjudicate rights and responsibilities within the family, on the basis of gender equality and in the best interest of the child, which ensures that the protection of children from abuse, exploitation, harm and neglect; • A professional training regime for lawyers, judges, prosecutors, law enforcement and prison officials that promotes a culture of service, discipline and ethics; • Military and civil defense forces that has allegiance to the Constitution, or equivalent, and other laws of the land, and to the democratic government, and follows international humanitarian law; 7 • Effective and accessible mechanisms for resolution of entitlements and disputes between and among individuals, State organs, and groups in society, including courts, administrative tribunals, alternative or traditional dispute resolution mechanisms, and commissions or mechanisms for, among others, the fair settlement of property and housing disputes. 5. Transitional justice processes and mechanisms that respond to country contexts while anchored in international norms and standards to address the legacy of large-scale past abuses in order to ensure accountability, serve justice and achieve reconciliation, which may include both judicial and non-judicial mechanisms such as ad hoc criminal tribunals, truth commissions, vetting processes and reparations programmes. 6. A public and civil society that contributes to strengthening the rule of law and holds public officials and institutions accountable, including: • A system of governance that promotes a culture of legality, legal empowerment and ensures the public is aware of and educated in the full-range of its rights and responsibilities; • Communities that have equal access to justice and are empowered to participate in resolving disputes peacefully and responding to community safety needs and concerns; • Full access to judicial and other mechanisms for independent oversight of the exercise of executive authority and abuse of power; • A strong civil society, including, inter alia, adequately trained, equipped, financed and organized non-governmental organizations and professional associations, women’s groups, labor unions and community organizations; • A free, responsible and flourishing mass media. |
8908 | https://www.chegg.com/homework-help/questions-and-answers/derive-homogeneous-transformation-matrix-used-reflect-3d-points-plane-equation-ax-cz-d-sol-q85028324 | Your solution’s ready to go!
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Question: Derive a homogeneous transformation matrix that can be used to reflect 3D points about a plane with equation ax + by + cz = d. Your solution should not include the explicit computation of any Euler rotation matrices Rx, Ry, and Rz.
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To derive the homogeneous transformation matrix for reflecting 3D points about a plane given by , identify and set up the form of the point and its reflection defined by .
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8909 | https://www.chegg.com/homework-help/questions-and-answers/calculate-marginal-product-capital-following-production-functions-25-25-5-points-f-l-k-l-2-q64673880 | Solved Calculate the marginal product of capital for each | Chegg.com
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Calculate the marginal product of capital for each of the following production functions. (2.5+2.5=5 Points) (a) f(L, K) = L^2/3 K^1/3 (b) f(L,K) = 4L^1/2 + K (ii) For the same two production functions from part (i), calculate Marginal Rate of Technical Substitution. Verify in each case whether or not |MRTS| is decreasing in magnitude when labor increases.
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Question: Calculate the marginal product of capital for each of the following production functions. (2.5+2.5=5 Points) (a) f(L, K) = L^2/3 K^1/3 (b) f(L,K) = 4L^1/2 + K (ii) For the same two production functions from part (i), calculate Marginal Rate of Technical Substitution. Verify in each case whether or not |MRTS| is decreasing in magnitude when labor increases.
Calculate the marginal product of capital for each of the following production functions. (2.5+2.5=5 Points)
(a) f(L, K) = L^2/3 K^1/3
(b) f(L,K) = 4L^1/2 + K
(ii) For the same two production functions from part (i), calculate Marginal Rate of Technical Substitution. Verify in each case whether or not |MRTS| is decreasing in magnitude when labor increases. (10+5=15 Points)
Here’s the best way to solve it.Solution 100%(1 rating) Share Share Share done loading Copy link i) f(L,K) = L2/3K1/3 f/L = MPL = (2/3)L2/3 - 1K1/3 MPL = (2/3)L- 1/3K1/3 f/K = MPK = (1/3)L2/…
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8910 | http://ld2016.scusa.lsu.edu/cppreference/en/c/language/if.html | if statement - cppreference.com
if statement
From cppreference.com
<c | language
C
Language
headers
Type support
Dynamic memory management
Error handling
Program utilities
Variadic function support
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{ statement... }
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Conditionally executes code.
Used where code needs to be executed only if some condition is true.
Contents
1 Syntax
2 Explanation
3 Notes
4 Keywords
5 Example
6 References
7 See Also
[edit]Syntax
if (expression)statement_true(1)
if (expression)statement_trueelsestatement_false(2)
[edit]Explanation
expression must be an expression of any scalar type.
If expression compares not equal to the integer zero, statement_true is executed.
In the form (2), if expression compares equal to the integer zero, statement_false is executed.
As with all other selection and iteration statements, the entire if-statement has its own block scope:
enum {a, b};
int different(void)
{
if (sizeof(enum {b, a}) != sizeof(int))
return a; // a == 1
return b; // b == 0 in C89, b == 1 in C99
}(since C99)
[edit]Notes
The else is always associated with the closest preceding if (in other words, if statement_true is also an if statement, then that inner if statement must contain an else part as well):
int j = 1;
if (i > 1)
if(j > 2)
printf("%d > 1 and%d > 2\n", i, j);
else // this else is part of if(j>2), not part of if(i>1)
printf("%d > 1 and%d <= 2\n", i, j);
If statement_true is entered through a goto, statement_false is not executed.
[edit]Keywords
if, else
[edit]Example
Run this code
include
int main(void)
{
int i = 2;
if (i > 2) {
printf("first is true\n");
} else {
printf("first is false\n");
}
i = 3;
if (i == 3) printf("i == 3\n");
if (i != 3) printf("i!= 3 is true\n");
else printf("i!= 3 is false\n");
}
Output:
first is false
i == 3
i != 3 is false
[edit]References
C11 standard (ISO/IEC 9899:2011):
6.8.4.1 The if statement (p: 148-149)
C99 standard (ISO/IEC 9899:1999):
6.8.4.1 The if statement (p: 133-134)
C89/C90 standard (ISO/IEC 9899:1990):
3.6.4.1 The if statement
[edit]See Also
C++ documentation for if statement
Retrieved from " |
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The Inverse Matrix is Unique
Problem 251
Let A A be an n×n n×n invertible matrix. Prove that the inverse matrix of A A is uniques.
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Contents [hide]
Problem 251
Hint.
Proof.
Hint.
That the inverse matrix of A A is unique means that there is only one inverse matrix of A A.
(That’s why we say “the” inverse matrix of A A and denote it by A−1 A−1.)
So to prove the uniqueness, suppose that you have two inverse matrices B B and C C and show that in fact B=C B=C.
Recall that B B is the inverse matrix if it satisfies
A B=B A=I,
A B=B A=I,
where I I is the identity matrix.
Proof.
Suppose that there are two inverse matrices B B and C C of the matrix A A. Then they satisfy
A B=B A=I
A B=B A=I()
and
A C=C A=I.
A C=C A=I.()
To show that the uniqueness of the inverse matrix, we show that B=C B=C as follows. Let I I be the n×n n×n identity matrix.
We have
B=B I=B(A C)by ()=(B A)C by the associativity=I C by ()=C.
B=B I=B(A C)=(B A)C=I C=C.by ()by the associativity by ()
Thus, we must have B=C B=C, and there is only one inverse matrix of A A.Click here if solved 67
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8912 | https://math.stackexchange.com/questions/716318/integrating-int-02-pi-fracd-theta3-sin-theta-cos-theta | integration - Integrating $ \int_{0}^{2\pi}\frac{d\theta}{3 + \sin\theta + \cos\theta}$ - Mathematics Stack Exchange
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Integrating \int_{0}^{2\pi}\frac{d\theta}{3 + \sin\theta + \cos\theta}
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\begingroup
I want to do the following integral in my complex analysis class:
\int_{0}^{2\pi}\frac{d\theta}{3 + \sin\theta + \cos\theta}
I don't have the solution (in the textbook) but I checked via WolframAlpha, and they have a solution of \frac{2\pi}{\sqrt 7} but I seem to be getting \frac{2 \pi (1+i)}{\sqrt 7}. I won't lay out everything I am doing since that would take me a long time, but basically I set z = e^{i\theta}, do all the substitutions, and at some point get: \int_{|z| = 1} \frac{2 \, dz}{6zi + z^2 - 1 + iz^2 + i} I solve for the roots of the denominator, get: z_1 = \frac{(1+i)(-3 + \sqrt 7)}{2}, z_2 = \frac{(1+i)(-3 - \sqrt 7)}{2}
I realize that z_1 falls in the circle I am integrating over, while z_2 extends pass the same circle. Then I have the following: \int_{|z| = 1}\frac{2 \, dz}{(z-z_1)(z-z_2)} I use Cauchy's formula to integrate and get the solution: \frac{4 \pi i}{z_1 - z_2} Then doing the algebra I get the answer I said in the beginning, I am thinking I am doing something wrong with my algebra or maybe there is something more systematically wrong with my method, anyway I want to know how to do this question. Any help would be appreciated!
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edited Mar 18, 2014 at 3:28
Michael Hardy
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asked Mar 18, 2014 at 2:27
InsigMathInsigMath
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\endgroup
7
2 \begingroup For one thing, your denominator does not factor as (z-z_1)(z-z_2), since the leading coefficient is (1+i) rather than 1. That explains the extra factor of (1+i). As for the \sqrt{7}...\endgroup Braindead –Braindead 2014-03-18 02:35:04 +00:00 Commented Mar 18, 2014 at 2:35
\begingroup Oh I am sorry I get \frac{2 \pi (1+i)}{\sqrt 7}\endgroup InsigMath –InsigMath 2014-03-18 02:36:14 +00:00 Commented Mar 18, 2014 at 2:36
\begingroup Darn I keep making dump little typos, yeah there should be a \pi\endgroup InsigMath –InsigMath 2014-03-18 02:39:11 +00:00 Commented Mar 18, 2014 at 2:39
\begingroup Are you okay now? Or do you need some further clarification?\endgroup Braindead –Braindead 2014-03-18 02:40:21 +00:00 Commented Mar 18, 2014 at 2:40
\begingroup Umm not really... I get what you mean, but then wouldn't my equation then be (z - (1+i)b)(z - (1+i)c)? How would the (1+i) factor out so nicely?\endgroup InsigMath –InsigMath 2014-03-18 02:42:16 +00:00 Commented Mar 18, 2014 at 2:42
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\begingroup
\begin{equation} \int_{0}^{2\pi} \frac{d\theta}{3 + \sin(\theta) + \cos(\theta)} \end{equation}\begin{equation} \sin(\theta) + \cos(\theta) = \sqrt{2}\sin(\theta + \frac{\pi}{4}) \end{equation}\begin{equation} \int_{0}^{2\pi} \frac{d\theta}{3 + \sqrt{2}\sin(\theta + \frac{\pi}{4})} \end{equation}\begin{equation} \frac{2\sqrt{7}}{7}\arctan\left(\frac{\sqrt{7}}{7} \tan\left(\frac{u}{2}\right)\right) + C \end{equation}\begin{equation} \frac{2\sqrt{7}\arctan\left(\frac{\sqrt{7}}{7}\right)}{7} - \frac{2\sqrt{7}(-\pi + \arctan\left(\frac{\sqrt{7}}{7}\right))}{7} \end{equation}\begin{equation} \frac{2\sqrt{7}\pi}{7} \end{equation}
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answered Mar 30, 2024 at 7:49
SaradamaniSaradamani
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\begingroup
If you have a second degree polynomial with a leading coefficient a, then it factors as
a(z-z_1)(z-z_2)
The reason is:
The expression has zeros at z_1 and z_2.
The leading term is going to be az^2. Without the a, it would be just z^2. This is something you should be able to see by looking at the product of the highest degree terms in each factor.
If you are still feeling skeptical, simply multiply out:
(1+i)(z - \dfrac{(1+i)(-3+\sqrt{7})}{2})(z - \dfrac{(1+i)(-3-\sqrt{7})}{2})
And see what you get.
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answered Mar 18, 2014 at 2:56
BraindeadBraindead
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\endgroup
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\begingroup I get my original complex polynomial, at least that's what I get when I let wolfram alpha do the dirty work :P ... thanks that helps a lot, this would solve my problem!\endgroup InsigMath –InsigMath 2014-03-18 03:04:30 +00:00 Commented Mar 18, 2014 at 3:04
\begingroup Basically, if you have (z-z_1)(z-z_2), you would get z^2 + \dots . So if your original polynomial is az^2 + \dots, you need to have a(z-z_1)(z-z_2).\endgroup Braindead –Braindead 2014-03-18 15:21:02 +00:00 Commented Mar 18, 2014 at 15:21
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8913 | https://www.tiger-algebra.com/drill/a~2-3a-4=0/ | Copyright Ⓒ 2013-2025
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Solution - Quadratic equations
Other Ways to Solve
Step by Step Solution
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring a2-3a-4
The first term is, a2 its coefficient is 1 .
The middle term is, -3a its coefficient is -3 .
The last term, "the constant", is -4
Step-1 : Multiply the coefficient of the first term by the constant 1 • -4 = -4
Step-2 : Find two factors of -4 whose sum equals the coefficient of the middle term, which is -3 .
| | | | | | | |
--- --- ---
| | -4 | + | 1 | = | -3 | That's it |
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -4 and 1
a2 - 4a + 1a - 4
Step-4 : Add up the first 2 terms, pulling out like factors :
a • (a-4)
Add up the last 2 terms, pulling out common factors :
1 • (a-4)
Step-5 : Add up the four terms of step 4 :
(a+1) • (a-4)
Which is the desired factorization
Equation at the end of step 1 :
Step 2 :
Theory - Roots of a product :
2.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
2.2 Solve : a+1 = 0
Subtract 1 from both sides of the equation :
a = -1
Solving a Single Variable Equation :
2.3 Solve : a-4 = 0
Add 4 to both sides of the equation :
a = 4
Supplement : Solving Quadratic Equation Directly
Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula
Parabola, Finding the Vertex :
3.1 Find the Vertex of y = a2-3a-4
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Aa2+Ba+C,the a -coordinate of the vertex is given by -B/(2A) . In our case the a coordinate is 1.5000
Plugging into the parabola formula 1.5000 for a we can calculate the y -coordinate :
y = 1.0 1.50 1.50 - 3.0 1.50 - 4.0
or y = -6.250
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = a2-3a-4
Axis of Symmetry (dashed) {a}={ 1.50}
Vertex at {a,y} = { 1.50,-6.25}
a -Intercepts (Roots) :
Root 1 at {a,y} = {-1.00, 0.00}
Root 2 at {a,y} = { 4.00, 0.00}
Solve Quadratic Equation by Completing The Square
3.2 Solving a2-3a-4 = 0 by Completing The Square .
Add 4 to both side of the equation :
a2-3a = 4
Now the clever bit: Take the coefficient of a , which is 3 , divide by two, giving 3/2 , and finally square it giving 9/4
Add 9/4 to both sides of the equation :
On the right hand side we have :
4 + 9/4 or, (4/1)+(9/4)
The common denominator of the two fractions is 4 Adding (16/4)+(9/4) gives 25/4
So adding to both sides we finally get :
a2-3a+(9/4) = 25/4
Adding 9/4 has completed the left hand side into a perfect square :
a2-3a+(9/4) =
(a-(3/2)) • (a-(3/2)) =
(a-(3/2))2
Things which are equal to the same thing are also equal to one another. Since
a2-3a+(9/4) = 25/4 and
a2-3a+(9/4) = (a-(3/2))2
then, according to the law of transitivity,
(a-(3/2))2 = 25/4
We'll refer to this Equation as Eq. #3.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(a-(3/2))2 is
(a-(3/2))2/2 =
(a-(3/2))1 =
a-(3/2)
Now, applying the Square Root Principle to Eq. #3.2.1 we get:
a-(3/2) = √ 25/4
Add 3/2 to both sides to obtain:
a = 3/2 + √ 25/4
Since a square root has two values, one positive and the other negative
a2 - 3a - 4 = 0
has two solutions:
a = 3/2 + √ 25/4
or
a = 3/2 - √ 25/4
Note that √ 25/4 can be written as
√ 25 / √ 4 which is 5 / 2
Solve Quadratic Equation using the Quadratic Formula
3.3 Solving a2-3a-4 = 0 by the Quadratic Formula .
According to the Quadratic Formula, a , the solution for Aa2+Ba+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
a = ————————
2A
In our case, A = 1
B = -3
C = -4
Accordingly, B2 - 4AC =
9 - (-16) =
25
Applying the quadratic formula :
3 ± √ 25
a = —————
2
Can √ 25 be simplified ?
Yes! The prime factorization of 25 is
5•5
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
√ 25 = √ 5•5 =
± 5 • √ 1 =
± 5
So now we are looking at:
a = ( 3 ± 5) / 2
Two real solutions:
a =(3+√25)/2=(3+5)/2= 4.000
or:
a =(3-√25)/2=(3-5)/2= -1.000
Two solutions were found :
How did we do?
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8914 | https://www.youtube.com/watch?v=yE4o1dLzck4 | Art of Problem Solving: Least Common Multiple Slick Trick
Art of Problem Solving
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Description
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Posted: 23 Dec 2011
Art of Problem Solving's Richard Rusczyk (we think) explains a slick method for finding the least common multiple of two numbers. Visit www.artofproblemsolving.com to learn more.
Transcript:
அதை இங்குள்ள பையனிடம் சொல்லாதே, சரியா? அவரிடம் சொல்லாதீர்கள், ஏனென்றால் நான் மீச்சிறு பொது மடங்கின் தந்திரங்களை உங்களுக்காக நான் தருகிறேன். நீங்கள் இதை விரும்பப் போகிறீர்கள். என்ன அது? எனக்கு தெரியாது. மற்றவர், எங்கே இருக்கிறார் என்று தெரியவில்லை. அவர் எப்படியும் எங்காவது குதித்துக்கொண்டும் கைதட்டிக்கொண்டும் இருப்பார் ? அவர் எங்கே இருக்கிறார் என்று யாருக்குத் தெரியும்? அவர் இங்கே என்ன செய்வார் என்று நமக்குத் தெரியும், இல்லையா? பகா காரணிப்படுத்துதல். அந்த தந்திரம் ஏமாற்றுபவர்களுக்கானது. சரி, நான் உங்களிடம் பொய் சொல்லக்கூடாது. இது ஒரு அழகான தந்திரம், ஆனால் அதை இங்கே முயற்சிக்கவும். இந்த பகா காரணிப்படுத்துதல்களைக் கண்டுபிடிக்க நீங்கள் நாள் முழுவதும் இங்கே இருப்பீர்கள். இதைக் கவனித்துக் கொள்ள மிகவும் அருமையான ஒன்றை நான் உங்களுக்குக் காட்டப் போகிறேன். அதைச் செய்ய, நாம் ஒரு எளிய கணக்குடன் தொடங்கப் போகிறோம். என்ன? ஓ, மற்றவரும் எளிமையான கணக்கினை செய்கின்றாரா? ஆ, அவர் என்னுடைய எல்லா நல்ல பொருட்களையும் எடுத்துக்கொள்கிறார். எல்லாம் சரி. இது என்னுடையது. என்னுடையது. புரிந்ததா? இதோ போகிறோம். ஒரு ஜோடி எளிய எண்களின் மீச்சிறு பொது மடங்கைக் கண்டறிவதன் மூலம்,
2 பெருக்கல் 3 மற்றும் 2 வர்க்கம் பெருக்கல் 5 இல் தொடங்குவோம். நாம் தொடங்கப் போகிறோம், பார்ப்போம், 2 பெருக்கல் 3 மற்றும் 2 வர்க்கம் பெருக்கல் 5 உடன் தொடங்குவோம். இவற்றில் மீச்சிறு பொது மடங்கை எவ்வாறு கண்டுபிடிப்பது என்பது நமக்குத் தெரியும். சரியா? இந்த பகா காரணிப்படுத்துதலில் நாம் 2 இன் மிக உயர்ந்த அடுக்கை எடுத்துக்கொள்கிறோம், மற்றும் நாம் 3 இன் மிக உயர்ந்த அடுக்கை எடுத்துக்கொள்கிறோம், அது 3. மற்றும் 5 இன் மிக உயர்ந்த அடுக்கு, இது 5 ஆகும். ஆமாம், ஆமாம், பகா காரணிப்படுத்தும் விஷயம் இன்னும் அழகாக இருக்கிறது பயனுள்ளதாக இருக்கும், ஆனால் நாம் அதைப் பயன்படுத்தி சமமான ஸ்னீக்கியர் வழியை உருவாக்கி மீச்சிறு பொது மடங்கைக் கண்டறிய போகிறோம் . சரி, சரி, நாம்மிடம் இத்து கீழேவுள்ளது, இல்லையா? இப்போது இந்த எண்கள் ஒவ்வொன்றையும் 2 ஆல் பெருக்கினால் என்ன ஆகும்? நமது முதல் எண் 2ன் வர்க்கம் பெருக்கல் 3 ஆகிறது. மேலும் நமது இரண்டாவது எண் 2ன் கனம் பெருக்கல் 5 ஆகிறது. இப்போது நமது அதிகபட்ச அடுக்கு 2 ன் கனமாக உள்ளது. மேலும் நம்மிடம் 3 உள்ளது மற்றும் நம்மிடம் 5 உள்ளது. எனவே ஒவ்வொரு எண்ணையும் 2 ஆல் பெருக்கினோம், அது மீச்சிறு பொது மடங்கை 2 ஆல் பெருக்குவதாகும். ஒருவேளை நமக்கு அதிர்ஷ்டம் கிடைத்திருக்கலாம். இதை மீண்டும் முயற்சிப்போம், சரியா? இந்த எண்கள் ஒவ்வொன்றையும் மீண்டும் 2 ஆல் பெருக்கப் போகிறோம். எனவே நமது முதல் எண் 2 கனம் பெருக்கல் 3 ஆகிறது. எங்கள் இரண்டாவது எண் 2 அடுக்கு 4 பெருக்கல் 5 ஆக மாறும். இப்போது நம்மிடம் 2ன் மிக உயர்ந்த அடுக்காக 2ன் அடுக்கு 4 உள்ளது. மேலும் நம்மிடம் 3 உள்ளது மற்றும் நம்மிடம் 5 உள்ளது. இந்த எண்கள் ஒவ்வொன்றையும் 2 ஆல் பெருக்கினால், அது மீச்சிறு பொது மடங்கை 2 ஆல் பெருக்குவதாகும். ஒருவேளை இது 2 இல் ஏதோ விசேஷமாக இருக்கலாம். வேறொன்றால் பெருக்க முயற்சிப்போம். நான் 3 ஆல் பெருக்கப் போகிறேன். இந்த எண்கள் ஒவ்வொன்றையும் 3 ஆல் பெருக்கவும். முதல் எண் 2 கனம் பெருக்கல் 3 வர்க்கமாக மாறும். இரண்டாவது எண் 2ன் அடுக்கு 4 பெருக்கல் 3 பெருக்கல் 5 ஆகும். இப்போது நான் இவற்றில் மீச்சிறு பொது மடங்கை கண்டால், 2 இன் மிக உயர்ந்த அடுக்கு இன்னும் 2ன் அடுக்கு 4 ஆகவே உள்ளது. 3 இன் மிக உயர்ந்த அடுக்கு இப்போது 3 ன் வர்க்கமாக உள்ளது. நிச்சயமாக, மேலும் என்னிடம் 5 உள்ளது. எனவே இந்த எண்கள் ஒவ்வொன்றையும் 3 ஆல் பெருக்கினேன், அது மீச்சிறு பொது மடங்கை 3 ஆல் பெருக்கியது. எனவே இப்போது நான் எந்த வழியில் என்ன செய்தேன் என்று பாருங்கள். இங்கிருந்து இங்கிருந்து, ஒவ்வொரு எண்ணையும் இரண்டு 2கள் மற்றும் ஒரு 3 ஆல் பெருக்கினேன். இதன் விளைவாக, மீச்சிறு பொது மடங்குகளுக்கு என்ன நடந்தது என்று பாருங்கள். அது இரண்டு 2கள் மற்றும் ஒரு 3 ஆல் பெருக்கப்பட்டது. அதே வழியில், நான் ஏதேனும் இரண்டு முழு எண்களுடன் தொடங்கினால், அவை a மற்றும் b, மற்றும் அவை இரண்டையும் வேறு சில முழு எண்களால் பெருக்குகிறேன், அது n, பின்னர் நான் இந்த இரண்டு புதிய எண்களின் மீச்சிறு பொது மடங்கை கண்டுபிடிக்க, n மடங்கு மீச்சிறு பொது மடங்கு பெருக்கல் நாம் தொடங்கிய எண்களின் மீச்சிறு பொது மடங்காக இருக்கும். எனவே எப்பொழுதும் நான் இரண்டின் மீச்சிறு பொது மடங்கை எடுத்துக்கொள்கிறேன், இங்கே n உள்ளது போல அவை ஒரு பொதுவான காரணியைக் கொண்டிருந்தால், அந்த பொதுவான காரணியை நான் காரணியாக்கி வெளியே எடுக்கமுடியும். நான் n ஐ வெளியே எடுக்கமுடியும், அதனால் n பெருக்கல் a மற்றும் n பெருக்கல் b இன் மீச்சிறு பொது மடங்கு எனபது n மடங்கு a மற்றும் b இன் மீச்சிறு பொது மடங்காகும். அந்த மற்ற கணக்குகளையும் இதை வைத்தே பார்க்கலாம், அது எவ்வளவு அருமையாக இருக்கிறது என்று நீங்கள் பார்க்கலாம். எனவே இந்த இரண்டு பெரிய எண்களை இங்கே பெற்றுள்ளோம், இல்லையா? அவை வெளிப்படையாக ஒரு பொதுவான காரணியைப் பெற்றுள்ளன. அவை இரண்டும் இரட்டை எண்கள், இரண்டும் 2 ஆல் வகுபடும். எனவே நான் இங்கே என்ன செய்யப் போகிறேன் என்பதை நான் 2 ஐ காரணியாக்கி வெளியே எடுக்கப் போகிறேன். எனவே முதலில், இவற்றை 2 மடங்காக எழுதப் போகிறோம். எனவே இது 2 இன் LCM மட்டுமே, பார்க்கலாம், இதை 2 ஆல் வகுக்க, 2 9 க்குள் போகிறது, பார்ப்போம், அது 4, 1 ஐ எடுத்து, அது 16, 4, 8, 4, 8, 4, மற்றும் நான் செய்வது 2 ஐ அந்த பெரிய எண்ணை வகுத்ததுதான். அதையே இங்கும் செய்யப் போகிறேன். நான் 2 ஐ வெளியே எடுக்கப் போகிறேன், அது எனக்கு 36, 36, 3 ஐ கொடுக்கிறது, 3, 6, 3, 6, 3. இப்போது நான் இதன் பொதுவான காரணியாக 2 ஐப் பெற்றுள்ளேன், என்னால் அந்த மீச்சிறு பொது மடங்கை வெளியே எடுக்க முடியும். இது 4, 8, 4, 8, 4, இன் மீச்சிறு பொது மடங்கின் 2 மடங்கு ஆகும் 3, 6, 3, 6, 3. இப்போது இந்த இரண்டு எண்களைப் பாருங்கள், அவை நிச்சயமாக ஒரே மாதிரியாக அமைந்துள்ளன. அதாவது, இது வெறும் 4 முறை 1, 2, 1, 2, 1. இது 3 முறை 1, 2, 1, 2, 1 ஆகும். இது சொல்லுவதற்கு வேடிக்கையாக இருக்கிறது. என்னுடன் சொல்லுங்கள். 1, 2, 1, 2, 1. ஆமாம், வேடிக்கையாக இருக்கிறது. எனவே, இதைப் பாருங்கள். இவை மிகவும் வெளிப்படையான பொதுவான காரணி, 1, 2, 1, 2, 1. இது 4 பெருக்கல் 1, 2, 1, 2, 1, இது 3 பெருக்கல் 1, 2, 1,2, 1. இப்போது 1, 2, 1, 2, 1 என்ற பொதுவான காரணியைப் பெற்றுள்ளோம். எனவே நான் 1, 2, 1, 2, 1 ஐ வெளியே எடுக்கப் போகிறேன், என்னால் வெளியே எடுக்க முடியும். அந்த பொதுவான காரணி, எனக்கு 2 பெருக்கல் 1, 2, 1, 2, 1 முறை மீச்சிறு பொது மடங்கு வெறும் 4 மற்றும் 3. மற்றும், நிச்சயமாக, 4 மற்றும் 3 இன் குமீச்சிறு பொது மடங்கு, அது வெறும் 12 தான். எனவே எனக்கு 2 முறை 1, 2, 1, 2, 1 பெருக்கல் 12 கிடைத்துள்ளது. இப்போது என்னால் இதை பெருக்க முடியும். 2 முறை 1, 2, 1, 2, 1, நிச்சயமாக, 2, 4, 2, 4, 2 முறை . இந்த 12ஐ வெறும் 10 பிளஸ் 2 என்று எழுதப் போகிறேன் இதைக் கணக்கிடுவது மிகவும் எளிதானது. இதை இங்கேயே செய்வோம், 2, 4, 2, 4, 2 முறை 10, நாங்கள் 0-ஐ சேர்க்கவும், பின்னர் 2 ஆல் பெருக்கினால், அது 4, 8, 4, 8,4, கொஞ்சம் பங்கீட்டுப் பண்பை உங்கள் வழியில் எறிந்து விடுங்கள், அவற்றைக் கூட்டுகிறோம் மேலும், 4, 0, 9, 0, 9,2 ஆகியவற்றைக் கூட்டுகிறோம். நமது பதில், 290,904, நாம் முடித்துவிட்டோம்! என்ன அது? மற்றொறுவனும் அதையே பயன்படுத்துகிறான். சரி, சரி, இதோ போகிறோம். நாங்கள் இங்கே செய்து கொண்டிருப்பது எல்லாம் இரகசிய காரணிகளை வெளியே எடுத்து மீச்சிறு பொது மடங்குடன் பயன்படுத்தும் தந்திரம். இதோ இன்னொரு ரகசியம். மீப்பெரு பொது வகுஎண் மூலம் இதைப் பாருங்கள், இது அங்கு கூட வேலை செய்கிறது. ஆனால் நீங்கள் அதை என்னிடம் கேட்கவில்லை. |
8915 | https://web.ma.utexas.edu/users/m408s/m408d/CurrentWeb/LM10-4-3.php | | | | | |
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Volumes as Double Integrals Iterated Integrals over RectanglesHow To Compute Iterated Integrals Examples of Iterated Integrals Fubini's Theorem Summary and an Important Example Double Integrals over General RegionsType I and Type II regions Examples 1-4 Examples 5-7 Swapping the Order of Integration Area and Volume Revisited Double integrals in polar coordinatesdA = r dr (d theta) Examples Multiple integrals in physicsDouble integrals in physics Triple integrals in physics Integrals in Probability and StatisticsSingle integrals in probability Double integrals in probability Change of VariablesReview: Change of variables in 1 dimension Mappings in 2 dimensions Jacobians Examples Bonus: Cylindrical and spherical coordinates | | To get the area between the polar curve r=f(θ) and the polar curve r=g(θ), we just subtract the area inside the inner curve from the area inside the outer curve. If f(θ)≥g(θ), this means 12∫baf(θ)2−g(θ)2dθ. Note that this is NOT 12∫ba[f(θ)−g(θ)]2dθ!! You first square and then subtract, not the other way around. As with most ``area between two curves'' problems, the tricky thing is figuring out the beginning and ending angles. This is typically where f(θ)=g(θ). In the following video, we compute the area inside the cardioid r=1+sin(θ) and outside the circle r=12. | | | Example: Find the area inside the circle r=2cos(θ) and outside the unit circle. Solution: Here f(θ)=2cos(θ) and g(θ)=1. These intersect when cos(θ)=1/2, i.e. at θ=±π/3. That is, f(θ) is only bigger than g(θ) when −π/3<θ<π/3. Our area is then 12∫π/3−π/3[(2cos(θ))2−12]dθ===12∫π/3−π/31+2cos(2θ)dθ12(θ+sin(2θ))∣∣π/3−π/3π3+3–√2. | << Prev Next >> | |
8916 | http://mathcentral.uregina.ca/QQ/database/QQ.09.14/h/vir1.pdf | Math Workshop October 2010 Fractions and Repeating Decimals This evening we will investigate the patterns that arise when converting fractions to decimals. As an example of what we will be looking at, 1. Write 3/7 as a repeating decimal by dividing 7 into 3.0000... . How many digits are in the repetend (i.e. in the repeating block of digits)? Add the number formed by the first half of the repetend to the number formed by the second half. We start by reviewing sequences. Adding arithmetic and geometric sequences. There are two types of sequences that appear so often that we give them names. To get any term after the first in an arithmetic sequence, you add a fixed constant to the previous term; for example, 7, 10, 13, 16, 19, 22, where the difference between consecutive terms is d = 3. For a geometric sequence you multiply the previous term by a fixed constant; for example, 7, 21, 63, 189, 567. Here the ratio of consecutive terms (one term divided by the previous term) is r = 3. There are simple tricks for adding the terms of these sequences. To easily compute the arithmetic sum S = 7 + 10 + 13 + 16 + 19 + 22, write the sum forwards and backwards, then add vertically: S = 7 + 10 + 13 + 16 + 19 + 22 S = 22 + 19 + 16 + 13 + 10 + 7 2S = 29 + 29 + 29 + 29 + 29 + 29 = 6 · 29; Since 2S = 6·29, we deduce that S = 3·29 = 87, which can easily be verified by adding the given six numbers. And here is the trick for geometric sums. To compute a geometric sum such as S = 7 + 21 + 63 + 189 + 567, subtract S from 3S (where the number we multiply S by is the common ratio of any two consecutive terms, 3 = 21/7 = 63/21 = 189/63 = ...): 3S = 21 + 63 + 189 + 567 + 1701 – S = – (7 + 21 + 63 + 189 + 567) 2S = – 7 + 1701; thus, 2S = 1694, whence S = 847, which again is easily checked directly. Here are four practice problems. Although there are formulas that give you the sum from the initial and final elements and the size of the jump, understanding the trick is much more important than mastering the formulas. 2. Compute the following four sums. a. Sum the odd numbers between 10 and 1000. b. Sum the powers of 4 from 4 to 1024. c. ! 7 10 + 7 100 + 7 1000 + 7 10000 + 7 100000. (Suggestion. Use 10S instead of S/10.) d. ! 7 10 + 7 100 + 7 1000 + 7 10000 + 7 100000 + … (where the three dots mean that you should continue the sum to infinity). Workshop, October 2010 page 2 Fractions and repeating decimals. Parts (c) and (d) of Exercise 2 suggest how geometric sums might be related to rational numbers, which is the main topic of today's workshop. We define a rational number to be any real number that can be written as a fraction; a fraction is any number that can be written in the form ! a b , where a and b are integers and b ≠ 0. Here are some examples: 2/5; 5/2; .8 = 8/10; 41/3 = 13/3; –18 = –18/1; – 4/3 = –4/3 = 4/–3; 0 = 0/1; 11.37 = 1137/100 . Every integer is a rational number since the integer n can be written as n/1. It is obvious that ! 81 is rational because it equals 9 = 9/1. On the other hand, it is entirely unobvious that ! 2 is not rational: ! 2 cannot be written as a fraction, a discovery that came as quite a surprise to the ancient Greeks. The proof of the irrationality of ! 2 is not hard, but we will leave it for another time. Tonight we are interested in exploring another unobvious fact: the rationals are precisely those numbers that can be written as repeating decimals. A repeating decimal is a number written in decimal notation whose digits to the right of the decimal point eventually consist of endless repetitions of a block of numbers called the repetend. The number of digits in the smallest repeating block is called the period of the repeating decimal. For example, 1/3 = .3333 ... has period 1, and 2779/110 = 25.2636363... has period 2. We use a bar over the repeating part of the decimal as shorthand; thus 1/3 is written as 0.3 (or just ! .3), while 2779/110 becomes 25.263 . ¼ = .250000000 … is also a repeating decimal, which we usually write as .25 (or 0.25). (Numbers where the repeating part is a zero are called "terminating decimals" by schoolteachers, but we have no need here for more terminology.) We can write any fraction as a repeating decimal by using long division. An example will make this clear. Let's write 345/22 as a decimal: ! 22 345.000 15.681 ) 22 125 110 150 132 180 176 40 22 18 We stopped when 18 repeated as the remainder — we would just bring down the zero again, find that 22 goes into 180 eight times with a remainder of 4, bring down the zero and get 22 goes into 40 one time with a remainder of 18, and continue getting remainders of 18 and 4 forever. We conclude that ! 345 22 =15.6818181… =15.681. We knew that eventually the remainders would start to repeat because there are only 22 possible remainders when dividing by 22, namely the integers from 0 up to 21. It just happens that the period is 2 and not something longer. In fact it is easy to see that a/b will have period 2 if b = 2m ·5n ·11 and a is any integer that is not divisible by 11; furthermore, the sum of the digits of the repeating part will be 9. Workshop, October 2010 page 3 3. Verify this for yourself — without a calculator — by showing that x = 7/440 has period 2, but save yourself time and effort by first multiplying x by 103 (because 440 = 23 ·5·11, so 3 is the highest power to which 2 or 5 has been raised), then divide top and bottom of 103x by 40 to reduce the fraction to an integer over 11, do the division, then solve for x. In the other direction we recognize that 15.6818181… is just 15.6 plus a geometric sequence! 15.6 + .0818181… = 15.6 + .081 + .00081 + .0000081 + …, where each term is 1/100 times the previous term: 4. Subtract x = 15.68181... from 100x, then solve for x. Reduce x to a fraction in lowest terms. (Remember that a fraction is the quotient of integers; you might have to multiply top and bottom by a power of 10 to get it in the proper form.) For a fraction a/b in lowest terms, the longest its period could possibly be is b – 1 (because there are b – 1 nonzero possible remainders). To simplify matters, we will restrict ourselves to the case where b is a prime number, and a < b. When the period of 1/b is b – 1, we will call b a full-period prime. The smallest full-period prime is 7. 5. a. Write 1/7 as a repeating decimal. b. Use part (a) to guess what the decimal expansion of 2/7 must be — in particular, you should note where the remainder of 2 shows up in your division of 7 into 1. Observe the evident patterns in Figure 1. Each fraction a/7 (in the box on the left) has the same cycle of digits in its decimal expansion, but with different starting points. Thus, to compute a/7, don't think of multiplying the decimal for 1/7 by a; rather, apply a cyclic permutation to the digits in the repeating block 142857. In this context it is natural to arrange the digits around a circle as shown in the figure. An arrow in the middle reminds us which way to order the digits, and each fraction a/7 is placed outside the circle beside the starting point for its decimal. Thus, looking at the circle diagram, we can read off the decimal expansion of 3/7 as follows. The first digit, 4, is inside the circle next to the label 3/7. Starting at that point, proceed around the circle reading off digits 4285714285714285… . Did you notice that the diametrically opposed digits inside the circle add to 9? Or that the diametrically opposed fractions outside the circle add to 1? Are these just coincidences, or do these patterns show up for denominators other than 7? It turns out that these questions are easily answered for full-period primes, so we will focus on them. Workshop, October 2010 page 4 It is not at all clear which primes have a full period. On the last page we include a list of the values of 1/b for integers b ≤ 50. The full-period primes on that list are 7, 17, 19, 23, 29, and 47. (There are many larger full-period primes — 59, 61, 97, 109, …— but nobody knows whether or not the list is infinite!) Here is one simple method for determining the period corresponding to a prime number: Theorem 1. For any prime number p larger than 5, the number of ones in the first element of the sequence 11, 111, 1111, … that is divisible by p equals the period of 1/p. 6. Test the theorem for the primes 7, 11, and 13 (with periods 6, 2, and 6, respectively). You can easily prove it by applying the method used in problem 4 above to convert 1/p to its decimal form: If k is the period, then (10k – 1)/p will be an integer, call it n; that is, 9·111…1/p = n. Since p is a prime larger than 5, it cannot divide evenly into 9, so it must divide the factor 111…1, as claimed. Go through this proof using 7 and 11 to better understand the argument—what does n equal when p = 7? p = 11? Consider the circle diagram for b = 17. 7. Your calculator won't help much in this exercise because the period of 1/17 is 16, but the figure has all the information you need. You should check (by long division) that the decimal expansion of 1/17 begins .05882...; that the first remainder after the decimal point is 10 (that is 17 goes into 1.0 zero times with a remainder of 10); that 17 goes into 100 five times with a remainder of 15 (Look where the 5 and 15 appear on the circle in Figure 2.); that 17 goes into 150 eight times with remainder 14. Furthermore, check that 15/17 begins .88235... . It should be clear to you that the decimal expansion of 15/17 would contain the same digits in the same order as 1/17, but it begins where the long division of 17 into 1 has a remainder of 15 (so that the first step is to divide 17 into 150). It should be clear from these examples that when b is a full-period prime, the b – 1 distinct fractions a/b with 1 ≤ a ≤ b – 1 have repetends that are cyclic permutations of a common sequence of digits d1, d2, d3,…, db–1. 8. Explain why b must be prime when 1/b has period b – 1. (Hint. If, to the contrary, b = pq had period b – 1, what would you know about the period of p/b = 1/q?) Workshop, October 2010 page 5 Theorem 2. Let p be a full-period prime. If the repetend of 1/p is the sequence of digits d1, d2, d3,…, d2k, then d1 + dk+1 = d2 + dk+2 = … dk + d2k = 9. 9. We expect older students who are good at algebra to provide a justification for each claim of my argument below. Younger students should just verify each claim using both p = 7 (with period 6) and p = 19 (with period 18). a. If p is a full period prime, then its repetend has length p – 1. b. We can denote the period by 2k (where k is a positive integer). c. p divides (102k – 1)/9 = ! 111…1 2k 1 2 3 , but fails to divide any shorter string of 1s. In particular, p does not divide ! 111…1 k 1 2 3 . d. ! 111…1 2k 1 2 3 = ! 10k +1 ( ) 10k "1 ( ) 9 = 10k +1 ( )#111…1 k 1 2 3 , so that p divides 10k + 1, whence 10k/p + 1/p is an integer. e. The decimal expansion of the sum 10k/p + 1/p may be represented this way: + ! d1d2…dk.dk+1dk+2…d2k d1 d2 … .d1 d2 …dk dk+1dk+2… f. Therefore, di + dk+i = 9 for each i between 1 and k. You may have noticed that in the proof of Theorem 2 we made little use of the condition that p be a full period prime. In fact, we used only the fact that the period is even. An 1836 theorem of the French mathematician E. Midy states that the property of 1/p in Theorem 2 holds for the fraction a/p, where p is an arbitrary prime larger than 5 and a an integer that is not a multiple of p, as long as the period of p is even. Observe in the table on the last page that the prime numbers from 31 to 43 all have odd periods so that Midy's theorem does not apply to them. It does apply to the other primes on the list. When p is not a full period prime, the period of 1/p is a divisor of p – 1, and for each a that is not a multiple of p, a/p has the same period as 1/p. 10. Check these claims out for a/11, where a < 11. Note that there will be five circle diagrams modeled after Figures 1 and 2, each with only one pair of opposite fractions. 11. Draw the circle diagrams for a/13, a < 13. Here the circle diagrams come with six fractions each; one circle can start with 1/13, the other with 2/13. Mod 10 arithmetic. Consider the long-division method you used with Exercise 11. At each step we bring down a zero, divide by 13, and subtract to obtain the new remainder. But "bring down a zero" really means multiply by 10. That is, if r is the remainder at a particular step in the division, the next remainder will be 10r (mod 13). So when expanding 1/13, starting with the first remainder, 10, the sequence of remainders is 10 (mod 13), 102 ≡ 9 (mod 13), 102 ≡ 9 (mod 13), 103 ≡ 12 (mod 13), 104 ≡ 3 (mod 13), 105 ≡ 4 (mod 13). (Please note how these numbers correspond to the fractions 10/13, 9/13, 12/13, 3/13, and 4/13 about the first circle you made in Exercise 11.) The sequence repeats when we reach a power of 10 that returns to the starting point: 106 ≡ 1 (mod 13); that is, 13 divides evenly into 106 – 1. Since 106 – 1 = (103 + 1)(103 – 1), and 13 does not divide 103 – 1 (because if it did, we would have found the period of 1/13 to be shorter than 6), we Workshop, October 2010 page 6 deduce that 13 must divide 103 + 1. (Of course, it's easy to check that 1001 = 7·11·13.) This approach gives us a second way to recognize full-period primes: Theorem 3. 1/p has period p – 1 if and only if every positive integer less than p is congruent to 10j (mod p) for some j < p. Table of the decimal expansions of 1/x for x ≤ 50. |
8917 | https://sahussaintu.files.wordpress.com/2014/03/spherical_harmonics.pdf | 1 1 Properties of Spherical Harmonics 1.1 Repetition In the lecture the spherical harmonics were introduced as the eigenfunctions of angular momentum operators and in spherical coordinates.
We found that [1.1] and .
[1.2] The spherical harmonics can be defined as [1.3] where is the quantum number of the orbital angular momentum and the magnetic quantum number. There are analytical definitions for the normalization factor and the associated Legendre Polynomials that allow the calculations of the spherical harmonics.
The spherical harmonics for = 0, 1, and 2 are given by [1.4] [1.5] Y, m , q f , ( ) l ˆz l ˆ2 l ˆzY, m , q f , ( ) "mY, m , q f , ( ) = l ˆ2Y, m , q f , ( ) ", , 1 + ( )Y, m , q f , ( ) = Y, m , q f , ( ) N, m , P, m q cos ( ) eimf ⋅ ⋅ = , m N, m , P, m q cos ( ) , Y0 0 , q f , ( ) 1 4p -----------= Y1 0 , q f , ( ) 1 2 ---3 p ---q cos = Y1 1 ± , q f , ( ) 1 2 ---3 2p ------q sin e if ± + − = 2 Chapter 1 .
[1.6] Note, that the sign of the functions and is defined differently than in the script of the lecture. The definition here is in agreement with most of the literature on spherical harmonics.
1.2 Graphical Representation of Spherical Harmonics The spherical harmonics are often represented graphically since their linear combinations correspond to the angular functions of orbitals. Figure 1.1a shows a plot of the spherical harmonics where the phase is color coded. One can clearly see that is symmetric for a rotation about the z axis. The linear combinations , and are always real and have the form of typical atomic orbitals that are often shown.
1.3 Properties of Spherical Harmonics There are some important properties of spherical harmonics that simplify working with them.
1.3.1 Orthogonality and Normalization The spherical harmonics are normalized and orthogonal, i.e., [1.7] where the Kronecker delta is defined as Y2 0 , q f , ( ) 1 4 ---5 p --- 3cos2q 1 – ( ) = Y2 1 ± , q f , ( ) 1 2 --15 2p ------q sin qe if ± cos + − = Y2 2 ± , q f , ( ) 1 4 ---15 2p ------sin2qe 2if ± = Y1 1 ± , q f , ( ) Y2 1 ± , q f , ( ) Y, m , q f , ( ) 1 2 ⁄ Y, m , q f , ( ) 1 – ( )mY, m – , q f , ( ) + ( ) 2 Y, m , q f , ( ) mf ( ) cos = Y, 0 , q f , ( ) i – 2 ⁄ Y, m , q f , ( ) 1 – ( )mY, m – , q f , ( ) – ( ) 2 Y, m , q f , ( ) mf ( ) sin = Y,1 m1 , q f , ( )Y,2 m2 , q f , ( ) q sin q d ( ) f d 0 p ∫ 0 2p ∫ dm1 m2 , d,1 ,2 , = Properties of Spherical Harmonics 3 , 0 = , 1 = , 2 = m 0 = m 2 – = m 1 – = m 1 = m 2 = m 3 = m 4 = m 3 – = m 4 – = , 3 = , 4 = -π -π/2 0 π π/2 , 0 = , 1 = , 2 = m 0 = m 2 = m 1 = m 1 = m 2 = m 3 = m 4 = m 3 = m 4 = , 3 = , 4 = -π -π/2 0 π π/2 a) b) Figure 1.1: Graphical Representation of the Spherical Harmonics a) Plot of the spherical harmonics where the phase of the function is color coded.
Note that is always axially symmetric with respect to a rotation about the z axis since it depends only on the angle . The phase of the function changes with a periodicity of . b) The linear combinations , , and are all real and show only a phase of 0 (positive) and (negative) and correspond to the typical orbital shapes.
Y, m , q f , ( ) Y, m , q f , ( ) q m 1 2 ⁄ Y, m , q f , ( ) 1 – ( )mY, m – , q f , ( ) + ( ) 2 Y, m , q f , ( ) mf ( ) cos = Y, 0 , q f , ( ) i – 2 ⁄ Y, m , q f , ( ) 1 – ( )mY, m – , q f , ( ) – ( ) 2 Y, m , q f , ( ) mf ( ) sin = p 4 Chapter 1 .
[1.8] They form a complete basis set of the Hilbert space of square-integrable functions, i.e., every such function can be expressed as a linear combination of spherical harmonics .
[1.9] The coefficients can be calculated as .
[1.10] 1.3.2 Product of Two Spherical Harmonics Since the spherical harmonics form a orthonormal basis set, the product of two spherical harmonics can again be expressed in spherical harmonics. Let us first look at a simple example .
[1.11] Comparing this to the spherical harmonics of Eqs. [1.4]-[1.6] it is immediately clear that we need the functions and to express the product. We can make an Ansatz [1.12] which leads to .
[1.13] From this it is immediately clear that da b , 0 a b ≠ 1 a b = ⎩ ⎨ ⎧ = f q f , ( ) f , m , Y, m , q f , ( ) m , – = , ∑ , 0 = ∞ ∑ = f , m , f , m , f q f , ( )Y, m , q f , ( ) q sin q d ( ) f d 0 p ∫ 0 2p ∫ = Y1 0 , q f , ( ) Y1 0 , q f , ( ) ⋅ 1 2 ---3 p ---q cos 1 2 --3 p ---q cos ⋅ 3 4p ------cos2q = = Y2 0 , q f , ( ) Y0 0 , q f , ( ) Y1 0 , q f , ( ) Y1 0 , q f , ( ) ⋅ c0 0 , Y0 0 , q f , ( ) c2 0 , Y2 0 , q f , ( ) + = 3 4p ------cos2q c0 0 , 1 4p -----------c2 0 , 1 4 --5 p --- 3cos2q 1 – ( ) + = c0 0 , 1 4p -----------c2 0 , 1 4 ---5 p ---– c2 0 , 3 4 ---5 p ---cos2q + = Properties of Spherical Harmonics 5 [1.14] and .
[1.15] For a general product this is of course more complicated but there are a few simple rules for the general product . Since the dependence on is always given by , it is immediately clear that the product function has to have the magnetic quantum number . Using similar arguments, the orbital angular quantum number can be limited to the range . In principle, it is not important to know these restrictions since the Clebsch-Gordan coefficients (or the Wigner 3j symbols) will do the selection automatically.
The product can in general be written as the following linear combination [1.16] where the Wigner 3j symbols are related to the Racah or Clebsch-Gordan coefficients by .
[1.17] The Wigner 3j symbols or the Clebsch-Gordan coefficients can be found in tables in books about angular momentum or calculated using programs like Matlab, Macsyma, or Mathematica. Written with Clebsch-Gordan coefficients we obtain for Eq. [1.16] c2 0 , 3 4p ------ 4 3 --- p 5 ---⋅ 1 5p -----------= = c0 0 , 1 5p -----------1 4 --5 p ---4p ⋅ 1 4p -----------= = Y,1 m1 , q f , ( ) Y,2 m2 , q f , ( ) ⋅ f imf ( ) exp M m1 m2 + = ,1 ,2 – L ,1 ,2 + ≤ ≤ Y,1 m1 , q f , ( ) Y,2 m2 , q f , ( ) ⋅ 2,1 1 + ( ) 2,2 1 + ( ) 2L 1 + ( ) 4p -------------------------------------------------------------------L M , ∑ = ,1 ,2 L m1 m2 M ⎝ ⎠ ⎜ ⎟ ⎛ ⎞YL M , q f , ( ) ,1 ,2 L 0 0 0 ⎝ ⎠ ⎜ ⎟ ⎛ ⎞ × ,1 ,2 L m1 m2 M ⎝ ⎠ ⎜ ⎟ ⎛ ⎞ 1 – ( ),1 ,2 – M – 1 2L 1 + ( ) -------------------------c ,1 m1 ,2 m2 L M – , , , , , ( ) = 6 Chapter 1 .
[1.18] To calculate the coefficients for the above example, we need the Wigner 3j symbols for [1.19] and obtain .
[1.20] In the same way more complex products can be calculated and decomposed in the spherical harmonics. This is an iterative way to calculate the functional form of higher-order spherical harmonics from the lower-order ones. We will discuss this in more detail in an exercise.
1.3.3 Addition Theorem of Spherical Harmonics The spherical harmonics obey an addition theorem that can often be used to simplify expressions [1.21] where omega describes the angle between two unit vectors oriented at the polar coordinates and with .
[1.22] Y,1 m1 , q f , ( ) Y,2 m2 , q f , ( ) ⋅ 2,1 1 + ( ) 2,2 1 + ( ) 4p 2L 1 + ( ) ----------------------------------------------YL M , q f , ( ) L M , ∑ = c ,1 m1 ,2 m2 L M , , , , , ( )c ,1 0 ,2 0 L 0 , , , , , ( ) × 1 1 0 0 0 0 ⎝ ⎠ ⎜ ⎟ ⎛ ⎞ 1 3 -------– = 1 1 1 0 0 0 ⎝ ⎠ ⎜ ⎟ ⎛ ⎞ 0 = 1 1 2 0 0 0 ⎝ ⎠ ⎜ ⎟ ⎛ ⎞ 2 15 ------= Y1 0 , q f , ( ) Y1 0 , q f , ( ) ⋅ 3 2L 1 + ( ) 4p --------------------1 1 L 0 0 0 ⎝ ⎠ ⎜ ⎟ ⎛ ⎞2 YL 0 , q f , ( ) L 0 = 2 ∑ = 1 4p -----------Y0 0 , q f , ( ) 0Y1 0 , q f , ( ) 1 5p -----------Y2 0 , q f , ( ) + + = Y, m , q1 f1 , ( )Y, m , q2 f2 , ( ) m , – = , ∑ 2, 1 + 4p ---------------P, w cos ( ) = w q1 f1 , ( ) q2 f2 , ( ) w cos q1 cos q2 cos q1 sin q2 sin f1 f2 – ( ) cos + = Properties of Spherical Harmonics 7 1.3.4 Integrals Over Spherical Harmonics The integration over the product of three spherical harmonics can be simplified using the product rule of Eq. [1.16] and the orthogonality of Eq. [1.7]. This leads to [1.23] a simple expression involving only a normalization constant and two Wigner 3j symbols.
1.4 Literature (1) D. M. Brink, G. R. Satchler, Angular Momentum, third edition, Clarendon Press, 1993.
(2) A. R. Edmonds, Angular Momentum in Quantum Mechanics, Princeton University Press, 1960.
(3) M. E. Rose, Elementary Theory of Angular Momentum, John Wiley & Sons Inc., New York, 1957.
Y,1 m1 , q f , ( )Y,2 m2 , q f , ( )Y,3 m3 , q f , ( ) q sin q d ( ) f d 0 p ∫ 0 2p ∫ 2,1 1 + ( ) 2,2 1 + ( ) 2L 1 + ( ) 4p -------------------------------------------------------------------,1 ,2 L m1 m2 M ⎝ ⎠ ⎜ ⎟ ⎛ ⎞ ,1 ,2 L 0 0 0 ⎝ ⎠ ⎜ ⎟ ⎛ ⎞ L M , ∑ = YL M , q f , ( )Y,3 m3 , q f , ( ) q sin q d ( ) f d 0 p ∫ 0 2p ∫ × 2,1 1 + ( ) 2,2 1 + ( ) 2,3 1 + ( ) 4p ---------------------------------------------------------------------,1 ,2 ,3 m1 m2 m3 ⎝ ⎠ ⎜ ⎟ ⎛ ⎞ ,1 ,2 ,3 0 0 0 ⎝ ⎠ ⎜ ⎟ ⎛ ⎞ = 8 Chapter 1 |
8918 | https://chem.libretexts.org/Courses/Madera_Community_College/MacArthur_Chemistry_3A_v_1.2/05%3A_Chemical_Reactions/5.04%3A_Types_of_Reactions/5.4.03%3A_Combustion_Reactions | 5.4.3: Combustion Reactions - Chemistry LibreTexts
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5.4: Types of Reactions
5: Chemical Reactions
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5.4.3: Combustion Reactions
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Contents
1. Home
2. Campus Bookshelves
3. Madera Community College
4. MacArthur Chemistry 3A v 1.2
5. 5: Chemical Reactions
6. 5.4: Types of Reactions
7. 5.4.3: Combustion Reactions
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MacArthur Chemistry 3A v 1.2
Front Matter
1: Basics of Measurement
2: Matter and Energy
3: Elements and Compounds
4: The Mole Concept
5: Chemical Reactions
6: Introduction to Stoichiometry
7: Electrons and Chemical Bonding
8: Gases
9: Attractive Forces
10: Aqueous Solutions
11: Acids and Bases
12: Stoichiometry Applications
13: Oxidation and Reduction
14: Radioactivity and Nuclear Chemistry
Back Matter
5.4.3: Combustion Reactions
Last updated Feb 14, 2022
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5.4.2: Decomposition Reactions
5.4.4: Single Replacement Reactions
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Table of contents
1. How do you cook the perfect marshmallow?
2. Combustion Reactions
1. Example 5.4.3.1: Combustion Reactions
1. Solution
2. Step 3: Think about your result.
Summary
Review
Figure 5.4.3.1 (CC BY 2.0; Nina Hale via Wikipedia)
How do you cook the perfect marshmallow?
Roasting marshmallows over an open fire is a favorite past-time for campers, outdoor cook-outs, and just gathering around a fire in the back yard. The trick is to get the marshmallow a nice golden brown without catching it on fire. Too often we are not successful and we see the marshmallow burning on the stick – a combustion reaction taking place right in front of us.
Combustion Reactions
A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve OA 2 as one reactant. The combustion of hydrogen gas produces water vapor:
2HA 2(g)+OA 2(g)→2HA 2O(g)
Notice that this reaction also qualifies as a combination reaction.
Figure 5.4.3.2: Explosion of the Hindenberg. (Public Domain; Gus Pasquerella/US Navy via Wikipedia)
The Hindenberg was a hydrogen-filled airship that suffered an accident upon its attempted landing in New Jersey in 1937. The hydrogen immediately combusted in a huge fireball, destroying the airship and killing 36 people. The chemical reaction was a simple one: hydrogen combining with oxygen to produce water.
Many combustion reactions occur with a hydrocarbon, a compound made up solely of carbon and hydrogen. The products of the combustion of hydrocarbons are carbon dioxide and water. Many hydrocarbons are used as fuel because their combustion releases very large amounts of heat energy. Propane (CA 3HA 8) is a gaseous hydrocarbon that is commonly used as the fuel source in gas grills.
CA 3HA 8(g)+5OA 2(g)→3COA 2(g)+4HA 2O(g)
Example 5.4.3.1: Combustion Reactions
Ethanol can be used as a fuel source in an alcohol lamp. The formula for ethanol is CA 2HA 5OH. Write the balanced equation for the combustion of ethanol.
Solution
Step 1: Plan the problem.
Ethanol and oxygen are the reactants. As with a hydrocarbon, the products of the combustion of an alcohol are carbon dioxide and water.
Step 2: Solve.
Write the skeleton equation:
CA 2HA 5OH(l)+OA 2(g)→COA 2(g)+HA 2O(g)
Balance the equation.
CA 2HA 5OH(l)+3OA 2(g)→2COA 2(g)+3HA 2O(g)
Step 3: Think about your result.
Combustion reactions must have oxygen as a reactant. Note that the water produced is in the gas state, rather than the liquid state, because of the high temperatures that accompany a combustion reaction.
Summary
Combustion reaction is defined and examples are given.
Review
What is needed for a combustion reaction to take place?
What is formed in any combustion reaction?
Mercury reacts with oxygen to form mercuric oxide. Is this a combustion reaction?
What are the products of any combustion reaction involving a hydrocarbon?
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8919 | https://labuladong.online/algo/en/data-structure/prefix-sum/ | Skip to main content
Prefix Sum Array Technique
labuladongOriginal
Solve 2 Problems
303. Range Sum Query - Immutable
304. Range Sum Query 2D - Immutable
This article will resolve
| LeetCode | Difficulty |
--- |
| 303. Range Sum Query - Immutable | 🟢 |
| 304. Range Sum Query 2D - Immutable | 🟠 |
Prerequisite Knowledge
Before reading this article, you need to learn:
Array Basics
The prefix sum technique is suitable for quickly and frequently calculating the sum of elements within a given index range.
Prefix Sum in One-Dimensional Arrays
Let's look at a sample problem, LeetCode Problem 303: "Range Sum Query - Immutable", which asks you to calculate the sum of elements within an array range. This is a standard prefix sum problem:
303. Range Sum Query - Immutable | LeetCode | 🟢
Given an integer array nums, handle multiple queries of the following type:
Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.
Implement the NumArray class:
NumArray(int[] nums) Initializes the object with the integer array nums.
int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + ... + nums[right]).
Example 1:
```
Input
["NumArray", "sumRange", "sumRange", "sumRange"]
, [0, 2], [2, 5], [0, 5]]
Output
[null, 1, -1, -3]
Explanation
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return (-2) + 0 + 3 = 1
numArray.sumRange(2, 5); // return 3 + (-5) + 2 + (-1) = -1
numArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3
```
Constraints:
1 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= left <= right < nums.length
At most 104 calls will be made to sumRange.
The problem is from LeetCode 303. Range Sum Query - Immutable.
java
```
// The problem requires you to implement such a class
class NumArray {
public NumArray(int[] nums) {}
// Query the cumulative sum of the closed interval [left, right]
public int sumRange(int left, int right) {}
}
```
cpp
```
// The problem requires you to implement such a class
class NumArray {
public:
NumArray(vector& nums) {}
// Query the accumulated sum of the closed interval [left, right]
int sumRange(int left, int right) {}
};
```
python
```
The problem requires you to implement such a class
class NumArray:
def __init__(self, nums: List[int]):
pass
# Query the accumulated sum of the closed interval [left, right]
def sumRange(self, left: int, right: int) -> int:
pass
```
go
```
// The problem requires you to implement such a class
type NumArray struct {}
func Constructor(nums []int) NumArray {}
// Query the cumulative sum of the closed interval [left, right]
func (this NumArray) SumRange(left int, right int) int {}
```
javascript
```
// The problem requires you to implement such a class
var NumArray = function(nums) {
this.nums = nums;
// Query the accumulated sum of the closed interval [left, right]
this.sumRange = function(left, right) {}
};
```
The sumRange function needs to calculate and return the sum of elements within an index range. Those unfamiliar with prefix sums might write code like this:
```
class NumArray {
private int[] nums;
public NumArray(int[] nums) {
this.nums = nums;
}
public int sumRange(int left, int right) {
// use for loop to sum
int sum = 0;
for (int i = left; i <= right; i++) {
sum += nums[i];
}
return sum;
}
}
```
This solution requires a for-loop traversal each time the sumRange function is called, resulting in a time complexity of O(N). Since the sumRange function may be called very frequently, this makes the algorithm inefficient.
The correct approach is to use the prefix sum technique for optimization, reducing the time complexity of the sumRange function to O(1):
java
```
class NumArray {
// prefix sum array
private int[] preSum;
// input an array to construct the prefix sum
public NumArray(int[] nums) {
// preSum = 0, to facilitate the calculation of accumulated sums
preSum = new int[nums.length + 1];
// calculate the accumulated sums of nums
for (int i = 1; i < preSum.length; i++) {
preSum[i] = preSum[i - 1] + nums[i - 1];
}
}
// query the sum of the closed interval [left, right]
public int sumRange(int left, int right) {
return preSum[right + 1] - preSum[left];
}
}
```
cpp
```
include
class NumArray {
// prefix sum array
std::vector preSum;
// input an array to construct the prefix sum
public:
NumArray(std::vector<int>& nums) {
// preSum = 0, to facilitate the calculation of accumulated sums
preSum.resize(nums.size() + 1);
// calculate the accumulated sums of nums
for (int i = 1; i < preSum.size(); i++) {
preSum[i] = preSum[i - 1] + nums[i - 1];
}
}
// query the sum of the closed interval [left, right]
int sumRange(int left, int right) {
return preSum[right + 1] - preSum[left];
}
};
```
python
```
class NumArray:
# prefix sum array
def init(self, nums: List[int]):
# input an array to construct the prefix sum
# preSum = 0, to facilitate the calculation of accumulated sums
self.preSum = (len(nums) + 1)
# calculate the accumulated sums of nums
for i in range(1, len(self.preSum)):
self.preSum[i] = self.preSum[i - 1] + nums[i - 1]
# query the sum of the closed interval [left, right]
def sumRange(self, left: int, right: int) -> int:
return self.preSum[right + 1] - self.preSum[left]
```
go
```
type NumArray struct {
// prefix sum array
PreSum []int
}
// input an array to construct the prefix sum
func Constructor(nums []int) NumArray {
// PreSum = 0, to facilitate the calculation of accumulated sums
preSum := make([]int, len(nums)+1)
// calculate the accumulated sums of nums
for i := 1; i < len(preSum); i++ {
preSum[i] = preSum[i-1] + nums[i-1]
}
return NumArray{PreSum: preSum}
}
// query the sum of the closed interval [left, right]
func (this NumArray) SumRange(left int, right int) int {
// The following line includes the missing comment:
// PreSum = 0, to facilitate the calculation of accumulated sums
return this.PreSum[right+1] - this.PreSum[left] // Here we are using the prefix sum property, no need to repeat the comment here.
}
```
javascript
```
var NumArray = function(nums) {
// prefix sum array
let preSum = new Array(nums.length + 1).fill(0);
// preSum = 0, to facilitate the calculation of accumulated sums
preSum = 0;
// input an array to construct the prefix sum
for (let i = 1; i < preSum.length; i++) {
// calculate the accumulated sums of nums
preSum[i] = preSum[i - 1] + nums[i - 1];
}
this.preSum = preSum;
};
// query the sum of the closed interval [left, right]
NumArray.prototype.sumRange = function(left, right) {
return this.preSum[right + 1] - this.preSum[left];
};
```
The core idea is to create a new array preSum, where preSum[i] stores the cumulative sum of nums[0..i-1]. For example, 10=3+5+2:
Using this preSum array, if we want to find the sum of all elements in the index range [1, 4], we can calculate it as preSum - preSum.
Thus, the sumRange function only needs to perform one subtraction operation, avoiding the for-loop calls each time, with a worst-case time complexity of O(1).
You can open the visualization below, click on the line preSum[i] = preSum[i - 1] + nums[i - 1] to see the calculation of the preSum array. Click on console.log multiple times to see the calls to the sumRange function:
Algorithm visualize Algorithm VisualizationLink copied!
This technique is widely used in real life. For example, if your class has several students, each with a final exam score (out of 100), you can implement an API that returns the number of students whose scores fall within a given range.
You can first use counting sort to calculate how many students have each specific score, then use the prefix sum technique to implement the score range query API:
```
// stores all the students' scores
int[] scores = new int[]{...};
// the full mark of the test paper is 100 points
int[] count = new int[100 + 1];
// record the number of students for each score
for (int score : scores) {
count[score]++;
}
// construct the prefix sum array
for (int i = 1; i < count.length; i++) {
count[i] = count[i] + count[i-1];
}
// use the prefix sum array 'count' to query score segments
// query how many students scored between 80 and 90
int result = count - count;
```
Next, let's see how the prefix sum approach is applied in a two-dimensional array.
Prefix Sum in a 2D Matrix
This is LeetCode problem 304, "Range Sum Query 2D - Immutable", which is similar to the previous problem. The previous problem asked you to calculate the sum of elements in a subarray, while this one requires you to calculate the sum of elements in a submatrix of a 2D matrix:
304. Range Sum Query 2D - Immutable | LeetCode | 🟠
Given a 2D matrix matrix, handle multiple queries of the following type:
Calculate the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).
Implement the NumMatrix class:
NumMatrix(int[][] matrix) Initializes the object with the integer matrix matrix.
int sumRegion(int row1, int col1, int row2, int col2) Returns the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).
You must design an algorithm where sumRegion works on O(1) time complexity.
Example 1:
```
Input
["NumMatrix", "sumRegion", "sumRegion", "sumRegion"]
], [2, 1, 4, 3], [1, 1, 2, 2], [1, 2, 2, 4]]
Output
[null, 8, 11, 12]
Explanation
NumMatrix numMatrix = new NumMatrix();
numMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangle)
numMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)
numMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)
```
Constraints:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 200
-104 <= matrix[i][j] <= 104
0 <= row1 <= row2 < m
0 <= col1 <= col2 < n
At most 104 calls will be made to sumRegion.
The problem is from LeetCode 304. Range Sum Query 2D - Immutable.
Of course, you could use a nested for loop to traverse the matrix, but this would increase the time complexity of the sumRegion function, and reduce the efficiency of your algorithm.
Note that the sum of elements in any submatrix can be transformed into the sum of elements in several larger matrices surrounding it:
These four larger matrices all share a common characteristic: their top-left corners are at the origin (0, 0).
The better approach to this problem, similar to the prefix sum in one-dimensional arrays, is to maintain a two-dimensional preSum array that records the sum of elements in matrices with the origin as the vertex. Thus, the sum of elements in any submatrix can be calculated using several addition and subtraction operations:
java
```
class NumMatrix {
// preSum[i][j] records the sum of elements in the matrix [0, 0, i-1, j-1]
private int[][] preSum;
public NumMatrix(int[][] matrix) {
int m = matrix.length, n = matrix.length;
if (m == 0 || n == 0) return;
// construct the prefix sum matrix
preSum = new int[m + 1][n + 1];
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
// calculate the sum of elements for each matrix [0, 0, i, j]
preSum[i][j] = preSum[i-1][j] + preSum[i][j-1] + matrix[i - 1][j - 1] - preSum[i-1][j-1];
}
}
}
// calculate the sum of elements in the submatrix [x1, y1, x2, y2]
public int sumRegion(int x1, int y1, int x2, int y2) {
// the sum of the target matrix is obtained by operations on four adjacent matrices
return preSum[x2+1][y2+1] - preSum[x1][y2+1] - preSum[x2+1][y1] + preSum[x1][y1];
}
}
```
cpp
```
include
class NumMatrix {
// preSum[i][j] records the sum of elements in the matrix [0, 0, i-1, j-1]
std::vector> preSum;
public:
NumMatrix(std::vector>& matrix) {
int m = matrix.size(), n = matrix.size();
if (m == 0 || n == 0) return;
// construct the prefix sum matrix
preSum.resize(m + 1, std::vector(n + 1, 0));
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
// calculate the sum of elements for each matrix [0, 0, i, j]
preSum[i][j] = preSum[i-1][j] + preSum[i][j-1] + matrix[i - 1][j - 1] - preSum[i-1][j-1];
}
}
}
// calculate the sum of elements in the submatrix [x1, y1, x2, y2]
int sumRegion(int x1, int y1, int x2, int y2) {
// the sum of the target matrix is obtained by operations on four adjacent matrices
return preSum[x2+1][y2+1] - preSum[x1][y2+1] - preSum[x2+1][y1] + preSum[x1][y1];
}
};
```
python
```
class NumMatrix:
# preSum[i][j] records the sum of elements in the matrix [0, 0, i-1, j-1]
def init(self, matrix: List[List[int]]):
m = len(matrix)
n = len(matrix)
if m == 0 or n == 0:
return
# construct the prefix sum matrix
self.preSum = [ (n + 1) for _ in range(m + 1)]
for i in range(1, m + 1):
for j in range(1, n + 1):
# calculate the sum of elements for each matrix [0, 0, i, j]
self.preSum[i][j] = (self.preSum[i - 1][j] + self.preSum[i][j - 1] +
matrix[i - 1][j - 1] - self.preSum[i - 1][j - 1])
# calculate the sum of elements in the submatrix [x1, y1, x2, y2]
def sumRegion(self, x1: int, y1: int, x2: int, y2: int) -> int:
# the sum of the target matrix is obtained by operations on four adjacent matrices
return (self.preSum[x2 + 1][y2 + 1] - self.preSum[x1][y2 + 1] -
self.preSum[x2 + 1][y1] + self.preSum[x1][y1])
```
go
```
type NumMatrix struct {
// preSum[i][j] records the sum of elements in the matrix [0, 0, i-1, j-1]
preSum [][]int
}
func Constructor(matrix [][]int) NumMatrix {
m := len(matrix)
if m == 0 {
return NumMatrix{}
}
n := len(matrix)
if n == 0 {
return NumMatrix{}
}
// construct the prefix sum matrix
preSum := make([][]int, m+1)
for i := range preSum {
preSum[i] = make([]int, n+1)
}
for i := 1; i <= m; i++ {
for j := 1; j <= n; j++ {
// calculate the sum of elements for each matrix [0, 0, i, j]
preSum[i][j] = preSum[i-1][j] + preSum[i][j-1] + matrix[i-1][j-1] - preSum[i-1][j-1]
}
}
return NumMatrix{preSum: preSum}
}
// calculate the sum of elements in the submatrix [x1, y1, x2, y2]
func (this NumMatrix) SumRegion(x1, y1, x2, y2 int) int {
// the sum of the target matrix is obtained by operations on four adjacent matrices
return this.preSum[x2+1][y2+1] - this.preSum[x1][y2+1] - this.preSum[x2+1][y1] + this.preSum[x1][y1]
}
```
javascript
```
var NumMatrix = function(matrix) {
let m = matrix.length, n = matrix.length;
// preSum[i][j] records the sum of elements in the matrix [0, 0, i-1, j-1]
this.preSum = Array.from({length: m + 1}, () => Array(n + 1).fill(0));
if (m == 0 || n == 0) return;
// construct the prefix sum matrix
for (let i = 1; i <= m; i++) {
for (let j = 1; j <= n; j++) {
// calculate the sum of elements for each matrix [0, 0, i, j]
this.preSum[i][j] = this.preSum[i-1][j] + this.preSum[i][j-1] + matrix[i - 1][j - 1] - this.preSum[i-1][j-1];
}
}
};
NumMatrix.prototype.sumRegion = function(x1, y1, x2, y2) {
// calculate the sum of elements in the submatrix [x1, y1, x2, y2]
// the sum of the target matrix is obtained by operations on four adjacent matrices
return this.preSum[x2+1][y2+1] - this.preSum[x1][y2+1] - this.preSum[x2+1][y1] + this.preSum[x1][y1];
};
```
In this way, the time complexity of the sumRegion function is optimized to O(1) using the prefix sum technique, which is a typical "space for time" trade-off.
You can open the visualization below and click multiple times on the line preSum[i][j] = ... to see the calculation process of the preSum array. By clicking multiple times on the line console.log, you can see the calls to the sumRegion function:
Algorithm visualize Algorithm VisualizationLink copied!
The prefix sum technique is explained here. This algorithm technique is easy for those who understand it and difficult for those who don't. In practical applications, you need to cultivate your mental flexibility to quickly identify a prefix sum problem when you see one.
Further Expansion
The prefix sum technique explained in this article uses a precomputed preSum array to quickly calculate the sum of elements within a given index range. However, it is not limited to summation; it can also be used for quickly computing products and other scenarios.
Moreover, prefix sum arrays are often combined with other data structures or algorithmic techniques. I will explain these combinations in High-Frequency Prefix Sum Exercises along with relevant exercises.
However, the prefix sum technique has several limitations.
First Limitation: The prerequisite for using the prefix sum technique is that the original array nums does not change.
If an element in the original array changes, the values in the preSum array after that element become invalid, requiring O(n) time to recalculate the preSum array, which is similar to the brute-force method.
Second Limitation: The prefix sum technique is only applicable in scenarios with inverse operations.
For example, in summation scenarios, if you know x+6=10, you can deduce x=10−6=4. Similarly, in product scenarios, if you know x∗6=12, you can deduce x=12/6=2. This is known as having an inverse operation, which allows the use of the prefix sum technique.
However, some scenarios do not have inverse operations. For example, in maximum value scenarios, if you know max(x,8)=8, you cannot deduce the value of x.
To address both issues simultaneously, more advanced data structures are required. The most general solution is the Segment Tree, which will be explained in detail in the data structure design chapter.
Related Problems
You can install my Chrome extension then open the link.
| LeetCode | Difficulty |
--- |
| 1314. Matrix Block Sum | 🟠 |
| 1352. Product of the Last K Numbers | 🟠 |
| 238. Product of Array Except Self | 🟠 |
| 325. Maximum Size Subarray Sum Equals k🔒 | 🟠 |
| 327. Count of Range Sum | 🔴 |
| 437. Path Sum III | 🟠 |
| 523. Continuous Subarray Sum | 🟠 |
| 525. Contiguous Array | 🟠 |
| 560. Subarray Sum Equals K | 🟠 |
| 724. Find Pivot Index | 🟢 |
| 862. Shortest Subarray with Sum at Least K | 🔴 |
| 918. Maximum Sum Circular Subarray | 🟠 |
| 974. Subarray Sums Divisible by K | 🟠 |
Please login to view/post comments |
8920 | https://www.johndcook.com/blog/2022/10/20/directrix-of-a-conic/ | (832) 422-8646
Contact
Directrix of a conic
Posted on by John
The most common way to define an ellipse geometrically is as the set of points whose distances to two foci sum to a constant. There is another way, however, to define an ellipse that generalizes to include the two other conic sections, parabolas and hyperbolas.
You can define a conic section as the set of points whose distance to a focus equals a constant multiple of the distance to a line called the directrix. This multiple is denoted e for eccentricity. If 0 < e < 1 you get an ellipse. If e = 1 you get a parabola, and if e > 1 you get a hyperbola .
The previous post defined the latus rectum for an ellipse. More generally, the latus rectum is the chord through the focus and parallel to the directrix.
And with this definition, you can define the latus rectum for each of the conic sections.
Ellipse
Given an ellipse with equation
x²/a² + y²/b² = 1
where a > b, the foci are at ± c where c = √(a² − b²). The eccentricity is e = c/a.
Consider the focus at −c. If the directrix is the line x = −d then the distance from the focus (-c, 0) to left vertex of the ellipse at (-a, 0) equals e times the distance from (-a, 0) to the point (-d, 0) on the directrix, and so
a − c = (c/a)(d − a)
and we find d = a²/c.
In the plot below, the green vertical line to the left of the ellipse is the directrix. The red vertical line through the focus is the latus rectum. The length of the orange segment from the focus to the ellipse should be e times the length of the segment from the point on the ellipse to the directrix.
Parabola
For a parabola with equation
y² = 4ax
with a > 0, the focus is at (a, 0) and the directrix is the line x = −a. The eccentricity of a parabola is 1.
Here again the green vertical line to the left of the parabola is the directrix and the red vertical line through the focus is the latus rectum. The two dashed orange lines have the same length.
Hyperbola
Given a hyperbola with equation
x²/a² − y²/b² = 1
where a > b, the foci are at ± ae where the eccentricity is e = √(1 + b² / a²).
Let d = a²/c. Then either the line x = d or the line x = −d could be used as the directrix.
As before the green vertical line is the directrix and the red vertical line through the focus is the latus rectum. The lengths of the two dashed orange lines are proportional, and the proportionality constant is the eccentricity e.
Related posts
Latus rectum of an ellipse
Best approximation of a catenary by a parabola
Trig in hyperbolic geometry
The case e = 0 gives a circle, but now the definition doesn’t apply directly. The circle is the limit as e goes to 0 and the directrix moves further and further away. Projective geometry gives a way to rigorously say the directrix is a line at infinity without taking limits. |
8921 | https://owlcalculator.com/area/unit-of-area-barn/square-centimeter | Convert From Barn To Square Centimeter
Barn
Barn (designation: b) is an off-system unit of measurement of area, used in nuclear physics to measure the effective cross section of nuclear reactions, as well as the quadrupole moment. 1 barn equals 10⁻²⁸ m² = 10⁻²⁴ cm² = 100 fm² (the approximate size of an atomic nucleus). Multiples and sub-multiples are also determined; of which are used:megabarn (Mbn, Mb, 10⁻¹⁸ cm²),kilobarn (kbn, kb, 10⁻²¹ cm²),millibarn (mbn, mb, 10⁻²⁷ cm²),microbarn (mkbn, mkb, 10-30 cm²),nanobarn (nbn, nb, 10−33 cm²),picobarn (pbn, pb, 10−36 cm²),femtobarn (fbn, fb, 10⁻³⁹ cm²),attobarn (abn, ab, 10⁻⁴² cm²),While the barn never was an SI unit, the SI standards body acknowledged it in the 8th SI Brochure (superseded in 2019) due to its use in particle physics.
Square Centimeter
A unit of area measurement equal to a square measuring one centimeter on each side.
Area
Using our calculator, you can easily convert units of area from one dimension to another. Area is a numerical characteristic of a two-dimensional (flat or curved) geometric figure, informally speaking, showing the size of this figure. Historically, the calculation of the area was called quadrature.
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convert from Barn to Square Centimeter conversion of Barn to Square Centimeter Barn to Square Centimeter Barn into Square Centimeter Barn in Square Centimeter conversion from Barn to Square Centimeter b to cm²
Acre
1 Acre (ac) =
##### 0.404686 Hectare (ha)
##### 4.04686 x 107 Square Centimeter (cm²)
##### 43560 Square Feet (ft²)
##### 0.00404686 Square Kilometer (km²)
##### 4046.86 Square Meter (m²)
##### 0.0015625 Square Miles (mi²)
##### 4840 Square Yards (yd²)
##### 404686 Are (а)
##### 40.4686 Square decimeter (dm²)
##### 4.04686 x 10-28 Barn (b)
##### 4.04686 x 106 Decar (daa)
##### 4.09435 x 106 Rood (ro)
Hectare
1 Hectare (ha) =
##### 2.47105 Acre (ac)
##### 1.0 x 108 Square Centimeter (cm²)
##### 107639 Square Feet (ft²)
##### 0.01 Square Kilometer (km²)
##### 10000 Square Meter (m²)
##### 0.00386102 Square Miles (mi²)
##### 11959.9 Square Yards (yd²)
##### 1.0 x 106 Are (а)
##### 100 Square decimeter (dm²)
##### 1.0 x 10-27 Barn (b)
##### 1.0 x 107 Decar (daa)
##### 1.01174 x 107 Rood (ro)
Square Centimeter
1 Square Centimeter (cm²) =
##### 2.47105 x 10-8 Acre (ac)
##### 1.0 x 10-8 Hectare (ha)
##### 0.00107639 Square Feet (ft²)
##### 1.0 x 10-10 Square Kilometer (km²)
##### 0.0001 Square Meter (m²)
##### 3.86102 x 10-11 Square Miles (mi²)
##### 0.000119599 Square Yards (yd²)
##### 0.01 Are (а)
##### 1.0 x 10-6 Square decimeter (dm²)
##### 1.0 x 10-35 Barn (b)
##### 0.1 Decar (daa)
##### 0.101174 Rood (ro)
Square Feet
1 Square Feet (ft²) =
##### 2.29568 x 10-5 Acre (ac)
##### 9.2903 x 10-6 Hectare (ha)
##### 929.03 Square Centimeter (cm²)
##### 9.2903 x 10-8 Square Kilometer (km²)
##### 0.092903 Square Meter (m²)
##### 3.58701 x 10-8 Square Miles (mi²)
##### 0.111111 Square Yards (yd²)
##### 9.2903 Are (а)
##### 0.00092903 Square decimeter (dm²)
##### 9.2903 x 10-33 Barn (b)
##### 92.903 Decar (daa)
##### 93.9934 Rood (ro)
Square Kilometer
1 Square Kilometer (km²) =
##### 247.105 Acre (ac)
##### 100 Hectare (ha)
##### 1.0 x 1010 Square Centimeter (cm²)
##### 1.07639 x 107 Square Feet (ft²)
##### 1.0 x 106 Square Meter (m²)
##### 0.386102 Square Miles (mi²)
##### 1.19599 x 106 Square Yards (yd²)
##### 1.0 x 108 Are (а)
##### 10000 Square decimeter (dm²)
##### 1.0 x 10-25 Barn (b)
##### 1.0 x 109 Decar (daa)
##### 1.01174 x 109 Rood (ro)
Square Meter
1 Square Meter (m²) =
##### 0.000247105 Acre (ac)
##### 0.0001 Hectare (ha)
##### 10000 Square Centimeter (cm²)
##### 10.7639 Square Feet (ft²)
##### 1.0 x 10-6 Square Kilometer (km²)
##### 3.86102 x 10-7 Square Miles (mi²)
##### 1.19599 Square Yards (yd²)
##### 100 Are (а)
##### 0.01 Square decimeter (dm²)
##### 1.0 x 10-31 Barn (b)
##### 1000 Decar (daa)
##### 1011.74 Rood (ro)
Square Miles
1 Square Miles (mi²) =
##### 640 Acre (ac)
##### 258.999 Hectare (ha)
##### 2.58999 x 1010 Square Centimeter (cm²)
##### 2.78784 x 107 Square Feet (ft²)
##### 2.58999 Square Kilometer (km²)
##### 2.58999 x 106 Square Meter (m²)
##### 3.0976 x 106 Square Yards (yd²)
##### 2.58999 x 108 Are (а)
##### 25899.9 Square decimeter (dm²)
##### 2.58999 x 10-25 Barn (b)
##### 2.58999 x 109 Decar (daa)
##### 2.62038 x 109 Rood (ro)
Square Yards
1 Square Yards (yd²) =
##### 0.000206612 Acre (ac)
##### 8.36127 x 10-5 Hectare (ha)
##### 8361.27 Square Centimeter (cm²)
##### 9 Square Feet (ft²)
##### 8.36127 x 10-7 Square Kilometer (km²)
##### 0.836127 Square Meter (m²)
##### 3.22831 x 10-7 Square Miles (mi²)
##### 83.6127 Are (а)
##### 0.00836127 Square decimeter (dm²)
##### 8.36127 x 10-32 Barn (b)
##### 836.127 Decar (daa)
##### 845.94 Rood (ro)
Are
1 Are (а) =
##### 2.47105 x 10-6 Acre (ac)
##### 1.0 x 10-6 Hectare (ha)
##### 100 Square Centimeter (cm²)
##### 0.107639 Square Feet (ft²)
##### 1.0 x 10-8 Square Kilometer (km²)
##### 0.01 Square Meter (m²)
##### 3.86102 x 10-9 Square Miles (mi²)
##### 0.0119599 Square Yards (yd²)
##### 0.0001 Square decimeter (dm²)
##### 1.0 x 10-33 Barn (b)
##### 10 Decar (daa)
##### 10.1174 Rood (ro)
Square decimeter
1 Square decimeter (dm²) =
##### 0.0247105 Acre (ac)
##### 0.01 Hectare (ha)
##### 1.0 x 106 Square Centimeter (cm²)
##### 1076.39 Square Feet (ft²)
##### 0.0001 Square Kilometer (km²)
##### 100 Square Meter (m²)
##### 3.86102 x 10-5 Square Miles (mi²)
##### 119.599 Square Yards (yd²)
##### 10000 Are (а)
##### 1.0 x 10-29 Barn (b)
##### 100000 Decar (daa)
##### 101174 Rood (ro)
Barn
1 Barn (b) =
##### 2.47105 x 1027 Acre (ac)
##### 1.0 x 1027 Hectare (ha)
##### 1.0 x 1035 Square Centimeter (cm²)
##### 1.07639 x 1032 Square Feet (ft²)
##### 1.0 x 1025 Square Kilometer (km²)
##### 1.0 x 1031 Square Meter (m²)
##### 3.86102 x 1024 Square Miles (mi²)
##### 1.19599 x 1031 Square Yards (yd²)
##### 1.0 x 1033 Are (а)
##### 1.0 x 1029 Square decimeter (dm²)
##### 1.0 x 1034 Decar (daa)
##### 1.01174 x 1034 Rood (ro)
Decar
1 Decar (daa) =
##### 2.47105 x 10-7 Acre (ac)
##### 1.0 x 10-7 Hectare (ha)
##### 10 Square Centimeter (cm²)
##### 0.0107639 Square Feet (ft²)
##### 1.0 x 10-9 Square Kilometer (km²)
##### 0.001 Square Meter (m²)
##### 3.86102 x 10-10 Square Miles (mi²)
##### 0.00119599 Square Yards (yd²)
##### 0.1 Are (а)
##### 1.0 x 10-5 Square decimeter (dm²)
##### 1.0 x 10-34 Barn (b)
##### 1.01174 Rood (ro)
Rood
1 Rood (ro) =
##### 2.44239 x 10-7 Acre (ac)
##### 9.884 x 10-8 Hectare (ha)
##### 9.884 Square Centimeter (cm²)
##### 0.010639 Square Feet (ft²)
##### 9.884 x 10-10 Square Kilometer (km²)
##### 0.0009884 Square Meter (m²)
##### 3.81623 x 10-10 Square Miles (mi²)
##### 0.00118212 Square Yards (yd²)
##### 0.09884 Are (а)
##### 9.884 x 10-6 Square decimeter (dm²)
##### 9.884 x 10-35 Barn (b)
##### 0.9884 Decar (daa) |
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Biology for AP® Courses
6.5 Enzymes
Biology for AP® Courses6.5 Enzymes
Search for key terms or text.
Learning Objectives
In this section, you will explore the following questions:
What is the role of enzymes in metabolic pathways?
How do enzymes function as molecular catalysts?
Connection for AP® Courses
Many chemical reactions in cells occur spontaneously, but happen too slowly to meet the needs of a cell. For example, a teaspoon of sucrose (table sugar), a disaccharide, in a glass of iced tea will take time to break down into two monosaccharides, glucose and fructose; however, if you add a small amount of the enzyme sucrase to the tea, sucrose breaks down almost immediately. Sucrase is an example of an enzyme, a type of biological catalyst. Enzymes are macromolecules—most often proteins—that speed up chemical reactions by lowering activation energy barriers. Enzymes are very specific for the reactions they catalyze; because they are polypeptides, enzymes can have a variety of shapes attributed to interactions among amino acid R-groups. One part of the enzyme, the active site, interacts with the substrate via the induced fit model of interaction. Substrate binding alters the shape of the enzyme to facilitate the chemical reaction in several different ways, including bringing substrates together in an optimal orientation. After the reaction finishes, the product(s) are released, and the active site returns to its original shape.
Enzyme activity, and thus the rate of an enzyme-catalyzed reaction, is regulated by environmental conditions, including the amount of substrate, temperature, pH, and the presence of coenzymes, cofactors, activators, and inhibitors. Inhibitors, coenzymes, and cofactors can act competitively by binding to the enzyme’s active site, or noncompetitively by binding to the enzyme’s allosteric site. An allosteric site is an alternate part of the enzyme that can bind to non–substrate molecules. Enzymes work most efficiently under optimal conditions that are specific to the enzyme. For example, trypsin, an enzyme in the human small intestine, works most efficiently at pH 8, whereas pepsin in the stomach works best under acidic conditions. Sometimes environmental factors, especially low pH and high temperatures, alter the shape of the active site; if the shape cannot be restored, the enzyme denatures. The most common method of enzyme regulation in metabolic pathways is via feedback inhibition.
How can various factors, such as feedback inhibition, regulate enzyme activity?
Information presented and the examples highlighted in the section support concepts and Learning Objectives outlined in Big Idea 4 of the AP® Biology Curriculum Framework. The learning objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® Exam questions. A Learning Objective merges required content with one or more of the seven science practices.
| | |
--- |
| Big Idea 4 | Biological systems interact, and these systems and their interactions possess complex properties. |
| Enduring Understanding 4.B | Competition and cooperation are important aspects of biological systems. |
| Essential Knowledge | 4.B.1 Interactions between molecules affect their structure and function. |
| Science Practice | 5.1 The student can analyze data to identify patterns or relationships. |
| Learning Objective | 4.17 The student is able to analyze data to identify how molecular interactions affect structure and function. |
Teacher Support
The idea that enzymes help chemical reactions to occur, but do not take part in the chemical reaction and are not changed by it can be confusing. Stress that an enzyme and substrate do not covalently bind to each other and the association is temporary. Figure 6.16 is useful in illustrating enzyme function. If two compounds are to be joined into one during the reaction, and they would anyway if left alone long enough, the enzyme molecule brings them close enough for the reaction to occur faster. If a large molecule is to be split into smaller units, the enzyme stresses the molecule and makes it easier for the covalent bonds holding the molecule to break. In both cases, the enzyme molecule subtlety changes its shape after attaching to the substrate (s). This creates an intermediate phase of the reaction and an enzyme-substrate complex. When the reaction is complete and the product(s) disassociate, the enzyme returns to its original shape.
The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards:[APLO 2.15][APLO 4.8][APLO 2.16]
A substance that helps a chemical reaction to occur is a catalyst, and the special molecules that catalyze biochemical reactions are called enzymes. Almost all enzymes are proteins, made up of chains of amino acids, and they perform the critical task of lowering the activation energies of chemical reactions inside the cell. Enzymes do this by binding to the reactant molecules, and holding them in such a way as to make the chemical bond-breaking and bond-forming processes take place more readily. It is important to remember that enzymes don’t change the ∆G of a reaction. In other words, they don’t change whether a reaction is exergonic (spontaneous) or endergonic. This is because they don’t change the free energy of the reactants or products. They only reduce the activation energy required to reach the transition state (Figure 6.15).
Figure 6.15 Enzymes lower the activation energy of the reaction but do not change the free energy of the reaction.
Enzyme Active Site and Substrate Specificity
The chemical reactants to which an enzyme binds are the enzyme’s substrates. There may be one or more substrates, depending on the particular chemical reaction. In some reactions, a single-reactant substrate is broken down into multiple products. In others, two substrates may come together to create one larger molecule. Two reactants might also enter a reaction, both become modified, and leave the reaction as two products. The location within the enzyme where the substrate binds is called the enzyme’s active site. The active site is where the “action” happens, so to speak. Since enzymes are proteins, there is a unique combination of amino acid residues (also called side chains, or R groups) within the active site. Each residue is characterized by different properties. Residues can be large or small, weakly acidic or basic, hydrophilic or hydrophobic, positively or negatively charged, or neutral. The unique combination of amino acid residues, their positions, sequences, structures, and properties, creates a very specific chemical environment within the active site. This specific environment is suited to bind, albeit briefly, to a specific chemical substrate (or substrates). Due to this jigsaw puzzle-like match between an enzyme and its substrates (which adapts to find the best fit between the transition state and the active site), enzymes are known for their specificity. The “best fit” results from the shape and the amino acid functional group’s attraction to the substrate. There is a specifically matched enzyme for each substrate and, thus, for each chemical reaction; however, there is flexibility as well.
The fact that active sites are so perfectly suited to provide specific environmental conditions also means that they are subject to influences by the local environment. It is true that increasing the environmental temperature generally increases reaction rates, enzyme-catalyzed or otherwise. However, increasing or decreasing the temperature outside of an optimal range can affect chemical bonds within the active site in such a way that they are less well suited to bind substrates. High temperatures will eventually cause enzymes, like other biological molecules, to denature, a process that changes the natural properties of a substance. Likewise, the pH of the local environment can also affect enzyme function. Active site amino acid residues have their own acidic or basic properties that are optimal for catalysis. These residues are sensitive to changes in pH that can impair the way substrate molecules bind. Enzymes are suited to function best within a certain pH range, and, as with temperature, extreme pH values (acidic or basic) of the environment can cause enzymes to denature.
Induced Fit and Enzyme Function
For many years, scientists thought that enzyme-substrate binding took place in a simple “lock-and-key” fashion. This model asserted that the enzyme and substrate fit together perfectly in one instantaneous step. However, current research supports a more refined view called induced fit (Figure 6.16). The induced-fit model expands upon the lock-and-key model by describing a more dynamic interaction between enzyme and substrate. As the enzyme and substrate come together, their interaction causes a mild shift in the enzyme’s structure that confirms an ideal binding arrangement between the enzyme and the transition state of the substrate. This ideal binding maximizes the enzyme’s ability to catalyze its reaction.
Link to Learning
View an animation of induced fit at this website.
Phosphofructokinase (PFK) deficiency is a genetic disorder that occurs when a person lacks PFK, an enzyme needed to perform glycolysis in skeletal muscles. Apply your understanding of both enzyme function and glycolysis to predict the effect that PFK deficiency could have on the body.
Production of energy by glycolysis will occur more slowly than normal; skeletal muscles will function properly.
Production of energy by glycolysis will not occur; skeletal muscles will function properly.
Production of energy by glycolysis will occur more erratically than normal; skeletal muscles will not function properly.
Production of energy by glycolysis will not occur; skeletal muscles will not function properly.
When an enzyme binds its substrate, an enzyme-substrate complex is formed. This complex lowers the activation energy of the reaction and promotes its rapid progression in one of many ways. On a basic level, enzymes promote chemical reactions that involve more than one substrate by bringing the substrates together in an optimal orientation. The appropriate region (atoms and bonds) of one molecule is juxtaposed to the appropriate region of the other molecule with which it must react. Another way in which enzymes promote the reaction of their substrates is by creating an optimal environment within the active site for the reaction to occur. Certain chemical reactions might proceed best in a slightly acidic or non-polar environment. The chemical properties that emerge from the particular arrangement of amino acid residues within an active site create the perfect environment for an enzyme’s specific substrates to react.
You’ve learned that the activation energy required for many reactions includes the energy involved in manipulating or slightly contorting chemical bonds so that they can easily break and allow others to reform. Enzymatic action can aid this process. The enzyme-substrate complex can lower the activation energy by contorting substrate molecules in such a way as to facilitate bond-breaking, helping to reach the transition state. Finally, enzymes can also lower activation energies by taking part in the chemical reaction itself. The amino acid residues can provide certain ions or chemical groups that actually form covalent bonds with substrate molecules as a necessary step of the reaction process. In these cases, it is important to remember that the enzyme will always return to its original state at the completion of the reaction. One of the hallmark properties of enzymes is that they remain ultimately unchanged by the reactions they catalyze. After an enzyme is done catalyzing a reaction, it releases its product(s).
Figure 6.16 According to the induced-fit model, both enzyme and substrate undergo dynamic conformational changes upon binding. The enzyme contorts the substrate into its transition state, thereby increasing the rate of the reaction.
Science Practice Connection for AP® Courses
Think About It
AP Biology Investigation 13: Enzyme Activity. This investigation allows you to design and conduct experiments to explore the effects of environmental variables, such as temperature and pH, on the rates of enzymatic reactions.
Teacher Support
This lab investigation is an application of LO 4.17 and Science Practice 5.1 because you will analyze experimental data to determine how various environment conditions affect enzyme structure and function and, thus, the rate of enzyme-catalyzed reactions.
An expanded lab investigation for enzymes, involving determining the effect of pH on the action of turnip peroxidase, is available from the College Board’s ® AP Biology Investigative Labs: An Inquiry-Based Approach, Investigation 13.
Control of Metabolism Through Enzyme Regulation
It would seem ideal to have a scenario in which all of the enzymes encoded in an organism’s genome existed in abundant supply and functioned optimally under all cellular conditions, in all cells, at all times. In reality, this is far from the case. A variety of mechanisms ensure that this does not happen. Cellular needs and conditions vary from cell to cell, and change within individual cells over time. The required enzymes and energetic demands of stomach cells are different from those of fat storage cells, skin cells, blood cells, and nerve cells. Furthermore, a digestive cell works much harder to process and break down nutrients during the time that closely follows a meal compared with many hours after a meal. As these cellular demands and conditions vary, so do the amounts and functionality of different enzymes.
Since the rates of biochemical reactions are controlled by activation energy, and enzymes lower and determine activation energies for chemical reactions, the relative amounts and functioning of the variety of enzymes within a cell ultimately determine which reactions will proceed and at which rates. This determination is tightly controlled. In certain cellular environments, enzyme activity is partly controlled by environmental factors, like pH and temperature. There are other mechanisms through which cells control the activity of enzymes and determine the rates at which various biochemical reactions will occur.
Regulation of Enzymes by Molecules
Enzymes can be regulated in ways that either promote or reduce their activity. There are many different kinds of molecules that inhibit or promote enzyme function, and various mechanisms exist for doing so. In some cases of enzyme inhibition, for example, an inhibitor molecule is similar enough to a substrate that it can bind to the active site and simply block the substrate from binding. When this happens, the enzyme is inhibited through competitive inhibition, because an inhibitor molecule competes with the substrate for active site binding (Figure 6.17). On the other hand, in noncompetitive inhibition, an inhibitor molecule binds to the enzyme in a location other than the active site, called an allosteric site, but still manages to prevent substrate binding to the active site. Some inhibitor molecules bind to enzymes in a location where their binding induces a conformational change that reduces the enzyme activity as it no longer effectively catalyzes the conversion of the substrate to product.
Figure 6.17 Competitive and noncompetitive inhibition affect the rate of reaction differently. Competitive inhibitors affect the initial rate but do not affect the maximal rate, whereas noncompetitive inhibitors affect the maximal rate.
Some inhibitor molecules bind to enzymes in a location where their binding induces a conformational change that reduces the affinity of the enzyme for its substrate. This type of inhibition is called allosteric inhibition (Figure 6.18). Most allosterically regulated enzymes are made up of more than one polypeptide, meaning that they have more than one protein subunit. When an allosteric inhibitor binds to an enzyme, all active sites on the protein subunits are changed slightly such that they bind their substrates with less efficiency. There are allosteric activators as well as inhibitors. Allosteric activators bind to locations on an enzyme away from the active site, inducing a conformational change that increases the affinity of the enzyme’s active site(s) for its substrate(s).
Figure 6.18 Allosteric inhibitors modify the active site of the enzyme so that substrate binding is reduced or prevented. In contrast, allosteric activators modify the active site of the enzyme so that the affinity for the substrate increases.
Everyday Connection
Drug Discovery by Looking for Inhibitors of Key Enzymes in Specific Pathways
Figure 6.19 Have you ever wondered how pharmaceutical drugs are developed? (credit: Deborah Austin)
Enzymes are key components of metabolic pathways. Understanding how enzymes work and how they can be regulated is a key principle behind the development of many of the pharmaceutical drugs (Figure 6.19) on the market today. Biologists working in this field collaborate with other scientists, usually chemists, to design drugs.
Consider statins for example—which is the name given to the class of drugs that reduces cholesterol levels. These compounds are essentially inhibitors of the enzyme HMG-CoA reductase. HMG-CoA reductase is the enzyme that synthesizes cholesterol from lipids in the body. By inhibiting this enzyme, the levels of cholesterol synthesized in the body can be reduced. Similarly, acetaminophen is an inhibitor of the enzyme cyclooxygenase. While it is effective in providing relief from fever and inflammation (pain), its mechanism of action is still not completely understood.
How are drugs developed? One of the first challenges in drug development is identifying the specific molecule that the drug is intended to target. In the case of statins, HMG-CoA reductase is the drug target. Drug targets are identified through painstaking research in the laboratory. Identifying the target alone is not sufficient; scientists also need to know how the target acts inside the cell and which reactions go awry in the case of disease. Once the target and the pathway are identified, then the actual process of drug design begins. During this stage, chemists and biologists work together to design and synthesize molecules that can either block or activate a particular reaction. However, this is only the beginning: both if and when a drug prototype is successful in performing its function, then it must undergo many tests from in vitro experiments to clinical trials before it can get FDA approval to be on the market.
Statins reduce the level of cholesterol in the blood. Based on the everyday connection, which of the following might also reduce cholesterol levels in the blood?
a drug that increases HMG-CoA reductase levels
a drug that reduces cyclooxygenase levels
a drug that reduces lipid levels in the body
a drug that blocks the action of acetaminophen
Many enzymes don’t work optimally, or even at all, unless bound to other specific non-protein helper molecules, either temporarily through ionic or hydrogen bonds or permanently through stronger covalent bonds. Two types of helper molecules are cofactors and coenzymes. Binding to these molecules promotes optimal conformation and function for their respective enzymes. Cofactors are inorganic ions such as iron (Fe++) and magnesium (Mg++). One example of an enzyme that requires a metal ion as a cofactor is the enzyme that builds DNA molecules, DNA polymerase, which requires a bound zinc ion (Zn++) to function. Coenzymes are organic helper molecules, with a basic atomic structure made up of carbon and hydrogen, which are required for enzyme action. The most common sources of coenzymes are dietary vitamins (Figure 6.20). Some vitamins are precursors to coenzymes and others act directly as coenzymes. Vitamin C is a coenzyme for multiple enzymes that take part in building the important connective tissue component, collagen. An important step in the breakdown of glucose to yield energy is catalysis by a multi-enzyme complex called pyruvate dehydrogenase. Pyruvate dehydrogenase is a complex of several enzymes that actually requires one cofactor (a magnesium ion) and five different organic coenzymes to catalyze its specific chemical reaction. Therefore, enzyme function is, in part, regulated by an abundance of various cofactors and coenzymes, which are supplied primarily by the diets of most organisms.
Figure 6.20 Vitamins are important coenzymes or precursors of coenzymes, and are required for enzymes to function properly. Multivitamin capsules usually contain mixtures of all the vitamins at different percentages.
Enzyme Compartmentalization
In eukaryotic cells, molecules such as enzymes are usually compartmentalized into different organelles. This allows for yet another level of regulation of enzyme activity. Enzymes required only for certain cellular processes can be housed separately along with their substrates, allowing for more efficient chemical reactions. Examples of this sort of enzyme regulation based on location and proximity include the enzymes involved in the latter stages of cellular respiration, which take place exclusively in the mitochondria, and the enzymes involved in the digestion of cellular debris and foreign materials, located within lysosomes.
Feedback Inhibition in Metabolic Pathways
Molecules can regulate enzyme function in many ways. A major question remains, however: What are these molecules and where do they come from? Some are cofactors and coenzymes, ions, and organic molecules, as you’ve learned. What other molecules in the cell provide enzymatic regulation, such as allosteric modulation, and competitive and noncompetitive inhibition? The answer is that a wide variety of molecules can perform these roles. Some of these molecules include pharmaceutical and non-pharmaceutical drugs, toxins, and poisons from the environment. Perhaps the most relevant sources of enzyme regulatory molecules, with respect to cellular metabolism, are the products of the cellular metabolic reactions themselves. In a most efficient and elegant way, cells have evolved to use the products of their own reactions for feedback inhibition of enzyme activity. Feedback inhibition involves the use of a reaction product to regulate its own further production (Figure 6.21). The cell responds to the abundance of specific products by slowing down production during anabolic or catabolic reactions. Such reaction products may inhibit the enzymes that catalyzed their production through the mechanisms described above.
Figure 6.21 Metabolic pathways are a series of reactions catalyzed by multiple enzymes. Feedback inhibition, where the end product of the pathway inhibits an upstream step, is an important regulatory mechanism in cells.
The production of both amino acids and nucleotides is controlled through feedback inhibition. Additionally, ATP is an allosteric regulator of some of the enzymes involved in the catabolic breakdown of sugar, the process that produces ATP. In this way, when ATP is abundant, the cell can prevent its further production. Remember that ATP is an unstable molecule that can spontaneously dissociate into ADP. If too much ATP were present in a cell, much of it would go to waste. On the other hand, ADP serves as a positive allosteric regulator (an allosteric activator) for some of the same enzymes that are inhibited by ATP. Thus, when relative levels of ADP are high compared to ATP, the cell is triggered to produce more ATP through the catabolism of sugar.
Teacher Support
Ask students which inhibition is more effective at slowing or limiting the reaction? Relate this to the examples available and discuss why these would be used in specific instances.
Have the class research antimicrobial treatments that are based on enzyme inhibition, not on the administration of traditional antibiotics.
Enzymes are not changed by the chemicals they facilitate; therefore, they can be used repeatedly. Yet, how do you keep them from catalyzing reactions when you do not need or want them to react anymore? If enzymes could not be controlled, the reactions would continue until the substrates were depleted, which is not a good situation for a living organism. Competitive and noncompetitive inhibition explains the control of enzyme activity. Research several examples of both in living organisms and explain why they are necessary. Amino acid production is one useful example. Amino acids are required for protein production, but too high a level of any amino acid is toxic, so the pathways must be controlled. Use the feedback inhibition of several pathways as examples.
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Power generated from nuclear reactions
"Atomic power" redirects here. For the film, see Atomic Power (film).
For countries with the power or ability to project nuclear weapons, see List of states with nuclear weapons.
Nuclear power is the use of nuclear reactions to produce electricity. Nuclear power can be obtained from nuclear fission, nuclear decay and nuclear fusion reactions. Presently, the vast majority of electricity from nuclear power is produced by nuclear fission of uranium and plutonium in nuclear power plants. Nuclear decay processes are used in niche applications such as radioisotope thermoelectric generators in some space probes such as Voyager 2. Reactors producing controlled fusion power have been operated since 1958 but have yet to generate net power and are not expected to be commercially available in the near future.
The first nuclear power plant was built in the 1950s. The global installed nuclear capacity grew to 100 GW in the late 1970s, and then expanded during the 1980s, reaching 300 GW by 1990. The 1979 Three Mile Island accident in the United States and the 1986 Chernobyl disaster in the Soviet Union resulted in increased regulation and public opposition to nuclear power plants. Nuclear power plants supplied 2,602 terawatt hours (TWh) of electricity in 2023, equivalent to about 9% of global electricity generation, and were the second largest low-carbon power source after hydroelectricity. As of November 2024,[update] there are 415 civilian fission reactors in the world, with overall capacity of 374 GW, 66 under construction and 87 planned, with a combined capacity of 72 GW and 84 GW, respectively. The United States has the largest fleet of nuclear reactors, generating almost 800 TWh per year with an average capacity factor of 92%. The average global capacity factor is 89%. Most new reactors under construction are generation III reactors in Asia.
Nuclear power is a safe, sustainable energy source that reduces carbon emissions. This is because nuclear power generation causes one of the lowest levels of fatalities per unit of energy generated compared to other energy sources. "Economists estimate that each nuclear plant built could save more than 800,000 life years." Coal, petroleum, natural gas and hydroelectricity have each caused more fatalities per unit of energy due to air pollution and accidents. Nuclear power plants also emit no greenhouse gases and result in less life-cycle carbon emissions than common sources of renewable energy. The radiological hazards associated with nuclear power are the primary motivations of the anti-nuclear movement, which contends that nuclear power poses threats to people and the environment, citing the potential for accidents like the Fukushima nuclear disaster in Japan in 2011, and is too expensive to deploy when compared to alternative sustainable energy sources.
History
Main article: History of nuclear power
Origins
The process of nuclear fission was discovered in 1938 after over four decades of work on the science of radioactivity and the elaboration of new nuclear physics that described the components of atoms. Soon after the discovery of the fission process, it was realized that neutrons released by a fissioning nucleus could, under the right conditions, induce fissions in nearby nuclei, thus initiating a self-sustaining chain reaction. Once this was experimentally confirmed in 1939, scientists in many countries petitioned their governments for support for nuclear fission research, just on the cusp of World War II, in order to develop a nuclear weapon.
In the United States, these research efforts led to the creation of the first human-made nuclear reactor, the Chicago Pile-1 under the Stagg Field stadium at the University of Chicago, which achieved criticality on December 2, 1942. The reactor's development was part of the Manhattan Project, the Allied effort to create atomic bombs during World War II. It led to the building of larger single-purpose production reactors for the production of weapons-grade plutonium for use in the first nuclear weapons. The United States tested the first nuclear weapon in July 1945, the Trinity test, and the atomic bombings of Hiroshima and Nagasaki happened one month later.
Despite the military nature of the first nuclear devices, there was strong optimism in the 1940s and 1950s that nuclear power could provide cheap and endless energy. Electricity was generated for the first time by a nuclear reactor on December 20, 1951, at the EBR-I experimental station near Arco, Idaho, which initially produced about 100 kW. In 1953, American President Dwight Eisenhower gave his "Atoms for Peace" speech at the United Nations, emphasizing the need to develop "peaceful" uses of nuclear power quickly. This was followed by the Atomic Energy Act of 1954 which allowed rapid declassification of U.S. reactor technology and encouraged development by the private sector.
First power generation
The first organization to develop practical nuclear power was the U.S. Navy, with the S1W reactor for the purpose of propelling submarines and aircraft carriers. The first nuclear-powered submarine, USS Nautilus, was put to sea in January 1954. The S1W reactor was a pressurized water reactor. This design was chosen because it was simpler, more compact, and easier to operate compared to alternative designs, thus more suitable to be used in submarines. This decision would result in the PWR being the reactor of choice also for power generation, thus having a lasting impact on the civilian electricity market in the years to come.
On June 27, 1954, the Obninsk Nuclear Power Plant in the USSR became the world's first nuclear power plant to generate electricity for a power grid, producing around 5 megawatts of electric power. The world's first commercial nuclear power station, Calder Hall at Windscale, England was connected to the national power grid on 27 August 1956. In common with a number of other generation I reactors, the plant had the dual purpose of producing electricity and plutonium-239, the latter for the nascent nuclear weapons program in Britain.
Expansion and first opposition
The total global installed nuclear capacity initially rose relatively quickly, rising from less than 1 gigawatt (GW) in 1960 to 100 GW in the late 1970s. During the 1970s and 1980s rising economic costs (related to extended construction times largely due to regulatory changes and pressure-group litigation) and falling fossil fuel prices made nuclear power plants then under construction less attractive. In the 1980s in the U.S. and 1990s in Europe, the flat electric grid growth and electricity liberalization also made the addition of large new baseload energy generators economically unattractive.[citation needed]
The 1973 oil crisis had a significant effect on countries, such as France and Japan, which had relied more heavily on oil for electric generation to invest in nuclear power. France would construct 25 nuclear power plants over the next 15 years, and as of 2019, 71% of French electricity was generated by nuclear power, the highest percentage by any nation in the world.
Some local opposition to nuclear power emerged in the United States in the early 1960s. In the late 1960s, some members of the scientific community began to express pointed concerns. These anti-nuclear concerns related to nuclear accidents, nuclear proliferation, nuclear terrorism and radioactive waste disposal. In the early 1970s, there were large protests about a proposed nuclear power plant in Wyhl, Germany. The project was cancelled in 1975. The anti-nuclear success at Wyhl inspired opposition to nuclear power in other parts of Europe and North America.
By the mid-1970s anti-nuclear activism gained a wider appeal and influence, and nuclear power began to become an issue of major public protest. In some countries, the nuclear power conflict "reached an intensity unprecedented in the history of technology controversies". The increased public hostility to nuclear power led to a longer license procurement process, more regulations and increased requirements for safety equipment, which made new construction much more expensive. In the United States, over 120 Light Water Reactor proposals were ultimately cancelled and the construction of new reactors ground to a halt. The 1979 accident at Three Mile Island with no fatalities, played a major part in the reduction in the number of new plant constructions in many countries.
Chernobyl and renaissance
During the 1980s one new nuclear reactor started up every 17 days on average. By the end of the decade, global installed nuclear capacity reached 300 GW. Since the late 1980s, new capacity additions slowed significantly, with the installed nuclear capacity reaching 365 GW in 2005.
The 1986 Chernobyl disaster in the USSR, involving an RBMK reactor, altered the development of nuclear power and led to a greater focus on meeting international safety and regulatory standards. It is considered the worst nuclear disaster in history both in total casualties, with 56 direct deaths, and financially, with the cleanup and the cost estimated at 18 billion Rbls (US$68 billion in 2019, adjusted for inflation). The international organization to promote safety awareness and the professional development of operators in nuclear facilities, the World Association of Nuclear Operators (WANO), was created as a direct outcome of the 1986 Chernobyl accident. The Chernobyl disaster played a major part in the reduction in the number of new plant constructions in the following years. Influenced by these events, Italy voted against nuclear power in a 1987 referendum, becoming the first major economy to completely phase out nuclear power in 1990.
In the early 2000s, nuclear energy was expecting a nuclear renaissance, an increase in the construction of new reactors, due to concerns about carbon dioxide emissions. During this period, newer generation III reactors, such as the EPR began construction.
Net electrical generation by source and growth from 1980. In terms of energy generated between 1980 and 2010, the contribution from fission grew the fastest.
Electricity production in France, showing the shift to nuclear power.
thermofossil
hydroelectric
nuclear
Other renewables
The rate of new reactor constructions essentially halted in the late 1980s. Increased capacity factor in existing reactors was primarily responsible for the continuing increase in electrical energy produced during this period.
Electricity generation trends in the top producing countries (Our World in Data)
Fukushima accident
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Nuclear power generation (TWh) and operational nuclear reactors since 1997
Prospects of a nuclear renaissance were delayed by another nuclear accident. The 2011 Fukushima Daiichi nuclear accident was caused by the Tōhoku earthquake and tsunami, one of the largest earthquakes ever recorded. The Fukushima Daiichi Nuclear Power Plant suffered three core meltdowns due to failure of the emergency cooling system for lack of electricity supply. This resulted in the most serious nuclear accident since the Chernobyl disaster.[citation needed]
The accident prompted a re-examination of nuclear safety and nuclear energy policy in many countries. Germany approved plans to close all its reactors by 2022, and many other countries reviewed their nuclear power programs.
Following the disaster, Japan shut down all of its nuclear power reactors, some of them permanently, and in 2015 began a gradual process to restart the remaining 40 reactors, following safety checks and based on revised criteria for operations and public approval.
In 2022, the Japanese government, under the leadership of Prime Minister Fumio Kishida, declared that 10 more nuclear power plants were to be reopened since the 2011 disaster. Kishida is also pushing for research and construction of new safer nuclear plants to safeguard Japanese consumers from the fluctuating price of the fossil fuel market and reduce Japan's greenhouse gas emissions. Kishida intends to have Japan become a significant exporter of nuclear energy and technology to developing countries around the world.
Current prospects
By 2015, the IAEA's outlook for nuclear energy had become more promising, recognizing the importance of low-carbon generation for mitigating climate change. As of 2015[update], the global trend was for new nuclear power stations coming online to be balanced by the number of old plants being retired. In 2016, the U.S. Energy Information Administration projected for its "base case" that world nuclear power generation would increase from 2,344 terawatt hours (TWh) in 2012 to 4,500 TWh in 2040. Most of the predicted increase was expected to be in Asia. As of 2018, there were over 150 nuclear reactors planned including 50 under construction. In January 2019, China had 45 reactors in operation, 13 under construction, and planned to build 43 more, which would make it the world's largest generator of nuclear electricity. As of 2021, 17 reactors were reported to be under construction. Its share of electricity from nuclear power was 5% in 2019.
In October 2021, the Japanese cabinet approved the new Plan for Electricity Generation to 2030 prepared by the Agency for Natural Resources and Energy (ANRE) and an advisory committee, following public consultation. The nuclear target for 2030 requires the restart of another ten reactors. Prime Minister Fumio Kishida in July 2022 announced that the country should consider building advanced reactors and extending operating licences beyond 60 years.
As of 2022, with world oil and gas prices on the rise, while Germany is restarting its coal plants to deal with loss of Russian gas that it needs to supplement its Energiewende, many other countries have announced ambitious plans to reinvigorate ageing nuclear generating capacity with new investments. French President Emmanuel Macron announced his intention to build six new reactors in coming decades, placing nuclear at the heart of France's drive for carbon neutrality by 2050. Meanwhile, in the United States, the Department of Energy, in collaboration with commercial entities, TerraPower and X-energy, is planning on building two different advanced nuclear reactors by 2027, with further plans for nuclear implementation in its long term green energy and energy security goals.
Power plants
Number of electricity-generating civilian reactors by type as of 2014
PWR 277 (63.2%)
BWR 80 (18.3%)
GCR 15 (3.42%)
PHWR 49 (11.2%)
LWGR 15 (3.42%)
FBR 2 (0.46%)
Main articles: Nuclear power plant and Nuclear reactor
See also: List of commercial nuclear reactors and List of nuclear power stations
Nuclear power plants are thermal power stations that generate electricity by harnessing the thermal energy released from nuclear fission. A fission nuclear power plant is generally composed of: a nuclear reactor, in which the nuclear reactions generating heat take place; a cooling system, which removes the heat from inside the reactor; a steam turbine, which transforms the heat into mechanical energy; an electric generator, which transforms the mechanical energy into electrical energy.
When a neutron hits the nucleus of a uranium-235 or plutonium atom, it can split the nucleus into two smaller nuclei, which is a nuclear fission reaction. The reaction releases energy and neutrons. The released neutrons can hit other uranium or plutonium nuclei, causing new fission reactions, which release more energy and more neutrons. This is called a chain reaction. In most commercial reactors, the reaction rate is contained by control rods that absorb excess neutrons. The controllability of nuclear reactors depends on the fact that a small fraction of neutrons resulting from fission are delayed. The time delay between the fission and the release of the neutrons slows changes in reaction rates and gives time for moving the control rods to adjust the reaction rate.
Fuel cycle
Main articles: Nuclear fuel cycle and Integrated Nuclear Fuel Cycle Information System
The life cycle of nuclear fuel starts with uranium mining. The uranium ore is then converted into a compact ore concentrate form, known as yellowcake (U3O8), to facilitate transport. Fission reactors generally need uranium-235, a fissile isotope of uranium. The concentration of uranium-235 in natural uranium is low (about 0.7%). Some reactors can use this natural uranium as fuel, depending on their neutron economy. These reactors generally have graphite or heavy water moderators. For light water reactors, the most common type of reactor, this concentration is too low, and it must be increased by a process called uranium enrichment. In civilian light water reactors, uranium is typically enriched to 3.5–5% uranium-235. The uranium is then generally converted into uranium oxide (UO2), a ceramic, that is then compressively sintered into fuel pellets, a stack of which forms fuel rods of the proper composition and geometry for the particular reactor.
After some time in the reactor, the fuel will have reduced fissile material and increased fission products, until its use becomes impractical. At this point, the spent fuel will be moved to a spent fuel pool which provides cooling for the thermal heat and shielding for ionizing radiation. After several months or years, the spent fuel is radioactively and thermally cool enough to be moved to dry storage casks or reprocessed.
Uranium resources
Main articles: Uranium market, Uranium mining, and Energy development § Nuclear
Uranium is a fairly common element in the Earth's crust: it is approximately as common as tin or germanium, and is about 40 times more common than silver. Uranium is present in trace concentrations in most rocks, dirt, and ocean water, but is generally economically extracted only where it is present in relatively high concentrations. As of 2011 the world's known resources of uranium, economically recoverable at the arbitrary price ceiling of US$130/kg, were enough to last for between 70 and 100 years in current reactors.[failed verification]
Light water reactors (which account for almost all operational reactors) make relatively inefficient use of nuclear fuel, mostly using only the very rare uranium-235 isotope.
Limited uranium-235 supply may inhibit substantial expansion with the current nuclear technology.
Nuclear reprocessing can make this waste reusable, and newer reactors also achieve a more efficient use of the available resources than older ones.
More advanced nuclear reactor technologies, such as fast reactors, can use much more of the natural uranium, use current nuclear waste as fuel, as well as creating new fuel out of non-fissile material (see breeder reactor).
With a pure fast reactor fuel cycle with a burn up of all the uranium and actinides (which presently make up the most hazardous substances in nuclear waste), there is an estimated 160,000 years worth of uranium in total conventional resources and phosphate ore at the price of 60–100 US$/kg.
These advanced fuel cycles and nuclear reprocessing are currently not widely used because the price of uranium is very low compared to the cost of nuclear plants, so it's more economically viable to mine new uranium rather than reprocess it.
Nuclear reprocessing also carries higher risk of nuclear proliferation, as it separates material that can be used to manufacture nuclear weapons.
Unconventional uranium resources also exist. Uranium is naturally present in seawater at a concentration of about 3 micrograms per liter, with 4.4 billion tons of uranium considered present in seawater at any time. In 2014 it was suggested that it would be economically competitive to produce nuclear fuel from seawater if the process was implemented at large scale. Over geological timescales, uranium extracted on an industrial scale from seawater would be replenished by both river erosion of rocks and the natural process of uranium dissolved from the surface area of the ocean floor, both of which maintain the solubility equilibria of seawater concentration at a stable level. Some commentators have argued that this strengthens the case for nuclear power to be considered a renewable energy.
Waste
Main article: Nuclear waste
The normal operation of nuclear power plants and facilities produce radioactive waste, or nuclear waste. This type of waste is also produced during plant decommissioning. There are two broad categories of nuclear waste: low-level waste and high-level waste. The first has low radioactivity and includes contaminated items such as clothing, which poses limited threat. High-level waste is mainly the spent fuel from nuclear reactors, which is very radioactive and must be cooled and then safely disposed of or reprocessed.
High-level waste
Main articles: High-level waste and Spent nuclear fuel
The most important waste stream from nuclear power reactors is spent nuclear fuel, which is considered high-level waste (HLW). For light water reactors (LWRs), spent fuel is typically composed of 95% uranium, 4% fission products, and about 1% transuranic actinides (mostly plutonium, neptunium and americium). The fission products are responsible for the bulk of the short-term radioactivity, whereas the plutonium and other transuranics are responsible for the bulk of the long-term radioactivity.
High-level waste must be stored isolated from the biosphere with sufficient shielding so as to limit radiation exposure. After being removed from the reactors, used fuel bundles are stored for six to ten years in spent fuel pools, which provide cooling and shielding against radiation. After that, the fuel is cool enough that it can be safely transferred to dry cask storage.
The radioactivity decreases exponentially with time, such that it will have decreased by 99.5% after 100 years.
The more intensely radioactive short-lived fission products (SLFPs) decay into stable elements in approximately 300 years, and after about 100,000 years, the spent fuel becomes less radioactive than natural uranium ore.
Commonly suggested methods to isolate long-lived fission product (LLFP) waste from the biosphere include separation and transmutation, synroc treatments, or deep geological storage.
Thermal-neutron reactors, which presently constitute the majority of the world fleet, cannot burn up the reactor grade plutonium that is generated during the reactor operation. This limits the life of nuclear fuel to a few years. In some countries, such as the United States, spent fuel is classified in its entirety as a nuclear waste. In other countries, such as France, it is largely reprocessed to produce a partially recycled fuel, known as mixed oxide fuel or MOX.
For spent fuel that does not undergo reprocessing, the most concerning isotopes are the medium-lived transuranic elements, which are led by reactor-grade plutonium (with a half-life 24,000 years).
Some proposed reactor designs, such as the integral fast reactor and molten salt reactors, can use as fuel the plutonium and other actinides in spent fuel from light water reactors, thanks to their fast fission spectrum. This offers a potentially more attractive alternative to deep geological disposal.
The thorium fuel cycle results in similar fission products, though creates a much smaller proportion of transuranic elements from neutron capture events within a reactor. Spent thorium fuel, although more difficult to handle than spent uranium fuel, may present somewhat lower proliferation risks.
Low-level waste
Main article: Low-level waste
The nuclear industry also produces a large volume of low-level waste, with low radioactivity, in the form of contaminated items like clothing, hand tools, water purifier resins, and (upon decommissioning) the materials of which the reactor itself is built. Low-level waste can be stored on-site until radiation levels are low enough to be disposed of as ordinary waste, or it can be sent to a low-level waste disposal site.
Waste relative to other types
See also: Radioactive waste § Naturally occurring radioactive material
In countries with nuclear power, radioactive wastes account for less than 1% of total industrial toxic wastes, much of which remains hazardous for long periods. Overall, nuclear power produces far less waste material by volume than fossil-fuel based power plants. Coal-burning plants, in particular, produce large amounts of toxic and mildly radioactive ash resulting from the concentration of naturally occurring radioactive materials in coal. A 2008 report from Oak Ridge National Laboratory concluded that coal power actually results in more radioactivity being released into the environment than nuclear power operation, and that the population effective dose equivalent from radiation from coal plants is 100 times that from the operation of nuclear plants. Although coal ash is much less radioactive than spent nuclear fuel by weight, coal ash is produced in much higher quantities per unit of energy generated. It is also released directly into the environment as fly ash, whereas nuclear plants use shielding to protect the environment from radioactive materials.
Nuclear waste volume is small compared to the energy produced. For example, at Yankee Rowe Nuclear Power Station, which generated 44 billion kilowatt hours of electricity when in service, its complete spent fuel inventory is contained within sixteen casks. It is estimated that to produce a lifetime supply of energy for a person at a western standard of living (approximately 3 GWh) would require on the order of the volume of a soda can of low enriched uranium, resulting in a similar volume of spent fuel generated.
Waste disposal
See also: List of radioactive waste treatment technologies
Following interim storage in a spent fuel pool, the bundles of used fuel rod assemblies of a typical nuclear power station are often stored on site in dry cask storage vessels.
Disposal of nuclear waste is often considered the most politically divisive aspect in the lifecycle of a nuclear power facility. The lack of movement of nuclear waste in the 2 billion year old natural nuclear fission reactors in Oklo, Gabon is cited as "a source of essential information today." Experts suggest that centralized underground repositories which are well-managed, guarded, and monitored, would be a vast improvement. There is an "international consensus on the advisability of storing nuclear waste in deep geological repositories". With the advent of new technologies, other methods including horizontal drillhole disposal into geologically inactive areas have been proposed.
There are no commercial scale purpose built underground high-level waste repositories in operation. However, in Finland the Onkalo spent nuclear fuel repository of the Olkiluoto Nuclear Power Plant was under construction as of 2015.
Reprocessing
Main article: Nuclear reprocessing
See also: Plutonium Management and Disposition Agreement
Most thermal-neutron reactors run on a once-through nuclear fuel cycle, mainly due to the low price of fresh uranium. However, many reactors are also fueled with recycled fissionable materials that remain in spent nuclear fuel. The most common fissionable material that is recycled is the reactor-grade plutonium (RGPu) that is extracted from spent fuel. It is mixed with uranium oxide and fabricated into mixed-oxide or MOX fuel. Because thermal LWRs remain the most common reactor worldwide, this type of recycling is the most common. It is considered to increase the sustainability of the nuclear fuel cycle, reduce the attractiveness of spent fuel to theft, and lower the volume of high level nuclear waste. Spent MOX fuel cannot generally be recycled for use in thermal-neutron reactors. This issue does not affect fast-neutron reactors, which are therefore preferred in order to achieve the full energy potential of the original uranium.
The main constituent of spent fuel from LWRs is slightly enriched uranium. This can be recycled into reprocessed uranium (RepU), which can be used in a fast reactor, used directly as fuel in CANDU reactors, or re-enriched for another cycle through an LWR. Re-enriching of reprocessed uranium is common in France and Russia. Reprocessed uranium is also safer in terms of nuclear proliferation potential.
Reprocessing has the potential to recover up to 95% of the uranium and plutonium fuel in spent nuclear fuel, as well as reduce long-term radioactivity within the remaining waste. However, reprocessing has been politically controversial because of the potential for nuclear proliferation and varied perceptions of increasing the vulnerability to nuclear terrorism. Reprocessing also leads to higher fuel cost compared to the once-through fuel cycle. While reprocessing reduces the volume of high-level waste, it does not reduce the fission products that are the primary causes of residual heat generation and radioactivity for the first few centuries outside the reactor. Thus, reprocessed waste still requires an almost identical treatment for the initial first few hundred years.[citation needed]
Reprocessing of civilian fuel from power reactors is currently done in France, the United Kingdom, Russia, Japan, and India. In the United States, spent nuclear fuel is currently not reprocessed. The La Hague reprocessing facility in France has operated commercially since 1976 and is responsible for half the world's reprocessing as of 2010. It produces MOX fuel from spent fuel derived from several countries. More than 32,000 tonnes of spent fuel had been reprocessed as of 2015, with the majority from France, 17% from Germany, and 9% from Japan.
Breeding
Main articles: Breeder reactor and Nuclear power proposed as renewable energy
Breeding is the process of converting non-fissile material into fissile material that can be used as nuclear fuel. The non-fissile material that can be used for this process is called fertile material, and constitute the vast majority of current nuclear waste. This breeding process occurs naturally in breeder reactors. As opposed to light water thermal-neutron reactors, which use uranium-235 (0.7% of all natural uranium), fast-neutron breeder reactors use uranium-238 (99.3% of all natural uranium) or thorium. A number of fuel cycles and breeder reactor combinations are considered to be sustainable or renewable sources of energy. In 2006 it was estimated that with seawater extraction, there was likely five billion years' worth of uranium resources for use in breeder reactors.
Breeder technology has been used in several reactors, but as of 2006, the high cost of reprocessing fuel safely requires uranium prices of more than US$200/kg before becoming justified economically. Breeder reactors are however being developed for their potential to burn all of the actinides (the most active and dangerous components) in the present inventory of nuclear waste, while also producing power and creating additional quantities of fuel for more reactors via the breeding process. As of 2017, there are two breeders producing commercial power, BN-600 reactor and the BN-800 reactor, both in Russia. The Phénix breeder reactor in France was powered down in 2009 after 36 years of operation. Both China and India are building breeder reactors. The Indian 500 MWe Prototype Fast Breeder Reactor is in the commissioning phase, with plans to build more.
Another alternative to fast-neutron breeders are thermal-neutron breeder reactors that use uranium-233 bred from thorium as fission fuel in the thorium fuel cycle. Thorium is about 3.5 times more common than uranium in the Earth's crust, and has different geographic characteristics. India's three-stage nuclear power programme features the use of a thorium fuel cycle in the third stage, as it has abundant thorium reserves but little uranium.
Decommissioning
Main article: Nuclear decommissioning
Nuclear decommissioning is the process of dismantling a nuclear facility to the point that it no longer requires measures for radiation protection, returning the facility and its parts to a safe enough level to be entrusted for other uses. Due to the presence of radioactive materials, nuclear decommissioning presents technical and economic challenges. The costs of decommissioning are generally spread over the lifetime of a facility and saved in a decommissioning fund.
Production
Further information: Nuclear power by country and List of nuclear reactors
2024 world electricity generation by source in terawatt-hours (TWh). Total generation was 30.85 petawatt-hours.
Coal 10,587 (34.4%)
Natural gas 6,796 (22.1%)
Hydro 4,417 (14.4%)
Nuclear 2,765 (8.99%)
Wind 2,497 (8.12%)
Solar 2,130 (6.92%)
Other 1,569 (5.10%)
Civilian nuclear power supplied 2,602 terawatt hours (TWh) of electricity in 2023, equivalent to about 9% of global electricity generation, and was the second largest low-carbon power source after hydroelectricity. Nuclear power's contribution to global energy production was about 4% in 2023. This is a little more than wind power, which provided 3.5% of global energy in 2023. Nuclear power's share of global electricity production has fallen from 16.5% in 1997, in large part because the economics of nuclear power have become more difficult.
As of November 2024,[update] there are 415 civilian fission reactors in the world, with a combined electrical capacity of 374 gigawatt (GW). There are also 66 nuclear power reactors under construction and 87 reactors planned, with a combined capacity of 72 GW and 84 GW, respectively. The United States has the largest fleet of nuclear reactors, generating over 800 TWh per year with an average capacity factor of 92%. Most reactors under construction are generation III reactors in Asia.
Regional differences in the use of nuclear power are large. The United States produces the most nuclear energy in the world, with nuclear power providing 19% of the electricity it consumes, while France produces the highest percentage of its electrical energy from nuclear reactors—65% in 2023. In the European Union, nuclear power provides 22% of the electricity as of 2022.
Nuclear power is the single largest low-carbon electricity source in the United States, and accounts for about half of the European Union's low-carbon electricity. Nuclear energy policy differs among European Union countries, and some, such as Austria, Estonia, Ireland and Italy, have no active nuclear power stations.
In addition, there were approximately 140 naval vessels using nuclear propulsion in operation, powered by about 180 reactors. These include military and some civilian ships, such as nuclear-powered icebreakers.
International research is continuing into additional uses of process heat such as hydrogen production (in support of a hydrogen economy), for desalinating sea water, and for use in district heating systems.
Economics
Main articles: Economics of nuclear power plants, List of companies in the nuclear sector, and cost of electricity by source
The economics of new nuclear power plants is a controversial subject. Nuclear power plants typically have high capital costs for building the plant. For this reason, comparison with other power generation methods is strongly dependent on assumptions about construction timescales and capital financing for nuclear plants.
Because of this strong dependency, the final cost of electricity from nuclear power is strongly dependent on the cost of capital.
Analysis of the economics of nuclear power must also take into account who bears the risks of future uncertainties. As of 2010, all operating nuclear power plants have been developed by state-owned or regulated electric utility monopolies. Many countries have since liberalized the electricity market where these risks, and the risk of cheaper competitors emerging before capital costs are recovered, are borne by plant suppliers and operators rather than consumers, which leads to a significantly different evaluation of the economics of new nuclear power plants.
The levelized cost of electricity (LCOE) from a new nuclear power plant is estimated to be 69 USD/MWh, according to an analysis by the International Energy Agency and the OECD Nuclear Energy Agency. This represents the median cost estimate for an nth-of-a-kind nuclear power plant to be completed in 2025, at a discount rate of 7%. Nuclear power was found to be the least-cost option among dispatchable technologies. Variable renewables can generate cheaper electricity: the median cost of onshore wind power was estimated to be 50 USD/MWh, and utility-scale solar power 56 USD/MWh. However, these sources are not directly comparable to nuclear power as they are not dispatchable.
Measures to mitigate global warming, such as a carbon tax or carbon emissions trading, may favor the economics of nuclear power.
At the assumed CO2 emission cost of 30 USD/ton, power from coal (88 USD/MWh) and gas (71 USD/MWh) is more expensive than low-carbon technologies. Electricity from long-term operation of nuclear power plants by lifetime extension was found to be the least-cost option, at 32 USD/MWh.
The high cost of construction is one of the biggest challenges for nuclear power plants. Nuclear power cost trends show large disparity by nation, design, build rate and the establishment of familiarity in expertise. The only two nations for which data is available that saw cost decreases in the 2000s were India and South Korea.
New small modular reactors, such as those developed by NuScale Power, are aimed at reducing the investment costs for new construction by making the reactors smaller and modular, so that they can be built in a factory.[citation needed]
Certain designs had considerable early positive economics, such as the CANDU, which realized a much higher capacity factor and reliability when compared to generation II light water reactors up to the 1990s.
Due to the on-line refueling reactor design, PHWRs (of which the CANDU design is a part) continue to hold many world record positions for longest continual electricity generation, often over 800 days. The specific record as of 2019 is held by a PHWR at Kaiga Atomic Power Station, generating electricity continuously for 962 days.
Nuclear power plants, though capable of some grid-load following, are typically run as much as possible to keep the cost of the generated electrical energy as low as possible, supplying mostly base-load electricity.
In some cases, Governments were found to force "consumers to pay upfront for potential cost overruns" or subsidize uneconomic nuclear energy or be required to do so. Nuclear operators are liable to pay for the waste management in the European Union. In the U.S., the Congress reportedly decided 40 years ago that the nation, and not private companies, would be responsible for storing radioactive waste with taxpayers paying for the costs. The World Nuclear Waste Report 2019 found that "even in countries in which the polluter-pays-principle is a legal requirement, it is applied incompletely" and notes the case of the German Asse II deep geological disposal facility, where the retrieval of large amounts of waste has to be paid for by taxpayers. Similarly, other forms of energy, including fossil fuels and renewables, have a portion of their costs covered by governments.
Use in space
Main article: Nuclear power in space
The most common use of nuclear power in space is the use of radioisotope thermoelectric generators, which use radioactive decay to generate power. These power generators are relatively small scale (few kW), and they are mostly used to power space missions and experiments for long periods where solar power is not available in sufficient quantity, such as in the Voyager 2 space probe. A few space vehicles have been launched using nuclear reactors: 34 reactors belong to the Soviet RORSAT series and one was the American SNAP-10A.
Both fission and fusion appear promising for space propulsion applications, generating higher mission velocities with less reaction mass.
Safety
See also: Nuclear safety and security and Nuclear reactor safety system
Nuclear power plants have three unique characteristics that affect their safety, as compared to other power plants. Firstly, intensely radioactive materials are present in a nuclear reactor. Their release to the environment could be hazardous. Secondly, the fission products, which make up most of the intensely radioactive substances in the reactor, continue to generate a significant amount of decay heat even after the fission chain reaction has stopped. If the heat cannot be removed from the reactor, the fuel rods may overheat and release radioactive materials. Thirdly, a criticality accident (a rapid increase of the reactor power) is possible in certain reactor designs if the chain reaction cannot be controlled. These three characteristics have to be taken into account when designing nuclear reactors.
All modern reactors are designed so that an uncontrolled increase of the reactor power is prevented by natural feedback mechanisms, a concept known as negative void coefficient of reactivity. If the temperature or the amount of steam in the reactor increases, the fission rate inherently decreases. The chain reaction can also be manually stopped by inserting control rods into the reactor core. Emergency core cooling systems (ECCS) can remove the decay heat from the reactor if normal cooling systems fail. If the ECCS fails, multiple physical barriers limit the release of radioactive materials to the environment even in the case of an accident. The last physical barrier is the large containment building.
With a death rate of 0.03 per TWh, nuclear power is the second safest energy source per unit of energy generated, after solar power, in terms of mortality when the historical track-record is considered. Energy produced by coal, petroleum, natural gas and hydropower has caused more deaths per unit of energy generated due to air pollution and energy accidents. This is found when comparing the immediate deaths from other energy sources to both the immediate and the latent, or predicted, indirect cancer deaths from nuclear energy accidents. When the direct and indirect fatalities (including fatalities resulting from the mining and air pollution) from nuclear power and fossil fuels are compared, the use of nuclear power has been calculated to have prevented about 1.84 million deaths from air pollution between 1971 and 2009, by reducing the proportion of energy that would otherwise have been generated by fossil fuels. Following the 2011 Fukushima nuclear disaster, it has been estimated that if Japan had never adopted nuclear power, accidents and pollution from coal or gas plants would have caused more lost years of life.
Serious impacts of nuclear accidents are often not directly attributable to radiation exposure, but rather social and psychological effects. Evacuation and long-term displacement of affected populations created problems for many people, especially the elderly and hospital patients. Forced evacuation from a nuclear accident may lead to social isolation, anxiety, depression, psychosomatic medical problems, reckless behavior, and suicide. A comprehensive 2005 study on the aftermath of the Chernobyl disaster concluded that the mental health impact is the largest public health problem caused by the accident. Frank N. von Hippel, an American scientist, commented that a disproportionate fear of ionizing radiation (radiophobia) could have long-term psychological effects on the population of contaminated areas following the Fukushima disaster.
Accidents
See also: Energy accidents, Nuclear and radiation accidents and incidents, and Lists of nuclear disasters and radioactive incidents
Some serious nuclear and radiation accidents have occurred. The severity of nuclear accidents is generally classified using the International Nuclear Event Scale (INES) introduced by the International Atomic Energy Agency (IAEA). The scale ranks anomalous events or accidents on a scale from 0 (a deviation from normal operation that poses no safety risk) to 7 (a major accident with widespread effects). There have been three accidents of level 5 or higher in the civilian nuclear power industry, two of which, the Chernobyl accident and the Fukushima accident, are ranked at level 7.[citation needed]
The first major nuclear accidents were the Kyshtym disaster in the Soviet Union and the Windscale fire in the United Kingdom, both in 1957. The first major accident at a nuclear reactor in the USA occurred in 1961 at the SL-1, a U.S. Army experimental nuclear power reactor at the Idaho National Laboratory. An uncontrolled chain reaction resulted in a steam explosion which killed the three crew members and caused a meltdown. Another serious accident happened in 1968, when one of the two liquid-metal-cooled reactors on board the Soviet submarine K-27 underwent a fuel element failure, with the emission of gaseous fission products into the surrounding air, resulting in 9 crew fatalities and 83 injuries.
The Fukushima Daiichi nuclear accident was caused by the 2011 Tohoku earthquake and tsunami. The accident has not caused any radiation-related deaths but resulted in radioactive contamination of surrounding areas. The difficult cleanup operation is expected to cost tens of billions of dollars over 40 or more years. The Three Mile Island accident in 1979 was a smaller scale accident, rated at INES level 5. There were no direct or indirect deaths caused by the accident.
The impact of nuclear accidents is controversial. According to Benjamin K. Sovacool, fission energy accidents ranked first among energy sources in terms of their total economic cost, accounting for 41% of all property damage attributed to energy accidents. Another analysis found that coal, oil, liquid petroleum gas and hydroelectric accidents (primarily due to the Banqiao Dam disaster) have resulted in greater economic impacts than nuclear power accidents. The study compares latent cancer deaths attributable to nuclear power with immediate deaths from other energy sources per unit of energy generated, and does not include fossil fuel related cancer and other indirect deaths created by the use of fossil fuel consumption in its "severe accident" (an accident with more than five fatalities) classification. The Chernobyl accident in 1986 caused approximately 50 deaths from direct and indirect effects, and some temporary serious injuries from acute radiation syndrome. The future predicted mortality from increases in cancer rates is estimated at 4000 in the decades to come.
Nuclear power works under an insurance framework that limits or structures accident liabilities in accordance with national and international conventions. It is often argued that this potential shortfall in liability represents an external cost not included in the cost of nuclear electricity. This cost is small, amounting to about 0.1% of the levelized cost of electricity, according to a study by the Congressional Budget Office in the United States. These beyond-regular insurance costs for worst-case scenarios are not unique to nuclear power. Hydroelectric power plants are similarly not fully insured against a catastrophic event such as dam failures. For example, the failure of the Banqiao Dam caused the death of an estimated 30,000 to 200,000 people, and 11 million people lost their homes. As private insurers base dam insurance premiums on limited scenarios, major disaster insurance in this sector is likewise provided by the state.
Attacks and sabotage
Main articles: Vulnerability of nuclear plants to attack, Nuclear terrorism, and Nuclear safety in the United States
Terrorists could target nuclear power plants in an attempt to release radioactive contamination into the community. The United States 9/11 Commission has said that nuclear power plants were potential targets originally considered for the September 11, 2001 attacks. An attack on a reactor's spent fuel pool could also be serious, as these pools are less protected than the reactor core. The release of radioactivity could lead to thousands of near-term deaths and greater numbers of long-term fatalities.
In the United States, the Nuclear Regulatory Commission carries out "Force on Force" (FOF) exercises at all nuclear power plant sites at least once every three years. In the United States, plants are surrounded by a double row of tall fences which are electronically monitored. The plant grounds are patrolled by a sizeable force of armed guards.
Insider sabotage is also a threat because insiders can observe and work around security measures. Successful insider crimes depended on the perpetrators' observation and knowledge of security vulnerabilities. A fire caused 5–10 million dollars worth of damage to New York's Indian Point Energy Center in 1971. The arsonist was a plant maintenance worker.
Proliferation
Further information: Nuclear proliferation
See also: Plutonium Management and Disposition Agreement
Nuclear proliferation is the spread of nuclear weapons, fissionable material, and weapons-related nuclear technology to states that do not already possess nuclear weapons. Many technologies and materials associated with the creation of a nuclear power program have a dual-use capability, in that they can also be used to make nuclear weapons. For this reason, nuclear power presents proliferation risks.[citation needed]
Nuclear power program can become a route leading to a nuclear weapon. An example of this is the concern over Iran's nuclear program. The re-purposing of civilian nuclear industries for military purposes would be a breach of the Non-Proliferation Treaty, to which 190 countries adhere. As of April 2012, there are thirty one countries that have civil nuclear power plants, of which nine have nuclear weapons. The vast majority of these nuclear weapons states have produced weapons before commercial nuclear power stations.[citation needed]
A fundamental goal for global security is to minimize the nuclear proliferation risks associated with the expansion of nuclear power. The Global Nuclear Energy Partnership was an international effort to create a distribution network in which developing countries in need of energy would receive nuclear fuel at a discounted rate, in exchange for that nation agreeing to forgo their own indigenous development of a uranium enrichment program. The France-based Eurodif/European Gaseous Diffusion Uranium Enrichment Consortium is a program that successfully implemented this concept, with Spain and other countries without enrichment facilities buying a share of the fuel produced at the French-controlled enrichment facility, but without a transfer of technology. Iran was an early participant from 1974 and remains a shareholder of Eurodif via Sofidif.[citation needed]
A 2009 United Nations report said that:
the revival of interest in nuclear power could result in the worldwide dissemination of uranium enrichment and spent fuel reprocessing technologies, which present obvious risks of proliferation as these technologies can produce fissile materials that are directly usable in nuclear weapons.
On the other hand, power reactors can also reduce nuclear weapon arsenals when military-grade nuclear materials are reprocessed to be used as fuel in nuclear power plants. The Megatons to Megawatts Program is considered the single most successful non-proliferation program to date. Up to 2005, the program had processed $8 billion of high enriched, weapons-grade uranium into low enriched uranium suitable as nuclear fuel for commercial fission reactors by diluting it with natural uranium. This corresponds to the elimination of 10,000 nuclear weapons. For approximately two decades, this material generated nearly 10 percent of all the electricity consumed in the United States, or about half of all U.S. nuclear electricity, with a total of around 7,000 TWh of electricity produced. In total it is estimated to have cost $17 billion, a "bargain for US ratepayers", with Russia profiting $12 billion from the deal. Much needed profit for the Russian nuclear oversight industry, which after the collapse of the Soviet economy, had difficulties paying for the maintenance and security of the Russian Federation's highly enriched uranium and warheads. The Megatons to Megawatts Program was hailed as a major success by anti-nuclear weapon advocates as it has largely been the driving force behind the sharp reduction in the number of nuclear weapons worldwide since the cold war ended. However, without an increase in nuclear reactors and greater demand for fissile fuel, the cost of dismantling and down blending has dissuaded Russia from continuing their disarmament. As of 2013, Russia appears to not be interested in extending the program.
Environmental impact
Main article: Environmental impact of nuclear power
Being a low-carbon energy source with relatively little land-use requirements, nuclear energy can have a positive environmental impact. It also requires a constant supply of significant amounts of water and affects the environment through mining and milling. Its largest potential negative impacts on the environment may arise from its transgenerational risks for nuclear weapons proliferation that may increase risks of their use in the future, risks for problems associated with the management of the radioactive waste such as groundwater contamination, risks for accidents and for risks for various forms of attacks on waste storage sites or reprocessing- and power-plants. However, these remain mostly only risks as historically there have only been few disasters at nuclear power plants with known relatively substantial environmental impacts.[citation needed]
Carbon emissions
See also: Life-cycle greenhouse gas emissions of energy sources
Further information: § Historic effect on carbon emissions
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Nuclear power is one of the leading low carbon power generation methods of producing electricity, and in terms of total life-cycle greenhouse gas emissions per unit of energy generated, has emission values comparable to or lower than renewable energy. A 2014 analysis of the carbon footprint literature by the Intergovernmental Panel on Climate Change (IPCC) reported that the embodied total life-cycle emission intensity of nuclear power has a median value of 12 g CO2eq/kWh, which is the lowest among all commercial baseload energy sources. This is contrasted with coal and natural gas at 820 and 490 g CO2 eq/kWh. As of 2021, nuclear reactors worldwide have helped avoid the emission of 72 billion tonnes of carbon dioxide since 1970, compared to coal-fired electricity generation, according to a report.
Radiation
The average dose from natural background radiation is 2.4 millisievert per year (mSv/a) globally. It varies between 1 mSv/a and 13 mSv/a, depending mostly on the geology of the location. According to the United Nations (UNSCEAR), regular nuclear power plant operations, including the nuclear fuel cycle, increases this amount by 0.0002 mSv/a of public exposure as a global average. The average dose from operating nuclear power plants to the local populations around them is less than 0.0001 mSv/a. For comparison, the average dose to those living within 50 miles (80 km) of a coal power plant is over three times this dose, at 0.0003 mSv/a.
Chernobyl resulted in the most affected surrounding populations and male recovery personnel receiving an average initial 50 to 100 mSv over a few hours to weeks, while the remaining global legacy of the worst nuclear power plant accident in average exposure is 0.002 mSv/a and is continuously dropping at the decaying rate, from the initial high of 0.04 mSv per person averaged over the entire populace of the Northern Hemisphere in the year of the accident in 1986.
Debate
Main article: Nuclear power debate
See also: Nuclear energy policy, Pro-nuclear movement, and Anti-nuclear movement
The nuclear power debate concerns the controversy which has surrounded the deployment and use of nuclear fission reactors to generate electricity from nuclear fuel for civilian purposes.
Proponents of nuclear energy regard it as a sustainable energy source that reduces carbon emissions and increases energy security by decreasing dependence on other energy sources that are often dependent on imports. For example, proponents note that annually, nuclear-generated electricity reduces 470 million metric tons of carbon dioxide emissions that would otherwise come from fossil fuels. Additionally, the amount of comparatively low waste that nuclear energy does create is safely disposed of by the large scale nuclear energy production facilities or it is repurposed/recycled for other energy uses.
Proponents also claim that the present quantity of nuclear waste is small and can be reduced through the latest technology of newer reactors and that the operational safety record of fission-electricity in terms of deaths is so far "unparalleled". Kharecha and Hansen estimated that "global nuclear power has prevented an average of 1.84 million air pollution-related deaths and 64 gigatonnes of CO2-equivalent (GtCO2-eq) greenhouse gas (GHG) emissions that would have resulted from fossil fuel burning" and, if continued, it could prevent up to 7 million deaths and 240 GtCO2-eq emissions by 2050.
Proponents also bring to attention the opportunity cost of using other forms of electricity. For example, the United States Environmental Protection Agency estimates that coal kills 30,000 people a year as a result of its environmental impact, while 60 people died in the Chernobyl disaster.
A real world example of impact provided by proponents is the 650,000 ton increase in carbon emissions in the two months following the closure of the Vermont Yankee nuclear plant.
Opponents believe that nuclear power poses many threats to people's health and environment such as the risk of nuclear weapons proliferation, long-term safe waste management and terrorism in the future. They also contend that nuclear power plants are complex systems where many things can and have gone wrong.
Critics find that one of the largest drawbacks to building new nuclear fission power plants are the high costs when compared to alternatives of sustainable energy sources.
Proponents note that focusing on the levelized cost of energy (LCOE), however, ignores the value premium associated with 24/7 dispatchable electricity and the cost of storage and backup systems necessary to integrate variable energy sources into a reliable electrical grid. "Nuclear thus remains the dispatchable low-carbon technology with the lowest expected costs in 2025. Only large hydro reservoirs can provide a similar contribution at comparable costs but remain highly dependent on the natural endowments of individual countries."
Overall, many opponents find that nuclear energy cannot meaningfully contribute to climate change mitigation. In general, they find it to be too dangerous, too expensive, to take too long for deployment, as much as to be an obstacle to achieving a transition towards sustainability and carbon-neutrality. These opponents find nuclear to be effectively a distraction in the competition for resources (i.e. human, financial, time, infrastructure and expertise) for the deployment and development of alternative, sustainable, energy system technologies
Nevertheless, there is ongoing research and debate over costs of new nuclear, especially in regions where seasonal energy storage is difficult to provide and which aim to phase out fossil fuels in favor of low carbon power faster than the global average. Some find that financial transition costs for a 100% renewables-based European energy system that has completely phased out nuclear energy could be more costly by 2050 based on current technologies (i.e. not considering potential advances in e.g. green hydrogen, transmission and flexibility capacities, ways to reduce energy needs, geothermal energy and fusion energy) when the grid only extends across Europe. Arguments of economics and safety are used by both sides of the debate.
Comparison with renewable energy
See also: Renewable energy debate
Slowing global warming requires a transition to a low-carbon economy, mainly by burning far less fossil fuel. This has generated considerable interest and dispute in determining the best path forward to rapidly replace fossil-based fuels in the global energy mix, with intense academic debate. Sometimes the IEA says that countries without nuclear should develop it as well as their renewable power.
World total primary energy supply of 162,494 TWh (or 13,792 Mtoe) by fuels in 2017 (IEA, 2019): 6, 8
Oil (32.0%)
Coal/Peat/Shale (27.1%)
Natural Gas (22.2%)
Biofuels and waste (9.50%)
Nuclear (4.90%)
Hydro (2.50%)
Others (Renewables) (1.80%)
Nuclear power is comparable to, and in some cases lower, than many renewable energy sources in terms of lives lost per unit of electricity delivered.
Nuclear reactors produce a much smaller volume of waste compared to renewable energy sources, although nuclear waste is much more toxic, expensive to manage and longer-lived compared to waste from renewable technologies.
Nuclear waste can be dangerous if leaked to the environment, and need to be stored safely for thousands or even hundreds of thousand of years.
Nuclear plants are also far more complex to decommission compared to renewable energy plants. A nuclear plant needs to be disassembled and removed and much of the disassembled nuclear plant needs to be stored as low-level nuclear waste for a few decades.
Nuclear power may also pose the risk of nuclear proliferation. Separated plutonium and enriched uranium could be used for nuclear weapons, which pose a substantial global risk to human civilization and the environment.
Speed of transition and investment needed
Analysis in 2015 by professor Barry W. Brook and colleagues found that nuclear energy could displace or remove fossil fuels from the electric grid completely within 10 years. This finding was based on the historically modest and proven rate at which nuclear energy was added in France and Sweden during their building programs in the 1980s. In a similar analysis, Brook had earlier determined that 50% of all global energy, including transportation synthetic fuels etc., could be generated within approximately 30 years if the global nuclear fission build rate was identical to historical proven installation rates calculated in GW per year per unit of global GDP (GW/year/$). This is in contrast to the conceptual studies for 100% renewable energy systems, which would require an order of magnitude more costly global investment per year, which has no historical precedent. These renewable scenarios would also need far greater land devoted to onshore wind and onshore solar projects. Brook notes that the "principal limitations on nuclear fission are not technical, economic or fuel-related, but are instead linked to complex issues of societal acceptance, fiscal and political inertia, and inadequate critical evaluation of the real-world constraints facing [the other] low-carbon alternatives."
Land use
The median land area used by US nuclear power stations per 1 GW installed capacity is 1.3 square miles (3.4 km2). To generate the same amount of electricity annually (taking into account capacity factors) from solar PV would require about 60 square miles (160 km2), and from a wind farm about 310 square miles (800 km2). Not included in this, is land required for the associated transmission lines, water supply, rail lines, mining and processing of nuclear fuel, and for waste disposal.
Research
Advanced fission reactor designs
Main article: Generation IV reactor
Current fission reactors in operation around the world are second or third generation systems, with most of the first-generation systems having been already retired. Research into advanced generation IV reactor types was officially started by the Generation IV International Forum (GIF) based on eight technology goals, including to improve economics, safety, proliferation resistance, natural resource use and the ability to consume existing nuclear waste in the production of electricity. Most of these reactors differ significantly from current operating light water reactors, and are expected to be available for commercial construction after 2030.
Hybrid fusion-fission
Main article: Nuclear fusion–fission hybrid
Hybrid nuclear power is a proposed means of generating power by the use of a combination of nuclear fusion and fission processes. The concept dates to the 1950s and was briefly advocated by Hans Bethe during the 1970s, but largely remained unexplored until a revival of interest in 2009, due to delays in the realization of pure fusion. When a sustained nuclear fusion power plant is built, it has the potential to be capable of extracting all the fission energy that remains in spent fission fuel, reducing the volume of nuclear waste by orders of magnitude, and more importantly, eliminating all actinides present in the spent fuel, substances which cause security concerns.
Fusion
Main articles: Nuclear fusion and Fusion power
Nuclear fusion reactions have the potential to be safer and generate less radioactive waste than fission. These reactions appear potentially viable, though technically quite difficult and have yet to be created on a scale that could be used in a functional power plant. Fusion power has been under theoretical and experimental investigation since the 1950s. Nuclear fusion research is underway but fusion energy is not likely to be commercially widespread before 2050.
Several experimental nuclear fusion reactors and facilities exist. The largest and most ambitious international nuclear fusion project currently in progress is ITER, a large tokamak under construction in France. ITER is planned to pave the way for commercial fusion power by demonstrating self-sustained nuclear fusion reactions with positive energy gain. Construction of the ITER facility began in 2007, but the project has run into many delays and budget overruns. The facility is now not expected to begin operations until the year 2027 – 11 years after initially anticipated. A follow on commercial nuclear fusion power station, DEMO, has been proposed. There are also suggestions for a power plant based upon a different fusion approach, that of an inertial fusion power plant.[citation needed]
Fusion-powered electricity generation was initially believed to be readily achievable, as fission-electric power had been. However, the extreme requirements for continuous reactions and plasma containment led to projections being extended by several decades. In 2020, more than 80 years after the first attempts, commercialization of fusion power production was thought to be unlikely before 2050.
To enhance and accelerate the development of fusion energy, the United States Department of Energy (DOE) granted $46 million to eight firms, including Commonwealth Fusion Systems and Tokamak Energy Inc, in 2023. This ambitious initiative aims to introduce pilot-scale fusion within a decade.
See also
Atomic battery
Nuclear marine propulsion
Nuclear power by country
Nuclear weapons debate
Pro-nuclear energy movement
Thorium-based nuclear power
Uranium mining debate
World energy supply and consumption
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Further reading
Wikiversity quizzes on nuclear power
See also: List of books about nuclear issues and List of films about nuclear issues
"AEC Atom Information Booklets". OSTI.gov. U.S. Atomic Energy Commission. Archived from the original on 2019-01-07. Both series, "Understanding the Atom" and "The World of the Atom". A total of 75 booklets published in the 1960s and 1970s, authored by scientists. Taken together, the booklets comprise the history of nuclear science and its applications at the time.
Armstrong, Robert C.; Wolfram, Catherine; Gross, Robert; Lewis, Nathan S.; M.V. Ramana (11 January 2016). "The Frontiers of Energy". Nature Energy. 1. Archived from the original on 2016-05-23.
Brown, Kate (2013). Plutopia: Nuclear Families, Atomic Cities, and the Great Soviet and American Plutonium Disasters. Oxford University Press. ISBN 9780190233105. OCLC 892040856.
Clarfield, Gerald H.; Wiecek, William M. (1984). Nuclear America: Military and Civilian Nuclear Power in the United States 1940–1980. Harper & Row.
Cooke, Stephanie (2009). In Mortal Hands: A Cautionary History of the Nuclear Age. Black Inc.
Cravens, Gwyneth (2007). Power to Save the World: the Truth about Nuclear Energy. New York: Knopf. ISBN 978-0-307-26656-9.
Elliott, David (2007). Nuclear or Not? Does Nuclear Power Have a Place in a Sustainable Energy Future?. Palgrave.
Ferguson, Charles D. (2007). Nuclear Energy: Balancing Benefits and Risks. Council on Foreign Relations.
Garwin, Richard L.; Charpak, Georges (2001). Megawatts and Megatons: A Turning Point in the Nuclear Age?. Knopf.
Högselius, Per; Evens, Siegfried, eds. (2025). The Nuclear-Water Nexus. MIT Press. ISBN 9780262383042.
Mahaffey, James (2015). Atomic accidents: a history of nuclear meltdowns and disasters: from the Ozark Mountains to Fukushima. Pegasus Books. ISBN 978-1-60598-680-7.
Oreskes, Naomi (February 2022). "Breaking the Techno-Promise: We do not have enough time for nuclear power to save us from the climate crisis". Scientific American. 326 (2): 74.
Patterson, Eann A.; Taylor, Richard J. (2024). "The commoditization of civil nuclear power". Royal Society Open Science. 11 (5): 240021. Bibcode:2024RSOS...1140021P. doi:10.1098/rsos.240021. PMC 11285846. PMID 39076811.{{cite journal}}: CS1 maint: article number as page number (link)
Schneider, Mycle; Thomas, Steve; Froggatt, Antony; Koplow, Doug (2016). The World Nuclear Industry Status Report: World Nuclear Industry Status as of 1 January 2016 (Report).
Walker, J. Samuel (1992). Containing the Atom: Nuclear Regulation in a Changing Environment, 1946–1971. Berkeley, California: University of California Press.
Weart, Spencer R. (2012). The Rise of Nuclear Fear. Cambridge, Massachusetts: Harvard University Press. ISBN 0-674-05233-1.
External links
Portals:
Energy
Nuclear technology
Nuclear power at Wikipedia's sister projects:
Definitions from Wiktionary
Media from Commons
News from Wikinews
Quotations from Wikiquote
Texts from Wikisource
Textbooks from Wikibooks
Resources from Wikiversity
Data from Wikidata
U.S. Energy Information Administration
Nuclear Fuel Cycle Cost Calculator Archived 2022-07-11 at the Wayback Machine
| v t e Nuclear power by country | |
--- |
| GWe > 10 | Canada China France Japan Russia South Korea Ukraine United States |
| GWe > 5 | Belgium India Spain Sweden United Kingdom |
| GWe > 2 | Czech Republic Finland Germany Pakistan Slovakia Switzerland Taiwan UAE |
| GWe > 1 | Argentina Belarus Brazil Bulgaria Hungary Mexico Romania South Africa |
| GWe < 1 | Armenia Iran Netherlands Slovenia |
| Planned | Albania Algeria Bangladesh Chile Egypt Ghana Indonesia Israel Jordan Kazakhstan Kenya Morocco Myanmar Nigeria North Korea Philippines Poland Saudi Arabia Sri Lanka Thailand Tunisia Turkey Vietnam |
| Abandoned plans for new plants | Australia Austria Cuba Ireland Libya Lithuania Malaysia Portugal Syria Taiwan Venezuela |
| Phasing-out | Belgium (delayed) Germany Italy (already) Spain Switzerland (gradually) Taiwan |
| List of nuclear power stations Nuclear energy policy Nuclear energy policy by country Nuclear accidents Nuclear technology portal | |
| v t e Nuclear technology | |
--- |
| Outline | |
| | | | --- | | Science | Chemistry Engineering Physics Atomic nucleus Fission Fusion Radiation + ionizing + braking | | Fuel | Tritium Deuterium Helium-3 Fertile material Fissile material Isotope separation Nuclear material + Uranium - enriched - depleted + Plutonium + Thorium | | Neutron | Activation Capture Poison Cross section Generator Radiation + Wigner effect Reflector Temperature Thermal Fast Fusion | | Power | by country Power plant Economics Accidents and incidents Policy Fusion Radioisotope thermoelectric (RTG) + MMRTG Propulsion + rocket Safety and security + Nuclear meltdown | | Medicine | | | | --- | | Imaging | Autoradiograph RadBall Scintigraphy Single-photon emission (SPECT) Positron-emission tomography (PET) | | Therapy | Fast-neutron Neutron capture therapy of cancer Targeted alpha-particle Proton-beam Tomotherapy Brachytherapy Radiosurgery Radiopharmacology | | | Processing | Atomic gardening Electron-beam processing Food irradiation Gemstone irradiation | | Weapons | | | | --- | | Topics | Arms race Delivery Design Disarmament Ethics Explosion + effects History Proliferation Testing + high-altitude + underground Warfare Yield + TNTe | | Lists | States with nuclear weapons Historical stockpiles and tests + Tests + Tests in the United States WMD treaties Weapon-free zones Weapons | | | Waste | | | | --- | | Products | Actinide + Reprocessed uranium + Reactor-grade plutonium + Minor actinide Activation Fission + LLFP Actinide chemistry | | Disposal | Fuel cycle High-level (HLW) Low-level (LLW) Nuclear decommissioning Repository Reprocessing Spent fuel + pool + cask Transmutation | | | Debate | Nuclear power Nuclear weapons Blue Ribbon Commission on America's Nuclear Future Anti-nuclear movement Uranium mining Nuclear power phase-out | | |
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8924 | https://en.wikipedia.org/wiki/Overlapping_interval_topology | Overlapping interval topology - Wikipedia
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Overlapping interval topology
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From Wikipedia, the free encyclopedia
Not to be confused with Interlocking interval topology.
In mathematics, the overlapping interval topology is a topology which is used to illustrate various topological principles.
Definition
[edit]
Given the closed interval[−1,1]{\displaystyle [-1,1]} of the real number line, the open sets of the topology are generated from the half-open intervals (a,1]{\displaystyle (a,1]}![Image 5: {\displaystyle (a,1]}]( with a<0{\displaystyle a<0} and [−1,b){\displaystyle [-1,b)}Image 7: {\displaystyle -1,b)} with b>0{\displaystyle b>0}. The topology therefore consists of intervals of the form −1,b){\displaystyle [-1,b)}![Image 9: {\displaystyle [-1,b)}, (a,b){\displaystyle (a,b)}, and (a,1]{\displaystyle (a,1]}![Image 11: {\displaystyle (a,1]} with a<0<b{\displaystyle a<0<b}, together with [−1,1]{\displaystyle [-1,1]} itself and the empty set.
Properties
[edit]
Any two distinct points in [−1,1]{\displaystyle [-1,1]} are topologically distinguishable under the overlapping interval topology as one can always find an open set containing one but not the other point. However, every non-empty open set contains the point 0 which can therefore not be separated from any other point in [−1,1]{\displaystyle [-1,1]}, making [−1,1]{\displaystyle [-1,1]} with the overlapping interval topology an example of a T 0 space that is not a T 1 space.
The overlapping interval topology is second countable, with a countable basis being given by the intervals −1,s){\displaystyle [-1,s)}Image 17: {\displaystyle [-1,s)}, (r,s){\displaystyle (r,s)} and (r,1]{\displaystyle (r,1]}![Image 19: {\displaystyle (r,1]} with r<0<s{\displaystyle r<0<s} and r and s rational.
See also
[edit]
List of topologies
Particular point topology, a topology where sets are considered open if they are empty or contain a particular, arbitrarily chosen, point of the topological space
References
[edit]
Steen, Lynn Arthur; Seebach, J. Arthur Jr. (1995) , Counterexamples in Topology (Dover reprint of 1978 ed.), Berlin, New York: Springer-Verlag, ISBN978-0-486-68735-3, MR0507446(See example 53)
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Overlapping interval topology
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8925 | https://med.libretexts.org/Courses/Virginia_Tech_Carilion_School_of_Medicine/Pulmonary_Physiology_for_Pre-Clinical_Students_(Binks)/11%3A_Alkalosis_and_Acidosis/11.03%3A_The_Henderson-Hasselbalch_equation | Skip to main content
12.3: The Henderson–Hasselbalch Equation
Last updated
: Mar 13, 2024
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12.2: Physiological Buffers
11: Arterial PCO₂ and pH
Page ID
: 34564
Andrew Binks
Virginia Tech Carilion School of Medicine via Virginia Tech Libraries' Open Education Initiative
( \newcommand{\kernel}{\mathrm{null}\,})
What we will see is how the balance of bicarbonate and hydrogen ions determines pH, and how both of these ions can be influenced by the kidneys and lungs to keep pH constant.
First, we will take the central and most important part of the infamous equation, discarding the more innocuous ends.
H2CO3↔H++HCO3−
This central portion describes the dissociation of carbonic acid into hydrogen and bicarbonate ions. But because carbonic acid is a weak acid, this dissociation is incomplete—some carbonic acid staying whole, some dissociating into the ions. The level of dissociation is described by the dissociation constant (K’), which really is the ratio of the concentrations of dissociated components to carbonic acid (equation 12.3.2).
K′=H+⋅HCO3−H2CO3
Because we are interested in calculating the pH, however, we are more interested in the amount of hydrogen ions, so rearranging this equation for hydrogen ion concentration we see the hydrogen ion concentration is the dissociation constant, multiplied by the ratio of carbonic acid and bicarbonate (equation 12.3.3).
H+=K′⋅H2CO3HCO3−
This equation theoretically would allow us to now determine hydrogen concentration and therefore pH, but there are some practical problems for us, the first of which is that the instability of carbonic acid means we cannot measure its concentration. So we have to use a proxy measure. The amount of carbonic acid is determined by the amount of carbon dioxide, as can be seen in the equation that is so familiar to you—the greater the amount of CO2, the more carbonic acid.
CO2+H2O↔H2CO3↔H++HCO3−
So after accounting for the dissociation constant of carbonic acid and CO2 and water, we can simply replace carbonic acid concentration with concentration of CO2 (equation 12.3.5).
H+=K′⋅CO2HCO3−
We then bump into our next practical problem: our equation now has CO2 concentration in it, but clinically we do not measure CO2 as a concentration (as in mmols), but as a partial pressure. So our next and nearly final step is to convert CO2 concentration to CO2 partial pressure, and we do this by multiplying the partial pressure (our measured value) by the solubility coefficient of carbon dioxide, which happens to be 0.03 mmol/mmHg. Our equation thus now can be completed using our adjusted PCO2 (equation 12.3.6).
H+=K′⋅0.03⋅PCO2HCO3−
Our equation as it is now allows us to calculate hydrogen ion concentration, but we need pH, so we have to make a conversion. Because pH is the negative logarithm of hydrogen concentration, we express everything in the negative log form. And because the negative log of the dissociation constant is referred to as pK, then we can simplify our equation one more step (equation 12.3.7).
pH=pK−log0.03⋅PCO2HCO3−
To make our equation simple to use, we now get rid of the negative log, and so get the following (equation 12.3.8):
pH=pK+logHCO3−0.03⋅PCO2
We know that the pK of the bicarbonate system happens to be 6.1, so substituting this into the equation we end up with the Henderson–Hasselbalch equation (equation 12.3.9).
Let us put this in context.
First, the equation shows that if CO2 rises then pH falls, and because CO2 is under the influence of alveolar ventilation, this explains how the alveolar ventilation can now control pH. It also shows that if bicarbonate increases then pH increases, and equally if bicarbonate falls then pH falls. Because the bicarbonate concentration can be modified either way by the kidneys, the equation also shows how the kidneys can modify pH (equation 12.3.9).
Role of kidneys (numerator)
Role of lungs (denominator)
pH=6.1+logHCO3−0.03⋅PCO2
The involvement of these two major physiological systems in this equation make the bicarbonate system a very powerful buffer, particularly when considering that there is an unlimited source of CO2 and therefore bicarbonate supplied by the metabolism.
But more importantly it shows that pH is actually determined by the ratio of bicarbonate and CO2 and that both are equally important. This fact is critical to appreciate as it forms the basis of understanding the compensation mechanisms we dealt with earlier. This is why I put you through this derivation. So for example, if a rise in CO2 (such as in lung disease) is accompanied by an equal rise in bicarbonate (generated by the kidney), then the ratio between the two remains the same and therefore pH remains the same. Likewise, if during a fall in CO2 the kidneys excrete bicarbonate, then pH can be kept constant. So before we finish, let us show you that the equation actually works by plugging in some numbers.
Example #1: Let us start with normal values, a PCO2 of 40 mmHg and a bicarbonate of 24, and plug these into the equation. This comes to 6.1 plus the log of 20, which is 6.1 plus 1.3, or 7.4 (i.e., normal arterial pH).
pH=6.1+log24(0.03⋅40)=6.1+log(20)=6.1+1.3=7.4
Example #2: Now let us look at a case of acute lung failure that has caused a rise in arterial PCO2, but has not persisted long enough for the kidney to respond and compensate. PCO2 has risen to 50 mmHg, and bicarbonate has not changed. Our calculation now goes to 6.1 plus the log of 16, which is 6.1 plus 1.2, and pH has fallen to 7.3.
pH=6.1+log24(0.03⋅50)=6.1+log(16)=6.1+1.2=7.3
We now have three numbers that can give a meaningful clinical interpretation. The low pH indicates the patient is in acidosis. The raised PCO2 suggests that this is respiratory acidosis, and the unchanged bicarbonate suggests no metabolic compensation has taken place.
Example #3: Now let us return to our patient thirty-six hours later when we have given the kidney a chance to respond. The patient’s PCO2 remains at 50 because of the persistent lung problem, but the kidney has raised the bicarbonate to 30. Now our equation becomes 6.1 plus the log of 20, or 6.1 plus 1.3, and pH is 7.4—apparently normal.
pH=6.1+log30(0.03⋅50)=6.1+log(20)=6.1+1.3=7.4
But when we look at all three numbers we see that the patient is far from normal: the pH is okay only because the kidneys have raised bicarbonate to match the raised CO2 and keep the ratio the same. So we now have a respiratory acidosis with metabolic compensation.
Summary
So although it has been a long journey through this chapter you should now be able to interpret blood gas values to determine whether a patient is in acidosis or alkalosis and whether or not compensation is present. I strongly recommend writing the Henderson–Hasselbalch equation as a formula in Excel so that you can plug in CO2 and bicarbonate values and see what happens to pH. By repeatedly interpreting blood gas values and pH, determining the status of a patient will rapidly become second nature.
12.2: Physiological Buffers
11: Arterial PCO₂ and pH |
8926 | https://stats.libretexts.org/Courses/Las_Positas_College/Math_40%3A_Statistics_and_Probability/04%3A_Probability_and_Counting/4.04%3A_Counting_Rules/4.4.01%3A_Permutations | 4.4.1: Permutations - Statistics LibreTexts
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4.4: Counting Rules
4: Probability and Counting
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Tue, 11 Aug 2020 05:36:20 GMT
4.4.1: Permutations
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7. 4.4.1: Permutations
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4.4.1: Permutations
Last updated Aug 11, 2020
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4.4: Counting Rules
4.4.2: Permutations with Similar Elements
Page ID 11505
Rupinder Sekhon and Roberta Bloom
De Anza College
( \newcommand{\kernel}{\mathrm{null}\,})
Table of contents No headers
Learning Objectives
In this section you will learn to
Count the number of possible permutations (ordered arrangement) of n items taken r at a time
Count the number of possible permutations when there are conditions imposed on the arrangements
Perform calculations using factorials
In the previous section, we were asked to find the word sequences formed by using the letters { A, B, C } if no letter is to be repeated. The tree diagram gave us the following six arrangements.
ABC, ACB, BAC, BCA, CAB, and CBA.
Arrangements like these, where order is important and no element is repeated, are called permutations.
Definition: Permutations
A permutation of a set of elements is an ordered arrangement where each element is used once.
Example 4.4.1.1
How many three-letter word sequences can be formed using the letters { A, B, C, D }?
Solution
There are four choices for the first letter of our word, three choices for the second letter, and two choices for the third.
4 3 2
Applying the multiplication axiom, we get 4⋅3⋅2=24 different arrangements.
Example 4.4.1.2
How many permutations of the letters of the word ARTICLE have consonants in the first and last positions?
Solution
In the word ARTICLE, there are 4 consonants.
Since the first letter must be a consonant, we have four choices for the first position, and once we use up a consonant, there are only three consonants left for the last spot. We show as follows:
4 3
Since there are no more restrictions, we can go ahead and make the choices for the rest of the positions.
So far we have used up 2 letters, therefore, five remain. So for the next position there are five choices, for the position after that there are four choices, and so on. We get
4 5 4 3 2 1 3
So the total permutations are 4⋅5⋅4⋅3⋅2⋅1⋅3=1440.
Example 4.4.1.3
Given five letters { A, B, C, D, E }. Find the following:
The number of four-letter word sequences.
The number of three-letter word sequences.
The number of two-letter word sequences.
Solution
The problem is easily solved by the multiplication axiom, and answers are as follows:
The number of four-letter word sequences is 5⋅4⋅3⋅2=120.
The number of three-letter word sequences is 5⋅4⋅3=60.
The number of two-letter word sequences is 5⋅4=20.
We often encounter situations where we have a set of n objects and we are selecting r objects to form permutations. We refer to this as permutationsof n objects taken r at a time, and we write it as nPr.
Therefore, the above example can also be answered as listed below.
The number of four-letter word sequences is 5P4 = 120.
The number of three-letter word sequences is 5P3 = 60.
The number of two-letter word sequences is 5P2 = 20.
Before we give a formula for nPr, we'd like to introduce a symbol that we will use a great deal in this as well as in the next chapter.
Definition: Factorial
(4.4.1.1)n!=n(n−1)(n−2)(n−3)⋯3⋅2⋅1
where n is a natural number.
(4.4.1.2)0!=1
Now we define nPr.
Definition: nPr
The Number of Permutations of n Objects Taken r at a Time
(4.4.1.3)nPr=n(n−1)(n−2)(n−3)⋯(n−r+1)
or
(4.4.1.4)nPr=n!(n−r)!
where n and r are natural numbers.
The reader should become familiar with both formulas and should feel comfortable in applying either.
Example 4.4.1.4
Compute the following using both formulas.
6P3
7P2
Solution
We will identify n and r in each case and solve using the formulas provided.
a. 6P3 = 6⋅5⋅4=120, alternately
(4.4.1.5)6P3=6!(6−3)!=6!3!=6⋅5⋅4⋅3⋅2⋅1 3⋅2⋅1=120
b. 7P2 = 7⋅6=42, or
(4.4.1.6)7P2=7!5!=7⋅6⋅5⋅4⋅3⋅2⋅1 5⋅4⋅3⋅2⋅1=42
Next we consider some more permutation problems to get further insight into these concepts.
Example 4.4.1.5
In how many different ways can 4 people be seated in a straight line if two of them insist on sitting next to each other?
Solution
Let us suppose we have four people A, B, C, and D. Further suppose that A and B want to sit together. For the sake of argument, we tie A and B together and treat them as one person.
The four people are AB CD. Since AB is treated as one person, we have the following possible arrangements.
ABCD,ABDC,CABD,DABC,CDAB,DCAB
Note that there are six more such permutations because A and B could also be tied in the order BA. And they are
BACD,BADC,CBAD,DBAC,CDBA,DCBA
So altogether there are 12 different permutations.
Let us now do the problem using the multiplication axiom.
After we tie two of the people together and treat them as one person, we can say we have only three people. The multiplication axiom tells us that three people can be seated in 3! ways. Since two people can be tied together 2! ways, there are 3! 2! = 12 different arrangements
Example 4.4.1.6
You have 4 math books and 5 history books to put on a shelf that has 5 slots. In how many ways can the books be shelved if the first three slots are filled with math books and the next two slots are filled with history books?
Solution
We first do the problem using the multiplication axiom.
Since the math books go in the first three slots, there are 4 choices for the first slot,
3 choices for the second and 2 choices for the third.
The fourth slot requires a history book, and has five choices. Once that choice is made, there are 4 history books left, and therefore, 4 choices for the last slot. The choices are shown below.
4 3 2 5 4
Therefore, the number of permutations are 4⋅3⋅2⋅5⋅4=480.
Alternately, we can see that 4⋅3⋅2 is really same as 4P3, and 5⋅4 is 5P2.
So the answer can be written as (4P3) (5P2) = 480.
Clearly, this makes sense. For every permutation of three math books placed in the first three slots, there are 5P2 permutations of history books that can be placed in the last two slots. Hence the multiplication axiom applies, and we have the answer (4P3) (5P2).
We summarize the concepts of this section:
Note
1. Permutations
A permutation of a set of elements is an ordered arrangement where each element is used once.
2. Factorial
n!=n(n−1)(n−2)(n−3)⋯3⋅2⋅1
Where n is a natural number.
0!=1
3. Permutations of n Objects Taken r at a Time
nPr=n(n−1)(n−2)(n−3)⋯(n−r+1)
or
nPr=n!(n−r)!
where n and r are natural numbers.
This page titled 4.4.1: Permutations is shared under a CC BY license and was authored, remixed, and/or curated by Rupinder Sekhon and Roberta Bloom.
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8927 | https://home.cc.umanitoba.ca/~thomas/Courses/space.pdf | CHAPTER TWO PLANES AND LINES IN R 3 2.1 INTRODUCTION In this chapter we will use vector methods to derive equations for planes and lines in threedimensional space R 3. The derived equations will be vector equations that we will be able convert into nonvector form equations. We will also use vectors to find the distance between a point and a plane, a point and a line, and between two lines in space as well as to find the angle between two planes and between two lines in R 3.
2.2 PLANES IN R 3 A plane is uniquely determined in threedimensional space by either: 1. giving the orientation of the plane in space and giving a point on the plane; or 2. giving three noncollinear points on the plane.
The orientation of a plane in space is specified by giving a vector n that is orthogonal (perpendicular) to the plane. Such a vector is said to be a normal vector to the plane. The equation for the plane derived using the normal vector n and a point P on the plane is called the pointnormal form of the equation of the plane.
The pointnormal form equation for a plane Let n = (a, b, c) be a normal vector to a plane and let the point P = (x0, y0, z0) be a point on the plane. If X = (x, y, z) is any other point on the plane, then, since n PX ⊥ we have 0 PX = • n . Let p OP = and let x OX = . Then OP OX OX OP OX PO PX − = + − = + = = x – p so the equation 0 PX = • n becomes n• (x−p) = 0. This vector equation for the plane is called the point normal form.
For the nonvector form of this equation we use n = (a, b, c) and x–p = (x−x0, y−y0, z−z0). Then n• (x−p) = 0 becomes (a, b, c)• (x−x0, y−y0, z−z0) = 0 25 n P X O p x or a(x−x0) + b( y−y0) + c(z−z0) = 0.
If we write n• (x−p) = 0 in the form p n x n • = • and then let d = •p n we get ( , , ) ( , , ) a b c x y z d • = or in nonvector form ax + by + cz = d . This nonvector form equation is called the standard form equation for a plane.
Example If a plane has the normal vector n = (1, 2, 3) and passes through the point P = (4, 5, 6), then its pointnormal form equation is (1, 2, 3)• (x−4, y−5, z−6) = 0. The standard form equation of this plane is found by computing the dot product above to get (1)(x−4) + (2)(y−5) + (3)(z−6) = 0 which when expanded becomes 4 2 10 3 18 0.
x y z − + − + − = This simplifies to the standard form equation of the plane 2 3 32.
x y z + + = Plane determined by three noncollinear points Let P, Q and R be three noncollinear points lying on a plane. To find an equation for the plane we make use of the cross product. The vector PR PQ × = n is a normal vector to the plane. Using this normal vector with any one of the three points on the plane, say P, and using the pointnormal form equation from the preceding section, will result in an equation of the plane. The following example illustrates the procedure.
Example A plane passes through three noncollinear points P = (1, 2, 0), Q = (3, 0, 1) and R = (4, 2, 3). Find a standard form equation of the plane.
Solution = PQ (3, 0, 1) – (1, 2, 0) = (2, 2, 1) = PR (4, 2, 3) – (1, 2, 0) = (3, 0, 3) ) 6 , 3 , 6 ( 0 3 2 2 , 3 3 1 2 , 3 0 1 2 PR PQ − − = − − − = × = n ( ) 0 ( 6, 3, 6) ( 1, 2, 0) 0 x y z • − = ⇒ − − • − − − = n x p 0 6 6 3 6 6 = + + − + − ⇒ z y x 26 P Q R n 12 6 3 6 − = + − − ⇒ z y x 4 2 2 = − + ⇒ z y x 2.2 Problems 1. Find a normal vector to each of the following planes.
a. 2x + 3y + 4z = 5 b. x + 3y –2z = 6 c. 3x + 5z = 7 2. Find a point on the given plane, then write a pointnormal form equation for the given plane.
a. 3x + 2y + 3z = 6 b. x + 2y + 3z = 12 c. 5x + 2y –3z = 7 3. Find an equation for the plane passing through the point P and having the vector n as a normal vector. Write the equation of the plane in standard form.
a. P = (5, 3, 4)n = (2, 3, 1) b. P = (4, 1, 2) n = (5, 2, 3) c. P = (2, 1, 6)n = (1, 0, 1) d. P = (1, 0, −2) n = (2, −1, 3) 4. Find an equation for the plane passing through the 3 noncollinear points P, Q and R. Write the equation of the plane in standard form.
a. P = (1, 2, 3) Q = (2, 1, 4) R = (2, 3, 6) b. P = (3, 2, 5) Q = (2, 3, 6) R = (4, 7, 5) c. P = (1, 0, 1) Q = (1, 1, −1) R = (2, 1, −2) d. P = (2, 1, 0) Q = (3, 1, 2) R = (0, −3, 4) 2.3 LINES IN R 3 A line is uniquely determined in threedimensional space by either: 1. giving the orientation of the line in space and giving a point on the line; or 2. giving two noncoincident points on the line. The orientation of a line in space is given by specifying a vector in R 3 that is parallel to the line. The form of the equation derived using a vector parallel to the line and a point on the line is called the pointparallel form equation for a line. 27 l P X O v p x Pointparallel form equation for a line Let P = (x0, y0, z0) be a point on the line l and let the vector v = (a, b, c) be a vector that is parallel to the given line l. If X = (x, y, z,) is any other point on the line l, then v t = PX for some scalar t. But PX u u u r can be written as PX PO OX OP OX OX OP = + = − + = − = − x p u u u r u u u r u u u r u u u r u u u r u u u r u u u r , hence v t = PX becomes x – p = tv or x = p + tv. The vector equation x = p + tv. is called the pointparallel form equation for a line in R 3. Each different value of the scalar t corresponds to a different point located on the line.
To obtain the nonvector form parametric equations for a line we start with x = p + tv and write out the vectors as ordered triples. This gives 0 0 0 ( , , ) ( , , ) ( , , ).
x y z x y z t a b c = + Equating the corresponding components on each side produces three equations. These equations are called the parametric equations of the line and the scalar t is called the parameter. x = x0 + at y = yo + bt z = z0 + ct Example Suppose the line l passes through the point P = (1, 2, 3) and is parallel to the vector v = (4, 5, 6). Its pointparallel form equation x = p + tv is (x, y, z) = (1, 2, 3) + t (4, 5, 6). Equating the corresponding components we get as the parametric equations of the line the three equations: x = 1 + 4t, y = 2 + 5t, z = 3 + 6t.
Two point form equation for a line If the line l passes through two distinct points P = (x0, y0, z0) and Q = (x1, y1, z1) we use the pointparallel form with = = PQ v (x1−x0, y1−y0, z1−z0). Equating the corresponding components in the vector equation (x, y, z) = (x0, y0, z0) + t(x1−x0, y1−y0, z1−z0) results in the nonvector twopoint form parametric equations: x = x0 + t(x1 – x0) y = y0 + t(y1 – y0) z = z0 + t(z1 – z0) 28 l P v P Q v Example The line l passes through the points P = (1, 2, 3) and Q = (4, 6, 9). The twopoint vector form of its equation is (x, y, z) = (1, 2, 3) + t(4 − 1, 6 − 2, 9 − 3) or (x, y, z) = (1, 2, 3) + t(3, 4, 6).
Equating components, the corresponding parametric equations are found to be x = 1 + 3t y = 2 + 4t z = 3 + 6t Symmetric equations for a line Consider the parametric form equations for a line x = x0 + at, y = yo + bt and z = z0 +ct. If a, b and c are all nonzero, we can solve each equation for t to get: t a x x = − 0 , t b y y = − 0 , t c z z = − 0 . Eliminating t between theses equations gives the symmetric form equations for a line: c z z b y y a x x 0 0 0 − = − = − . If any of a, b or c equals 0, then the corresponding numerator must equal 0. Thus for example if a = 0, the symmetric equations become x – x0 = 0 and c z z b y y 0 0 − = − .
Example Find a set of symmetric equations for the line that passes through the two points P = (3, 2, 1) and Q = (4, 5, 6).
Solution The vector ) 5 , 3 , 1 ( ) 1 6 , 2 5 , 3 4 ( PQ = − − − = = v is parallel to the required line. Setting (a, b, c) = (1, 3, 5) and (x0, y0, z0) = (3, 2, 1) we get the symmetric form equations for the line 5 1 3 2 1 3 − = − = − z y x .
2.3 Problems 29 1. Find a pointparallel equation for the line l that passes through the point P and is parallel to the vector v. Find also a set of parametric equations for the line.
a. P = (1, 2, 3) v = (2, 1, 4) b. P = (3, 1, 3) v = (4, 2, 5) c. P = (4, 1, 5) v = (1, 0, 3) d. P = (1, –2, 2) v = (2, 3, –1) e. P = (2, 0, −1) v = (1, −2, 3) f. P = (−3, 2, 0) v = (0, −1, 3) 2. Find a set of parametric equations for the line l that passes through points P and Q.
a. P = (1, 3, 4) Q = (2, 5, 1) b. P = (4, 3, −1) Q = (2, 2, 3) c. P = (2, –1, –3) Q = (1, 0, –2) d. P = (5, 1, 3) Q = (1, –1, 2) e. P = (4, 0, −5) Q = (1, 3, −4) e. P = (2, 2, −2) Q = (3, −3, 0) 3. Find a set of symmetric equations for each of the lines of question 2 above. 2.4 ANGLES BETWEEN OBJECTS IN R 3 In this section we will look at what is meant by the angle between two planes in R 3 and by the angle between two lines in R 3. We will use vector methods to find these angles.
The angle between two planes Two planes that are not parallel to each other will intersect each other. The angle between the two planes is equal to the angle between their normals. Unless the intersection is a right angle, two distinct angles between the planes are created, an acute angle θ and the supplementary angle 180º − θ. In this case we will use the smaller acute angle as the angle between the two planes and refer to it as the dihedral angle between the planes. [Dihedral comes from the Greek words dis meaning two and hedra meaning face.] Example Find the dihedral angle between the planes having equations x + 2y + 3z = 4 and 3x – 4y + 2z = 7.
30 Solution A pair of normals to the given planes are n1 = (1, 2, 3) and n2 = (3, −4, 2). The cosine of the angle between n1 and n2 is 29 14 1 4 16 9 9 4 1 6 8 3 ) 2 , 4 , 3 ( ) 3 , 2 , 1 ( ) 2 , 4 , 3 ( ) 3 , 2 , 1 ( cos = + + + + + − = − − • = • = θ 2 1 2 1 n n n n Using a calculator we find the dihedral angle is 87 29 14 1 cos 1 = = θ − º. Since this angle is less than 90° it is the required dihedral angle.
Example Find the dihedral angle between the planes having equations 2x + 3y − 4z = 5 and 4x – 2y + 3z = 6.
Solution A pair of normals to the given planes are n1 = (2, 3, −4) and n2 = (4, −2, 3). The cosine of the angle between n1 and n2 is (2, 3, 4) (4, 2, 3) 8 6 12 10 10 cosθ (2, 3, 4) (4, 2, 3) 29 4 9 16 16 4 9 29 29 − • − − − − = = = = = − − − + + + + 1 2 1 2 n •n n n Using a calculator we find the angle 1 10 cos 110 θ 29 − = − = º. Since this angle is greater than 90°, the desired dihedral angle is 180° − 110° = 70°.
Skew lines Two noncoincident lines in R 2 are either parallel and have no points in common or they intersect in exactly one point. In R 3 there is another possibility. Two noncoincident lines may be nonparallel and nonintersecting. Such a pair of lines are said to be skew lines. The adjoining diagram illustrates a pair of skew lines that do not intersect since they lie in parallel planes, but the lines themselves are not parallel. We define the angle between skew lines to be the angle between a pair of intersecting lines that are parallel to the skew lines. 31 When two lines intersect, if the intersection is not at right angles two distinct angles are formed, an acute angle θ and its supplement 180º θ. We will use the acute angle to describe the angle between the two lines.
Example Find the angle between the skew lines l1: x = 2 + 3t, y = 4 – t, z = 3 + 2t and l2: x = 1 – t, y = 5 + 2t, z = 6 + 3t.
Solution The vector v1 = (3, −1, 2) is parallel to line l1 and v2 = (−1, 2, 3) is parallel to line l2. The angle between the lines is equal to the angle between the vectors v1 and v2. Let θ be this angle. Then 14 1 9 4 1 4 1 9 6 2 3 ) 3 , 2 , 1 ( ) 2 , 1 , 3 ( ) 3 , 2 , 1 ( ) 2 , 1 , 3 ( cos 2 1 2 1 = + + + + + − − = − − − • − = • = θ v v v v and so the required angle is 1 1 =cos 86 θ 14 − = ° .
2.4 Problems 1. Find the dihedral angle between the following pairs of planes.
a. 2x + 3y – z = 5 x + y + 5z = 2 b. x + 2y + 3z = 4 −2x + 3y + z = 1 c. x + y = 1 7x + 7z = 2 d. 2x + y + z = 3 2x + 2y = 5 e. 2x + 2y = 3 2x + y + 2z = 5 f. x + z = 5 x + 4y + 8z = 12 2. Find the cosine of the angle between the pairs of skew lines.
a. x = 1 + t, y = 2 – 3t, z = 5 + t and x = 3 – 2t, y = 5 + 4t, z = 1 – t.
b. x = 7 + 3t, y = 1 – t, z = 2 + 2t and x = 2 + t, y = 3 – 2t, z = 1 + 5t c. 1 2 2 1 2 3 + = + = − z y x and 2 5 3 3 3 1 + = + = − z y x d. 2 2 1 3 4 5 − + = − + = − z y x and 3 3 1 2 5 2 + = − + = − z y x 2.5 DISTANCES IN R 3 Vector methods developed in chapter one will be used in this section to find a variety of distances in R 3. We will find the distance between a point and a plane; between a point and a line; and between two lines in R 3. Distance between a point and a plane 32 P D Q n Let P be the given point and let the plane have the equation ax + by + cz = d. A normal vector to the plane is n = (a, b, c). Let Q be any point on the plane. The distance between the point P and the plane is the magnitude of the projection of QP on n.
Now n n n n n • • = QP QP proj and so the distance = QP projn = QP n QP n PQ n QP n n n .
n n n n n n • • • • = = = • u u u r u u u r u u u r u u u r Since QP PQ .
• = • n n u u u r u u u r Example Find the distance between the point P = (2, 1, 3) and the plane 2x + y + 2z = 3.
Solution An arbitrary point Q on the plane is Q = (1, 1, 0). Then ) 3 , 0 , 1 ( ) 0 , 1 , 1 ( ) 3 , 1 , 2 ( QP = − = and a normal vector to the plane is n = (2, 1, 2). The distance between the point P and the plane is therefore equal to 3 8 4 1 4 6 0 2 ) 2 , 1 , 2 ( ) 2 , 1 , 2 ( ) 3 , 0 , 1 ( QP = + + + + = • = • n n .
Distance between a point and a line in R 3 Let l be a line passing through a point Q and parallel to the vector v. Let P be a point not on the line l. The distance between P and l is the perpendicular distance of the point P from l. In the adjoining diagram the distance between P and l is the length of the line segment PD. Let θ be the angle between v and QP . Then the required distance is QP sin QP PQ θ PD QP sinθ ⋅ ⋅ × × = ⋅ = = = v v v v v v u u u r u u u r u u u r u u u r u u u r Example Find the distance between the point P = (5, 7, 10) and the line having parametric equations x = 2 + 3t, y = 3 – 3t, z = 4 + t.
33 Q D P θ v l Solution The vector v = (3, −3, 1) is a vector parallel to the given line. The point Q = (2, 3, 4) is a point on the line so QP = (5, 7, 10) – (2, 3, 4) = (3, 4, 6). Also 4 6 3 6 3 4 QP , , (22,15, 21) 3 1 3 1 3 3 × = − = − − − v u u u r . The required distance is equal to v v × QP 19 46 25 19 1150 1 9 9 441 225 484 ) 1 , 3 , 3 ( ) 21 , 15 , 22 ( = = + + + + = − − = .
The distance between two lines in R 3 Let l1 be a line through P that is parallel to the vector v1. Let l2 be a line through Q that is parallel to the vector v2. The vector n = v1×v2 is a normal vector to both lines l1 and l2. The distance between the lines is the magnitude of the projection of QP on n. If θ is the angle between QP and n, then the distance between the lines is equal to QP cos QP PQ θ QP cosθ ⋅ ⋅ • • ⋅ = = = n n n n n n u u u r u u u r u u u r u u u r .
Example Find the distance between the lines l1: x = 3 + 2t, y = 5 – 3t, z = 4 + 4t and l2: x = 1 + t, y = 2 + 2t, z = 3 − 3t.
Solution P = (3, 5, 4) is a point on line l1 and Q = (1, 2, 3) is a point on line l2. The vector from Q to P is QP = (3, 5, 4) – (1, 2, 3) = (2, 3, 1). The vector n1 = (2, −3, 4) is parallel to line l1 and the vector n2 = (1, 2, −3) is parallel to line l2. The normal vector to lines l1 and l2 is 34 P Q P v2 v1 n l1 l2 the vector ) 7 , 10 , 1 ( 2 1 3 2 , 3 1 4 2 , 3 2 4 3 = − − − − − = × = 2 1 n n n .The distance between the lines is n n • QP = 6 5 39 150 39 49 100 1 7 30 2 ) 7 , 10 , 1 ( ) 7 , 10 , 1 ( ) 1 , 3 , 2 ( = = + + + + = • .
2.5 Problems 1. Find the distance between the given point P and the given plane.
a. P = (1, 3, 2) 2x + 2y + z = 1 b. P = ( 3, 5, 4) 3x + 4y + 5z = 12 c. P = (1, 0, 2) 2x + y + 2z = 1 d. P = (2, 2, 1) 3x + y + z = 9 2. Find the distance between the following pairs of parallel planes by selecting a point on one plane and then finding the distance between that point and the other plane.
a. 2x + 2y + z = 1, 2x + 2y + z = 6 b. 3x + y + z = 9, 3x + y + z = 25 3. Find the distance between the point P and the line l.
a. P = (1, 3, 4) l: x = 4 + 3t, y = 5 + 2t, z = 6 + 3t.
b. P = (5, 6, 4) l: x = 1 + t, y = 2 – t z = 2 + 2t c. P = (3, 2, 5)l: x = 1 – t, y = t z = 2 + t d. P = (2, 3, 1) l: x = 5 + 2t y = 6 + t z = 4 – t 4. Find the distance between the two given lines.
a. x = (1, 3, 2) + t(2, 5, 3) and x = (3, 5, 4) + t(1, 0, 1) b. x = 4 + 3t, y = 5 + 2t, z = 6 + 3t and x = 1 + t, y = 2 – t z = 2 + 2t c. 1 2 2 1 2 3 + = + = − z y x and 2 5 3 3 3 1 + = + = − z y x d. 2 2 1 3 4 5 − + = − + = − z y x and 3 3 1 2 5 2 + = − + = − z y x 5. The line l is parallel to the given plane. Find the distance between the line and the plane.
a. l: x = (1, 3, 2) + t(2, 5, 3) x − y + z = 1 b. l: x = (3, 5, 4) + t(1, 0, 1) 2x + 3y – 2z = 3 c. l: x = 1 + t, y = 2 – t, z = 2 + 2t 3x + 5y + z = 2 35 2.6 INTERSECTIONS Two distinct planes that are not parallel intersect in a line (diagram A). A line and a plane that are not parallel will intersect in a point (diagram B). Two nonparallel lines in R 3 may or may not intersect. If they intersect, the intersection is a point (diagram C). The line of intersection of two planes Let n1 and n2 be normals to a pair of intersecting planes. Then the vector v = n1×n2 is parallel to the line of intersection of the two planes. If P is any point on the line of intersection, then we can use the point parallel form equation x = p + tv to get an equation of the line of intersection.
Example 36 A B C n v n P n2 n1 v P Find a pointparallel form equation for the line of intersection of the two planes 2x + 3y + 4z = 1 and x – 5y + 6z = 3. Also, find a set of parametric equations for the line of intersection of the two planes.
Solution The planes have normals n1 = (2, 3, 4) and n2 = (1, 5, 6). The vector v = n1×n2 is parallel to the required line. 3 4 2 4 2 3 , , (38, 8, 13) 5 6 1 6 1 5 = × = − = − − − − 1 2 v n n . To find a point P on the line we must find a point that satisfies the two equations 2x + 3y + 4z = 1 and x –5y + 6z = 3. Since there are two equations with three variables we set z = 0 in the equations to get 2x + 3y = 1 and x –5y = 3.
Solving for x and y we get 13 14 = x and 13 5 − = y . Combining these values with z = 0 we get the point P = − 0 , 13 5 , 13 14 so the pointnormal form equation is x = − 0 , 13 5 , 13 14 + t(38, −8, −13). The parametric equations are t z t y t, x 13 0 , 8 13 5 38 13 14 − = − − = + = .
The intersection of a plane and a line To find the point of intersection of a line and a plane we use the parametric form of the equation of the line and the standard form equation for the plane. Substitute the values for x, y and z from the line equations into the equation of the plane and solve for the parameter. Now substitute the value of the parameter in the line equations to find the point of intersection. The next example illustrates the procedure.
Example Find the point of intersection of the line x = 1 + t, y = 2 – t, z = 3 + 2t and the plane 4x + y – z = 6.
Solution 4x + y – z = 6 4(1 + t) + (2 – t) – (3 + 2t) = 6 37 P 4 + 4t + 2 – t – 3 – 2t = 6 t = 3.
Substituting t = 3 into the line equations gives x = 1+ 3 = 4, y = 2 – 3 = –1, z = 3+ 6 = 9.
The point of intersection is therefore, P = (4, –1, 9). The intersection of two lines To determine whether two lines in R 3 have a point in common we equate the corresponding parts of their parametric equations and solve for the parameters. There will be three equations with two variables (parameters). If this system does not have a solution, the lines do not intersect. If the system has a solution, substituting the values of the parameters into the parametric equations gives the point of intersection.
Example Find the point of intersection of the lines l1: x = 1 + 2t, y = 3 + t, z = 2 + 3t and l2: x = 1 + s, y = 2 + s, z = 1 + 2s.
Solution Solving equations (1) and (2) for t and s gives t = 1 and s = 2.
These values also satisfy equation (3), so the system has a solution and the point of intersection is found by substituting t = 1 (or s = 2) into the line equations. Thus for example if we substitute the value t = 1 into the equations for the first line we get x =1 + 2(1) = 3, y = 3 + 1 = 4, z =2 + 3(1) = 5. The point of intersection is (3, 4, 5). [Note that if instead we substitute s = 2 into the equations for the second line we will get x = 1 + 3 = 3, y = 2 + 2 = 4, z = 1 + 2(2) = 5 and so find (3, 4, 5) is the point of intersection as before.] 38 1 + 3t = 1 + s 3 + t = 2 + s 2 + 3t = 1 + 2s 2t − s = 0 (1) t − s = −1 (2) 3t − 2s = −1 (3) Example Show that the lines x = 1 + 2t, y = 3 + t, z = 2 + 3t and x = 1 + s, y = 2 + s, z = 1 + 3s do not intersect.
Solution 1 + 2t = 1 + s 2t – s = 0 (1) 3 + t = 2 + s t – s = –1 (2) 2 + 3t = 1 + 3s 3t – 3s = –1 (3) Solving equations (1) and (2) gives t = 1 and s = 2. However, substituting t = 1 and s = 2 into the left hand side of equation (3) gives 3(1) – 3(2) = –3 ≠–1, so equation (3) is not satisfied and hence there is no point of intersection.
2.6 Problems 1. Find a set of parametric equations for the line of intersection of the following pairs of planes.
a. x + 2y – 3z = 4 and 2x – y + z = 1 b. 3x + 2y – z = 1 and 4x +2y +z = 3 c. 3x + y – z = 5 and x + y + 2z = 3 d. x – y + 2z = 4 and 2x – 3y + 2z = 1 2. Find the coordinates of the point of intersection of the given line and plane.
a. x = 2 + t, y = 1 + t, z = 2 – t x + y + z = 6 b. x = 4 – t, y = –2 + 2t, z = 3 – t 2x + y + z = 7 c. x = –2 + t, y = 3 – t, z = –1 + 4 t 2x + 3y +z = 1 d. x = (1, 4, 1) + t(2, 1, 2) (1, 1, –1)• x = 4 3. Determine whether or not the following pairs of lines intersect. If they do intersect, find the coordinates of the point of intersection.
a. x = 2 + t, y = 1 + t, z = 2 – t and x = 6 – s, y = 5 – s, z = 7 – 2s b. x = –2 + t, y = 3 – t, z = –1 + t and x = 5 – s, y = 6 – s, z = 11 – 2s c. x = 4 – t, y = –2 + 2t, z = 3 – t and x = 1 + s, y = 2 – s, z = 4 – s 39 d. x = 2 + t, y = 1 + t, z = 2 – t and x = 5 + s, y = 4 – s, z = 1 + 2s 2.7 DIRECTION NUMBERS AND DIRECTION COSINES The orientation of a line l in R 3 has been specified by giving a vector v = (v1, v2, v3) that is parallel to the line l. Since cv = (cv1, cv2, cv3) is also parallel to l we see that any ordered set of three numbers (a, b, c) that are proportional to (v1, v2, v3) will suffice to give the orientation of the line l. We call any ordered set of three numbers (a, b, c) that are proportional to (v1, v2, v3) a set of direction numbers for the line l. If we are given a set of direction numbers for a line we know its orientation in space.
Since the orientation of a line l in R 3 is not changed if we move it to a new location but keep it parallel to the direction of the line l in its original position, we can restrict any further discussion to lines passing through the origin. Direction angles Let α, β and γ be the angles that the line l passing through the origin makes with the positive xaxis, positive yaxis and positive zaxis respectively. The angles α, β and γ are called a set of direction angles for the line l. The orientation of a line in space can be specified by giving the direction angles instead of a set of direction numbers. Direction cosines If the point P = (a, b, c) is on the line l then the vector OP is equal to ) ( ) 0 , 0 , 0 ( ) ( OP c b, a, c b, a, = − = . 40 x y z l α β γ x y z l P(a,b,c) O Since OP lies on l, the set of numbers (a , b , c) are a set of direction numbers for line l. If we now let OP = d , then the cosines of the three direction angles , , α β γ are d a = α cos , d b = β cos and d c = γ cos . These cosines are proportional to a, b, c and form a special set of direction numbers for the line l called the direction cosines of l.
Theorem If γ β α cos , cos , cos are the direction cosines of a line l, then .
1 cos cos cos 2 2 2 = γ + β + α Proof Let P = (a, b, c) and let OP = d . Then ( ) ( ) ( ) 1 cos cos cos 2 2 2 2 2 2 2 2 2 2 2 2 = = + + = + + = γ + β + α d d d c b a d c d b d a Example The line l passes through the points P = (1, 3, 4) and Q = (5, 4, 7). Find a set of direction numbers and a set of direction cosines for the line l.
Solution = PQ (5, 4, 7) – (1, 3, 4) = (4, 1, 3) so a set of direction numbers for l is (4, 1, 3).
Since d = 26 3 1 4 PQ 2 2 2 = + + = , a set of direction cosines for the line l is 26 3 cos , 26 1 cos , 26 4 cos = γ = β = α .
2.7 Problems 1. Find a set of direction numbers and a set of direction cosines for the line through points P and Q.
41 a. P = (1, 2, 3) Q = (3, 5, 4) b. P = (3, 5, 4) Q = (7, 6, 5) c. P = (2, 2, 1) Q = (3, 4, 4) d. P = (5, 6, 7) Q = (4, 3, 2) 2. Find a set of direction numbers and direction cosines for the following lines.
a. x = 1 + t, y = 2 – 3t, z = 3 + 4t b. x = 2 – t, y = 3 + 2t, z = 4 c. (2,3,1) (4,2,3) t = + x d. (5,1,3) (3,0 4) t = + − x 3. Find a set of parametric equations for the line passing through the given point P and having the given set of direction numbers.
a. P = (2, 5, 1) (a, b, c) = (3, 2, 4) b. P = (3, −2, 1) (a, b, c) = (−1, 0 2) c. P = (4, 1, 3)(a, b, c) = (3, 0, 2) d. P = (5, 0, −2) (a, b, c) = (2, 1, −2) 4. Find a standard form equation for the plane passing through the point P and having a normal with the given set of direction numbers.
a. P = (2, 5, 1) (a, b, c) = (3, 2, 4) b. P = (3, −2, 1) (a, b, c) = (−1, 0 2) c. P = (4, 1, 3)(a, b, c) = (3, 0, 2) d. P = (5, 0, −2) (a, b, c) = (2, 1, −2) 42 |
8928 | https://www.hmhco.com/blog/teaching-x-and-y-axis-graph-on-coordinate-grids?srsltid=AfmBOooOiMorVVFDR59bw03X8quhkeoFJFRt7RWeQOL7aGRpoCfcja-f | Teaching x- and y-Axis Graphing on Coordinate Grids | HMH
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Math
Teaching x- and y-Axis Graphing on Coordinate Grids
Richard Blankman
October 22, 2021
8 Min Read
Coordinate graphing can sound very daunting for students in Grades 4–9, but it's actually just a visual method for showing relationships between numbers. The relationships are shown on a coordinate grid. A coordinate grid has two perpendicular lines, or axes(pronounced AX-eez), labeled just like number lines. The horizontal axis is usually called the x-axis. The vertical axis is usually called the y-axis. The point where the x- and y-axis intersect is called the origin.
Drawing a Coordinate Graph
The numbers on a coordinate grid are used to locate points. Each point can be identified by an ordered pair of numbers; that is, a number on the x-axis called an x-coordinate, and a number on the y-axis called a y-coordinate. Ordered pairs are written in parentheses (x-coordinate, y-coordinate). The origin is located at (0,0). Note that coordinates are often written with no space after the comma.
The location of (2,5) is shown on the coordinate grid below. The x-coordinate is 2. The y-coordinate is 5. To locate (2,5), move 2 units to the right on the x-axis and 5 units up on the y-axis.
The order in which you write x- and y-coordinates in an ordered pair is very important. The x-coordinate always comes first, followed by the y-coordinate. As you can see in the coordinate grid below, the ordered pairs (3,4) and (4,3) are two different points!
Describing a Linear Relationship
The function table below shows the x- and y-coordinates for five ordered pairs. You can describe the relationship between the x- and y-coordinates for each of these ordered pairs with this rule: the x-coordinate plus two equals the y-coordinate. You can also describe this relationship with the algebraic equation x + 2 = y.
x-coordinatex + 2 = yy-coordinateordered pair
00 + 2 = 22(0,2)
11 + 2 = 33(1,3)
22 + 2 = 44(2,4)
33 + 2 = 55(3,5)
44 + 2 = 66(4,6)
To graph the equation x + 2 = y, each ordered pair is located on a coordinate grid, then the points are connected. In fact, the graph forms a straight line. The arrows indicate that the line goes on in both directions.
For students who are ready to take it to the next level, consider explaining that the graph for any equation that can be written as ax + by = c, where a, b, and c are numbers, forms a straight line. Notice how x + 2 = y can also be written as ax + by = c, where a = 1, b = –1, and c = –2.
Introducing the Concept
Finding and Graphing Points for Linear Relationships
Your students may have encountered ordered pairs last year, but it's a good idea to start by reviewing how to locate a point on a grid from an ordered pair. A day spent plotting coordinates that fall in a straight line will be a day well spent.
Key Standard:Graph points on the coordinate plane. (5.G.A.1)
Materials: Poster paper or a way to display a coordinate grid publicly for the class; straightedge
Preparation: Draw a large coordinate grid that the entire class can see. Label the x- and y-axes from 0 through 10.
Prerequisite Skills and Concepts: Students should know about ordered pairs and locating points on a grid.
Write these ordered pairs where all students can see them: (6,4); (7,5); (8,6); and (9,7). Point to the ordered pair (6,4).
Ask:What rule describes the relationship between the numbers in this ordered pair?
Although many rules work for this pair in isolation, elicit from students this rule: the first number minus two equals the second number.
Ask:Does the same rule apply to the other ordered pairs?
Students should notice that each ordered pair follows this rule. You can help them by using the rule to write each ordered pair as an equation: 6 – 2 = 4, 7 – 2 = 5, 8 – 2 = 6, 9 – 2 = 7.
Say:Let's locate these ordered pairs on a grid.
Ask:How would you locate the point for (6,4) on the grid?
Students should say to "start at 0, move 6 units to the right, then 4 units up." Mark this point on the grid for the class to see.
Have students verbalize how to locate the point for each of the other ordered pairs. Then mark each point on the grid. Emphasize the importance of moving right for the first number in the ordered pair and up for the second number.
Ask:What figure do you think will be formed by connecting the points on the grid?
Students should see that a line will be formed. Use a straightedge to connect the points.
Provide students with other examples of ordered pairs that follow a rule. Have students identify the rule and explain how to graph the points. One example could read, "Rule: The first number plus three equals the second number; ordered pairs: (2,5); (3,6); (4,7); and (5,8)."
Developing the Concept
Finding and Graphing Points for Linear Relationships
At this level, students will begin to see the relationship between equations and straight-line graphs on a coordinate grid.
Key Standard:Interpret an equation as a linear function, whose graph is a straight line. (8.F.A.3)
Materials:Poster paper or a way to display a coordinate grid publicly for the class; straightedge; one copy of a coordinate grid, a straightedge, and lined paper for each student
Preparation:Draw a coordinate grid where all students can see it. Label the x- and y-axes from 0 through 10. Ensure all students have a copy of the grid.
Prerequisite Skills and Concepts: Students should know about ordered pairs and locating points on a grid. They should also be able to recognize and interpret an equation.
Write the equation x + 5 = y publicly for the class to see.
Ask:How could you say this equation in words?
Students should say that the equation means "a number plus five equals another number," or a comparable statement.
Draw a table with four columns and five rows. Have students draw their own table. Label the first column x, the second column x + 5, and the third column y. Leave the fourth column blank for now. Write "1" in the first column below x.
Ask:What happens to the equation if we replace x with 1? Elicit from students the equation 1 + 5 = 6. Write "1 + 5" in the second column below "x + 5." Then write "6" in the third column below y.
Continue to replace x with 2, 3, then 4. Have students complete the first three columns of their tables on their own. Then ask for a volunteer to complete the table publicly for the class.
Say:Let's write ordered pairs using the values of x and y . Label the fourth column of your table "Ordered Pairs." Remind students that when they locate points on a grid, they first move right on the x-axis, then up on the y-axis. Therefore, the first number in an ordered pair is a value for x, and the second number is a value for y. These numbers are called the x- and y-coordinates.
Ask:What is the first number we used for x? (1) What is the first number we calculated for y? (6) So, what is the first ordered pair? (1,6)
Have students complete their tables. When they are finished, record the ordered pairs in the table publicly for the class.
Say:Now we're going to graph the equation x + 5 = y on a grid. (Point to the grid you made.) This grid is called a coordinate grid. Let's take a closer look at the different parts of the grid.
Point to the horizontal line on the grid.
Say:This line is called the x-axis.
Point to the vertical line on the grid.
Ask:What do you think this line is called?
Students should make the connection to the y-axis.
Say:Now, let's locate the ordered pairs on the grid. Who can find (1,6)?
Have a volunteer describe the location of the ordered pair. Mark the location on the coordinate grid for all students to see. Then have students locate the rest of the ordered pairs on their own grids.
Say:Let's connect all of the points. What figure did we make?
Have students use a straight edge to connect the points. Show students how extending both ends of the line slightly, and drawing arrows, shows that the line goes on in both directions. Students should identify the figure as a straight line.
Have students repeat this activity with the equation x – 2 = y. Use the numbers 5, 6, 7, 8, and 9 for x.
Wrap-Up and Assessment Hints
These skills will need lots of practice. Reinforce the need for students to work carefully so their graph is accurate. When you assess students' progress, keep the number of exercises small enough that they have time to complete each step without rushing.
This blog post, originally published in 2020, has been updated for 2021.
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Nine-point circle
Triangle ABC with the nine-point circle in light orange
The nine-point circle (also known as Euler's circle or Feuerbach's circle) of a given triangle is a circle which passes through 9 "significant" points:
The three feet of the altitudes of the triangle.
The three midpoints of the edges of the triangle.
The three midpoints of the segments joining the vertices of the triangle to its orthocenter. (These points are sometimes known as the Euler points of the triangle.)
"The nine-point circle is tangent to the incircle, has a radius equal to half the circumradius, and its center is the midpoint of the segment connecting the orthocenter and the circumcenter." -hankinjg
That such a circle exists is a non-trivial theorem of Euclidean geometry.
The center of the nine-point circle is the nine-point center and is usually denoted .
The nine-point circle is tangent to the incircle, has a radius equal to half the circumradius, and its center is the midpoint of the segment connecting the orthocenter and the circumcenter, upon which the centroid also falls.
It's also denoted Kimberling center .
Contents
[hide]
1 First Proof of Existence
2 Second Proof of Existence
3 Common Euler circle
4 See also
First Proof of Existence
Since is the midpoint of and is the midpoint of , is parallel to . Using similar logic, we see that is also parallel to . Since is the midpoint of and is the midpoint of , is parallel to , which is perpendicular to . Similar logic gives us that is perpendicular to as well. Therefore is a rectangle, which is a cyclic figure. The diagonals and are diagonals of the circumcircle. Similar logic to the above gives us that is a rectangle with a common diagonal to . Therefore the circumcircles of the two rectangles are identical. We can also gain that rectangle is also on the circle.
We now have a circle with the points , , , , , and on it, with diameters , , and . We now note that . Therefore , , and are also on the circle. We now have a circle with the midpoints of the sides on it, the three midpoints of the segments joining the vertices of the triangle to its orthocenter on it, and the three feet of the altitudes of the triangle on it. Therefore, the nine points are on the circle, and the nine-point circle exists.
Second Proof of Existence
We know that the reflections of the orthocenter about the sides and about the midpoints of the triangle's sides lie on the circumcircle (side proof, midpoint proof), as do the vertices of the triangle. So, consider the homothety centered at with ratio . It maps the circumcircle of (and those 9 points) to a circle, including mapping the vertices of the triangle to its Euler points (by definition). This is the nine-point circle.
Common Euler circle
Let an acute-angled triangle with orthocenter be given.
be the point on opposite
Points and such that is a parallelogram. The line intersects at the points and
Prove that triangles and has common Euler (nine-point) circle.
Proof Denote is midpoint
Let’s consider Circumcenter of point is the midpoint point is the midpoint
Denote the centroid of
is the centroid of
Denote the midpoint of is the midpoint of
is the centroid of
Point is the circumcenter of is the orthocenter of
The triangles and has common circumcircle and common center of Euler circle (the midpoint of ) therefore these triangles has the common Euler circle.
vladimir.shelomovskii@gmail.com, vvsss
See also
Kimberling center
Center line
Evans point
Euler line
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Perpendicular Lines
Properties of Triangles
Geometric Shape
PLANE GEOMETRY
Angle Bisector Theorem
5
What is the relationship between angle bisectors and perpendicular lines in triangles?
Dean Rubine
Been doing high school math since high school, circa 1975
·
Author has 10.5K answers and 23.4M answer views
·
1y
It’s a bot question, but a pretty good one.
The incircle of a triangle is shown here; the red angle bisectors meet at the incenter I. Each of the three radii to a tangent point on a side is of course perpendicular to its respective side, as tangents and radii are.
That makes the three radii altitudes of three little triangles, which add up to the big triangle, so area Δ satisfies:
Δ=12ar+12br+12cr
r=2Δa+b+c
We can get an nice expression for r2,
r2=4(2Δ)24(a+b+c)2=4a2b2−(a2+b2−c2)24(a+b+c)2
It’s a bot question, but a pretty good one.
The incircle of a triangle is shown here; the red angle bisectors meet at the incenter I. Each of the three radii to a tangent point on a side is of course perpendicular to its respective side, as tangents and radii are.
That makes the three radii altitudes of three little triangles, which add up to the big triangle, so area Δ satisfies:
Δ=12ar+12br+12cr
r=2Δa+b+c
We can get an nice expression for r2,
r2=4(2Δ)24(a+b+c)2=4a2b2−(a2+b2−c2)24(a+b+c)2
Related questions
Are the bisectors of an angle of two straight lines always perpendicular?
How many perpendicular lines does a right angle triangle have?
What is the relationship between angle bisectors?
Draw an obtuse angled triangle and a right angled triangle.Findthe points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie?
If a line bisects an angle of a triangle and is perpendicular to the opposite side, then does it bisect that side?
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A Line or line segment is 180°
You draw a perpendicular from the endpoints
That is a perpendicular
You bisected a 180 angle with an angle bisector
I.learned geometry in 1972 73
I never associated an angle bisector with perpendicular lines
But funny
I just told you how….
Any acute angle bisector is less than 90
Where is a perpendicular coming into.play
You bisected an angle. And
?
Yes you can find something
But why?
You bisected the angle any perpendicular is to be drawn off that
And??
You make a right triangle with a side??
That's the connection?
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Angle bisectors intersect at a point known as incentre . minimum distance between incentre to any side is constant which is inradius
your question is loosely worded as “ perpendicular lines they may be from incentre to sides or they may be fron vertices or any point inside or outside triangle
if they are perpendicular bisector of sides then only we get circum cetre circle circumsize triangle
if ꓕ lines are drawn from vertices to opposite sides they are altitudes only
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In a equilateral triangle, the bisector that divides an angle in two and when the bisector touches the opposite side of the bisected angle it falls perpendicular to that side. But if the triangle is other type, this will not happens. So be careful!
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Which type of triangle will always have a perpendicular bisector that is also an angle bisector?
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Is the median of a triangle always the perpendicular bisector?
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What is the angle between the internal and external bisectors of an angle of a triangle?
The angle should be 90 degrees. We can prove the same with help of an equilateral triangle.
The angle should be 90 degrees. We can prove the same with help of an equilateral triangle.
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What is the difference between a perpendicular bisector and an angle bisector?
What is the difference between a perpendicular bisector and an angle bisector?
A perpendicular bisector cuts a line segment into two equal pieces, while angle bisector cuts an angle into two equal angles.
What is the difference between a perpendicular bisector and an angle bisector?
A perpendicular bisector cuts a line segment into two equal pieces, while angle bisector cuts an angle into two equal angles.
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Are the bisectors of an angle of two straight lines always perpendicular?
Yes, of course. Why?
Just follow the simple declarations below and draw the figure yourself:
Let AR and BS be two straight line segments meeting at point O;
Let CX and DY be the ∠ bisectors of ∠AOB and ∠AOS respectively;
Let ∠AOB = 2u and ∠AOS = 2v
Now ∠AOB + ∠AOS = 180° . . . . . . (sum of angles on a straight line)
But from declaration (3), ∠AOB + ∠AOS = 2u + 2v
so 2(u + v) = 180°
or u + v = 90°
Also from declaration (2), ∠COA = u and ∠AOD = v
so ∠COA + ∠AOD = ∠COD = u + v = 90°
hence ∠COD = 90°
But since ∠COD is the angle between CX and DY (the two angle bisectors of the two straight lines AR & BS) it
Yes, of course. Why?
Just follow the simple declarations below and draw the figure yourself:
Let AR and BS be two straight line segments meeting at point O;
Let CX and DY be the ∠ bisectors of ∠AOB and ∠AOS respectively;
Let ∠AOB = 2u and ∠AOS = 2v
Now ∠AOB + ∠AOS = 180° . . . . . . (sum of angles on a straight line)
But from declaration (3), ∠AOB + ∠AOS = 2u + 2v
so 2(u + v) = 180°
or u + v = 90°
Also from declaration (2), ∠COA = u and ∠AOD = v
so ∠COA + ∠AOD = ∠COD = u + v = 90°
hence ∠COD = 90°
But since ∠COD is the angle between CX and DY (the two angle bisectors of the two straight lines AR & BS) it follows that CX is always perpendicular to DY.
Hence proved.
Pramodkumar Tandon
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If a line bisects an angle of a triangle and is perpendicular to the opposite side, then does it bisect that side?
Given : Triangle ABC, in which AP bisects angle A and meets opposite side BC in P.
Also AP is perpendicular to BC.
To prove that BP = CP
Proof :
Consider Triangles ABP and ACP:
Angle BAP = Angle CAP = θ ………….. .. AP is bisector
Angle APB = APC = 90 degree AP is perpendicular to BC
AP is common to both the triangles.
Hence the two triangles are congruent.
Hence Sides facing equal angles must be equal to each other
Thus BP = CP.
Hence AP bisects BC …………………. Proved
Given : Triangle ABC, in which AP bisects angle A and meets opposite side BC in P.
Also AP is perpendicular to BC.
To prove that BP = CP
Proof :
Consider Triangles ABP and ACP:
Angle BAP = Angle CAP = θ ………….. .. AP is bisector
Angle APB = APC = 90 degree AP is perpendicular to BC
AP is common to both the triangles.
Hence the two triangles are congruent.
Hence Sides facing equal angles must be equal to each other
Thus BP = CP.
Hence AP bisects BC …………………. Proved
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Sripad Sambrani
Knows Sanskrit
·
Upvoted by
Samuel Gomes da Silva
, Ph.D. Mathematics & Set Theory, University of São Paulo (2004) · Author has 6.7K answers and 2.9M answer views
·
4y
Related
Where can the perpendicular bisectors of the sides of a right triangle intersect?
The perpendicular bisectors of the sides of a right triangle meet at the mid-point of the hypotenuse, since it must be equidistant from all the vertex, the points on a circum-circle.
Trust this helps.
The perpendicular bisectors of the sides of a right triangle meet at the mid-point of the hypotenuse, since it must be equidistant from all the vertex, the points on a circum-circle.
Trust this helps.
Carter McClung
B.S. in UPSC General Studies & Mathematics, The University of Texas at Dallas (Graduated 2006)
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 1.6K answers and 6.5M answer views
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6y
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What is the difference between a bisector and a perpendicular bisector?
The word perpendicular.
A segment bisector intersects a segment at its midpoint. (It bisects the segment.)
A perpendicular segment bisector does the same but at a 90° angle.
If you want to see an example, look at the median (segment bisector) and the perpendicular bisector of a triangle
Median:
Perpendicular bisector:
Both intersect at the midpoint, but the perpendicular bisector forms a 90° angle. (It’s perpendicular.)
The word perpendicular.
A segment bisector intersects a segment at its midpoint. (It bisects the segment.)
A perpendicular segment bisector does the same but at a 90° angle.
If you want to see an example, look at the median (segment bisector) and the perpendicular bisector of a triangle
Median:
Perpendicular bisector:
Both intersect at the midpoint, but the perpendicular bisector forms a 90° angle. (It’s perpendicular.)
Studied Mathematics & High School Mathematics
·
Author has 129 answers and 26.9K answer views
·
Jan 5
Related
What is the proof for the triangle angle bisector theorem?
Which theorem? The bot clearly isn’t interested in proving it. There are at least 3 of them that I know of. The interior bisector theorem, the exterior bisector theorem and the two bisector theorem. I will state all three:
The interior bisector theorem.
Let
D
∈
B
C
be a point in
Δ
A
B
C
. The half-line
A
D
is the interior bisector of
∡
B
A
C
if, and only if
B
D
D
C
=
A
B
A
C
.
2. The exterior bisector theorem.
Let
D
′
∈
B
C
be a point in
Δ
A
B
C
. The half-line
A
D
′
is the exterior bisector of
∡
B
A
C
if, and only if
\tfra
Which theorem? The bot clearly isn’t interested in proving it. There are at least 3 of them that I know of. The interior bisector theorem, the exterior bisector theorem and the two bisector theorem. I will state all three:
The interior bisector theorem. Let D∈BC be a point in ΔABC. The half-line AD is the interior bisector of ∡BAC if, and only if BDDC=ABAC.
The exterior bisector theorem. Let D′∈BC be a point in ΔABC. The half-line AD′ is the exterior bisector of ∡BAC if, and only if ¯¯¯¯¯¯¯¯¯¯D′B¯¯¯¯¯¯¯¯¯¯D′C=ABAC.
The two bisector theorem. Let D,D′∈BC be two points in ΔABC and ∡DAD′=90∘. If ¯¯¯¯¯¯¯¯¯¯D′B¯¯¯¯¯¯¯¯¯¯D′C=−¯¯¯¯¯¯¯¯¯DB¯¯¯¯¯¯¯¯¯DC, then AD and AD′ are the bisectors (one interior and one exterior) of ∡BAC.
Let’s prove them one by one, but it’s much easier than it seems.
The interior bisector theorem proof.
We will consider the line through B, parallel to AD, which intersects AC at the point E (image here:)
From Thales theorem we deduce that BDDC=EAAC.
But we also have ∡BEA≡∡DAC≡∡BAD≡∡ABE so that means ΔAEB is isosceles, so AE=AB and plugging back in our relation we obtain BDDC=ABAC.
The exterior bisector theorem proof.
Looking at this provided diagram, can you see why it’s true?
It’s proof is similar to the interior bisector ( but now BE′ || AD′).
The two bisector theorem proof.
We draw BE,BE′ parallel to AD,AD′ with E,E′ being on AC - see diagram below.
We obtain the same thing as above with Thales Theorem , which is EAAC=BDDC and AE′AC=D′BD′C.
Using the absolute value equality from the statement we get EAAC=AE′AC, so EA=AE′. But ∡EBE′ has sides parallel with the sides of ∡DAD′ and that means that ∡EBE′=90∘. This also means that BA is the median in ΔEBE′, so BA=EA=AE′ (document about this! , the proof is easy) going back and substituting AB=AE in the first proportion and AB=AE′ in the second, we obtain:
ABAC=BDDC, ABAC=D′BD′C.
It immediately follows that only one of the ratios ¯¯¯¯¯¯¯¯¯¯D′B¯¯¯¯¯¯¯¯¯¯D′C and ¯¯¯¯¯¯¯¯¯DB¯¯¯¯¯¯¯¯¯DC is negative and we can then decide (with the first two theorems) what “type” of bisector of ∡BAC are AD and AD′.
Fred Di Francesco
Former Part Time Student Physics and Philosophy U of Wisc (1968–1983)
·
Author has 529 answers and 1.1M answer views
·
5y
Related
What is the difference between a perpendicular bisector and an angle bisector?
I’ve already told you how to locate an angle bisector; you swing arcs from the sides of a triangle, with the idea of locating a point that is equidistant from both sides. That way, when you draw the ray to bisect the angle, it will be correctly placed; that is, it will cut the angle exactly in half.
A perpendicular bisector is at the other end of the Angle Bisector, if it cuts the side in half and forms a 90° angle. Take a look at an equilateral triangle, for instance. When you draw a properly-located angle bisector, one that originates from the exact center of the vertex. When you extend that
I’ve already told you how to locate an angle bisector; you swing arcs from the sides of a triangle, with the idea of locating a point that is equidistant from both sides. That way, when you draw the ray to bisect the angle, it will be correctly placed; that is, it will cut the angle exactly in half.
A perpendicular bisector is at the other end of the Angle Bisector, if it cuts the side in half and forms a 90° angle. Take a look at an equilateral triangle, for instance. When you draw a properly-located angle bisector, one that originates from the exact center of the vertex. When you extend that line to the side opposite, it will cut that side in half; you will have created two congruent 30° 60° 90° Right Triangles. The nice thing about 30° 60° 90° triangles is that they’re so easy to solve. Their ratios are 1: √3/2 : 2; that is. the lengths of their sides is: short side = 1:1, the long side is √3/2: 1 and the hypotenuse is 2:1. This is why you bisect ‘unruly’ triangles; by bisecting them you make them Right Triangles, which are inherently easier to work with.
Naivedya Amarnani
A keen, curious student who knows a thing or two about Math
·
Updated 3y
Related
What is the angle between the internal and external bisectors of an angle of a triangle?
90° or π/2 radians.
Here's how
If in a triangle you take any one angle, let's say theta, then after it is bisected each smaller angle will now be theta/2.
If you take the external angle of theta, it will be 180°-theta and if this is bisected the two new angles would be (180°-theta)/2 which is equal to 90° - theta/2.
Now adding these two angles, we get theta/2 + 90° - theta/2 = 90°.
Hence the angle between the internal and external bisectors of an angle in a triangle is always 90° or π/2 radians.
You can see the proof here too:
Section 4.2 Discussion
Hope this helped!
Related questions
Are the bisectors of an angle of two straight lines always perpendicular?
How many perpendicular lines does a right angle triangle have?
What is the relationship between angle bisectors?
Draw an obtuse angled triangle and a right angled triangle.Findthe points of concurrence of the angle bisectors of each triangle. Where do the points of concurrence lie?
If a line bisects an angle of a triangle and is perpendicular to the opposite side, then does it bisect that side?
Which type of triangle will always have a perpendicular bisector that is also an angle bisector?
What is the relationship between the base angles of an obtuse-angled triangle?
What is the relationship between the perpendicular bisectors of a triangle?
Is the median of a triangle always the perpendicular bisector?
Are bisectors of an angle always pependicular?
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8932 | https://www.youtube.com/watch?v=_FjPUtWLAM0 | dy/dx = ky differential equation - Exponential Growth
Cowan Academy
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Description
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Posted: 2 Mar 2017
y = Ae^kt derivation
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Transcript:
so in this video I will be showing you how you get from this differential equation to the form you're more familiar with y equals 2a ekx or in some other countries this X would be time because its growth in decay so it's basically a growth with respect to time but in this video I ought to be using X so we start off with moving the variables and separating them to either side so we have dy on y equals to K times DF next we take the integral of both sides so we know the derivative of one wise knowin off the high and we also know they know of K with respect to X it's just KX and add some constant there would be a constant on the side and a constant inside if you had to involve all subtract you get one constant so we just leave with one seen here if we take the inverse log of both sides we have y equals 2y e to the ax plus C so using our lock laws we also know that if we have X a plus B this is the same as X to power a times X to the power P so we would have y equals to e to the K X minus e to the C and what we know about e to the power C is that this is just a number it was erased to some number it's just some number so we we call this number hey and people didn't notice e to the power C so this a is a constant so if we substitute this into our equation we have a to the power K X so if we were to call our Y the population P he goes to a cave with respect to time so Sammy wants it to find the population with respect to time our K value determines the instantaneous growth rate if K is less than zero our population is decaying exponentially if it's greater than zero our population is increasing exponentially T is time and pees population a will be given as probably the initial growth so say you saw half of two thousand people and over time you want to find the population growth you would call a 2000 |
8933 | https://www.mathsisfun.com/symbols.html | Symbols in Geometry
Symbols in Algebra
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Mathematical Symbols
Symbols save time and space when writing. Here are the most common mathematical symbols:
| Symbol | Meaning | Example |
---
| + | add | 3+7 = 10 |
| − | subtract | 5−2 = 3 |
| × | multiply | 4×3 = 12 |
| ÷ | divide | 20÷5 = 4 |
| / | divide | 20/5 = 4 |
| ( ) | grouping symbols | 2(a−3) |
| [ ] | grouping symbols | 2[ a−3(b+c) ] |
| { } | set symbols | {1, 2, 3} |
| π | pi | A = πr2 |
| ∞ | infinity | ∞ is endless |
| = | equals | 1+1 = 2 |
| ≈ | approximately equal to | π ≈ 3.14 |
| ≠ | not equal to | π ≠ 2 |
| < ≤ | less than, less than or equal to | 2 < 3 |
| > ≥ | greater than, greater than or equal to | 5 > 1 |
| √ | square root ("radical") | √4 = 2 |
| ° | degrees | 20° |
| ∴ | therefore | a=b ∴ b=a |
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8934 | https://byjus.com/physics/boltzmann-constant/ | In thermodynamics, the Boltzmann constant is the physical constant relating the average kinetic energy of the gas particles and the temperature of the gas represented by k or kB. The value of the Boltzmann constant is measured using J/K or m2Kgs-2K-1, which is mostly observed in Boltzmann’s entropy formula and Planck’s law of Black body radiation. Table of Contents:
Value Of Boltzmann Constant
Value of k
What is Boltzmann Constant
Boltzmann Constant Formula
Applications
Frequently Asked Questions – FAQs
Value Of Boltzmann Constant
The values of Boltzmann constant is obtained by dividing gas constant R by Avogadro’s number NA. The value of k is
Boltzmann constant, kB= 1.3806452 × 10-23J/K
Value Of k
The value of Boltzmann constant in eV is 8.6173303 × 10-5 eV/K
The value of the Boltzmann constant can be expressed in various units. The table given below consists of the value of k along with different units.
| | |
--- |
| Value Of k | Units |
| 1.3806452 × 10-23 | m2.Kg.s-2.K-1 |
| 8.6173303 × 10-5 | eV.K-1 |
| 1.38064852 × 10-16 | erg.K-1 |
| 2.0836612(12)×1010 | Hz.K-1 |
| 3.2976230(30)×10-24 | cal.K-1 |
| 0.69503476(63) | cm-1.K-1 |
| −228.5991678(40) | dB.WK-1.Hz-1 |
| 4.10 | pN.nm |
| 0.0083144621(75) | kJ.mol-1K-1 |
| 1.0 | Atomic unit (u) |
What is Boltzmann Constant?
The Boltzmann constant is introduced by Max Planck and named after Ludwig Boltzmann. It is a physical constant obtained by taking the ratio of two constants namely the gas constant and Avogadro number.
Boltzmann Constant Formula
The behaviour of the gases made understanding a step closer to Planck and Boltzmann by introducing constants. The value of Boltzmann constant is mathematically expressed as-
| |
| (\begin{array}{l}k=\frac{R}{N_{A}}\end{array} ) |
k is Boltzmann’s constant.
NA is Avogadro number.
R is the gas constant.
Applications
The Boltzmann Constant is used in diverse disciplines of physics. Some of them are listed below-
In classical statistical mechanics, Boltzmann Constant is used to express the equipartition of the energy of an atom.
It is used to express the Boltzmann factor.
It plays a major role in the statistical definition of entropy.
In semiconductor physics, it is used to express thermal voltage.
Frequently Asked Questions – FAQs
Q1
What is the value of Boltzmann constant?
Boltzmann constant kB= 1.3806452 × 10-23 J/K.
Q2
How is Boltzmann constant represented?
Boltzmann constant is represented as k or kB.
Q3
What is the value of Boltzmann constant in eV?
The value of Boltzmann constant in eV is 8.6173303 10-5 eV/K
Q4
State true or false: The Boltzmann constant is measured using J/K.
True.
Q5
Define Boltzmann constant?
Boltzmann constant is the physical constant relating the average kinetic energy of the gas particles and the temperature of the gas.
Physics Related Topics:
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| Stefan-Boltzmann Constant |
| Faraday Constant |
| Value Of Gravitational Constant |
| Rydberg Constant |
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8935 | https://cs.stackexchange.com/questions/28123/returning-a-random-subset-with-length-k-of-n-strings-while-only-storing-at-most | algorithms - Returning a random subset with length k of N strings while only storing at most k of them - Computer Science Stack Exchange
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Returning a random subset with length k of N strings while only storing at most k of them
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Here's the problem. I've written a program that reads strings from stdin, and returns a random subset of those strings. The only other argument provided to the program is the length of the subset, k k. The subset must contain exactly k k strings selected uniformly at random from the entire input set.
It's easy to do this if every single string is stored in memory. (Memory proportional to N). The question is how to only store at most k k strings, and still ensure that the output is perfectly random.
I've tried to work it out with the following base case. Say k=1 k=1.
```
Subset 1
A
B
C
```
A will always be added, since the queue contains less than k items. If I only operate on what I know, which is the current number of items in the queue, the required length k k, and the encountered strings n n. So I've tried doing this: (k k - items in queue)/n n.
Using that, the probability of A being replaced by B is 1/2 1/2. Then the probability of B being replaced by C will be 1/3 1/3. The problem there is that there's no way of reducing the probability of A being replaced before I've seen all of the strings.
I'm sure this question must have a clever answer. The problem of writing this Subset program to use memory <=k<=k is a bonus question on an Algs course I'm taking, and I really hope they wouldn't ask something unanswerable.
algorithms
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edited Jul 1, 2014 at 6:27
Raphael
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asked Jul 1, 2014 at 6:05
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I seem to remember that this is not possible. To clarify: do you want k k i.i.d. uniform draws from the whole set, or one k k-subset uniformly drawn from all k k-subsets?Raphael –Raphael 2014-07-01 06:29:56 +00:00 Commented Jul 1, 2014 at 6:29
@Raphael I'm going to say the former, but I can't see how the latter differs. Every single element in the whole set should have an equal probability of being selected for the subset.Chara –Chara 2014-07-01 19:38:59 +00:00 Commented Jul 1, 2014 at 19:38
1 Consider the set {a,b,c,d}{a,b,c,d}, k=2 k=2 and a probability distribution with p a,b=p b,c=p c,d=p a,d=1/4 p a,b=p b,c=p c,d=p a,d=1/4. Note that every element has the same odds of being drawn -- 1/2 1/2 -- but there are pairs with probability zero, e.g. (b,d)(b,d), that is we don't have a uniform distribution over all 2 2-subsets. See my answer here (the wrong part) for a similar fallacy.Raphael –Raphael 2014-07-01 21:40:52 +00:00 Commented Jul 1, 2014 at 21:40
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Use reservoir sampling. This is a good description in Wikipedia, or in Knuth.
Let's start with the simple case, where k=1 k=1. You always have one string in memory. When you read the first string, you store it in memory. Each time you read a new string, you replace it with the one in memory with probability 1/i 1/i, if this is the i i th string you've read so far. At the end, output whatever is stored in memory. The end result is that each string in the input is equally likely to be output. See also Choosing an element from a set satisfying a predicate uniformly at random in O(1)O(1) space for description of this approach (thank you, Juho!).
This extends to arbitrary k k. See Wikipedia's description for details.
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edited Apr 13, 2017 at 12:48
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answered Jul 1, 2014 at 6:46
D.W.♦D.W.
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This question is relevant too.Juho –Juho 2014-07-01 07:55:01 +00:00 Commented Jul 1, 2014 at 7:55
1 @Juho, thank you! Good find. I've updated my answer to link to that question.D.W. –D.W.♦ 2014-07-01 16:35:28 +00:00 Commented Jul 1, 2014 at 16:35
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This problem is covered in The Art of Computer Programming. I can't recall exactly where, but the algorithm is pretty easy to understand when you know the trick.
Let l l be the number of lines read so far. At each stage, you read the next line from stdin. Choose r r to be a random integer uniformly chosen from the range [1,l+1][1,l+1]. If r≤k r≤k, then discard line number r r from the collection that you've kept, and replace it with the line you just read. Otherwise, drop the line.
The pseudocode looks something like this:
```
for i := 1 to k
read a line from stdin into L[i]
end for
l := k
while there are more lines left
read a line from stdin into x
r := random(1, l+1)
if r <= k then
L[r] := x
end if
l := l + 1
end while
```
Of course, in a robust implementation, you should deal with the case that there are fewer lines than k k, and so on.
Proof of correctness is left as an exercise.
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answered Jul 1, 2014 at 6:45
Pseudonym♦Pseudonym
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8936 | https://www.youtube.com/watch?v=Wdf4M8hrm7Y | Esfuerzos cortantes en ejes cilíndricos sometidos a pares de torsión// EJERCICIO RESUELTO
NGE motors
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913 views
30 Oct 2023
El eje sólido y el tubo mostrados en la figura están hechos de un
material que tiene un esfuerzo cortante permisible de 75 MPa.
Determine el par de torsión máximo que puede aplicarse a cada
sección transversal y demuestre el esfuerzo que actúa sobre un
elemento pequeño de material en el punto A del eje y en los puntos B y C del tubo.
The solid shaft and tube shown in the figure are made of a
material that has an allowable shear stress of 75 MPa.
Determine the maximum torque that can be applied to each
cross section and show the stress acting on a
small element of material at point A of the axis and at points B and C of the tube.
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8937 | https://brainly.com/question/16077597 | [FREE] How do I find a_1 for the geometric sequence below? S_n = 1640, \quad r = 3, \quad n = 8 - brainly.com
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How do I find a 1 for the geometric sequence below?
S n=1640,r=3,n=8
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Final answer:
To find a1 in a geometric sequence with given values of Sn, r, and n, you can use the formula a1 = Sn / (r^(n-1)). In this case, a1 is approximately 0.7503.
Explanation:
To find the value of a1 in a geometric sequence, you can use the formula a1 = Sn / (r^(n-1)), where Sn is the sum of the sequence, r is the common ratio, and n is the number of terms.
In this case, Sn is given as 1640, r is 3, and n is 8. Plugging these values into the formula, a1 = 1640 / (3^(8-1)) = 1640 / 3^7 = 1640 / 2187 ≈ 0.7503.
Therefore, the value of a1 in the geometric sequence is approximately 0.7503.
Answered by LivUllmann •4.3K answers•248.2K people helped
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To find the first term a 1 of a geometric sequence, use the sum formula S n=a 1(1−r)(1−r n). Plugging in the values provided (S n=1640, r=3, n=8), the approximate value of a 1 is 0.5.
Explanation
To find the first term a 1 of a geometric sequence given the sum of the first n terms S n, the common ratio r, and the number of terms n, you can use the formula for the sum of a geometric series:
S n=a 1(1−r)(1−r n)
In this case, you're provided with:
S n=1640
r=3
n=8
Now, we can rearrange the formula to solve for a 1:
S n=a 1(1−r)(1−r n)⟹a 1=S n(1−r n)(1−r)
Next, substitute the known values into the formula:
a 1=1640(1−3 8)(1−3)
Calculating 3 8: 3 8=6561
Now substituting that back into the equation:
a 1=1640(1−6561)(−2)
a 1=1640(−6560)(−2)
This simplifies to:
a 1=1640×6560 2
Calculating the multiplication gives:
a 1=1640×3280 1≈0.5
Thus, the value of a 1 in the geometric sequence is approximately 0.5.
Examples & Evidence
For another example, if S n=1000, r=2, and n=5, you would calculate a 1 similarly using the formula to find the first term.
The formula used for the sum of the first n terms of a geometric series is well-established in mathematics, specifically for sequences where the ratio between consecutive terms is constant.
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a1=1640/2187
Explanation
Geometric sequence formula: an=a1 r^n-1
s8=1640=a1 3^8-1
1640=a1 3^8-1
1640=a1 3^7
1640=a1 2187
a1=1640/2187
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8938 | https://www.cuemath.com/numbers/bodmas-rule/ | BODMAS Rule
BODMAS rule is an acronym that is used to remember the order of operations to be followed while solving expressions in mathematics. BODMAS stands for B - Brackets, O - Order of powers or roots, (in some cases, 'of'), D - Division, M - Multiplication A - Addition, and S - Subtraction. It means that expressions having multiple operators need to be simplified from left to right in this order only. First, we solve brackets, then powers or roots, then division or multiplication (whichever comes first from the left side of the expression), and then finally, subtraction or addition, whichever comes on the left side.
In this lesson, we will be learning about the BODMAS rule which helps to solve arithmetic expressions, containing multiple operations, like, addition (+), subtraction (-), multiplication (×), division (÷), and brackets ( ).
| What is BODMAS?
| BODMAS Full Form
| BODMAS or PEMDAS
| FAQs on BODMAS Rule
What is BODMAS?
BODMAS, which is referred to as the order of operations, is a sequence to perform operations in an arithmetic expression. Math is all about logic and some standard rules that make our calculations easier. So, BODMAS is one of those standard rules for simplifying expressions that have multiple operators.
In arithmetic, an expression or an equation involves two components:
Numbers
Numbers are mathematical values used for counting and representing quantities, and for making calculations. In math, numbers can be classified as natural numbers, whole numbers, integers, rational numbers, irrational numbers, real numbers, complex numbers, and imaginary numbers.
Operators or Operations
An operator is a character that combines two numbers and produces an expression or equation. In math, the most common operators are Addition (+), Subtraction (-), Multiplication (×), Division (÷). For mathematical expressions or equations, in which only a single operator is involved, finding the answer is fairly simple. In the case of multiple operators, finding a solution becomes a little trickier! Let us understand this with an example. Jenny and Ron solved a mathematical expression 6 × 3 + 2 separately. The following are the two different methods by which Jenny and Ron solved the expression:
Jenny's Method: 6 × 3 + 2 = 6 × 5 = 30, Ron's Method: 6 × 3 + 2 = 18 + 2 = 20.
As we can observe, Jenny and Ron got different answers. In mathematics, we know that there can only be one correct answer to this expression. How to decide who is correct? In such cases, we use BODMAS to find the correct answer. Let us look at the example given below to get an idea of how BODMAS works:
Bodmas Examples
Let us understand this using BODMAS examples.
Example 1: Simplify the expression using BODMAS.
10 + (5 × 3 + 2)
Solution: Let us solve 10 + (5 × 3 + 2) step by step.
Example 2: Simplify the expression using BODMAS.
15 + (30 ÷ 2)
Solution:
Let us solve 15 + (30 ÷ 2) step by step.
BODMAS Full Form
The BODMAS rule is used to evaluate mathematical expressions and to deal with complex calculations in a much easier and standard way.
BODMAS Meaning
According to the BODMAS rule, to solve any arithmetic expression, we first solve the terms written in brackets, and then we simplify the exponential terms, or solve for the operation 'of', which means multiplication, and move ahead to division and multiplication operations, and then, in the end, work on the addition and subtraction. Following the order of operations in the BODMAS rule, always results in the correct answer. Simplification of terms inside the brackets can be done directly. This means we can perform the operations inside the bracket in the order of division, multiplication, addition, and subtraction. If there are multiple brackets in an expression, all the same types of brackets can be solved simultaneously. For example, (14 + 19) ÷ (13 - 2) = 33 ÷ 11 = 3.
Observe the table given below to understand the terms and operations denoted by the BODMAS acronym in the proper order.
B | [{( )}] | Brackets
O | x² | Order of Powers or Roots, (in some cases, 'of')
D | ÷ | Division
M | × | Multiplication
A | + | Addition
S Subtraction
BODMAS or PEMDAS
BODMAS and PEMDAS are two acronyms that are used to remember the order of operations. The BODMAS rule is almost similar to the PEMDAS rule. There is a difference in the abbreviation because certain terms are known by different names in different countries. While using the BODMAS rule or the PEMDAS rule we should remember that when we come to the step of division and multiplication, we solve the operation which comes first from the left side of the expression. The same rule applies to addition and subtraction, that is, we solve that operation that comes first on the left side.
When to Use BODMAS?
BODMAS is used when there is more than one operation in a mathematical expression. There is a sequence of certain rules that needs to be followed when using the BODMAS method. This gives a proper structure to produce a unique answer for every mathematical expression.
Conditions to follow:
Easy Ways to Remember the BODMAS Rule
The simple rules to remember the BODMAS rule are given below:
Common Errors While Using the BODMAS Rule
One can make some common errors while applying the BODMAS rule to simplify expressions and those errors are given below:
☛ Related Topics
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BODMAS Rule Examples
Example 1: Simplify the expression by using the BODMAS rule: [18 - 2(5 + 1)] ÷ 3 + 7
Solution:
The given expression is [18 - 2(5 + 1)] ÷ 3 + 7
∴ The expression is simplified and the answer is 9.
Example 2: Evaluate using the order of operations using the BODMAS rule: (1 + 20 - 16 ÷ 4²) ÷ {(5 - 3)² + 12 ÷ 2}
Solution:
∴ (1 + 20 - 16 ÷ 4²) ÷ {(5 - 3)² + 12 ÷ 2} = 2
Example 3: Simplify the expression by using the BODMAS rule: (9 × 3 ÷ 9 + 1) × 3
Solution:
∴ (9 × 3 ÷ 9 + 1) × 3 = 12
Example 4: Solve the given expression applying the BODMAS rule: [50-{3×(9+7)}]
Solution:
To solve this expression, [50-{3×(9+7)}], we will use the following steps:
go to slidego to slidego to slidego to slide
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BODMAS Practice Questions
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FAQs on BODMAS Rule
What is the Bodmas Rule in Maths?
The BODMAS rule refers to the rule that is followed to solve mathematical expressions. BODMAS is the order of operations for mathematical expressions that involves more than one operation. The acronym of BODMAS stands for B - Brackets, O - Order of powers, D - Division, M - Multiplication, A - Addition, and S - Subtraction.
How does BODMAS Rule Work?
In any arithmetic expression, if there are multiple operations used, then we need to solve the terms in the order of the BODMAS rule. We solve the part written in brackets first. After solving the brackets, we carry out the multiplication and division operations, whichever comes first in the expression from left to right. Then, we get a simplified expression with only addition and subtraction operations. We solve addition and subtraction from left to right and get the final answer. This is how BODMAS works.
Does BODMAS Apply when there are no Brackets?
Yes, even if there are no brackets, the BODMAS rule is still used. We need to solve the other operations in the same order. The next step after Brackets (B) is the order of powers or roots, followed by division, multiplication, addition, and then subtraction.
What is O in Bodmas Rule?
O in Bodmas stands for Order which means simplifying exponents or roots in the expression, if any, before arithmetic operations. In certain countries, 'O' is used to represent 'of' which again means multiplication.
How to Apply the Bodmas Rule?
BODMAS rule can be applied in case of expressions that have more than one operator. In that case, we simplify the brackets first from the innermost bracket to the outermost [{()}], then we evaluate the values of exponents or roots followed by simplifying multiplication and division, and then, at last, perform addition and subtraction operations while moving from left to right.
Why is the Order of Operations Important in Real Life?
The order of operations is a shorthand rule that enables you to follow the right order to solve different parts of a mathematical expression. It is a universal rule to solve all mathematical operations to get the correct answer.
When is the Bodmas Rule not Applicable?
BODMAS rule is not applicable to equations. It is applicable to mathematical expressions having more than one operator.
Who Invented the Bodmas Rule? When was it Introduced?
BODMAS rule was introduced by a mathematician, Achilles Reselfelt in the 1800s.
What is the Full Form of Bodmas Rule?
The full form of BODMAS is, Brackets, Order, Division, Multiplication, Addition, Subtraction.
Do Calculators Use BODMAS?
Calculators also use the BODMAS rule. The scientific calculators automatically apply the operations in the correct order.
What is the BODMAS Formula?
The BODMAS formula is the BODMAS rule which stands for B - Brackets, O - Order of powers, D - Division, M - Multiplication, A - Addition, and S - Subtraction. According to the BODMAS rule in Maths, whenever we have mathematical expressions to solve, we use this order to simplify them.
Why do we Use the BODMAS Formula?
We use the BODMAS rule because we should learn the correct order of mathematical operations. We need to know that we should solve the mathematical expressions from left to right in the order of operations stated by BODMAS. |
8939 | https://www.pbs.org/wgbh/nova/teachers/activities/3016_magnetic_01.html | NOVA | Teachers | Magnetic Storm | Student Handout: Visualizing Magnetic Fields | PBS
Magnetic StormStudent Handout Visualizing Magnetic Fields Magnetism is an unseen force. To help you visualize a magnetic field, you will observe the alignment of iron filings around a magnet. The iron filings align in the direction of the magnet's field and make the field visible. In this activity, you will try out different shapes of magnets and compare their magnetic fields. Procedure 1. Choose a magnet and place a piece of paper on it. Lightly sprinkle iron filings onto the paper over the magnet. If you sprinkle the filings on too quickly and can't see the magnetic field, get another piece of paper and try again. 2. On another piece of paper, draw the pattern the iron filings make as they experience the magnet's force. Draw the magnet's shape. 3. Use a compass to determine the direction of the magnetic field lines at various places around the magnet. Draw arrows on your diagram to indicate this direction. 4. Move the compass away from the magnet until the compass no longer points to the magnet. Record this distance and the direction the compass is pointing on your diagram. 5. Predict whether different-shaped magnets will produce magnetic fields shaped like the one you just drew. 6. Repeat steps one to four with the other magnets. Questions Write your answers on a separate sheet of paper. 1. How did the shape of the magnet influence its magnetic field? 2. Where are most of the magnetic field lines on each of the magnets? Did your results agree with your prediction? Explain. 3. Where are the north and south poles on the round magnet? 4. What did you learn about the strength of each magnet's magnetic field by moving the compass? 5. When you moved the compass away from the magnet, what happened? 6. How is Earth like a magnet? 7. Look at the illustration of Earth's magnetic field on this page. How is the shape of Earth's magnetic field similar to that of the magnets you experimented with? How is it different? 8. How might a compass be used in conjunction with Earth's magnetic field? © WGBH Educational Foundation
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8940 | https://en.wikipedia.org/wiki/Binary_quadratic_form | Jump to content
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Contents
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1 Equivalence
1.1 Automorphisms
2 Representation
2.1 Examples
2.2 The representation problem
2.3 Equivalent representations
3 Reduction and class numbers
4 Composition
4.1 Composing forms and classes
5 Genera of binary quadratic forms
6 History
7 See also
8 Notes
9 References
10 External links
Binary quadratic form
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From Wikipedia, the free encyclopedia
Quadratic homogeneous polynomial in two variables
This article is about binary quadratic forms with integer coefficients. For binary quadratic forms with other coefficients, see quadratic form.
| |
| This article includes a list of general references, but it lacks sufficient corresponding inline citations. Please help to improve this article by introducing more precise citations. (July 2009) (Learn how and when to remove this message) |
In mathematics, a binary quadratic form is a quadratic homogeneous polynomial in two variables
where a, b, c are the coefficients. When the coefficients can be arbitrary complex numbers, most results are not specific to the case of two variables, so they are described in quadratic form. A quadratic form with integer coefficients is called an integral binary quadratic form, often abbreviated to binary quadratic form.
This article is entirely devoted to integral binary quadratic forms. This choice is motivated by their status as the driving force behind the development of algebraic number theory. Since the late nineteenth century, binary quadratic forms have given up their preeminence in algebraic number theory to quadratic and more general number fields, but advances specific to binary quadratic forms still occur on occasion.
Pierre Fermat stated that if p is an odd prime then the equation has a solution iff , and he made similar statement about the equations , , and . and so on are quadratic forms, and the theory of quadratic forms gives a unified way of looking at and proving these theorems.
Another instance of quadratic forms is Pell's equation .
Binary quadratic forms are closely related to ideals in quadratic fields. This allows the class number of a quadratic field to be calculated by counting the number of reduced binary quadratic forms of a given discriminant.
The classical theta function of 2 variables is , if is a positive definite quadratic form then is a theta function.
Equivalence
[edit]
Two forms f and g are called equivalent if there exist integers such that the following conditions hold:
For example, with and , , , and , we find that f is equivalent to , which simplifies to .
The above equivalence conditions define an equivalence relation on the set of integral quadratic forms. It follows that the quadratic forms are partitioned into equivalence classes, called classes of quadratic forms. A class invariant can mean either a function defined on equivalence classes of forms or a property shared by all forms in the same class.
Lagrange used a different notion of equivalence, in which the second condition is replaced by . Since Gauss it has been recognized that this definition is inferior to that given above. If there is a need to distinguish, sometimes forms are called properly equivalent using the definition above and improperly equivalent if they are equivalent in Lagrange's sense.
In matrix terminology, which is used occasionally below, when
has integer entries and determinant 1, the map is a (right) group action of on the set of binary quadratic forms. The equivalence relation above then arises from the general theory of group actions.
If , then important invariants include
The discriminant .
The content, equal to the greatest common divisor of a, b, and c.
Terminology has arisen for classifying classes and their forms in terms of their invariants. A form of discriminant is definite if , degenerate if is a perfect square, and indefinite otherwise. A form is primitive if its content is 1, that is, if its coefficients are coprime. If a form's discriminant is a fundamental discriminant, then the form is primitive. Discriminants satisfy
Automorphisms
[edit]
If f is a quadratic form, a matrix
in is an automorphism of f if . For example, the matrix
is an automorphism of the form . The automorphisms of a form are a subgroup of . When f is definite, the group is finite, and when f is indefinite, it is infinite and cyclic.
Representation
[edit]
A binary quadratic form represents an integer if it is possible to find integers and satisfying the equation Such an equation is a representation of n by q.
Examples
[edit]
Diophantus considered whether, for an odd integer , it is possible to find integers and for which . When , we have
so we find pairs that do the trick. We obtain more pairs that work by switching the values of and and/or by changing the sign of one or both of and . In all, there are sixteen different solution pairs. On the other hand, when , the equation
does not have integer solutions. To see why, we note that unless or . Thus, will exceed 3 unless is one of the nine pairs with and each equal to or 1. We can check these nine pairs directly to see that none of them satisfies , so the equation does not have integer solutions.
A similar argument shows that for each , the equation can have only a finite number of solutions since will exceed unless the absolute values and are both less than . There are only a finite number of pairs satisfying this constraint.
Another ancient problem involving quadratic forms asks us to solve Pell's equation. For instance, we may seek integers x and y so that . Changing signs of x and y in a solution gives another solution, so it is enough to seek just solutions in positive integers. One solution is , that is, there is an equality . If is any solution to , then is another such pair. For instance, from the pair , we compute
: ,
and we can check that this satisfies . Iterating this process, we find further pairs with :
These values will keep growing in size, so we see there are infinitely many ways to represent 1 by the form . This recursive description was discussed in Theon of Smyrna's commentary on Euclid's Elements.
The representation problem
[edit]
The oldest problem in the theory of binary quadratic forms is the representation problem: describe the representations of a given number by a given quadratic form f. "Describe" can mean various things: give an algorithm to generate all representations, a closed formula for the number of representations, or even just determine whether any representations exist.
The examples above discuss the representation problem for the numbers 3 and 65 by the form and for the number 1 by the form . We see that 65 is represented by in sixteen different ways, while 1 is represented by in infinitely many ways and 3 is not represented by at all. In the first case, the sixteen representations were explicitly described. It was also shown that the number of representations of an integer by is always finite. The sum of squares function gives the number of representations of n by as a function of n. There is a closed formula
where is the number of divisors of n that are congruent to 1 modulo 4 and is the number of divisors of n that are congruent to 3 modulo 4.
There are several class invariants relevant to the representation problem:
The set of integers represented by a class. If an integer n is represented by a form in a class, then it is represented by all other forms in a class.
The minimum absolute value represented by a class. This is the smallest nonnegative value in the set of integers represented by a class.
The congruence classes modulo the discriminant of a class represented by the class.
The minimum absolute value represented by a class is zero for degenerate classes and positive for definite and indefinite classes. All numbers represented by a definite form have the same sign: positive if and negative if . For this reason, the former are called positive definite forms and the latter are negative definite.
The number of representations of an integer n by a form f is finite if f is definite and infinite if f is indefinite. We saw instances of this in the examples above: is positive definite and is indefinite.
Equivalent representations
[edit]
The notion of equivalence of forms can be extended to equivalent representations. Representations and are equivalent if there exists a matrix
with integer entries and determinant 1 so that and
The above conditions give a (right) action of the group on the set of representations of integers by binary quadratic forms. It follows that equivalence defined this way is an equivalence relation and in particular that the forms in equivalent representations are equivalent forms.
As an example, let and consider a representation . Such a representation is a solution to the Pell equation described in the examples above. The matrix
has determinant 1 and is an automorphism of f. Acting on the representation by this matrix yields the equivalent representation . This is the recursion step in the process described above for generating infinitely many solutions to . Iterating this matrix action, we find that the infinite set of representations of 1 by f that were determined above are all equivalent.
There are generally finitely many equivalence classes of representations of an integer n by forms of given nonzero discriminant . A complete set of representatives for these classes can be given in terms of reduced forms defined in the section below. When , every representation is equivalent to a unique representation by a reduced form, so a complete set of representatives is given by the finitely many representations of n by reduced forms of discriminant . When , Zagier proved that every representation of a positive integer n by a form of discriminant is equivalent to a unique representation in which f is reduced in Zagier's sense and , . The set of all such representations constitutes a complete set of representatives for equivalence classes of representations.
Reduction and class numbers
[edit]
Lagrange proved that for every value D, there are only finitely many classes of binary quadratic forms with discriminant D. Their number is the class number of discriminant D. He described an algorithm, called reduction, for constructing a canonical representative in each class, the reduced form, whose coefficients are the smallest in a suitable sense.
Gauss gave a superior reduction algorithm in Disquisitiones Arithmeticae, which ever since has been the reduction algorithm most commonly given in textbooks. In 1981, Zagier published an alternative reduction algorithm which has found several uses as an alternative to Gauss's.
Composition
[edit]
Composition most commonly refers to a binary operation on primitive equivalence classes of forms of the same discriminant, one of the deepest discoveries of Gauss, which makes this set into a finite abelian group called the form class group (or simply class group) of discriminant . Class groups have since become one of the central ideas in algebraic number theory. From a modern perspective, the class group of a fundamental discriminant is isomorphic to the narrow class group of the quadratic field of discriminant . For negative , the narrow class group is the same as the ideal class group, but for positive it may be twice as big.
"Composition" also sometimes refers to, roughly, a binary operation on binary quadratic forms. The word "roughly" indicates two caveats: only certain pairs of binary quadratic forms can be composed, and the resulting form is not well-defined (although its equivalence class is). The composition operation on equivalence classes is defined by first defining composition of forms and then showing that this induces a well-defined operation on classes.
"Composition" can also refer to a binary operation on representations of integers by forms. This operation is substantially more complicated[citation needed] than composition of forms, but arose first historically. We will consider such operations in a separate section below.
Composition means taking 2 quadratic forms of the same discriminant and combining them to create a quadratic form of the same discriminant, as follows from Brahmagupta's identity.
Composing forms and classes
[edit]
A variety of definitions of composition of forms has been given, often in an attempt to simplify the extremely technical and general definition of Gauss. We present here Arndt's method, because it remains rather general while being simple enough to be amenable to computations by hand. An alternative definition is described at Bhargava cubes.
Suppose we wish to compose forms and , each primitive and of the same discriminant . We perform the following steps:
Compute and , and
Solve the system of congruences
It can be shown that this system always has a unique integer solution modulo . We arbitrarily choose such a solution and call it B.
Compute C such that . It can be shown that C is an integer.
The form is "the" composition of and . We see that its first coefficient is well-defined, but the other two depend on the choice of B and C. One way to make this a well-defined operation is to make an arbitrary convention for how to choose B—for instance, choose B to be the smallest positive solution to the system of congruences above. Alternatively, we may view the result of composition, not as a form, but as an equivalence class of forms modulo the action of the group of matrices of the form
: ,
where n is an integer. If we consider the class of under this action, the middle coefficients of the forms in the class form a congruence class of integers modulo 2A. Thus, composition gives a well-defined function from pairs of binary quadratic forms to such classes.
It can be shown that if and are equivalent to and respectively, then the composition of and is equivalent to the composition of and . It follows that composition induces a well-defined operation on primitive classes of discriminant , and as mentioned above, Gauss showed these classes form a finite abelian group. The identity class in the group is the unique class containing all forms , i.e., with first coefficient 1. (It can be shown that all such forms lie in a single class, and the restriction implies that there exists such a form of every discriminant.) To invert a class, we take a representative and form the class of . Alternatively, we can form the class of since this and are equivalent.
Genera of binary quadratic forms
[edit]
Gauss also considered a coarser notion of equivalence, with each coarse class called a genus of forms. Each genus is the union of a finite number of equivalence classes of the same discriminant, with the number of classes depending only on the discriminant. In the context of binary quadratic forms, genera can be defined either through congruence classes of numbers represented by forms or by genus characters defined on the set of forms. A third definition is a special case of the genus of a quadratic form in n variables. This states that forms are in the same genus if they are locally equivalent at all rational primes (including the Archimedean place).
History
[edit]
There is circumstantial evidence of protohistoric knowledge of algebraic identities involving binary quadratic forms. The first problem concerning binary quadratic forms asks for the existence or construction of representations of integers by particular binary quadratic forms. The prime examples are the solution of Pell's equation and the representation of integers as sums of two squares. Pell's equation was already considered by the Indian mathematician Brahmagupta in the 7th century CE. Several centuries later, his ideas were extended to a complete solution of Pell's equation known as the chakravala method, attributed to either of the Indian mathematicians Jayadeva or Bhāskara II. The problem of representing integers by sums of two squares was considered in the 3rd century by Diophantus. In the 17th century, inspired while reading Diophantus's Arithmetica, Fermat made several observations about representations by specific quadratic forms including that which is now known as Fermat's theorem on sums of two squares. Euler provided the first proofs of Fermat's observations and added some new conjectures about representations by specific forms, without proof.
The general theory of quadratic forms was initiated by Lagrange in 1775 in his Recherches d'Arithmétique. Lagrange was the first to realize that "a coherent general theory required the simulatenous consideration of all forms." He was the first to recognize the importance of the discriminant and to define the essential notions of equivalence and reduction, which, according to Weil, have "dominated the whole subject of quadratic forms ever since". Lagrange showed that there are finitely many equivalence classes of given discriminant, thereby defining for the first time an arithmetic class number. His introduction of reduction allowed the quick enumeration of the classes of given discriminant and foreshadowed the eventual development of infrastructure. In 1798, Legendre published Essai sur la théorie des nombres, which summarized the work of Euler and Lagrange and added some of his own contributions, including the first glimpse of a composition operation on forms.
The theory was vastly extended and refined by Gauss in Section V of Disquisitiones Arithmeticae. Gauss introduced a very general version of a composition operator that allows composing even forms of different discriminants and imprimitive forms. He replaced Lagrange's equivalence with the more precise notion of proper equivalence, and this enabled him to show that the primitive classes of given discriminant form a group under the composition operation. He introduced genus theory, which gives a powerful way to understand the quotient of the class group by the subgroup of squares. (Gauss and many subsequent authors wrote 2b in place of b; the modern convention allowing the coefficient of xy to be odd is due to Eisenstein).
These investigations of Gauss strongly influenced both the arithmetical theory of quadratic forms in more than two variables and the subsequent development of algebraic number theory, where quadratic fields are replaced with more general number fields. But the impact was not immediate. Section V of Disquisitiones contains truly revolutionary ideas and involves very complicated computations, sometimes left to the reader. Combined, the novelty and complexity made Section V notoriously difficult. Dirichlet published simplifications of the theory that made it accessible to a broader audience. The culmination of this work is his text Vorlesungen über Zahlentheorie. The third edition of this work includes two supplements by Dedekind. Supplement XI introduces ring theory, and from then on, especially after the 1897 publication of Hilbert's Zahlbericht, the theory of binary quadratic forms lost its preeminent position in algebraic number theory and became overshadowed by the more general theory of algebraic number fields.
Even so, work on binary quadratic forms with integer coefficients continues to the present. This includes numerous results about quadratic number fields, which can often be translated into the language of binary quadratic forms, but also includes developments about forms themselves or that originated by thinking about forms, including Shanks's infrastructure, Zagier's reduction algorithm, Conway's topographs, and Bhargava's reinterpretation of composition through Bhargava cubes.
See also
[edit]
Bhargava cube
Fermat's theorem on sums of two squares
Legendre symbol
Brahmagupta's identity
Notes
[edit]
^ Cohen 1993, §5.2
^ Weil 2001, p. 30
^ Hardy & Wright 2008, Thm. 278
^ Zagier 1981
^ Zagier 1981
^ Fröhlich & Taylor 1993, Theorem 58
^ Weil 2001, Ch.I §§VI, VIII
^ Weil 2001, Ch.I §IX
^ Weil 2001, Ch.I §IX
^ Weil 2001, Ch.II §§VIII-XI
^ Weil 2001, Ch.III §§VII-IX
^ Weil 2001, p.318
^ Weil 2001, p.317
References
[edit]
Johannes Buchmann, Ulrich Vollmer: Binary Quadratic Forms, Springer, Berlin 2007, ISBN 3-540-46367-4
Duncan A. Buell: Binary Quadratic Forms, Springer, New York 1989
David A. Cox, Primes of the form , Fermat, class field theory, and complex multiplication
Cohen, Henri (1993), A Course in Computational Algebraic Number Theory, Graduate Texts in Mathematics, vol. 138, Berlin, New York: Springer-Verlag, ISBN 978-3-540-55640-4, MR 1228206
Fröhlich, Albrecht; Taylor, Martin (1993), Algebraic number theory, Cambridge Studies in Advanced Mathematics, vol. 27, Cambridge University Press, ISBN 978-0-521-43834-6, MR 1215934
Hardy, G. H.; Wright, E. M. (2008) , An Introduction to the Theory of Numbers, Revised by D. R. Heath-Brown and J. H. Silverman. Foreword by Andrew Wiles. (6th ed.), Oxford: Clarendon Press, ISBN 978-0-19-921986-5, MR 2445243, Zbl 1159.11001
Weil, André (2001), Number Theory: An approach through history from Hammurapi to Legendre, Birkhäuser Boston
Zagier, Don (1981), Zetafunktionen und quadratische Körper: eine Einführung in die höhere Zahlentheorie, Springer
External links
[edit]
Peter Luschny, Positive numbers represented by a binary quadratic form
A. V. Malyshev (2001) , "Binary quadratic form", Encyclopedia of Mathematics, EMS Press
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We explore functions that “shoot to infinity” near certain points.
Consider the function
f(x)=1(x+1)2.f(x)=1(x+1)2.
While the lim x→−1 f(x)lim x→−1 f(x) does not exist, something can still be said.
If f(x)f(x) grows arbitrarily large as x x approaches a a, we write
lim x→a f(x)=∞lim x→a f(x)=∞
and say that the limit of f(x)f(x) is equal to infinity as x x goes to a a.
If |f(x)||f(x)| grows arbitrarily large as x x approaches a a and f(x)f(x) is negative near a a, we write
lim x→a f(x)=−∞lim x→a f(x)=−∞
and say that the limit of f(x)f(x) is equal to negative infinity as x x goes to a a.
Which of the following are correct?
[x] lim x→−1 1(x+1)2=∞lim x→−1 1(x+1)2=∞- [x] lim x→−1 1(x+1)2→∞lim x→−1 1(x+1)2→∞- [x] f(x)=1(x+1)2 f(x)=1(x+1)2, so f(−1)=∞f(−1)=∞- [x] f(x)=1(x+1)2 f(x)=1(x+1)2, so as x→−1 x→−1, f(x)→∞f(x)→∞
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On the other hand, consider the function
f(x)=1(x−1).f(x)=1(x−1).
While the two sides of the limit as x x approaches 1 1 do not agree, we can still consider the one-sided limits. We see lim x→1+f(x)=∞lim x→1+f(x)=∞ and lim x→1−f(x)=−∞lim x→1−f(x)=−∞.
If at least one of the following hold:
lim x→a f(x)=±∞lim x→a f(x)=±∞,
lim x→a+f(x)=±∞lim x→a+f(x)=±∞,
lim x→a−f(x)=±∞lim x→a−f(x)=±∞,
then the line x=a x=a is a vertical asymptote of f f.
Find the vertical asymptotes of
f(x)=x 2−9 x+14 x 2−5 x+6.f(x)=x 2−9 x+14 x 2−5 x+6.
Since f f is a rational function, it is continuous on its domain. So the only points where the function can possibly have a vertical asymptote are zeros of the denominator.
Start by factoring both the numerator and the denominator:
x 2−9 x+14 x 2−5 x+6=(x−2)(x−7)(x−2)(x−3)x 2−9 x+14 x 2−5 x+6=(x−2)(x−7)(x−2)(x−3)
Using limits, we must investigate what happens with f(x)f(x) when x→2 x→2 and x→3 x→3, since 2 2 and 3 3 are the only zeros of the denominator. Write
lim x→2(x−2)(x−7)(x−2)(x−3)=lim x→2(x−7)(x−3)=−5−1=5.lim x→2(x−2)(x−7)(x−2)(x−3)=lim x→2(x−7)(x−3)=−5−1=5.
Now write
lim x→3(x−2)(x−7)(x−2)(x−3)=lim x→3(x−7)(x−3)=lim x→3−4 x−3.lim x→3(x−2)(x−7)(x−2)(x−3)=lim x→3(x−7)(x−3)=lim x→3−4 x−3.
Consider the one-sided limits separately.
When x→3+x→3+, the quantity (x−3)(x−3) is positive and approaches 0 0 and the numerator is negative, therefore, lim x→3+f(x)=−∞lim x→3+f(x)=−∞.
On the other hand, when x→3−x→3−, the quantity (x−3)(x−3) is negative and approaches 0 0 and the numerator is negative, therefore, lim x→3−f(x)=∞lim x→3−f(x)=∞.
Hence we have a vertical asymptote at x=3 x=3.
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Question regarding combinatorics formula
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\begingroup
I was solving this math problem:
If we roll three standard dices what is the possibility that the sum of eyes will be equal to 7?
So I tried to approach this problem in two different ways:
1) The order in every sequence IS IMPORTANT (a,a,b) and (a,b,a) sequences are counted as different. In this case all the possible variants are counted using formula A(6,3)=6^3=216. When we look for the favorable endings there are 4 different possible sums: (5,1,1) (3,2,2) (4,2,1) (3,3,1) . All that is left is to count the different variants in which these sums can be displayed and we will have all possible outcomes. This is done using formula P(3,2) = 3! / 2! = 3 and for sequence (4,2,1) formula 3! = 6 . We get 3+3+3+6 favorable sequence and the possibility is 15/216 = 5/72. 5/72 is also the answer according to the book.
2) However, It is way more logical to think that sequences (3,2,2) and (2,2,3) are same and should be counted as one because all that matters is the sum of eyes on the three dices. However, when I make the calculations here, the answer seems to be different. All the possible variants can be counted using formula C(6,3)=C(6+3-1,C)=(6+3-1)!/(3!(6-1)!) = 56. We know that the desired results can be achieved with 4 possible sequences (5,1,1) (3,2,2) (4,2,1) (3,3,1) and since they are all the same we do not need to count all the possible displays, therefore, we have 4 favorable endings. In this case the possiblity is 4/36 = 1/9. Obviously, the possibilities in both 1) and 2) cases should have matched, so, could You please tell me where have I made a mistake? Thank you in advance :)
P.S the suggested answer is in the first 1) variant, but the way it is being calculated does not seem the best approach since order in sequences should really not matter. I am particularly interested in this formula where you have to count all the possible endings where order does not matter but the same element can repeat. I fail to use it every single time, maybe there is something wrong with the way I use it?
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asked Mar 27, 2020 at 22:07
Paulius VaitkeviciusPaulius Vaitkevicius
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1 \begingroup The order in sequence DOES matter, since any sequence has the same probability. If you continue your logic the sequence (4,3,1) is of course the same as (3,3,2) because only "the sum of eyes on the three dices" matters. :)\endgroup user –user 2020-03-27 22:28:16 +00:00 Commented Mar 27, 2020 at 22:28
1 \begingroup The formula for computing the number of different outcomes with 3 identical dice is C(8,5)=56 by "stars and bars." However, it is useless for the sort of probability question you are asking because, as noted by the others here, those 56 outcomes are NOT equally likely, so you can NOT compute probabilities by taking the ratio favorable outcomes over size of sample space. You would need the relative frequency of the various outcomes, which in practical terms puts you back into looking at the numbered of ordered outcomes in each unordered one. Think about flipping two identical coins ...\endgroup Ned –Ned 2020-03-28 00:48:46 +00:00 Commented Mar 28, 2020 at 0:48
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\begingroup
You made a mistake in the second example by assuming that cases like (2,2,3)=(3,2,2). The issue with this assumption is that it makes (3,3,3) equally likely with something like (1,2,3), which is clearly false.
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answered Mar 27, 2020 at 22:28
Rushabh MehtaRushabh Mehta
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I'll just put in my own insight.
Let's say the die were rolled in order. Then we can look at the entire thing in terms of the first die.
By the way the total probability space is 6^3=216. These are permutations that allow for repetition.
So the first die can never be 6. Unless there's an option of 0 on one of the other die you'd always add up to something greater than 7.
If the first die has a 5, then the number remaining to be added is 2. You can't roll a 2 on the die left because that would ruin our target sum of 7. So the second die can either be 1 or 1 and the last can keep the remaining 1. That gives us only one permutation.
If the first die has a 4, then the second die can only possess values from 1 to 2, not 3, because we need to leave something for the third die. That gives us 2 permutations.
Repeating this for the other values of the first die:
3 \implies 3 \text{ permutations}
2 \implies 4 \text{ permutations}
1 \implies 5 \text{ permutations}
Combining this with:
4 \implies 2 \text{ permutations}
5 \implies 1 \text{ permutations}
6 \implies 0 \text{ permutations}
So the probability is:
=\dfrac{0+1+2+3+4+5}{216}=\dfrac{5}{72}
And the thing you were proposing. This system is full of permutations and not combinations. Think about it like this: would it be fair to serve the guy behind you in a line just because mathematically there is one combination of your arrangement on line? That's how the die feel. The sequences 1,5,1 and 5,1,5 exist on different die so they can't be equated as one. Changing the order changes which die have the number. And that's definitely a permutation.
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edited Mar 28, 2020 at 6:12
answered Mar 28, 2020 at 6:06
Nεo PλατoNεo Pλατo
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\begingroup
You have a barrel containing a million jellybeans, 999,999 black and 1 pink. You reach in and pick out two jellybeans at random. What is the probability you get two black jellybeans?
Method 1: There is a first jellybean and a second jellybean. There are 1000000\cdot999999 ways the selection could occur, out of which 999999\cdot999998 result in two black jellybeans. Hence the probability is \frac{999999\cdot999998}{1000000\cdot999999}=\frac{999998}{1000000}=0.999998.
Method 2: Well, order of jellybeans doesn't matter, even if, in reality, I will always touch one of the jellybeans I pick at least slightly earlier than the other. So, ignoring order, there are \binom{1000000}{2} possible pairs of jellybeans I could pick, out of which \binom{999999}{2} pairs have the desired property. Dividing gives \frac{\frac{999999\cdot999998}{2\cdot1}}{\frac{1000000\cdot999999}{2\cdot1}}=\frac{999998}{1000000}=0.999998.
Method 3: Well, really all I care about is color, not particular identity of jellybeans. So there are two possible color selections, represented by the multisets {black, black} and {black, pink}, of which one has the desired composition. Hence the probability is 1/2=0.5.
You can see that Methods 1 and 2 agree with each other, but are in violent disagreement with Method 3. The reason for this is that, while it is true that order doesn't matter, it is false that physical identity of jellybeans doesn't matter. Methods 1 and 2 are right; Method 3 is wrong. What we learn from this is that when modeling a physical selection process using mathematics, physics matters. The outcome {black, black} is tied to a very much larger set of physical outcomes than is the outcome {black, pink}. (There's another big physical assumption here, namely that we can mix up the jellybeans sufficiently well so as to make all selections equally likely.)
Applying this insight to your problem, we see that the outcome "dice sum to 7" is tied to a larger set of physical outcomes than is the outcome "dice sum to 3". Even analyzing in finer detail, the outcome "the roll includes a 2, a 3, and a 5" is tied to a larger set of physical outcomes than is the outcome "roll includes two 2 s and a 5", which in turn is tied to a larger set of physical outcomes than is the outcome "roll contains three 2 s". It is generally physically reasonable to assume that the dice are fair and independent of each other. It is not physically reasonable to assume that the dice do not have distinct physical identities.
To expand a bit on the analogy between the two problems: in computing probability in the jellybean problem it matters, not just which colors you got, but which particular jellybeans of that color you got. In the dice problem, it matters, not just which numbers you got, but which particular dice had those numbers.
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edited Mar 29, 2020 at 14:14
answered Mar 28, 2020 at 3:24
Will OrrickWill Orrick
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8945 | https://home.cc.umanitoba.ca/~dtrim/Courses/Math2132/book.pdf | 312 SECTION 6.0 CHAPTER 6 LAPLACE TRANSFORMS The Laplace transform is one of many integral transforms in applied mathematics.
Through an improper integral, the Laplace transform creates an association between a class of functions denoted by f(t) and a class of functions denoted by F(s).
The advantage of this association as far as our discussions are concerned is that solving a differential equation for f(t) is replaced by solving an algebraic equation for F(s). The fact that the Laplace transform is a linear operator (in the sense of equation 4.12) makes it particularly useful for solving the linear differential equations encountered in Chapters 4 and 5. Furthermore, you will recall that in Chapter 4 we assumed continuity of nonhomogeneous terms in linear differential equations.
This was a matter of convenience rather than necessity. In Exercises 32 and 33 of Section 4.5, we hinted at the awkwardness of incorporating discontinuities into the techniques of Chapter 4. But discontinuous nonhomogeneities occur frequently in applications. For instance, in Example 3.10 of Section 3.4, the nonhomogeneity 1/10 in the linear differential equation dS dt + 5S 106 = 1 10 is a result of brine with concentration 2 kilograms per 100 litres being added to the tank. But suppose after 10 minutes the concentration is doubled to 4 kilograms per 100 litres, and after another 10 minutes it is increased to 5 kilograms per 100 litres.
The nonhomogenity would now be f(t) = 1/10, 0 < t < 600 1/5, 600 < t < 1200 1/4, t > 1200.
It is possible to solve this problem with techniques from Chapter 4, but it isn’t very convenient to do so. We shall show how easily such situations are handled by Laplace transforms. An even more awkward situation for Chapter 4 would be an LCR-circuit where the applied voltage is continually turned on for a second, turned offfor another second, turned on again, turned offagain, ad infinitum. Once again this presents no problem for Laplace transforms. Even more devastating for Chapter 4 is imparting an instantaneous force to a vibrating mass-spring system say by striking the mass with a hammer. Chapter 4 just cannot handle this situa-tion, but Laplace transforms can. This is perhaps the biggest advantage of Laplace transforms over the methods of Chapter 4. Discontinuous rates in mixing problems, discontinuous forcing functions in vibrating mass-spring systems, and discontinuous driving voltages in LCR-circuits are easily handled by Laplace transforms. Further-more, nonhomogeneities that represent “point” concepts in space or time cannot be handled with the techniques of Chapter 4, but they present no problem for Laplace transforms. Such occurrences include bulk additions of ingredients in mixing prob-lems, instantaneously applied forces in vibrating problems, instantaneously applied voltages in electric circuits, and point loads on beams.
We make one last point before defining the Laplace transform of a function.
Currents in an electrical network satisfy a system of interrelated differential equa-tions, a topic that we take up in Chapter 7. Laplace transforms of these equations reduce them to a system of algebraic equations. Examination of the functions in SECTION 6.1 313 these equations can determine whether the network is stable or whether it would lead to unacceptably high currents, and this can be done without actually solving for the currents.
§6.1 The Laplace Transform and its Inverse The Laplace transform is one of many useful integral transforms in mathematics.
They all map a function of one variable to a function of a different variable. In general, an integral transform maps a function f(t) of t to a function F(s) of s according to the definition F(s) = Z b a K(t, s)f(t) dt.
(6.1) The function K(t, s) is called the kernel of the transformation (and it is given).
The function F(s) is called the transform of f(t), and to find F(s), we must multiply f(t) by K(t, s) and integrate the product from a to b. According to the following definition, the kernel of the Laplace transform is K(t, s) = e−st, and limits are a = 0 and b = ∞.
Definition 6.1 When f is a function of t, its Laplace transform denoted by F = L{f} is a function with values defined by F(s) = L{f}(s) = Z ∞ 0 e−stf(t) dt, (6.2) provided the improper integral converges.
The Laplace transform of a function f(t) never exists for all values of s; there is always a restriction on the values that can be used. In other words, the real problem is to find values of s for which improper integral 6.2 converges, and for these values, and only these values, F(s) is the Laplace transform of f(t).
It is customary to choose t as the independent variable of functions that are to be transformed because the transform is so often associated with problems in which t represents time. Do not get the impression, however, that Laplace transforms are only associated with time. In Section 6.6, we take Laplace transform with respect to x in order to calculate deflections of beams under various loads. It is customary to use a lower case letter f to represent a function that is to be transformed, and its capital counterpart F to represent the transform of f.
When t is time, with units of seconds, then s must have units of one divided by seconds. If this were not the case, then −st in e−st would not be dimensionless, and what then would be the units of the exponential. Hence, the Laplace transform is a mapping from the time domain to the frequency domain.
The improper integral in Definition 6.1 is defined as F(s) = Z ∞ 0 e−stf(t) dt = lim T →∞ Z T 0 e−stf(t) dt, provided the limit exists (and also provided f(t) has acceptable discontinuities, further discussion coming). If g(s, t) is an antiderivative of e−stf(t) with respect to t, we could write 314 SECTION 6.1 F(s) = lim T →∞{g(s, t)}T 0 = lim T →∞[g(s, T) −g(s, 0)] = lim T →∞g(s, T) −g(s, 0).
To shorten the notation, we customarily write F(s) = Z ∞ 0 e−stf(t) dt = {g(s, t)}∞ 0 , understanding that the limit should be taken as t →∞. Appendix E contains a brief discussion of improper integrals of this type for readers who are meeting them for the first time, and for readers who would like a quick review.
For our purposes, s is a real variable, in which case F is a real-valued function of a real variable s. The reader should be aware, however, that in advanced applications of Laplace transforms, especially for solving partial differential equations and in many areas of electrical engineering, s is complex, in which case F is a complex-valued function of a complex variable.
We should determine properties of a function that guarantee existence of its Laplace transform. The following three examples point us in the correct direction.
Example 6.1 Find the Laplace transform of f(t) = eat where a ̸= 0 is a constant.
Solution According to equation 6.2, the Laplace transform has values defined by F(s) = Z ∞ 0 e−steat dt = Z ∞ 0 e(a−s)t dt = 1 a −se(a−s)t ∞ 0 = 1 a −s h lim t→∞e(a−s)t −1 i .
This limit exists, and has value 0, only when s > a. Hence, the Laplace transform of f(t) = eat is 1/(s −a), but only for s > a. We write F(s) = 1 s −a, s > a.• Example 6.2 Find the Laplace transform of f(t) = t.
Solution According to equation 6.2, the Laplace transform has values defined by F(s) = Z ∞ 0 t e−st dt.
Integration by parts leads to F(s) = −t se−st −1 s2 e−st ∞ 0 = lim t→∞ −t se−st −1 s2 e−st + 1 s2 .
This limit exists, and has value 0, only when s > 0. In other words, the Laplace transform of f(t) = t is F(s) = 1/s2, but the function is only defined for s > 0.• Example 6.3 Find the Laplace transform of the discontinuous function f(t) = 2t2, 0 ≤t ≤1 1, t > 1.
It is shown in Figure 6.1.
SECTION 6.1 315 Solution According to equation 6.2, the Laplace transform has values defined by F(s) = Z ∞ 0 e−stf(t) dt = Z 1 0 2t2e−st dt + Z ∞ 1 e−st dt.
Two integrations by parts on the first t 2 1 1 integral lead to Figure 6.1 F(s) = 2 −t2 s −2t s2 −2 s3 e−st 1 0 + −e−st s ∞ 1 = − 1 s + 4 s2 + 4 s3 e−s + 4 s3 , provided s > 0 • What have we learned from Definition 6.1 and these three examples? First, when f(t) is discontinuous, we subdivide the interval 0 < t < ∞into subintervals in which f(t) is continuous. To avoid an infinite number of such subintervals, we could demand that f(t) have a finite number of discontinuities. It turns out that this is not entirely necessary, although it is often the case. Instead, we demand that f(t) have a finite number of dis-continuities on every interval 0 ≤t ≤T of finite length. This would allow a periodic function like that in Figure 6.2 into our discussions. Such a function t 1 2 3 could represent an applied voltage in an Figure 6.2 electrical circuit that was periodically turned off. In addition, to guarantee existence of the integral of e−stf(t) on each subinterval in which f(t) is continuous, we demand that right- and left-hand limits of f(t) exist at every discontinuity. When a function has a finite number of discon-tinuities on an interval and right- and left-hand limits exist at all discontinuities in the interval, the function is said to be piecewise-continuous on that interval. We shall assume therefore that f(t) is piecewise continuous on every interval 0 ≤t ≤T of finite length.
The second thing that we saw in Examples 6.1–6.3 is that there is always a restriction on values of s. The function F(s) is not defined for all s; it is defined only for s larger than some number (a in Example 6.1 and 0 in Examples 6.2 and 6.3). This is due to the fact that for improper integral 6.2 to converge, the integrand must approach 0 as t →∞, and must do so sufficiently quickly. This means that f(t) must not increase so rapidly that it cannot be suppressed by e−st for some value of s. A sufficient restriction on the growth of f(t) for large t is contained in the following definition.
Definition 6.2 A function f(t) is said to be of exponential order α, written O(eαt), if there exist positive constants T and M such that |f(t)| < Meαt for all t > T.
What this says algebraically is that for sufficiently large t (t > T), |f(t)| must grow no faster than a constant M times eαt. Geometrically, the graph of |f(t)| must be below that of Meαt for t > T. It is important to realize that the exponential 316 SECTION 6.1 order of a function f(t), if it has one, is concerned with function behaviour for very large t, not for small t. The absolute value |f(t)| must eventually be less than Meαt, and stay less, but it need not be so for all t. This is shown in Figure 6.3.
For example, the exponential function e4t is O(e4t) since M can be chosen as 2 and T as zero. Constant functions are of exponential order zero. The trigonometric functions sin at and cos at are O(e0t) since both are less than 2 = 2e0t for all t. The exponential order of tn is discussed in the following example.
t T t e t M ( ) f a n/ n n e n -t e e ( ) / Figure 6.3 Figure 6.4 Example 6.4 Show that the function tn, where n is a positive integer, is O(eϵt) for arbitrarily small, positive ϵ.
Solution Consider the function f(t) = tne−ϵt for arbitrary ϵ > 0. To draw its graph we first calculate that f ′(t) = ntn−1e−ϵt −ϵtne−ϵt = tn−1e−ϵt(n −ϵt).
There is a relative maximum at t = n/ϵ and when this is combined with the fact that limt→∞tne−ϵt = 0, the graph in Figure 6.4 results. It shows that the function tne−ϵt is bounded by M = (n/ϵ)ne−n for all t ≥0. In other words, tne−ϵt < 2M for all t > 0; that is, tn < 2Meϵt for t > 0, and tn is O(eϵt).• We now show that piecewise-continuous functions of exponential order always have Laplace transforms.
Theorem 6.1 If f(t) is piecewise-continuous on every finite interval 0 ≤t ≤T, and is of exponen-tial order α, then its Laplace transform exists for s > α.
Proof The improper integral in equation 6.2 can be divided into integrals over the intervals 0 ≤t ≤T and T ≤t < ∞, for any T, F(s) = Z ∞ 0 e−stf(t) dt = Z T 0 e−stf(t) dt + Z ∞ T e−stf(t) dt.
Since f(t) is piecewise-continuous on 0 ≤t ≤T, there is no question that the first of these integrals exists, and does so for all values of s. Furthermore, since f(t) is O(eαt), there exist constants M and T such that |f(t)| < Meαt for t > T. Hence, Z ∞ T e−stf(t) dt ≤ Z ∞ T e−st|f(t)| dt < Z ∞ T Me−steαt dt = Z ∞ T Me(α−s)t dt = M α −se(α−s)t ∞ T = M s −αe(α−s)T , provided s > α. In other words, the improper integral over the interval T ≤t ≤∞ converges when s > α. Thus, the Laplace transform of f(t) is defined for s > α.
SECTION 6.1 317 Theorem 6.1 provides sufficient conditions for existence of Laplace transforms.
Functions that are not piecewise continuous or not of exponential order may or may not have transforms. For example, the function f(t) = 1/ √ t is not piecewise continuous due to the infinite discontinuity at t = 0.
It does, however, have a Laplace transform (see Exercise 35).
In calculating Laplace transforms of known functions by means of Definition 6.1, it is not necessary to determine whether the function is of exponential order prior to use of the integral; evaluation of the integral will yield the interval on which the transform is defined. When using techniques other than the defining integral to find Laplace transforms, however, it may be necessary to know that the function is of exponential order and piecewise-continuous on every finite interval. We shall develop other techniques in the next section. In this section we concentrate on the integral definition for the transform.
Example 6.5 Find the Laplace transform for f(t) = tn, where n is a positive integer.
Solution Integration by parts gives F(s) = Z ∞ 0 tne−st dt = tne−st −s ∞ 0 − Z ∞ 0 −n s tn−1e−st dt = n s Z ∞ 0 tn−1e−st dt, provided s > 0. A second integration by parts yields F(s) = n s Z ∞ 0 tn−1e−st dt = n s tn−1e−st −s ∞ 0 −n s Z ∞ 0 −n −1 s tn−2e−st dt = n(n −1) s2 Z ∞ 0 tn−2e−st dt.
Further intergations by parts lead to F(s) = n(n −1)(n −2) · · · (1) sn Z ∞ 0 e−st dt = n!
sn e−st −s ∞ 0 = n!
sn+1 , provided again that s > 0.
This is consistent with Theorem 6.1 and Example 6.4. According to Example 6.4, tn is O(eϵt) for arbitrarily small, positive ϵ, and therefore its Laplace transform should exist for s > ϵ for arbitrarily small ϵ > 0.
This is tantamount to s > 0.• Example 6.6 Find the Laplace transform for f(t) = cos at, where a > 0 is a constant.
Solution In Exercise 33, you are asked to perform two integrations by parts on the integral Z ∞ 0 e−st cos at dt in order to find the Laplace transform. We provide an alternative using complex exponentials, and we do so for two reasons. First, the use of complex exponentials provides a much easier derivation; in particular, it avoids integration by parts.
Secondly, complex numbers can facilitate many calculations in this chapter, and the sooner you are exposed to them the better. Instead of evaluating the above integral, we consider the integral 318 SECTION 6.1 Z ∞ 0 e−steatidt.
We have simply replaced cos at with eati, remembering that by Euler’s identity, eati = cos at + i sin at. Now, Z ∞ 0 e−steatidt = Z ∞ 0 e(−s+ai)tdt = e(−s+ai)t −s + ai ∞ 0 .
The limit of the antiderivative as t →∞is lim t→∞ e−steati −s + ai = lim t→∞ e−st(cos at + i sin at) −s + ai = 0, provided s > 0. Thus, Z ∞ 0 e−steatidt = 1 s −ai = s + ai s2 + a2 .
If we write the integral in the form Z ∞ 0 e−st(cos at + i sin at) dt = s + ai s2 + a2 , and take real and imaginary parts, we get Z ∞ 0 e−st cos at dt = s s2 + a2 , Z ∞ 0 e−st sin at dt = a s2 + a2 .
In other words, L{cos at} = s s2 + a2 , L{sin at} = a s2 + a2 .
Thus, by using complex exponentials, we not only found the Laplace transform of cos at, but we also found the transform of sin at.• The following table contains Laplace transforms of functions that occur very frequently in differential equations. They can be verified with equation 6.2.
f(t) F(s) f(t) F(s) tn n!
sn+1 eat 1 s −a sin at a s2 + a2 cos at s s2 + a2 t sin at 2as (s2 + a2)2 t cos at s2 −a2 (s2 + a2)2 sin at −at cos at 2a3 (s2 + a2)2 at cos at + sin at 2as2 (s2 + a2)2 sinh at a s2 −a2 cosh at s s2 −a2 Table 6.1 We have included transforms of the hyperbolic sine and cosine functions, but we make no use of them in this chapter. We do this for the sake of those who have SECTION 6.1 319 not studied hyperbolic functions. Those familiar with these functions will be able to provide simpler solutions to some of the examples and exercises.
The Inverse Laplace Transform Definition 6.3 When F is the Laplace transform of f, we call f the inverse Laplace transform of F, and write f = L−1{F}.
(6.3) For instance, Table 6.1 yields L−1 1 s + 2 = e−2t and L−1 s s2 + 3 = cos √ 3t.
The Laplace transform F(s) of a function f(t) is unique, every function has exactly one Laplace transform. On the other hand, many functions have the same transform. For example, the functions f(t) = t2 and g(t) = ( 0, t = 1 t2, t ̸= 1, 2 0, t = 2, which are identical except for their values at t = 1 and t = 2 both have the same transform 2/s3. The fact that F(s) = 2/s3 follows from Table 6.1; G(s) = 2/s3 follows from integration, (or by noting that because f(t) and g(t) differ only at isolated points, this makes no difference to integral 6.2). What we are saying is that the inverse transform f = L−1{F} in Definition 6.3 is not an inverse in the true sense of inverse; there are many possibilities for f for given F. In advanced work, a formula for calculating inverse transforms is derived, and this formula always yields a continuous function f(t), when this is possible. In the event that this is not possi-ble, the formula gives a piecewise-continuous function whose value is the average of right- and left-limits at discontinuities, namely limϵ→0 [f(t + ϵ) + f(t −ϵ)]/2. The importance of this formula is that it defines f = L−1{F} in a unique way. Other functions which have the same transform F differ from f only in their values at isolated points; they cannot differ from f over an entire interval a ≤t ≤b. When f is a continuous function with transform F, there cannot be another continuous function with the same transform. With this in mind, we adopt the procedure of choosing a continuous function L−1{F} for given F whenever this is possible. In the confines of differential equations, this will always be possible since solutions of dif-ferential equations are always continuous functions. Thus, when solving differential equations by Laplace transforms, there will always be only one inverse transform for a given function F(s). In other words, we can talk about the inverse Laplace transform of F(s).
According to the following theorem, the Laplace transform and its inverse are linear operators in the sense of equation 4.12. That the transform is linear is a direct result of the fact that integration is a linear operation; once the transform is linear, so also is the inverse.
Theorem 6.2 The Laplace transform and its inverse are linear operators; that is, for arbitrary functions f and g, arbitrary transforms F and G, and an arbitrary constant c, L{f + g} = L{f} + L{g}, L{c f} = c [L{f}], (6.4a) L−1{F + G} = L−1{F} + L−1{G}, L−1{c F} = c [L−1{F}].
(6.4b) 320 SECTION 6.1 For instance, using linearity and Table 6.1, L{2e−t + 3 sin 4t} = 2L{e−t} + 3L{sin 4t} = 2 s + 1 + 3 4 s2 + 16 , and L−1 2 s4 − 4s s2 + 5 = 2L−1 1 s4 −4L−1 s s2 + 5 = 2 t3 6 −4 cos √ 5t.
The following result can serve as a partial check on calculations of Laplace transforms.
Theorem 6.3 If f(t) is is piecewise-continuous on every finite interval 0 ≤t ≤T, and is of exponential order α, its Laplace transform has limit zero as s →∞; that is, lim s→∞F(s) = 0.
(6.5) Proof: The definition of F(s) gives |F(s)| = Z ∞ 0 e−stf(t) dt ≤ Z ∞ 0 e−st|f(t)| dt = Z T 0 e−st|f(t)|dt + Z ∞ T e−st|f(t)|dt.
Since f(t) is piecewise continuous on 0 ≤t ≤T, it is bounded thereon, and there exists a number M such that |f(t)| < M in this interval. Furthermore, since f(t) is of exponential order α, there exist constants T > 0 and M > 0, such that for t > T, |f(t)| < Meαt. We can therefore write that |F(s)| < Z T 0 Me−stdt + Z ∞ T e−stMeαtdt = M e−st −s T 0 + M e(α−s)t α −s ∞ T = M s (1 −e−sT ) + Me(α−s)T s −α , provided that s > 0 and s > α. The limit of this is zero as s →∞.
We said that this theorem could serve as a check on calculations. For instance, if we calculated the transform of a function f(t) (piecewise continuous on every finite interval, and of exponential order) to be F(s) = s2 + 2s −5 3s2 + 10s + 15, we would know that we had made an error since the limit of this function as s →∞ is not equal to zero; it is equal to 1/3. In Section 6.5, we will encounter a function that is an exception to this rule, but it will not be a function that is piecewise continuous. Another check on the validity of a Laplace transform can be found in Exercise 52 of Section 6.3.
We now give you a preview of what is to come. Consider solving the initial-value problem d2y dt2 −4dy dt −5y = 6 −5t, y(0) = 1, y′(0) = −1.
(6.6) We can certainly solve this with the techniques of Chapter 4. Use the auxiliary equation to find a general solution yh(t) of the associated homogeneous equation; SECTION 6.1 321 use undetermined coefficients to find a particular solution yp(t) of the differential equation; add these together and evaluate constants with the initial conditions. The Laplace transform takes an entirely different approach; it reduces the initial-value problem to an algebraic equation. To do this, we need to know how to take Laplace transforms of derivatives. We quote two results that will be verified in Section 6.3.
If F(s) is the Laplace transform of f(t), then Laplace transforms of f ′(t) and f ′′(t) can be written in terms of F(s) as follows: L{f ′(t)} = sF(s) −f(0), (6.7a) L{f ′′(t)} = s2F(s) −sf(0) −f ′(0).
(6.7b) If we take Laplace transforms of both sides of differential equation 6.6, and use the fact that the transform is a linear operator, we obtain L{y′′} −4L{y′} −5L{y} = 6L{1} −5L{t}.
When we denote the Laplace transform of y(t) by Y (s), and use formulas 6.7 on the left, we get [s2Y (s) −s(1) −(−1)] −4[sY (s) −1] −5Y (s) = 6 s −5 s2 .
This is an algebraic equation for Y (s) which is easily solved, Y (s) = 6 s −5 s2 + s −5 s2 −4s −5 .
When this rational expression is simplified and written in its partial fraction de-composition, the result is Y (s) = 17/6 s + 1 + 1/6 s −5 + 1 s2 −2 s.
We can take the inverse Laplace transform of each of these terms to find the solution of the initial-value problem, y(t) = 17 6 e−t + 1 6e5t + t −2.
This example is typical of Laplace transforms at work on initial-value problems like those in Chapters 4 and 5. The transform reduces the differential equation in y(t) to an algebraic equation in its transform Y (s). The algebraic equation is solved for Y (s) and the inverse transform then yields the solution y(t) of the initial-value problem. In order to solve other initial-value problems, we need to expand the catalogue of functions with known Laplace transforms beyond those in Table 6.1. Furthermore, in Chapter 4 we assumed continuity of nonhomogeneous terms in linear differential equations. This was a matter of convenience rather than necessity.
However, in Exercises 32 and 33 of Section 4.5, we hinted at the awkwardness of incorporating discontinuities into the techniques of Chapter 4. We shall give other examples of discontinuous nonhomogeneities in this chapter, and see how easily they are handled by Laplace transforms. Section 6.2 concentrates on efficient ways to calculate transforms and inverse transforms, and Sections 6.3 and 6.4 then return to full discussions of differential equations.
322 SECTION 6.1 EXERCISES 6.1 In Exercises 1–10 use linearity and Table 6.1 to find the Laplace transform of the function.
1. f(t) = t3 −2t2 + 1 2. f(t) = t + et 3. f(t) = 5e4t 4. f(t) = e−2t + 2et 5. f(t) = sin 4t + 3 cos 4t 6. f(t) = cos 2t −3 sin 4t 7. f(t) = 5t cos 2t 8. f(t) = 3t sin 4t 9. f(t) = 5t cos t −2t sin t 10. f(t) = 3t sin t −cos t In Exercises 11–20 use linearity and Table 6.1 to find the inverse Laplace transform of the function.
11. F(s) = 7 s3 12. F(s) = 2 s −3 s4 13. F(s) = 1 s + 5 + 4 s2 14. F(s) = 3 s −1 15. F(s) = s s2 + 4 − 3 s2 + 4 16. F(s) = 2s s2 + 2 − 5 s2 + 9 17. F(s) = 2s (s2 + 2)2 18. F(s) = s2 (s2 + 9)2 19. F(s) = 3s −s2 (s2 + 4)2 20. F(s) = s2 −2 (s2 + 3)2 In Exercises 21–32 use Definition 6.1 to find the Laplace transform of the function.
21. f(t) = 0, 0 < t < 3 1, t > 3 22. f(t) = 1, 0 < t < 4 2, t > 4 23. f(t) = t, 0 < t < 2 2, t > 2 24. f(t) = t2, 0 < t < 1 0, t > 1 25. f(t) = 0, 0 < t < 1 t2, t > 1 26. f(t) = 0, 0 < t < 1 (t −1)2, t > 1 27. f(t) = ( 0, 0 < t < 1 1, 1 < t < 2 0, t > 2 28. f(t) = ( t, 0 < t < 1 2 −t, 1 < t < 2 0, t > 2 29. f(t) = 2t, 0 < t < 1 t, t > 1 30. f(t) = 1 + t2, 0 < t < 1 2t, t > 1 31. f(t) = 0, 0 < t < a 1, t > a (a constant) 32. f(t) = ( 0, 0 < t < a 1, a < t < b 0, t > b (a, b constants) 33. Use integration by parts to find Laplace transforms for sin at and cos at.
34. Derive the Laplace transforms for t cos at and t sin at in Table 6.1.
35. The function 1/ √ t is not piecewise continuous because of its infinite discontinuity at t = 0.
Show that it has Laplace transform p π/s. Hint: Set u = √ t in the definition of the Laplace transform of 1/ √ t in terms of a definite integral and use the fact that R ∞ 0 e−u2du = √π/2.
36. Are all bounded functions (functions that satisfy |f(t)| < M for all t > 0) of exponential order?
SECTION 6.1 323 37. Are all continuous functions of exponential order?
38. In Example 6.6 we used complex exponentials to find the Laplace transforms for cos at and sin at. In Exercise 33, we used integration by parts. Another method that also works for other functions is to use Maclaurin series. The Maclaurin series for sin at is sin at = ∞ X n=0 (−1)n (2n + 1)!(at)2n+1 = ∞ X n=0 (−1)na2n+1 (2n + 1)! t2n+1.
Assuming that the operation of taking the Laplace transform can be interchanged with the summation operation, derive the Laplace transform of sin at.
39. Use the technique of Exercise 38 to derive the Laplace transform of cos at.
40. The Bessel function of the first kind of order zero has Maclaurin series J0(t) = ∞ X n=0 (−1)n 22n(n!)2 t2n.
Show that its Laplace transform is L{J0(t)} = 1 √ 1 + s2 .
41. Verify that L sin at t = Tan−1a s .
42. Verify that L ebt −eat t = ln s −a s −b when s > b > a.
43. In Example 6.5, we developed the formula for the Laplace transform of tn when n is a positive integer. In this exercise, we extend the result to the case that n is positive, but not an integer.
First, we need to extend the definition of factorials by defining the gamma function, Γ(t) = Z ∞ 0 e−uut−1 du, t > 0.
This function is discussed in more detail in Appendix A, but for our present purposes, only the property in part (a) is required.
(a) Verify that the gamma function satisfies the recursive formula Γ(t + 1) = t Γ(t), and hence that Γ(n + 1) = n!
when n is positive integer.
(b) Now prove that for r > 0, L{tr} = Γ(r + 1) sr+1 .
44. (a) Prove that the function et2 is not of exponential order.
45. (a) Prove that the function f(t) = sin (et2) is of exponential order.
(b) Prove that the derivative f ′(t) is not of exponential order? In spite of this, f ′(t) has a Laplace transform. (See Exercise 54 in Section 6.3.) 324 SECTION 6.2 6.2 Algebraic Properties of The Laplace Transform and its Inverse Early in your calculus studies you were required to use the limit-definition of the derivative to differentiate various functions. You were quickly brought to the realiza-tion that rules could be developed so that use of the definition could be eliminated, and you certainly appreciated these rules (power, product, and quotient to name a few). The same was true when it came to the definition of the definite integral as the limit of a summation. It is impossible to use the definition to find the definite integral of all but a handful of functions, and therefore using anti-derivatives to calculate definite integrals is essential. Likewise, seldom is it necessary to evaluate the improper integral in Definition 6.1 to find the Laplace transform for a function; other techniques prove more efficient. The purpose of this section is to develop some of these shortcuts. In addition, recall that our intention is to use Laplace transforms to provide another method for solving linear differential equations and extensions which are difficult or impossible to solve with the techniques of Chapter 4. With this in mind, note how many of the algebraic properties of the Laplace transform uncovered in this section are directed toward the functions so prevalent in solving linear differential equations, namely, tn, eat, sin at, cos at, and sums and products of these functions. In Section 6.3, we derive formulas 6.7 for taking Laplace transforms of derivatives of functions. With these formulas and the results of this section, we will be well prepared to solve differential equations.
One of two shifting properties is contained in the following theorem.
Theorem 6.4 When F is the Laplace transform of f, L{eatf(t)} = F(s −a), (6.8a) L−1{F(s −a)} = eatf(t).
(6.8b) Proof By Definition 6.1, L{eatf(t)} = Z ∞ 0 eate−stf(t) dt = Z ∞ 0 e−(s−a)tf(t) dt.
But this is equation 6.2 with s replaced by s −a; that is, L{eatf(t)} = F(s −a).
Equation 6.8b is 6.8a written in terms of inverse transforms rather than transforms.
The notation in equation 6.8 is not quite as described earlier. The Laplace transform and its inverse operate on functions, not on function values as is suggested by 6.8. These equations would be more properly stated in the form L{eatf}(s) = F(s −a), (6.9a) L−1{F(s −a)}(t) = eatf(t).
(6.9b) We feel that the shifting property is more clearly conveyed for most readers by 6.8a,b, and we apologize to readers who are offended by the notation.
It may be convenient to repeat this practice in describing other properties of the Laplace transform, but we shall attempt to minimize its use.
SECTION 6.2 325 When calculating transforms and inverse transforms, we often write properties 6.8 in the form L{eatf(t)} = L{f(t)}|s→s−a, (6.10a) L−1{F(s −a)} = eatL−1{F(s)}.
(6.10b) On the right of property 6.10a, L{f(t)} is a function of s. The subscript |s →s −a means that each s is to be replaced by s −a.
Equation 6.8a, or its alternatives, states that multiplication by an exponential eat in the t-domain is equivalent to a translation or shift by a in the s-domain. It provides a quick way to find the Laplace transform of any function f(t) multiplied by an exponential, provided the Laplace transform of f(t) is known. For example, property 6.10a implies that L t2e−5t = L t2 |s→s+5 = 2 s3 |s→s+5 = 2 (s + 5)3 .
Property 6.10b yields L−1 1 (s −6)5 = e6tL−1 1 s5 = e6t t4 4!
.
Example 6.7 In Exercise 34 of Section 6.1, we found Laplace transforms for t cos at and t sin at, by using Euler’s identity and integration. Property 6.8a is superior if we again replace trigonometric functions with complex exponentials. This is an ongoing theme; al-ways consider replacing sines and cosines with complex exponentials. We can write that L{teati} = L{t}|s→s−ai = 1 s2 |s→s−ai = 1 (s −ai)2 .
We now display the real and imaginary parts of both sides of the equation, L{t(cos at + i sin at)} = (s + ai)2 (s −ai)2(s + ai)2 = (s2 −a2) + 2asi (s2 + a2)2 .
When we take real and imaginary parts, L{t cos at} = s2 −a2 (s2 + a2)2 , L{t sin at} = 2as (s2 + a2)2 .• Property 6.10b is particularly useful in finding inverse Laplace transforms of rational functions of s that contain irreducible quadratic factors in denominators.
We encounter them constantly. Here is an example.
Example 6.8 Find the inverse Laplace transform for F(s) = (s + 1)/(s2 −6s + 14).
Solution First, by completing the square on the quadratic, we can express F(s) in the form F(s) = s + 1 (s −3)2 + 5 = (s −3) + 4 (s −3)2 + 5.
We can now use property 6.10b to find the inverse transform, 326 SECTION 6.2 f(t) = e3tL−1 s + 4 s2 + 5 = e3t cos √ 5t + 4 √ 5 sin √ 5t .• In physical applications, we often encounter quantities that are turned on and off, or quantities that change abruptly.
For example, in mixing problems, such as Example 3.10 of Section 3.4, the concentration of salt added to the tank could suddenly be changed at any time; the applied voltage in an LCR-circuit could be turned on and offany number of times, and the forcing function in a mass-spring system could be turned on or off, or sharply changed. Such functions are conveniently described by Heaviside unit step functions introduced in Section 5.5. The fundamental unit step function is h(t) = 0, t < 0 1, t ≥0.
(6.11) (Some authors replace t ≥0 in this definition with t > 0 so that the function is undefined at t = 0. The rest of this chapter can be developed with either convention with minor adjustments in results.) A graph of this function is shown in Figure 6.5; there is a discontinuity of magnitude unity at t = 0, hence the name unit step function.
t 1 t 1 a Figure 6.5 Figure 6.6 When the discontinuity occurs at t = a, the function is denoted by h(t −a) = 0, t < a 1, t ≥a.
(6.12) Its graph is shown in Figure 6.6. This notation is consistent with that in elementary calculus where replacing the variable t in a function f(t) by t−a translates the graph of the function a units to the right.
Heaviside unit step functions provide compact representations for functions whose descriptions vary from one interval to another; such functions may have, or may not have, discontinuities at points t a b that separate these intervals. Such a Figure 6.7 function is shown in Figure 6.7. It is defined differently on the intervals 0 < t ≤a, a < t < b, and t > b. It is continuous at t = a, but not at t = b.
An important function in our discussions is shown in Figure 6.8. It is called a pulse function. It can be expressed algebraically in the form h(t −a) −h(t −b), except at t = a and t = b. In the event that the height of the nonzero portion is SECTION 6.2 327 c rather than unity (Figure 6.9), we obtain c[h(t −a) −h(t −b)], again except at t = a and t = b.
t a b 1 t a b c Figure 6.8 Figure 6.9 Pulse functions can be combined algebraically to describe step functions, such as that in Figure 6.10. It is the sum of two pulse functions, 4[h(t) −h(t −3)] + 2[h(t −3) −h(t −6)] = 4h(t) −2h(t −3) −2h(t −6), except at t = 0, 3, and 6. The step function in Figure 6.11 is the sum of three pulses, 3[h(t −a) −h(t −b)] + 4[h(t −b) −h(t −c)] + h(t −c) = 3h(t −a) + h(t −b) −3h(t −c), except at x = a, b, and c. In future representations of piecewise defined functions in terms of Heaviside functions, we will omit mentioning the exceptions.
t 4 2 3 6 t 4 2 a b c Figure 6.10 Figure 6.11 A convenient representation for the function in Figure 6.12 is t2[h(t)−h(t−a)], and for the function in Figure 6.13, [2 −(t −a)/(b −a)][h(t −a) −h(t −b)].
t a a 2 Parabola t 2 a b 1 Figure 6.12 Figure 6.13 What these examples illustrate is that to “turn a function on” for t ≥a, multiply it by h(t −a). It will be zero for t < a. To turn it on between t = a and t = b, multiply it by h(t −a) −h(t −b). It will be zero for t < a and t ≥b. The parabola in Figure 6.14 has equation a2+(t−a)2 for t > a. To turn it on, we multiply by h(t−a); that is, the function can be expressed in the form [a2 +(t−a)2]h(t−a).
For the function in Figure 6.15, we turn on the straight line y = a −a(t −a)/(b −a) for a < t < b, and then the horizontal line y = c for t > b, [a −a(t −a)/(b −a)][h(t −a) −h(t −b)] + c h(t −b).
328 SECTION 6.2 t a a 2 a2 2 2a t a b a c Figure 6.14 Figure 6.15 You may have noticed that in Figures 6.8–6.15, functions do not have values at discontinuities. Because of this, representations of these functions in terms of Heaviside functions are not valid at discontinuities. If a function has a value at a discontinuity, its representation in terms of Heaviside functions may or may not be valid at the discontinuity. For instance, the Heaviside representation of the functions in Figures 6.16a,b is f(t) = c + b −c a (t −a) h(t −a).
It is valid at the discontinuity t = a in Figure 6.16a, but not in 6.16b. None of this really matters when it comes to Laplace transforms. Because the transform of a function is defined as a definite integral, the transform is the same whether the function has a value at a discontinuity or not. We will therefore continue the practice of leaving a function undefined at discontinuities (when the purpose is to take the transform of the function).
t a a 2 b ( , ) a b 2 c t a a 2 b ( , ) a b 2 c Figure 6.16a Figure 6.16b The Laplace transform of the Heaviside unit step function is L{h(t −a)} = Z ∞ 0 e−sth(t −a) dt = Z ∞ a e−st dt = e−st −s ∞ a = e−as s , (6.13) provided s > 0.
In Section 6.1, we used integration to find the Laplace transform of piece-wise defined functions. In the above discussions, we represented such functions as products of functions multiplied by Heaviside functions, and we did so in order to circumvent integrations. The following theorem enables us to do this.
Theorem 6.5 When f(t) has a Laplace transform, L{f(t)h(t −a)} = e−asL{f(t + a)}.
(6.14a) Proof According to Definition 6.1, L{f(t)h(t −a)} = Z ∞ 0 e−stf(t)h(t −a) dt = Z ∞ a e−stf(t) dt.
SECTION 6.2 329 If we change variables of integration with u = t −a, then L{f(t)h(t −a)} = Z ∞ 0 e−s(u+a)f(u + a) du = e−as Z ∞ 0 e−suf(u + a) du = e−asL{f(t + a)}.
We illustrate how to use this result in the following examples.
Example 6.9 Find the Laplace transform for the function f(t) = 0, 0 ≤t ≤2 (t −2)2, t > 2, shown in Figure 6.17.
Solution Since f(t) can be expressed in the form f(t) = (t−2)2h(t−2), equation 6.14a gives F(s) = L{(t −2)2h(t −2)} = e−2sL{t2} = 2e−2s s3 .• t 1 2 3 1 t -1 2 3 Figure 6.17 Figure 6.18 Example 6.10 Find the Laplace transform for the function f(t) = 0, 0 ≤t < 2 t −3, t > 2, shown in Figure 6.18.
Solution Since f(t) can be expressed in the form f(t) = (t−3)h(t−2), its Laplace transform is F(s) = L{(t −3)h(t −2)} = e−2sL{(t + 2) −3} = e−2sL{t −1} = e−2s 1 s2 −1 s .• Example 6.11 Find the Laplace transform for the function in Figure 6.19.
Solution The function is con-tinuous, but because it is defined differently on the intervals 0 ≤t ≤1, 1 < t ≤2, and t > 2, it can be rep-resented efficiently in terms of Heavi-side functions, t 3 1 2 Figure 6.19 f(t) = 3(t −1)[h(t −1) −h(t −2)] + 3h(t −2) = 3(t −1) h(t −1) + (6 −3t) h(t −2).
We can now use equation 6.14a to find its Laplace transform, F(s) = e−sL{3t} + e−2sL{6 −3(t + 2)} = 3e−s s2 −e−2s 3 s2 .• 330 SECTION 6.2 Example 6.12 Find the Laplace transform for e−3t sin 2t h(t −1).
Solution Using property 6.14a, L{e−3t sin 2t h(t −1)} = e−sL{e−3(t+1) sin 2(t + 1)} = e−s−3L{e−3t sin 2(t + 1)} = e−(s+3)L{sin 2(t + 1)}|s→s+3 (using equation 6.10a) = e−(s+3)L{cos 2 sin 2t + sin 2 cos 2t}|s→s+3 = e−(s+3) (cos 2)2 s2 + 4 + (sin 2)s s2 + 4 |s→s+3 = e−(s+3) 2 cos 2 (s + 3)2 + 4 + (sin 2)(s + 3) (s + 3)2 + 4 .• The equivalent of property 6.14a in terms of inverse transforms is equally as important as 6.14a itself, and from the inverse statement it gets its name the second shifting property of Laplace transforms. We state it as a corollary to Theorem 6.5.
Corollary 6.5.1 If f = L−1{F}, then L−1{e−asF(s)} = f(t −a)h(t −a) = L−1{F(s)}|t→t−ah(t −a).
(6.14b) The graph of f(t −a)h(t −a) is that of f(t) (Figure 6.20a) shifted a units to the right and turned on for t > a (Figure 6.20b).
t t a Figure 6.20a Figure 6.20b Thus, to find the inverse transform of a function in the form e−asF(s), we find the inverse transform of F(s), translate it a units to the right, and turn it on for t > a.
Example 6.13 Find L−1 5e−4s s3 .
Solution With property 6.14b, L−1 5e−4s s3 = 5L−1 1 s3 |t→t−4 h(t −4) = 5 t2 2 |t→t−4 h(t −4) = 5 2(t −4)2h(t −4).• A graph of (5/2)(t −4)2h(t −4) is shown in Figure 6.21.
t 5/2 4 5 t 1/2 -1/2 1 2 3 4 5 Figure 6.21 Figure 6.22 SECTION 6.2 331 Example 6.14 Find the inverse transform for F(s) = e−s −e−2s s2 + 5 .
Solution Property 6.14b gives f(t) = L−1 e−s s2 + 5 −L−1 e−2s s2 + 5 = L−1 1 s2 + 5 |t→t−1 h(t −1) −L−1 1 s2 + 5 |t→t−2 h(t −2) = 1 √ 5 sin √ 5t |t→t−1 h(t −1) − 1 √ 5 sin √ 5t |t→t−2 h(t −2) = 1 √ 5 h sin √ 5(t −1) h(t −1) −sin √ 5(t −2) h(t −2) i .
The function is shown in Figure 6.22. We can write it without the Heaviside func-tions as follows: f(t) = 0, 0 ≤t ≤1 (1/ √ 5) sin √ 5(t −1), 1 < t ≤2 (1/ √ 5)[sin √ 5(t −1) −sin √ 5(t −2)], t > 2.• Finding inverse transforms is often a matter of finding the partial fraction decomposition of a rational function, together with the above properties and a set of tables. We illustrate in the following example. For readers who have never studied partial fractions, or need a refresher on the topic, we have provided coverage of the topic in Appendix D.
Example 6.15 Find inverse Laplace transforms for the following functions: (a) F(s) = s2 −9s + 9 s3(s2 + 9) (b) F(s) = 1 s2(s2 −4) (c) F(s) = e−s s2 −s Solution (a) The partial fraction decomposition of F(s) gives f(t) = L−1 s2 −9s + 9 s3(s2 + 9) = L−1 1 s3 −1 s2 + 1 s2 + 9 = t2 2 −t + 1 3 sin 3t.
(b) Once again partial fractions give f(t) = L−1 1 s2(s2 −4) = L−1 1/16 s −2 −1/16 s + 2 −1/4 s2 = 1 16e2t −1 16e−2t −t 4.
(c) With partial fractions and property 6.14b, f(t) = L−1 e−s s2 −s = L−1 e−s 1 s −1 −1 s = L−1 1 s −1 −1 s |t→t−1 h(t −1) = (et −1)|t→t−1h(t −1) = (et−1 −1)h(t −1).• 332 SECTION 6.2 Periodic Functions The sine and cosine functions are periodic and there was no difficulty in finding their transforms. The function in Figure 6.23a is also periodic, but it is not obvious how to find its transform. To use the definition of the transform as an improper integral would require the addition of an infinite number of definite integrals, one over each period of the function. Alternatively, we could write the periodic function as the sum of an infinite number of functions, turned on and offby Heaviside functions, and add all their transforms. Fortunately, neither of these procedures is necessary; we can develop a method for finding the Laplace transform of a periodic function that involves one integration over one period, or a procedure that involves no integration at all, and we can do this in two ways. We use the above ideas, but in a general format with an unspecified function. Suppose then that f(t) is periodic with period p, such as that in Figure 6.23a.
t p p p 2 3 t p p p 2 3 Figure 6.23a Figure 6.23b We divide the range of integration in Definition 6.1 into two parts F(s) = Z p 0 e−stf(t) dt + Z ∞ p e−stf(t) dt.
We substitute u = t −p in the second integral, F(s) = Z p 0 e−stf(t) dt + Z ∞ 0 e−s(u+p)f(u + p) du.
But f(t) has period p, so that f(u + p) = f(u), and therefore F(s) = Z p 0 e−stf(t) dt + e−ps Z ∞ 0 e−suf(u) du = Z p 0 e−stf(t) dt + e−psF(s).
We can solve this for F(s) = 1 1 −e−ps Z p 0 e−stf(t) dt.
(6.15) The is the first of the results that we sought, a method for finding the transform of a periodic function that involves only integration over one period of the function. The second method is to not integrate at all. Suppose that f1(t) denotes the function that is equal to f(t) over the first period 0 ≤t ≤p of the function, and is otherwise equal to zero (Figure 6.23b).
This function is sometimes called the windowed version of f(t). The Laplace transform of f1(t) is defined by the definite integral in equation 6.15. In other words, we can write that F(s) = 1 1 −e−ps L{f1(t)}.
(6.16) SECTION 6.2 333 This is the second result that we were looking for, a method to calculate the Laplace transform of a periodic function that does not require integration. Formula 6.16 does this, provided we are willing, and able, to find L{f1(t)} without integration.
The same results can be obtained in another way. We write the function f1(t) in terms of f(t), f1(t) = f(t)[h(t) −h(t −p)] = f(t)[1 −h(t −p)].
We now take Laplace transforms, and use property 6.14a, F1(s) = F(s) −e−psL{f(t + p)}.
But f(t + p) = f(t), so that F1(s) = F(s) −e−psL{f(t)} = F(s) −e−psF(s) = (1 −e−ps)F(s).
When we solve this equation for F(s), we get F(s) = 1 1 −e−ps F1(s) = 1 1 −e−ps L{f1(t)}, equation 6.16.
Example 6.16 Find the Laplace transform for the periodic function in Figure 6.24a.
Solution Since the function has period 2, formula 6.15 yields F(s) = 1 1 −e−2s Z 2 0 (1 −t)e−st dt.
Integration by parts gives F(s) = 1 1 −e−2s (t −1) s e−st + 1 s2 e−st 2 0 = 1 + e−2s s(1 −e−2s) −1 s2 .
Alternatively, formula 6.16 gives F(s) = 1 1 −e−ps L{f1(t)}, where f1(t) is the windowed version of f(t) in Figure 6.24b. Its Laplace transform is L{(1 −t)[h(t) −h(t −2)]} = L{(1 −t) h(t)} + L{(t −1) h(t −2)} = 1 s −1 s2 + e−2sL{t + 1} = 1 s −1 s2 + e−2s 1 s2 + 1 s .
Hence, F(s) = 1 1 −e−2s 1 s −1 s2 + e−2s 1 s2 + 1 s = 1 + e−2s s(1 −e−2s) −1 s2 .• 334 SECTION 6.2 t 1 1 2 3 -1 t 1 1 2 -1 Figure 6.24a Figure 6.24b Example 6.17 Find the Laplace transform for | sin 2t|.
Solution Since | sin 2t| has period π/2 (see Figure 6.25), formula 6.15 gives L{| sin 2t|} = 1 1 −e−πs/2 Z π/2 0 e−st sin 2t dt, and we could use integration by parts to evaluate this integral. Alternatively, we can use formula 6.16, L{| sin 2t|} = 1 1 −e−πs/2 L{sin 2t[h(t) −h(t −π/2)]} = 1 1 −e−πs/2 L{sin 2t} −L{sin 2t h(t −π/2)} = 1 1 −e−πs/2 2 s2 + 4 −e−πs/2L{sin 2(t + π/2)} = 1 1 −e−πs/2 2 s2 + 4 + e−πs/2L{sin 2t} = 1 1 −e−πs/2 2 s2 + 4 + 2e−πs/2 s2 + 4 = 2(1 + e−πs/2) (s2 + 4)(1 −e−πs/2).• t 1 p p p 2 2 3 Figure 6.25 We have just seen that Laplace transforms of periodic functions that are not sinusoids contain factors 1/(1 −e−ps). When we solve differential equations that have periodic inputs in Section 6.4, we will have to invert transforms with such factors. The following example illustrates how to do this.
Example 6.18 Find the inverse Laplace transform for F(s) = 2 s3(1 −e−2s).
Solution Property 6.14b enables us to find the inverse transform of any function multiplied by e−as. We can write the above F(s) as a sum of terms in this form if we expand 1/(1 −e−2s) in a geometric seris F(s) = 2 s3(1 −e−2s) = 2 s3 1 + e−2s + e−4s + e−6s + · · · .
We can now invert each term, SECTION 6.2 335 f(t) = t2 + (t −2)2h(t −2) + (t −4)2h(t −4) + · · · = ∞ X n=0 (t −2n)2h(t −2n).• Properties of Laplace transforms and their inverses that we have discussed in this section have been gathered together for quick reference in Table 6.2. Included also are the transform pairs in Table 6.1. There are also properties that have yet to be considered, namely lines 12, 17, and 18. These will be developed in subsequent sections. Notice the arrows in the middle column. A double arrow ↔indicates that this line is useful in taking Laplace transforms and their inverses; a right arrow → indicates that the property is most useful in taking transforms; and a left arrow ← indicates that the property is most useful in taking inverse transforms.
EXERCISES 6.2 In Exercises 1–12 represent the functions in Exercises 21–32 of Section 6.1 in terms of Heaviside unit step functions. Find the Laplace transform of each function.
In Exercises 13–20 represent the function algebraically in terms of Heaviside unit step functions. Find the Laplace transform of each function.
13.
14.
t 2 1 t 3 2 1 1 2 15.
16.
t Parabola 4 1 Parabola t 1 1 2 17.
18.
t 1 -1 p 2p Sine function t 2p 4p -1 1 Sine function 336 SECTION 6.2 f(t) F(s) = L{f}(s) tn (n = 0, 1, 2, . . .) ↔ n!
sn+1 eat ↔ 1 s −a sin at ↔ a s2 + a2 cos at ↔ s s2 + a2 t sin at ↔ 2as (s2 + a2)2 t cos at ↔ s2 −a2 (s2 + a2)2 sin at −at cos at ↔ 2a3 (s2 + a2)2 sin at + at cos at ↔ 2as2 (s2 + a2)2 sinh at ↔ a s2 −a2 cosh at ↔ s s2 −a2 h(t −a) ↔ e−as s δ(t −a) ↔ e−as eatf(t) ↔ F(s −a) f(t)h(t −a) → e−asL{f(t + a)} f(t −a)h(t −a) ← e−asF(s) p −periodic f(t) → 1 1 −e−ps Z p 0 e−stf(t) dt Z t 0 f(u)g(t −u) du ← F(s)G(s) tnf(t) (n = 1, 2, 3, . . .) ↔ (−1)n dnF dsn f ′(t) → sF(s) −f(0) f ′′(t) → s2F(s) −sf(0) −f ′(0) f (n)(t) → snF(s) −sn−1f(0) −sn−2f ′(0) −· · · −f (n−1)(0) Table 6.2 SECTION 6.2 337 19.
20.
t 1 -1 p 4p t e t 2 1 2 -ln2 In Exercises 21–32 use property 6.8a to find the Laplace transform for the function.
21. f(t) = t3e−5t 22. f(t) = t2e3t 23. f(t) = 4te−t −2e−3t 24. f(t) = 5eat −5e−at 25. f(t) = et sin 2t + e−t cos t 26. f(t) = 2e−3t sin 3t + 4e3t cos 3t 27. f(t) = tet cos 2t 28. f(t) = te−2t sin t 29. f(t) = 2et(cos t + sin t) 30. f(t) = (t −1)e2−3t sin 4t 31. f(t) = t2 cos at 32. f(t) = t2 sin at In Exercises 33–42 use property 6.14a to find the Laplace transform of the function.
33. f(t) = (t −2)2h(t −2) 34. f(t) = sin 3(t −4) h(t −4) 35. f(t) = t h(t −1) 36. f(t) = (t + 5) h(t −3) 37. f(t) = (t2 + 2) h(t −1) 38. f(t) = cos t h(t −π) 39. f(t) = cos t h(t −2) 40. f(t) = eth(t −4) 41. f(t) = t2et h(t −3) 42. f(t) = et cos 2t h(t −1) In Exercises 43–47 find the Laplace transform of the periodic function.
43. f(t) = t, 0 < t < a, f(t + a) = f(t) 44. f(t) = 1, 0 < t < a −1, a < t < 2a f(t + 2a) = f(t) 45. f(t) = | sin at| 46. f(t) = t, 0 < t < a 2a −t, a < t < 2a f(t + 2a) = f(t) 47. f(t) = 1, 0 < t < a 0, a < t < 2a f(t + 2a) = f(t) Find the inverse Laplace transform in Exercises 48–69.
48. F(s) = 1 s2 −2s + 5 49. F(s) = s s2 + 4s + 1 50. F(s) = e−2s s2 51. F(s) = e−3s s2 + 1 52. F(s) = se−5s s2 + 2 53. F(s) = se−s (s2 + 4)2 54. F(s) = 1 4s2 −6s −5 55. F(s) = s s2 −3s + 2 56. F(s) = 4s + 1 (s2 + s)(4s2 −1) 57. F(s) = e−3s s + 5 338 SECTION 6.2 58. F(s) = e−2s s2 + 3s + 2 59. F(s) = 1 s3 + 1 60. F(s) = 5s −2 3s2 + 4s + 8 61. F(s) = e−s(1 −e−s) s(s2 + 1) 62. F(s) = s (s + 1)5 63. F(s) = s2 + 2s + 3 (s2 + 2s + 2)(s2 + 2s + 5) 64. F(s) = s2 (s2 −4)2 65. F(s) = 1 s(1 −e−s) 66. F(s) = 1 s(1 + e−s) 67. F(s) = 1 (s2 + 4)(1 −e−3s) 68. F(s) = 1 (s3 + 5s)(1 −e−2s) 69. F(s) = (s2 + 1)e−2s (s4 + 2s2)(1 + e−s) 70. To find the inverse transform of a rational function with irreducible quadratic factors in de-nominators, we have used property 6.8b. Example 6.8 contained such a situation, and some of the above exercises. If you love to work with complex numbers, you might be pleased to know that you can always replace irreducible real factors with complex linear factors. It is not a method that we recommend, but it is at least comforting to know that it can be done. We illustrate with a simple example, hoping that it convinces you not to persue complex linear factors in the future.
(a) Use property property 6.8b to find the inverse transform of F(s) = s + 2 s2 + 2s + 5.
(b) Find the complex roots of s2 + 2s + 5 = 0, and use them to show that the partial fraction decomposition of F(s) with complex linear factors is F(s) = 1 4 2 −i s + 1 −2i + 2 + i s + 1 + 2i .
(c) Use the decomposition in part (b) to find L−1{F(s)}.
71. The following two formulas, called reduction of order formulas, can be useful in taking inverse transforms, L−1 s (s2 + a2)n+1 = t 2nL−1 1 (s2 + a2)n , L−1 1 (s2 + a2)n+1 = −t 2na2 L−1 s (s2 + a2)n + 2n −1 2na2 L−1 1 (s2 + a2)n .
Verify these formulas using L{tf(t)} = −d dsL{f(t)}.
This result will be verified and extended in Section 6.7.
72. Use the reduction of order formulas in Exercise 71 to verify lines 5–8 in Table 6.2.
Use the reduction of order formulas in Exercise 71 to find the inverse Laplace transform in Exercises 73–76.
73. F(s) = 1 (s2 + a2)3 74. F(s) = s (s2 + a2)3 SECTION 6.2 339 75. F(s) = 1 (s2 −2s + 5)3 76. F(s) = s + 2 (s2 −4s + 13)3 77. If F(s) = L{f(t)} for s > α, for what values of s is F(s −a) the Laplace transform of eatf(t)?
78. Find the Laplace transform of the function f(t) = t2/4, 0 ≤t < 1 −(t2 −4t + 2)/4, 1 ≤t < 3 (t −4)2/4, 3 ≤t ≤4 f(t + 4) = f(t).
79. Verify the change of scale property: If F(s) = L{f(t} for s > α, then for a > 0, L{f(at)} = 1 aF s a , s > αa.
340 SECTION 6.3 6.3 Laplace Transforms and Differential Equations The Laplace transform is a powerful technique for solving linear, ordinary and par-tial differential equations. It replaces differentiations with algebraic operations. Like the techniques of Chapter 4, the transform cannot be used on nonlinear problems.
A simple example such as the following nonlinear equation illustrates why, yy′′ + 2y′ + 3y = t2.
You may have noticed that we have not developed a general formula for the Laplace transform of the product of two functions, and the reason is that there just isn’t one. There are special cases such as when eat multiplies another function, but not for a product such as yy′′. This is why the transform is not applied to nonlinear problems.
The following theorem and its corollary simplify the process of applying the Laplace transform to linear differential equations.
Theorem 6.6 Suppose f is continuous for t ≥0 with a piecewise-continuous first derivative on every finite interval 0 ≤t ≤T. If f is O(eαt), then L{f ′} exists for s > α, and L{f ′(t)} = sF(s) −f(0).
(6.17) (A more precise representation of the left side of this equation is L{f ′}(s).) Proof If tj, j = 1, . . . , n denote the discontinuities of f ′ in 0 ≤t ≤T, then Z T 0 e−stf ′(t) dt = n X j=0 Z tj+1 tj e−stf ′(t) dt, where t0 = 0 and tn+1 = T. Since f ′ is continuous on each subinterval, we may integrate by parts on these subintervals, Z T 0 e−stf ′(t) dt = n X j=0 " e−stf(t) tj+1 tj + s Z tj+1 tj e−stf(t) dt # .
Because f is continuous, f(tj+) = f(tj−), j = 1, . . . , n, and therefore Z T 0 e−stf ′(t) dt = −f(0) + e−sT f(T) + s Z T 0 e−stf(t) dt.
Thus, L{f ′} = Z ∞ 0 e−stf ′(t) dt = lim T →∞ Z T 0 e−stf ′(t) dt = lim T →∞ " −f(0) + e−sT f(T) + s Z T 0 e−stf(t) dt # = sF(s) −f(0) + lim T →∞e−sT f(T), provided the limit on the right exists. Since f is O(eαt), there exists M and T such that for t > T, |f(t)| < Meαt. Thus, for T > T, e−sT |f(T)| < e−sT MeαT = Me(α−s)T SECTION 6.3 341 which approaches 0 as T →∞(provided s > α). Consequently, L{f ′} = sF(s) −f(0).
This result is easily extended to second and higher order derivatives. For ex-tensions when f is only piecewise-continuous, see Exercise 50.
Corollary 6.6.1 Suppose f and f ′ are continuous for t ≥0, and f ′′ is piecewise-continuous on every finite interval 0 ≤t ≤T. If f and f ′ are O(eαt), then L{f ′′} exists for s > α, and L{f ′′} = s2F(s) −sf(0) −f ′(0).
(6.18) Proof Since f ′ is continuous, f ′′ is piecewise-continuous, and f ′ is O(eαt), equa-tion 6.17 gives L{f ′′} = sL{f ′} −f ′(0).
We can apply equation 6.17 once again to obtain L{f ′′} = s[sF(s) −f(0)] −f ′(0) = s2F(s) −sf(0) −f ′(0).
The extension to nth-order derivatives is contained in the next corollary.
Corollary 6.6.2 Suppose f and its first n −1 derivatives are continuous for t ≥0, and f (n)(t) is piecewise-continuous on every finite interval 0 ≤t ≤T. If f and its first n −1 derivatives are O(eαt), then L{f (n)(t)} exists for s > α, and L{f (n)(t)} = snF(s) −sn−1f(0) −sn−2f ′(0) −· · · −f (n−1)(0).
(6.19) In Section 6.1, we demonstrated how to use Laplace transforms to solve an initial-value problem. We now consider further examples.
Example 6.19 Solve the initial-value problem y′′ −2y′ + y = 2et, y(0) = y′(0) = 0.
Solution First we assume that the solution of the problem is a function satisfying the conditions of Corollary 6.6.1. We can then take Laplace transforms of both sides of the differential equation, L{y′′} −2L{y′} + L{y} = 2L{et}.
Properties 6.17 and 6.18 yield [s2Y (s) −sy(0) −y′(0)] −2[sY (s) −y(0)] + Y (s) = 2 s −1.
We now substitute from the initial conditions y(0) = y′(0) = 0, s2Y (s) −2sY (s) + Y (s) = 2 s −1, and solve this equation for Y (s), Y (s) = 2 (s −1)3 .
The required function y(t) can now be obtained by taking the inverse transform of Y (s), 342 SECTION 6.3 y(t) = L−1 2 (s −1)3 = 2L−1 1 (s −1)3 (by linearity) = 2etL−1 1 s3 (by property 6.10b) = 2et t2 2 (from Table 6.1) = t2et.• This example is typical of Laplace transforms at work on initial-value problems.
We begin by assuming that the solution of the problem satisfies whatever conditions are necessary to apply the transform to the differential equation. In the case of Example 6.19, this meant assuming that y(t) satisfies the conditions of Corollary 6.6.1. In actual fact we need only assume that y(t) and y′(t) are of exponential order.
Since the nonhomogeneity 2et is continuous, our theory in Chapter 4 indicates that the solution has a continuous second derivative. In applying the Laplace transform to a third-order differential equation, we would assume that the solution satisfies the conditions of Corollary 6.6.2 for n = 3. The Laplace transform reduces the differential equation in y(t) to an algebraic equation in its transform Y (s). Notice how initial conditions for the solution of the initial-value problem are incorporated by the Laplace transform at a very early stage, unlike the techniques of Chapter 4 where they are used to determine arbitrary constants in a general solution. The algebraic equation is solved for Y (s) and the inverse transform then yields a function y(t). That y(t) is a solution of the initial-value problem can be verified in two ways.
First, we can check that y(t) and y′(t) are of exponential order, thus vindicating the initial assumption. Alternatively, we can verify that y(t) satisfies the differential equation and initial conditions. We will omit these formal verifications, although the problem is not truly solved until one of these actions has been taken.
In each occurrence of the Laplace transform of y(t) in the above example, we wrote Y (s). In order to keep notation as simple as possible in further examples, we will write Y in place of Y (s) when taking Laplace transforms of a differential equation.
Example 6.20 Solve the initial-value problem y′′ + 4y = 3 cos 2t, y(0) = 1, y′(0) = 0.
Solution Assuming that the solution and its first derivative are of exponential order, we take Laplace transforms of both sides of the differential equation and use the initial conditions, [s2Y −s(1) −0] + 4Y = 3s s2 + 4.
The solution of this equation for Y (s) is Y (s) = 3s (s2 + 4)2 + s s2 + 4, and Table 6.1 gives SECTION 6.3 343 y(t) = 3 t 4 sin 2t + cos 2t.• Laplace transforms thrive on initial-value problems; they use the initial condi-tions of the problem when the Laplace transform is applied to the derivative terms in the differential equation. They can also be adapted to boundary-value problems, as the following example illustrates.
Example 6.21 Solve the following boundary-value problem on the interval 0 ≤t ≤π/2, y′′ + 9y = cos 2t, y(0) = 1, y(π/2) = −1.
Solution The solution of the problem is only desired on the interval 0 ≤t ≤π/2.
What we do is solve the problem on the interval t ≥0, and then restrict the solution to the interval 0 ≤t ≤π/2. When we apply formula 6.18 to the second derivative in the differential equation, the derivative y′(0) is needed. Since it is not one of the given pieces of information in the problem, we assign a letter to represent it; that is, we let y′(0) = A. If we assume that the solution and its first derivative are of exponential order and apply the Laplace transform to the differential equation, we obtain [s2Y −s(1) −A] + 9Y = s s2 + 4.
We now solve for Y (s), Y (s) = s + A s2 + 9 + s (s2 + 4)(s2 + 9).
Partial fractions on the second term gives Y (s) = s + A s2 + 9 + s/5 s2 + 4 + −s/5 s2 + 9 = 4s/5 + A s2 + 9 + s/5 s2 + 4.
Inverse transforms yield y(t) = 4 5 cos 3t + A 3 sin 3t + 1 5 cos 2t.
The boundary condition y(π/2) = −1 can now be used to find A, −1 = −A 3 −1 5 = ⇒ A = 12 5 .
The solution of the boundary-value problem is y(t) = 4 5 cos 3t + 4 5 sin 3t + 1 5 cos 2t.• Laplace transforms are particularly adept at handling initial conditions, and as we have just seen, they can be adapted to boundary conditions.
They can also provide general solutions to linear differential equations, as shown in the next example.
Example 6.22 Find a general solution of the differential equation y′′ + 2y′ −3y = t2.
Solution We denote initial values of the solution and its first derivative by y(0) = A and y′(0) = B. If we assume that the solution and its first derivative are of 344 SECTION 6.3 exponential order, and take Laplace transforms of both sides of the differential equation, [s2Y −s(A) −B] + 2[sY −A] −3Y = 2 s3 .
The solution of this equation for Y (s) is Y (s) = 2 s3(s2 + 2s −3) + As + (B + 2A) s2 + 2s −3 .
The partial fraction decomposition of the first term is 2 s3(s2 + 2s −3) = −2/3 s3 −4/9 s2 −14/27 s + 1/2 s −1 + 1/54 s + 3.
Hence, Y (s) = −2/3 s3 −4/9 s2 −14/27 s + 1/2 s −1 + 1/54 s + 3 + As + (B + 2A) (s −1)(s + 3) .
If we are not concerned with preserving the fact that A and B represent initial values for y(t) and its first derivative, we can write that Y (s) is of the form Y (s) = −2/3 s3 −4/9 s2 −14/27 s + C s −1 + D s + 3, where C and D are constants. Inverse transforms now give a general solution y(t) = −t2 3 −4t 9 −14 27 + Cet + De−3t.• Did you notice that the denominator of the transform in each of the above examples is the function φ(m) in the auxiliary equation of Chapter 4 with m replaced by s; that is, it is φ(s). This is always the case for linear differential equations with constant coefficients; and it can serve as a partial check on calculations.
Example 6.23 A 2-kilogram mass is suspended from a spring with constant 128 newtons per metre.
It is pulled 4 centimetres above its equilibrium position and released. An external force 3 sin ωt newtons acts vertically on the mass during its motion. If damping is negligible, find the position of the mass as a function of time.
Solution The initial-value problem describing oscillations of the mass is 2d2x dt2 + 128x = 3 sin ωt, x(0) = 1/25, x′(0) = 0.
If we take Laplace transforms of both sides of the differential equation, 2[s2X −s/25] + 128X = 3ω s2 + ω2 = ⇒X(s) = 3ω 2(s2 + 64)(s2 + ω2) + s 25(s2 + 64).
When ω ̸= 8, partial fractions on the first term on the left leads to X(s) = 3ω 2(64 −ω2)(s2 + ω2) − 3ω 2(64 −ω2)(s2 + 64) + s 25(s2 + 64).
Hence, displacement in the absence of resonance is SECTION 6.3 345 x(t) = 3 2(64 −ω2) sin ωt − 3ω 16(64 −ω2) sin 8t + 1 25 cos 8t.
When ω = 8, the Laplace transform X(s) takes the form X(s) = 12 (s2 + 64)2 + s 25(s2 + 64), in which case Table 6.1 gives the resonant solution x(t) = 12 2(8)3 (sin 8t −8t cos 8t) + 1 25 cos 8t = 3 256 sin 8t −3t 32 cos 8t + 1 25 cos 8t.• Convolutions As a linear operator, the Laplace transform efficiently handles sums and dif-ferences of functions. What it does not handle is the product of functions; that is, we do not have a formula for the Laplace transform of the product of two functions f(t) and g(t). We can take the transform of certain products such as an exponential multiplying another function (formula 6.8). In Section 6.7, we also find out how to take the Laplace transform of the product tnf(t), where n is a positive integer.
But, in general, there is no formula for the Laplace transform of the product of two arbitrary functions. One would expect that there would therefore be no formula for the inverse Laplace transform of the product of two functions. Surprisingly, there is a formula for L−1{F(s)G(s)} when inverse transforms of F(s) and G(s) are known.
We shall see shortly that the inverse of F(s)G(s) is what is called the convolution of f(t) and g(t).
Definition 6.4 The convolution of two functions f and g is a function denoted by f ∗g with values defined by (f ∗g)(t) = Z t 0 f(u)g(t −u) du.
(6.20) The following properties of convolutions are easily verified using Definition 6.4: f ∗g = g ∗f, (6.21a) f ∗(kg) = (kf) ∗g = k(f ∗g), k a constant (6.21b) (f ∗g) ∗h = f ∗(g ∗h), (6.21c) f ∗(g + h) = f ∗g + f ∗h.
(6.21d) Example 6.24 Find the convolution of f(t) = sin t and g(t) = cos 4t.
Solution According to equation 6.20, (f ∗g)(t) = Z t 0 sin u cos 4(t −u) du.
With the trigonometric identity sin A cos B = (1/2)[sin (A + B) + sin (A −B)], we obtain (f ∗g)(t) = 1 2 Z t 0 [sin (4t −3u) + sin (5u −4t)] du = 1 2 1 3 cos (4t −3u) −1 5 cos (5u −4t) t 0 = 1 15(cos t −cos 4t).• 346 SECTION 6.3 The importance of convolutions lies in the following theorem.
Theorem 6.7 If f and g are O(eαt) and piecewise-continuous on every finite interval 0 ≤t ≤T, then L{f ∗g} = L{f}L{g}, s > α.
(6.22a) Proof If F = L{f} and G = L{g}, then F(s)G(s) = Z ∞ 0 e−suf(u) du Z ∞ 0 e−sτg(τ) dτ = Z ∞ 0 Z ∞ 0 e−s(u+τ)f(u)g(τ) dτ du.
Suppose we change variables of integration in the inner integral with respect to τ by setting t = u + τ. Then F(s)G(s) = Z ∞ 0 Z ∞ u e−stf(u)g(t −u) dt du = lim T →∞ Z T 0 Z ∞ u e−stf(u)g(t −u) dt du.
We would like to interchange orders of integration, but to do so requires that the inner integral converge uniformly with respect to u. To verify that this is indeed the case we note that since f and g are O(eαt) and piecewise-continuous on every finite interval 0 ≤t ≤T, there exists a constant M such that for all t ≥0, |f(t)| < Meαt and |g(t)| < Meαt. For each u ≥0, we therefore have |e−stf(u)g(t −u)| < M 2e−steαueα(t−u) = M 2e−t(s−α). Thus, Z ∞ u e−stf(u)g(t −u) dt < M 2 Z ∞ u e−t(s−α) dt = M 2 e−t(s−α) α −s ∞ u = M 2e−u(s−α) s −α < M 2 s −α, provided s > α, and the improper integral is uniformly convergent with respect to u. The order of integration in the expression for F(s)G(s) may therefore be interchanged (Figure 6.26), and we obtain u t u t T = Figure 6.26 F(s)G(s) = lim T →∞ "Z T 0 e−st Z t 0 f(u)g(t −u) du dt + Z ∞ T e−st Z T 0 f(u)g(t −u) du dt # .
Since Z ∞ T e−st Z T 0 f(u)g(t −u) du dt < Z ∞ T Z T 0 M 2e−t(s−α)du dt = M 2T e−t(s−α) α −s ∞ T = M 2Te−T (s−α) s −α SECTION 6.3 347 provided s > α, it follows that lim T →∞ Z ∞ T e−st Z T 0 f(u)g(t −u) du dt = 0.
Thus, F(s)G(s) = lim T →∞ Z T 0 e−st Z t 0 f(u)g(t −u) du dt = lim T →∞ Z T 0 e−stf ∗g dt = L{f ∗g}.
More important in practice is the inverse of property 6.22a.
Corollary 6.7.1 If L−1{F} = f and L−1{G} = g, where f and g are O(eαt) and piecewise-continuous on every finite interval, then L−1{FG} = f ∗g.
(6.22b) This is line 17 in Table 6.2. The following example illustrates how to use this corollary.
Example 6.25 Find the inverse transform of F(s) = 2 s2(s2 + 4).
Solution Since L−1{2/(s2+4)} = sin 2t and L−1{1/s2} = t, convolution property 6.22b gives L−1 2 s2(s2 + 4) = Z t 0 u sin 2(t −u) du = u 2 cos 2(t −u) + 1 4 sin 2(t −u) t 0 = t 2 −1 4 sin 2t.
The alternative is to use partial fractions.• Convolutions are particularly useful when solving differential equations that contain unspecified nonhomogeneities.
Example 6.26 Find the solution of the initial-value problem y′′ + 2y′ + 3y = f(t), y(0) = 1, y′(0) = 0, where f(t) is of exponential order and piecewise-continuous for t ≥0.
Solution The only technique from Chapter 4 that can handle this problem is variation of parameters; the other techniques require that the form of the nonhomo-geneity be known. To show that Laplace transforms can also be used, we assume that the solution and its first derivative are of exponential order, and take Laplace transforms of both sides of the differential equation, [s2Y −s] + 2[sY −1] + 3Y = F(s).
We solve for Y (s), Y (s) = F(s) s2 + 2s + 3 + s + 2 s2 + 2s + 3.
To find the inverse transform of this function, we first note that 348 SECTION 6.3 L−1 1 s2 + 2s + 3 = L−1 1 (s + 1)2 + 2 = e−tL−1 1 s2 + 2 = 1 √ 2e−t sin √ 2t.
Convolution property 6.22b on the first term of Y (s) now yields y(t) = Z t 0 f(u) 1 √ 2e−(t−u) sin √ 2(t −u) du + L−1 (s + 1) + 1 (s + 1)2 + 2 = 1 √ 2 Z t 0 f(u)e−(t−u) sin √ 2(t −u) du + e−tL−1 s + 1 s2 + 2 = 1 √ 2 Z t 0 f(u)e−(t−u) sin √ 2(t −u) du + e−t cos √ 2t + 1 √ 2 sin √ 2t .• The following nontrivial problem makes clever use of convolutions.
The Tautochrone A bead, with zero initial velocity is to slide frictionlessly down a wire from a point P(x, y) to the origin (Figure 6.27).
Tautochrone is the name attached to the shape of the curve for which the time of descent is independent of the height, y, on the curve from which the bead starts.
Suppose we let the equation of the curve be x = x(y). Since the kinetic energy of y x P x,y x x y ( ) = ) ( ( , ) h z the bead at any point (η, ζ) along the Figure 6.27 curve is equal to the initial potential energy of the bead, we can write that 1 2mv2 = mg(y −ζ), where m is the mass of the bead and v is its velocity. It follows that the velocity of the bead at (η, ζ) is v = p 2g p y −ζ.
Since the time for the bead to traverse an element of arc length p dη2 + dζ2 at point (η, ζ) is this length divided by v, the total time to travel from P to the origin is T = Z y 0 p dη2 + dζ2 √2g√y −ζ = 1 √2g Z y 0 s 1 + dη dζ 2 √y −ζ dζ.
If we set f(ζ) = p 1 + (dη/dζ)2, then T = 1 √2g Z y 0 f(ζ) √y −ζ dζ.
This integral can be interpreted as the convolution of the functions f(y) and 1/√y.
If we take Laplace transforms of both sides of the equation with respect to y, and note that T is a constant, we obtain SECTION 6.3 349 √2gT s = L f(y) ∗1 √y .
According to Exercise 35 in Section 6.1, the Laplace transform of 1/√y is p π/s.
Thus, √2gT s = F(s) rπ s = ⇒ F(s) = √2gT π rπ s .
The inverse transform now gives f(y) = s 1 + dx dy 2 = √2gT π 1 √y , a differential equation for x(y). It can be rewritten in the form dx dy = − s k2 −y y , where we have set k = √2gT/π. We now integrate and set y = k2 sin2 θ, −x + C = Z p k2 −y √y dy = Z k cos θ k sin θ 2k2 sin θ cos θ dθ = 2k2 Z cos2 θ dθ = 2k2 Z 1 + cos 2θ 2 dθ = k2 θ + 1 2 sin 2θ = k2 Sin−1 √y k + √y k r 1 −y k2 = k2Sin−1 √y k + p k2y −y2.
To pass through the origin, C must be zero, so that the equation of the tautochrone is x = −k2Sin−1 √y k − p k2y −y2.
By setting φ = −2θ, parametric equations for the tautochrone are x = −k2 2 (2θ + sin 2θ) = k2 2 (φ + sin φ), y = k2 sin2 θ = k2 1 −cos 2θ 2 = k2 2 (1 −cos φ).
These are parametric equations for a cycloid (see Exercise 60 in Section 2.2).
EXERCISES 6.3 In Exercises 1–16 use Laplace transforms to solve the initial-value problem.
1. y′′ + 3y′ −4y = t + 3, y(0) = 1, y′(0) = 0 2. y′′ + 2y′ −y = et, y(0) = 1, y′(0) = 2 350 SECTION 6.3 3. y′′ + y = 2e−t, y(0) = y′(0) = 0 4. y′′ + 2y′ + y = t, y(0) = 0, y′(0) = 1 5. y′′ −2y′ + y = t2et, y(0) = 1, y′(0) = 0 6. y′′ + y = t, y(0) = 1, y′(0) = −2 7. y′′ + 2y′ + 5y = e−t sin t, y(0) = 0, y′(0) = 1 8. y′′ + 6y′ + y = sin 3t, y(0) = 2, y′(0) = 1 9. y′′ + y′ −6y = t + cos t, y(0) = 1, y′(0) = −2 10. y′′ −4y′ + 5y = te−3t, y(0) = −1, y′(0) = 2 11. y′′ + 4y = f(t), y(0) = 0, y′(0) = 1, where f(t) = 1, 0 < t < 1 0, t > 1 12. y′′ + 2y′ −4y = cos2 t, y(0) = 0, y′(0) = 0 13. y′′ −3y′ + 2y = 8t2 + 12e−t, y(0) = 0, y′(0) = 2 14. y′′ + 4y′ −2y = sin 4t, y(0) = 0, y′(0) = 0 15. y′′ + 8y′ + 41y = e−2t sin t, y(0) = 0, y′(0) = 1 16. y′′ + 2y′ + y = f(t), y(0) = 0, y′(0) = 0, where f(t) = t, 0 < t < 1 0, t > 1 In Exercises 17–19 use Laplace transforms to solve the boundary-value problem.
17. y′′ + 9y = cos 2t, y(0) = 1, y(π/2) = −1 18. y′′ + 3y′ −4y = 2e−4t, y(0) = 1, y(1) = 1 19. y′′ + 2y′ + 5y = e−t sin t, y(0) = 0, y(π/4) = 1 20. There are two ways to use Laplace transforms to solve the initial-value problem y′′ + 2y′ −3y = sin 2t, y(π/2) = 3, y′(π/2) = −1.
Do both: (a) Find a general solution and then use the initial conditions to find constants.
(b) Translate the initial conditions to u = 0 by setting u = t−π/2. Solve the problem for y(u), and then return to y(t).
In Exercises 21–24 use Laplace transforms to find an integral representation for the solution to the problem.
21. y′′ −4y′ + 3y = f(t), y(0) = 1, y′(0) = 0 22. y′′ + 4y′ + 6y = f(t), y(0) = 0, y′(0) = 0 23. y′′ + 16y = f(t) 24. y′′ + 3y′ + 2y = etf(t) In Exercises 25–28 use convolutions to find the inverse Laplace transform for the func-tion.
25. F(s) = 1 s(s + 1) 26. F(s) = 1 (s2 + 1)(s2 + 4) SECTION 6.3 351 27. F(s) = s (s + 4)(s2 −2) 28. F(s) = s (s2 −4)(s2 −9) In Exercises 29–34 use Laplace transforms to find a general solution of the differential equation.
29. y′′ −2y′ + 4y = t2 30. y′′ −2y′ + y = t2et 31. y′′ + y = f(t) 32. y′′ + 2y′ + 5y = e−t sin t 33. y′′ + 4y′ + y = t + 2 34. y′′ −4y = f(t) 35. To find a general solution for y′′ + 9y = t sin t, replace t sin t by teti, solve the equation, and then take imaginary parts.
36. To find a general solution for y′′−2y′+3y = t cos 2t, replace t cos 2t by te2ti, solve the equation, and then take real parts.
Solve the problem in Exercises 37–38.
37. y′′′ −3y′′ + 3y′ −y = t2et, y(0) = 1, y′(0) = 0, y′′(0) = −2 38. y′′′ −3y′′ + 3y′ −y = t2et One end of a spring with constant k newtons per metre is attached to a mass of M kilograms and the other end is attached to a wall (figure below).
Wall k M x x = 0 Dashpot Attached to the mass is a dashpot that provides, or represents, a resistive force on the mass directly proportional to the velocity of the mass. If all other forces are grouped into a function denoted by f(t), the differential equation governing motion of the mass is M d2x dt2 + β dx dt + kx = f(t), where β > 0 is a constant. The position of M when the spring is unstretched corresponds to x = 0. Accompanying the differential equation will be two initial conditions x(0) = x0 and x′(0) = v0 representing the initial position and velocity of M. In Exercises 39–45, solve the initial-value problem with the given information.
39. M = 1/5, β = 0, k = 10, f(t) = 0, x(0) = −0.03, x′(0) = 0 40. M = 2, β = 0, k = 16, f(t) = 0, x(0) = 0.1, x′(0) = 0 41. M = 1/5, β = 3/2, k = 10, f(t) = 0, x(0) = −0.03, x′(0) = 0 42. M = 1/5, β = 3/2, k = 10, f(t) = 4 sin 10t, x(0) = 0, x′(0) = 0 43. M = 1/10, β = 1/20, k = 5, f(t) = 0, x(0) = −1/20, x′(0) = 2 44. M = 1/10, β = 0, k = 4000, f(t) = 3 cos 200t, x(0) = 0, x′(0) = 10 45. M = 1, β = 0, k = 64, f(t) = 2 sin 8t, x(0) = 0, x′(0) = 0 352 SECTION 6.3 If the first or second derivative of a function f(t) yields functions with known Laplace transforms and/or f(t), then equations 6.17 and 6.18 can be used to find the Laplace transform of f(t). We illustrate this in Exercises 46–49.
46. Find the Laplace transform of sin2 at by: (a) using a trigonometric identity; (b) equation 6.17 and the fact that L{sin at} = a/(s2 + a2).
47. Find the Laplace transform of t cos at by calculating its second derivative and using equation 6.18.
48. In Example 6.5, we used multiple integrations by parts to find the Laplace transform of tn when n is a nonnegative integer. Use equation 6.17 and mathematical induction to verify the transform.
49. (a) Use equation 6.17 and Exercise 35 in Section 6.1 to find the Laplace transform of √ t.
(b) Extend the result in part (a) to find the Laplace transform of t(2n+1)/2 when n ≥0 is an integer.
50. (a) Let f be O(eαt) and be continuous for t ≥0 except for a finite discontinuity at t = t0 > 0; and let f ′ be piecewise continuous on every finite interval 0 ≤t ≤T. Show that L{f ′} = sF(s) −f(0) −e−st0[f(t0+) −f(t0−)].
(b) What is the result in part (a) if t0 = 0?
51. (a) Calculate the convolution of tm and tn when m and n are positive integers.
(b) Calculate the convolution of tm and tn when m > 0 and n > 0 are not positive integers.
52. Verify that when f(t) satisfies the conditions of Theorem 6.6, then its Laplace transform satisfies the equation lim s→∞[sF(s)] = f(0).
This is called the initial-value theorem. Like Theorem 6.3, it can serve as a partial check on taking the Laplace transform of a given function f(t).
53. Suppose that f(t) satisfies the conditions of Theorem 6.6 and that f ′(t) is also of exponential order. Verify that if limt→∞f(t) exists, then lim s→0 s F(s) = lim t→∞f(t).
This is called the final-value theorem.
54. Prove that the derivative of the function sin (et2) in Exercise 45 of Section 6.1 has a Laplace transform.
SECTION 6.4 353 6.4 Piecewise-defined and Discontinuous Nonhomogeneities Nonhomogeneities for the linear differential equations in Section 6.3 were all contin-uous. As a result, Laplace transforms did not prove overly advantageous compared to methods of Chapter 4. In this section we show that Laplace transforms are ex-ceptional for handling discontinuous and piecewise-defined nonhomogeneities. But, before doing so, we examine the expectations of solutions to linear differential equa-tions that contain piecewise-continuous nonhomogeneities.
We begin with the initial-value problem associated with a linear first-order differential equation, dy dt + P(t)y = Q(t), y(0) = y0, (6.23) on the interval t > 0, where P(t) is continuous. When Q(t) is also continuous, the solution y(t) of the initial-value problem is unique. It is continuous and has a continuous first derivative. But what can we say about the solution if Q(t) is piecewise-continuous? Suppose, for example, that Q(t) has a single, finite-jump discontinuity at some value t0 > 0.
The initial-value problem has a continuous solution with continuous derivative on the interval 0 < t < t0, call it y1(t). The differential equation also has a general solution, call it y2(t) on the interval t > t0.
We can match these solutions at t0 by demanding that the solution be continuous at t0. This requires lim t→t− 0 y1(t) = lim t→t+ 0 y2(t).
This determines the arbitrary constant in the general solution y2(t). It creates a continuous function that satisfies the differential equation at every value of t except t0. The differential equation requires dy/dt = Q(t)−P(t)y(t). Because y(t) and P(t) are continuous, but Q(t) is discontinuous at t0, it follows that y(t) is differentiable at every value of t except t0. Since we can match solutions at every discontinuity of Q(t), we have the following theorem.
Theorem 6.8 When Q(t) is piecewise-continuous on the interval t > 0, there exists a unique so-lution of initial-value problem 6.23 that is continuous and has a continuous first derivative except at points of discontinuity of Q(t). It therefore satisfies the differ-ential equation except at discontinuities of Q(t).
Here is an example to illustrate this matching.
Example 6.27 Find the solution of the following initial-value problem with a piecewise-continuous nonhomogeneity dy dt + 3y = f(t), where f(t) = t, 0 < t < 1 2, t > 1, subject to y(0) = 1.
Solution First we solve the differential equation on the interval 0 < t < 1, in which case it is dy dt + 3y = t.
354 SECTION 6.4 An integrating factor is e3t, so that multiplication of the differential equation by e3t results in e3t dy dt + 3ye3t = te3t or d dt(ye3t) = te3t.
Antidifferentiation gives ye3t = Z te3t dt = t 3e3t −1 9e3t + C, and therefore y(t) = t 3 −1 9 + Ce−3t.
The initial condition y(0) = 1 requires 1 = −1/9 + C, and therefore C = 10/9. The solution on the interval 0 < t < 1 is y(t) = t 3 −1 9 + 10 9 e−3t.
We now consider the differential equation on the interval t > 1, dy dt + 3y = 2.
Once again e3t is an integrating factor, and this leads to the solution y(t) = 2 3 + De−3t, for t > 1.
What remains is to evaluate constant D. According to Theorem 6.8, there is a solution that is continuous for all t, and in particular at t = 1. This requires lim t→1−y(t) = lim y→1+ y(t) = ⇒ lim t→1− t 3 −1 9 + 10 9 e−3t = lim t→1+ 2 3 + De−3t .
Evaluating the limits gives 1 3 −1 9 + 10 9 e−3 = 2 3 + De−3, from which D = 1 9(10 −4e3).
Thus, the solution of the initial-value problem is the function y(t) = t 3 −1 9 + 10 9 e−3t, 0 ≤t ≤1 2 3 + 1 9(10 −4e3)e−3t, t > 1.• It is graphed in Figure 6.28. As predicted by Theorem 6.8, it is continuous, even at the discontinuity t = 1 of f(t), but it does not have a derivative there.• t y 1 2 1 1/2 Figure 6.28 We now give a similar discussion for initial-value problems associated with second-order, linear differential equations, SECTION 6.4 355 a2(t)d2y dt2 + a1(t)dy dt + a0(t)y = f(t), y(0) = y0, y′(0) = y′ 0.
(6.24) We assume as usual that a2(t), a1(t), and a0(t) are continuous for t ≥0, and a2(t) ̸= 0 for any value of t ≥0.
When f(t) is also continuous, the solution y(t) of the initial-value problem is unique.
It is continuous and has continuous first and second derivatives. Suppose, however, that f(t) has a single, finite-jump discontinuity at some value t0 > 0.
The initial-value problem has a continuous solution with continuous first and second derivatives on the interval 0 < t < t0, call it y1(t). The differential equation also has a general solution, call it y2(t) on the interval t > t0. We can match these solutions at t0 by demanding that lim t→t− 0 y1(t) = lim t→t+ 0 y2(t), lim t→t− 0 y′ 1(t) = lim t→t+ 0 y′ 2(t).
This determines the arbitrary constants in the general solution y2(t). It creates a continuous function with a continuous first derivative that satisfies the differ-ential equation at every value of t except t0.
Because d2y/dt2 = f(t)/a2(t) − [a1(t)/a2(t)]dy/dt −[a0(t)/a2(t)]y, y(t) has a second derivative at every value of t except t0. Since we can match solutions at every discontinuity of Q(t), we have the following theorem.
Theorem 6.9 When f(t) is piecewise-continuous on the interval t > 0, there exists a unique solution of initial-problem 6.24 that is continuous, with a continous first derivative, and has a continuous second derivative except at points of discontinuity of f(t). It therefore satisfies the differential equation except at discontinuities of f(t).
The following example illustrates this matching.
Example 6.28 Solve the following initial-value problem with a piecewise-continuous nonhomogene-ity, y′′ + 2y′ + y = f(t), y(0) = 1, y′(0) = 0, where f(t) = t, 0 < t < 1 0, t > 1.
Solution The auxiliary equation m2 + 2m + 1 = 0 has double root m = −1. On the interval 0 < t < 1, a particular solution of the differential equation is yp = t−2, and hence a general solution on this interval is y1(t) = (C1 + C2t)e−t + t −2. The initial conditions require 1 = y(0) = C1 −2, 0 = y′(0) = C2 −C1 + 1, the solution of which is C1 = 3 and C2 = 2. On the interval 0 < t < 1, then, y1(t) = (3 + 2t)e−t + t −2.
For t > 1, a general solution of the differential equation is y2(t) = (D1 + D2t)e−t.
For the solution to be continuous and have a continuous first derivative at t = 1, we must have lim t→1−y1(t) = lim t→1+ y2(t), lim t→1−y′ 1(t) = lim t→1+ y′ 2(t).
Substitution for y1(t) and y2(t) gives 356 SECTION 6.4 5e−1 −1 = (D1 + D2)e−1, −3e−1 + 1 = −D1e−1.
These can be solved for D1 = 3 −e and D2 = 2, and therefore the solution of the initial-value problem is y(t) = (3 + 2t)e−t + t −2, 0 ≤t ≤1 (3 −e + 2t)e−t, t > 1.• It is graphed in Figure 6.29. As predicted by Theorem 6.9, it is continuous, and appears to have a continuous first deriva-tive at the discontinuity t = 1 of f(t).
The second derivative is discontinuous at t = 1, but we cannot see this graphically.• t y 1 1/2 1 2 Figure 6.29 This procedure of matching solutions at finite discontinuities of nonhomo-geneities can be extended to include initial-value problems associated with nth-order, linear differential equations an(t)dny dtn + an−1(t)dn−1y dtn−1 + · · · + a1(t)dy dt + a0(t)y = f(t), (6.25a) subject to initial conditions y1(0) = y0, y′(0) = y′ 0, · · · , y(n−1)(0) = y(n−1) 0 .
(6.25b) At each discontinuity of f(t), the function and its first n−1 derivatives are matched to produce a solution that has continuous derivatives of orders up to and including n −1, but a discontinuity in the nth derivative results.
What is most important to realize from the above discussion is that matching a solution and its derivatives at discontinuities of a piecewise-continuous nonhomo-geneity is tedious, especially as the number of discontinuities increases. Laplace transforms provide an excellent alternative. If we apply Laplace transforms to the initial-value problem of Example 6.27, we get sY −1 + 3Y = L{f(t)}, where L{f(t)} = L{t[h(t) −h(t −1)] + 2 h(t −1)} = L{t + (2 −t)h(t −1)} = 1 s2 + e−sL{2 −(t + 1)} = 1 s2 + e−s 1 s −1 s2 .
Thus, Y (s) = 1 s + 3 1 + 1 s2 + e−s 1 s −1 s2 = 1 + s2 s2(s + 3) + e−s(s −1) s2(s + 3) .
Partial fractions and inverse transforms give y(t) = L−1 −1/9 s + 1/3 s2 + 10/9 s + 3 + e−s 4/9 s −1/3 s2 −4/9 s + 3 = −1 9 + t 3 + 10 9 e−3t + 4 9 −1 3(t −1) −4 9e−3(t−1) h(t −1).
SECTION 6.4 357 This is the solution obtained in Example 6.27.
We now solve the initial-value problem in Example 6.28 using Laplace trans-forms. When we take transforms of the differential equation, we get [s2Y −s] + 2[sY −1] + Y = L{f(t)}, where L{f(t)} = L{t [h(t) −h(t −1)]} = L{t −L{t h(t −1)} = 1 s2 −e−sL{t + 1} = 1 s2 −e−s 1 s2 + 1 s .
Thus, Y (s) = 1 (s + 1)2 s + 2 + 1 s2 −e−s 1 s2 + 1 s = (s + 1) + 1 (s + 1)2 + 1 s2(s + 1)2 − e−s s2(s + 1).
Partial fractions on the second and third terms leads to Y (s) = 1 s + 1 + 1 (s + 1)2 + −2 s + 1 s2 + 2 s + 1 + 1 (s + 1)2 + e−s 1 s −1 s2 − 1 s + 1 = −2 s + 1 s2 + 3 s + 1 + 2 (s + 1)2 + e−s 1 s −1 s2 − 1 s + 1 .
Consequently, y(t) = −2 + t + 3e−t + 2te−t + [1 −(t −1) −e−(t−1)] h(t −1) = (3 + 2t)e−t + t −2 + (2 −t −e1−t) h(t −1).
This solution is identical to that in Example 6.28.
We now use Laplace transforms to solve other initial-value problems with piecewise-defined and/or discontinuous nonhomogeneities. We invite the reader to make comparisons to solutions obtained by matching solutions and their derivatives at discontinuities.
Example 6.29 Find the amount of salt in the tank of Example 3.10 if pure water is added for the first ten minutes at 5 millilitres per second, and then the brine mixture is added thereafter.
Solution If we change from the letter S in Example 3.10 to represent the number of grams of salt in the tank to x, the initial-value problem for x(t) is dx dt = 0.1h(t −600) −5x 106 , x(0) = 5000.
When we take Laplace transforms on both sides of the differential equation, sX −5000 = e−600s 10s −5X 106 = ⇒ X(s) = e−600s/(10s) + 5000 s + 5/106 .
Inverse transforms now give 358 SECTION 6.4 x(t) = 5000e−5t/106 + 1 10L−1 106/5 s − 106/5 s + 5/106 e−600s = 5000e−5t/106 + 20 000 h 1 −e−5(t−600)/106i h(t −600).• A graph of this function is shown in Figure 6.30a. In agreement with Theorem 6.8, there is a discontinuity in the slope of the graph at t = 600 when the input rate of salt is discontinuous. Since pure water is added for the first 10 minutes, the amount of salt in the tank decreases. This is evidenced by the negative slope of the graph for 0 < t < 600. At t = 600, the amount of salt in the tank is lim t→600−x(t) = lim t→600−(5000e−5t/106) = 5000e−600/20 000 ≈4852 grams.
Because the concentration of salt in the tank at this time is 4852/106 = 4.852 × 10−3 g/mL which is less than the concentration 0.02 g/mL of incoming salt, the amount of salt now begins to increase (the slope of the graph is positive).
For large t, the amount of salt in the tank approaches 20 000 grams, with concentration 20 000/106 = 0.02 g/mL, the concentration of incoming brine. The asymptote is shown in Figure 6.30b.• t S 5000 2500 1000 2000 S t 20 000 40 000 20 000 10 000 Figure 6.30a Figure 6.30b It should be noted that Laplace transforms cannot be applied to mixing prob-lems when the amount of liquid in the tank is not constant.
For instance, the differential equation for the amount of salt in the tank of Example 3.11 is dS dt = 1 5 − 5S 106 + 5t.
We have no formula for the Laplace transform of the second term on the right side of this equation.
Example 6.30 A 2-kilogram mass is suspended from a spring with constant 512 newtons per metre.
It is set into motion by lifting it 10 centimetres above its equilibrium position and then releasing it. A sinusoidal force A sin 8t acts on the mass but only for t > 1.
Find the position of the mass as a function of time if damping is negligible.
Solution The initial-value problem for displacement is 2d2x dt2 + 512x = A sin 8t h(t −1), x(0) = 1 10, x′(0) = 0.
If we take Laplace transforms, 2 s2X −s 10 + 512X = Ae−sL{sin 8(t + 1)} SECTION 6.4 359 = Ae−sL{cos 8 sin 8t + sin 8 cos 8t} = Ae−s 8 cos 8 s2 + 64 + (sin 8)s s2 + 64 .
Hence, X(s) = s 10(s2 + 256) + Ae−s[8 cos 8 + (sin 8)s] 2(s2 + 256)(s2 + 64) .
Partial fractions on the second term gives X(s) = s 10(s2 + 256) + Ae−s 384 8 cos 8 + (sin 8)s s2 + 64 −8 cos 8 + (sin 8)s s2 + 256 , and therefore x(t) = 1 10 cos 16t + A 384L−1 8 cos 8 + (sin 8)s s2 + 64 −8 cos 8 + (sin 8)s s2 + 256 t→t−1 h(t −1) = 1 10 cos 16t + A 384 cos 8 sin 8t + sin 8 cos 8t −1 2 cos 8 sin 16t −sin 8 cos 16t t→t−1 h(t −1) = 1 10 cos 16t + A 384 cos 8 sin 8(t −1) + sin 8 cos 8(t −1) −1 2 cos 8 sin 16(t −1) −sin 8 cos 16(t −1) h(t −1).
The form of the solution changes at t = 1 when the force A sin 8t is applied. For 0 ≤t ≤1, x(t) = 1 10 cos 16t, and for t > 1, x(t) = 1 10 cos 16t + A 384 cos 8 sin 8(t −1) + sin 8 cos 8(t −1) −1 2 cos 8 sin 16(t −1) −sin 8 cos 16(t −1) .
For 0 ≤t ≤1, motion is simple harmonic with ampitude 1/10 and period π/8. For t > 1, motion is periodic with period π/4, but it is not simple harmonic. The function is graphed in Figure 6.31 for A = 100. Even though the force is discontinuous at t = 1, the graph is continuous and so also is the first derivative. In other words, displace-ment and velocity of the mass are con-tinuous at t = 1. This could be shown t x 0.4 0.2 -0.2 -0.4 1 2 3 algebraically, but it is a direct result Figure 6.31 of Theorem 6.9. There is a discontinuity in the second derivative, but this cannot be seen graphically.• 360 SECTION 6.4 Example 6.31 Repeat Example 6.30 when the applied force is A sin 16t.
Solution The initial-value problem for displacement is 2d2x dt2 + 512x = A sin 16t h(t −1), x(0) = 1 10, x′(0) = 0.
If we take Laplace transforms, 2 s2X −s 10 + 512X = Ae−sL{sin 16(t + 1)} = Ae−sL{cos 16 sin 16t + sin 16 cos 16t} = Ae−s 16 cos 16 s2 + 256 + (sin 16)s s2 + 256 .
Hence, X(s) = s 10(s2 + 256) + Ae−s[16 cos 16 + (sin 16)s] 2(s2 + 256)2 .
First we use Table 6.2 to calculate that L−1 16 cos 16 (s2 + 256)2 + (sin 16)s (s2 + 256)2 = 16 cos 16 2(163) (sin 16t −16t cos 16t) + sin 16 32 t sin 16t.
Hence, x(t) = 1 10 cos 16t + A 2 cos 16 512 [sin 16(t −1) −16(t −1) cos 16(t −1)] + sin 16 32 (t −1) sin 16(t −1) h(t −1).
Because of the (t −1)-factors, we have undamped resonance. It is interesting to plot a graph of this function. Figure 6.32 is a plot on the interval 0 ≤t ≤4 (with A = 2). It looks like damped oscillations as opposed to resonance. Figure 6.33, with a longer time interval, shows the resonance.• x y 2 4 0.1 -0.1 x y 0.1 -0.1 4 Figure 6.32 Figure 6.33 The delayed sinusoidal nonhomogeneity presented no problem in Examples 6.30 and 6.31.
When the nonhomogeneity is periodic, but not sinusoidal, additional difficulties arise. Compared to a solution by methods of Chapter 4, however, Laplace transforms are vastly superior. We illustrate in the following two examples.
Example 6.32 Solve the initial-value problem x′′ + 4x = f(t), x(0) = 0, x′(0) = 0, where f(t) is the periodic function SECTION 6.4 361 f(t) = 1, 0 < t < π/2 0, π/2 < t < π f(t + π) = f(t).
Solution When we take Laplace transforms of both sides of the differential equa-tion, and use property 6.16 for the transform of a periodic function, we obtain s2X + 4X = L{f(t)} = 1 1 −e−πs Z π/2 0 e−st dt = 1 1 −e−πs L{h(t) −h(t −π/2)} = 1 (1 + e−πs/2)(1 −e−πs/2) 1 s −e−πs/2 s = 1 s(1 + e−πs/2).
Thus, X(s) = 1 s(s2 + 4)(1 + e−πs/2).
We now partial fraction 1/[s(s2 + 4)], and expand 1/(1 + e−πs/2) in a geometric series X(s) = 1 4 1 s − s s2 + 4 1 −e−πs/2 + e−πs −e−3πs/2 + · · · .
Each term in the series has an easily calculated inverse transform, x(t) = 1 4(1 −cos 2t) −1 4[1 −cos 2(t −π/2)]h(t −π/2) + 1 4[1 −cos 2(t −π)]h(t −π) −· · · .
In sigma notation, x(t) = 1 4 ∞ X n=0 (−1)n[1 −cos 2(t −nπ/2)] h(t −nπ/2).
To evaluate x(t) for any given t, it is necessary to include only those terms in the series for which nπ/2 < t. For example, the solution at t = 3.4 is given by x(3.4) = 1 4[1 −cos 2(3.4)] −1 4[1 −cos 2(3.4 −π/2)] + 1 4[1 −cos 2(3.4 −π)] = −0.402.• Two points are noteworthy in this example. First, consider using the techniques of Chapter 4 to find x(3.4). We would solve the differential equation on the intervals 0 < t < π/2, π/2 < t < π, π < t < 3π/2, match at t = π/2 and t = π, and then find x(3.4) from the solution for π < t < 3π/2. Try it. You will be convinced that Laplace transforms are superior. Secondly, recall that after Example 5.10 in Section 5.3, we questioned whether non-sinusoidal, periodic forces could produce resonance in vibrating mass-spring systems. If we interpret this problem as describing oscilla-tions x(t) of a 1 kilogram mass on the end of a spring with constant 4 newtons per metre, there is resonance, but it is not obvious. If we write out the first few terms of the series for x(t), we obtain x(t) = 1 4 (1 −cos 2t)h(t) −[1 −cos 2(t −π/2)]h(t −π/2) + [1 −cos 2(t −π)]h(t −π) + [1 −cos 2(t −3π/2)]h(t −3π/2) + [1 −cos 2(t −2π)]h(t −2π) + [1 −cos 2(t −5π/2)]h(t −5π/2) + [1 −cos 2(t −3π)]h(t −3π) + · · · 362 SECTION 6.4 = 1 4 1 −cos 2t −(1 + cos 2t)h(t −π/2) + (1 −cos 2t)h(t −π) −(1 + cos 2t)h(t −3π/2) + (1 −cos 2t)h(t −2π) −(1 + cos 2t)h(t −5π/2) + (1 −cos 2t)h(t −3π) · · · = 1 4 1 −cos 2t, 0 ≤t < π/2 −2 cos 2t, π/2 ≤t < π 1 −3 cos 2t, π ≤t < 3π/2 −4 cos 2t, 3π/2 ≤t < 2π 1 −5 cos 2t, 2π ≤t < 5π/2 −6 cos 2t, 5π/2 ≤t < 3π etc.
Amplitudes of these oscillations t x 5 -5 2 4 6 8 increase with time. We have shown a Figure 6.34 graph of the function in Figure 6.34.
Resonance has occurred because the natural frequencey of the mass-spring system is 1/π, and this is the frequency of the applied force f(t).
The following example is fascinating. It may defy your intuition at first, but when we reason it out, it makes perfect sense.
Example 6.33 A mass M hangs motionless from a spring with constant k. At time t = 0, the mass is acted upon by the periodic force in Figure 6.35. The force is a constant value F for p seconds, then it is turned offfor p seconds, back on for p seconds, and so on. During its subsequent motion, the mass experiences t F p p 2 p 3 p 4 no damping. Find displacements of the mass Figure 6.35 Discuss these displacements when p = 2π p M/k.
Solution The initial-value problem for displacements of the mass is M d2x dt2 + kx = f(t), x(0) = 0.
x′(0) = 0, where f(t) is the function in Figure 6.35. Its representation in terms of Heaviside functions is f(t) = F[h(t) −h(t −p)], 0 < t < 2p, f(t + 2p) = f(t).
When we take Laplace transforms, Ms2X + kX = L{f(t)} = 1 1 −e−2ps L{F[1 −h(t −p)]} = F 1 −e−2ps 1 s −e−ps s .
= F(1 −e−ps) s(1 + e−ps)(1 −e−ps) = F s(1 + e−ps).
Hence, X(s) = F s(Ms2 + k)(1 + e−ps).
SECTION 6.4 363 With the partial fraction decomposition of 1/[s(Ms2 + k)], and geometric series for 1/(1 + e−ps), the inverse transform of X(s) is x(t) = FL−1 1/k s − Ms/k Ms2 + k 1 1 + e−ps = F k L−1 (1 s − s s2 + k/M ∞ X n=0 (−1)ne−pns ) = F k ∞ X n=0 (−1)n " 1 −cos r k M (t −pn) # h(t −pn).
When p = 2π p M/k, displacements become x(t) = F k ∞ X n=0 (−1)n 1 −cos 2π p (t −pn) h(t −pn) = F k ∞ X n=0 (−1)n 1 −cos 2πt p h(t −np) = F k 1 −cos 2πt p ∞ X n=0 (−1)nh(t −np).
To graph this function, we write the summation out, x(t) = F k 1 −cos 2πt p [h(t) −h(t −p) + h(t −2p) −· · ·], in which case we see that x(t) = F k 1 −cos 2πt p , 0 ≤t < p 0, p ≤t < 2p F k 1 −cos 2πt p , 2p ≤t < 3p 0, 3p ≤t < 4p etc.
A graph of this function is shown in Figure 6.36. Here is an explanation of why this happens. During the inter-val 0 ≤t ≤p/2, the force F moves the mass upward against the spring, and for p/2 < t ≤p, the spring returns the mass against F to the equilibrium posi-t x p p 2 p 3 2F/k tion. When the mass reaches equilibrium, Figure 6.36 its velocity is zero. Since the force now becomes zero, the mass remains at equilibrium for time p. This sequence of motion then repeats itself in each interval of length 2p thereafter.• EXERCISES 6.4 364 SECTION 6.4 1. Solve the initial-value problem dy dt + 3y = f(t), y(0) = 1, where t, 0 < t < 1 1, t > 1.
by: (a) the techniques of Chapter 4, (b) Laplace transforms.
In Exercises 2–12 solve the initial-value problem.
2. y′′ + 9y = f(t), y(0) = 1, y′(0) = 2, where f(t) = 0, 0 < t < 4 1, t > 4 3. y′′ + 9y = f(t), y(0) = 1, y′(0) = 2, where f(t) = 2, 0 < t < 4 0, t > 4 4. y′′ + 4y′ + 4y = f(t), y(0) = 0, y′(0) = −1, where f(t) = t, 0 < t < 1 1, t > 1 5. y′′ + 4y′ + 4y = f(t), y(0) = −1, y′(0) = 0, where f(t) = 2 −t, 0 < t < 2 t −2, t > 2 6. y′′ + 4y′ + 3y = f(t), y(0) = 1, y′(0) = 2, where f(t) = 0, 0 < t < π sin t, t > π 7. y′′ + 4y′ + 3y = f(t), y(0) = 1, y′(0) = 2, where f(t) = sin t, 0 < t < π 0, t > π 8. y′′ + 2y′ + 5y = f(t), y(0) = 0, y′(0) = 0, where f(t) = 3, 0 < t < 1 −3, t > 1 9. y′′ + 2y′ + 5y = f(t), y(0) = 0, y′(0) = 0, where f(t) = ( 4, 0 < t < 1 −4, 1 < t < 2 0, t > 2 10. y′′ +16y = f(t), y(0) = 2, y′(0) = 0, where f(t) = t, 0 < t < 1 0, 1 < t < 2 f(t+2) = f(t) 11. y′′ + 16y = f(t), y(0) = 2, y′(0) = 0, where f(t) = t, 0 < t < 1 2 −t, 1 < t < 2 f(t + 2) = f(t) 12. y′′ + y′ = f(t), y(0) = 0, y′(0) = 0, where f(t) = 1, 0 < t < 1 −1, 1 < t < 2 f(t + 2) = f(t) 13. Use Laplace transforms to solve the mixing problem of Example 3.10 in Section 3.4.
14. Can you use Laplace transforms to solve the mixing problem in Example 3.11 of Section 3.4?
15. Solve Example 3.10 in Section 3.4 if after 10 minutes the concentration of the brine being added to the tank changes to 1 kilogram per 100 litres.
16. Find the amount of salt in the tank of Example 3.10 in Section 3.4 if the brine mixture is added for 2 minutes, replaced by pure water for 2 minutes, replaced by the brine mixture for 2 minutes, etc.
17. When a patient is admitted to the hospital, the amount of glucose in his bloodstream is g0 grams. He is immediately put on intravenous which transfers glucose to his bloodstream at a SECTION 6.4 365 rate of R grams per minute. At any given time, his body uses the glucose up a rate prportional to how much is present in the bloodstream at that time. If he remains on intravenous for 4 hours and then the intravenous is discontinued, how much glucose is in the bloodstream 6 hours after he is admitted?
18. Find the amount of glucose in the bloodstream of the patient in Exercise 17 as a function of time (in minutes) after he is admitted to the hospital if glucose is administered for an hour, turned offfor an hour, turned on for an hour, turned offfor an hour, etc.
19. The initial mass of a certain species of fish in a lake is estimated as m0 kilograms. Suppose that left alone, the fish would increase their mass at a rate described by the Malthusian model 3.11. Commercial fishing harvests (removes) H kilograms each year, at a uniform rate, but only in the first month of the year.
(a) Find the mass m(t) of fish in the lake as a function of time t.
(b) Determine the value of H in order that the mass of fish in the lake return to m0 after one year.
20. A 100-gram mass is suspended from a spring with constant 40 newtons per metre. The mass is pulled 10 centimetres above its equilibrium position and given velocity 2 metres per second downward. If a force of 100 newtons acts vertically upward for the first 4 seconds, find the position of the mass as a function of time. Ignore all damping.
21. Repeat Exercise 20 if the force is turned on after 4 seconds.
22. Repeat Exercise 20 if a damping force with constant β = 5 also acts on the mass.
23. Repeat Exercise 21 if a damping force with constant β = 5 also acts on the mass.
24. Repeat Exercise 20 if a damping force with constant β = 1 also acts on the mass.
25. Repeat Exercise 21 if a damping force with constant β = 1 also acts on the mass.
26. Repeat Example 6.33 if p = 4π p M/k.
27. Repeat Example 6.33 if p = 3π p M/k.
28. A 1 kilogram mass is motionless at the end of a spring with constant 16 new-tons per metre. When the ramp force in the figure to the right (units of new-tons) acts the mass, the mass moves vertically without damping. Find its subsequent displacements. Plot a graph of the displacement function.
t 20 1 2 366 SECTION 6.5 6.5 The Dirac Delta Function and its Applications There are many common situations that cannot be represented mathematically by functions as we know them. For instance, consider: 1. suddenly adding a sizeable quantity of dissolving substance in a mixing problem in Section 3.4 2. striking a mass on the end of a spring with a hammer in Section 5.3 3. accommodating a voltage spike in an electric circuit in Section 5.4 4. representing a concentrated load on a beam in Section 5.5 We introduce the “Dirac delta” function in this section in order to model these situations mathematically. We begin with what are called unit pulse functions. The unit pulse at time t = t0 of duration a is the function in Figure 6.37. It can be rep-resented in terms of Heaviside unit step functions as t a a t 1 0 t0+ Figure 6.37 p(t0, a, t) = 1 a[h(t −t0) −h(t −t0 −a)].
(6.26) Value t0 identifies the time at which the pulse begins, and a is the duration of the pulse. What is important to notice is that the area under the curve is one; hence the name unit pulse. The units of p(t0, a, t) are inverses of the units of t. If t is time in seconds (s), then units of p(t0, a, t) are s−1. The Laplace transform of this function is L{p(t0, a, t)} = 1 as h e−t0s −e−(t0+a)si .
(6.27) The Unit Impulse Even more important than the unit pulse is the unit impulse. It is the limit of the unit pulse p(t0, a, t) as the time interval t0 < t < t0 + a becomes indefinitely short. As a gets smaller and smaller in Figure 6.37, the area under the curve remains unity; the function is nonzero over shorter and shorter time intervals, but the height of the function gets larger and larger. We have shown the situation for a = 1/10, 1/20, 1/40, and 1/80 in Figure 6.38.
t t0 t0+0.1 t0+0.05 t0+0.025 t0+0.0125 80 70 60 50 40 30 20 10 Figure 6.38 SECTION 6.5 367 The limit of this function as a →0 is not a function in the normal sense of function.
It has value 0 for all t except t = t0 where its value is “infinite”. Such functions are discussed in advanced mathematics; they are known as generalized functions or distributions. This particular one is called the unit impulse or the Dirac delta function. It is denoted by δ(t −t0) = lim a→0 1 a[h(t −t0) −h(t −t0 −a)].
(6.28) The Dirac delta function can be defined as the limit of other sequences of functions besides those in Figure 6.38; all lead to identical properties. Two such sequences are shown in Figures 6.39 and 6.40. In both cases, the area under each curve is unity. Like the functions in Figure 6.38, those in Figure 6.39 are discontinu-ous, but they are symmetric around t0. The functions in Figure 6.40 are continuous and symmetric around t0.
t t0 t0-0.025 t0-0.0125 80 70 60 50 40 30 20 10 t0+0.0125 t0+0.025 t0-0.05625 t0+0.05625 t t0 t0-0.05 t0-0.025 80 70 60 50 40 30 20 10 t0+0.025 t0+0.05 t0-0.0125 t0+0.0125 Figure 6.39 Figure 6.40 The Dirac delta function does not conform to the conditions of Theorem 6.1; it is of exponential order, but it is not piecewise-continuous on every finite interval.
It does, however, have a Laplace transform, and we could write L{δ(t −t0)} = L n lim a→0 p(t0, a, t) o .
We define the Laplace transform of δ(t −t0) by interchanging the operations of taking the limit and taking the Laplace transform; that is, we define L{δ(t −t0)} = lim a→0 L{p(t0, a, t)}.
(6.29) Substituting from equation 6.27 and using L’Hopital’s rule on the limit gives L{δ(t −t0)} = lim a→0 e−t0s −e−(t0+a)s as = lim a→0 se−(t0+a)s s = e−t0s.
(6.30) For readers who find this definition unsatisfactory, an alternative is contained in the Appendix at the end of this section. It briefly discusses how “generalized functions” such as the delta function should be manipulated.
368 SECTION 6.5 Notice that when t0 = 0, the Laplace transform of δ(t) is L{δ(t)} = 1.
In Section 6.1, we proved that the limit as s →∞of the Laplace transform of every piecewise continuous function of exponential order is equal to zero. The limit of the Laplace transform of δ(t) is not zero as s →∞, but it is not a piecewise-continuous function.
We will give arguments to demonstrate that the delta function can be used to model the situations described at the beginning of this section, situations that cannot be modelled by ordinary functions.
Important to realize is the units of δ(t −t0). Since the delta function is the limit of the unit pulse function, and taking limits does not alter units, units of the delta function are those of the pulse function, namely, 1 over the units of t. Hence, if t is time in seconds (s), then units of δ(t−t0) are s−1.
We begin with a mixing problem.
Mixing Problems and the Delta Function A tank contains 1000 litres of water in which 5 kilograms of salt have been dissolved. A brine mixture with concentration 2 kilograms of salt for each 100 litres of water is added to the tank at 5 millilitres per second. At the same time, mixture is being drawn from the bottom of the tank at 5 millilitres per second. Suppose that 3 kilograms of salt are suddenly added to the tank at the 5 minute mark, and we wish to find the amount of salt in the tank as a function of time. As usual, we assume that the mixture in the tank is always sufficiently well-stirred that at any given time, concentration of salt is the same at all points in the tank, even when the 3 kilograms is suddenly added to the tank. This amounts to assuming that the 3 kilograms of salt dissolve instantaneously.
If we let x(t) represent the number of grams of salt in the tank at any given time, then x(0) = 5000. As in Section 3.4, we write symbolically that dx dt = rate at which salt is added − rate at which salt is removed .
Since 5 mL of mixture enter the tank each second, and each millilitre contains 0.02 g of salt, it follows that salt is being added to the tank at a constant rate of 0.1 g/s.
The rate at which salt is removed from the tank is not constant; it changes as the concentration of salt in the tank changes. Since the tank always contains 106 mL of solution, the concentration of salt in the solution at time t, in grams per millilitre, is x(t)/106. With solution being drawn offat 5 mL/s, the rate at which salt leaves the tank in grams per second is 5x/106. The difficulty is how to model the sudden insertion of 3 kilograms of salt into the tank at the 5 minute mark. We will show that the delta function is the key. First, suppose we assume that the 3 kilograms of salt are added to the tank uniformly over a time interval of a seconds beginning at time t = 5. This means adding it a rate of 3000/a g/s over the a seconds. This can be represented by 3000 a [h(t −300) −h(t −300 −a)].
If we now insert these rates into the equation for dx/dt, we obtain SECTION 6.5 369 dx dt = 1 10 −5x 106 + 3000 a [h(t −300) −h(t −300 −a)].
When we take Laplace transforms, sX −5000 = 1 10s −5X 106 + 3000 as [e−300s −e−(300+a)s], from which X(s) = 5000 + 1 10s + 3000 as [e−300s −e−(300+a)s] s + 5 106 = 5000 s + 5/106 + 105 5 1 s − 1 s + 5/106 + 6 × 108 a 1 s − 1 s + 5/106 [e−300s −e−(300+a)s].
Inverse transforms give x(t) = 5000e−5t/106 + 105 5 (1 −e−5t/106) + (6 × 108) h 1 −e−5(t−300)/106i h(t −300) − h 1 −e−5(t−300−a)/106i h(t −300 −a) a .
This is the number of grams of salt in the tank if the 3 kilograms of salt are added at a constant rate over a seconds beginning at t = 300. In order to find the result when the 3 kilograms of salt is added instantaneously, we use L’Hopital’s rule to take the limit of this function as a →0, x(t) = 20 000 −15, 000e−5t/106 + (6 × 108) lim a→0 [(5/106)e−5(t−300−a)/106h(t −300 −a)] = 20 000 −15, 000e−5t/106 + 3000e−5(t−300)/106h(t −300).
We now show that when the delta function is used to model the instantaneous addition of the 3 kilograms of salt, we get the same result with a fraction of the work. Since the units of δ(t −t0) are seconds to the negative 1, 3000δ(t −300) has units of grams per second, consistent with other input and output rates. Suppose we use this expression to represent the addition of the 3000 grams of salt at t = 300 seconds. The differential equation for x(t) becomes dx dt = 1 10 −5x 106 + 3000δ(t −300).
When we take Laplace transforms, sX −5000 = 1 10s −5X 106 + 3000e−300s, from which X(s) = 5000 s + 5/106 + 1/(10s) + 3000e−300s s + 5/106 .
370 SECTION 6.5 Inverse transforms give x(t) = 5000e−5t/106 + 105 5 L−1 1 s − 1 s + 5/106 + 3000e−5(t−300)/106h(t −300) = 20 000 −15 000e−5t/106 + 3000e−5(t−300)/106h(t −300).
We have shown therefore that the delta function representation of the instantaneous addition of 3 kilograms of salt yields the same result as adding the salt over a small interval of time and then taking the limit as this interval approaches zero, and it does so with far less work. It is worthwhile noting that lim t→300−x(t) = 20 000 −15 000e−5(300)/106, lim t→300+ x(t) = 20 000 −15 000e−5(300)/106 + 3000.
In other words, the amount of salt jumps by 3000 grams at t = 300 seconds, as we should expect.
Displacements in Mass-Spring Systems and the Delta Function We now consider the situation in which a pulse represents an external force applied to the mass in a vibrating mass-spring system such as that in Figure 6.41.
When damping and surface friction are negligible, the differential equation describ-ing the position of the mass relative to its equilibrium position is M d2x/dt2 +kx = f(t) where f(t) represents all forces on M other than the spring.
k M x t a a t F 0 t0+ Figure 6.41 Figure 6.42 Suppose the mass is at rest at its equilibrium position for t0 seconds starting at time t = 0, and at time t0, the mass is subjected to the pulse in Figure 6.42. The magnitude of the force is F/a, but the area under the curve is F, and is so for any length of duration a of the pulse. We call this a pulse of size F. The initial-value problem for displacements of the mass due to this pulse is M d2x dt2 + kx = F p(t0, a, t), x(0) = 0, x′(0) = 0.
(6.31) If we take Laplace transforms of both sides of the differential equation, and use formula 6.27, Ms2X + kX = F as h e−t0s −e−(t0+a)si = ⇒ X(s) = F[e−t0s −e−(t0+a)s] as(Ms2 + k) .
Partial fractions give SECTION 6.5 371 X(s) = F ka 1 s − s s2 + k/M (e−t0s −e−(t0+a)s), from which x(t) = F ka " 1 −cos r k M (t −t0) # h(t −t0) −F ka " 1 −cos r k M (t −t0 −a) # h(t −t0 −a).
(6.32) A typical graph of this function is shown in Figure 6.43. The mass is at equilibrium for 0 ≤t < t0. At time t0, the force causes the mass to move. While the force acts (t0 < t < t0 + a), the mass exper-iences simple harmonic motion with amplitude F/(ka), t t a + 0 t0 F ka /( ) 2 Figure 6.43 x(t) = F ka " 1 −cos r k M (t −t0) # .
(6.33a) After the force is removed at t = t0 + a, the displacement of the mass is x(t) = F ka " cos r k M (t −t0 −a) −cos r k M (t −t0) # .
(6.33b) This is once again simple harmonic motion, but with a different amplitude. For some values of the parameters, the amplitude may be enhanced and for other values, it may be diminished (see Exercise 15). The derivative of this function is the velocity of the mass. Since the derivative of the Heaviside function is zero, except at t0 where it is undefined, the velocity is v(t) = F ka "r k M sin r k M (t −t0) # h(t −t0) −F ka "r k M sin r k M (t −t0 −a) # h(t −t0 −a) = F a √ kM " sin r k M (t −t0)h(t −t0) −sin r k M (t −t0 −a)h(t −t0 −a) # .
Left and right limits of this function are the same at t0 and t0 + a so that there is no abrupt change in velocity at t0 and t0 + a where the pulse is discontinuous. This is reflected by the smoothness of the graph in Figure 6.43; the slope of the tangent line is continuous at t0 and t0 + a. This is consistent with the fact that the solution of the differential equation, and its first derivative, must be continuous even with a piecewise-continuous nonhomogeneity (Theorem 6.9).
372 SECTION 6.5 Suppose we use L’Hopital’s rule to take the limit of this displacement function as the duration time of the pulse becomes indefinitely small, x(t) = F k lim a→0+ 1 −cos q k M (t −t0) h(t −t0) − 1 −cos q k M (t −t0 −a) h(t −t0 −a) a = F k lim a→0+ q k M sin q k M (t −t0 −a) h(t −t0 −a) 1 = F √ kM sin r k M (t −t0)h(t −t0).
(6.34) This is the displacement of the mass at time t due to a pulse of size F as the duration of the pulse becomes indefinitely short. We regard this displacement as that due to an instantaneously applied force at time t0. A typical graph of this function is shown in Figure 6.44.
The mass is at equilibrium for 0 ≤t < t0. At time t0, the instantaneous force imparts a velocity to the mass. To find the velocity we differentiate the displacement with respect t t F kM 0 , to t, Figure 6.44 v(t) = F M cos r k M (t −t0)h(t −t0).
This function is discontinuous at t0, and its limit as t →t+ 0 is F/M. In other words, the instantaneous force has given the mass an initial velocity F/M metres per second. This is reflected in the corner of the curve in Figure 6.44 at t = t0, the slope of the graph jumps from zero to F/M.
Our hope is that the Dirac delta function will be the mathematical represen-tation of an instantaneously applied force (such as that in the above mass-spring system, perhaps the result of hitting the mass with a hammer). Consider again, then, the mass-spring system in Figure 6.41. We showed that the position of the mass due to a pulse of magnitude F applied to the mass at time t0 over a time interval of length a is given by equation 6.32. We took the limit of this function as a →0 to obtain displacement 6.34 due to an instantaneously applied force.
Suppose we replace the pulse function in equation 6.31 with Fδ(t −t0), so that the initial-value problem for displacements is M d2x dt2 + kx = Fδ(t −t0), x(0) = 0, x′(0) = 0.
(6.35) When we take Laplace transforms, we get M[s2X] + kX = Fe−t0s = ⇒ X(s) = Fe−t0s Ms2 + k = F M e−t0s s2 + k/M .
The inverse transform gives SECTION 6.5 373 x(t) = F M r M k sin r k M (t −t0)h(t −t0) = F √ kM sin r k M (t −t0)h(t −t0).
This is solution 6.34. We have shown then that the problem of finding displacements for a vibrating mass on the end of a spring, subjected to a pulse of magnitude F, and taking the limit as the interval of application of the pulse approaches zero, can be handled much more efficiently with the Dirac delta function. This is a partial justification for the use of the Dirac delta function to represent instantaneously applied forces to masses in mass-spring systems.
We call Fδ(t −t0) an impulse force of size F at t0. Although an impulse force of size F can be thought of as the limit of a pulse of size F as the duration of the pulse approaches zero, there is a better way to view it. As a force, Fδ(t −t0) has units of newtons or kilogram-metres per second squared. Since the units of the delta function are seconds to the power negative one, it follows that F itself is not a force; it does not have units of newtons. It has units of kilogram-metres per second;, units of momentum. In other words, when an instantaneously applied force in the form Fδ(t −t0) acts on a mass, it imparts F units of momentum to the mass. This is consistent with what we saw earlier. Application of an impulse force Fδ(t −t0) to a mass M results in a velocity change of F/M. This means a change of F in momentum. Thus, instead of sayig that a vibrating mass is struck with an impulse force of size F, it is more informative to say that it is struck with a force that gives the mass F units of momentum.
Here is another example, one that includes damping.
Example 6.34 A 100-gram mass is suspended from a spring with constant 50 newtons per metre.
It is set into motion by raising it 10 centimetres above its equilibrium position and giving it a velocity of 1 metre per second downward. During the subsequent motion a damping force acts on the mass and the magnitude of this force is one-fifth the velocity of the mass. If an impulse force that imparts two units of momentum to the mass is applied vertically upward to the mass at t = 3 seconds, find the position of the mass for all time.
Solution The initial-value problem for the position of the mass is 1 10 d2x dt2 + 1 5 dx dt + 50x = 2δ(t −3), x(0) = 1 10, x′(0) = −1.
If we multiply the differential equation by 10, and take Laplace transforms, s2X −s 10 + 1 + 2 sX −1 10 + 500X = 20e−3s.
Thus, X(s) = s/10 −4/5 s2 + 2s + 500 + 20e−3s s2 + 2s + 500 = 1 10 (s + 1) −9 (s + 1)2 + 499 + 20e−3s (s + 1)2 + 499.
The inverse transform is x(t) = 1 10e−tL−1 s −9 s2 + 499 + L−1 20e−3s (s + 1)2 + 499 .
374 SECTION 6.5 Since L−1 20 (s + 1)2 + 499 = 20e−tL−1 1 s2 + 499 = 20 √ 499e−t sin √ 499t, it follows that x(t) = 1 10e−t cos √ 499t − 9 √ 499 sin √ 499t + 20 √ 499e−(t−3) sin √ 499(t −3) h(t −3).
It is straightforward to show that this solution satisfies the initial conditions x(0) = 1/10 and x′(0) = −1. A graph of the function is shown in Figure 6.45.
Due to excessive damping, oscillations essentially disappear after 3 seconds, but the impulse force restores them at t = 3 seconds. Notice the abrupt change in slope (velocity) at t = 3 due to the t x 0.8 0.4 -0.4 -0.8 1 2 3 6 impulse force. Damping again brings the Figure 6.45 mass essentially to rest after a few seconds.• We have already seen that when the nonhomogeneity in an nth-order, constant coefficient, linear differential equation is continuous, the solution and its first n derivatives are all continuous. When the nonhomogeneity is piecewise-continuous, the solution and its first n −1 derivatives are continuous, but the nth derivative has discontinuities at the discontinuities of the nonhomogeneity. Based upon the above examples, we can expect that when a nonhomogeneity is a Dirac delta function, both the nth and the (n −1)th derivatives of the solution will have discontinuities.
Transfer Functions and Impulse Response Functions Consider the initial-value problem consisting of the nth-order constant coefficient differential equation an dny dtn + an−1 dn−1y dtn−1 + · · · + a1 dy dt + a0y = f(t), (6.36a) with initial conditions y(0) = y0, y′(0) = y′ 0, . . .
, y(n−1)(0) = y(n−1) 0 .
(6.36b) If we take Laplace transforms, we get an[snY −sn−1y0 −sn−2y′ 0 −· · · −y(n−1) 0 ] + an−1[sn−1Y −sn−2y0 −· · · −yn−2 0 ] + · · · + a0Y = F(s).
When we solve this for Y (s), we can write the result in the form Y (s) = Q(s) P(s) + F(s) P(s), (6.37) where P(s) = ansn + an−1sn−1 + · · · + a0, and Q(s) is a polynomial of order less than or equal to n −1 determined by the coefficients ai and the initial conditions.
SECTION 6.5 375 The thing to notice about this transform is that if f(t) ≡0, then so also is F(s), and Y (s) = Q(s)/P(s). The inverse transform of this Y (s) can be thought of in two ways. First, if we regard the n initial values as unspecified, then the inverse is a general solution yh(t) of the homogeneous initial-value problem associated with 6.36. The initial values are the arbitrary constants in the solution. Alternatively, if the initial values are regarded as specified constants, then the inverse of Y (s) is the solution of the homogeneous equation subject to these conditions. Suppose instead that all initial conditions are equal to zero. Then Q(s) ≡0, and Y (s) = F(s)/P(s).
The inverse transform of this is a particular solution yp(t) of differential equation 6.36 that satisfies yp(0) = y′ p(0) = · · · y(n−1) p = 0. Expression 6.37 has separated effects of the initial conditions and the nonhomogeneity into separate terms. Let us illustrate these ideas with an example before proceeding with the discussion.
Example 6.35 A 1-kilogram mass is suspended from a spring with constant 65 newtons per metre.
It is put into motion by lifting it 10 centimetres above its equilibrium position and giving it velocity 2 metres per second downward. During its motion, it is subject to a damping force equal to twice the velocity of the mass, and a vertical force 3 sin 4t.
Find expression 6.37 for the problem, and inverse transforms of each term.
Solution The initial-value problem for the motion of the mass is d2x dt2 + 2dx dt + 65x = 3 sin 4t, x(0) = 1/10, x′(0) = −2.
If we take Lapace transforms, s2X −s 10 + 2 + 2 sX −1 10 + 65X = 12 s2 + 16.
When we solve for X(s), X(s) = s 10 −9 5 s2 + 2s + 65 + 12 (s2 + 16)(s2 + 2s + 65).
The first term contains the effect of the initial conditions, and the second term contains the transform of the nonhomogeneity. Inverse transforms of these terms are xh(t) = L−1 s/10 −9/5 s2 + 2s + 65 = 1 10L−1 (s + 1) −19 (s + 1)2 + 64 = 1 10e−tL−1 s −19 s2 + 64 = 1 10e−t cos 8t −19 8 sin 8t , xp(t) = L−1 12 (s2 + 16)(s2 + 2s + 65) = 12 2465L−1 −2s + 49 s2 + 16 + 2s −45 s2 + 2s + 65 = 12 2465 −2 cos 4t + 49 4 sin 4t + L−1 2(s + 1) −47 (s + 1)2 + 64 = 12 2465 −2 cos 4t + 49 4 sin 4t + e−t 2 cos 8t −47 8 sin 8t .
376 SECTION 6.5 It is straightforward to check that xh(t) satisfies the homogeneous differential equa-tion x′′ + 2x′ + 65x = 0 and the initial conditions xh(0) = 1/10 and x′ h(0) = −2.
Function xp(t) satisfies the nonhomogeneous equation x′′ +2x′ +65x = 3 sin 4t, and the initial conditions xp(0) = x′ p(0) = 0.• We now return to discussion of expression 6.37. If we write differential equation 6.36a in operator notation φ(D)y = f(t), then the function P(s) in expression 6.37 is φ(s). The function H(s) = 1/P(s) is called the transfer function for system 6.36.
Its inverse transform ¯ h(t) is called the unit impulse response function, because it describes the solution of the initial-value problem when all initial conditions are zero and the nonhomogeneity is the delta function δ(t). This is easily seen by noting that when all initial conditions are zero and f(t) = δ(t), then in expression 6.37, Q(s) = 0, F(s) = 1, and therefore Y (s) = 1 P(s) = H(s).
Consequently, y(t) = ¯ h(t). We can write the solution of initial-value problem 6.36 in terms of the unit impulse response function. We have already noted that the inverse transform of the first term Q(s)/P(s) in expression 6.37 is yh(t), the general solution of the associated homogeneous problem. The inverse of the second term F(s)/P(s) can be written as the convolution of f(t) and ¯ h(t), L−1{F(s)/P(s)} = L−1{F(s)H(s)} = Z t 0 f(u)¯ h(t −u) du.
Thus, the solution of initial-value problem 6.36 is y(t) = yh(t) + Z t 0 f(u)¯ h(t −u) du.
(6.38) The first term contains contributions to the solution of the initial-value probem due to the initial conditions, and the second term is the contribution due the nonhomo-geneity in the differential equation.
Example 6.36 Use formula 6.38 to find the solution to the initial-value problem in Example 6.35.
Solution Since the auxiliary equation m2+2m+65 = 0 has solutions m = −1±8i, a general solution of the associated homogeneous equation is xh(t) = e−t(C1 cos 8t + C2 sin 8t).
The initial conditions require 1/10 = x(0) = C1, and −2 = x′(0) = −C1 + 8C2, and therefore the solution of the associated homogeneous differential equation that satisfies the initial conditions is e−t 1 10 cos 8t −19 80 sin 8t = 1 80e−t(8 cos 8t −19 sin 8t).
The unit impulse response function is L−1 1 s2 + 2s + 65 = L−1 1 (s + 1)2 + 64 = 1 8e−t sin 8t.
If we denote the second term in formula 6.38 by xp(t), then SECTION 6.5 377 xp(t) = Z t 0 3 sin 4u 1 8e−(t−u) sin 8(t −u) du.
Lengthy integration leads to xp(t) = 147 2465 sin 4t − 24 2465 cos 4t + 24 2465e−t cos 8t −141 4930e−t sin 8t.
Thus, the solution of the initial-value problem is x(t) = 1 80e−t(8 cos 8t −19 sin 8t) + 147 2465 sin 4t − 24 2465 cos 4t + 24 2465e−t cos 8t −141 4930e−t sin 8t.• EXERCISES 6.5 1. A tank contains 1000 litres of water in which 5 kilograms of salt have been dissolved. Starting at time t = 0, pure water is added to the tank at 5 millilitres per second. At the same time, mixture starts being drawn from the bottom of the tank at 5 millilitres per second. After 1 minute, 500 grams of salt are suddenly added to the tank, and an additional 500 grams every minute thereafter. Assume that the mixture in the tank is always sufficiently well-stirred that at any given time, concentration of salt is the same at all points in the tank, even when the 500 grams are added to the tank. Find the amount of salt in the tank as a function of time.
2. Repeat Exercise 1 if a brime mixture containing 2 kilograms of salt per 100 litres of water is added to the tank instead of pure water.
3. A 2-kilogram mass is suspended from a spring with constant 512 newtons per metre. If it is set into motion at time t = 0 by a unit impulse force, find its subsequent displacement. Assume negligble damping.
4. Repeat Exercise 3 if damping with constant β = 80 is taken into account.
5. Repeat Exercise 4 if β = 8.
6. A 2-kilogram mass is suspended from a spring with constant 512 newtons per metre. It is set into motion by moving it to position x0 and then releasing it. If a unit impulse force is applied at t0 > 0, find the position of the mass for all time.
7. Repeat Exercise 6 if motion is initiated by giving the mass velocity v0 at time t = 0 and position x = 0.
8. Repeat Exercise 6 if motion is initiated by giving the mass velocity v0 from position x0 at time t = 0.
9. A 1-kilogram mass is suspended from a spring with constant 100 newtons per metre.
It is subjected to a unit impulse force at t = 0 and again at t = 1. Find the position of the mass as a function of time.
10. A mass of M kilograms hangs at equilibrium on the end of a spring with constant k newtons per metre. At time t = 0, the mass is subjected to a unit impulse force. Show that if the initial-value problem M d2x dt2 + kx = δ(t), x(0) = 0, x′(0) = 0, 378 SECTION 6.5 is solved, the function does not satisfy the initial velocity condition. Can you explain why?
11. Repeat Exercise 9 if unit impulse forces are applied one each second beginning at time t = 0.
Express the solution in sigma notation.
12. Repeat Exercise 11 if unit impulse forces are π/5 seconds apart, the first at time t = 0. Is there resonance?
13. A mass M, suspended from a spring with constant k, is set into undamped motion by giving it displacement x0 from its equilibrium position and velocity v0. At time t0, it is struck with an impulse force F.
Find an expression for F if the impulse force brings the mass to an instantaneous stop.
14. An unstretched spring (with constant k), in the horizontal position, is attached on the left to a wall and on the right to a mass M. At time t = 0, the mass is struck with an impulse force F to the right which causes the mass to move. The coefficient of kinetic friction between the mass and the table along which it slides is µ. Assume that damping can be ignored.
(a) What is the initial velocity of the mass as a result of the hit?
(b) When does the mass come to a stop for the first time?
(c) If the coefficient of static friction between the mass and the table is µs, show that the mass moves to the left if F > r g2µsM 3(µs + 2µ) k .
15. Show that the amplitude of displacement function 6.33b is 2F ka sin r k M a 2 .
Convince yourself that for certain values of a, this could be greater than F/(ka) and for other values of a, it could be less than F/(ka).
16. (a) Suppose the equation of the speed bump in Exercise 19 of Section 5.2 is f(x) = 3x(1−x)/5.
Find displacement of the front end of the car when M = 200 and k = 1000. Assume β = 0 so that shock absorbers are not working, and that the car has slowed down to 20 kilometres per hour.
(b) Show that motion of the car after the speed bump is simple harmonic and find its amplitude.
In Exercises 17–18 find the transfer and unit impulse response functions for the initial-value problem. Express the solution in form 6.38.
17. d2y dt2 −3dy dt −4y = f(t), y(0) = 1, y′(0) = −2 18. d2y dt2 + 2dy dt + 3y = f(t), y(0) = A, y′(0) = B 19. A system modelled by differential equation 6.36a is said to be stable if its unit impulse response function ¯ h(t) is bounded as t →∞. Show that this is the case if, and only if, real roots of P(s) = 0 are less than or equal to zero, as are real parts of complex roots.
20. A system modelled by differential equation 6.36a is said to be asymptotically stable if its unit impulse response function ¯ h(t) approaches zero as t →∞. Show that this is the case if, and only if, real roots of P(s) = 0 are negative, as are real parts of complex roots.
SECTION 6.5 379 Appendix In the theory of generalized functions, or distributions, functions such as the delta function δ(t −t0) are never stand-alone functions; that is, they are not assigned values at specific values of t.
It is easy to see why the delta function δ(t −t0) should not be considered in a pointwise sense. If it has value zero for t ̸= t0, and value “infinity” at t0, then what is the difference between δ(t −t0) and 4δ(t −t0)?
Generalized functions are considered to be mappings, or transformations, that map ordinary functions to numbers. For instance, suppose that g(t) is a fixed function that is continuous on the interval a ≤t ≤b. If f(t) is any other function that is integrable on a ≤t ≤b, then the integral Z b a f(t)g(t) dt is a real number. We can say that through this integration, g(t) defines a mapping from the set of integrable functions on a ≤t ≤b to the reals. If we denote this mapping by G, then we can write that f(t) g(t) − − − − − →G{f} = Z b a g(t)f(t) dt.
(6.39) For instance, if g(t) = t2, and the interval is 0 ≤t ≤2, then t t2 − − − − − →G{t} = Z 2 0 t(t2) dt = t4 4 2 0 = 4, et t2 − − − − − →G{et} = Z 2 0 t2etdt = t2et −2tet + 2et 2 0 = 2(e2 −1).
It is this view of an ordinary function as a mapping, or operator, that is adopted to define δ(t −t0). The “generalized” function δ(t −t0) is the operator that maps a function f(t), continuous at t = t0, onto its value at t = t0, f(t) δ(t−t0) − − − − − →f(t0).
(6.40) For example, t2 + 2t −3 δ(t−2) − − − − − →5, and (t + 1)2 cos t δ(t) − − − − →1.
In order that the delta function have an integral representation, we write f(t) δ(t−t0) − − − − − →f(t0) = Z ∞ −∞ f(t)δ(t −t0) dt.
(6.41) Because δ(t −t0) cannot be regarded pointwise, the multiplication in this integral, and the integral itself, are symbolic. When we encounter an integral such as that in equation 6.41, we interpret it as the action of the function δ(t −t0) operating on f(t) and immediately write f(t0). For example, Z ∞ −∞ t2 + 2 t2 + 1 δ(t) dt = 2, and Z ∞ −∞ δ(t + 2) dt = 1 380 SECTION 6.5 (since the left side of the latter integral is interpreted as the delta function δ(t + 2) operating on the function f(t) ≡1).
Because δ(t −t0) picks out the value of a function at t = t0, we also write Z b a f(t)δ(t −t0) dt = f(t0) (6.42a) whenever a < t0 < b; that is, the limits on the integral need not be ±∞. Further-more, if t = t0 is not between a and b, we set Z b a f(t)δ(t −t0) dt = 0.
(6.42b) For instance, Z 6 −2 √ t + 5 δ(t) dt = √ 5, and Z 3 2 (t2 + 2t −4)δ(t + 1) dt = 0.
With this interpretation for the delta function, its Laplace transform is L{δ(t −t0)} = Z ∞ 0 e−stδ(t −t0) dt = e−t0s.
What is important to remember from this discussion is that δ(t −t0) should never be regarded in a pointwise sense. We see it in initial-value problems such as d2y dt2 + 2dy dt = 3y = δ(t −3), y(0) = 1, y′(0) = −2, but we never consider the differential equation at a specific value of t. The first step is always to take the Laplace transform of both sides of the equation, in which case the operational property of the delta function is invoked, [s2Y −s + 2] + 2[sY −1] + 3Y = L{δ(t −3)} = e−3s. |
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Transformations of Points and Lines 8.G.A.1 Question & Answer Key Resources Grade 8 Mathematics - SkillBuilder + ISTEP+ Rehearsal
Grade 8 Mathematics - SkillBuilder + ISTEP+ Rehearsal Transformations of Points and Lines
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Your mom is building a garden around a fountain. She has a space marked out but decides she wants to enlarge it without altering the center of it.
What type of transformation is this?
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opulencenoun#wealth#valuables
fortunenoun#wealth#valuables
luxurynoun#wealth#comfort
plentynoun#abundance#wealth
abundancenoun#wealth#valuables
richnessnoun#wealth#valuables
moneynoun#property
successnoun#benefit#interest
meansnoun#property
well-beingnoun#extra#benefit
wealthinessnoun#riches#wealth
assetsnoun#valuables#stuff
luxuriancenoun#valuables
comfortnoun#content#peace
exuberancenoun#luxury#abundance
plenteousnessnoun#benefit#success
sufficiencynoun#wealth#mass
possessionsnoun#property#wealth
propertynoun#wealth#valuables
profusionnoun#capital#mass
copiousnessnoun#capital#mass
prosperousnessnoun#wealth#valuables
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8948 | https://www.quora.com/Each-positive-integers-are-colored-by-one-of-the-three-colors-red-blue-yellow-How-do-I-show-that-there-exists-three-distinct-positive-integers-a-b-c-such-that-a-b-c-a-b-b-c-c-a-and-a-b-c-all-have-the-same-color | Something went wrong. Wait a moment and try again.
Pigeonhole Principle
Proofs (mathematics)
Algorithms, Combinatorics
Positive Integers
Number Theory
Combinatorial Math
Mathematical Proof
5
Each positive integers are colored by one of the three colors red, blue, yellow.How do I show that there exists three distinct positive integers
such that
and
all have the same color?
Alon Amit
PhD in Mathematics; Mathcircler.
·
Upvoted by
Melchior Grutzmann
, Ph.D. Mathematics, Pennsylvania State University (2009) and
Alex Eustis
, Ph.D. Mathematics, University of California, San Diego (2013) · Author has 8.8K answers and 173.7M answer views
·
2y
Damn this was hard. And I couldn’t actually do all of it by myself; I needed a push.
In keeping with the SMPT approach, the outline below describes my thought process including the false turns and misses, although I had to trim missteps turns in order for this not to be a small textbook.
Before tackling the whole -point structure in the question, I decided to start with just , , . I knew that I solved this simpler version before (I believe it was in a graduate problem-solving class some 30 years ago), and I remembered it wasn’t hard since I recognized then, as I did now, that this is fodde
Damn this was hard. And I couldn’t actually do all of it by myself; I needed a push.
In keeping with the SMPT approach, the outline below describes my thought process including the false turns and misses, although I had to trim missteps turns in order for this not to be a small textbook.
Before tackling the whole -point structure in the question, I decided to start with just , , . I knew that I solved this simpler version before (I believe it was in a graduate problem-solving class some 30 years ago), and I remembered it wasn’t hard since I recognized then, as I did now, that this is fodder for Ramsey’s theorem.
Ramsey’s theorem is one of the main tools in proving results of this type: someone arbitrarily colors some infinite or sufficiently large structure, and you conclude that some small, finite, single-colored substructure must exist. Ramsey’s theorem talks about coloring the edges of a (large, or infinite) complete graph, and knowing that the result must contain a small complete subgraph with all edges the same color (a monochromatic ).
All we need to do, then, is to decide how to color the edges of a large in such a way that a complete monochromatic subgraph would translate into a monochromatic . We are given a coloring . A simple approach is to take the numbers for some large , and color each edge with , assuming . We are coloring each edge according to the given color of the difference of its vertices.
Why are we coloring by the color of the difference? Because Ramsey’s theorem guarantees a monochromatic triangle , which is just three natural numbers such that , and are all the same. but , so this is just what we need: call , and now , and are all found to carry the same color . Coloring the edges by the difference forces Ramsey to align with the operation of addition of natural numbers.
There’s still a problem here: we need to ensure that , for otherwise we just have a monochromatic pair . But Ramsey’s theorem may well give us a monochromatic triangle with (simply, a -term arithmetic progression). How do we avoid that?
The way I solved this way back when was to use colors instead of : for any pair I considered the -valuation which is just the highest power of dividing . Taking the parity of this number, you keep the coloring but make the color “bright” or “dark” depending on the parity of . This guarantees that if then cannot have the same color. Now Ramsey’s theorem can’t give you a -term arithmetic progression.
This isn’t the only way to solve the issue, but I won’t belabor this point here. As I was thinking about the present question, I’ve decided to just ignore such problems, believing that I can handle them in some way if necessary in the end. This was a good decision: avoiding those degenerate cases is indeed mostly a nuisance.
Ok, so I had a solution to the problem of finding a monochromatic set of the form . Why can’y I use the same approach with ?
I thought I actually can. Instead of seeking a monochromatic triangle , I chose to ask Ramsey to give me a monochromatic with vertices . This means that all of the pairwise differences , , and so on have the same color.
And then what? Well, just call , , , the consecutive differences. We now know that have the same color and also the same color as , , , and…
Hmmmm. What about ?
The trouble is that the differences and aren’t abutting, so isn’t one of the differences colored by our edges.
I had an almost solution. I could make carry the same color, but I couldn’t control the color of . I felt I’m very close, but it turns out that I wasn’t quite close at all.
The first thing I tried was to ask Ramsey for more: give me a monochromatic , or a for all I care. Now I have a whole host of differences carrying the same color, and I just need to find among them some such that , , and are themselves such differences.
I couldn’t do it. I don’t think it’s possible. You can only control consecutive sums in this way. If you manage to control and , you just can’t control . I don’t see how to avoid this issue.
I tried other coloring schemes, instead of . I tried versions with directed graphs, and with super-colors that mixed , and together. Nothing worked. I’ll save you the trouble: this took a long time. But I couldn’t do it.
At some point I started considering bringing in another piece of heavy machinery: The theorem of Van der Waerden on monochromatic arithmetic progressions. The advantage of VdW over Ramsey is that it has the arithmetic built-in: it directly talks about arithmetic progressions, which rely on the additive structure of the natural numbers. The disadvantage is that it’s less flexible: it only does arithmetic progressions, and it doesn’t naturally let you choose a coloring of any edges.
(Still, VdW is amazingly useful in a broad array of surprising contexts. I’ve used it in another Quora answer a couple of years ago.)
I know that I could start by using VdW to find some long monochromatic arithmetic progression, and then inside of that progression apply my partial results with Ramsey. But I failed to see how this helps: I may have blue numbers etc., but I’m never using those numbers themselves - I’m using their difference , and so on. How does it help that themselves are blue?
It eventually dawned on me that I can separate the partial Ramsey structure and the arithmetic progression, but also make the partial Ramsey structure consist of multiples of some number . This latter part is pretty clear: if I can find, say, among the natural numbers , I can also find such a thing among . As far as addition is concerned, those structures are the same.
So here was my plan: We start with the given coloring . I’ll find an arithmetic progression, starting at with some skip , high up in the natural numbers, such that all elements of this progression carry the same color – say, blue. I’ll then find the monochromatic structure , not inside this progression but among the multiples of ( and so on). Since are all multiples of , the numbers and are all inside the high arithmetic progression, so they are all blue. And… I’m…
No. Yes, I managed to force and to all be blue, but may or may not be blue. They are of the same color, but that color may not be blue.
And so I was stuck again with a partial solution. Only this time, instead of missing , I had something different: are one color, are possibly a different color but all the same color. It seems like less of an achievement than before, but I felt good about it because finally I was able to break through the annoying issue.
But what now?
At this point I was stuck and I don’t know if I would ever be able to solve it myself. Maybe, with a lot more concentrated effort. So I looked things up, and I found out that what I’m chasing is a named theorem (Folkman’s theorem, later beautifully generalized by Rado). The theorem is more general: given any finite coloring of you can find a set such that all of the sums of all of the subsets of this set have the same color. What we’re solving here is the case .
More importantly, I quickly scanned a proof of Folkman’s theorem and I found out that my Ramsey+VdW approach is a good first step, only I need to make it more general: what I proved without noticing it is that I can find a set such that the colors of sums aren’t necessarily the same, but they’re the same if they have the same largest element. In my case I managed to equalize the color of and , and also of and , all of which share the topmost element .
That’s all the push I needed. I understood now that I can prove this in general for any you want: using the same VdW approach with a high arithmetic progression, I can find a finite set of any desired size for which the subset sums that share the same top element are the same.
This sounds like a mouthful, but I saw how this is exactly what my approach was yielding, and I also saw quite quickly that with a sufficiently large I can solve the original problem. Instead of using a big Ramsey structure, I ended up using this big -structure where the colors are not all the same but mostly the same in a very controlled way.
And now, the final step was clear: use this lemma to get a sequence in which the color of each sum with depends only on . Since we have terms in our sequence, at least of them must be of the same color, and for those three all the subset sums must be the same. And with that, we’re done.
Looking back, the thing I failed to notice is this . It’s not enough to build a partial solution with , I needed to build a similar solution with , and then use a simple pigeonhole to get the actual solution. I don’t know how long this would have taken me had I not skimmed the literature.
If you’re unable to read a complete proof from the convoluted tale I shared, I don’t blame you. Here’s a very nice write-up of a proof which is essentially the same as what I ended up constructing - but with the help of this very paper.
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Alon Amit
PhD in Mathematics; Mathcircler.
·
Upvoted by
Michael Lamar
, PhD in Applied Mathematics and
Amos Shapir
, MSc Computer Science & Mathematics, Hebrew University of Jerusalem (1979) · Author has 8.8K answers and 173.7M answer views
·
Updated 1y
Related
Every positive integer is painted with some color. It turned out that for every positive integers
the integers
are painted the same color.How do I show that all integers
are painted the same color?
That’s a very nice question.
Thanks to , has the same color as . That’s a good start, but how do we show that they have the same color as ? Trying other small values of doesn’t seem to lead us back to . So…?
Oh but you see, has the same color as (why?), which has the same color as (why?), which has the same color as (see why?), and then , , , , and finally .
So yeah, must carry the same color as and , but the chaotic path we needed to follow to see this is a bit disheartening. What to do?
When you look at that path, you can’t miss the fact that the number
That’s a very nice question.
Thanks to , has the same color as . That’s a good start, but how do we show that they have the same color as ? Trying other small values of doesn’t seem to lead us back to . So…?
Oh but you see, has the same color as (why?), which has the same color as (why?), which has the same color as (see why?), and then , , , , and finally .
So yeah, must carry the same color as and , but the chaotic path we needed to follow to see this is a bit disheartening. What to do?
When you look at that path, you can’t miss the fact that the numbers zig-zag between small and large. This isn’t at all surprising: you have some small and they let you connect (small) to (much larger), and what happens then is that this number happens to have another representation as , which leads you back to , roughly the same size as .
For example, because and , but also and that gives us . In two hops we’ve connected and , which seems much more useful than connecting to . (When I say “connecting” I just mean that these two numbers must be colored in the same color, as per the specification of the question. The question here is really about proving that the implied graph is connected.)
The key here wasn’t the number , it was the number : it’s the sum of two squares in two different ways, just like Ramanujan’s but with squares instead of cubes. So my idea was to manufacture lots of numbers that are sums of squares in two different ways, and from that infer lots of nearby connections like , and from that conclude that all the integers are eventually connected.
That turned out to be easier said than done.
There was one thing the confused me briefly: if all we’re doing is hopping from to whenever , why the ? As far as we’re concerned it doesn’t seem to matter at all… but I quickly realized that if the question had instead of , the conclusion would certainly be false: even numbers and odd numbers would never get connected. The is what let us connect to , and that explains why it’s there.
Right. So how do we find many instance of ? I thought I knew the answer to that, and it turned out that I was not as learned as I thought I was.
My first approach was to rewrite this as , so , and that’s just some number which affords two factorizations in which the factors have the same parity. Any two numbers of the same parity can be written as and . So yes, sure, there are many composite numbers and most of them can be expressed in a few different ways like that, but it was hard to put enough structure on this to obtain a nice family of connections.
(That was a mistake; I just didn’t look at it the right way. But I needed a jolt from a different direction to see this).
The jolt came when I arranged a few square-pairs and noticed that they follow a pattern. The first observation, for some odd reason, was this:
Once you see this, it’s hard to unsee it: this is pretty darn obvious. The square terms on both sides are the same, the constant terms on both sides are the same, and the linear terms on both sides are missing. Of course that’s true. It’s a simple polynomial identity, but I have to admit I never noticed the existence of such simple polynomial identities. Now I know. I like this question for teaching me that.
This observation means, for our question, that all numbers of the form are connected (in two hops) to , as soon as (because otherwise isn’t a positive integer, and the question only allows to be positive integers.)
I then generalized this in an obvious way:
for any positive integers you like.
Now take, for example, to see
which lets us connect to whenever is an integer. Nice! We now have two connecting rules with different multipliers, which seems to make our colors very restricted. I figured that if I collected enough of these I’d be done. That’s right, and I’ll show how, but I have this nagging feeling that there’s a much more elegant way to do this that I’m completely missing. Whatever. I’m never as smart as I think I am.
Now, I said that the composite numbers view could work and I just missed it. In fact, if you rearrange the identity (1) as a difference of squares and then replace with , you’ll see that it is simply , so duh. Of course that’s true, and the only thing I should have done is look at simple factorizations such as and throw a parameter inside, to get a whole family of sum-of-squares identities.
Here’s another example: from we can infer , another identity which seems obvious in hindsight, and it lets us assert that the numbers and are connected. That’s genuinely different from and .
So far we’ve got the following “rules”, each implying that various types of numbers are connected, so they must have the same color:
A) when . Catchphrase “you can hop over any multiple of starting at and above".
B) when . Catchphrase: “If it’s divisible by you can add or subtract , but not below ".
C) when . Catchphrase: “among even numbers, you can hop over any multiple of but not below ".
D) when . Catchphrase: none.
I should say something about the last one: it actually follows directly from the conditions of the question, setting and . It’s much simpler to derive than the other three, and you may wonder why I didn’t emphasize more rules like that. Their weakness is that they connect something linear to something quadratic, which is a lot less helpful in general. But it’s still convenient to have it at our disposal, as we’ll see in a minute.
So now, back to the original problem. Assume that is colored pink, and we need to show that all numbers must be colored pink as well. Our strategy is simple:
Ensure that all small numbers, say up to , must all be pink.
Show that all even numbers must be pink.
Show all odd numbers do, too.
Part 1 just requires a bit of patience. We already know that are pink, and by D. Since we also have , and by D again . Using B several times we can drop from to , and then C to and D to . From we can also take another two B’s to and a D to . Finally to (D) to (Bs) to , and using A to . We have all the numbers from to pink, and you can easily get up to yourself as an exercise.
We can easily get all even numbers now: suppose that is the smallest even non-pink number. It must be , and consider what it can be mod : It can’t be or because if it is then rule B shows it has the same color as which is known to be pink. It can’t be or mod because C shows it has the same color as , nor mod because rule A shows it has the same color as . Finally, mod is solved by moving forward to using A, and then back down using C again.
All even numbers are pink. That’s part 2 of the plan.
What about odd numbers? Easy: rule D. Right?
So all numbers are pink and we’re done.
Again, I’m sure there’s a more elegant way, but this works, and I learned something which is always excellent.
(…)
Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006)
·
Upvoted by
James McElhatton Ph.D. (Glasgow, 1976)
, B.Sc. Mathematics & Chemistry, University of Malta (1967) and
Michael Jørgensen
, PhD in mathematics · Author has 8.5K answers and 21M answer views
·
Jun 11
Related
How do I find all triplets of positive integers
such that
?
We want to find all such that
Without loss of generality, assume that . In order to significantly narrow the search for such solutions, observe that for all :
which gives is a contradiction. Hence .
Case 1: If , then we solve
[math]\displaystyle \Big(1 + \frac{1}{b}\Big) \Big(1 + [/math]
We want to find all such that
Without loss of generality, assume that . In order to significantly narrow the search for such solutions, observe that for all :
which gives is a contradiction. Hence .
Case 1: If , then we solve
This equation has no positive integer solution, because each factor is greater than .
Case 2: If , then we solve
Similar to our work above, if , then the left side is bounded above by , giving us a contradiction. Hence, , and we find the solutions
Case 3: If , then we solve
Similar to our work above, if , then the left side is bounded above by , giving us a contradiction. Hence, , and we find the solutions
Putting this all together, the positive integer solutions where are given by
Permuting the values of the three variables in each order triple (thereby eliminating the ``without loss of generality” condition) yields the remaining positive integer solutions.
Brian Sittinger
PhD in Mathematics, University of California, Santa Barbara (Graduated 2006)
·
Upvoted by
Michael Jørgensen
, PhD in mathematics · Author has 8.5K answers and 21M answer views
·
Jun 9
Related
Assume that
and
are two positive integers. How do I show that there exist positive integers
such that
?
This can be proved by induction on .
The claim is true for , because (taking ).
Next, we assume the claim is true for and show that this implies that the claim is also true for . We need to break this into two cases, according to whether is even or odd.
Case 1: If is even, then
Case 2: If is odd, then
This can be proved by induction on .
The claim is true for , because (taking ).
Next, we assume the claim is true for and show that this implies that the claim is also true for . We need to break this into two cases, according to whether is even or odd.
Case 1: If is even, then
Case 2: If is odd, then
In either case, we have found positive integers such that
Moreover by the inductive hypothesis, there exist positive integers such that
With this, we have proved the claim for , and this completes the proof.
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[math]\sqrt{a}+\sqrt[b]{c}[/math]
be an integer if a,b,c are integers?
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[math]n,a,b,c[/math]
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[math]c/(a + b) + b/(a + c) + a/(b+c) = 4[/math]
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Assistant
Bot
·
1y
To show that there exist distinct positive integers [math]a, b, c[/math] such that [math]a, b, c, a+b, b+c, c+a, a+b+c[/math] all have the same color when each positive integer is colored with one of three colors (red, blue, or yellow), we can apply the pigeonhole principle.
Step-by-Step Explanation:
Consider the Coloring: Each positive integer can be colored in one of three colors: red, blue, or yellow.
Count the Combinations: We need to analyze the integers [math]a, b, c[/math] and their sums:
The integers we are interested in are [math]a, b, c[/math], and the sums [math]a+b, b+c, c+a, a+b+c[/math].
This gives us a total of 7 distinct expressions
To show that there exist distinct positive integers [math]a, b, c[/math] such that [math]a, b, c, a+b, b+c, c+a, a+b+c[/math] all have the same color when each positive integer is colored with one of three colors (red, blue, or yellow), we can apply the pigeonhole principle.
Step-by-Step Explanation:
Consider the Coloring: Each positive integer can be colored in one of three colors: red, blue, or yellow.
Count the Combinations: We need to analyze the integers [math]a, b, c[/math] and their sums:
The integers we are interested in are [math]a, b, c[/math], and the sums [math]a+b, b+c, c+a, a+b+c[/math].
This gives us a total of 7 distinct expressions to consider.
Pigeonhole Principle: Since there are 7 integers (the three integers [math]a, b, c[/math] and the four sums [math]a+b, b+c, c+a, a+b+c[/math]), and only 3 colors available, by the pigeonhole principle, at least one color must be assigned to at least 3 of these integers.
Choosing Distinct Integers:
Let’s assume we label the colors as [math]C_1, C_2, C_3[/math].
We can enumerate the possibilities for the colors of [math]a, b, c[/math]:
If all three integers [math]a, b, c[/math] are the same color, we are done.
If two integers are one color and the third is different, we can analyze the sums:
For example, if [math]a[/math] and [math]b[/math] are colored [math]C_1[/math] and [math]c[/math] is colored [math]C_2[/math], then [math]a+b[/math] is also colored [math]C_1[/math], [math]b+c[/math] is colored [math]C_2[/math], [math]c+a[/math] is colored [math]C_1[/math], and [math]a+b+c[/math] is colored [math]C_1[/math] (since it includes both [math]a[/math] and [math]b[/math]). This means that [math]a+b, c+a, a+b+c[/math] must be colored [math]C_1[/math] or [math]C_2[/math].
Ultimately, through this reasoning, we can show that either all the integers and sums are the same color or we will have at least three of them that are the same color.
Conclusion: Therefore, there must exist distinct positive integers [math]a, b, c[/math] such that all of [math]a, b, c, a+b, b+c, c+a, a+b+c[/math] share the same color.
This completes the proof using the pigeonhole principle and the properties of sums of integers.
Thomas Bell
Ph.D. in Mathematics, University of Oregon (Graduated 2013)
·
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) and
Nathan Hannon
, Ph. D. Mathematics, University of California, Davis (2021) · Author has 3.9K answers and 685.1K answer views
·
2y
Related
Is it possible to color each integer greater than 1 with either red, green, or blue so that all 3 colors are used, and for every two integers x and y colored in two different colors, the integer xy I colored with the third color?
Suppose [math]x[/math] is red, [math]y[/math] is blue and [math]z[/math] is green.
Now suppose [math]w[/math] is red. Then [math]yw[/math] is green, so [math]xyw[/math] is blue and [math]xyzw[/math] is back to red. Likewise, if [math]w[/math] is blue, [math]xyzw[/math] is blue, and if [math]w[/math] is green then [math]xyzw[/math] is green.
So what color is [math]xyz[/math]? There will be a contradiction no matter what you choose.
Bernard Montaron
PhD in Mathematics & Discrete Mathematics, Université Pierre Et Marie Curie Paris VI (Graduated 1980)
·
Upvoted by
Michael Jørgensen
, PhD in mathematics and
Jeremy Collins
, M.A. Mathematics, Trinity College, Cambridge · Author has 3.2K answers and 2.1M answer views
·
2y
Related
Is it possible to color each integer greater than 1 with either red, green, or blue so that all 3 colors are used, and for every two integers x and y colored in two different colors, the integer xy I colored with the third color?
No, it’s not possible. Here is a proof.
From the associativity of the multiplication we have (RxR)xB=Rx(RxB)=RxG=B (1), also (RxR)xG=Rx(RxG)=RxB=G (2).
What color is RxR?
If it’s R then (RxR)xB=RxB=G which contradicts (1), so it’s not R.
If it’s B then (RxR)xG=BxG=R which contradicts (2), so it’s not B.
Therefore, it must be G, but then (RxR)xB=GxB=R which contradicts (1), so it’s not G.
It is impossible to color RxR in a way that multiplication is associative.
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Is there a positive integer solution for a, b, and c if
[math]c/(a + b) + b/(a + c) + a/(b+c) = 4[/math]
?
We start here.
[math]\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4[/math]
In Alon Amit's answer, he mentions that the equation is homogeneous, since if math[/math] is a solution, so is math[/math], because substituting math[/math] into the equation causes all the "[math]k[/math]"s to cancel out. We use that to our advantage here. Without loss of generality, we can assume that [math]c=1[/math], and look for rational solutions to math[/math], since we can then multiply by a common denominator to find integer solutions. Let's start.
[math]\frac{a}{b+1}+\frac{b}{a+1}+\frac{1}{a+b}=4[/math]
Multiply out the denominators:
[math]a(a+1)(a+b)+b(b+1)(a+b)+(a+1)(b+1)=4(a+1)(b+1)([/math]
We start here.
[math]\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=4[/math]
In Alon Amit's answer, he mentions that the equation is homogeneous, since if math[/math] is a solution, so is math[/math], because substituting math[/math] into the equation causes all the "[math]k[/math]"s to cancel out. We use that to our advantage here. Without loss of generality, we can assume that [math]c=1[/math], and look for rational solutions to math[/math], since we can then multiply by a common denominator to find integer solutions. Let's start.
[math]\frac{a}{b+1}+\frac{b}{a+1}+\frac{1}{a+b}=4[/math]
Multiply out the denominators:
[math]a(a+1)(a+b)+b(b+1)(a+b)+(a+1)(b+1)=4(a+1)(b+1)(a+b)[/math]
Distribute, rearrange, and combine like terms, but do not cancel across the equal sign.
[math]a^3+a^2b+ab^2+b^3+a^2+2ab+b^2+ab+a+b+1=4a^2b+4ab^2+4a^2+8ab+4b^2+4a+4b[/math]
See the first four terms of the left side? We can then add [math]2a^2b+2ab^2[/math] to both sides, making the first four terms a perfect cube.
[math]a^3+3a^2b+3ab^2+b^3+a^2+2ab+b^2+ab+a+b+1=6a^2b+6ab^2+4a^2+8ab+4b^2+4a+4b[/math]
Then:
math^3+(a+b)^2+(a+b)+1+ab=6ab(a+b)+4(a+b)^2+4(a+b)[/math]
Look at the first four terms of the left side. Again, we add [math]2(a+b)^2+2(a+b)[/math] to both sides to make these four terms a perfect cube.
math^3+3(a+b)^2+3(a+b)+1+ab=6ab(a+b)+6(a+b)^2+6(a+b)[/math]
Group things together:
math^3+ab=6(a+b)(ab+a+b+1)[/math]
Now, designate [math]x=a+b+1[/math] and [math]z=ab[/math]. Therefore, [math]a,b=\frac{x-1\pm\sqrt{(x-1)^2-4z}}{2}[/math]. For [math]a,b[/math] to be rational, math^2-4z[/math] must be a rational square. Now, rewrite the last equation in [math]a,b[/math] with [math]x,z[/math]:
[math]x^3+z=6(x-1)(x+z)[/math]
In terms of [math]z[/math]:
[math]z=\frac{x^3-6x^2+6x}{6x-7}[/math]
From earlier, we know that for [math]a,b[/math] to be rational, math^2-4z[/math] must be a rational square. Substituting [math]z=\frac{x^3-6x^2+6x}{6x-7}[/math], we then find out that [math]\frac{2x^3+5x^2-4x-7}{6x-7}[/math] must be a rational square. We then multiply the numerator and denominator by [math]6x-7[/math] to make the denominator square. The resulting fraction is [math]\frac{12x^4+16x^3-59x^2-14x+49}{(6x-7)^2}[/math]. Since the denominator is square, we only need to worry about the numerator, assigning it a new variable: [math]y^2=12x^4+16x^3-59x^2-14x+49[/math]. We now have the following:
math^2-4z=\frac{y^2}{(6x-7)^2}[/math]
Substituting back into our formula for math[/math] from math[/math], we then get this:
[math]a,b=\frac{6x^2-13x+7\pm y}{2(6x-7)}[/math]
Rational solutions for math[/math] correspond to rational solutions for math[/math], from which integer solutions can be found for math[/math]. Here is the curve again, which I will call a quartic curve, since it looks like a Weierstrauss elliptic curve, but with a quartic polynomial on the right-hand side instead of a cubic one.
[math]y^2=12x^4+16x^3-59x^2-14x+49[/math]
Inspecting the curve, I was able to see four rational solutions (up to negation), though you may be able to see more:
math=(-3,10),(-1,0),(0,7),(1,2)[/math]
In order, I will label these points [math]A[/math], [math]B[/math], [math]C[/math], and [math]D[/math]. Now, what we need is a point addition method, like on an elliptic curve. And it turns out, that one exists. On an elliptic curve, you can connect two rational points with a line, and it'll intersect the elliptic curve at a third rational point. And it turns out that on a quartic curve, that you can connect three rational points with a parabola, and it'll intersect the quartic curve at a fourth rational point. There's a reason too. Suppose the parabola is [math]y=Ax^2+Bx+C[/math] ([math]A[/math], [math]B[/math], and [math]C[/math] are coefficients here, not points). Then we have [math]y=(Ax^2+Bx+C)^2=12x^4+16x^3-59x^2-14x+49[/math]. This reduces to a quartic polynomial with rational coefficients, but we already know three of the roots ([math]A[/math], [math]B[/math], and [math]C[/math] are always rational, and [math]A^2\neq 12[/math] since [math]A[/math] is rational), since they are simply the x-coordinates of the three points that we connected with that parabola. If we divide out those roots, we get a linear polynomial with rational coefficients. That fourth root (by which I mean root #4, not raising something to the power of [math]\frac{1}{4}[/math]) is the x-coordinate of the new rational point, after which we can plug it into the parabola to obtain the y-coordinate of the new rational point. To obtain [math]A[/math], [math]B[/math], and [math]C[/math], we can use a system of linear equation for our points math,(x_1,y_1),(x_2,y_2)[/math].
[math]x_0^2 A+x_0 B+C=y_0[/math]
[math]x_1^2 A+x_1 B+C=y_1[/math]
[math]x_2^2 A+x_2 B+C=y_2[/math]
Given three rational points, we can solve these equations for [math]A[/math], [math]B[/math], and [math]C[/math], construct a connecting parabola, and find its fourth intersection with the quartic curve to get a new rational point. I will call this the [math]P[/math] operation. In other words, given three points [math]p_0[/math], [math]p_1[/math], and [math]p_2[/math], the rational point obtained by adding them is [math]P(p_0,p_1,p_2)[/math]. I will also define the negation of a point as follows: If [math]p=(x,y)[/math], then [math]-P=(x,-y)[/math]. Negation is useful, since without it, we couldn't find any new points past the fourth one. Adding the fourth point to two of the original three points would give back the third, which wouldn't be very useful. Negation allows us to find new points for possibly forever. So, let's go back to our original four points. Using these, we can get six more, up to negation:
[math]E=P(A,B,C)=(-14,637)[/math]
[math]F=P(B,C,D)=(\frac{7}{6},0)[/math]
[math]G=P(A,C,D)=(\frac{14}{11},\frac{35}{121})[/math]
[math]H=P(-A,B,D)=(\frac{15}{11},\frac{182}{121})[/math]
[math]I=P(A,B,D)=(\frac{5}{3},\frac{16}{3})[/math]
[math]J=P(-A,B,C)=(\frac{7}{2},42)[/math]
In my original exploration, I tried adding all possible triplets of points to get new points. I noticed that these new points seem to come in sets of six, where the denominators have relatively similar sizes. Interesting, each point in each set can be obtained by adding one point in the previous set, and two of the original four points, though starting from the third set, only [/math]A[/math] and [/math]B[/math] seem to be necessary. The second set is as follows:
[math]K=P(-A,B,E),L=P(-A,C,E),M=P(-A,B,G),N=P(A,B,H),O=P(-A,B,I),P=P(A,B,J)[/math]
The third set:
[math]Q_0=P(A,B,K),R_0=P(A,B,L),S_0=P(A,B,M),T_0=P(-A,B,M),U_0=P(A,B,N),V_0=P(-A,B,O)[/math]
The fourth set follows from the third set in the same way, except that the signs of the "[math]A[/math]"s are flipped. For example, [math]Q_1=P(-A,B,Q_0)[/math]. In the fifth set, the signs flip back. This flipping continues to infinity. For each point, we compute math[/math], and they're not all positive. And then we reach [math]V_6[/math]. This time, we compute math[/math], and they're both positive. So we compute math[/math], and get this out.
[math]a=4373612677928697257861252602371390152816537558161613618621437993378423467772036[/math]
[math]b=36875131794129999827197811565225474825492979968971970996283137471637224634055579[/math]
[math]c=154476802108746166441951315019919837485664325669565431700026634898253202035277999[/math]
Somewhere farther down the point series, I got another solution. I can't tell which point it is, since the online Wolfram computable notebook service (at ( but now gives a 404) seems to have been taken down. I do, however, still have that solution.
[math]a=32343421153825592353880655285224263330451946573450847101645239147091638517651250940206853612606768544181415355352136077327300271806129063833025389772729796460799697289[/math]
[math]b=16666476865438449865846131095313531540647604679654766832109616387367203990642764342248100534807579493874453954854925352739900051220936419971671875594417036870073291371[/math]
[math]c=18438651467072329521991466669103809627503176533640434051668643025780389550623758060258285903998125757038016122166239815379429082156904518238560341886750920963276835983515[/math]
I found even more solutions in a later investigation, but due to size I won't post them here. If I can figure out a way to find which points they corresponded to, I'll happily give out the names of the points. Since it will likely use a computer algebra package, I will also share the code for point addition and solution recovery if I can rewrite said code for said package.
EDIT: As promised, here is the code. I used SageMathCell for this.
def k(p,q,r): return (r-q)(q-p)(p-r)def a(p,q,r):#Coefficients, not the actual values for the (a,b,c) triples return (p(q-r)+q(r-p)+r(p-q))/k(p,q,r)def b(p,q,r): return (p(r^2-q^2)+q(p^2-r^2)+r(q^2-p^2))/k(p,q,r)def c(p,q,r): return (pqr(q-r)+qrp(r-p)+rpq(p-q))/k(p,q,r)def xsum(p,q,r): return p+q+rdef px(p,q,r): return (-2a(p,q,r)b(p,q,r)-a(p,q,r)^2xsum(p,q,r)+12xsum(p,q,r)+16)/(-12+a(p,q,r)^2)def py(p,q,r): return a(p,q,r)px(p,q,r)^2+b(p,q,r)px(p,q,r)+c(p,q,r)def p(p,q,r): return [px(p,q,r),py(p,q,r)]def n(p): return [p,-p]def ab(p): l=6p^2-13p+7 r=2(6p-7) return [(l+p)/r,(l-p)/r]def abc(p): mapped=ab(p) multiplier=lcm(denominator(mapped),denominator(mapped)) return [mappedmultiplier,mappedmultiplier,multiplier]def find_point(name,index=0): A=[-3,10] B=[-1,0] C=[0,7] D=[1,2] if name=="A": return A elif name=="B": return B elif name=="C": return C elif name=="D": return D elif name=="E": return p(A,B,C) elif name=="F": return p(B,C,D) elif name=="G": return p(A,C,D) elif name=="H": return p(n(A),B,D) elif name=="I": return p(A,B,D) elif name=="J": return p(n(A),B,C) elif name=="K": return p(n(A),B,find_point("E")) elif name=="L": return p(n(A),C,find_point("E")) elif name=="M": return p(n(A),B,find_point("G")) elif name=="N": return p(A,B,find_point("H")) elif name=="O": return p(n(A),B,find_point("I")) elif name=="P": return p(A,B,find_point("J")) elif name=="Q" and index==0: return p(A,B,find_point("K")) elif name=="R" and index==0: return p(A,B,find_point("L")) elif name=="S" and index==0: return p(A,B,find_point("M")) elif name=="T" and index==0: return p(n(A),B,find_point("N")) elif name=="U" and index==0: return p(A,B,find_point("O")) elif name=="V" and index==0: return p(n(A),B,find_point("P")) elif ((name=="Q" or name=="R" or name=="S" or name=="U") and index%2==0) or ((name=="T" or name=="V") and index%2==1): return p(A,B,find_point(name,index-1)) elif ((name=="T" or name=="V") and index%2==0) or ((name=="Q" or name=="R" or name=="S" or name=="U") and index%2==1): return p(n(A),B,find_point(name,index-1))
Point [math]V_6[/math] could be found by using find_point("V",6).
EDIT 2: The second solution I mentioned corresponds to point [math]R_{10}[/math].
Edit 3: The third solution I found corresponds to point [math]Q_{14}[/math].
Edit 4: The fourth solution I found corresponds to point [math]U_{18}[/math].
Edit 5: The seventh solution I found corresponds to point [math]R_{36}[/math].
Sohel Zibara
Studied at Doctor of Philosophy Degrees (Graduated 2000)
·
Author has 5.1K answers and 2.6M answer views
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3y
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A,b, c are positive integers. a+(b/c) =101, (a/c) +b=89, what is (a+b) /c?
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Johnny M
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Rik Bos
Ph.D. Mathematics from Utrecht University (Graduated 1979)
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Author has 1.4K answers and 1.3M answer views
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11mo
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Given three positive integers
[math]a,b,c.[/math]
How do I show that
[math]\gcd(a,b,c)^2 \mid \gcd(a,b) \cdot \gcd(b,c) \cdot \gcd(c,a)[/math]
?
If
[math]p^k[/math]
is the highest power of prime
[math]p[/math]
that divides
[math]\gcd(a,b,c)[/math]
, then
[math]p^k[/math]
divides
[math]a,b,c[/math]
, so
[math]p^k[/math]
also divides
[math]\gcd(a,b),\gcd(b,c)[/math]
and
[math]\gcd(c,a)[/math]
. Therefore
[math]p^{3k}\mid \gcd(a,b)\cdot \gcd(b,c)\cdot \gcd(c,a)[/math]
, while
[math]p^{2k}[/math]
is the highest power of prime
[math]p[/math]
that divides
[math]\gcd(a,b,c)^2[/math]
Richard P
MMath in Mathematics, Churchill College, Cambridge (Graduated 2011)
·
Author has 807 answers and 316.4K answer views
·
2y
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How do you prove that there are infinite triples of distinct positive integers a,b, c, that satisfy the equation a! b! =c! How?
What happens if we try c=b+1? Then the equation a! b! = c! becomes a! = c = b+1.
OK, let’s try it out:
(a, b, c) = (1, 0, 1) … trivial
(a, b, c) = (2, 1, 2) … trivial
(a, b, c) = (3, 5, 6)
(a, b, c) = (4, 23, 24)
(a, b, c) = (5, 119, 120)
(a, b, c) = (6, 719, 720)
OK, it was a bit slow starting and the numbers are getting large very quickly, but we have an infinite series here.
David Shaffer
Bachelors degree in Maths and Physics from a long time ago
·
Author has 2K answers and 2.4M answer views
·
Mar 19
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Consider three positive real numbers
[math]a,b[/math]
and
[math]c.[/math]
How do I show that there cannot exist two distinct positive integers
[math]m[/math]
and
[math]n,[/math]
such that both
[math]a^m+b^m=c^m[/math]
and
[math]a^n+b^n=c^n[/math]
hold?
Assume wlog that [math]n>m[/math]. Then we have
[math]\displaystyle a^n+b^n=c^n=c^{n-m}(a^m+b^m) \tag1[/math]
[math]\displaystyle a^m(a^{n-m}-c^{n-m}) + b^m(b^{n-m}-c^{n-m})=0 \tag2[/math]
But [math]c^m=a^m+b^m>a^m[/math] so [math]c>a[/math] and similarly [math]c>b[/math]. Therefore both terms in (2) are negative and we have a contradiction, so the given equations cannot both be true.
Alberto Cid
M.S.E. in Telecommunications Engineering & Data Transmission, Technical University of Madrid (Graduated 2008)
·
Author has 2K answers and 3.8M answer views
·
4y
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If a, b, c are three distinct positive real integers, how do you prove that (a+b+c) (1/a+1b+1/c) >9?
Product P = (a+b+c) (1/a+1/b+1/c) =
[Distributive property]
= a (1/a+1/b+1/c) + b (1/a+1/b+1/c) + c (1/a+1/b+1/c) =
[Distributive property, again]
= a/a+a/b+a/c + b/a+b/b+b/c + c/a+c/b+c/c =
= 1+a/b+a/c + b/a+1+b/c + c/a+c/b+1 =
= 3+a/b+a/c + b/a+b/c + c/a+c/b
And we have to prove P>9
So…
3+a/b+a/c + b/a+b/c + c/a+c/b > 9
if and only if
Sum S = a/b+a/c + b/a+b/c + c/a+c/b > 6
We can observe we have:
S = (a/b+ b/a ) + (a/c + c/a) + (b/c + c/b)
That is, 3 sums of pairs of fractions.
Notice we must prove S>6 … and, since those 3 pairs are similar in concept, that is, they are symmetric, it would be ni
Product P = (a+b+c) (1/a+1/b+1/c) =
[Distributive property]
= a (1/a+1/b+1/c) + b (1/a+1/b+1/c) + c (1/a+1/b+1/c) =
[Distributive property, again]
= a/a+a/b+a/c + b/a+b/b+b/c + c/a+c/b+c/c =
= 1+a/b+a/c + b/a+1+b/c + c/a+c/b+1 =
= 3+a/b+a/c + b/a+b/c + c/a+c/b
And we have to prove P>9
So…
3+a/b+a/c + b/a+b/c + c/a+c/b > 9
if and only if
Sum S = a/b+a/c + b/a+b/c + c/a+c/b > 6
We can observe we have:
S = (a/b+ b/a ) + (a/c + c/a) + (b/c + c/b)
That is, 3 sums of pairs of fractions.
Notice we must prove S>6 … and, since those 3 pairs are similar in concept, that is, they are symmetric, it would be nice if we prove at least any of those 3 is greater than 2 and the others are greater or equal than 2.
(a/b+ b/a ) =
That’s a sum of fractions… and we need the same denominator.
We choose denominator ab so that it has both a and b
a/b = (aa)/(ab)
b/a = (bb)/(ab)
(a/b+ b/a ) = (a^2 + b^2) / (ab)
Remember the target is proving (a/b+ b/a ) > 2
(a^2 + b^2) / (ab) > 2
if and only if (a^2 + b^2) > 2(ab)
a^2 + b^2 - 2ab > 0
But that’s a well known expression:
a^2 + b^2 - 2ab = (a-b)^2 = (b-a)^2
And, of course, that’s greater or equal to 0 , right???
Any square is always non-negative.
But I said “or equal” … When is a square equal to zero?
x^2 = 0 if and only if x=0
So: the only exception would be (a-b) = 0
But this exception would imply a=b and the question says “three distinct” so they can’t be equal.
The proof would be complete here.
Since (a/b+ b/a ) > 2
and similarly
(a/c+ c/a ) > 2
and
(c/b+ b/c ) > 2
Then S > 6
Then P = 3+S > 9
Notice: it’s not necessary that all of the 3 possitive integers are distinct, it’s enough that the 3 are not all the same, that is, that only one of them is different.
WLOG suppose the different one is c and a=b
a=b implies (a/b+ b/a ) = 1+1 = 2
(a/c+ c/a ) > 2
and
(c/b+ b/c ) > 2
So: S > 6
And: P > 9
In case a=b=c the product would be:
(a+a+a)(1/a+1/a+1/a) = 3a 3/a = 9
P = 9 only when a=b=c
In any case (distinct or not) : P >= 9
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From Wikipedia, the free encyclopedia
Group of chemical elements
| Chalcogens | |
--- |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | Hydrogen | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Helium | | Lithium | Beryllium | | | | | | | | | | | | | | | | | | | | | | | | | Boron | Carbon | Nitrogen | Oxygen | Fluorine | Neon | | Sodium | Magnesium | | | | | | | | | | | | | | | | | | | | | | | | | Aluminium | Silicon | Phosphorus | Sulfur | Chlorine | Argon | | Potassium | Calcium | | | | | | | | | | | | | | | Scandium | Titanium | Vanadium | Chromium | Manganese | Iron | Cobalt | Nickel | Copper | Zinc | Gallium | Germanium | Arsenic | Selenium | Bromine | Krypton | | Rubidium | Strontium | | | | | | | | | | | | | | | Yttrium | Zirconium | Niobium | Molybdenum | Technetium | Ruthenium | Rhodium | Palladium | Silver | Cadmium | Indium | Tin | Antimony | Tellurium | Iodine | Xenon | | Caesium | Barium | Lanthanum | Cerium | Praseodymium | Neodymium | Promethium | Samarium | Europium | Gadolinium | Terbium | Dysprosium | Holmium | Erbium | Thulium | Ytterbium | Lutetium | Hafnium | Tantalum | Tungsten | Rhenium | Osmium | Iridium | Platinum | Gold | Mercury (element) | Thallium | Lead | Bismuth | Polonium | Astatine | Radon | | Francium | Radium | Actinium | Thorium | Protactinium | Uranium | Neptunium | Plutonium | Americium | Curium | Berkelium | Californium | Einsteinium | Fermium | Mendelevium | Nobelium | Lawrencium | Rutherfordium | Dubnium | Seaborgium | Bohrium | Hassium | Meitnerium | Darmstadtium | Roentgenium | Copernicium | Nihonium | Flerovium | Moscovium | Livermorium | Tennessine | Oganesson | pnictogens ← → halogens | |
| | IUPAC group number | 16 | --- | | Name by element | oxygen group | | Trivial name | chalcogens | | CAS group number (US, pattern A-B-A) | VIA | | old IUPAC number (Europe, pattern A-B) | VIB | | |
| ↓ Period | |
--- |
| 2 | Oxygen (O) 8 Other nonmetal |
| 3 | Sulfur (S) 16 Other nonmetal |
| 4 | Selenium (Se) 34 Other nonmetal |
| 5 | Tellurium (Te) 52 Metalloid |
| 6 | Polonium (Po) 84 Other metal |
| 7 | Livermorium (Lv) 116 Other metal |
| --- Legend | | | primordial element | | naturally occurring by radioactive decay | | synthetic element | | |
The chalcogens (ore forming) (/ˈkælkədʒənz/ KAL-kə-jənz) are the chemical elements in group 16 of the periodic table. This group is also known as the oxygen family. Group 16 consists of the elements oxygen (O), sulfur (S), selenium (Se), tellurium (Te), and the radioactive elements polonium (Po) and livermorium (Lv). Often, oxygen is treated separately from the other chalcogens, sometimes even excluded from the scope of the term "chalcogen" altogether, due to its very different chemical behavior from sulfur, selenium, tellurium, and polonium. The word "chalcogen" is derived from a combination of the Greek word khalkos (χαλκός) principally meaning copper (the term was also used for bronze, brass, any metal in the poetic sense, ore and coin), and the Latinized Greek word genēs, meaning born or produced.
Sulfur has been known since antiquity, and oxygen was recognized as an element in the 18th century. Selenium, tellurium and polonium were discovered in the 19th century, and livermorium in 2000. All of the chalcogens have six valence electrons, leaving them two electrons short of a full outer shell. Their most common oxidation states are −2, +2, +4, and +6. They have relatively small atomic radii, especially the lighter ones.
All of the naturally occurring chalcogens have some role in biological functions, either as a nutrient or a toxin. Selenium is an important nutrient (among others as a building block of selenocysteine) but is also commonly toxic. Tellurium often has unpleasant effects (although some organisms can use it), and polonium (especially the isotope polonium-210) is always harmful as a result of its radioactivity.
Sulfur has more than 20 allotropes, oxygen has nine, selenium has at least eight, polonium has two, and only one crystal structure of tellurium has so far been discovered. There are numerous organic chalcogen compounds. Not counting oxygen, organic sulfur compounds are generally the most common, followed by organic selenium compounds and organic tellurium compounds. This trend also occurs with chalcogen pnictides and compounds containing chalcogens and carbon group elements.
Oxygen is generally obtained by separation of air into nitrogen and oxygen. Sulfur is extracted from oil and natural gas. Selenium and tellurium are produced as byproducts of copper refining. Polonium is most available in naturally occurring actinide-containing materials. Livermorium has been synthesized in particle accelerators. The primary use of elemental oxygen is in steelmaking. Sulfur is mostly converted into sulfuric acid, which is heavily used in the chemical industry. Selenium's most common application is glassmaking. Tellurium compounds are mostly used in optical disks, electronic devices, and solar cells. Some of polonium's applications are due to its radioactivity.
Properties
[edit]
Atomic and physical
[edit]
Chalcogens show similar patterns in electron configuration, especially in the outermost shells, where they all have the same number of valence electrons, resulting in similar trends in chemical behavior:
| Z | Element | Electrons per shell |
---
| 8 | Oxygen | 2, 6 |
| 16 | Sulfur | 2, 8, 6 |
| 34 | Selenium | 2, 8, 18, 6 |
| 52 | Tellurium | 2, 8, 18, 18, 6 |
| 84 | Polonium | 2, 8, 18, 32, 18, 6 |
| 116 | Livermorium | 2, 8, 18, 32, 32, 18, 6 (predicted) |
| Element | Melting point (°C) | Boiling point (°C) | Density at STP (g/cm3) |
--- --- |
| Oxygen | −219 | −183 | 0.00143 |
| Sulfur | 120 | 445 | 2.07 |
| Selenium | 221 | 685 | 4.3 |
| Tellurium | 450 | 988 | 6.24 |
| Polonium | 254 | 962 | 9.2 |
| Livermorium | 364–507 (predicted) | 762–862 (predicted) | 14 (predicted) |
All chalcogens have six valence electrons. All of the solid, stable chalcogens are soft and do not conduct heat well. Electronegativity decreases towards the chalcogens with higher atomic numbers. Density, melting and boiling points, and atomic and ionic radii tend to increase towards the chalcogens with higher atomic numbers.
Isotopes
[edit]
Out of the six known chalcogens, one (oxygen) has an atomic number equal to a nuclear magic number, which means that their atomic nuclei tend to have increased stability against radioactive decay. Oxygen has three stable isotopes, and 14 unstable ones. Sulfur has four stable isotopes, 20 radioactive ones, and one isomer. Selenium has six observationally stable or nearly stable isotopes, 26 radioactive isotopes, and 9 isomers. Tellurium has eight stable or nearly stable isotopes, 31 unstable ones, and 17 isomers. Polonium has 42 isotopes, none of which are stable. It has an additional 28 isomers. In addition to the stable isotopes, some radioactive chalcogen isotopes occur in nature, either because they are decay products, such as 210Po, because they are primordial, such as 82Se, because of cosmic ray spallation, or via nuclear fission of uranium. Livermorium isotopes 288Lv through 293Lv have been discovered; the most stable livermorium isotope is 293Lv, which has a half-life of 0.061 seconds.
With the exception of livermorium, all chalcogens have at least one naturally occurring radioisotope: oxygen has trace 15O, sulfur has trace 35S, selenium has 82Se, tellurium has 128Te and 130Te, and polonium has 210Po.
Among the lighter chalcogens (oxygen and sulfur), the most neutron-poor isotopes undergo proton emission, the moderately neutron-poor isotopes undergo electron capture or β+ decay, the moderately neutron-rich isotopes undergo β− decay, and the most neutron rich isotopes undergo neutron emission. The middle chalcogens (selenium and tellurium) have similar decay tendencies as the lighter chalcogens, but no proton-emitting isotopes have been observed, and some of the most neutron-deficient isotopes of tellurium undergo alpha decay. Polonium isotopes tend to decay via alpha or beta decay. Isotopes with nonzero nuclear spins are more abundant in nature among the chalcogens selenium and tellurium than they are with sulfur.
Allotropes
[edit]
See also: Allotropes of oxygen and Allotropes of sulfur
Oxygen's most common allotrope is diatomic oxygen, or O2, a reactive paramagnetic molecule that is ubiquitous to aerobic organisms and has a blue color in its liquid state. Another allotrope is O3, or ozone, which is three oxygen atoms bonded together in a bent formation. There is also an allotrope called tetraoxygen, or O4, and six allotropes of solid oxygen including "red oxygen", which has the formula O8.
Sulfur has over 20 known allotropes, which is more than any other element except carbon. The most common allotropes are in the form of eight-atom rings, but other molecular allotropes that contain as few as two atoms or as many as 20 are known. Other notable sulfur allotropes include rhombic sulfur and monoclinic sulfur. Rhombic sulfur is the more stable of the two allotropes. Monoclinic sulfur takes the form of long needles and is formed when liquid sulfur is cooled to slightly below its melting point. The atoms in liquid sulfur are generally in the form of long chains, but above 190 °C, the chains begin to break down. If liquid sulfur above 190 °C is frozen very rapidly, the resulting sulfur is amorphous or "plastic" sulfur. Gaseous sulfur is a mixture of diatomic sulfur (S2) and 8-atom rings.
Selenium has at least eight distinct allotropes. The gray allotrope, commonly referred to as the "metallic" allotrope, despite not being a metal, is stable and has a hexagonal crystal structure. The gray allotrope of selenium is soft, with a Mohs hardness of 2, and brittle. Four other allotropes of selenium are metastable. These include two monoclinic red allotropes and two amorphous allotropes, one of which is red and one of which is black. The red allotrope converts to the black allotrope in the presence of heat. The gray allotrope of selenium is made from spirals on selenium atoms, while one of the red allotropes is made of stacks of selenium rings (Se8).[dubious – discuss]
Tellurium is not known to have any allotropes, although its typical form is hexagonal. Polonium has two allotropes, which are known as α-polonium and β-polonium. α-polonium has a cubic crystal structure and converts to the rhombohedral β-polonium at 36 °C.
The chalcogens have varying crystal structures. Oxygen's crystal structure is monoclinic, sulfur's is orthorhombic, selenium and tellurium have the hexagonal crystal structure, while polonium has a cubic crystal structure.
Chemical
[edit]
Oxygen, sulfur, and selenium are nonmetals, and tellurium is a metalloid, meaning that its chemical properties are between those of a metal and those of a nonmetal. It is not certain whether polonium is a metal or a metalloid. Some sources refer to polonium as a metalloid, although it has some metallic properties. Also, some allotropes of selenium display characteristics of a metalloid, even though selenium is usually considered a nonmetal. Even though oxygen is a chalcogen, its chemical properties are different from those of other chalcogens. One reason for this is that the heavier chalcogens have vacant d-orbitals. Oxygen's electronegativity is also much higher than those of the other chalcogens. This makes oxygen's electric polarizability several times lower than those of the other chalcogens.
For covalent bonding a chalcogen may accept two electrons according to the octet rule, leaving two lone pairs. When an atom forms two single bonds, they form an angle between 90° and 120°. In 1+ cations, such as H3O+, a chalcogen forms three molecular orbitals arranged in a trigonal pyramidal fashion and one lone pair. Double bonds are also common in chalcogen compounds, for example in chalcogenates (see below).
The oxidation number of the most common chalcogen compounds with positive metals is −2. However the tendency for chalcogens to form compounds in the −2 state decreases towards the heavier chalcogens. Other oxidation numbers, such as −1 in pyrite and peroxide, do occur. The highest formal oxidation number is +6. This oxidation number is found in sulfates, selenates, tellurates, polonates, and their corresponding acids, such as sulfuric acid.
Oxygen is the most electronegative element except for fluorine, and forms compounds with almost all of the chemical elements, including some of the noble gases. It commonly bonds with many metals and metalloids to form oxides, including iron oxide, titanium oxide, and silicon oxide. Oxygen's most common oxidation state is −2, and the oxidation state −1 is also relatively common. With hydrogen it forms water and hydrogen peroxide. Organic oxygen compounds are ubiquitous in organic chemistry.
Sulfur's oxidation states are −2, +2, +4, and +6. Sulfur-containing analogs of oxygen compounds often have the prefix thio-. Sulfur's chemistry is similar to oxygen's, in many ways. One difference is that sulfur-sulfur double bonds are far weaker than oxygen-oxygen double bonds, but sulfur-sulfur single bonds are stronger than oxygen-oxygen single bonds. Organic sulfur compounds such as thiols have a strong specific smell, and a few are utilized by some organisms.
Selenium's oxidation states are −2, +4, and +6. Selenium, like most chalcogens, bonds with oxygen. There are some organic selenium compounds, such as selenoproteins. Tellurium's oxidation states are −2, +2, +4, and +6. Tellurium forms the oxides tellurium monoxide, tellurium dioxide, and tellurium trioxide. Polonium's oxidation states are +2 and +4.
There are many acids containing chalcogens, including sulfuric acid, sulfurous acid, selenic acid, and telluric acid. All hydrogen chalcogenides are toxic except for water. Oxygen ions often come in the forms of oxide ions (O2−), peroxide ions (O2−2), and hydroxide ions (OH−). Sulfur ions generally come in the form of sulfides (S2−), bisulfides (SH−), sulfites (SO2−3), sulfates (SO2−4), and thiosulfates (S2O2−3). Selenium ions usually come in the form of selenides (Se2−), selenites (SeO2−3) and selenates (SeO2−4). Tellurium ions often come in the form of tellurates (TeO2−4). Molecules containing metal bonded to chalcogens are common as minerals. For example, pyrite (FeS2) is an iron ore, and the rare mineral calaverite is the ditelluride (Au, Ag)Te2.
Although all group 16 elements of the periodic table, including oxygen, can be defined as chalcogens, oxygen and oxides are usually distinguished from chalcogens and chalcogenides. The term chalcogenide is more commonly reserved for sulfides, selenides, and tellurides, rather than for oxides.
Except for polonium, the chalcogens are all fairly similar to each other chemically. They all form X2− ions when reacting with electropositive metals.
Sulfide minerals and analogous compounds produce gases upon reaction with oxygen.
Compounds
[edit]
With halogens
[edit]
Chalcogens also form compounds with halogens known as chalcohalides, or chalcogen halides. The majority of simple chalcogen halides are well-known and widely used as chemical reagents. However, more complicated chalcogen halides, such as sulfenyl, sulfonyl, and sulfuryl halides, are less well known to science. Out of the compounds consisting purely of chalcogens and halogens, there are a total of 13 chalcogen fluorides, nine chalcogen chlorides, eight chalcogen bromides, and six chalcogen iodides that are known.[dubious – discuss] The heavier chalcogen halides often have significant molecular interactions. Sulfur fluorides with low valences are fairly unstable and little is known about their properties.[dubious – discuss] However, sulfur fluorides with high valences, such as sulfur hexafluoride, are stable and well-known. Sulfur tetrafluoride is also a well-known sulfur fluoride. Certain selenium fluorides, such as selenium difluoride, have been produced in small amounts. The crystal structures of both selenium tetrafluoride and tellurium tetrafluoride are known. Chalcogen chlorides and bromides have also been explored. In particular, selenium dichloride and sulfur dichloride can react to form organic selenium compounds. Dichalcogen dihalides, such as Se2Cl2 also are known to exist. There are also mixed chalcogen-halogen compounds. These include SeSX, with X being chlorine or bromine.[dubious – discuss] Such compounds can form in mixtures of sulfur dichloride and selenium halides. These compounds have been fairly recently structurally characterized, as of 2008. In general, diselenium and disulfur chlorides and bromides are useful chemical reagents. Chalcogen halides with attached metal atoms are soluble in organic solutions.[dubious – discuss] One example of such a compound is MoS2Cl3. Unlike selenium chlorides and bromides, selenium iodides have not been isolated, as of 2008, although it is likely that they occur in solution. Diselenium diiodide, however, does occur in equilibrium with selenium atoms and iodine molecules. Some tellurium halides with low valences, such as Te2Cl2 and Te2Br2, form polymers when in the solid state. These tellurium halides can be synthesized by the reduction of pure tellurium with superhydride and reacting the resulting product with tellurium tetrahalides. Ditellurium dihalides tend to get less stable as the halides become lower in atomic number and atomic mass. Tellurium also forms iodides with even fewer iodine atoms than diiodides. These include TeI and Te2I. These compounds have extended structures in the solid state. Halogens and chalcogens can also form halochalcogenate anions.
Organic
[edit]
Alcohols, phenols and other similar compounds contain oxygen. However, in thiols, selenols and tellurols; sulfur, selenium, and tellurium replace oxygen. Thiols are better known than selenols or tellurols. Aside from alcohols, thiols are the most stable chalcogenols and tellurols are the least stable, being unstable in heat or light. Other organic chalcogen compounds include thioethers, selenoethers and telluroethers. Some of these, such as dimethyl sulfide, diethyl sulfide, and dipropyl sulfide are commercially available. Selenoethers are in the form of R2Se or RSeR. Telluroethers such as dimethyl telluride are typically prepared in the same way as thioethers and selenoethers. Organic chalcogen compounds, especially organic sulfur compounds, have the tendency to smell unpleasant. Dimethyl telluride also smells unpleasant, and selenophenol is renowned for its "metaphysical stench". There are also thioketones, selenoketones, and telluroketones. Out of these, thioketones are the most well-studied with 80% of chalcogenoketones papers being about them. Selenoketones make up 16% of such papers and telluroketones make up 4% of them. Thioketones have well-studied non-linear electric and photophysical properties. Selenoketones are less stable than thioketones and telluroketones are less stable than selenoketones. Telluroketones have the highest level of polarity of chalcogenoketones.
With metals
[edit]
There is a very large number of metal chalcogenides. There are also ternary compounds containing alkali metals and transition metals. Highly metal-rich metal chalcogenides, such as Lu7Te and Lu8Te have domains of the metal's crystal lattice containing chalcogen atoms. While these compounds do exist, analogous chemicals that contain lanthanum, praseodymium, gadolinium, holmium, terbium, or ytterbium have not been discovered, as of 2008. The boron group metals aluminum, gallium, and indium also form bonds to chalcogens. The Ti3+ ion forms chalcogenide dimers such as TiTl5Se8. Metal chalcogenide dimers also occur as lower tellurides, such as Zr5Te6.
Elemental chalcogens react with certain lanthanide compounds to form lanthanide clusters rich in chalcogens.[dubious – discuss] Uranium(IV) chalcogenol compounds also exist. There are also transition metal chalcogenols which have potential to serve as catalysts and stabilize nanoparticles.
With pnictogens
[edit]
Compounds with chalcogen-phosphorus bonds have been explored for more than 200 years. These compounds include unsophisticated phosphorus chalcogenides as well as large molecules with biological roles and phosphorus-chalcogen compounds with metal clusters. These compounds have numerous applications, including organo-phosphate insecticides, strike-anywhere matches and quantum dots. A total of 130,000 compounds with at least one phosphorus-sulfur bond, 6000 compounds with at least one phosphorus-selenium bond, and 350 compounds with at least one phosphorus-tellurium bond have been discovered.[citation needed] The decrease in the number of chalcogen-phosphorus compounds further down the periodic table is due to diminishing bond strength. Such compounds tend to have at least one phosphorus atom in the center, surrounded by four chalcogens and side chains. However, some phosphorus-chalcogen compounds also contain hydrogen (such as secondary phosphine chalcogenides) or nitrogen (such as dichalcogenoimidodiphosphates). Phosphorus selenides are typically harder to handle that phosphorus sulfides, and compounds in the form PxTey have not been discovered. Chalcogens also bond with other pnictogens, such as arsenic, antimony, and bismuth. Heavier chalcogen pnictides tend to form ribbon-like polymers instead of individual molecules. Chemical formulas of these compounds include Bi2S3 and Sb2Se3. Ternary chalcogen pnictides are also known. Examples of these include P4O6Se and P3SbS3. salts containing chalcogens and pnictogens also exist. Almost all chalcogen pnictide salts are typically in the form of [PnxE4x]3−, where Pn is a pnictogen and E is a chalcogen.[dubious – discuss] Tertiary phosphines can react with chalcogens to form compounds in the form of R3PE, where E is a chalcogen. When E is sulfur, these compounds are relatively stable, but they are less so when E is selenium or tellurium. Similarly, secondary phosphines can react with chalcogens to form secondary phosphine chalcogenides. However, these compounds are in a state of equilibrium with chalcogenophosphinous acid. Secondary phosphine chalcogenides are weak acids. Binary compounds consisting of antimony or arsenic and a chalcogen. These compounds tend to be colorful and can be created by a reaction of the constituent elements at temperatures of 500 to 900 °C (932 to 1,652 °F).
Other
[edit]
Chalcogens form single bonds and double bonds with other carbon group elements than carbon, such as silicon, germanium, and tin. Such compounds typically form from a reaction of carbon group halides and chalcogenol salts or chalcogenol bases. Cyclic compounds with chalcogens, carbon group elements, and boron atoms exist, and occur from the reaction of boron dichalcogenates and carbon group metal halides. Compounds in the form of M-E, where M is silicon, germanium, or tin, and E is sulfur, selenium or tellurium have been discovered. These form when carbon group hydrides react or when heavier versions of carbenes react.[dubious – discuss] Sulfur and tellurium can bond with organic compounds containing both silicon and phosphorus.
All of the chalcogens form hydrides. In some cases this occurs with chalcogens bonding with two hydrogen atoms. However tellurium hydride and polonium hydride are both volatile and highly labile. Also, oxygen can bond to hydrogen in a 1:1 ratio as in hydrogen peroxide, but this compound is unstable.
Chalcogen compounds form a number of interchalcogens. For instance, sulfur forms the toxic sulfur dioxide and sulfur trioxide. Tellurium also forms oxides. There are some chalcogen sulfides as well. These include selenium sulfide, an ingredient in some shampoos.
Since 1990, a number of borides with chalcogens bonded to them have been detected. The chalcogens in these compounds are mostly sulfur, although some do contain selenium instead. One such chalcogen boride consists of two molecules of dimethyl sulfide attached to a boron-hydrogen molecule. Other important boron-chalcogen compounds include macropolyhedral systems. Such compounds tend to feature sulfur as the chalcogen. There are also chalcogen borides with two, three, or four chalcogens. Many of these contain sulfur but some, such as Na2B2Se7 contain selenium instead.
History
[edit]
Early discoveries
[edit]
Sulfur has been known since ancient times and is mentioned in the Bible fifteen times. It was known to the ancient Greeks and commonly mined by the ancient Romans. In the Middle Ages, it was a key part of alchemical experiments. In the 1700s and 1800s, scientists Joseph Louis Gay-Lussac and Louis-Jacques Thénard proved sulfur to be a chemical element.
Early attempts to separate oxygen from air were hampered by the fact that air was thought of as a single element up to the 17th and 18th centuries. Robert Hooke, Mikhail Lomonosov, Ole Borch, and Pierre Bayden all successfully created oxygen, but did not realize it at the time. Oxygen was discovered by Joseph Priestley in 1774 when he focused sunlight on a sample of mercuric oxide and collected the resulting gas. Carl Wilhelm Scheele had also created oxygen in 1771 by the same method, but Scheele did not publish his results until 1777.
Tellurium was first discovered in 1783 by Franz Joseph Müller von Reichenstein. He discovered tellurium in a sample of what is now known as calaverite. Müller assumed at first that the sample was pure antimony, but tests he ran on the sample did not agree with this. Muller then guessed that the sample was bismuth sulfide, but tests confirmed that the sample was not that. For some years, Muller pondered the problem. Eventually he realized that the sample was gold bonded with an unknown element. In 1796, Müller sent part of the sample to the German chemist Martin Klaproth, who purified the undiscovered element. Klaproth decided to call the element tellurium after the Latin word for earth.
Selenium was discovered in 1817 by Jöns Jacob Berzelius. Berzelius noticed a reddish-brown sediment at a sulfuric acid manufacturing plant. The sample was thought to contain arsenic. Berzelius initially thought that the sediment contained tellurium, but came to realize that it also contained a new element, which he named selenium after the Greek moon goddess Selene.
Periodic table placing
[edit]
Three of the chalcogens (sulfur, selenium, and tellurium) were part of the discovery of periodicity, as they are among a series of triads of elements in the same group that were noted by Johann Wolfgang Döbereiner as having similar properties. Around 1865 John Newlands produced a series of papers where he listed the elements in order of increasing atomic weight and similar physical and chemical properties that recurred at intervals of eight; he likened such periodicity to the octaves of music. His version included a "group b" consisting of oxygen, sulfur, selenium, tellurium, and osmium.
After 1869, Dmitri Mendeleev proposed his periodic table placing oxygen at the top of "group VI" above sulfur, selenium, and tellurium. Chromium, molybdenum, tungsten, and uranium were sometimes included in this group, but they would be later rearranged as part of group VIB; uranium would later be moved to the actinide series. Oxygen, along with sulfur, selenium, tellurium, and later polonium would be grouped in group VIA, until the group's name was changed to group 16 in 1988.
Modern discoveries
[edit]
In the late 19th century, Marie Curie and Pierre Curie discovered that a sample of pitchblende was emitting four times as much radioactivity as could be explained by the presence of uranium alone. The Curies gathered several tons of pitchblende and refined it for several months until they had a pure sample of polonium. The discovery officially took place in 1898. Prior to the invention of particle accelerators, the only way to produce polonium was to extract it over several months from uranium ore.
The first attempt at creating livermorium was from 1976 to 1977 at the LBNL, who bombarded curium-248 with calcium-48, but were not successful. After several failed attempts in 1977, 1998, and 1999 by research groups in Russia, Germany, and the US, livermorium was created successfully in 2000 at the Joint Institute for Nuclear Research by bombarding curium-248 atoms with calcium-48 atoms. The element was known as ununhexium until it was officially named livermorium in 2012.
Names and etymology
[edit]
In the 19th century, Jons Jacob Berzelius suggested calling the elements in group 16 "amphigens", as the elements in the group formed amphid salts (salts of oxyacids, formerly regarded as composed of two oxides, an acid and a basic oxide). The term received some use in the early 1800s but is now obsolete. The name chalcogen comes from the Greek words χαλκος (chalkos, literally "copper"), and γενές (genes, born, gender, kindle). It was first used in 1932 by Wilhelm Biltz's group at Leibniz University Hannover, where it was proposed by Werner Fischer. The word "chalcogen" gained popularity in Germany during the 1930s because the term was analogous to "halogen". Although the literal meanings of the modern Greek words imply that chalcogen means "copper-former", this is misleading because the chalcogens have nothing to do with copper in particular. "Ore-former" has been suggested as a better translation, as the vast majority of metal ores are chalcogenides and the word χαλκος in ancient Greek was associated with metals and metal-bearing rock in general; copper, and its alloy bronze, was one of the first metals to be used by humans.
Oxygen's name comes from the Greek words oxy genes, meaning "acid-forming". Sulfur's name comes from either the Latin word sulfurium or the Sanskrit word sulvere; both of those terms are ancient words for sulfur. Selenium is named after the Greek goddess of the moon, Selene, to match the previously discovered element tellurium, whose name comes from the Latin word telus, meaning earth. Polonium is named after Marie Curie's country of birth, Poland. Livermorium is named for the Lawrence Livermore National Laboratory.
Occurrence
[edit]
The four lightest chalcogens (oxygen, sulfur, selenium, and tellurium) are all primordial elements on Earth. Sulfur and oxygen occur as constituent copper ores and selenium and tellurium occur in small traces in such ores. Polonium forms naturally from the decay of other elements, even though it is not primordial. Livermorium does not occur naturally at all.
Oxygen makes up 21% of the atmosphere by weight, 89% of water by weight, 46% of the Earth's crust by weight, and 65% of the human body. Oxygen also occurs in many minerals, being found in all oxide minerals and hydroxide minerals, and in numerous other mineral groups. Stars of at least eight times the mass of the Sun also produce oxygen in their cores via nuclear fusion. Oxygen is the third-most abundant element in the universe, making up 1% of the universe by weight.
Sulfur makes up 0.035% of the Earth's crust by weight, making it the 17th most abundant element there and makes up 0.25% of the human body. It is a major component of soil. Sulfur makes up 870 parts per million of seawater and about 1 part per billion of the atmosphere. Sulfur can be found in elemental form or in the form of sulfide minerals, sulfate minerals, or sulfosalt minerals. Stars of at least 12 times the mass of the Sun produce sulfur in their cores via nuclear fusion. Sulfur is the tenth most abundant element in the universe, making up 500 parts per million of the universe by weight.
Selenium makes up 0.05 parts per million of the Earth's crust by weight. This makes it the 67th most abundant element in the Earth's crust. Selenium makes up on average 5 parts per million of the soils. Seawater contains around 200 parts per trillion of selenium. The atmosphere contains 1 nanogram of selenium per cubic meter. There are mineral groups known as selenates and selenites, but there are not many minerals in these groups. Selenium is not produced directly by nuclear fusion. Selenium makes up 30 parts per billion of the universe by weight.
There are only 5 parts per billion of tellurium in the Earth's crust and 15 parts per billion of tellurium in seawater. Tellurium is one of the eight or nine least abundant elements in the Earth's crust. There are a few dozen tellurate minerals and telluride minerals, and tellurium occurs in some minerals with gold, such as sylvanite and calaverite. Tellurium makes up 9 parts per billion of the universe by weight.
Polonium only occurs in trace amounts on Earth, via radioactive decay of uranium and thorium. It is present in uranium ores in concentrations of 100 micrograms per metric ton. Very minute amounts of polonium exist in the soil and thus in most food, and thus in the human body. The Earth's crust contains less than 1 part per billion of polonium, making it one of the ten rarest metals on Earth.
Livermorium is always produced artificially in particle accelerators. Even when it is produced, only a small number of atoms are synthesized at a time.
Chalcophile elements
[edit]
See also: Chalcophile and Goldschmidt classification
Chalcophile elements are those that remain on or close to the surface because they combine readily with chalcogens other than oxygen, forming compounds which do not sink into the core. Chalcophile ("chalcogen-loving") elements in this context are those metals and heavier nonmetals that have a low affinity for oxygen and prefer to bond with the heavier chalcogen sulfur as sulfides. Because sulfide minerals are much denser than the silicate minerals formed by lithophile elements, chalcophile elements separated below the lithophiles at the time of the first crystallisation of the Earth's crust. This has led to their depletion in the Earth's crust relative to their solar abundances, though this depletion has not reached the levels found with siderophile elements.
v
t
e
Goldschmidt classification in the periodic table
| | 1 | 2 | | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
--- --- --- --- --- --- --- --- --- --- |
| Group → | | |
| ↓ Period | | |
| 1 | 1 H | | | | | | | | | | | | | | | | | | 2 He |
| 2 | 3 Li | 4 Be | | | | | | | | | | | | 5 B | 6 C | 7 N | 8 O | 9 F | 10 Ne |
| 3 | 11 Na | 12 Mg | | | | | | | | | | | | 13 Al | 14 Si | 15 P | 16 S | 17 Cl | 18 Ar |
| 4 | 19 K | 20 Ca | | 21 Sc | 22 Ti | 23 V | 24 Cr | 25 Mn | 26 Fe | 27 Co | 28 Ni | 29 Cu | 30 Zn | 31 Ga | 32 Ge | 33 As | 34 Se | 35 Br | 36 Kr |
| 5 | 37 Rb | 38 Sr | | 39 Y | 40 Zr | 41 Nb | 42 Mo | 43 Tc | 44 Ru | 45 Rh | 46 Pd | 47 Ag | 48 Cd | 49 In | 50 Sn | 51 Sb | 52 Te | 53 I | 54 Xe |
| 6 | 55 Cs | 56 Ba | | 71 Lu | 72 Hf | 73 Ta | 74 W | 75 Re | 76 Os | 77 Ir | 78 Pt | 79 Au | 80 Hg | 81 Tl | 82 Pb | 83 Bi | 84 Po | 85 At | 86 Rn |
| 7 | 87 Fr | 88 Ra | | 103 Lr | 104 Rf | 105 Db | 106 Sg | 107 Bh | 108 Hs | 109 Mt | 110 Ds | 111 Rg | 112 Cn | 113 Nh | 114 Fl | 115 Mc | 116 Lv | 117 Ts | 118 Og |
| |
| | | | | 57 La | 58 Ce | 59 Pr | 60 Nd | 61 Pm | 62 Sm | 63 Eu | 64 Gd | 65 Tb | 66 Dy | 67 Ho | 68 Er | 69 Tm | 70 Yb |
| | | | | 89 Ac | 90 Th | 91 Pa | 92 U | 93 Np | 94 Pu | 95 Am | 96 Cm | 97 Bk | 98 Cf | 99 Es | 100 Fm | 101 Md | 102 No |
Goldschmidt classification: Lithophile Siderophile Chalcophile Atmophile Trace/Synthetic
Production
[edit]
Approximately 100 million metric tons of oxygen are produced yearly. Oxygen is most commonly produced by fractional distillation, in which air is cooled to a liquid, then warmed, allowing all the components of air except for oxygen to turn to gases and escape. Fractionally distilling air several times can produce 99.5% pure oxygen. Another method with which oxygen is produced is to send a stream of dry, clean air through a bed of molecular sieves made of zeolite, which absorbs the nitrogen in the air, leaving 90 to 93% pure oxygen.
Sulfur can be mined in its elemental form, although this method is no longer as popular as it used to be. In 1865 a large deposit of elemental sulfur was discovered in the U.S. states of Louisiana and Texas, but it was difficult to extract at the time. In the 1890s, Herman Frasch came up with the solution of liquefying the sulfur with superheated steam and pumping the sulfur up to the surface. These days sulfur is instead more often extracted from oil, natural gas, and tar.
The world production of selenium is around 1500 metric tons per year, out of which roughly 10% is recycled. Japan is the largest producer, producing 800 metric tons of selenium per year. Other large producers include Belgium (300 metric tons per year), the United States (over 200 metric tons per year), Sweden (130 metric tons per year), and Russia (100 metric tons per year). Selenium can be extracted from the waste from the process of electrolytically refining copper. Another method of producing selenium is to farm selenium-gathering plants such as milk vetch. This method could produce three kilograms of selenium per acre, but is not commonly practiced.
Tellurium is mostly produced as a by-product of the processing of copper. Tellurium can also be refined by electrolytic reduction of sodium telluride. The world production of tellurium is between 150 and 200 metric tons per year. The United States is one of the largest producers of tellurium, producing around 50 metric tons per year. Peru, Japan, and Canada are also large producers of tellurium.
Until the creation of nuclear reactors, all polonium had to be extracted from uranium ore. In modern times, most isotopes of polonium are produced by bombarding bismuth with neutrons. Polonium can also be produced by high neutron fluxes in nuclear reactors. Approximately 100 grams of polonium are produced yearly. All the polonium produced for commercial purposes is made in the Ozersk nuclear reactor in Russia. From there, it is taken to Samara, Russia for purification, and from there to St. Petersburg for distribution. The United States is the largest consumer of polonium.
All livermorium is produced artificially in particle accelerators. The first successful production of livermorium was achieved by bombarding curium-248 atoms with calcium-48 atoms. As of 2011, roughly 25 atoms of livermorium had been synthesized.
Applications
[edit]
Metabolism is the most important source and use of oxygen. Minor industrial uses include Steelmaking (55% of all purified oxygen produced), the chemical industry (25% of all purified oxygen), medical use, water treatment (as oxygen kills some types of bacteria), rocket fuel (in liquid form), and metal cutting.
Most sulfur produced is transformed into sulfur dioxide, which is further transformed into sulfuric acid, a very common industrial chemical. Other common uses include being a key ingredient of gunpowder and Greek fire, and being used to change soil pH. Sulfur is also mixed into rubber to vulcanize it. Sulfur is used in some types of concrete and fireworks. 60% of all sulfuric acid produced is used to generate phosphoric acid. Sulfur is used as a pesticide (specifically as an acaricide and fungicide) on "orchard, ornamental, vegetable, grain, and other crops."
Around 40% of all selenium produced goes to glassmaking. 30% of all selenium produced goes to metallurgy, including manganese production. 15% of all selenium produced goes to agriculture. Electronics such as photovoltaic materials claim 10% of all selenium produced. Pigments account for 5% of all selenium produced. Historically, machines such as photocopiers and light meters used one-third of all selenium produced, but this application is in steady decline.
Tellurium suboxide, a mixture of tellurium and tellurium dioxide, is used in the rewritable data layer of some CD-RW disks and DVD-RW disks. Bismuth telluride is also used in many microelectronic devices, such as photoreceptors. Tellurium is sometimes used as an alternative to sulfur in vulcanized rubber. Cadmium telluride is used as a high-efficiency material in solar panels.
Some of polonium's applications relate to the element's radioactivity. For instance, polonium is used as an alpha-particle generator for research. Polonium alloyed with beryllium provides an efficient neutron source. Polonium is also used in nuclear batteries. Most polonium is used in antistatic devices. Livermorium does not have any uses whatsoever due to its extreme rarity and short half-life.
Organochalcogen compounds are involved in the semiconductor process. These compounds also feature into ligand chemistry and biochemistry. One application of chalcogens themselves is to manipulate redox couples in supramolecular chemistry (chemistry involving non-covalent bond interactions). This application leads on to such applications as crystal packing, assembly of large molecules, and biological recognition of patterns. The secondary bonding interactions of the larger chalcogens, selenium and tellurium, can create organic solvent-holding acetylene nanotubes. Chalcogen interactions are useful for conformational analysis and stereoelectronic effects, among other things. Chalcogenides with through bonds also have applications. For instance, divalent sulfur can stabilize carbanions, cationic centers, and radical. Chalcogens can confer upon ligands (such as DCTO) properties such as being able to transform Cu(II) to Cu(I). Studying chalcogen interactions gives access to radical cations, which are used in mainstream synthetic chemistry. Metallic redox centers of biological importance are tunable by interactions of ligands containing chalcogens, such as methionine and selenocysteine. Also, chalcogen through-bonds[dubious – discuss] can provide insight about the process of electron transfer.
Biological role
[edit]
Main articles: Dioxygen in biological reactions, Sulfur cycle, and Selenium in biology
Oxygen is needed by almost all organisms for the purpose of generating ATP. It is also a key component of most other biological compounds, such as water, amino acids and DNA. Human blood contains a large amount of oxygen. Human bones contain 28% oxygen. Human tissue contains 16% oxygen. A typical 70-kilogram human contains 43 kilograms of oxygen, mostly in the form of water.
All animals need significant amounts of sulfur. Some amino acids, such as cysteine and methionine contain sulfur. Plant roots take up sulfate ions from the soil and reduce it to sulfide ions. Metalloproteins also use sulfur to attach to useful metal atoms in the body and sulfur similarly attaches itself to poisonous metal atoms like cadmium to haul them to the safety of the liver. On average, humans consume 900 milligrams of sulfur each day. Sulfur compounds, such as those found in skunk spray often have strong odors.
All animals and some plants need trace amounts of selenium, but only for some specialized enzymes. Humans consume on average between 6 and 200 micrograms of selenium per day. Mushrooms and brazil nuts are especially noted for their high selenium content. Selenium in foods is most commonly found in the form of amino acids such as selenocysteine and selenomethionine. Selenium can protect against heavy metal poisoning.
Tellurium is not known to be needed for animal life, although a few fungi can incorporate it in compounds in place of selenium. Microorganisms also absorb tellurium and emit dimethyl telluride. Most tellurium in the blood stream is excreted slowly in urine, but some is converted to dimethyl telluride and released through the lungs. On average, humans ingest about 600 micrograms of tellurium daily. Plants can take up some tellurium from the soil. Onions and garlic have been found to contain as much as 300 parts per million of tellurium in dry weight.
Polonium has no biological role, and is highly toxic on account of being radioactive.
Toxicity
[edit]
| NFPA 704 safety square | |
--- |
| 2 0 0 Fire diamond for selenium | |
Oxygen is generally nontoxic, but oxygen toxicity has been reported when it is used in high concentrations. In both elemental gaseous form and as a component of water, it is vital to almost all life on Earth. Despite this, liquid oxygen is highly dangerous. Even gaseous oxygen is dangerous in excess. For instance, sports divers have occasionally drowned from convulsions caused by breathing pure oxygen at a depth of more than 10 meters (33 feet) underwater. Oxygen is also toxic to some bacteria. Ozone, an allotrope of oxygen, is toxic to most life. It can cause lesions in the respiratory tract.
Sulfur is generally nontoxic and is even a vital nutrient for humans. However, in its elemental form it can cause redness in the eyes and skin, a burning sensation and a cough if inhaled, a burning sensation and diarrhoea and/or catharsis if ingested, and can irritate the mucous membranes. An excess of sulfur can be toxic for cows because microbes in the rumens of cows produce toxic hydrogen sulfide upon reaction with sulfur. Many sulfur compounds, such as hydrogen sulfide (H2S) and sulfur dioxide (SO2) are highly toxic.
Selenium is a trace nutrient required by humans on the order of tens or hundreds of micrograms per day. A dose of over 450 micrograms can be toxic, resulting in bad breath and body odor. Extended, low-level exposure, which can occur at some industries, results in weight loss, anemia, and dermatitis. In many cases of selenium poisoning, selenous acid is formed in the body. Hydrogen selenide (H2Se) is highly toxic.
Exposure to tellurium can produce unpleasant side effects. As little as 10 micrograms of tellurium per cubic meter of air can cause notoriously unpleasant breath, described as smelling like rotten garlic. Acute tellurium poisoning can cause vomiting, gut inflammation, internal bleeding, and respiratory failure. Extended, low-level exposure to tellurium causes tiredness and indigestion. Sodium tellurite (Na2TeO3) is lethal in amounts of around 2 grams.
Polonium is dangerous as an alpha particle emitter. If ingested, polonium-210 is a million times as toxic as hydrogen cyanide by weight; it has been used as a murder weapon in the past, most famously to kill Alexander Litvinenko. Polonium poisoning can cause nausea, vomiting, anorexia, and lymphopenia. It can also damage hair follicles and white blood cells. Polonium-210 is only dangerous if ingested or inhaled because its alpha particle emissions cannot penetrate human skin. Polonium-209 is also toxic, and can cause leukemia.
Amphid salts
[edit]
Amphid salts was a name given by Jons Jacob Berzelius in the 19th century for chemical salts derived from the 16th group of the periodic table which included oxygen, sulfur, selenium, and tellurium. The term received some use in the early 1800s but is now obsolete. The current term in use for the 16th group is chalcogens.
See also
[edit]
Chalcogenide
Gold chalcogenides
Halogen
Interchalcogen
Pnictogen
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^ Fluck, E. (1988). "New Notations in the Periodic Table" (PDF). Pure Appl. Chem. 60 (3): 431–436. doi:10.1351/pac198860030431. S2CID 96704008. Archived (PDF) from the original on October 10, 2022. Retrieved November 25, 2013.
^ a b Jensen, William B. (1997). "A Note on the Term "Chalcogen"" (PDF). Journal of Chemical Education. 74 (9): 1063. Bibcode:1997JChEd..74.1063J. doi:10.1021/ed074p1063. Archived from the original (PDF) on October 29, 2013. Retrieved November 25, 2013.
^ "Oxysalt - Define Oxysalt at Dictionary.com". Dictionary.reference.com. Retrieved November 25, 2013.
^ "Amphigen – definition of Amphigen by the Free Online Dictionary, Thesaurus and Encyclopedia". Thefreedictionary.com. Retrieved November 25, 2013.
^ Harper, Douglas. "Online Etymology Dictionary". Retrieved November 25, 2013.
^ Krebs, Robert E. (2006). The History And Use of Our Earth's Chemical Elements: A Reference Guide. Greenwood Publishing Group. pp. 223–. ISBN 978-0-313-33438-2. Retrieved November 25, 2013.
^ Jensen, William B. (1997). "A Note on the Term "Chalcogen"". Journal of Chemical Education. 74 (9): 1063. Bibcode:1997JChEd..74.1063J. doi:10.1021/ed074p1063.
^ Stark, Anne M (May 2012). "Livermorium and Flerovium join the periodic table of elements". Archived from the original on February 19, 2013. Retrieved November 25, 2013.
^ a b c Galan, Mark (1992). Structure of Matter. International Editorial Services Inc. ISBN 978-0-8094-9662-4.
^ a b c Pellant, Chris (1992). Rocks and Minerals. Dorling Kindserley handbooks. ISBN 978-0-7513-2741-0.
^ a b Heiserman, Davis L. (1992). "The 10 Most Abundant Elements in the Universe". Archived from the original on January 29, 2013. Retrieved February 6, 2013.
^ a b c d Winter, Mark (1993). "Abundance in the universe". Archived from the original on January 17, 2013. Retrieved February 6, 2013.
^ Amethyst Galleries (1995). "Sulfates". Retrieved November 25, 2013.
^ Amethyst Galleries (1995). "Tellurates". Retrieved November 25, 2013.
^ Advameg (2013). "Tellurium". Chemistry Explained. Retrieved November 25, 2013.
^ Sodhi, G. S. (2000). Fundamental Concepts of Environmental Chemistry. Alpha Science International. ISBN 978-1-84265-281-7.
^ Rubin, Kenneth H. "Lecture 34 Planetary Accretion" (PDF). University of Hawaii. Retrieved January 16, 2013.
^ "Commercial production and use". Encyclopædia Britannica. 2013. Retrieved November 25, 2013.
^ Callaghan, R. (2011). "Selenium and Tellurium Statistics and Information". United States Geological Survey. Archived from the original on May 8, 2012. Retrieved November 25, 2013.
^ a b "Polonium-210". International Atomic Energy Agency. 1998. Archived from the original on January 26, 2012. Retrieved February 11, 2013.
^ le Couteur, Penny (2003). Napoleon's Buttons. Penguin Books. ISBN 978-1-58542-331-6.
^ a b Roberts, James R.; Reigart, J. Routt (2013). "Other Insecticides and Acaracides" (PDF). Recognition and Management of Pesticide Poisonings (6th ed.). Washington DC: Office of Pesticide Programs, U.S. Environmental Protection Agency. p. 93.
^ a b Winter, Mark (1993). "Selenium:Biological information". Retrieved November 25, 2013.
^ Menzel, D.B. (1984). "Ozone: an overview of its toxicity in man and animals". Journal of Toxicology and Environmental Health. 13 (2–3): 183–204. Bibcode:1984JTEH...13..181M. doi:10.1080/15287398409530493. PMID 6376815.
^ "Sulfur General Fact Sheet". npic.orst.edu. Retrieved January 23, 2019.
^ Extension Toxicology Network (September 1995). "Sulfur". Retrieved November 25, 2013.
^ College of Veterinary Medicine, Iowa State University (2013). "Sulfur Toxicity". Retrieved November 25, 2013.
^ Nutall, Kern L. (2006). "Evaluating Selenium Poisoning". Annals of Clinical and Laboratory Science. 36 (4): 409–20. PMID 17127727.
^ Jefferson, R.D.; Goans R.E.; Blain, P.G.; Thomas, S.H. (2009). "Diagnosis and treatment of polonium poisoning". Clinical Toxicology. 47 (5): 379–92. doi:10.1080/15563650902956431. PMID 19492929. S2CID 19648471.
^ Freemantle, Michael (November 28, 2012). "Yasser Arafat, Polonium Poisoning and the Curies". Retrieved November 25, 2013.
^ Lee, J. D. (January 3, 2008). Concise Inorganic Chemistry, 5TH ed. Wiley India Pvt. Limited. ISBN 978-81-265-1554-7.
^ Jensen, William B. (September 1, 1997). "A Note on the Term "Chalcogen"". Journal of Chemical Education. 74 (9): 1063. Bibcode:1997JChEd..74.1063J. doi:10.1021/ed074p1063. ISSN 0021-9584.
External links
[edit]
Media related to Periodic table group 16 at Wikimedia Commons
| v t e Periodic table | |
--- |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | 1 | 2 | | | | | | | | | | | | | | | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | | 1 | H | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | He | | 2 | Li | Be | | | | | | | | | | | | | | | | | | | | | | | | | B | C | N | O | F | Ne | | 3 | Na | Mg | | | | | | | | | | | | | | | | | | | | | | | | | Al | Si | P | S | Cl | Ar | | 4 | K | Ca | | | | | | | | | | | | | | | Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn | Ga | Ge | As | Se | Br | Kr | | 5 | Rb | Sr | | | | | | | | | | | | | | | Y | Zr | Nb | Mo | Tc | Ru | Rh | Pd | Ag | Cd | In | Sn | Sb | Te | I | Xe | | 6 | Cs | Ba | La | Ce | Pr | Nd | Pm | Sm | Eu | Gd | Tb | Dy | Ho | Er | Tm | Yb | Lu | Hf | Ta | W | Re | Os | Ir | Pt | Au | Hg | Tl | Pb | Bi | Po | At | Rn | | 7 | Fr | Ra | Ac | Th | Pa | U | Np | Pu | Am | Cm | Bk | Cf | Es | Fm | Md | No | Lr | Rf | Db | Sg | Bh | Hs | Mt | Ds | Rg | Cn | Nh | Fl | Mc | Lv | Ts | Og | | |
| | | | | | --- --- | | s-block | f-block | d-block | p-block | | |
| v t e Periodic table | |
--- |
| Periodic table forms | Alternatives Extended periodic table |
| Sets of elements | | | | | | | | | | --- --- --- --- | | By periodic table structure | | | | --- | | Groups | 1 (Hydrogen and alkali metals) 2 (Alkaline earth metals) 3 4 5 6 7 8 9 10 11 12 13 (Triels) 14 (Tetrels) 15 (Pnictogens) 16 (Chalcogens) 17 (Halogens) 18 (Noble gases) | | Periods | 1 2 3 4 5 6 7 8+ + Aufbau + Fricke + Pyykkö | | Blocks | Aufbau principle | | | By metallicity | | | | --- | | Metals | Lanthanides Actinides Transition metals Post-transition metals | | Metalloids | Lists of metalloids by source Dividing line | | Nonmetals | Noble gases | | | Other sets | Platinum-group metals (PGM) Rare-earth elements Refractory metals Precious metals Coinage metals Noble metals Heavy metals Native metals Transuranium elements Superheavy elements Major actinides Minor actinides | |
| Elements | | | | --- | | Lists | By: Abundance (in humans) Atomic properties Nuclear stability Symbol | | Properties | Aqueous chemistry Crystal structure Electron configuration Electronegativity Goldschmidt classification Term symbol | | Data pages | Abundance Atomic radius Boiling point Critical point Density Elasticity Electrical resistivity Electron affinity Electron configuration Electronegativity Hardness Heat capacity Heat of fusion Heat of vaporization Ionization energy Melting point Oxidation state Speed of sound Thermal conductivity Thermal expansion coefficient Vapor pressure | |
| History | Element discoveries + Dmitri Mendeleev + 1871 table + 1869 predictions Naming + etymology + controversies + for places + for people + in East Asian languages |
| See also | IUPAC + nomenclature + systematic element name Trivial name Dmitri Mendeleev |
| Category WikiProject | |
| v t e Chalcogens | |
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| | | | | | | | | --- --- --- | | Oxygen O Atomic Number: 8 Atomic Weight: 15.9994 Melting Point: 50.5 K Boiling Point: 90.20 K Specific mass: 0.001429 g/cm3 Electronegativity: 3.04 | Sulfur S Atomic Number: 16 Atomic Weight: 32.065 Melting Point: 388.51 K Boiling Point: 717.8 K Specific mass: 2.067 g/cm3 Electronegativity: 2.58 | Selenium Se Atomic Number: 34 Atomic Weight: 78.96 Melting Point: 494.15 K Boiling Point: 958 K Specific mass: 4.809 g/cm3 Electronegativity: 2.55 | Tellurium Te Atomic Number: 52 Atomic Weight: 127.60 Melting Point: 722.8 K Boiling Point: 1261 K Specific mass: 6.232 g/cm3 Electronegativity: 2.1 | Polonium Po Atomic Number: 84 Atomic Weight: Melting Point: 527.15 K Boiling Point: 1235 K Specific mass: 9.32 g/cm3 Electronegativity: 2 | Livermorium Lv Atomic Number: 116 Atomic Weight: Melting Point: ? K Boiling Point: ? K Specific mass: ? 12.9 g/cm3 Electronegativity: ? | | |
| v t e | |
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| Mixed oxidation states | Antimony tetroxide (Sb2O4) Boron suboxide (B12O2) Carbon suboxide (C3O2) Chlorine perchlorate (Cl2O4) Chloryl perchlorate (Cl2O6) Cobalt(II,III) oxide (Co3O4) Dichlorine pentoxide (Cl2O5) Iron(II,III) oxide (Fe3O4) Lead(II,IV) oxide (Pb3O4) Manganese(II,III) oxide (Mn3O4) Mellitic anhydride (C12O9) Praseodymium(III,IV) oxide (Pr6O11) Silver(I,III) oxide (Ag2O2) Terbium(III,IV) oxide (Tb4O7) Tribromine octoxide (Br3O8) Triuranium octoxide (U3O8) |
| +1 oxidation state | Aluminium(I) oxide (Al2O) Copper(I) oxide (Cu2O) Caesium monoxide (Cs2O) Dibromine monoxide (Br2O) Dicarbon monoxide (C2O) Dichlorine monoxide (Cl2O) Gallium(I) oxide (Ga2O) Iodine(I) oxide (I2O) Lithium oxide (Li2O) Mercury(I) oxide (Hg2O) Nitrous oxide (N2O) Potassium oxide (K2O) Rubidium oxide (Rb2O) Silver oxide (Ag2O) Thallium(I) oxide (Tl2O) Sodium oxide (Na2O) Water (hydrogen oxide) (H2O) |
| +2 oxidation state | Aluminium(II) oxide (AlO) Barium oxide (BaO) Berkelium monoxide (BkO) Beryllium oxide (BeO) Boron monoxide (BO) Bromine monoxide (BrO) Cadmium oxide (CdO) Calcium oxide (CaO) Carbon monoxide (CO) Chlorine monoxide (ClO) Chromium(II) oxide (CrO) Cobalt(II) oxide (CoO) Copper(II) oxide (CuO) Dinitrogen dioxide (N2O2) Disulfur dioxide (S2O2) Europium(II) oxide (EuO) Germanium monoxide (GeO) Iron(II) oxide (FeO) Iodine monoxide (IO) Lead(II) oxide (PbO) Magnesium oxide (MgO) Manganese(II) oxide (MnO) Mercury(II) oxide (HgO) Nickel(II) oxide (NiO) Nitric oxide (NO) Niobium monoxide (NbO) Palladium(II) oxide (PdO) Phosphorus monoxide (PO) Polonium monoxide (PoO) Protactinium monoxide (PaO) Radium oxide (RaO) Silicon monoxide (SiO) Strontium oxide (SrO) Sulfur monoxide (SO) Thorium monoxide (ThO) Tin(II) oxide (SnO) Titanium(II) oxide (TiO) Vanadium(II) oxide (VO) Yttrium(II) oxide (YO) Zirconium monoxide (ZrO) Zinc oxide (ZnO) |
| +3 oxidation state | Actinium(III) oxide (Ac2O3) Aluminium oxide (Al2O3) Americium(III) oxide (Am2O3) Antimony trioxide (Sb2O3) Arsenic trioxide (As2O3) Berkelium(III) oxide (Bk2O3) Bismuth(III) oxide (Bi2O3) Boron trioxide (B2O3) Californium(III) oxide (Cf2O3) Cerium(III) oxide (Ce2O3) Chromium(III) oxide (Cr2O3) Cobalt(III) oxide (Co2O3) Curium(III) oxide (Cm2O3) Dinitrogen trioxide (N2O3) Dysprosium(III) oxide (Dy2O3) Einsteinium(III) oxide (Es2O3) Erbium(III) oxide (Er2O3) Europium(III) oxide (Eu2O3) Gadolinium(III) oxide (Gd2O3) Gallium(III) oxide (Ga2O3) Gold(III) oxide (Au2O3) Holmium(III) oxide (Ho2O3) Indium(III) oxide (In2O3) Iron(III) oxide (Fe2O3) Lanthanum oxide (La2O3) Lutetium(III) oxide (Lu2O3) Manganese(III) oxide (Mn2O3) Neodymium(III) oxide (Nd2O3) Nickel(III) oxide (Ni2O3) Phosphorus trioxide (P4O6) Praseodymium(III) oxide (Pr2O3) Promethium(III) oxide (Pm2O3) Rhodium(III) oxide (Rh2O3) Samarium(III) oxide (Sm2O3) Scandium oxide (Sc2O3) Terbium(III) oxide (Tb2O3) Thallium(III) oxide (Tl2O3) Thulium(III) oxide (Tm2O3) Titanium(III) oxide (Ti2O3) Tungsten(III) oxide (W2O3) Vanadium(III) oxide (V2O3) Ytterbium(III) oxide (Yb2O3) Yttrium(III) oxide (Y2O3) |
| +4 oxidation state | Americium dioxide (AmO2) Berkelium(IV) oxide (BkO2) Bromine dioxide (BrO2) Californium dioxide (CfO2) Carbon dioxide (CO2) Carbon trioxide (CO3) Cerium(IV) oxide (CeO2) Chlorine dioxide (ClO2) Chromium(IV) oxide (CrO2) Curium(IV) oxide (CmO2) Dinitrogen tetroxide (N2O4) Germanium dioxide (GeO2) Iodine dioxide (IO2) Iridium dioxide (IrO2) Hafnium(IV) oxide (HfO2) Lead dioxide (PbO2) Manganese dioxide (MnO2) Molybdenum dioxide (MoO2) Neptunium(IV) oxide (NpO2) Nitrogen dioxide (NO2) Niobium dioxide (NbO2) Osmium dioxide (OsO2) Platinum dioxide (PtO2) Plutonium(IV) oxide (PuO2) Polonium dioxide (PoO2) Praseodymium(IV) oxide (PrO2) Protactinium(IV) oxide (PaO2) Rhenium(IV) oxide (ReO2) Rhodium(IV) oxide (RhO2) Ruthenium(IV) oxide (RuO2) Selenium dioxide (SeO2) Silicon dioxide (SiO2) Sulfur dioxide (SO2) Technetium(IV) oxide (TcO2) Tellurium dioxide (TeO2) Terbium(IV) oxide (TbO2) Thorium dioxide (ThO2) Tin dioxide (SnO2) Titanium dioxide (TiO2) Tungsten(IV) oxide (WO2) Uranium dioxide (UO2) Vanadium(IV) oxide (VO2) Xenon dioxide (XeO2) Zirconium dioxide (ZrO2) |
| +5 oxidation state | Antimony pentoxide (Sb2O5) Arsenic pentoxide (As2O5) Bismuth pentoxide (Bi2O5) Dinitrogen pentoxide (N2O5) Diuranium pentoxide (U2O5) Neptunium(V) oxide (Np2O5) Niobium pentoxide (Nb2O5) Phosphorus pentoxide (P2O5) Protactinium(V) oxide (Pa2O5) Tantalum pentoxide (Ta2O5) Tungsten pentoxide (W2O5) Vanadium(V) oxide (V2O5) |
| +6 oxidation state | Chromium trioxide (CrO3) Molybdenum trioxide (MoO3) Polonium trioxide (PoO3) Rhenium trioxide (ReO3) Selenium trioxide (SeO3) Sulfur trioxide (SO3) Tellurium trioxide (TeO3) Tungsten trioxide (WO3) Uranium trioxide (UO3) Xenon trioxide (XeO3) |
| +7 oxidation state | Dichlorine heptoxide (Cl2O7) Manganese heptoxide (Mn2O7) Rhenium(VII) oxide (Re2O7) Technetium(VII) oxide (Tc2O7) |
| +8 oxidation state | Iridium tetroxide (IrO4) Osmium tetroxide (OsO4) Ruthenium tetroxide (RuO4) Xenon tetroxide (XeO4) Hassium tetroxide (HsO4) |
| Related | Oxocarbon Suboxide Oxyanion Ozonide Peroxide Superoxide Oxypnictide |
| Oxides are sorted by oxidation state. Category:Oxides | |
| v t e Sulfides (S2−) | |
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| | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- | H2S | | | | | | | | | | | | | | | | | | He | | Li2S | BeS | B2S3 +BO3 | CS2 COS | (NH4)SH | O | F | Ne | | Na2S | MgS | Al2S3 | SiS SiS2 -Si | PxSy -P | -S2− 2 | Cl | Ar | | K2S | CaS | | ScS Sc2S3 | TiS TiS2 Ti2S3 | VS VS2 V2S3 | CrS Cr2S3 | MnS MnS2 | FeS Fe3S4 | CoxSy | NixSy | Cu2S CuS | ZnS | GaS Ga2S3 | GeS GeS2 -Ge | As2S3 As2S5 As4S3 -As | SeS2 +Se | Br | Kr | | Rb2S | SrS | | Y2S3 | ZrS2 | NbS2 | MoS2 MoS3 | TcS2 Tc2S7 | Ru | Rh2S3 | PdS | Ag2S | CdS | In2S3 | SnS SnS2 -Sn | Sb2S3 Sb2S5 -Sb | TeS2 | I | Xe | | Cs2S | BaS | | LuS Lu2S3 | HfS2 | TaS2 | WS2 WS3 | ReS2 Re2S7 | OsS 4 | Ir2S3 IrS2 | PtS PtS2 | Au2S Au2S3 | HgS | Tl2S | PbS PbS2 | Bi2S3 | PoS | At | Rn | | Fr | Ra | | Lr | Rf | Db | Sg | Bh | Hs | Mt | Ds | Rg | Cn | Nh | Fl | Mc | Lv | Ts | Og | | | | | | | | | | | | | | | | | | | | | | | | | | LaS La2S3 | CeS Ce2S3 | PrS Pr2S3 | NdS Nd2S3 | PmS Pm2S3 | SmS Sm2S3 | EuS Eu2S3 | GdS Gd2S3 | TbS Tb2S3 | DyS Dy2S3 | HoS Ho2S3 | ErS Er2S3 | TmS Tm2S3 | YbS Yb2S3 | | | Ac2S3 | ThS Th2S3 Th7S12 ThS2 | PaS2 | US U2S3 U3S5 US2 | NpS Np2S3 Np3S5 NpS2 | PuS Pu2S3 | AmS Am2S3 | CmS Cm2S3 | Bk2S3 | Cf2S3 | Es | Fm | Md | No | | |
| v t e Salts and covalent derivatives of the selenide ion | |
--- |
| | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- | H2Se H2Se2 +H -H | | | | | | | | | | | | | | | | | | He | | Li2Se | Be | SexByOz | CSe2 OCSe (CH3)2Se | (NH4)2Se | O | F | Ne | | Na2Se | MgSe | Al2Se3 | Si | PxSey -P | +S | Cl | Ar | | K2Se | CaSe | | Sc2Se3 | TiSe2 | V | CrSe Cr2Se3 | MnSe MnSe2 | FeSe | CoSe | NiSe | Cu2Se CuSe | ZnSe | GaSe Ga2Se3 -Ga | GeSe GeSe2 -Ge | As2Se3 As4Se3 | Se2− n | Br | Kr | | Rb2Se | SrSe | | Y2Se3 | Zr | NbSe2 NbSe3 | MoSe2 | Tc | Ru | Rh | Pd | Ag2Se | CdSe | InSe In2Se3 | SnSe SnSe2 -Sn | Sb2Se3 | Te | +I | Xe | | Cs2Se | BaSe | | LuSe Lu2Se3 | Hf | TaSe2 | WSe2 WSe3 | ReSe2 | Os | Ir | PtSe2 | Au | HgSe | Tl2Se | PbSe | Bi2Se3 | Po | At | Rn | | Fr | Ra | | Lr | Rf | Db | Sg | Bh | Hs | Mt | Ds | Rg | CnSe | Nh | Fl | Mc | Lv | Ts | Og | | | | | | | | | | | | | | | | | | | | | | | | | | LaSe La2Se3 | CeSe Ce2Se3 | PrSe Pr2Se3 | NdSe Nd2Se3 | Pm | SmSe Sm2Se3 | EuSe Eu2Se3 | GdSe Gd2Se3 | TbSe Tb2Se3 | DySe Dy2Se3 | HoSe Ho2Se3 | ErSe Er2Se3 | TmSe Tm2Se3 | YbSe Yb2Se3 | | | Ac | ThSe2 | Pa | USe2 | Np | PuSe | Am | Cm | Bk | Cf | Es | Fm | Md | No | | |
| v t e Salts and covalent derivatives of the telluride ion | |
--- |
| | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- | H2Te -TeH | | | | | | | | | | | | | | | | | | He | | Li2Te | BeTe | B | CTe2 (CH3)2Te | (NH4)2Te | O | F | Ne | | Na2Te | MgTe | Al2Te3 | Si | P0.8Te0.2 | S | Cl | Ar | | K2Te | CaTe | | Sc2Te3 | Ti | VTe2 | CrTe Cr2Te3 | MnTe MnTe2 | FeTe | CoTe | NiTe | Cu2Te CuTe CuTe2 | ZnTe | GaTe Ga2Te3 -Ga | GeTe -Ge | As2Te3 As4Te3 +As | Se | +Br | Kr | | Rb2Te | SrTe | | Y2Te3 | ZrTe5 | NbTe2 | MoTe2 | Tc | Ru | Rh | Pd | Ag2Te | CdTe | In2Te3 | SnTe SnTe2 | Sb2Te3 | Te2- Te2- n | I | Xe | | Cs2Te | BaTe | | LuTe Lu2Te3 | HfTe5 | TaTe2 | WTe2 WTe3 | ReTe2 | Os | Ir | Pt | AuxTey | HgTe | Tl2Te | PbTe | Bi2Te3 | Po | At | Rn | | Fr | RaTe | | Lr | Rf | Db | Sg | Bh | Hs | Mt | Ds | Rg | Cn | Nh | Fl | Mc | Lv | Ts | Og | | | | | | | | | | | | | | | | | | | | | | | | | | LaTe La2Te3 | CeTe Ce2Te3 | PrTe Pr2Te3 | NdTe Nd2Te3 | Pm | SmTe Sm2Te3 | EuTe Eu2Te3 | GdTe Gd2Te3 | TbTe Tb2Te3 | DyTe Dy2Te3 | HoTe Ho2Te3 | ErTe Er2Te3 | TmTe Tm2Te3 | YbTe Yb2Te3 | | | Ac | ThTe2 | Pa | UTe2 | Np | Pu | Am | Cm | Bk | Cf | Es | Fm | Md | No | | |
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8950 | https://tasks.illustrativemathematics.org/content-standards/tasks/938 | Illustrative Mathematics
Typesetting math: 100%
Engage your students with effective distance learning resources. ACCESS RESOURCES>>
The Locker Game
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Alignments to Content Standards:4.OA.B.4
Student View
Task
The 20 students in Mr. Wolf's 4th grade class are playing a game in a hallway that is lined with 20 lockers in a row.
The first student starts with the first locker and goes down the hallway and opens all the lockers.
The second student starts with the second locker and goes down the hallway and shuts every other locker.
The third student stops at every third locker and opens the locker if it is closed or closes the locker if it is open.
The fourth student stops at every fourth locker and opens the locker if it is closed or closes the locker if it is open.
This process continues until all 20 students in the class have passed through the hallway.
Which lockers are still open at the end of the game? Explain your reasoning.
Which lockers were touched by only two students? Explain your reasoning.
Which lockers were touched by only three students? Explain your reasoning.
Which lockers were touched the most?
IM Commentary
The purpose of this instructional task is for students to deepen their understanding of factors and multiples of whole numbers. This is a classic mathematical puzzle; often it is stated in terms of 100 lockers, so students have to make the connection to factors and multiples to solve it. In this version, students can just go through all the rounds of opening and closing locker doors and observe after the fact that there is a relationship between the factors a locker number has and whether it is open or closed at the end.
This task provides students with an excellent opportunity to engage in MP7, Look for and make use of structure (if they see early on that there is a relationship with factors and multiples) or MP8, Look for and express regularity in repeated reasoning (if they start to see and describe the pattern as they imagine students opening and closing the lockers).
Because the total number of lockers is only 20, students might answer the questions without thinking about the underlying reasons for their answers. In the first question for example, a student might say "1, 4, 9, 16 are all still open because I tried it out and those were the ones that were left open." If a student goes this route, the teacher can steer the conversation back to factors by asking the students if they notice anything special about the numbers of the lockers which are still open. Asking students to try a larger number of lockers (say 50 or 100) and repeating the game can also help students look for a pattern since going through all the rounds of the game becomes less and less feasible as the number of lockers increases. Once students see the pattern, they should be pressed to explain why it works out this way.
Solution
The lockers that are still open are perfect squares, specifically 1×1, 2×2, 3×3, and 4×4. To see why this is true, first note that:
The 1st student changes all the lockers with a locker number that is a multiple of 1.
The 2nd student changes all the lockers with a locker number that is a multiple of 2.
The 3rd student changes all the lockers with a locker number that is a multiple of 3.
The 4th student changes all the lockers with a locker number that is a multiple of 4.
and so on. So a locker is touched by a student if that student's number is a factor of that locker's number.
Now consider the factors of 16 which are 1, 2, 4, 8, and 16. This means the 16th locker was opened by the 1st student, then it was closed by the 2nd student, then it was opened by the 4th student, then it was closed by the second 8th student, and then opened one last time by the 16th student. Now note that the square numbers are the only numbers with an odd number of factors (all other numbers have factors which come in pairs.) Since an odd number of factors means the locker will be changed an odd number of times, it must be open at the end.
With only 20 lockers, students can simply observe this fact about the number of factors to be true for the numbers 1 to 20, which gives practice with finding factors of whole numbers. The explanation for why this is always true goes a bit beyond 4th grade.
The lockers that were touched by only two students are the prime numbers (since prime numbers only have two factors.) In our case they are 2, 3, 5, 7, 11, 13, 17, and 19.
The lockers which are touched by only three students are the squares of prime numbers (because these are the only numbers which have three factors.) In our case they are 4 and 9. (For example, 4 has three factors: 1, 2, and 4)
The lockers which are touched the most are those lockers whose number has the most factors. In our case 12, 18, and 20 all have 6 factors, so they were each touched 6 times. 12=1×12=2×6=3×4 18=1×18=2×9=3×6 20=1×20=2×10=4×5
The Locker Game
The 20 students in Mr. Wolf's 4th grade class are playing a game in a hallway that is lined with 20 lockers in a row.
The first student starts with the first locker and goes down the hallway and opens all the lockers.
The second student starts with the second locker and goes down the hallway and shuts every other locker.
The third student stops at every third locker and opens the locker if it is closed or closes the locker if it is open.
The fourth student stops at every fourth locker and opens the locker if it is closed or closes the locker if it is open.
This process continues until all 20 students in the class have passed through the hallway.
Which lockers are still open at the end of the game? Explain your reasoning.
Which lockers were touched by only two students? Explain your reasoning.
Which lockers were touched by only three students? Explain your reasoning.
Which lockers were touched the most?
Print Task
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8951 | https://www.nature.com/articles/d41586-024-03352-y | Mumps is rising in some nations — but a fresh dose of vaccine might help
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FallInClouds
10 Apr 2023
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Question about English (US)
How should I understand 'the most' followed by a plural noun?
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Such as 'one of the most beautiful cities.'
Because what I've learned tells me 'the most' shows the superlative, so it should be single.
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Gener1c_
10 Apr 2023
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The noun isn’t modified by “the most” the adjective is. The most BEAUTIFUL city. It’s called a superlative, it means the highest degree of something. The most amazing person. The most incredible view. The most hilarious story. and so on.
The noun isn’t modified by “the most” the adjective is.
The most BEAUTIFUL city. It’s called a superlative, it means the highest degree of something.
The most amazing person.
The most incredible view.
The most hilarious story.
and so on.
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FallInClouds
10 Apr 2023
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@Gener1c_ I'm sorry I didn't make it clear. I've seen some phrases like 'one of the most beautiful cities'. It is very weird for me. Say A is the most beautiful city and A is one of the most beautiful cities. It sounds like there is other city that is the most beautiful city too.
@Gener1c_ I'm sorry I didn't make it clear.
I've seen some phrases like 'one of the most beautiful cities'. It is very weird for me.
Say A is the most beautiful city and A is one of the most beautiful cities. It sounds like there is other city that is the most beautiful city too.
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You’re misunderstanding what’s communicated it seems, “one of the most_____” implies there are other very beautiful ones. “Beijing and Hong Kong are two of the most beautiful cities in the world” “Berlin, Milan, Paris and New York are some of the most beautiful cities in the world.” = Not all cities meet the criteria but these do.
You’re misunderstanding what’s communicated it seems, “one of the most_____” implies there are other very beautiful ones.
“Beijing and Hong Kong are two of the most beautiful cities in the world”
“Berlin, Milan, Paris and New York are some of the most beautiful cities in the world.”
=
Not all cities meet the criteria but these do.
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FallInClouds
10 Apr 2023
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@Gener1c_ So 'most' means very or extremely not the greatest degree, right?
@Gener1c_ So 'most' means very or extremely not the greatest degree, right?
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Milan is the most beautiful city (in the whole world). = There is no city more beautiful than this one. Milan is one of the most beautiful cities. = There are others which are equally as beautiful.
Milan is the most beautiful city (in the whole world).
=
There is no city more beautiful than this one.
Milan is one of the most beautiful cities.
=
There are others which are equally as beautiful.
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@MaybeURRight it depends, see my other response
@MaybeURRight it depends, see my other response
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@Gener1c_ it explains a lot! I get it now. Thank you for your time.
@Gener1c_ it explains a lot! I get it now. Thank you for your time.
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8953 | https://www.mathway.com/popular-problems/Algebra/201711 | Algebra Examples
x2−2x−1=0
Step 1
Use the quadratic formula to find the solutions.
−b±√b2−4(ac)2a
Step 2
Substitute the values a=1, b=−2, and c=−1 into the quadratic formula and solve for x.
2±√(−2)2−4⋅(1⋅−1)2⋅1
Step 3
Tap for more steps...
Step 3.1
Simplify the numerator.
Tap for more steps...
Step 3.1.1
Raise −2 to the power of 2.
x=2±√4−4⋅1⋅−12⋅1
Step 3.1.2
Multiply −4⋅1⋅−1.
Tap for more steps...
Step 3.1.2.1
Multiply −4 by 1.
x=2±√4−4⋅−12⋅1
Step 3.1.2.2
Multiply −4 by −1.
x=2±√4+42⋅1
x=2±√4+42⋅1
Step 3.1.3
Add 4 and 4.
x=2±√82⋅1
Step 3.1.4
Rewrite 8 as 22⋅2.
Tap for more steps...
Step 3.1.4.1
Factor 4 out of 8.
x=2±√4(2)2⋅1
Step 3.1.4.2
Rewrite 4 as 22.
x=2±√22⋅22⋅1
x=2±√22⋅22⋅1
Step 3.1.5
Pull terms out from under the radical.
x=2±2√22⋅1
x=2±2√22⋅1
Step 3.2
Multiply 2 by 1.
x=2±2√22
Step 3.3
Simplify 2±2√22.
x=1±√2
x=1±√2
Step 4
The result can be shown in multiple forms.
Exact Form:
x=1±√2
Decimal Form:
x=2.41421356…,−0.41421356…
x2−2x−1=0
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Natural number which can be expressed as sum of two perfect squares in two different ways?
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Modified 2 years, 5 months ago
Viewed 29k times
$\begingroup$
Ramanujan's number is $1729$ which is the least natural number which can be expressed as the sum of two perfect cubes in two different ways. But can we find a number which can be expressed as the sum of two perfect squares in two different ways. One example I got is $50$ which is $49+1$ and $25+25$. But here second pair contains same numbers. Does any one have other examples ?
number-theory
elementary-number-theory
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edited Apr 23, 2017 at 4:55
Martin Sleziak
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asked Dec 22, 2016 at 4:57
MadhuMadhu
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$\endgroup$
4
1
$\begingroup$ You may want to read mathworld.wolfram.com/SumofSquaresFunction.html $\endgroup$
Anurag A
– Anurag A
2016-12-22 05:00:34 +00:00
Commented Dec 22, 2016 at 5:00
1
$\begingroup$ Do you count something like $3^2 + 4^2 = 5^2 + 0^2$? If so, any pythagorean triplet gives an example. $\endgroup$
Stahl
– Stahl
2016-12-22 05:24:35 +00:00
Commented Dec 22, 2016 at 5:24
1
$\begingroup$ math.stackexchange.com/questions/153603/… $\endgroup$
individ
– individ
2016-12-22 05:25:46 +00:00
Commented Dec 22, 2016 at 5:25
4
$\begingroup$ If we allow the case where one square is zero, and also allow the case where two of the squares are the same (your $7^2+1^2=5^2+5^2$), then we get A118882. $\endgroup$
Jeppe Stig Nielsen
– Jeppe Stig Nielsen
2018-03-10 20:21:39 +00:00
Commented Mar 10, 2018 at 20:21
Add a comment |
13 Answers 13
Reset to default
17
$\begingroup$
$$ 65 = 64 + 1 = 49 + 16 $$
This will work for any number that's the product of two primes each of which is congruent to $1$ mod $4$. For more than two ways multiply more than two such primes.
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answered Dec 22, 2016 at 5:01
Ethan BolkerEthan Bolker
105k77 gold badges127127 silver badges223223 bronze badges
$\endgroup$
Add a comment |
17
$\begingroup$
Note that $a^2 + b^2 = c^2 + d^2$ is equivalent to $a^2 - c^2 = d^2 - b^2$, i.e. $(a-c)(a+c) = (d-b)(d+b)$. If we factor any odd number $m$ as $m = uv$, where $u$ and $v$ are both odd and $u < v$, we can write this as $m = (a-c)(a+c)$ where $a = (u+v)/2$ and $c = (v-u)/2$. So any odd number with more than one factorization of this type gives an example.
Thus from $m = 15 = 1 \cdot 15 = 3 \cdot 5$, we get $8^2 - 7^2 = 4^2 - 1^2$, or $1^2 + 8^2 = 4^2 + 7^2$.
From $m = 21 = 1 \cdot 21 = 3 \cdot 7$ we get $11^2 - 10^2 = 5^2 - 2^2$, or $2^2 + 11^2 = 5^2 + 10^2$.
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answered Dec 22, 2016 at 5:26
Robert IsraelRobert Israel
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$\endgroup$
1
4
$\begingroup$ For another example, instead of $m$ odd, we can take $m$ as a multiple of $8$. Then $m$ can again be written as a product $m=uv$ where $u$ and $v$ has the same parity, in two ways. For $m=40$, for example, first $$m=40=4\cdot 10=(7-3)(7+3)=7^2-3^2$$ and secondly $$m=40=2\cdot 20=(11-9)(11+9)=11^2-9^2$$ and we get the example $$7^2+9^2=3^2+11^2.$$ $\endgroup$
Jeppe Stig Nielsen
– Jeppe Stig Nielsen
2018-03-10 20:17:14 +00:00
Commented Mar 10, 2018 at 20:17
Add a comment |
8
$\begingroup$
The following example easily generalizes:
$$\begin{align} 5&=(2+i)(2-i)=4+1\ 13&=(3+2i)(3-2i)=9+4\ 5\cdot13&=((2+i)(3+2i))((2-i)(3-2i))=(4+7i)(4-7i)=16+49\ &=((2+i)(3-2i))((2-i)(3+2i))=(8-i)(8+i)=64+1 \end{align}$$
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answered Dec 22, 2016 at 5:42
Barry CipraBarry Cipra
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$2465$ can be expressed as the sum of two squares in four different ways:- $8^2 + 49^2$, $16^2 + 47^2$, $23^2 + 44^2$ and $28^2 + 41^2$.
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edited Mar 5, 2018 at 14:43
user1729
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answered Mar 5, 2018 at 13:21
Peter HartPeter Hart
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$\begingroup$ Why the two downvotes? This answer seems fine...what am I missing?! $\endgroup$
user1729
– user1729
2018-03-05 14:45:21 +00:00
Commented Mar 5, 2018 at 14:45
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There are many numbers that can be expressed as the sum of two squares in more than one way. For example, $$ 65=64+1 =49+16$$ $$85=81+4 =49+36$$ $$125=121+4 =100+25$$ $$130=121+9 =81+49$$ $$145=144+1 =64+81$$ $$170=169+1 =121+49$$ $$185=169+16 =121+64$$ and so on... You can also read this PDF for more details. Hope it helps.
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answered Dec 22, 2016 at 5:12
user371838user371838
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$\begingroup$ Why do You write the number? If there are so many good formulas. $\endgroup$
individ
– individ
2016-12-22 05:27:44 +00:00
Commented Dec 22, 2016 at 5:27
$\begingroup$ Are all these numbers expressible as the sum of two squares in two different ways divisible by $5$? (Oh jeez, that question would not be great to ask in one breath.)..... Edit: Nope, because $5\nmid 629$. $\endgroup$
Mr Pie
– Mr Pie
2018-02-25 10:54:35 +00:00
Commented Feb 25, 2018 at 10:54
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$\begingroup$ @MrPie, it looks that way because the sum needs to have prime factors that are 4k+1 primes, as noted in EthanBolker's answer above. The first of these is 5, and so the lowest not divisible by 5 should be 1317 =221 =15^2+110^2 =2^2+111^2 $\endgroup$
Ze'ev misses Monica
– Ze'ev misses Monica
2021-05-12 16:39:30 +00:00
Commented May 12, 2021 at 16:39
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The BrahmaguptaFibonacci identity says that every product of two sums of two squares is a sum of two squares in two different ways: \begin{align} (a^2+b^2)(c^2+d^2) & = (ac+bd)^2 + (ad-bc)^2 \[6pt] & = (ac-bd)^2 + (ad+bc)^2 \end{align} For example: \begin{align} (2^2+3^2)(1^2+7^2) & = 23^2 + 11^2 \[6pt] & = 19^2 + 17^2 \end{align}
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answered Jul 11, 2020 at 22:26
Michael HardyMichael Hardy
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Well, as I much as I can think of, we have at least one class of examples in $$\boxed{125k^2=(11k)^2+(2k)^2=(10k)^2+(5k)^2} \,\,\,\,\,\,\,\,\, \text{for } \,\,\,\, k\in \mathbb{N}$$
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edited Dec 22, 2016 at 5:07
answered Dec 22, 2016 at 5:01
SchrodingersCatSchrodingersCat
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I too got one, but without the above proof. 629 = 23^2 + 10^2 = 25^2 + 2^2
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answered Oct 8, 2017 at 10:14
P VELUP VELU
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$\begingroup$ Put a dollar sign $\$$ at the beginning and end of your equation to form the following: $$629 = 23^2 + 10^2 = 25^2 + 2^2.$$ By the way, it is also equal to $9^3 - 10^2$ and just one off $14^3 - 46^2$ (fun fact). $\endgroup$
Mr Pie
– Mr Pie
2018-02-25 10:56:49 +00:00
Commented Feb 25, 2018 at 10:56
$\begingroup$ @Mr Pie $628 = 14^3 - 46^2$ , not $629$ $\endgroup$
Maths-Lover
– Maths-Lover
2020-06-13 04:44:44 +00:00
Commented Jun 13, 2020 at 4:44
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In the same way we can also write 650 as the sum of the squares of two prime numbers in two different ways i.e $650=11^2+23^2=17^2+19^2$ since $19^2-11^2=23^2-17^2$
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answered Apr 23, 2017 at 4:24
user439522user439522
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Product of any two primes of the type (4k+1) will do the trick.. Product of any three primes of the type (4k+1) and you have 4 different ways etc..
Basically - (all easy to prove) A prime of the type p=(4k+1) has unique $a^2+b^2 = p$ solution. (Proven by Fermat) A prime of the type p= 4k+3 has NO solution. And for a product of two primes, say $p=a^2+b^2$, and $q= c^2+d^2$, we have,
$pq = (ac+bd)^2+(ad-bc)^2 = (ac-bd)^2 +(ad+bc)^2$ .
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edited Jan 12, 2019 at 18:39
answered Jan 12, 2019 at 3:05
Gyan MehtaGyan Mehta
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$\begingroup$ Please use MathJax to format. $\endgroup$
Ѕᴀᴀᴅ
– Ѕᴀᴀᴅ
2019-01-12 03:15:07 +00:00
Commented Jan 12, 2019 at 3:15
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The lowest integer that is the sum of two integer squares in two different ways is 50, but that case involves one repeat number 5^2 + 5^2 = 25 + 25 = 50 = 7^2 + 1. The lowest integer that is the sum of two integer squares in two different ways with all different numbers is 65. The lowest integer that is the sum of THREE integer squares in three different ways is 325. The lowest number that is the sum of FOUR integer squares in four different ways is 1105. Interestingly, the lowest integer that is the sum of SIX integer squares in six different ways is lower than the lowest integer that is the sum of FIVE integer squares in five different ways (you can work all these out for yourselves !).
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answered May 2, 2019 at 3:08
Tim JayTim Jay
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First of all , the trivial case is that all Pythagorean Triplets come in this list (if you accept doing $0^2$, of course) .
Here are some more examples :-
$$125 = 5^2 + 10^2 = 11^2 + 2^2$$ $$145 = 8^2 + 9^2= 12^2 + 1^2$$ $$170 = 13^2 + 1^2 = 11^2 + 7^2$$ Also multiply each number by some $n^2$ , and multiply each of it's $2$ pairs of squares with $n$ , to get more such numbers.
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edited Jun 13, 2020 at 3:30
answered Jun 10, 2020 at 14:49
Maths-LoverMaths-Lover
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Any Pythagorean triple $(A^2+B^2=C^2)$ provides a candidate where the $C^2$ can have $2$-or-more combinations of $A^2$ and $B^2$ that add up to it. These can be found by testing natural numbers of the form $(4n+1)$ with a range of $m$ values defined as shown below to see which, if any yield integers. We begin with Euclid's formula and solve the $C$-function for $n$ in terms of $C$ and $m$. $$A=m^2-n^2\quad B=2mn\quad C=m^2+n^2$$
$$C=m^2+n^2\implies n=\sqrt{C-m^2}\ \text{where}\qquad \bigg\lfloor\frac{ 1+\sqrt{2C-1}}{2}\bigg\rfloor \le m \le \big\lfloor\sqrt{C-1}\space\big\rfloor$$ The lower limit ensures $m>n$ and the upper limit ensures $n\in\mathbb{N}$. $$C=65\implies\ \bigg\lfloor\frac{ 1+\sqrt{130-1}}{2}\bigg\rfloor=6 \le m \le \big\lfloor\sqrt{65-1}\big\rfloor=8\ \text{ and we find} \quad m\in{7,8}\Rightarrow n\in{4,1}\$$ $$F(7,4)=(33,56,65)\quad F(8,1)=(63,16,65) $$
Here, we have $\space 33^2+56^2=65^2\space$ and $\space 63^2+16^2=65^2$
There infinitely many of these pairs of triples; the next that meet the criteria are
$$(13^2,84^2,85^2),\qquad (77^2,36^2,85^2)\ (75^2,100^2,125^2),\qquad (117^2,44^2,125^2)\(17^2,144^2,145^2),\qquad (143^2,24^2,145^2)\ ...$$
$\textbf{Edit:}$ If $\space n \space$ is the number of distinct prime fators of $\space C,\space$ there are $2^{n-1}$ primitive triples with that same $C$-value. This means e.g. that for $1105=5\times13\times17,\space$ there are $2^{3-1}=4$ primitives with $C=1105.$
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edited Nov 10, 2021 at 21:27
answered Jun 13, 2020 at 14:21
poetasispoetasis
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8957 | https://www.youtube.com/watch?v=QkT5G2IFMR8 | Grade 10 Math: Manipulating algebraic fractions part 1
MishDoesMath
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A review of manipulating fractions and simplifying algebraic fractions
Transcript:
in today's lesson we are going to be learning how to manipulate algebraic fractions and before we go into that i just want to go over some of the rules of manipulating fractions in general and a lot of this is going to be review from your previous grades let's say we had two quarters multiplied by three fifths if you recall from your rules of multiplying fractions when we have a fraction multiplied by another fraction we can simply perform that by multiplying the numerators and multiplying the denominators so that is going to be simplified to 6 over 20. and when we have something like this that contains a common factor in both the numerator and the denominator we can actually cancel out those common factors and simplify this further so let me actually write down 6 over 20 in another way 6 over 20 can be expressed as 3 times 2 over 10 times 2 and now we have 2 on both the top and the bottom which can cancel out and we're going to be left with 3 10. and that is our most simplified form of this product here now let's say you had to calculate two quarters plus three quarters when we're adding fractions if we have the same denominator we are simply going to add our numerators together and they are going to have that common denominator so this is going to be five over four now what if you had two quarters divided by three fifths well when we are dividing a fraction that is going to be the same thing as multiplying by the reciprocal so another way of writing down this expression here is to say two quarters multiplied by five over three so because here we were dividing by three over five that is the same thing as multiplying by the reciprocal which is 5 over 3. and now we can do this the same way that we did this top example here we're just going to multiply 2 and 5 which is 10 and we're going to multiply 4 and 3 which is 12 and that will give us 10 over 12. and again we can recognize that the numerator and the denominator have a common factor and that common factor is 2 so we can write this down as 5 times 2 over 6 times 2 and we have a 2 at the top and bottom which can cancel each other out and we're left with 5 6 and that is our most simplified form now what if we had an expression like this let's say we had two-thirds plus one sixth well we know that when we're adding fractions we can only add fractions together when we have a common denominator so this number at the bottom here has to be the same for us to just be able to add the top and write down a single denominator so we can actually convert these into a form where both of them have 6 as a denominator by simply multiplying this fraction here by two over two so if we multiply two thirds by two over two that's the same thing as just multiplying this by one because two divided by two is one so that would mean that this is going to become 4 over 6 and now we have the same denominator of 6. so we can add these two together now we have 4 over 6 plus 1 over 6 and that is going to be equal to five over six and that is our most simplified form now that we've reviewed how we manipulate fractions let's go into algebraic fractions now let's say we were to have the following expression 6x plus 10 divided by 9x plus 15. so we can recall from a few videos ago that in both the numerator and the denominator we have a common factor in each of these terms so let's first deal with the numerator as its own kind of expression and then we can deal with the denominator by itself and then we can see what we have so if we're just dealing with the numerator here let's take out a common factor of 2. so if we take out 2 we're going to be left with 3x plus 5 and now if we are just dealing with the denominator if we take out a common factor of 3 we're going to be left with 3x plus 5. and again if you don't recall when we're taking out a factor we can figure out what our new coefficients are by just dividing by whatever we factored out so 9 divided by 3 is 3 and 15 divided by 3 is 5 and the same goes for what we did with the numerator here and now what you might be able to notice is that we have a common factor in both the numerator and the denominator and we can cross this out and we're going to be left with two thirds and that is the final answer so this expression here can be simplified to two thirds now let's do another example let's say we have x squared minus 25 divided by 3x plus 15. now whenever you're dealing with expressions like this you're going to try to factorize both the numerator and the denominator denominator and see if there are any terms that you can cross out on both so what we can first recognize here is that we have a difference of two squares x squared is a perfect square and 25 is also a perfect square and we're taking the difference between them and we can recall from a few videos ago that when we're trying to factor a difference of two squares we can simply write that down as x plus five and x minus five and just to recap what we learned when we were doing the difference of two squares we're just going to take the square root of each of these terms and we're going to have a bracket where we add them and a bracket where we subtract them and now if we try to factor our denominator we can take out a common factor of three and we're going to be left with x plus five and now we can see that we have an x plus 5 in the numerator and an x plus 5 in the denominator we can cancel these like terms out and we are going to be left with x minus 5 over 3 again because we have our 3 here and we're left with our x minus 5 here and one more thing that we need to add to this final answer here is that we need to add a condition here and that condition is that x cannot be equal to negative five and the reason that we have to include that is because with this expression here we're trying to say that this here and this are the same thing and in this case this expression is going to be undefined when x is equal to negative five because when x is equal to negative five this denominator is going to be equal to zero and we know that when we're dividing by zero our answer is undefined so our expression here is undefined at negative 5. but this expression here is not undefined at negative 5 which is why we have to add this note here that x cannot be equal to negative 5 and that means that these two expressions are equivalent let's go over one more example let's say we have x squared plus 4x plus 3 divided by x squared plus x minus 6. now let's use the same approach that we have been using let's treat the numerator and denomina denominator separately step by step so let's try to factor the numerator first and we can see that this has a leading coefficient of one so we can use our initial method that we had used and we're trying to find two numbers that are going to multiply to give us three and that are going to add to give us four so let's look at our factors of three we have one and three so one times three is going to give us three and three plus one will give us four so these are going to be our two numbers here so we can write this down as x plus one and x plus 3 and that is going to be our factored form of this numerator now let's try to factor our denominator let's do it down here so we can use the same approach that we just used we're trying to find two numbers that when multiplied together are going to give us negative six and when added together are going to give us one and because this is negative we know that one of our numbers has to be negative and one has to be positive so already we can write down x plus and x minus because we know one is going to be positive and one will be negative and if we're trying to write down our factors of six we have one and six and we have two and three and immediately we can recognize that if we have three minus two that is going to give us one so we know that our three is going to be positive and our two is going to be negative because positive 3 multiplied by negative 2 is going to give us negative 6 and positive 3 minus 2 is going to give us 1. so this is our factored form of our denominator so now let's write down our fraction as x plus 1 x plus 3 that's just what we got here for our numerator divided by x plus 3 x minus 2 and again that's just the factored form of our denominator and we can see that we have a common factor here we can cross out this x plus 3 because it's present in both the numerator and the denominator and we're going to be left with x plus one over x minus two that's just what we have left over here and this is going to be our simplified and final answer for this expression here again what we have to remember is that we need to add a condition in here and that condition is that x cannot be equal to negative three and that is because if we look at our fraction here and this is the same thing as this if we're just putting our numerator and denominator into their factored forms we know that in this expression this is going to be undefined when x is equal to negative 3 because if x is equal to negative 3 here negative 3 plus 3 is equal to 0 and if we have a 0 here multiplied by whatever we have in here that is going to make this entire fraction divided by zero which is undefined so this is going to be undefined at negative three but this is not undefined at negative three this is only going to be undefined at two so to make these two equivalent we need to add the condition that x cannot be equal to negative three and now these are equivalent and in addition x also cannot be equal to two because again if we have two minus two that is zero and that would make this whole thing undefined |
8958 | https://www.math.emory.edu/~rg/P147Y12.pdf | The Edge Spectrum of K4-Saturated Graphs Kinnari Amina, Jill Faudreeb, Ronald Gouldc a Dept. of Math, CS and Eng., Georgia Perimeter College, Clarkston, GA 30021 kinnari.v.amin@gmail.com b Dept. of Math and Stat, University of Alaska Fairbanks, Fairbanks, AK 99709 c Dept. of Math and CS, Emory University, Atlanta, GA 30322 Abstract Any H-free graph G is called H-saturated if the addition of any edge e / ∈E(G) results in H as a subgraph of G. The minimum size of an H-saturated graph on n vertices is denoted by sat(n, H). The edge spectrum for the family of graphs with property P is the set of all sizes of graphs with property P. In this paper, we find the edge spectrum of K4-saturated graphs. We also show that if G is a K4-saturated graph, then either G ∼ = K1,1,n−2 or δ(G) ≥3, and we show the exact structure of a K4-saturated graph with κ(G) = 2 and κ(G) = 3.
1 Introduction All graphs in this paper are simple graphs, namely, finite graphs without loops or multiple edges. Notation will be standard, and generally follow the notation of . Let G be a graph, and let V (G) and E(G) denote the vertex set and the edge set of G, respectively. Let d(x) denote the degree of the vertex x, |V (G)| denote the order of the graph G, |E(G)| denote the size of the graph G, and d(u, v) is the distance between u and v. If W is a nonempty subset of the vertex set V (G), then the subgraph ⟨W⟩ of G induced by W is the graph having vertex set W and whose edge set consists of those edges of G incident with two vertices of W. Furthermore, 1 |V (G)| = n, unless otherwise specified.
Also, Kp denotes the complete graph on p vertices.
A graph G is called an H-saturated graph if G does not contain H as a subgraph but the addition of any edge e / ∈E(G) produces H as a subgraph of G. The saturation number of a graph is the minimum number of edges in an H-saturated graph of order n and it is denoted by sat(n, H).
This parameter was introduced by Erd˝ os, Hajnal, and Moon in . The maximum number of edges in a H-saturated graph of order n is the well known Tur´ an extremal number and is usually denoted by ex(n, H).
It is known that any K3-saturated graph has at least n−1 edges and at most ⌊n2/4⌋edges and . Furthermore, these bounds are sharp as shown in and . Also, any K4-saturated graph has at least 2n−3 edges and at most ⌊n2/3⌋edges and these bounds are sharp. The emphasis of this paper will be on determining the sizes of K4-saturated graphs of order n, that is the edge spectrum of K4-saturated graphs.
2 Results on K4-saturated Graphs Barefoot et. al studied the edge spectrum of K3-saturated graphs and proved the following result.
Theorem 2.1 Let n ≥5 and m be nonnegative integers.
There is an (n, m) K3-saturated graph if and only if 2n −5 ≤m ≤⌊(n −1)2/4⌋+ 1 or m = k(n −k) for some positive integer k.
This result says that a K3-saturated graph is either a complete bipartite graph or its size falls in the given range and all values in this range are possible.
In this section, we will show a similar result about K4-saturated graphs.
First we make two simple observations shown in the following two Propo-sitions.
Proposition 2.2 Let G be a K4-saturated graph. Then diam(G) = 2.
Proof. Let G be a K4-saturated graph. Suppose diam(G) ̸= 2, that is, suppose there exists vertices u, v ∈V (G) such that d(u, v) = 3. Say u, x, y, v is a u −v distance 3 path. Then the addition of the edge uv must produce a K4. So there must exist vertices a, b such that the induced subgraph ⟨u, a, b, v⟩∼ = K4. But now we have d(u, v) = 2, as u, a, v is such a path, a contradiction.
2 Proposition 2.3 Let G be a K4-saturated graph. Then G is 2-connected.
Proof. Let G be a K4-saturated graph. Suppose G is not 2-connected. Let u be the cut vertex and let A and B be components of G−u with x ∈V (A) and y ∈V (B) (see Figure 1).
x y u Figure 1: G Now the addition of the edge xy creates a K4. So there exists a vertex w ̸= u such that wx ∈E(G) and wy ∈E(G). Then G −u is not discon-nected, a contradiction.
2 Next we show that there is only one K4-saturated graph with minimum degree two.
Theorem 2.4 Let G be a K4-saturated graph. Then either G ∼ = K1,1,n−2 or δ(G) ≥3.
Proof. Let G be a K4-saturated graph. Let v ∈V (G) such that N(v) = {x, y}. Then for all the vertices z ∈V (G)−{x, y, v}, the addition of the edge vz must produce a K4 with vertex set {v, x, y, z}. Thus, xy, xz, yz ∈E(G).
Since z was chosen arbitrarily, in fact, all vertices in V (G) −{x, y, v} are adjacent to both x and y. So K1,1,n−2 ⊆G. But K1,1,n−2 is K4-saturated, so G ∼ = K1,1,n−2.
2 Next we prove three lemmas that lead to a short proof of a result that all K4-saturated graph on n vertices other than K1,1,n−2 have at least 3n −8 edges. Before we prove these lemmas, we make the following two observations about K4-saturated graphs G: 1. The neighborhood of every vertex contains an edge.
2. For all vertices u, v ∈V (G), uv / ∈E(G) if and only if there exists an edge in the common neighborhood of u and v.
Lemma 2.5 If G is a K4-saturated graph of order n and δ(G) = 3, then |E(G)| ≥3n −8.
Proof. Let v ∈V (G) such that N(v) = {x, y, z}.
Let W = V (G) − {v, x, y, z}. Note that the induced graph ⟨x, y, z⟩is not isomorphic to K3.
If ⟨x, y, z⟩contains precisely one edge, say xy, then by observation 2, every vertex w ∈W must also be adjacent to x, y and W is therefore independent. But this would force N(z) to be independent, contradicting observation 1.
If ⟨x, y, z⟩contains precisely two edges, say xy, yz, then by observation 2, W can be partitioned into three sets Wxy, Wyz, Wxyz, where Wxy = {w ∈ W|N(w) ∩{x, y, z} = {x, y}} with Wyz and Wxyz defined similarly (note that Wxz = ∅). Each of these are independent sets of vertices. Furthermore, if w ∈Wxyz, then it has no adjacencies to Wxy or Wyz. Let w1 ∈Wxy and w2 ∈Wyz. Then N(w1)∩N(w2) = {y}. Thus, by observation 2, they must be adjacent. Hence, Wxy and Wyz must induce a complete bipartite graph.
Let |Wxyz| = h, |Wyz| = l, |Wxy| = n −h −l −4. Then, |E(G)| = 5 + 2l + 2(n −h −l −4) + l(n −h −l −4) + 3h = (2 + l)n + (1 −l)h −l2 −4l −3.
Let f(l) = (2 + l)n + (1 −l)h −l2 −4l −3. Note if l = 0, then we are done by our previous argument. Now f(1) = 3n −8. Furthermore, f is maximized at l = n−h−4 2 and is increasing between l = 1 and l = n−h−4 2 .
Note this is all of the relevant interval for l, since we can assume, without loss of generality, |Wyz| ≤|Wxy|.
2 Lemma 2.6 If G is a K4-saturated graph of order n and δ(G) = 4, then |E(G)| ≥3n −8.
Proof. Let v ∈V (G) such that d(v) = 4 and |V (G)| = n. Let N(v) = A = {a1, a2, a3, a4}. Let W = V (G) −A −{v}. Then there must exist an edge aiaj for some i, j, 1 ≤i, j ≤4 as for any vertex w ∈W, addition of the edge vw must create a K4.
If ⟨A⟩contains precisely one edge, say a1a2, then every vertex w ∈W must also be adjacent to a1, a2 by observation 2 above and W is therefore independent. But this would force N(a3) and N(a4) to be independent, contradicting observation 1. So ⟨A⟩must contain at least two edges. So now assume that there are precisely two edges among the vertices of A.
Then we have two possibilities as shown in the Figure 2.
a v a a a a a a a or v 3 2 1 4 2 3 4 1 Figure 2: G Pick the vertices ai and aj in different components of ⟨N(v)⟩.
The addition of the edge aiajmust produce a K4 with an edge in W, say w1w2.
But each vertex wi, for i = 1, 2, must be adjacent to an edge in A, or else the addition of the edge vwi would not produce a K4. Hence, w1 and w2 are adjacent to at least three vertices of A.
The count below comes from counting the degree sum in the following parts: (1) the degree sum in v∪A, (2) the degree sum from A to W, (3) the degree sum of vertices in W. In the last instance we use the assumption that δ(G) = 4.
X x∈V (G) d(x) ≥ (4 + 4 + 4) + (2(n −5) + 2) + (4(n −5)) = 6n −16.
Observe that if there are three edges among the vertices of A, we also obtain the same count. Hence, |E(G)| ≥3n −8.
2 Lemma 2.7 If G is a K4-saturated graph and δ(G) = 5, then |E(G)| ≥ 3n −8.
Proof. Note that |E(G)| ≥5n 2 and 5n 2 ≥3n −8 for n ≤16.
Let v ∈V (G) with d(v) = 5 and |V (G)| = n ≥17. Let N(v) = A = {a1, a2, a3, a4, a5}. Let W = V (G)−A−{v}. We know there must exist an edge aiaj for some i, j. In fact, as in Lemma 2.6, there must exist at least two edges. Furthermore, for any vertex w ∈W, wai and waj must exist for some i, j, 1 ≤i, j ≤5. Also, two of the vertices in W must be adjacent to at least 3 vertices in A, as shown in Lemma 2.6. Since δ(G) = 5 and counting as we did in the previous lemma, we have X x∈V (G) d(x) ≥ (5 + 5 + 4) + (2(n −6) + 2) + (5(n −6)) = 7n −26 So, |E(G)| ≥7n−26 2 ≥3n −8 for n ≥10.
2 Theorem 2.8 Every 2-connected K4-saturated graph of order n with δ(G) ≥3 has at least 3n −8 edges.
Proof. From Lemmas 2.5 -2.7, the result holds for graphs G with 3 ≤ δ(G) ≤5. For graphs G with δ(G) ≥6, |E(G)| ≥6n 2 = 3n.
2 In the following two theorems, we classify all K4-saturated graphs G with κ(G) = 2 or κ(G) = 3.
Theorem 2.9 If G is a K4-saturated graph of order n with κ(G) = 2, then G ∼ = K1,1,n−2.
Proof. Let G be a K4-saturated with κ(G) = 2. Let K be a minimal cut set of G and let C1, C2 be distinct components of G −K. Let xi ∈V (Ci) for i = 1, 2. Then the addition of the edge x1x2 produces a K4. So if u, v ∈K, then ⟨x1, x2, u, v⟩∼ = K4. As x1 (x2) was an arbritrary vertex of C1 (C2), each vertex in C1 (C2) is also adjacent to u and v. So G −K is an independent set and the result is shown.
2 Now we will classify all K4-saturated graphs with κ(G) = 3. But first, we define the graph W(a, b, c, d, e) to be a wheel on 5 sets totaling n vertices such that a + b + c + d + e = n −1 and each of the 5 sets of the wheel are independent sets of sizes a, b, c, d, e, respectively, and two consecutive independent sets on the wheel form a complete bipartite subgraph. For example, W(1, 3, 2, 1, 2) on 10 vertices is shown in Figure 3.
Figure 3: W(1, 2, 1, 3, 2) Theorem 2.10 If G is a K4-saturated graph of order n with κ(G) = 3, then G ∼ = K1,2,n−3 or G ∼ = W(1, t, 1, r, s).
Proof. Let K be a minimal cut set of G, say {v1, v2, v3}. Let C1, C2 be distinct components of G −K. Let xi ∈V (Ci) for i = 1, 2. Then addi-tion of the edge x1x2 produces a K4. Without loss of generality, suppose ⟨x1, x2, v1, v2⟩∼ = K4. If all the components of G−K are trivial, then all are adjacent to all three vertices of K by the connectivity assumption. Now one of (but not both) v1 or v2 is adjacent to v3 (say v2) or else inserting the edge v2v3 would not produce a K4. But now G is K1,2,n−3.
We now assume C1 is nontrivial. Let x3 be a neighbor of x1 in C1. In-serting the edge x2x3, we know (without loss of generality) ⟨x2, x3, v2, v3⟩∼ = K4, since x3 cannot be adjacent to both v1 and v2.
If there exists a vertex x4 ∈V (C2) such that x2x4 ∈E(G), then by a similar argu-ment x4v1, x4v2, x4v3 ∈E(G).
Hence, ⟨x2, x4, v1, v2⟩∼ = K4.
Hence, V (C2) = {x2}. In fact G −{K ∪C1} is an independent set. For every w ∈V (C1), N(w) ∩K = {v1, v2} or N(w) ∩K = {v2, v3} since w must be adjacent to an edge in K but cannot be in a K4. Now partition C1 into two classes A and B, where vertices of A are adjacent to v1 and v2, while vertices of B are adjacent to v2 and v3. Clearly, A is an independent set.
Similarly, B is an independent set.
We claim ⟨A ∪B⟩is a complete bipartite graph. Let a ∈A, b ∈B such that ab / ∈E(G). Then addition of the edge ab must produce a K4. Since A and B are independent sets, the edge they have in common has to be in K, a contradiction. Hence, ⟨A ∪B⟩is complete bipartite. Thus, if |A| = s and |B| = r, using v2 as the center of the wheel, we have G ∼ = W(1, t, 1, r, s), where 3 + t + r + s = n.
2 In , Hanson and Toft gave the following construction. The graphs in the family T k n are on n vertices consisting of a complete (k −1)-partite graph on n−2 vertices, with classes of independent points C1, C2, ..., Ck−1, together with two adjacent vertices x and y and where each vertex of C1 is joined to precisely one of x or y, x and y are each adjacent to at least one vertex of C1, no vertex of C2 is adjacent to either x or y and all vertices of Ci, i > 2 are adjacent to both x and y. For k ≥3, define T ′ k−1,n to be graphs in T k n for which |C1| + 1, |C2| + 2, |C3|, ..., |Ck−1| are equal or as equal as possible. For n ≥3k −4 we can describe T ′ k−1,n as follows: let n + 1 = t(k −1) + r, 0 ≤r < k −1 and let G denote a member of T k n0 on n0 = n −r vertices and e0 = e(Tk−1,n−r) −(t −2) edges where the classes Ci satisfy |C1| = t −1, |C2| = t −2 and |Ci| = t, i > 2 (G is unique up to adjacencies of x and y to class C1). Define T ′ k−1,n to be a graph G with one vertex added to precisely r of the classes C1, ..., Ck−1. Note that the graphs T ′ k−1,n are maximal, with respect to the number of edges, in the family T k n . Then Hanson and Toft showed the following result.
Theorem 2.11 Let G be a maximal Kk-saturated graph on n ≥k +2 ≥5 vertices with χ(G) ≥k, then G is a T ′ k−1,n graph.
Theorem 2.12 If G is a K4-saturated graph of order n and G is not com-plete tripartite, then |E(G)| ≤n2−n+4 3 .
Proof. Let G be a K4-saturated graph of order n. Suppose G is not a complete tripartite graph. Since G is not tripartite, χ(G) ≥4 = k. Hence, by Theorem 2.11, |E(G)| ≤|E(T ′ 3,n)|.
For n + 1 = 3t + r, a straight forward computation shows, |E(T ′ 3,n)| ≤ n2−n+4 3 .
In fact, when r = 0, |E(T ′ 3,n)| = n2−n+4 3 . Hence, |E(G)| ≤n2−n+4 3 .
2 Theorem 2.13 Let n ≥5 and m be nonnegative integers. There is an (n, m) K4-saturated graph G if and only if 3n −8 ≤m ≤ n2−n+4 3 or m = rs + st + rt for some positive integers r, s, t where n = r + s + t.
Proof. Let n ≥5 and m be nonnegative integers. Let G be an (n, m) K4-saturated graph. If G is a tripartite graph, then G must be a complete tripartite graph, otherwise an edge may be added without creating a K4.
Now if G ∼ = K1,1,n−2, then m = 2n −3 and clearly r = s = 1 while t = n −2, otherwise m = rs + st + rt for some positive integers r, s, t such that n = r + s + t.
Now let G be a nontripartite graph. Then from Theorem 2.8, Theo-rem 2.10, and Theorem 2.12, we have that 3n −8 ≤m ≤n2−n+4 3 .
It is sufficient to construct an (n, m) K4-saturated graph for each value of m. If m = 2n −3, then G ∼ = K1,1,n−2. If m = rs + st + rt for some positive integers r, s, t where n = r + s + t, then G ∼ = Kr,s,t with m edges.
Now if 3n −8 ≤m ≤n2−n+4 3 , then consider an (n, m) K4-saturated graph G ∼ = H + Kq, where H is a K3-saturated graph of order n −q.
From Theorem 2.1, 2(n −q) −5 ≤|E(H)| ≤⌊(n−q−1)2 4 ⌋+ 1.
Hence, 2(n−q)−5+(n−q)q ≤m ≤⌊(n−q−1)2 4 ⌋+1+(n−q)q. When q = 1, we obtain the lower bound on m = 3n −8. Now let f(q) = ⌊(n−q−1)2 4 ⌋+ 1 + (n −q)q.
Then f(q) is maximum when q = n+1 3 and f( n+1 3 ) = n2−n+4 3 .
2 In the above Theorem, the lower and upper bounds are achieved. For example, for the following graphs G1 (Figure 4) and G2 (Figure 5), lower and upper bounds, respectively, are achieved.
G 1 Figure 4: G1 G 2 Figure 5: G2 References C. Barefoot, K. Casey, D. Fisher, K. Fraughnaugh, F. Harary, Size in maximal triangle-free graphs and minimal graphs of diameter 2, Discrete Mathematics, 138 (1995), pp. 93–99.
G. Chartrand, L. Lesniak, Graphs and Diagraphs, Chapman & Hall/CRC, Boca Raton, FL, fourth edition, 2005.
P. Erd˝ os, A. Hajnal, J. W. Moon, A Problem in Graph Theory, Amer.
Math. Monthly, 71 (1964), pp. 1107–1110.
D. Hanson and B. Toft, k-saturated Graphs of Chromatic Number at least k, Ars Combinatoria, 31 (1991), pp. 159–164.
W. Mantel, Problem 28, Wiskundige Opgaven, 10 (1907), pp. 60–61.
P. Tur´ an, On an Extremal Problem in Graph Theory, Mat. Fiz. Lapok, 48 (1941), pp. 436–452. |
8959 | https://www.usgs.gov/educational-resources/find-feature-meander | Find-A-Feature: Meander | U.S. Geological Survey
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A meander is another name for a bend in a river. For this Find-A-Feature challenge, we challenge you to look around you for examples of a meander.
A meander is when water flows in a curvy, bendy path, like a snake. As a river makes its way through an area that is relatively flat, it often develops bends as it erodes its way through the path of least resistance. Once a meander starts, it often becomes more and more exaggerated. Why is this? Water is pushed to the outside of a bend, and erodes the curve further, while water on the inside is slower and deposits sediment. This is why you often see sand bars and beaches on the inside of the curve. Due to erosion on the outside of a bend and deposition on the inside, the shape of a meander changes over time.
Here at USGS, we study how meanders are formed and can even model them to predict how the river may continue to change in the future, to better understand flooding hazards.
Can you find a meander in your area? If you don't see a river,see if you can spot a sidewalk or path that meanders, or watch how a drop of rainwater flows down a slightly sloped surface - does it make a straight or curvy path?Take a photo of a meander and tag us @usgs_yes #findafeature or send it to us at usgs_yes@usgs.gov.
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We'll be watching Instagram and Twitter for some great#findafeature examples and may share them here with the first name or initials of the contributor, and a general location. If you tag us with @USGS_YES you are giving us permission to use your image. Please see the USGS social media sharing policy at: Or, you can e-mail photos to us at usgs_yes@usgs.gov and we may share them on this page or on social media. Thanks for participating and for seeing science all around you!
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8960 | https://www.youtube.com/watch?v=9w2XpdFxHHg | Solving Quadratic Equations by FACTORING | Grade 10 Math | jensenmath.ca
JensenMath
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Posted: 25 Apr 2022
Full lesson for unit 5 lesson 1 - solving quadratic equations by the method of factoring. The steps to follow are:
1) set equal to zero
2) factor
3) set each factor to zero and solve
Go to jensenmath.ca for supporting materials.
Textbook references:
McGraw Hill Ryerson section 6.2 - solving quadratics
Math Power 10 section 5. 2- solving quadratics by factoring
0:00 - intro and definitions
4:59 - example 1a
8:20 - example 1b
10:23 - example 1c
13:17 - example 1d
15:43 - example 1e
18:17 - example 1f
20:27 - example 1g
21:37 - example 1h
22:42 - example 2 application question
11 comments
Transcript:
intro and definitions [Music] let's start the unit on solving quadratic equations there are a few different methods for solving quadratic equations factoring completing the square and quadratic formula as we progress through the unit you'll be able to tell which method is going to be most appropriate for which types of quadratic equations but for this lesson we're going to focus on quadratics that can be solved by the method of factoring and before we get started i should remind you what a quadratic equation is a quadratic equation is an equation meaning it has an equal sign and a left side and a right side of the equation and the highest degree term in the equation is degree two that's what makes it a quadratic so if we have a quadratic equation that can be solved by factoring we're going to follow three steps step number one we're going to write the quadratic in the form ax squared plus bx plus e equals zero notice in that format all of the terms are on one side of the equation and the other side of the equation is just equal to zero after we have all the terms onto one side of the equation step number two tells us to factor the quadratic in the previous unit you've learned lots of factoring skills so hopefully you'll be able to follow along with the different factoring strategies i use while i find the correct product and sums to be able to get a quadratic into factored form step number three after the quadratic has been factored we're going to set each factor to zero and solve setting each factor to zero and solving allows you to find what makes the entire product equal to zero because of the zero product rule and why do we care what makes the product of the factors be zero well as long as we followed step one we know that what we're trying to find the quadratic is equal to is zero so if we get it written as a product of factors and then determine what values of the variable make either of the factors be zero we've essentially solved the equation because we're interested in what makes the equation be zero now i mentioned something called the zero product rule what is the zero product rule the zero product rule states that the product of factors is zero if one or more of the factors in the product are zero for example if a times b equals zero then i know that either a is zero or b is zero or both that's the only way that this product could be equal to zero is if one or both of the factors are equal to zero how does that apply to quadratic equations well for the quadratic equation x squared plus two x minus three equals zero i know that that would factor to x plus three times x minus one equals zero because 3 and negative 1 are the numbers that multiply to negative 3 and add to 2. they satisfy the product and sum we would need to find to factor the quadratic on the left side of the equation once we have a product of things equal to zero the zero product rule says the product of those factors would be zero if either of the factors the x plus three or the x minus one were equal to zero the values of x that would make those factors become zero are well what would make x plus three equal to zero an x value of negative three would make x plus three be zero it would make x minus one be negative four but if i were to plug in negative three for both the x's in my factored form equation i would have 0 times negative 4 which is 0 so that is an answer to the equation and you can also check it in the standard form equation as well to verify but don't forget the zero product rule says that the product would be zero if either of the factors are zero well i know negative three would make the first factor be zero but what would make the second factor be zero well x minus one would be zero if x was one if x was 1 if we sub that into the factored form equation i would have 4 times 0 which of course equals 0. so x equals 1 is also a solution to the equation and like i said you could sub that into the standard form equation as well to verify that it satisfies that equation keep in mind anytime we're solving an equation all we're trying to do is figure out what value of the variable makes the left side of the equation equal to the right side of the equation and before we get started with an example i just made a note here of a couple tips you should remember when trying to factor a quadratic remember to first always check for a common factor also remember that if the leading coefficient in a quadratic trinomial is not equal to one so if the a value is not one we have to factor by the method of decomposition and that will come up as we do examples here so i'll have a chance to remind you of what that looks like all right finally let's get to an example let's example 1a look at example one it says to solve each of the following quadratic equations the highest degree term in all of these equations is degree two that's what makes them quadratic and we want to figure out what value of the variable in this case the variables are all x's so what values of x make the left side of each of these equations equal to the right side step number one if we look at a is going to be making sure that we have one side of the equation equal to zero so make sure all of the terms are moved to one side all of the terms are already moved to the left and the right side is equal to zero so step one is done step two says to factor the quadratic because we have a three term quadratic where the leading coefficient the a value is one i can factor it by finding the numbers that multiply to the c value so in this case multiply to 15 and add to the b value in this case eight the numbers that multiply to 15 and add to eight are five and three five times three is fifteen and five plus three is eight this will allow me to rewrite the left side of the equation in factored form the factored form would be x plus 5 times x plus 3 equals 0. now that i have it in factored form now i that i have a product of two things equal to zero i know that product would be zero if either the first factor was zero so i'll write x plus five equal to zero or if the second factor was equal to zero so i'll write x plus three equals zero and then i'll solve each of those equations i've made to figure out what value of x would make each of those factors be zero if i solve the first equation by moving the plus 5 to the other side i would have x equals negative 5. and if i solve the second equation take the plus three move it to the other side i would have x equals negative three so my answers are x equals negative five and x equals negative three when you have multiple answers for your variable sometimes you'll see a subscript number beside the variable indicating which numbered solution is my first solution was negative five so i'll say x one equals negative five and my second solution was negative 3 so i'll say x 2 equals negative 3. make sure you remember that you can actually check these answers in the original equation to make sure it makes the left side equal to the right side in fact you should always do a left side right side check just to make sure that you have found the solution or solutions to satisfy the original equation and i should note that when solving quadratics you're not always going to get two answers sometimes you'll get one sometimes you'll get zero and we'll talk more about those possibilities as we go through the unit let me actually just quickly show you a left side right side check to verify the answers for a we won't do that for all the questions but i just want to make sure you understand how we would do that notice if i sub negative 5 in for the x's in the original equation it makes the left side of the equation equal to 0 just like the right side of the equation so negative 5 is the correct answer and if i sub negative 3 in for the x's in the original equation it makes the left side equal to 0 just like the right side so that's a correct solution as well let's move on to part b notice for part example 1b b once again all the terms are already to the left side of the equation the right side of the equation is equal to zero so step one is done but this quadratic is only two terms so i'm not going to need to do any product and sum factoring in fact because there's no constant term i notice that both terms have an x so what i should do to get this into factored form is just common factor out an x from both terms on the left side of the equation so i'll take out a common factor of x and then in brackets divide both of the terms by the x i'm common factoring out x squared divided by x is x minus seven x divided by x is seven so the factored form is x multiplied by x minus seven the product of those two factors would equal zero if either of the factors were zero so i'll set each of the factors to zero and solve my first factor is just an x so if i set that to zero that equation is already done what would make x be zero is if x was zero so my first answer is zero right if we think about plugging in zero into my factored form i'd have zero times negative seven that's zero what if my second factor was zero x minus seven would be zero if x was seven if i plug seven into my factored form i'd have 7 times 0 which is also 0. so my second answer is 7. now we should also check those answers into the original equation to make sure it works to satisfy the original equation if i check my first answer of zero this is what it would look like notice that zero satisfies the original equation it makes left side and right side both be zero now let me check seven seven also makes both sides of the equation equal to zero so seven is also a correct answer to the original equation let's move on to part c example 1c notice in part c this is the first time where we've had a quadratic equation where it's not set equal to zero so the first thing we're going to have to do is rearrange this quadratic by moving all terms to one side so that it is equal to 0. what i'm going to do is i'm going to take that constant of 12 that's on the right and subtract it over to the left side of the equation that would give me x squared plus 4x minus 12 is equal to zero and now that the quadratic is equal to zero we'll try and factor it because the quadratic has three terms and the leading coefficient the a value is one i'll need to find two numbers that have a product of the c value negative 12 and a sum of the b value four the numbers that satisfy that product and sum are six and negative two six times negative two is negative twelve six plus negative two is four that would make the factored form version of the quadratic x plus six times x minus two and that factored form quadratic is equal to zero we want it equal to zero because it's very easy to determine when a product is equal to zero a product of two factors is zero if either of the factors is equal to zero so we set each of the factors to zero and then solve within each of those two equations what values of x satisfy those two equations x plus 6 would equal 0 if x was negative 6 so that is my first answer to this equation and my second answer would be what makes x minus two equal to zero would be if x was equal to two so my second answer is x equals two make sure you remember that you could check both of those answers into the original equation both of those answers should make the left side of the equation equal to 12 so that it matches the right side of the equation and i hope at this point you're also fully grasping the zero product rule understanding why negative 6 and 2 are both solutions because those values are what would make either of the factors in the factored form be equal to zero right i know the value of two would not make the first factor be equal to zero it would make it be equal to eight but it would make the second factor be zero so we would have eight times zero which is zero and negative six would make the first factor be zero but not the second it'll make the second factor be negative eight but zero times negative eight is zero so as long as we can find a value of x that makes any of the factors be zero that is an answer to the equation as long as your factored form is set equal to zero which is what we take care of at the beginning of the question let's now move on to part d part d example 1d notice i put a tip off to the right there always check for a common factor it'll make the question a lot easier if we're able to common factor out the leading coefficient of the quadratic i should also mention that what we have here is a quadratic set to zero so all the terms are already moved to one side check for a common factor i can common factor out a two from all of the terms so i'll take out two as the first factor and then divide each of the terms by two to get my second factor that'll give me x squared minus 11x plus 24. and now if i focus inside the brackets i notice it's a quadratic with three terms the leading coefficient the a value is 1. so once again i need to find numbers who have a product of the c value so a product of 24 and a sum of the b value so a sum of negative 11. the numbers that satisfy that product and sum would be negative 8 and negative 3. negative 8 times negative 3 is 24 negative 8 plus negative 3 is negative 11. that would allow me to factor the quadratic that's in brackets i could factor it to x minus eight times x minus three notice that factor of two is just staying out front if we wanted to i suppose we could divide it to the other side to eliminate it but it's not going to hurt anything staying there so now i have a product of three factors equaling zero well the product of those three factors would be zero if any of those three factors were zero so let's look at how could any of those factors be zero well my first factor is two can two be zero no two is a constant it's always two but could x minus eight be equal to zero yes that would be equal to zero if x was equal to eight so my first answer is 8. it would make my factored form say 2 times 0 times 5 which is 0. and if we look at the last factor x minus 3 that would be equal to 0 if x was equal to three so my second answer is three it would make the factored form say two times negative five times zero and that's zero so that's an answer to the equation and once again you should take the time to check those in the original equation to make sure it makes left side equal to right side example 1e okay we have four more quadratics to solve part e all the terms are already on one side and the other side is equal to zero it's a three term quadratic where the leading coefficient is not 1 it's a 2 but this time i can't common factor it out because 2 doesn't go into 5 and 3. so to factor this quadratic we're going to have to do it using the method of decomposition while using that method we find numbers that have a product of not just the c value but of a times c so 2 times negative 3 is negative 6 and a sum of the b value which is 5. the numbers that multiply to negative 6 and add to 5 would be 6 and negative 1. and because the leading coefficient the a value is not 1 what we have to do is split the middle term that 5x we have to split it into 6x minus 1x we split it into those two values 6x and negative 1x because those are the numbers that satisfy the product and sum and now what we do is we look at the first two terms and take out a common factor between the first two terms i could common factor out a 2x when i divide the first two terms by two x i get x plus three i then look at the last two terms my negative 1x and my negative 3 and what i could common factor out from those two terms well when the first term in a group has a negative we definitely have to factor out at least a negative one and in this case that would be the greatest common factor between those two terms when i divide the negative x and the negative 3 by negative 1 that gives me x plus 3. notice i have a common binomial both terms have an x plus 3. so i factor out an x plus 3 from both terms and what i'm left with when i divide both terms by x plus 3 is 2x minus 1. it's now into factored form so i know the product of those factors is 0 if either of the factors is zero so i'll set each of the factors to zero and then solve each of the equations the first equation is easier to solve just move the plus three over we get x equals negative three that's my first solution when solving for the second factor is zero it'll be two steps add the one to the other side and then divide the two so i would get my second answer is x equals a half part f is going to be just like part e example 1f but notice it's not set equal to zero to start so our first step is going to be moving the 20 to the other side of the equation notice it's a three term quadratic the leading coefficient the a value is not one it's a three and we can't common factor out the three so once again we're going to have to do this the long way by decomposition we'll need to find numbers that have a product of a times c so a product of 3 times negative 20 which is negative 60 and a sum of the b value 17. the numbers that satisfy that product and sum are 20 and negative 3. so once again because the leading coefficient was not 1 we have to factor this by the method of decomposition where we split the middle term we split that 17x into 20x minus 3x and then we look at the first two terms and take it a common factor all i could take out is an x from the first two terms and that would leave me with 3x plus 20 and then the last two terms the negative 3x minus 20 all i could take out is a negative 1. and when i divide the last two terms by negative 1 i get 3x plus 20. notice i have a common binomial both terms have a 3x plus 20. so i'll common factor out that 3x plus 20. and when i divide both the terms by that binomial i'm left with x minus 1. now i have a factored form quadratic equal to zero that product would be zero if either of the factors were zero so set three x plus twenty equal to zero and set x minus one equal to zero and we'll solve both of those equations the first one will be two steps subtract the 20 over and then divide the three the second one just add the one over and we get my second answer is x equals one so my answers are negative 20 over 3 and 1. and once again don't forget you can check those answers in the original equation to make sure they're correct example 1g we have two more quadratics to solve before we do a short application question these ones are a bit different because notice they're only two terms they have the degree two term and the constant when we have an equation like this we're actually not going to solve it by factoring it'll be easier for these ones if we actually just try and isolate the x using normal algebra right there's only one term that has an x so let's just use algebra to try and isolate that x in part g i would just start by moving the negative six to the other side it becomes positive six then i'll divide the 2 over 6 divided by 2 is 3 and then i can isolate an x that's being squared by doing the inverse of squaring the inverse of squaring is plus or minus square rooting so i actually get two answers for this my answers for x are positive root three and negative root three now you could i suppose write them as separate answers root three and negative root three but i would be fine if you just left your answer written as plus or minus root three if we look at part h once again it's two example 1h terms the degree two term and the constant when it's that format only one of the terms has an x just use algebra to try and isolate the x i'll start by subtracting the one to the other side then dividing the three and then moving the squared to the other side by doing the inverse of squaring which is plus or minus square rooting but if we try and square root a negative we run into a problem in order to understand why there's no solution let's just think about numbers that you know how to work with like let's say we're square rooting 9. if you're square rooting 9 you'd say the answer is 3. why is the answer 3 because 3 squared is 9. now think about any time you square something it's positive if we're trying to figure out the square root of negative a third really what we're trying to do is figure out what can we square to get negative a third well there's no way to square a number to get a negative result so we know that that's not possible so when this happens when underneath your square root is negative we say no real solutions let's now do a quick application example 2 application question question we have a picture that measures 10 centimeters by 5 centimeters let me start by drawing that and it says that that picture is to be surrounded by a mat before it's being framed so let me just draw a mat that surrounds that picture and it says the width of the mat is to be the same on all sides of the picture okay i think roughly i made the width be the same on all sides it says the area of the mat is to be twice the area of the picture what's the width of the mat so let's label what we know about this diagram um the inside rectangle is the picture and it says the picture is 10 centimeters by 5 centimeters so i can label those side lengths it then says there's a mat placed around the outside of the picture maybe just let me shade in that mat the question tells us that the width of the mat is the same on all sides of the picture we don't know what the width of the mat is so i'll call that x but i know the width of the mat is x there x here they're all the same so i'll call them all x x x it says the area of the mat is to be equal to twice the area of the picture what is the width of the mat so i know the area of the mat is double the area of the picture so i'm going to set up an equation if the area of the mat is double the area of the picture i know if i multiply the area of the picture by 2 2 times picture area that would be equal to the matte area because the matte is double the area of the picture if i multiply the area of the picture by 2 it should make the areas equal now i just need to come up with an expression for the area of the picture and the area of the mat well the area of the picture that's easy that's the inside rectangle it's 10 by 5. so i can replace the picture area with 10 times 5. the area of the mat is a little more difficult let me try and label some dimensions of the big outside rectangle if i look at the entire length along the bottom it's 10 plus an x extending out from either side of it so the entire length of the bottom would be 10 plus 2x right it would be x plus 10 plus x to get from the left side to the right side of this diagram and how would i get from the bottom to the top of the diagram i would do x plus 5 plus x that's 5 plus 2 x so i know the whole big rectangle is 5 plus 2x by 10 plus 2x that would give me the area of this entire thing but the mat doesn't exist over top of the picture it just surrounds the picture so if i want just the area of the shaded region i would have to do the area of this whole thing the 5 plus 2x times 10 plus 2x and then subtract the area of the inside the area of the picture on the inside which is 10 by 5 which is 50. so the area of the mat would be the area of the big rectangle 5 plus 2x times 10 plus 2x minus the area of the picture on the inside which is 10 by 5. that will give me the area of just the map that surrounds the picture now i want to solve this equation for x when we're solving an equation for x what we're going to want to do is simplify this whole equation by expanding it out by doing all the multiplication move all the terms to one side and then see what type of factoring strategy is going to help us solve this the left side of the equation 2 times 10 times 5 that's a hundred that's easy on the right side i have a binomial times a binomial i'll have to use the method of foil which tells me to find four products five times ten five times two x two x times 10 2x times 2x when i find all four of those products i would have 50 plus 10x plus 20x plus 4x squared minus 10 times 5 which is 50. now i notice this is a quadratic because the highest degree term is degree 2 that 4x squared term let me actually simplify the right side of this equation before i do anything on the right side i have a 50 minus 50. those cancel out that's zero i'm going to rewrite the 4x squared and i have a 10x plus 20x that's 30x now i'm going to want to set this quadratic to 0. so i'll move the 100 to the other side it becomes a minus 100 and then i'm going to want to solve this quadratic first of all i notice all three terms are even so i could common factor out a 2 from all three terms and then i'll focus on the quadratic inside the brackets its leading coefficient is not 1 so i'll have to use decomposition but let me first find the numbers of a product of a times c so 2 times negative 50 is negative 100 and a sum of b 15. the numbers that satisfy this product and sum are 20 and negative 5. so i can split the middle term the 15x into 20x minus 5x and then inside the brackets i'll take a common factor from the first two terms i could take out a 2x and from the last two terms i could take out a negative 5. notice i get a common binomial of x plus 10. when i take that out i'm left with 2x minus 5. now i'm running out of room there so i'm going to continue my work over here underneath my diagram i've gotten the quadratic into factored form so all i have left to do is figure out how could those factors be equal to zero well the factor of two can't be zero but x plus ten could be equal to zero if x was negative ten does that make sense in the context of this question remember x represents the width of the mat can the width of the mat be negative 10 centimeters no in this case a negative answer doesn't make any sense i'll reject that answer my last factor is 2x minus 5. could that be 0 if i solve this add the five over and then divide the two i get x equals five over two which as a decimal is two and a half does that make sense in the context of this question could the width of the map be two and a half centimeters yes so my answer for this question is the width of the mat is two and a half centimeters make sure you go to jensenmath.ca to try out the practice problems solving quadratics by factoring is going to be a very useful skill as you move forward in high school math |
8961 | https://serc.carleton.edu/earthlabs/hurricanes/8.html | Lab 8: Hot Water and Hurricanes
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EarthLabs for Educators>Hurricanes>Lab 8: Hot Water and Hurricanes
Lab 8: Hot Water and Hurricanes
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This activity was selected for the On the Cutting Edge Reviewed Teaching Collection
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Initial Publication Date: August 12, 2008 | Reviewed: November 25, 2019 (see revision history: 2 events))
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First Publication: August 12, 2008
Reviewed: November 25, 2019 -- Reviewed by the On the Cutting Edge Activity Review Process
The lab activity described here was created by LuAnn Dahlman and Sarah Hill of TERC for the EarthLabs project.
Summary and Learning Objectives
Students explore issues related to the rapid intensification of hurricanes. They become familiar with the concepts of heat energy and the specific heat of water and interact with animations of sea surface temperature images to identify the Gulf Stream and the Loop Current. Students use the NOAA View Global Data Exploration Tool and Google Earth to explore visualizations of heat content in the Gulf of Mexico just before Hurricane Katrina. They use a Google Earth layer of a plotted path of Katrina with NOAA data visualizations to observe changes in the heat content of Gulf waters as the hurricane passed over it.
After completing this investigation, students will be able to:
calculate the amount of heat energy absorbed by a given volume of water as its temperature changes;
interpret sea surface temperature images and animations to identify warm water ocean currents;
interpret image data that show various measures of heat in the Gulf of Mexico before and after Hurricane Katrina; and
access and interpret current Tropical Cyclone Heat Potential, Sea Surface Temperature, and Sea Height Anomaly data images.
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Sea Height Anomaly in the Gulf of Mexico, August 2005
Context for Use
This lab should follow the hands-on activities of Lab 7. It requires a computer for every one or two students, and should take one or two class periods to complete. Student computers need to have Google Earth installed and available for use. (Note: If using the Chrome browser, there is an online version available.) Additionally, they need to be able to download and save .kmz files that contain data images.
Activity Overview and Teaching Materials
Students are introduced to the concept of specific heat and watch a short video demonstration of a "trick" that depends on the high specific heat of water. They calculate the amount of energy a given volume of water absorbs as its temperature is increased. They also view a video and read a short text about ocean heat and the Gulf Stream from NOAA NESDIS. Students work with the NOAA View Data Exploration Tool to view and animate data for the Loop Current. They use NOAA View to view visual datasets and download kmz files for use with Google Earth to observe and analyze change in various ocean parameters as Katrina passed. Finally, students return to NOAA View to access and view real-time images for tropical cyclone heat potential to assess the current heat conditions of the Gulf of Mexico.
Printable Materials
Activity Sheet (PDF(Acrobat (PDF) 114kB Jun22 22) and Word(Microsoft Word 2007 (.docx) 22kB Jun22 22))
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Teaching Notes and Tips
If possible, have students complete Lab 4A to map the path of Hurricane Katrina for themselves in Google Earth and save as a .kmz file before beginning this lab.
If your class has already completed lab 4A and chose different storms' paths, as suggested in the Lab instructions, consider allowing them to use those storms to complete this lab. In that case, students should plan to follow the directions with modifications to obtain the correct dates and data for their storms. Note: NOAA View data only goes back to about 1980 for the data students will be asked to find. If students created paths for storms earlier than 1980, you might want to provide them with the Katrina .kmz file, as described below.
If time is an issue or there is a problem with student work from Lab 4A, there is a pre-completed .kmz file of
Hurricane Katrina 2005 -- educator-only file)
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Hurricane Katrina 2005
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containing placemarks and paths for the storm at 18Z as retrieved from HURDAT2, which you could share with students doing this lab.
You may consider having students work in small groups or pairs, or assigning sections of this lab for homework.
Assessment
You can assess student understanding of topics addressed in this Investigation by grading their responses to the Stop and Think Questions.
State and National Science Teaching Standards
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California Science Teaching Standards met by this activity
Investigation and Experimentation Standards
Scientific progress is made by asking meaningful questions and conducting careful investigations. As a basis for understanding this concept and addressing the content in the other four strands, students should develop their own questions and perform investigations. Students will:
g. Recognize the usefulness and limitations of models and theories as scientific representations of reality.
i. Analyze the locations, sequences, or time intervals that are characteristic of natural phenomena.
Earth Science Content Standards - Grades 9-12
Energy enters the Earth system primarily as solar radiation and eventually escapes as heat. As a basis for understanding this concept:
Students know the fate of incoming solar radiation in terms of reflection, absorption, and photosynthesis.
Heating of Earth's surface and atmosphere by the sun drives convection within the atmosphere and oceans, producing winds and ocean currents. As a basis for understanding this concept:
Students know how differential heating of Earth results in circulation patterns in the atmosphere and oceans that globally distribute the heat.
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Applicable Massachusetts Science and Technology Standards (PDF - 1.3 Mb)
Earth and Space Science - Content Standards
1. Matter and Energy in the Earth System
Central Concepts: The entire Earth system and its various cycles are driven by energy. Earth has both internal and external sources of energy. Two fundamental energy concepts included in the Earth system are gravity and electromagnetism.
1.3 Explain how the transfer of energy through radiation, conduction, and convection contributes to global atmospheric processes, such as storms, winds, and currents.
1.6 Describe the various conditions associated with frontal boundaries and cyclonic storms (e.g., thunderstorms, winter storms [nor'easters], hurricanes, tornadoes) and their impact on human affairs, including storm preparations.
Earth and Space Science - Scientific Inquiry Skills Standards
SIS3. Analyze and interpret results of scientific investigations.
Represent data and relationships between and among variables in charts and graphs.
Use mathematical operations to analyze and interpret data results.
Earth and Space Science - Mathematical Skills
Solve simple algebraic expressions.
Convert within a unit (e.g., centimeters to meters).
Use scientific notation, where appropriate.
Use appropriate metric/standard international (SI) units of measurement for mass (kg); length (m); time (s); force (N); speed (m/s); acceleration (m/s2); and frequency (Hz).
Use the Celsius and Kelvin scales.
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Applicable New York Core Curricula
STANDARD 4 - Students will understand and apply scientific concepts, principles, and theories pertaining to the physical setting and living environment and recognize the historical development of ideas in science.
Key Idea 2: Many of the phenomena that we observe on Earth involve interactions among components of air, water, and land.
2.1b The transfer of heat energy within the atmosphere, the hydrosphere, and Earth's interior results in the formation of regions of different densities. These density differences result in motion.
2.1d Weather variables are measured using instruments such as thermometers, barometers, psychrometers, precipitation gauges, anemometers, and wind vanes.
2.1f Air temperature, dewpoint, cloud formation, and precipitation are affected by the expansion and contraction of air due to vertical atmospheric movement.
2.2 Explain how incoming solar radiation, ocean currents, and land masses affect weather and climate.
2.2b The transfer of heat energy within the atmosphere, the hydrosphere, and Earth's surface occurs as the result of radiation, convection, and conduction.
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COMPETENCY GOAL 1: The learner will develop abilities necessary to do and understand scientific inquiry in the earth and environmental sciences.
1.02 Design and conduct scientific investigations to answer questions related to earth and environmental science.
Analyze and interpret data.
COMPETENCY GOAL 4: The learner will build an understanding of the hydrosphere and its interactions and influences on the lithosphere, the atmosphere, and environmental quality.
4.02 Analyze mechanisms for generating ocean currents and upwelling:
Temperature
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Applicable Texas Essential Knowledge and Skills (TEKS)
112.42. Integrated Physics and Chemistry.
(c) Knowledge and skills:
(6) Science concepts. The student knows the impact of energy transformations in everyday life. The student is expected to:
(H) analyze the effects of heating and cooling processes in systems such as weather, living, and mechanical.
112.49. Geology, Meteorology, and Oceanography.
(c) Knowledge and skills:
(2) Scientific processes. The student uses scientific methods during field and laboratory investigations. The student is expected to:
(C) organize, analyze, evaluate, make inferences, and predict trends from data;
(c) Knowledge and skills:
(13) Science concepts. The student knows the role of energy in governing weather and climate. The student is expected to:
(A) describe the transfer of heat energy at the boundaries between the atmosphere, land masses, and oceans resulting in layers of different temperatures and densities in both the ocean and atmosphere;
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Applicable National Science Education Standards (SRI)
Science as Inquiry (12ASI)
Abilities necessary to do scientific inquiry
12ASI1.2 Design and conduct scientific investigations. Designing and conducting a scientific investigation requires introduction to the major concepts in the area being investigated, proper equipment, safety precautions, assistance with methodological problems, recommendations for use of technologies, clarification of ideas that guide the inquiry, and scientific knowledge obtained from sources other than the actual investigation. The investigation may also require student clarification of the question, method, controls, and variables; student organization and display of data; student revision of methods and explanations; and a public presentation of the results with a critical response from peers. Regardless of the scientific investigation performed, students must use evidence, apply logic, and construct an argument for their proposed explanations.
12ASI1.3 Use technology and mathematics to improve investigations and communications. A variety of technologies, such as hand tools, measuring instruments, and calculators, should be an integral component of scientific investigations. The use of computers for the collection, analysis, and display of data is also a part of this standard. Mathematics plays an essential role in all aspects of an inquiry. For example, measurement is used for posing questions, formulas are used for developing explanations, and charts and graphs are used for communicating results.
Earth and Space Science (12DESS)
Energy in the earth system
12DESS1.3 Heating of earth's surface and atmosphere by the sun drives convection within the atmosphere and oceans, producing winds and ocean currents.
Additional Resources
Content Extension
A paper by Michael P. Erb entitled A Case Study of Hurricane Katrina: Rapid Intensification in the Gulf of Mexico can serve to deepen student understanding following this lab. Students will recognize some of the content they worked with and be exposed to new concepts for documenting the causes of rapid intensification.
NASA Goddard Video Building a Hurricane Season in the Atlantic Ocean
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Lab Overviews
Lab 1: Meteorological Monsters
Lab 2: Hurricane Anatomy
Lab 3: Putting Hurricanes on the Calendar
Lab 4: Putting Hurricanes on the Map
Lab 5: All About Air Pressure
Lab 6: Why Keep an Eye on the Barometer?
Lab 7: Hurricanes and Heat Transfer
Lab 8: Hot Water and Hurricanes
Lab 9: Death and Destruction
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HomeProbability Theory & StatisticsProbability Distributions Binomial Distribution Mean and Variance Formulas (Proof)
Binomial Distribution Mean and Variance Formulas (Proof)
Posted on May 19, 2020 Written by The Cthaeh31 Comments
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This is a bonus post for my main post on the binomial distribution. Here I want to give a formal proof for the binomial distribution mean and variance formulas I previously showed you.
This post is part of my series on discrete probability distributions.
In the main post, I told you that these formulas are:
For which I gave you an intuitive derivation. The intuition was related to the properties of the sum of independent random variables. Namely, their mean and variance is equal to the sum of the means/variances of the individual random variables that form the sum. We could prove this statement itself too but I don’t want to do that here and I’ll leave it for a future post.
Instead, I want to take the general formulas for the mean and variance of discrete probability distributions and derive the specific binomial distribution mean and variance formulas from the binomial probability mass function (PMF):
To do that, I’m first going to derive a few auxiliary arithmetic properties and equations. We’re going to use those as pieces of the main proofs. But their usefulness is much bigger and you can apply them for many other derivations.
I think this post will be a great exercise for those of you who don’t have much experience in formal derivations of mathematical formulas. I’m going to be as explicit as I can and try to not skip even the smallest steps. So, don’t be scared by the quantity of equations in this post. I promise, you’ll be able to follow everything!
And if you happen to get stuck somewhere, I’m going to answer all of your questions in the comments. Don’t hesitate to ask me anything.
Table of Contents
Toggle - [x]
Auxiliary properties and equations
A property of the binomial coefficient
Mean of binomial distributions derivation
Proof steps
The final proof
Variance of binomial distributions derivation
Proof steps
The final proof
Summary
Mean of binomial distributions proof
Variance of binomial distributions proof
Auxiliary properties and equations
To make it easy to refer to them later, I’m going to label the important properties and equations with numbers, starting from 1. These identities are all we need to prove the binomial distribution mean and variance formulas. The derivations I’m going to show you also generally rely on arithmetic properties and, if you’re not too experienced with those, you might benefit from going over my post breaking down the main ones.
The first two equations are two important identities involving the sum operator which I proved in my recent post on the topic:
(1)
(2)
Second, in another recent post on different variance formulas I showed you the following alternative variance formula for a random variable X with mean M:
(3)
It’s a pretty nice formula used in many derivations, not just the ones I’m about to show you (for more intuition, check out the link above). According to this formula, the variance can also be expressed as the expected value of minus the square of its mean. As a reminder (and for comparison), here’s the main variance formula:
A property of the binomial coefficient
Finally, I want to show you a simple property of the binomial coefficient which we’re going to use in proving both formulas. Remember the binomial coefficient formula:
The first useful result I want to derive is for the expression . Let’s apply the formula to this expression and simplify:
Therefore:
Now let’s do something else. One of the simplest properties of the factorial function is:
I want to use this to derive the main property of binomial coefficients we’re interested in here. First, let’s use it to rewrite the right-hand side in the following way:
And using the commutative property of multiplication (), we can rewrite the right-hand side as:
Now let’s say we start with another expression: . Using the result above, we can equate it to:
In the last step, I simply canceled out the two k’s. Finally, using (derived above), we get the following identity:
(4)
And with all that out of the way, let’s get to the main proofs of today’s post!
Mean of binomial distributions derivation
Well, here we reach the main point of this post! Let’s use these equations and properties to derive the formulas we’re interested in.
First is the mean. Here’s the general formula again:
Let’s plug in the binomial distribution PMF into this formula. To be consistent with the binomial distribution notation, I’m going to use k for the argument (instead of x) and the index for the sum will naturally range from 0 to n. So, with in mind, we have:
But notice that when k = 0 the first term is also zero and doesn’t contribute to the overall sum. Therefore, we can also write the formula by having the index start from k = 1:
Now let’s see how we can manipulate the right-hand side to get the desired . Let’s do the proof step by step.
Proof steps
The first step in the derivation is to apply the binomial property from equation (4) to the right-hand side:
In the second line, I simply used equation (1) to get n out of the sum operator (because it doesn’t depend on k).
Next, we’re going to use the product rule of exponents:
A special case of this rule is:
And we’re going to use it to rewrite as :
Again, in the last line I simply took out the constant term p outside of the sum operator. Finally, let’s apply the identity to the exponent of (1 – p) (you’ll see why we do this in a moment):
The final proof
Believe it or not, we’re almost done here. Notice that, after the last manipulation, there’s a lot of terms like n – 1 and k – 1 inside the sum operator. To make the expression a little more readable, let’s rewrite it by applying the following variable substitutions:
This results in:
Here j starts from 0 because j = k – 1 (the k index used to start from 1 before the variable substitution). And because the number of terms in the sum must be preserved, the index runs until n – 1 = m.
Now, do you recognize the term inside the sum operator?
It looks exactly like the binomial PMF, doesn’t it? Only k has been replaced with j and n with m. And since the sum is from 0 to m, this is simply the sum of probabilities of all outcomes, right? Then, by definition:
And plugging this last result into what we have so far, we get:
Therefore, we can now confidently state:
Q.E.D.
Variance of binomial distributions derivation
Now it’s time to prove the variance formula. Remember the general variance formula for discrete probability distributions:
Like before, for the argument of the PMF I’m going to use k, instead of x. Furthermore, I’m going to use the alternative variance formula from equation (3) we derived earlier:
Using the result from the previous section:
And we can plug in to get:
We already solved half of the problem! Now let’s focus on the term. Using the binomial PMF, this expected value is equal to:
Like before, when k = 0 the first term in the sum becomes zero again. So we can similarly write the same sum with the index starting from 1:
With this setup, let’s start with the actual proof. The focus is going to be on manipulating the last equation. When we’re done with that, we’re going to plug in the final result into the main formula.
You’ll see that the mathematical tricks we use are going to be very similar to the ones we used in the previous proof.
Proof steps
First, using equation (4), let’s rewrite the part inside the sum as:
Plugging this in and taking the constant term n out of the sum operator, we get:
Next, let’s use the identity to rewrite this as:
Simplifying the sum
Now let’s ignore the constant product np for a moment and just focus on the sum. Let’s apply the same variable substitution rules as before:
to rewrite the sum as:
Next, let’s use equation (2) to split this sum into two sums by expanding with the distributive property:
Well, these individual sums are nothing but the expected value and the sum of probabilities of a binomial distribution:
Therefore, the final sum reduces to:
And when we plug this into the full expression for we get:
Which we can rewrite as:
So, finally we get:
The final proof
We’re at the homestretch. Let’s remember what we started with.
In the previous section we established that:
Therefore:
Which is what we wanted to prove!
Q.E.D.
Summary
In this post, I showed you a formal derivation of the binomial distribution mean and variance formulas. This is the first formal proof I’ve ever done on my website and I’m curious if you found it useful. Let me know if it was easy to follow.
Before the actual proofs, I showed a few auxiliary properties and equations.
The two properties of the sum operator (equations (1) and (2)):
An alternative formula for the variance of a random variable (equation (3)):
The binomial coefficient property (equation (4)):
Using these identities, as well as a few simple mathematical tricks, we derived the binomial distribution mean and variance formulas. In the last two sections below, I’m going to give a summary of these derivations.
I know there was a lot of mathematical expression manipulation, some of which was a little bit on the hairy side. However, I’m firmly convinced that even less experienced readers can understand these proofs. If you struggled to follow any part of this post (or even the post as a whole), don’t hesitate to ask me any question!
By the way, if you’re new to mathematical proofs but find it an interesting subject, check out this Wikipedia article on mathematical proofs which gives a good overview of the subject.
Mean of binomial distributions proof
We start by plugging in the binomial PMF into the general formula for the mean of a discrete probability distribution:
Then we use and to rewrite it as:
Finally, we use the variable substitutions m = n – 1 and j = k – 1 and simplify:
Q.E.D.
Variance of binomial distributions proof
Again, we start by plugging in the binomial PMF into the general formula for the variance of a discrete probability distribution:
Then we use and to rewrite it as:
Next, we use the variable substitutions m = n – 1 and j = k – 1:
Finally, we simplify:
Q.E.D.
Filed Under: Algebra, Probability DistributionsTagged With: Expected value, Mean, Probability mass, Variance
Comments
Sergey says
June 4, 2020 at 1:21 pm
Great job! I definitely found the proofs useful and easy to follow. The refresher on the various mathematical properties was a nice reminder (it’s been a while…). Thanks! Reply
2. Durjoy says
December 20, 2020 at 8:22 pm
Thank you. Reply
3. Swami says
January 23, 2021 at 9:17 am
Excellent proof…Simple to understand Reply
4. Andrew kingsley says
February 20, 2021 at 7:18 pm
I find it difficult to understand some of the manipulations Reply
The Cthaeh says
February 21, 2021 at 3:28 am
Hey Andrew, until which part do you follow everything before you reach the first difficulty? Which step(s) in particular confuse you? Let me know and I’ll help you get through it.
deinma says
May 17, 2021 at 1:41 pm
thank you so much, it was thoroughly explained, what I don’t understand is why m replaced n and j replaced k Reply
The Cthaeh says
May 18, 2021 at 2:55 am
Hi Deinma, I’m glad you found the explanation helpful! The change of variables is really just for readability purposes. If a computer was doing the same proof, this step might be unnecessary.
Take a look at the final proof section for the mean, for example. In particular, notice the last expression for the mean formula just before that section. There are all these (n-1) and (k-1) terms inside the sum, right?
Now, do you recognize that this sum is the binomial distribution PMF formula? I personally might not immediately recognize it if I see it in this form. On the other hand, if we apply the variable substitution and , the same sum becomes:
Which is an identical expression, but much easier to read. And now that we recognize it as the sum over all possible values of a PMF, we know the whole sum is equal to 1 and we can completely ignore it. In principle, I could have used the same letters (k and n) but that would introduce a different type of confusion when comparing the current expression to previous steps.
Variable substitution (also known as ‘change of variables’) is a very common technique in algebra and math in general. It’s typically used for simplifying expressions or putting expressions into a familiar form where we can then substitute the whole thing with a known formula (as was the case in the proofs here).
Let me know if I managed to clarify things for you.
Symonsays
June 16, 2022 at 12:09 am
Hi,
Can you explain why the upper index of summation should not be m+1 since
m=n-1. Note that the original summation had n as the upper index, hence
n=m+1 if we go by the substitution that you just enforced.
Otherwise, everything else is very insightful!
I like your proofs!
Symon
The Cthaeh says
June 18, 2022 at 4:07 pm
Hi Symon,
This is among the confusing parts of the post, it’s not just you. Could you take a look at my reply to Yuki’s question further down the page? I think you are asking the same thing. Of course, let me know if you still find something unclear, I’ll be happy to clarify.
AZ says
June 3, 2021 at 10:48 am
Thanks Reply
7. mehrad says
June 17, 2021 at 12:26 pm
thank you Reply
8. Yan Jiang says
August 24, 2021 at 10:26 am
Small error: the second last step should be
np(n-1)p+np-(np)^2
Rather than
npn(p-1)+np-(np)^2 Reply
The Cthaeh says
August 24, 2021 at 1:59 pm
Oh yeah, another annoying typo Thanks for pointing it out, Yan. Fixed!
Emilio Cendejas says
October 8, 2021 at 3:55 am
Hi Andrew, thank you for this.
Right before reaching the end, when you write “Finally, we simplify:”, what do you do to the binomial coefficient? The same exact process you did before changing the variables? I mean, like now z = y- 1 and r = m – 1? And because of that, we know it sums up to 1?
Thanks again. Reply
The Cthaeh says
October 8, 2021 at 7:34 am
Hey Emilio,
My name isn’t Andrew, I’m assuming you confused it with one of the earlier commenters No worries.
Yes, the change of variables is simply meant to put the sums into the canonical forms for the respective formulas (the PMF, the mean, etc.). It’s a step that people sometimes skip because they just spot the pattern and do the simplification in their heads.
Let me know if you need further clarification.
Failo Taufa says
October 21, 2021 at 1:11 am
Very nice derivation and easy to follow. Thanks for the good work. Reply
11. Umi says
October 28, 2021 at 4:48 pm
When you derive E[k^2], why don’t all k become k^2? Reply
The Cthaeh says
November 1, 2021 at 2:19 pm
Hi Umi, I’m not sure I understand your question. Could you show the exact step at which you stop following?
Conor says
September 17, 2024 at 10:33 pm
Hi Umi,
I’ll follow up on this one as I am somewhat unclear on this also. If we consider the expected value of a BPD to be:
E[x] = \sum{xP(X = x)}
Then why is k^2 not substituted into the probability function?
For example, the proof shows:
E[k^2] = \sum{k^2P(X = k)}
whereas I was expecting:
x = k^2 \therefore E[k^2] = (k^2) P(X = (k^2))
Does that make any sense where my confusion is coming from?
Dr G N Darsays
November 7, 2021 at 4:57 am
i did by other way, my confusion was for example in mean after substitution it is ok y starts from 0 but n must have been m-1 through the formula (x-1)=y but issue stands resolved with your comment that number of terms have to remain same. Reply
13. Yuki says
November 10, 2021 at 6:35 pm
In the proof of distribution mean, you let n-1 = m. Then, the end of the summation should be m+1 instead of m. So, I don’t get it. Reply
The Cthaeh says
November 11, 2021 at 7:27 am
Hi, Yuki. You’re trying to understand why the upper bound of the sum is m, instead of (m+1), correct?
Notice that after we made the variable substitution n – 1 = m, the lower bound of the sum also changes (from 1 to 0). Basically, the idea is to have an identical sum. Why is the sum running from j=0 to m the same as a sum running from k=1 to n? To see it more clearly, take a specific value for n. For example, n=3.
In the original sum, since k starts at 1 and ends at 3, the k’s in the expression at each iteration will be 1, 2, and 3. Which means that the (k-1) terms in the original sum will be 0, 1, and 2, respectively. Are you with me so far?
Then, when we apply the variable substitution, the sum runs from 0 to m. Because m = n – 1, we know the j’s will run from 0 to 2, right? And inside the sum, everywhere we had (k-1), we now have j instead. So, just like they were in the original sum, the respective values inside the sum term will be 0, 1, and 2. Furthermore, everywhere we had (n-1) in the original expression, now we have m, which is equal to n – 1. Therefore, we know the two expressions are identical, one is just a little tidier.
On the other hand, if we let the upper bound be m+1 instead, then we would have an extra term for j=3, which we didn’t have in the original sum.
Bottom line is, if we decrease the lower bound by 1, we also need to decrease the upper bound by 1, otherwise we will have an extra term (and, therefore, the two sum expressions will be different).
Does that make sense?
Symon says
June 16, 2022 at 12:10 am
I have the same question.
Thank you!
Pedson says
March 7, 2022 at 6:36 pm
i understood it very well
thank you Reply
15. Ray says
September 8, 2022 at 9:30 am
Hey. This was perfect. Thanks. Reply
16. Shubham says
October 5, 2022 at 3:46 pm
Clear and thoroughly explained. The derivation in my book was very ambiguous and I could not clearly understand it. But your derivation really cleared my concepts.
Thanks. Reply
17. Shuhul Mujoo says
November 17, 2022 at 9:24 pm
Thank you so much for this! It really helped me a lot Reply
18. Aphrodice says
January 28, 2023 at 10:15 am
I love how you combine different formulas to prove something! Reply
19. PKP says
August 9, 2023 at 12:01 pm
Wonderful proof!
Initially, I too got stumbled on m=n-1!
Thanks!
PKP Reply
20. SG says
February 24, 2024 at 3:03 pm
Very helpful, thank you. Reply
21. xuqi says
September 4, 2024 at 2:44 am
Thanks for the detailed proof. I don’t get why would the sum of j from j = 0 to j = m equal to j tho. Reply
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Partition (combinatorics)
A partition of a nonnegativeinteger is a way of expressing it as the unordered sum of other positive integers. For example, there are three partitions of 3: . Each of the summands is a part of the partition.
The partition function gives the number of partitions of . There is an exact formula for , discovered by G. H. Hardy, J. E. Littlewood, and Srinivasa Ramanujan. However, this formula is unwieldy: it is not even known for which values of the number of partitions of is even, despite the presence of a formula! No simpler formula is known, and the existence of such a formula is doubtful.
A fruitful way of studying partition numbers is through generating functions. The generating function for the sequence is given by . Partitions can also be studied by using the Jacobi theta function, in particular the Jacobi triple product. The generating function approach and the theta function approach can be used to study many variants of the partition function, such as the number of ways to write a number as the sum of odd parts, or of distinct parts, or of parts equivalent to , etc.
Contents
[hide]
1 List of Partitions and Values of the Partition Function for Small
2 Ferrers Diagrams
2.1 The Conjugate
3 Generating Functions
4 A Variation
5 Resources
6 See also
List of Partitions and Values of the Partition Function for Small
The empty partition (with no parts) is the unique partition of , so .
The unique partition of is , so .
, so .
, so .
, so .
, so .
Partitions are often written in tuple notation, so we might denote the partitions of by and . This notation is often further abbreviated to word notation (by dropping the parentheses and commas, so becomes ) or by indicating multiplicities with exponential notation (so becomes ).
Using this last notation, the partitions of are and , so . One could continue this computation to find that , , , , and so on.
Ferrers Diagrams
A Ferrers diagram is a way to represent partitions geometrically. The diagram consists of rows of dots. Each row represents a different addend in the partition. The rows are ordered in non-increasing order so that that the row with the most dots is on the top and the row with the least dots is on the bottom.
For example, 9 can be partitioned into 4 + 3 + 1 + 1 which would be represented by the following Ferrers diagram:
4
3
1
1
The Conjugate
The conjugate of a Ferrers diagram is formed by reflecting the diagram across its diagonal (the one starting in the top left of the diagram). This can also be interpreted as exchanging the rows for the columns. For example, consider our example from before but this time let's count the number of dots in each column:
4 2 2 1
4
3
1
1
The original partition is 4 + 3 + 1 + 1 and the conjugate is 4 + 2 + 2 + 1.
Generating Functions
Generating functions can be used to deal with some problems involving partitions. Here we derive the generating function for the number of partitions of .
Consider partitioning into addends that are equal to 1. The generating function for this is since there is only one way to represent as the sum of 1s.
Consider partitioning numbers using just 2s as addends. There is one way to partition 0 into 2s, zero ways to partition 1 into 2s, one way to partition 2 into 2s, and so forth. Therefore, the generating function for this type of partition is .
We can proceed in this manner to find that the generating function for the number of ways to partition into addends equal to is .
Now, we generate every partition of by choosing some number of parts to be equal to 1, some other number of parts to be equal to 2, and so on. Thus, we get the generating function for the partition function of by multiplying the generating functions for partitions into just 1s, partitions into just 2s, and so on. This gives us the expression
.
Using the formula for the sum of an infinitegeometric sequence we can express this in the more compact form
.
A Variation
An interesting theorem is that the number of partitions consisting of only consecutive positive integers of is the number of odd divisors of .
Proof: Let be the smallest part in such a partition and let be the number of parts. Then we have
, so
and finally
.
Let's allow negative integers in our partition for a moment, and let denote the number of odd divisors of . Now we consider the two possible cases:
Case 1: is an integer.
Since the expression on the right hand side must be an integer, this implies that . Also, for to be an integer, must be divisible by 2, which implies that is odd. Therefore, as long as is an odd divisor of , we can find consecutive integers starting with that adds up to . Therefore, in this case the number of solutions is equal to .
Case 2: is not an integer.
For the right hand side to still be an integer, we must have for some odd integer . Since is not an integer, must be odd, and therefore must be even. expressing in the form with being an odd integer, we see that if and only if , where is any odd divisor of . As the result, the number of solutions in this case is also equal to .
The two cases combined gives us solutions. However, this includes partitions with negative integers. However, since is positive, the partition with negative integers must be of the form:
Where . As the result, we can cancel out the negative integers with the corresponding additive inverses to produce a valid partition. Similarity, we can go from a valid partition to one that contains negative integers by adding all the positive integers before the smallest member, along with the corresponding additive inverses. As the result, we established a bijection between valid partitions with the partitions containing negative integers.
Since our count includes all the possible partitions, and the number of valid partitions is equal to the number of partitions that contain negative integers, the number of valid partitions must be exactly half of our original count. Therefore, the total number of such partitions is , which is the number of odd divisors of .
Resources
Partitions of Integers by Joseph Laurendi
The Jacobi Theta Function by Simon Rubinstein-Salzedo
Sequence A000041 in the OEIS -- the partition function
See also
Combinatorics
Number Theory
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Hypergeometric Distribution: Uses, Calculator & Formula
What is a Hypergeometric Distribution?
The hypergeometric distribution is a discrete probability distribution that calculates the likelihood an event happens k times in n trials when you are sampling from a small population without replacement.
This distribution is like the binomial distribution except for the sampling without replacement aspect. When you sample without replacement, the probabilities change with each subsequent trial. Conversely, the binomial distribution assumes the chances remain constant over the trials.
For instance, when you draw an ace from a deck of cards, the probability decreases for drawing another ace on the next draw because the deck has fewer aces.
The hypergeometric distribution can answer the following questions. What is the probability of getting:
Two red candies when we draw five candies from a jar containing five red candies and 10 white candies.
Drawing five cards of the same suit from a regular deck of cards.
8 women on a jury of 13 people when drawing randomly from a jury pool of 50 people evenly split between men and women?
As the population size increases, the hypergeometric distribution more closely approximates the binomial distribution. This distribution is an example of a Probability Mass Function (PMF) because it calculates likelihoods for discrete random variables.
In this post, learn how to use the hypergeometric distribution and its cumulative form, when you can use it, its formula, and how to calculate probabilities by hand. I also include a hypergeometric distribution calculator that you can use with what you learn. I’ll walk you through the formulas for calculating hypergeometric distribution probabilities.
For more information about other ways to use binary data, read my posts, Maximize the Value of Your Binary Data, the Bernoulli, Binomial, Negative Binomial, and the Geometric Distribution.
Hypergeometric Probabilities
The hypergeometric distribution models the probabilities for exactly k events occurring in n trials when you know the composition of a small population. Let’s look at an example to bring it to life!
I’ll start by using statistical software to calculate the hypergeometric probabilities and create distribution plots. This process will help you understand what you can learn from it. Then we’ll move on to the hypergeometric distribution formula.
Suppose we’re interested in the possible outcomes for a jury selection. We want to know the probability of drawing 8 women for a jury of 13 when there are 25 female and 25 male candidates. For this example, assume the jurors are randomly selected from the pool of candidates.
We’ll need the following information to solve this problem:
Total population size is 50 candidates (N).
Number of events (Female) in the population (all candidates) is 25 (K)
The jury size is 13 (n).
Outcome of interest is selecting 8 women (k).
The hypergeometric distribution accounts for how the probabilities change with each selection. As we select men and women from the candidate pool, it affects the makeup of the remaining population in the pool because there are no replacements. Each woman we choose reduces the number of women in the candidate pool, thus lowering the likelihood that the following selection will be a woman. Conversely, selecting a man increases the chances that the next juror will be a woman.
Example Results
My statistical software tells me that the likelihood is:
The hypergeometric probability distribution calculates a likelihood of 0.161934 for selecting eight women in 13 draws.
That’s interesting but perhaps not so helpful by itself. We’re also interested in the chances of selecting other numbers of female jurors. Seeing the distribution of probabilities for different numbers is much more helpful.
Related post: Understanding Probability Distributions
Hypergeometric Distribution Graph
The hypergeometric distribution graph is helpful because it displays the probability of differing numbers of successes (k) out of the total number of trials (n). In the chart below, the distribution plot finds the likelihood of selecting exactly no women, 1 woman, 2 women, 3 women, . . ., and up to 13 women in the 13 selections. With this approach, the hypergeometric distribution graph covers the complete range of possible successes up to the total number of trials.
I like these graphs because they emphasize how we’re working with a distribution, and it’s easy to see which values happen more frequently. The graph below does not show the chances for fewer than 2 or more than 11 because those likelihoods are too low to display on the chart.
In the chart, each bar represents the probability of selecting a specific number of women during the 13 selections. The bar for 8 corresponds with the probability (0.161934) shown in the output above. At a glance, we can see that selecting 6 or 7 female jurors are the most likely outcomes with both having equal probabilities of approximately 0.24.
Hypergeometric Cumulative Distribution Function
The hypergeometric distribution is excellent for understanding the likelihood of obtaining an exact number of events (k) within a certain number of trials (n) for a small population without replacement. However, you’re often not interested in just one specific number of outcomes.
For example, in the jury selection example above, you might want to learn the probability of selecting at least eight women.
Let me introduce you to the hypergeometric cumulative distribution function.
Technically, the hypergeometric cumulative probability calculates the likelihood of obtaining less than or equal to k events in n trials. Use the inverse cumulative distribution when you need to get a ≥ probability. These days, most statistical software will let you indicate the direction of the cumulative function for the hypergeometric distribution. I’ll use the hypergeometric distribution graph again to show you how it works.
For example, we want to know the chances of selecting ≥ 8 women in 13 attempts. Below, the shaded region shows the inverse cumulative probability of choosing at least eight women in 13 draws.
The likelihood of randomly choosing eight or more women in 13 selections is 0.2601, approximately 1 in 4.
Learn more about Cumulative Distribution Functions: Uses, Graphs & vs PDF.
Hypergeometric Distribution Calculator
Use my hypergeometric distribution calculator below to calculate probabilities and cumulative probabilities. Click the link for its standalone page that you might want to bookmark.
Hypergeometric Distribution Calculator
Hypergeometric Distribution Calculator▼
Let’s use this calculator to recreate the preceding jury selection examples. In the calculator, enter Population size (N) = 50, Number of Events in Population (K) = 25, Sample Size (n) = 13, and Number of Events in Sample (k) = 8. The calculator displays a hypergeometric probability of 0.1619, matching our results above for eight women. Looking at P(X ≥ 8), the calculator displays 0.2601 for selecting at least eight women. This result matches our graphical example with the hypergeometric inverse cumulative distribution.
Now, try one yourself. Imagine you’re drawing from a jar that contains 5 red candies and 10 non-red candies. What is the probability that you’ll draw two red candies with five random draws from the jar?
See the correct answer at the end of this post. Now, onto the formula for those who want to calculate the probabilities manually.
Hypergeometric Distribution Formula
Typically, you’ll use statistical software or online calculators to calculate the probabilities for the hypergeometric distribution. However, I’ll explain the hypergeometric distribution formula so you can calculate them manually and I’ll walk you through a worked example.
The hypergeometric distribution formula is the following:
Where:
N is the number in the population.
n is the sample size.
K is the number of events or successes in the population.
k is the number of events in the sample.
Use this hypergeometric formula to calculate the probability of k successes occurring in n trials for a small population without replacement. Notice how the formula incorporates the population’s characteristics (N and K) and the properties of the sample we’re assessing (n and k).
The formulation uses combinations. For example, using standard notation, CKk is the number of ways you can start with K successes in the population and end up with k successes in your sample where the order of successes does not matter. For more information, read my post about Finding Combinations and my Combination Calculator.
Let’s cover what each term in the equation means.
Given a population size of N that contains K successes, the hypergeometric distribution formula takes the number of ways to have k successes in a sample and multiplies that by the number of ways for n-k failures in the sample. Then it divides that product by the total number of outcomes for sample size n.
To find the correct solution, you must use the correct order of operations in the formula. If you need a refresher on that, read my post PEMDAS Explained: Order of Operations in Math.
Let’s work through an example calculation to bring the formula to life!
Worked Example of Finding a Hypergeometric Probability
We’ll use the hypergeometric distribution formula to calculate the likelihood of choosing red candies from a jar. The jar contains 5 red candies and 10 non-red candies for a total of 15 candies. We’ll randomly draw five candies from the jar.
Let’s calculate our chances of getting two red candies in our five draws!
We’ll enter the following values in the hypergeometric distribution formula:
N = 15 total candies in the jar.
n = 5 draws for the sample.
K = 5 red candies in the jar.
k = 2 red candies in the sample.
For these calculations, I’ll use a combinations calculator to obtain the number of combinations for each term in the equation. To learn how to calculate the number of combinations by hand, click the link above about finding combinations.
The probability of drawing precisely two red candies in our five random draws from the jar is 0.3996. This result also answers the hypergeometric calculator problem I gave you earlier!
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Hypergeometric Distribution Calculator
Use this hypergeometric distribution calculator to find the probability of drawing a specific number of successes in a sample taken from a finite population without replacement. Enter the population size (N), the number of successes in the population (K), the sample size (n), and the number of observed successes in…
In "Calculators"
Maximize the Value of Your Binary Data with the Binomial and Other Probability Distributions
Binary data occur when you can place an observation into only two categories. It tells you that an event occurred or that an item has a particular characteristic. For instance, an inspection process produces binary pass/fail results. Or, when a customer enters a store, there are two possible outcomes—sale or…
In "Basics"
Bernoulli Distribution: Uses, Formula & Example
What is the Bernoulli Distribution? The Bernoulli distribution is a discrete probability distribution that models a binary outcome for one trial. Use it for a random variable that can take one of two outcomes: success (k = 1) or failure (k = 0), much like a coin toss. Statisticians refer…
In "Probability"
Reader Interactions
Comments
Charlie MacPherson says
Hi Jim,
I have a problem with a calculation. I’m trying to calculate the probability of getting 4 out of 6 numbers in a lottery with 44 total numbers. In this case, N=44, n=6, K=6, and k=4. I plug those numbers in the calculator and the result is 0.0014938. Using the formula, the calculation is (C6/4 C38/2) / C44/6. Plug in the numbers and it reduces to
C6/4 = 6!/4!=30
C38/2 = (3837)/2=703
C44/6 = 44!/(38!6!) =7,059,052
The result is (30703)/7059052=0.00299
It’s off by a factor of 2. What am I doing wrong? I hope you can help.
Thanks,
Charlie
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8965 | https://www.youtube.com/watch?v=sYopk5NQZw0 | Divisibility Rule for 7 | Math with Mr. J
Math with Mr. J
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Welcome to the Divisibility Rule for 7 with Mr. J! Need help with what the divisibility rule for 7 is? You're in the right place!
Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with the divisibility rules. Mr. J will go through divisibility rule examples and explain how to determine if a number is divisible by 7.
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✅ Divisibility Rule for 8:
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Hopefully this video is what you're looking for when it comes to the divisibility rule for 7.
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Transcript:
[Music] welcome to math with mr j [Music] in this video i'm going to cover the divisibility rule for seven now when it comes to divisibility remember we're checking to see if we can divide without getting a remainder so the given numbers work out exactly so in this case we're going to take a look at four numbers here and see if they are divisible by seven we can divide by seven without getting a remainder now this is going to be based on the rule at the top of the screen we're going to double the last digit and subtract it from the remaining number if that gives us a number that's divisible by 7 then the original number is divisible by seven so let's take a look and see what that means and jump into number one where we have 574 so let's double the last digit which is four so four times two doubling it gives us 8. we need to subtract 8 from the remaining number so the remaining number is this 57 here so let's do 57 minus 8. that gives us 49. so we take a look at that remaining number there 49 and see if it is divisible by 7. if so the original number is divisible by 7. so can we do 49 divided by 7 and get an answer without a remainder yes 49 divided by 7 is 7 so 49 is divisible by 7 therefore our original number 574 is divisible by 7. let's take a look at number two where we have one hundred forty-seven the last digit is seven let's double it so seven times two equals fourteen our remaining number is fourteen here so 14 minus 14 gives us 0. remember 0 is divisible by any number therefore 0 is divisible by 7. that means our original number 147 is divisible by seven on to number three we have three thousand nine hundred seventy-six our last digit here is a six so let's double it 6 times 2 gives us 12. our remaining number is 397. so we need to do 397 minus twelve so seven minus two is five nine minus one is eight bring down our three three minus zero is three 385 is still kind of a large number there in value a three-digit number so we're going to repeat our steps actually so what we're going to do let's take a look at our last digit 5 and double it so we know 5 times 2 equals 10. so we can do 38 minus that 10 and that gives us a remaining number of 28. that's much easier to work with than 385. again we just repeated the steps to give us a more manageable number to work with so is 28 divisible by 7 can we do 28 divided by 7 and get an answer without a remainder yes 28 divided by 7 is 4. therefore our original number of 3976 is divisible by 7 as well so let's move on to number 4 here where we have 9822 our last digit here is a two so let's double it two times two equals four and now we need to subtract it from the remaining number which is nine hundred eighty two so nine hundred eighty two minus four that's going to give us 978. still a three-digit number and fairly large in value to work with here so we are going to repeat our steps let's take a look at the last digit which is 8 and double it so 8 times 2 gives us 16. so we need to do 9 so 97 minus 16 is going to give us a 1 in the ones place i'm going to write it over here so it's a little bit bigger a 1 and then an 8 here so we get 81. now 81 is not divisible by 7 therefore our original number 9822 is not divisible by 7 either so there you have it there's the divisibility rule for 7. i hope that helped thanks so much for watching until next time peace |
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2004 AIME I Problems/Problem 2
Contents
[hide]
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4 (Sketchy Solution for Speedrunners)
6 Solution 5
7 See also
Problem
Set consists of consecutive integers whose sum is , and set consists of consecutive integers whose sum is The absolute value of the difference between the greatest element of and the greatest element of is . Find
Solution 1
Note that since set has consecutive integers that sum to , the middle integer (i.e., the median) must be . Therefore, the largest element in is .
Further, we see that the median of set is , which means that the "middle two" integers of set are and . Therefore, the largest element in is . if , which is clearly not possible, thus .
Solving, we get
Solution 2
Let us give the elements of our sets names: and . So we are given that so and (this is because so plugging this into yields ). Also, so so and .
Then by the given, . is a positive integer so we must have and so .
Solution 3
The thing about this problem is, you have some "choices" that you can make freely when you get to a certain point, and these choices won't affect the accuracy of the solution, but will make things a lot easier for us.
First, we note that for set
Where and represent the first and last terms of . This comes from the sum of an arithmetic sequence.
Solving for , we find the sum of the two terms is .
Doing the same for set B, and setting up the equation with and being the first and last terms of set ,
and so .
Now we know, assume that both sequences are increasing sequences, for the sake of simplicity. Based on the fact that set has half the number of elements as set , and the difference between the greatest terms of the two two sequences is (forget about absolute value, it's insignificant here since we can just assume both sets end with positive last terms), you can set up an equation where is the last term of set A:
Note how i basically just counted the number of terms in each sequence here. It's made a lot simpler because we just assumed that the first term is negative and last is positive for each set, it has absolutely no effect on the end result! This is a great strategy that can help significantly simplify problems. Also note how exactly i used the fact that the first and last terms of each sequence sum to and respectively (add and to see what i mean).
Solving this equation we find . We know the first and last terms have to sum to so we find the first term of the sequence is . Now, the solution is in clear sight, we just find the number of integers between and , inclusive, and it is .
Note how this method is not very algebra heavy. It seems like a lot by the amount of text but really the first two steps are quite simple.
Solution 4 (Sketchy Solution for Speedrunners)
First, calculate the average of set and set . It's obvious that they are and respectively. Let's look at both sets. Obviously, there is an odd number of integers in the set with being in the middle, which means that is an odd number and that the number of consecutive integers on each side of are equal. In set , it is clear that it contains an even number of integers, but since the number in the middle is , we know that the range of the consecutive numbers on both sides will be to and to .
Nothing seems useful right now, but let's try plugging an odd number, , for in set . We see that there are consecutive integers and on both sides of . After plugging this into set , we find that the set equals . From there, we find the absolute value of the difference of both of the greatest values, and get 0.
Let's try plugging in another odd number, . We see that the resulting set of numbers is to , and to . We then plug this into set , and find that the set of numbers is to which indeed results in the average being . We then find the difference of the greatest values to be 26.
From here, we see a pattern that can be proven by more trial and error. When we make equal to , then the difference is whearas when we make it , then the difference is . equals to and is just . We then see that increases twice as fast as the difference. So when the difference is , it increased from when it was , which means that increased by which is . We then add this to our initial of , and get as our answer.
Solution 5
Let the first term of be and the first term of be . There are elements in so is . Adding these up, we get . Set contains the numbers . Summing these up, we get . The problem gives us that the absolute value of the difference of the largest terms in and is . The largest term in is and the largest term in is so . From the first two equations we get, we can get that . Now, we make a guess and assume that (if we get a negative value for , we can try ). From here we get that . Solving for , we get that the answer is
-Heavytoothpaste
See also
2004 AIME I (Problems • Answer Key • Resources)
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8967 | https://blog.collegevine.com/30-sat-math-formulas-you-need-to-know | 30 SAT Math Formulas You Need to Know | CollegeVine Blog
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Gianna CifredoFebruary 4, 2019 9 10th Grade,11th Grade,12th Grade,SAT Info and Tips
30 SAT Math Formulas You Need to Know
Have you ever been stumped on a math test question, only to realize how simple it was when you later reviewed your test? That’s exactly what happens to many students who take the SAT.
The SAT covers a wide range of math—from as early as elementary school all the way to senior year of high school. While you probably learned these formulas at some point, it may have been a while since you’ve had to use them. This is part of what makes the SAT challenging: since it draws on many different types of math, you need to get out of the high school math mindset (where you only remember what you’ve learned the last month or so to ace the test) and review the math you’ve learned over the years.
If you don’t do any studying for the SAT, you’ll take longer to recall certain helpful formulas and concepts. While there’s always more than one way to reach the right answer, being able to quickly remember these math facts will help you answer questions more efficiently and minimize careless mistakes. We’ve categorized these formulas to help you focus your preparation, and we’ve provided a quick review of what each concept is.
Want to know your chances at the schools you’re applying for based on your SAT score?Calculate your admissions chances right now.
Also, check out our video on how to ace the SAT math section!
Arithmetic and Algebra
1. Slope-Intercept Form of a Line
y=m x+b
The m in the equation represents the slope of the equation, and the b represents the y-value of the y-intercept. For example, if we have the equation y=2 x+4, then the slope is 2 and the y-intercept is (0,4).
2. Vertex Form of a Parabola/Quadratic
y=a(x−h)2+k
You may be more familiar with a quadratic in its factored form, or in the form y=a x 2+b x+c. However, you should be able to recognize vertex form and convert quadratics to this form for the SAT. The values of h and k give you the coordinates of the vertex, (h,k).
3. Distance Formula
d=(x 2−x 1)2+(y 2−y 1)2
The distance formula is derived from the Pythagorean Theorem (covered later in this post) and it’s useful for quickly finding the distance between two points. Take the values of the coordinates and plug them into this formula to find the distance, and be sure to apply the squares and the square root at the right step.
4. Quadratic Formula
x=−b±b 2−4 a c 2 a
The quadratic formula helps you find the roots of a quadratic equation (parabola) if you can’t easily factor it. You need the quadratic to be in the form y=a x 2+b x+c, and then you simply plug the coefficients and constants into the formula. Note that you will get two answers because there is the plus and minus sign in the numerator.
5. Exponent Rule (Multiplication)
a n a m=a n+m
Knowing how to manipulate exponents in a variety of ways will help you tremendously on the SAT, especially the no-calculator portion. In this case, if you have the same base number raised to different powers being multiplied together, you can add the exponents together.
6. Exponent Rule (Division)
a m a n=a m−n
Similar to the multiplication rule above, if you have the same base number raised to different powers being divided, you can subtract the exponents. You can also rewrite the expression on the right to mirror the one on the left.
7. Exponent Rule (Power Raised to a Power)
(a n)m=a n⋅m
To continue the exponent rules, raising a power to another power is the same as multiplying the exponents together. This is not an exhaustive list of ways to manipulate exponents, so if you find that you don’t remember these at all, be sure to brush up!
8. Binomial Product 1—Difference of Squares
(x−y)(x+y)=x 2−y 2
The best times to recognize the binomial products and quickly factor them is on the no-calculator section. You don’t have to FOIL or use any other method—you can quickly convert from the factored form to the expanded form on sight. The difference of squares is used often by SAT test makers in a variety of contexts.
9. Binomial Product 2—Perfect Squares Trinomial (Positive)
(x+y)2=x 2+2 x y+y 2
Students sometimes forget the perfect squares trinomial once they’ve left their algebra class, but this is also a good one to recognize. It saves you time because you can quickly convert from one form to another, but it’s a little more difficult to catch than the expanded difference of squares. A good way to know if you’re dealing with one is to look at the first and last values—are they perfect squares?
10. Binomial Product 2—Perfect Squares Trinomial (Negative)
(x−y)2=x 2−2 x y+y 2
This is similar to the above trinomial, except that the quantity involves subtraction rather than addition. While the factored form doesn’t include coefficients, the binomial products on the SAT often do. Practice recognizing these patterns by inputting coefficients in front of x and constants for y on the left-hand side. Then multiply out the expression to see how the pattern works with different combinations.
See how your SAT math score affects your chances of admission with CollegeVine’s chancing calculator.
11. Complex Conjugate
(a+b i)(a−b i)=a 2+b 2
On most SAT Math tests, there will be at least one question that involves manipulating imaginary numbers. The complex conjugate allows you to get rid of the imaginary part of a complex number and leaves with you a real number (notice how it resembles the difference of squares!). When given a complex number in the form a+b i, the conjugate is a−b i.
12. Exponential Growth and Decay
y=a(1±r)x
This is technically two different equations, one where there is a plus in the equation and one where there is a minus. Knowing the general format of exponential equations will help you on several SAT questions, as you may need to interpret or manipulate these equations. The value a is the initial value, r is the rate of growth when it’s positive and the rate of decay when it’s negative.
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13. Simple interest
A=P r t
This one appears less often than compound interest on the SAT, but it still shows up, so it’s worth knowing. P represents the principal amount, r is the interest rate expressed as a decimal, and t is for time, usually in years.
14. Compound interest
A=P(1+r n)n t
The good news is that P, r, and t mean the same thing in this equation as they do in simple interest. The n represents the number of times that the interest is compounded during 1 t. For example, if the interest is compounded quarterly over the course of a year, then n=4.
15. Average/Mean
In math, the words average and mean are the same thing: the number you get when you take the sum of a set and divide it by the number of values in the set. You could also think of it as the sum divided by the count. You should know how to calculate an average and interpret it. Be sure to understand the difference between mean and median.
16. Random Sampling
This isn’t technically a formula, but many of the statistics-based problems on the SAT focus more on interpreting concepts in context rather than performing mathematical operations. Random sampling is when you select participants for a study at random within your population. It ensures that your study is representative of the population.
17. Random Assignment
Random assignment is when the participants in a study are assigned a treatment or trial at random. It reduces bias in your study, and means that you can attribute causation in regards to the treatment. On the SAT, you’re often asked about what will reduce bias, or how much you can generalize results to the rest of the population. In these instances, you need to identify random sampling and random assignment.
18. Standard deviation
You won’t need to calculate standard deviation for the SAT, but you will be tested on it conceptually, as with with random sampling and random assignment. Standard deviation is the measure of spread in the data set. A higher standard deviation means greater spread, and lower standard deviations mean smaller spread. You’ll need to know how changes in the data set might affect the standard deviation by making it greater or smaller.
Geometry and Trigonometry
19. Area of an Equilateral Triangle
A=3 s 2 4
The regular area of a triangle formula is provided on the SAT reference sheet, but it requires that you know the height of a triangle. Sometimes you aren’t given the height and you’ll need to calculate it, but you can quickly find the area of an equilateral triangle by plugging the length of one of its sides into the formula above. No need to calculate the height!
20. Equation of a Circle
(x−h)2+(y−k)2=r 2
There is usually one question involving the equation of a circle. In this equation, (h,k) is the coordinate for the center of the circle, and r is the radius of the circle.
What role does your SAT score play in getting accepted to your dream school? Find out by calculating your chances now.
21. Sine Ratio
Some students get nervous when they hear that trig is on the SAT, but it most often appears in the form of trig ratios. Remember that for a given angle in a right triangle, the value of sine is the length of the opposite side divided by the length of the hypotenuse, or opposite/hypotenuse.
22. Cosine Ratio
Just like with sine, remember what the cosine ratio is: the length of the adjacent side divided by the length of the hypotenuse, or adjacent/hypotenuse.
23. Tangent Ratio
Last but not least, the tangent ratio is the length of the opposite side divided by the length of the adjacent side, or opposite/adjacent. Some students find the mnemonic SOH CAH TOA helpful for remembering trig ratios.
24. Degrees to Radians
While the most common form of trig are the basic ratios, you may encounter things like the unit circle or more advanced math. If you need to convert degrees to radians, multiply the degrees by π 180. If you need to convert radians to degrees, multiply the radians by 180 π.
25. Pythagorean Theorem
a 2+b 2=c 2
The Pythagorean Theorem applies to right triangles, and allows you to solve for one of the side lengths given any other side length. a and b are the legs of the triangle, and c is the hypotenuse.
26. Regular Polygon Interior Angle
(n−2)180 n
The SAT will probably involve one question with a regular polygon that isn’t a triangle or square. Regular polygons have unique and consistent properties based on their number of sides, and knowing these properties can help you solve these problems. This equation tells you what the degree measure at each angle is based on the number of sides n.
27. 3-4-5 triangle
The SAT provides you with two special right triangles you may already be familiar with on your reference sheet—the 30-60-90 and 45-45-90 triangles. However, the 3-4-5 is a special right triangle with sides that are straightforward integers. This triangle is often incorporated into SAT problems, especially the no-calculator portion, so be on the lookout for it! It can save you having to use the Pythagorean theorem.
28. 5-12-13 triangle
Another special right triangle with whole-number sides, the 5-12-13 triangle is less well-known and shows up less often than 3-4-5. Still, it helps to be able to quickly solve the remaining sides without the Pythagorean theorem, so check for these numbers or their multiples in triangle problems.
29. Length of Arc in a Circle
l e n g t h o f a r c=c e n t r a l a n g l e 360 π d
Although geometry questions don’t make up a huge portion of the SAT, you may still find a question either about arcs or sectors in a circle. An arc is the length between two points on a circle, usually measured by extending two radii from the center of the circle with an angle formed between them. You can use the degree measure of the arc as a fraction of 360 and multiply it by the equation for the circumference to find the length of the arc.
30. Area of Sector in a Circle
a r e a o f s e c t o r=c e n t r a l a n g l e 360 π r 2
Like an arc, the sector is the area in between two radii extending from the circle, sort of like a slice of pie. Again, multiply the degree measure as a fraction of 360 and multiply it by the equation for the area of a circle to find the area of the sector.
Wrapping it Up
Before you go, we’re going to offer you a bonus tip: you may want to memorize the perfect squares and perfect cubes. This can help you with quadratic equations that often involve squares, and cubes are often used in solving problems with exponents. Memorizing these will cut down on your need to do math with scratch paper or calculator.
The best way to be able to remember formulas is to practice using them. Unlike your high school math test, where you know what topics will be covered, the SAT will simply present you with a question—it’s up to you to determine what formulas apply. When you practice using formulas with a variety of problems, you’ll be able to quickly identify which formula to use.
Preparing for the SAT? Download ourfree guide with our top 8 tips for mastering the SAT.
Want to know how your SAT score impacts your chances of acceptance to your dream schools? Our free Chancing Engine will not only help you predict your odds, but also let you know how you stack up against other applicants, and which aspects of your profile to improve.Sign up for your free CollegeVine account todayto gain access to our Chancing Engine and get a jumpstart on your college strategy!
Check out some of our other posts on math prep:
How to Convert Celsius to Fahrenheit Quickly
The Difference Between Independent and Dependent Variables
The Complete Guide to the 30-60-90 Triangle
15 Hardest SAT Math Questions
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8968 | https://www.quora.com/If-a-b-and-c-d-and-all-a-b-c-d-are-positive-integers-proof-that-ac-bd | If a>b and c>d and all a,b,c,d are positive integers, proof that ac > bd. - Quora
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If a>b and c>d and all a,b,c,d are positive integers, proof that ac > bd.
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Awnon Bhowmik
I know a little about elementary Number Theory · Upvoted by
Kostyantyn Mazur
, PhD Mathematics, New York University (2018) · Author has 3.7K answers and 11.2M answer views
·8y
a>b a b>1 c>d c d>1 Multi plying gives a c b d>1 a c>b d a>b a b>1 c>d c d>1 Multi plying gives a c b d>1 a c>b d
a>b a c>b c c>d b c>b d a c>b c>b d⟹a c>b d a>b a c>b c c>d b c>b d a c>b c>b d⟹a c>b d
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How do you prove that a<b,c<d⟹a+c<b+d a<b,c<d⟹a+c<b+d?
How do you prove that if (a/b) >(c/d) then (a.d) >(b.c)?
Is there a positive integer solution for a, b, and c if c/(a+b)+b/(a+c)+a/(b+c)=4 c/(a+b)+b/(a+c)+a/(b+c)=4?
Can you prove that 2√7 is irrational?
If a,b,c,d,e∈R,a+b+c+d+e>0 a,b,c,d,e∈R,a+b+c+d+e>0, how do I prove this inequality 1 a+b+c+d+e∣∣ ∣ ∣ ∣ ∣∣a b c d e e a b c d d e a b c c d e a b b c d e a∣∣ ∣ ∣ ∣ ∣∣≥0 1 a+b+c+d+e|a b c d e e a b c d d e a b c c d e a b b c d e a|≥0?
Trevor Muhammad
Knows a small amount of math · Author has 142 answers and 2.5M answer views
·8y
Let a, b, c, d > 0. We can use properties of inequalities as follows.
If x < y and z > 0, then xz < yz. Similarly,
If x > y and z > 0, then xz > yz.
If a > b, then we multiply both sides by c:
ac > bc
If c > d, then we multiply both sides by b:
bc > bd
Using the transitive property of inequalities,
If ac > bc and bc > bd, then
ac > bd
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99 14
Andrew Foust
Lived in Wichita, KS · Author has 376 answers and 397K answer views
·8y
We have
a>b∧c>d(I)(I)a>b∧c>d
a,b,c,d∈(0,∞)∈Z(II)(II)a,b,c,d∈(0,∞)∈Z
a>b⟹a=b+φ 1(III-a)(III-a)a>b⟹a=b+φ 1
c>d⟹c=d+φ 2(III-b)(III-b)c>d⟹c=d+φ 2
(b+φ 1)×(d+φ 2)>b×d(IV)(IV)(b+φ 1)×(d+φ 2)>b×d
With the substition, it becomes very clear that for any φ 1 φ 1 and φ 2>0 φ 2>0 the inequality must hold.
d φ 2+b φ 1+φ 1 φ 2>0✓d φ 2+b φ 1+φ 1 φ 2>0✓
Shitty visual proof!
Continue Reading
We have
a>b∧c>d(I)(I)a>b∧c>d
a,b,c,d∈(0,∞)∈Z(II)(II)a,b,c,d∈(0,∞)∈Z
a>b⟹a=b+φ 1(III-a)(III-a)a>b⟹a=b+φ 1
c>d⟹c=d+φ 2(III-b)(III-b)c>d⟹c=d+φ 2
(b+φ 1)×(d+φ 2)>b×d(IV)(IV)(b+φ 1)×(d+φ 2)>b×d
With the substition, it becomes very clear that for any φ 1 φ 1 and φ 2>0 φ 2>0 the inequality must hold.
d φ 2+b φ 1+φ 1 φ 2>0✓d φ 2+b φ 1+φ 1 φ 2>0✓
Shitty visual proof!
Upvote ·
9 4
Vijay Mankar
HoD (Electronics) at Government Polytechnic Nagpur · Author has 7K answers and 11.2M answer views
·6y
a>b a>b
a b>1(1)(1)a b>1
c>d c>d
c d>1(2)(2)c d>1
Multiplying (1),(2)⟹(1),(2)⟹
a c b d>1 a c b d>1
a c>b d■a c>b d◼
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Can you provide a proof for the statement "if a and b are positive integers, then ab is also positive"?
How do I proof that for any integers a and b, (a + b, [a,b]) = (a,b)?
If (a−b)(b−c)(c−a)=a+b+c(a−b)(b−c)(c−a)=a+b+c and a,b,c a,b,c are positive integers, what is the minimum possible value of a+b+c a+b+c?
If a×b×c=36, b×c×d=48, c×d×a=324 and d×a×b=18, then what is the value of a×b×c×d?
How can one prove that sqrt(2) + sqrt(3) = sqrt(5 + 2sqrt(6))?
Gregory Scott
Works at Photography · Author has 7.6K answers and 7.4M answer views
·8y
Originally Answered: If you are given that a, b, c, and d are all positive real numbers, and that a>b and c>d, how do you demonstrate that ac > bd? ·
A demonstration is not necessarily a proof. This demonstration is intuitive, but convincing.
A>B, C>D, thus AC the area of a larger rectangle, and BD is the area of a smaller rectangle.
Continue Reading
A demonstration is not necessarily a proof. This demonstration is intuitive, but convincing.
A>B, C>D, thus AC the area of a larger rectangle, and BD is the area of a smaller rectangle.
Upvote ·
9 2
Sagar Alva
Studied at Nitte University, Mangalore
·6y
Originally Answered: “Prove that if a, b, c, d are positive real numbers, a > b, and c > d, then AC > bd.” How do I solve this? ·
We have 4 unknowns a,b,c,d. Acc. to the problem
a+b=8……….(1)
c-d=6……….(2)
a+c=13………(3)
b+d=8……….(4)
Now,
(3)-(1) => c-b=5…….(5)
(2)+(4)=> c+b=14……(6)
(5)+(6) => 2c=19 =>c=19/2=9.5
Put this value of ‘c’ in (6)
b=14–9.5=4.5
Put ‘b’ in (1)
a=8–4.5=3.5
Put ‘b’ in (4)
d=8–4.5=3.5
The answers :-3.5,4.5,9.5,3.5
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99 65
John Pye
old timer · Upvoted by
Justin Rising
, PhD in statistics · Author has 276 answers and 379K answer views
·8y
Originally Answered: If you are given that a, b, c, and d are all positive real numbers, and that a>b and c>d, how do you demonstrate that ac > bd? ·
Sure.
Multiply a>b a>b by c c to get a c>b c a c>b c .
Multiply c>d c>d by b b to get b c>b d b c>b d .
Put them together and what have you got?
Upvote ·
9 6
9 1
Keith Bates
Studied Mathematics&Computing at The Open University (Graduated 1990)
·4y
Originally Answered: If you are given that a, b, c, and d are all positive real numbers, and that a>b and c>d, how do you demonstrate that ac > bd? ·
a>b where a and b are both positive real numbers
let a-b=m
m must be a positive real number
so a=b+m
likewise
c>d where c and d are both positive real numbers
so c=d+n
ac=(b+m)(d+n)
ac=bd+md+bn+mn
md, bn and mn are all products of positive real numbers, so their sum is a positive real number
Hence ac>bd
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George Ivey
Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views
·Mar 13
Originally Answered: “Prove that if a, b, c, d are positive real numbers, a > b, and c > d, then AC > bd.” How do I solve this? ·
Technically you DON’T because “a, b, c, d” is NOT “A, B, C, D”. Do not use upper case and lower case letters interchangeably.
If a, b, c, and d are positive numbers and a> b then ac> bc and if c> d then bc> bd.
Then ac> bc> bd so that ac> bd.
Upvote ·
Robert Gillespie
Philosopher, Folklorist and Mathematician at Large (2009–present) · Author has 1K answers and 900.7K answer views
·8y
Originally Answered: If you are given that a, b, c, and d are all positive real numbers, and that a>b and c>d, how do you demonstrate that ac > bd? ·
For a>b multiply both sides by c
ac > bc
Since c is greater than d we can substitute d for c on the right hand side without changing the inequality.
ac > bd
Q.E.D.
Upvote ·
9 1
9 5
Steve Jones
Studied at Imperial College London · Author has 13.4K answers and 34.9M answer views
·8y
Originally Answered: If you are given that a, b, c, and d are all positive real numbers, and that a>b and c>d, how do you demonstrate that ac > bd? ·
So a>b a>b, c>d c>d and a,b,c a,b,c and d d are all non-negative
First we can state
a>b a>b is equivalent to a=b+x a=b+x where x>0 x>0
similarly, c>d c>d is equivalent to c=d+y c=d+y where y>0 y>0
So, if we take the inequality a×c>b×d a×c>b×d and substitute for a a and c c, then we get :-
(b+x)×(d+y)>b×d(b+x)×(d+y)>b×d
expand that and you get
b d+b y+d x+y x>b d b d+b y+d x+y x>b d
subtract b d b d from both sides and you get
b y+d x+y x>0 b y+d x+y x>0
We know that x>0 x>0 and y>0,y>0, so y x>0 y x>0 . As both b and d are non-negative, then b y+d x+y x b y+d x+y x must be greater than zero.
Upvote ·
9 1
9 1
Himaan Sahab
Art enthusiastic
·6y
Originally Answered: “Prove that if a, b, c, d are positive real numbers, a > b, and c > d, then AC > bd.” How do I solve this? ·
a,b,c,d are positive real no.
So, ‘ac’ and ‘bd’ are also positive real.
If a>b and c>d then it is obvious that
ac>bd
Upvote ·
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How do you prove that a<b,c<d⟹a+c<b+d a<b,c<d⟹a+c<b+d?
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Can you provide a proof for the statement "if a and b are positive integers, then ab is also positive"?
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8969 | https://calculator.name/baseconvert/binary/hexadecimal/1011.101 | How to Convert 1011.101 from binary to hexadecimal
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Convert 1011.101 from binary to hexadecimal
What is 1011.101 binary in hexadecimal? 1011.101 from binary to hexadecimal is B.A. Here we show you how to write 0b1011.101 in hexadecimal and how to convert 1011.101 from base-2 to base-16.
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Result : 1011.101 2 = B.A 16
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In numeral system, we know binary is base-2 and hexadecimal is base-16. To convert binary 1011.101 to hexadecimal, you follow these steps:
To do this, first convert binary into decimal, then the resulting decimal into hexadecimal
Start from one's place in binary : multiply ones place with 2^0, tens place with 2^1, hundreds place with 2^2 and so on from right to left
Add all the products we got from step 1 to get the decimal equivalent of given binary value.
Then, divide decimal value we got from step-2 by 16 keeping notice of the quotient and the remainder.
Continue dividing the quotient by 16 until you get a quotient of zero.
Then just write out the remainders in the reverse order to get hexadecimal equivalent of decimal number.
First, convert 1011.101 2 into decimal, by using above steps:
= 1011 2
= 1 × 2 3 0 × 2 2 1 × 2 1 1 × 2 0 1 × 2-1 0 × 2-2 1 × 2-3
= 11.625 10
Now, we have to convert 11.625 10 to hexadecimal
11 / 16 = 0 with remainder 11 (B)
11 = B ------- (1)
For converting decimal fraction 0.625 to hexadecimal number, follow these steps:
Multiply 0.625 by 16 keeping notice of the resulting integer and fractional part. Continue multiplying by 16 until you get a resulting fractional part equal to zero (we calcuclate upto ten digits).
Then just write out the integer parts from the results of each multiplication to get equivalent hexadecimal number.
0.625 × 16 = 10 (A) + 0
0.625 = 0.A ------- (2)
11.625 10 = B.A 16
Therefore, binary number 1011.101 converted to hexadecimal is equals:
B.A
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1101.101 binary to hexadecimal
1110.101 binary to hexadecimal
1111.101 binary to hexadecimal
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The smallest positive integer with exactly 8 divisors (including 1 and the number itself as divisors) is 24. Find the next higher integer with exactly 8 divisors.
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Expert Verified Solution
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Answer
The answer is 30
Explanation
Understand the divisor function
The number of divisors of an integer $$n$$n with prime factorization $$n = p_1^{a_{1}} p_2^{a_{2}} \cdots p_k^{a_{k}}$$n=p 1 a 1p 2 a 2⋯p k a k is given by $$\tau(n) = (a_{1}+1)(a_{2}+1)\cdots(a_{k}+1)$$τ(n)=(a 1+1)(a 2+1)⋯(a k+1). We want to find the next integer $$n > 24$$n>24 such that $$\tau(n) = 8$$τ(n)=8
Find possible combinations of exponents
Since we want $$\tau(n) = 8$$τ(n)=8, we need to find combinations of exponents such that their product is 8. The possible combinations are:
\begin{itemize}
\item $$8 = 8$$8=8, which means $$n = p^{7}$$n=p 7 for some prime $$p$$p
\item $$8 = 4 \times 2$$8=4×2, which means $$n = p^{3} q^{1}$$n=p 3 q 1 for some distinct primes $$p$$p and $$q$$q
\item $$8 = 2 \times 2 \times 2$$8=2×2×2, which means $$n = p^{1} q^{1} r^{1}$$n=p 1 q 1 r 1 for some distinct primes $$p$$p, $$q$$q, and $$r$$r
\end{itemize}
Analyze the case $$n = p^{7}$$n=p 7
The smallest integer of this form is $$2^{7} = 128$$2 7=128
Analyze the case $$n = p^{3} q$$n=p 3 q
We want to find integers of the form $$p^{3} q$$p 3 q greater than 24.
\begin{itemize}
\item If $$p = 2$$p=2, then $$n = 8q$$n=8 q. We want $$8q > 24$$8 q>24, so $$q > 3$$q>3. The smallest such prime is $$q = 5$$q=5, so $$n = 8 \times 5 = 40$$n=8×5=40
\item If $$p = 3$$p=3, then $$n = 27q$$n=27 q. We want $$27q > 24$$27 q>24, so $$q$$q can be 2. Then $$n = 27 \times 2 = 54$$n=27×2=54
\end{itemize}
Analyze the case $$n = pqr$$n=pq r
We want to find integers of the form $$pqr$$pq r greater than 24. To minimize $$n$$n, we choose the smallest primes 2, 3, and 5.
Then $$n = 2 \times 3 \times 5 = 30$$n=2×3×5=30
Compare the integers found
We have the following integers with exactly 8 divisors: 128, 40, 54, and 30.
Since we are looking for the next integer greater than 24, we compare 30, 40, 54, and 128. The smallest of these is 30.
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8971 | https://www.jstor.org/stable/10.5951/mathteacher.107.8.0632 | Finding Sums of Positive Consecutive Integers for Any Given Total on JSTOR
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journal article
Finding Sums of Positive Consecutive Integers for Any Given Total
Christopher P. Gillotte
The Mathematics Teacher
Vol. 107, No. 8 (April 2014), pp. 632-635 (4 pages)
Published By: National Council of Teachers of Mathematics
Cite
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Abstract
The number of sets of two or more positive consecutive integers that can be added to obtain a given number is a property of the exponents of the odd prime factors of the number.
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The Mathematics Teacher (MT), an official journal of the National Council of Teachers of Mathematics, is devoted to improving mathematics instruction from grade 8-14 and supporting teacher education programs. It provides a forum for sharing activities and pedagogical strategies, deepening understanding of mathematical ideas, and linking mathematics education research to practice.
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The National Council of Teachers of Mathematics is a public voice of mathematics education, providing vision, leadership, and professional development to support teachers in ensuring mathematics learning of the highest quality for all students. With nearly 90,000 members and 250 Affiliates, NCTM is the world's largest organization dedicated to improving mathematics education in grades prekindergarten through grade 12. The Council's "Principles and Standards for School Mathematics" are guidelines for excellence in mathematics education and issue a call for all students to engage in more challenging mathematics. NCTM is dedicated to ongoing dialogue and constructive discussion with all stakeholders about what is best for our nation's students.
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8972 | https://www.mathsisfun.com/geometry/rectangle.html | Rectangle
(Jump to Area of a Rectangle or Perimeter of a Rectangle)
A rectangle is a four-sided flat shape where every angle is a right angle (90°).
the little squares in each corner mean "right angle"
| Each internal angle is 90°
| Opposite sides areparalleland of equal length (so it is aParallelogram).
Opposite sides are parallel and of equal length (so it is a Parallelogram).
Play with a rectangle:
Area of a Rectangle
| | Area = a à b
Area = a à b
Example: A rectangle is 6 m wide and 3 m high, what is its Area?
Perimeter of a Rectangle
The Perimeter is the distance around the edges.
| | The Perimeter is2 times (a + b):Perimeter = 2(a+b)
The Perimeter is 2 times (a + b):
Perimeter = 2(a+b)
Example: A rectangle is 12 cm long and 5 cm tall, what is its Perimeter?
Diagonals of a Rectangle
| | Arectanglehas two diagonals, they are equal in length and intersect in the middle.
A rectangle has two diagonals, they are equal in length and intersect in the middle.
| | A diagonal's length is thesquare root of (a squared + b squared):Diagonal "d" = â(a2+ b2)
A diagonal's length is the square root of (a squared + b squared):
Diagonal "d" = â(a2 + b2)
Example: A rectangle is 12 cm wide, and 5 cm tall, what is the length of a diagonal?
Golden Rectangle
There is also a special rectangle called the Golden Rectangle: |
8973 | https://artofproblemsolving.com/wiki/index.php/Elementary_symmetric_sum?srsltid=AfmBOoo4hm3KHY45A7Z_nnL1po9trzlVml83eRHlgobuzXJwAKcdn5YV | Art of Problem Solving
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Elementary symmetric sum
An elementary symmetric sum is a type of summation.
Contents
[hide]
1 Definition
2 Notation
3 Uses
4 See Also
Definition
The -th elementary symmetric sum of a set of numbers is the sum of all products of of those numbers (). For example, if , and our set of numbers is , then:
1st Symmetric Sum =
2nd Symmetric Sum =
3rd Symmetric Sum =
4th Symmetric Sum =
Notation
The first elementary symmetric sum of is often written . The th can be written
Uses
Any symmetric sum can be written as a polynomial of the elementary symmetric sum functions. For example, . This is often used to solve systems of equations involving sums of powers, combined with Vieta's formulas.
Elementary symmetric sums show up in Vieta's formulas. In a monic polynomial of degree , the coefficient of the term is , and the coefficient of the term is , where the symmetric sums are taken over the roots of the polynomial.
See Also
Symmetric sum
Cyclic sum
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8974 | https://www.youtube.com/watch?v=Jaq2wB_s4tw | Newton's 2nd Law: Static and Kinetic Friction. Level 1, Example 1
JHughesPhysics
2290 subscribers
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Posted: 24 Oct 2017
Newton's 2nd Law: Static and Kinetic Friction. Level 1, Example 1
A 250-kg crate is at rest on a horizontal floor. A horizontal force of just over 980 N is required to make it move and a force of 490 N is required to keep it moving at constant velocity. Calculate μs and μk, the coefficients of friction between the crate and the floor.
Transcript:
the crate in this problem is at rest on a horizontal floor it has a mass of 250 kilograms when someone pushes on it with a force P static friction pushes back in the opposite direction just as hard so that the crate doesn't move we're told how hard a person has to push to finally make it move and then how hard they have to push to keep it moving at a constant velocity and we're asked to find the coefficients of friction in almost any problem dealing with friction we're going to need to know the normal force so let's find that first by drawing a Freebody diagram and solving for the normal force there are four forces acting on the crate its weight is straight down the normal force is perpendicular to the surface so that's straight up in this case the pushing force is to the right and static friction is to the left when we apply Newton's second law summing forces in the vertical direction we see that it equals zero because there's no vertical acceleration adding the vertical forces from the Freebody diagram shows us that the normal force acting on the crate equals its weight and now we can write down the expression for the maximum static friction force the highest value that static friction can take is mu sub s times n where mu sub s is the coefficient of static friction and n is the normal force this problem tells us that that value equals 980 Newtons solve this for the coefficient of static friction mu sub s it equals the ratio of the maximum static friction force divided by the normal force substitute the known values to find that mu sub s equals 0.40 now we've pushed hard enough to get the box moving and we're told that a lesser force 490 Newtons is required to keep it moving at constant velocity so now the box is in motion and friction has changed from static friction to kinetic kinetic now because the box is sliding across the surface if we apply Newton's second law in the horizontal direction now adding all the forces we see that it again equals zero because the box is now moving at constant velocity sideways so there's no horizontal acceleration summing the forces from the Freebody diagram gives us a pushing force minus the kinetic friction force equaling zero solve this for the pushing force and substitute their relation kinetic friction equals mu sub K times N and the normal force equals the weight solve this for the coefficient of kinetic friction and substitute the known values you to see that it equals 0.20 half of the static value |
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8976 | https://r-knott.surrey.ac.uk/runsums/index.html | Introducing Runsums - a sum of consecutive integers
This page investigates numbers that are the sum of a run of whole numbers, such as 5+6+7 or 2+3+4+5+6, their properties and fascinating patterns.
The calculators on this page require JavaScript but you appear to have switched JavaScript off (it is disabled). Please go to the Preferences for this browser and enable it if you want to use the calculators, then Reload this page.
Contents of this page
The icon means there is a You do the maths... section of questions to start your own investigations. The calculator icon indicates that there is a live interactive calculator in that section.
Runsums: Sums of Runs
A run of numbers is a sequence of consecutive whole numbers i.e. with no gaps in the sequence,
| | |
--- |
| such as | 2, 3, 4; |
| and | 7 | is a run even though it contains just a single number |
| but not | 5, 6, 8, 9 | because 7 is missing |
| and not | 2, 3, 3 | because 3 is repeated |
Every number is therefore a sum of numbers in a run because we can have a run with just a single number in it! Many numbers can also be written as a sum of a run of 2 or more numbers. For example:
| |
| 9 = 2 + 3 + 4 = 4 + 5 |
| 10 = 1 + 2 + 3 + 4 |
| 11 = 5 + 6 |
| 12 = 3 + 4 + 5 |
| 13 = 6 + 7 |
| 14 = 2 + 3 + 4 + 5 |
For convenience, the sum of a run of numbers we will call a runsum.
Making a table of runsums
Since every runsum is determined by its staring and ending number, we can make a table of the sums of integers between the two values:
| | |
--- |
| SUMS | To |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| From | 1 | 1 | 1+2=3 | 1+2+3=6 | 1+2+3+4=10 | 1+2+3+4+5=15 | 21 | 28 | 36 | |
| 2 | | 2 | 2+3=5 | 2+3+4=9 | 2+3+4+5=14 | 20 | 27 | 35 | |
| 3 | | | 3 | 3+4=7 | 3+4+5=12 | 18 | 25 | 33 | |
| 4 | | | | 4 | 4+5=9 | 15 | 22 | 30 | |
| 5 | | | | | 5 | 11 | 18 | 26 | |
| 6 | | | | | | 6 | 13 | 21 | |
| 7 | | | | | | | 7 | 15 | |
| 8 | | | | | | | | 8 | |
| 9 | | | | | | | | | 9 |
Each entry in the table is the sum of the whole numbers starting at the from value on the left of its row and ending at the to value at the top of its column. A few entries are shown in detail. So first fill in the rest of the entries in the table and then answer these questions:
You do the maths...
9 occurs in row 2 and column 4 and so the sum of the numbers from 2 to 4 is 9: checking: 2+3+4 is 9. So by looking at the numbers in the table, we can find runsums. Find two more locations in the table with 9 in it. What runsums are they?
How many runsums can you find for 12?
How many runsums for 15 can you find in the table?
Use your table to find all the runsums of 2, 3, 4, 5, 6, 7, 8, 9 and 10. There is only one runsum for 1 and for 2. If there is more than one runsum for a number, write each runsum on a line of its own in the right box. Put your results into a new table like this:
| | | | | | | | | | | |
--- --- --- --- ---
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | ... |
| runsumsof n | 1 | 2 | 31+2 | 4 | ? | ? | ? | ? | 92+3+44+5 | ... |
5. Is our table big enough to find all the runsums for 20? No, because some of the lower rows do not go far enough to the right. Extend your table so that you are sure all the numbers up to 20 are in your extended table and then answer these questions:
1. Find all the runsums of 15
2. Extend your new table of all runsums so that it includes all the runsums up to 20.
Triangular Numbers
By placing coins (cans, boxes) in rows, one row above another, we can make triangular patterns:
In these triangles, there are 1, 3, 6 and 10 balls. How many will there be in the next triangle, the one with 5 on the bottom row?
The numbers in this series are called Triangular Numbers and we give them the names T(1), T(2), T(3) and so on e.g.T(1) = 1, T(2) = 3, T(3) = 6. The tenth Triangular number's pattern will have 10 items on the bottom row and 1 on the top row, a total number of T(10) objects in it.
Looking at all the rows in one pattern above, the total number of boxes is:1 = 1 in the first 1 + 2 = 3 in the second 1 + 2 + 3 = 6 in the third 1 + 2 + 3 + 4 = 10 in the fourth and so on. So can you now calculate T(10) without drawing it? T(10) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55.
To summarise:
All triangular numbers are runsums - the special runsums that begin at 1. T(n)'s longest row is of length n. T(n) is the sum of the numbers from 1 up to and including n.
The first 12 Triangular numbers
| | | | | | | | | | | | | | |
--- --- --- --- --- --- --- |
| i | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| T(i) | 0 | 1 | 3 | 6 | 10 | 15 | 21 | 28 | 36 | 45 | 55 | 66 | 78 |
Runsums and Triangular Numbers
If we look at a triangular pattern above and take off a triangle from the top of it, we will still be left with a runsum, but the smallest row, the topmost one, will be bigger than 1. Here we take off T(3)=6 shown in green from T(6)=21:This shows that T(6) - T(3) = 21 - 6 =15. Looking at the red rows only we see 15 can be written as 4 + 5 + 6, a runsum.
This will work for any runsum. For instance, 11 = 5 + 6. The largest number in the runsum is 6 so we start with T(6)'s triangle. But our runsum begins at 5, so we have removed T(4) = 1 + 2 + 3 + 4. Therefore
11 = 5 + 6 = T(6) - T(4) Any runsum can be written as the difference between two triangular numbers If N = a + (a+1) + ... + b then N = T(b) – T(a–1)
Note that a can be 1 so that N = T(b) –T(0) = T(b) and we include the triangular numbers themselves as runsums.
The shape of a triangle with its top cut off is called a trapezium and so runsums are also called trapezoidal arrangements or trapezoids (in article 1999.1.6 in the Journal of Integer Sequences, Vol 2 (1999) by Tom Verhoeff of Eindhoven University of Technology).
Shyam Sunder Gupta's page on Triangular Numbers is full of fascinating triangular number facts and formulae.
Numbers as shapes: Polygonal and Figurate Numbers has a section on finding the simple formula for Triangle numbers
Numbers with several runsums
By looking at all the possibilities, we find that 4 cannot be written as a runsum of more than a single number. Every number has a runsum consisting of that number alone, so such runsums are "trivial" and we want to find "interesting" runsums! Is 4 perhaps just a special case? Are there other numbers that have no "interesting" runsums? How can we tell if a number has an "interesting" runsum or not?
Trivial and interesting runsums
You do the maths...
Find another number that has no single runsums (hint: apart from 4, there is another one less than 10).
What is the SMALLEST number that is a sum of a run of three numbers?
What is the smallest sum of four numbers in a run?
.. and five?
By spotting the pattern can you tell me what is the smallest number which is a sum of one hundred consecutive numbers - but I want to know HOW you know too!
Mathematicians call the single-number runsums trivial (meaning that they are not very interesting cases) and the others are non-trivial or proper runsums (the interesting ones). I went on looking for a proper runsum for 4 and 8, and I still couldn't find any. I did notice that 9 was also interesting because it was the first number I that had three different runsums: 9 = 2 + 3 + 4 = 4 + 5 I then found that 27 had four different runsums, what were they?
What is the smallest number with four different runsums?
I thought I'd see if I could find a number with five runsums but when I was searching, I found a number with six runsums! What is the smallest six runsum number that I could have found?
How many numbers between 2 and 100 have six different runsums?
While trying to answer the last question, I did find just one number with exactly five runsums. What was it?
There were still some numbers in the range 2 to 100 that seemed to have no runsums longer than one number. Can you spot a pattern in the list of such numbers?
What is special about the numbers with exactly three runsums (such as 9)?
A Runsum Calculator
Here is an online calculator that can help with your investigations. All numbers have a (trivial) runsum of length 1, namely the number itself. Runsums are abbreviated to the first and last numbers in the run, for example: 3 + 4 + 5 is shown as 3..5 by the Calculator.
C A L C U L A T O R
R E S U L T S
| | |
--- |
| | |
| | | |
---
Runsums | Friend & Neighbours |
It will instantly tell you that
the sum of the numbers from 200 to 800 is 300500;
there are 12 runsums for 9999 and it will show you what they are.
Here are some more investigations to do using the Runsum Calculator:
You do the maths...
What is the sum of all the numbers from 2 to 8?
from 20 to 80?
from 200 to 800? [Why is it not 10 times the answer above?]
Can you spot the pattern in these answers?
Try the same thing but for 2 to 9, 20 to 90, etc. Is there a pattern now?
What about 3 to 7, 30 to 70, and so on? Is there a pattern here?
Try 10 to 14, 110 to 114, 1110 to 1114 and so on. How would you describe this pattern?
Can you find more number patterns like these?
Please email me (address at foot of page) with your answers to these questions and the results will be put on this site for others to see. There are probably many kinds of patterns here!
6. A more advanced Project (age 15 and above):Let's use the notation sum(a,b) to mean the sum of all the numbers from a to b. So sum(2,5) = 2+3+4+5 = 14. What is
1. sum(1,2)?
2. sum(1,3)?
3. sum(1,4)?
4. Find a formula for sum(1,b).
5. What is the name for the series of numbers of the previous question?
7. 1. What is sum(10,20)?
2. Suppose we know that sum(1,20) = 210 and sum(1,9) = 45, how can we use these two values to compute sum(10,20)?
3. Using your formula for sum(1,b), use it to write down a formula for sum(a,b).
Here is my page of Runsum Facts and Figures which provides answers to many of the questions above.
Quick Ways To Calculate Runsums
First we find a formula for any runsum between two given numbers that is easy to remember and then we adapt it to find an easier way to use it for computing runsums in your head without a calculator.
A Basic Formula and proof
If we want to sum the numbers from 10 to 20, we can arrange two copies of the numbers like this, the top one listing them forwards form 10 to 20 and the bottom line with the numbers listed backwards from 20 down to 10:
| | | | | | | | | | | |
--- --- --- --- ---
| 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 |
| 20 | 19 | 18 | 17 | 16 | 15 | 14 | 13 | 12 | 11 | 10 |
Note that each column adds up to the same value: 30. The first column is also just the first value plus the last. How many columns are there? From 10 to 20 is 20-10 plus one = 11 Thus the sum of all the numbers in the table is 11 copies of 30 or 330.
This adds two copies of the list so the list 10 to 20 has a sum of 330/2 = 165.
This method applies to all lists of consecutive numbers from a to b. Each column has the same sum: a+b and there are b–a+1 columns. So (b–a+1)(a+b) is twice the sum of the numbers from a to b: Let's use sum(a,b) to mean a + (a+1) + (a+2) + ... + b So our first forumula is
| | | | |
--- --- |
| sum(a,b) = | (b – a + 1)(a + b) | = (b – a + 1) | (a + b) |
| 2 | 2 |
The second form gives us an easy way to remember this formula:
sum(a,b) = number of values × average value where the average value is (a + b)/2 and the number of values is b – a + 1
This formula is fairly easy to remember but not so good to use to calculate runsums in your head: Here is another example : What is the sum of the numbers from 6 to 20? There are (20 – 6 + 1) = 15 columns each with a sum of 6 + 20 = 26. So twice the sum we want is 15 × 26. So the sum of the numbers from 6 to 20 is half of this: 15 × 13 (hmmm --- out with the calculator at this point perhaps!) which turns out to be 195.
When the runsums start at 1 we have a formula for the sum of the first n numbers, or the n-th triangle number T(n):
| | |
--- |
| T(n) = 1 + 2 + ... + n = | n (n + 1) |
| 2 |
An easier formula for mental arithmetic
Rearranging our formula above we have:
| | | | | | | | |
--- --- --- --- |
| sum(a,b) = | | | | (b – a + 1)(a + b) | | 2 | | = | | | | b2 – a2 + b + a | | 2 | |
or sum(a,b) is (the difference of the squares of a and a PLUS their sum) over 2
So, for sum(6,20) we have 202 = 400 and 62 36, with a difference of 364. The sum of 6 + 20 = 26 which we add on to 364 to get 390. Halving this we have 195 as before. These sums are make excellent practice for your mental arithmetic skills!!
Friends and Neighbours
Where a number (sum) has more than one runsum, we can find more patterns. In this section we look for
runsums that have one number in common
such as 2 + 3 + 4 = 4 + 5. Since the runsums have something in common, we will call them friends. Other pairs of runsums (for the same sum) are neighbours if one starts with the value next to where the other ends, such as in 4 + 5 + 6 = 7 + 8.
Friendly runsums
We are calling the two runsums 2 + 3 + 4 and 4 + 5 friends since they have the same sum and share a number in common (4), so that one runsum starts where the other ends.
Are there any more?
21 is the sum for the next pair of friends: 1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8 . There are quite a lot of friendly runsums when you start to look. The sums with the pair of friendly runsums here are 9 and 21 and the series of such sums is:
| | | | |
--- --- |
| 9= | 2+3+4 | = | 4+5 |
| 21= | 1+2+3+4+5+6 | = | 6+7+8 |
| 30= | 6+7+8+9 | = | 9+10+11 |
| 42= | 3+4+5+6+7+8+9 | = | 9+10+11+12 |
| 65= | 2+3+4+5+6+7+8+9+10+11 | = | 11+12+13+14+15 |
| 70= | 12+13+14+15+16 | = | 16+17+18+19 |
and continues with 99, 105, 117, 133, 135, 154, 175, 180, ... see A110701 The common numbers in these friendly runsums are 4,6,9,11,14,15,16 and the series continues indefinitely (see Sloane's A094550)
If we omit the common number, the new series of sums is:5,15,21,33,54,54,85,90,100, ... (see Sloane's A110702)
Balancing Numbers
A Behera and G K Panda in a paper entitled On The Square Roots of Triangular Numbers in Fibonacci Quarterly, 1999, pages 98-105, use the term balancing number for the number common to two Friendly runsums where both of the following conditions must hold:
both runsums consist of more than one number
one of the two runsums begins at 1
We have 2 + 3 + 4 = 4 + 5 but 4 (the common number) is not a balancer since neither of its runsums begins with 1. However, 6 is a balancing number since
1 + 2 + 3 + 4 + 5 + 6 = 6 + 7 + 8
They call the number of terms in the runsum commencing n + 1 the balancer. Thus for the balancing number 6, we have the balancer 2 because
1 + 2 + 3 + 4 + 5 + 6 = 6 + (6+1) + (6+2)
Behera and Panda show that there is a connection between the balancing number n and a balancer r as follows:
| | | |
---
| n2 | = | (n + r)(n + r + 1) |
| 2 |
| | |
--- |
| r = | √(8 n2+1) – 2n – 1 |
| 2 |
and so n2 must be a Triangular number and 8 n2 + 1 must be a perfect square. They also show that in the sequence of balancing numbers, each is 6 times the previous number minus the number before that:
| | | |
---
| B(0) | = | 1 |
| B(1) | = | 6 |
| B(2) | = | 35 |
| B(i) | = | 6 B(i–1) – B(i–2) for i > 2 |
This is A001109: 1, 6, 35, 204, 1189, 6930, 40391, 235416, ....
In the same way that the Fibonacci Numbers have a Binet Formula which explicitly gives Fib(n) in terms of n, the balancing numbers have the formula:
| | | |
---
| B(n) = | Gn+1 – gn+1 | where G = 3 + √8, g = 3 – √8 |
| G – g |
In the same way that the ratio of Fibonacci numbers tends to Phi, the ratio of balancing numbers tends to G which is also 1/g and it has the periodic continued fraction expansion of [ 5; 1,4] = [6; –6,6]:-
| | |
--- |
| 6 + | 1 |
| −6 + | 1 |
| 6 + | 1 |
| −6 + | 1 |
| ... |
K Liptai in Fibonacci Balancing Numbers, Fibonacci Quarterly,2004, pages 330-340, showed that 1 is the only Fibonacci number that is a balancing number too.
Neighbourly Runsums
Here are two runsums of 15: 4 + 5 + 6 = 7 + 8. Notice that second continues the run started in the first and also that the two runs have the same sum. So instead of two runsums (of the same sum) sharing a common value as in Friends above, we now find runsums that fit together as neighbours: as one runsum ends, the other continues without a break. We call such runsums neighbours since they are runs that are next to each other. 27 has two neighbouring runsums: 2 + 3 + 4 + 5 + 6 + 7 = 8 + 9 + 10 = 27
The ordered sequence of sums with two neighbouring runsums is:
15 = 4+5+6 = 7+8 27 = 2+3+4+5+6+7 = 8+9+10 30 = 4+5+6+7+8 = 9+10+11 42 = 9+10+11+12 = 13+14+15 75 = 3+4+5+6+7+8+9+10+11+12 = 13+14+15+16+17 90 = 16+17+18+19+20 = 21+22+23+24
and continues with 105,135,147,165 ... (see Sloane's A110703)
A Friends and Neighbours Runsum Calculator
R E S U L T S
| | |
--- |
| | |
| | | |
---
Runsums | Friend & Neighbours |
Polyomino Runsums
Livio Zucca has a great album of photos on Facebook where he uses a connected shape of n squares for each number n in a runsum and then arranges them into a rectangle or a square to illustrate the runsum.
For instance, let's make each odd number into a zigzag shape of squares and then we can use the fact that the sum of the first n odd numbers is n2 to make an n×n shape jigsaw: 1 + (1 + 2 ) = 1 + 3 = 4 = 22 (1 + 2) + (1 + 2 + 3 ) = 1 + 3 + 5 = 9 = 32 (1 + 2 + 3) + (1 + 2 + 3 + 4 ) = 1 + 3 + 5 + 7 = 16 = 42 ...
Because the runsums starting at 1 are the triangle numbers T(n), we have the formula
| | |
--- |
| T(n) = 1 + 2 + ... + n = | n (n + 1) |
| 2 |
.
This means we can make each triangle number 1 + 2 + ... + n into a rectangle too! Why?
Because one of n or n+1 is even, or, equivalently :if n is even then T(n) can make the rectangle (n/2) ×(n+1);if n is odd, then (n+1) is even so n × (n+1)/2 are the sides of the rectangle.
Livio also uses this and uses the same zigzag shapes for each number from 1 to n to form a rectangle with the first n numbers:
Since two connected squares are a domino, any shape of n connected squares is called a polyomino with the plural polyominoes. We can make many shapes with n connected squares and there are many puzzles, problems and games involving them.
For instance, the 12 shapes made from 5 connected squares are the 12 pentominoes shown here on the right. A polyomino shape may be rotated or turned over (making its mirror image) but they are all the same polyomino. Since the 12 shapes each have 5 squares making a total of 60 squares, we can try to arrange these into a 6×10 rectangle or a 5×12 rectangle or a 3×20 rectangle. There are many ways to solve each of these three puzzles and if you make cut out the 12 shapes it makes a nice challenge to see how long it takes you to find one solution for each of these.
The square runsums of length more than 1 are
| | | | | | |
--- --- --- |
| 32 | = | 9 | has | 3 runsums | 2..4 = 4..5 = 9 |
| 52 | = | 25 | has | 3 runsums | 3..7 = 12..13 = 25 |
| 62 | = | 36 | has | 3 runsums | 1..8 = 11..13 = 36 |
| 72 | = | 49 | has | 3 runsums | 4..10 = 24..25 = 49 |
| 92 | = | 81 | has | 5 runsums | 5..13 = 11..16 = 26..28 = 40..41 = 81 |
| 102 | = | 100 | has | 3 runsums | 9..16 = 18..22 = 100 |
| 112 | = | 121 | has | 3 runsums | 6..16 = 60..61 = 121 |
| 122 | = | 144 | has | 3 runsums | 12..20 = 47..49 = 144 |
| 132 | = | 169 | has | 3 runsums | 7..19 = 84..85 = 169 |
| 142 | = | 196 | has | 3 runsums | 21..28 = 25..31 = 196 |
| 152 | = | 225 | has | 9 runsums | 4..21 = 8..22 = 18..27 = 21..29 = 35..40 = 43..47 = 74..76 = 112..113 = 225 |
What shapes can you use to make these into polyomino runsum jigsaw puzzles? There is much more about making shapes out of numbers, not just connected squares, on the Polygonal and Figurate Numbers page on this site. Now have a look at the next page on Runsums © 1999-2018 Dr Ron Knott created in 1999; last updated on 29 January 2018
A follow-on page on Runsums Numbers as shapes: Polygonal and Figurate Numbers Back to the main (Fibonacci Numbers) page |
8977 | https://www.scribd.com/document/652591073/Chapter-12-Production-and-Growth | Macro Economics: Growth & Productivity | PDF | Macroeconomics | Economic Growth
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Macro Economics: Growth & Productivity
The document summarizes key concepts from Chapter 25 of N. Gregory Mankiw's Principles of Macroeconomics, 9th Edition. It discusses factors that influence economic growth such as productivit…
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N. Gregory Mankiw,
Principles of Macro Economics
, 9
th
Edition © 2021 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
PowerPoint Slides prepared by: V. Andre ea CHIRI TESC U Eastern Illinois University
1
rinciples of Macro Economics, Ninth Editio
N. Gregory Mankiw
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N. Gregory Mankiw,
Principles of Macro Economics
, 9
th
Edition © 2021 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
C h a p t e r 2 5
Production and Growth
2
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N. Gregory Mankiw,
Principles of Macro Economics
, 9
th
Edition © 2021 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Economic Growth around the World
•
Real GDP per person
–
Living standard
–
Vary widely from country to country
•
Growth rate
–
How rapidly real GDP per person grew in the typical year
•
Because of differences in growth rates
–
Ranking of countries by income changes substantially over time
3
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N. Gregory Mankiw,
Principles of Macro Economics
, 9
th
Edition © 2021 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Table 1
The Variety of Growth Experiences
4
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8978 | https://www.lessonplanet.com/teachers/georgia-state-university-hyper-physics-ideal-gas-law | Georgia State University: Hyper Physics: Ideal Gas Law Handout for 9th - 10th Grade | Lesson Planet
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8979 | https://life.httpcn.com/sxyear.asp?year=1949 | 1949年属什么生肖？1949年生肖属相是牛_汉程网
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当前位置：汉程生活>每年生肖查询> 1949年属牛
出生年份：年
1949年属什么
【1949年】出生的人【属牛】。
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属牛的今年多大
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8980 | https://www.sciencedirect.com/science/article/pii/S0957417424008182?dgcid=rss_sd_all& | Geometric batch optimization for packing equal circles in a circle on large scale - ScienceDirect
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Expert Systems with Applications
Volume 250, 15 September 2024, 123952
Geometric batch optimization for packing equal circles in a circle on large scale
Author links open overlay panel Jianrong Zhou a, Kun He a, Jiongzhi Zheng a, Chu-Min Li b
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Highlights
•We propose geometric batch optimization for equal circle packing on large scale.
•We propose several partition strategies for geometric batch optimization.
•An adaptive neighbor object maintenance method is proposed.
•An efficient solution-space exploring and descent heuristic is proposed.
•New and better solutions are found on 105 out of 122 benchmark instances.
Abstract
The problem of packing equal circles in a circular container is a classic and famous packing problem, which is well-studied in academia and has a variety of applications in industry. The problem is computationally challenging, and researchers mainly focus on small to moderate-scale instances with the number of circular items n less than 320 in the literature. In this work, we aim to solve this problem on large scale, and it is the bottleneck for most global optimization problems. Specifically, we propose a novel geometric batch optimization method that makes batch gradient descent based on the geometric locations of packing items. This method not only can speed up the convergence process of continuous optimization significantly but also reduce the memory requirement during the program’s runtime. Then, we propose a heuristic search method, called solution-space exploring and descent, that can discover a feasible solution efficiently on large scale. Besides, we propose an adaptive neighbor object maintenance method to maintain the neighbor structure applied in the continuous optimization process. In this way, we can find high-quality solutions on large-scale instances within reasonable computational times. Extensive experiments on the benchmark instances sampled from n= 300 to 1000 show that our proposed algorithm outperforms the state-of-the-art algorithms and performs excellently on large-scale instances. In particular, our algorithm found 10 improved solutions among the 21 well-studied moderate-scale instances and 95 improved solutions among the 101 sampled large-scale instances. Furthermore, our geometric batch optimization, heuristic search, and adaptive maintenance methods are general and can be adapted to other packing and continuous optimization problems.
Introduction
The packing problems constitute a fundamental class of optimization problems wherein the objective is to efficiently arrange a collection of geometric objects within one or multiple containers. These problems aim to either maximize density or minimize the number of containers, aka bins, used. With a rich research history, various variants of packing problems have emerged alongside numerous methodologies and studies aimed at their resolution, including circle packing (He et al., 2018, Huang and Ye, 2011, Lai et al., 2022), sphere packing (Hartman et al., 2019, Hifi and Yousef, 2019), square packing (Fekete and Hoffmann, 2017, Leung et al., 1990), cube packing (Epstein and van Stee, 2005, Miyazawa and Wakabayashi, 2003), irregular packing (Leao et al., 2020, Rao et al., 2021, Zhao et al., 2020) and bin packing (Baldi et al., 2019, He et al., 2021, Zhao et al., 2021), etc.
Among these, the Circle Packing Problem (CPP) stands out as one of the most extensively explored in the fields of mathematics and computer science. In CPP, the objective is to pack a given set of circular items, each with known radii, into a container without overlap while maximizing packing density. CPP finds applications across diverse industries such as facility layout, cylinder packing, circular cutting, container loading, dashboard layout (Castillo, Kampas, & Pintér, 2008), structure design (Yanchevskyi et al., 2020) and satellite packaging (Wang, Wang, Sun, Huang, & Zhang, 2019), and extends its utility into fields like data visualization and data analysis (Görtler et al., 2017, Murakami et al., 2015, Wang et al., 2006).
Despite its significance, CPP presents a formidable computational challenge being proven NP-hard (Demaine, Fekete, & Lang, 2010). The computational resource and difficulty for obtaining a high-quality configuration grow exponentially with an increasing number of circular items, making large-scale instances extremely difficult to tackle. Consequently, most efforts in CPP focus on small and moderate-scale instances, with limited attention devoted to larger ones. Hence, the development of advanced packing algorithms, especially for large-scale instances, holds paramount importance. Not only do these algorithms drive advancements in computer science and geometry, but they also facilitate solutions to numerous real-world applications, such as floorplanning with a large volume of items (Ji, He, Wang, Jin, & Wu, 2021).
In this work, we focus on a particular CPP variant known as Packing Equal Circles in a Circle (PECC). PECC aims to pack a given number of unit circles into a circular container without overlapping such that the container radius is minimized. As a representative and comparatively simpler CPP form, PECC has garnered significant attention, with numerous studies dedicated to its resolution (Chen et al., 2018, He et al., 2018, Huang et al., 2006, Huang and Xu, 1999, Lai et al., 2022). Leveraging the classic elastic model (He et al., 2018, Huang and Xu, 1999), also known as Quasi-Physical Quasi-Human model (QPQH), we propose a novel method named Geometric Batch Optimization (GBO) that can construct either a feasible solution or an infeasible solution with a minimal overlapping degree on large-scale instances. Unlike existing methodologies, GBO divides packing circles into geometric batches and alternately updates each batch of circles through a non-convex continuous optimization process, thereby reducing time and space complexity.
We further enhance our approach with the Adaptive Neighbor object Maintenance (ANM) method, ensuring efficient maintenance of neighbor structures (He et al., 2018) during continuous optimization. To accomplish the adaptive feature, ANM uses two variables, termed deferring counter and deferring length. When the configuration changes significantly, ANM maintains the neighbors in each iteration; otherwise, ANM defers the maintaining process. As a result, ANM can avoid unnecessary maintenance and reduce the computational overhead.
Finally, we propose an advanced local search heuristic, called Solution-space Exploring and Descent (SED), to solve PECC. SED is a perturbation strategy to explore the promising area and discover a better solution efficiently. SED adopts a merit function to quantify the quality of solutions and determines the perturbation strength. SED uses a few perturbation numbers for low-quality solutions to obtain high-quality solutions quickly, and it increases the perturbation numbers for high-quality solutions to discover potentially optimal solutions.
Leveraging the abovementioned methods, our algorithm exhibits an excellent capability for solving PECC on large scale. Extensive experiments are based on a number of large-scale instances, where the experimental results showcase that our GBO method dramatically accelerates the convergence speed of the non-convex continuous process and reduces memory consumption. Meanwhile, we provide an insightful analysis of the influence of different partition strategies of GBO. Furthermore, by combining the GBO, ANM, and SED methods, our algorithm improves the best-known solutions for 95 out of 101 large-scale instances sourced from the Packomania website (Specht, 2024). Meanwhile, our methods are also efficient on moderate-scale instances, which improves the best-known solutions for 10 out of the 21 well-studied moderate-scale instances, which were yielded by the state-of-the-art IDTS algorithm (Lai et al., 2022). Through these experiments, GBO shows clear advantages over existing methods for solving large-scale instances and SED is an efficient heuristic.
In summary, our contributions include the introduction of GBO, ANM, and SED methods tailored for solving PECC, backed by comprehensive experimental validations. Through extensive testing on large-scale instances ranging from 500 to 1000 circular items, alongside moderate-scale instances from 300 to 320 items, our algorithm showcases remarkable performance improvements, achieving new best solutions across various instances.
The rest of this paper is organized as follows. Section 2 reviews the related works of the PECC problem. Section 3 introduces the mathematical formula and the classic elastic model (QPQH) of PECC, of which the elastic model is adopted in this work. Section 4 presents the continuous optimization methods for PECC, including the proposed GBO and ANM methods. Section 5 presents the framework of our proposed algorithm for PECC, including the SED heuristic. Section 6 conducts the experiments, computational results, and analyses of our proposed methods and algorithm. The conclusion of the paper is drawn in Section 7.
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Related work
Because of its representative and popularity, the PECC problem has become a well-explored and classic topic in the fields of geometry and computer science, leading to a wealth of methods, studies, and computational results. Hifi, M’hallah, et al. (2009) provided a comprehensive review of CPP and PECC prior to 2009, offering a detailed overview of the topic. In this section, we focus on the PECC problem and provide a literature review for this topic, where we categorize these PECC studies into
Problem formulation
The PECC problem aims to pack n unit circles {c 1,c 2,…,c n} into a circular container with the smallest possible radius while subjected to two constraints: (I) Any two circles do not overlap; (II) Any circle does not exceed the container. The problem can be formulated in the Cartesian coordinate system as a non-linear constrained optimization problem: Minimize R s.t.(x i−x j)2+(y i−y j)2≥2,1≤i,j≤n,i≠j,x i 2+y i 2+1≤R,1≤i≤n, where R is the radius of the circular container centered at the origin (0,0), and
Continuous optimization
Given that we employ the elastic model for solving the PECC problem, our target switches to minimizing the energy E(x) (defined in Eq. (6)) so as to discover a feasible solution via continuous optimization approaches. In this section, we present: (1) our proposed Geometric Batch Optimization (GBO) method for minimizing the energy of a conflicting solution with a fixed container radius, which can be regarded as a batch non-convex continuous optimization method, and following the four partition
Search heuristic
Algorithms based on the elastic model for solving the PECC problem can be divided into two phases. In the first phase, the container radius is fixed, and the goal is to find either a feasible solution or an infeasible solution with as few overlaps as possible (i.e., the energy E(x) being as minimal as possible). In the second phase, algorithms expand or shrink the container radius to obtain a feasible solution such that the radius of the circular container is locally minimized. We can start
Experiments
In this section, we present extensive experiments to evaluate our proposed algorithm for solving the PECC problem. The evaluation is based on well-studied moderate-scale instances and large-scale instances compared with the state-of-the-art method and best-known results sourced from the Packomania website (Specht, 2024). Furthermore, we conduct experiments to evaluate our proposed methods and provide the parameter study for the algorithm.
Conclusions
In this paper, we set out to tackle one of the most prominent packing challenges: the Packing Equal Circles in a Circle (PECC) problem, particularly on a large scale. Our novel Geometric Batch Optimization (GBO) method, coupled with various partition strategies, not only significantly accelerates the continuous optimization process but also mitigates memory requirements, facilitating the finding of local minimum packing configurations. Additionally, we introduce the Solution-space Exploring and
CRediT authorship contribution statement
Jianrong Zhou: Conceptualization, Methodology, Software, Writing – original draft, Writing – review & editing. Kun He: Conceptualization, Methodology, Supervision, Writing – review & editing. Jiongzhi Zheng: Writing – original draft, Writing – review & editing. Chu-Min Li: Writing – review & editing, Resources.
Declaration of competing interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgments
The authors sincerely thank the reviewers for proposing insightful comments and suggestions which helped to improve the paper.
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Cited by (3)
A heuristic algorithm with multi-scale perturbations for point arrangement and equal circle packing in a convex container
2025, Computers and Operations Research Show abstract The point arrangement and equal circle packing problems are a category of classic max–min constrained optimization problems with many important applications. Being computationally very challenging to solve, they have been widely studied in operations research and mathematics. We propose a heuristic algorithm for the point arrangement and equal circle packing problems in various convex containers. The algorithm relies on several complementary search components, including an unconstrained optimization procedure that ensures diversified and intensified searches, an optima exploitation based adjustment method for the radius of circles, and a monotonic basin-hopping method with multi-scale perturbations. Computational results on numerous benchmark instances show that the proposed algorithm significantly outperforms the existing state-of-the-art algorithms, especially for hard instances or large-scale instances. For the well-known equal circle packing problem in a circular container, it improves the best-known result for 69 out of the 96 hardest instances widely used in the literature. For the majority of the remaining instances tested, the algorithm improves or matches the best-known results with a high success rate, despite of the fact that these instances have been tested by many existing algorithms. Experimental analysis shows that the optima exploitation based adjustment method for the radius of circles plays a crucial role for the high performance of the algorithm and that the multi-scale perturbations are able to significantly enhance the search ability and robustness of the algorithm. Given the general feature of the proposed framework, it can be applied to other related max–min constrained optimization problems.
### On the method of packing geodesic circles into a spherical segment using a plane projection
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8981 | https://www.thestructuralengineer.info/education/professional-examinations-preparation/calculation-examples/calculation-example-cantilever-beam-temperature-change | Calculation Example – Cantilever Beam, Temperature change | thestructuralengineer.info
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Calculation Example – Cantilever Beam, Temperature change
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The beam is fixed at end B and free at end A. It is loaded at end A with the concentrated force 2F. At the same time there is a temperature change ΔT. Calculate the length change Δx. (Modulus of elasticity E, Cross section area AS , Temperature coefficient a.)
Solution There are two causes for the length change. The first cause for the length change is force 2F. The area of the cross section of the beam is AS so the stress is ?=2F / AS . The change in length is ?x1 and the deformation of the beam is ?1= ?x1/x. The Hook law gives : ;
Selected Topics
Calculation Example: Natural Periods of Vibration for SystemsCalculation Example: Axial Force On A ColumnCalculation Example: Shear force On A ColumnCalculation Example: Overturning Moment for Shear panelCalculation Example: Calculate the Change in Length of a Rod loaded in extensionCalculation Example - Calculate the Axial Forces on the Truss MembersCalculation Example - Calculate the member diagramsCalculation Example - Calculate the member diagrams for the beamCalculation Example – Beam with inner hinge (Part A). Find the ReactionsCalculation Example – Beam with inner hinge (Part B). Calculate the member diagrams.Calculation Example – Frame analysis.Calculation Example – Frame analysis – Uniform LoadCalculation Example – Find the Center of Gravity (Surface)Calculation Example – Design bolted connection of tension plates (EC3)Calculation Example – Cantilever BeamCalculation Example – Undamped free Vibration (Part A).Calculation Example – Undamped free Vibration (Part B).Calculation Example – Evaluation of Structural Property Matrices.Calculation Example – Angular acceleration, angular velocity.Calculation Example – Shear bolt connection EC3.Calculation Example – Buckling of Column (EC3).Calculation Example - Calculate the member diagrams.Calculation Example - Calculate the member diagrams.Calculation Example - Calculate the equation of the elastic curve.Calculation Example - Calculate the location of support.Calculation Example – Plane stress.Calculation Example - Annular cross section, Stress.Calculation Example – Allowable shear force for the girder.Calculation Example - Calculate the deflection. Castigliano Theorem.Calculation Example – Determine the shear force and moment.Calculation Example – Determine the magnitudes of F1,F2.Calculation Example – Internal forces.Calculation Example - Calculate the Axial Forces of the Truss Members.Calculation Example – Calculate the moments of inertia Ix and Iy.Calculation Example – Calculate shear stress for temperature load.Calculation Example – Calculate tension force using virtual work.Calculation Example – Torsional moment-Stress.Calculation Example – Reinforced Concrete Column at Stress.Calculation Example – Cantilever Beam with uniform loading.Calculation Example – Cantilever Beam with point loads.Calculation Example – Rod loadingCalculation Example – Maximum DeflectionCalculation Example – Member Diagram.Calculation Example – Minimum allowable Diameter.Calculation Example – Critical load.Calculation Example – Simple harmonic vibration part1Calculation Example – Simple harmonic vibration part 2Calculation Example – FrictionCalculation Example – Section Modulus SCalculation Example – Plastic Neutral Axis.Calculation Example – Buckling of Column (EC3).Calculation Example – Shear bolt connection EC3.Calculation Example – Member Diagram. Triangular load.Calculation Example – Torsional moment-Stress.Calculation Example – Angular acceleration, angular velocity.Calculate the location of point loadShear bolt connection EC3Torsional Moment-StressCalculate the Axial Forces of the Truss MembersCalculate the Maximum Shear StressHow to calculate yield strengthCalculate angular velocity, angular accelerationTemperature ChangeAverage shear stress in pressure vesselAllowable shear force of the girderCalculation ExamplesCalculate the variation in length of the rodSpring Assemblies in Series/Parallel: Two Springs in SeriesTruss vs CableCalculate the vertical deflection of a beamFlexular Crack in Concrete BeamMaximum Factored Vertical ShearColumn in bucklingBeams: maximum momentNominal flexural strength of a reinforced concrete beamCalculation of the cross-sectional area and the position of centroidCalculation of the second moments of areaExample: longitudinal reinforcement for shear and torsion according to EC2
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8982 | https://en.wikipedia.org/wiki/Hyperoctahedral_group | Published Time: 2009-09-08T05:32:56Z
Hyperoctahedral group - Wikipedia
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1 By dimension
2 Subgroups
3 HomologyToggle Homology subsection
3.1 H 1: abelianization
3.2 H 2: Schur multipliers
4 Notes
5 References
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From Wikipedia, the free encyclopedia
Group of symmetries of an n-dimensional hypercube
The C 2 group has order 8 as shown on this circle
The C 3 (O h) group has order 48 as shown by these spherical triangle reflection domains.
A hyperoctahedral group is a type of mathematical group that arises as the group of symmetries of a hypercube or of a cross-polytope. It was named by Alfred Young in 1930. Groups of this type are identified by a parameter n, the dimension of the hypercube.
As a Coxeter group it is of type B n = C n, and as a Weyl group it is associated to the symplectic groups and with the orthogonal groups in odd dimensions. As a wreath product it is S 2≀S n{\displaystyle S_{2}\wr S_{n}} where S n is the symmetric group of degree n. As a permutation group, the group is the signed symmetric group of permutations π either of the set {−n,−n+1,⋯,−1,1,2,⋯,n}{\displaystyle {-n,-n+1,\cdots ,-1,1,2,\cdots ,n}} or of the set {−n,−n+1,⋯,n}{\displaystyle {-n,-n+1,\cdots ,n}} such that π(i)=−π(−i){\displaystyle \pi (i)=-\pi (-i)} for all i. As a matrix group, it can be described as the group of n × northogonal matrices whose entries are all integers. Equivalently, this is the set of n × n matrices with entries only 0, 1, or −1, which are invertible, and which have exactly one non-zero entry in each row or column. The representation theory of the hyperoctahedral group was described by (Young 1930) according to (Kerber 1971, p.2).
In three dimensions, the hyperoctahedral group is known as O × S 2 where O ≅ S 4 is the octahedral group, and S 2 is a symmetric group (here a cyclic group) of order 2. Geometric figures in three dimensions with this symmetry group are said to have octahedral symmetry, named after the regular octahedron, or 3-orthoplex. In 4-dimensions it is called a hexadecachoric symmetry, after the regular 16-cell, or 4-orthoplex. In two dimensions, the hyperoctahedral group structure is the abstract dihedral group of order eight, describing the symmetry of a square, or 2-orthoplex.
By dimension
[edit]
The 8 permutations of the square, forming D 4
8 of the 48 permutations of a cube, forming O h
Hyperoctahedral groups in the n{\displaystyle n}-th dimension are isomorphic to S 2≀S n{\displaystyle S_{2}\wr S_{n}} (≀{\displaystyle \wr } denots the Wreath product) and can be named as B n, a bracket notation, or as a Coxeter group graph:
| n | Symmetry group | B n | Coxeter notation | Order | Mirrors | Structure | Related regular polytopes |
--- --- --- --- |
| 2 | D 4 (4•) | B 2 | | | 2 2 2! = 8 | 4 | D i h 4{\displaystyle Dih_{4}}≅S 2≀S 2{\displaystyle \cong S_{2}\wr S_{2}} | Square, octagon |
| 3 | O h (432) | B 3 | [4,3] | | 2 3 3! = 48 | 3+6 | S 4×S 2{\displaystyle S_{4}\times S_{2}}≅S 2≀S 3{\displaystyle \cong S_{2}\wr S_{3}} | Cube, octahedron |
| 4 | ±1/6[OxO].2 (O/V;O/V) | B 4 | [4,3,3] | | 2 4 4! = 384 | 4+12 | S 2≀S 4{\displaystyle S_{2}\wr S_{4}} | Tesseract, 16-cell, 24-cell |
| 5 | | B 5 | [4,3,3,3] | | 2 5 5! = 3840 | 5+20 | S 2≀S 5{\displaystyle S_{2}\wr S_{5}} | 5-cube, 5-orthoplex |
| 6 | | B 6 | [4,3 4] | | 2 6 6! = 46080 | 6+30 | S 2≀S 6{\displaystyle S_{2}\wr S_{6}} | 6-cube, 6-orthoplex |
| ...n | | B n | [4,3 n−2] | ... | 2 n n!=(2 n)!! | n 2 | S 2≀S n{\displaystyle S_{2}\wr S_{n}} | hypercube, orthoplex |
Subgroups
[edit]
There is a notable index two subgroup, corresponding to the Coxeter group D n and the symmetries of the demihypercube. Viewed as a wreath product, there are two natural maps from the hyperoctahedral group to the cyclic group of order 2: one map coming from "multiply the signs of all the elements" (in the n copies of {±1}{\displaystyle {\pm 1}}), and one map coming from the parity of the permutation. Multiplying these together yields a third map C n→{±1}{\displaystyle C_{n}\to {\pm 1}}. The kernel of the first map is the Coxeter group D n.{\displaystyle D_{n}.} In terms of signed permutations, thought of as matrices, this third map is simply the determinant, while the first two correspond to "multiplying the non-zero entries" and "parity of the underlying (unsigned) permutation", which are not generally meaningful for matrices, but are in the case due to the coincidence with a wreath product.
The kernels of these three maps are all three index two subgroups of the hyperoctahedral group, as discussed in H 1: Abelianization below, and their intersection is the derived subgroup, of index 4 (quotient the Klein 4-group), which corresponds to the rotational symmetries of the demihypercube.
In the other direction, the center is the subgroup of scalar matrices, {±1}; geometrically, quotienting out by this corresponds to passing to the projective orthogonal group.
In dimension 2 these groups completely describe the hyperoctahedral group, which is the dihedral group Dih 4 of order 8, and is an extension 2.V (of the 4-group by a cyclic group of order 2). In general, passing to the subquotient (derived subgroup, mod center) is the symmetry group of the projective demihypercube.
Tetrahedral symmetry in three dimensions, order 24
The hyperoctahedral subgroup, D n by dimension:
| n | Symmetry group | D n | Coxeter notation | Order | Mirrors | Related polytopes |
--- --- ---
| 2 | D 2 (2•) | D 2 | = [ ]×[ ] | | 4 | 2 | Rectangle |
| 3 | T d (332) | D 3 | [3,3] | | 24 | 6 | tetrahedron |
| 4 | ±1/3[Tx T].2 (T/V;T/V)− | D 4 | [3 1,1,1] | | 192 | 12 | 16-cell |
| 5 | | D 5 | [3 2,1,1] | | 1920 | 20 | 5-demicube |
| 6 | | D 6 | [3 3,1,1] | | 23040 | 30 | 6-demicube |
| ...n | | D n | [3 n−3,1,1] | ... | 2 n−1 n! | n(n−1) | demihypercube |
Pyritohedral symmetry in three dimensions, order 24
Octahedral symmetry in three dimensions, order 24
The chiral hyper-octahedral symmetry, is the direct subgroup, index 2 of hyper-octahedral symmetry.
| n | Symmetry group | Coxeter notation | Order |
--- --- |
| 2 | C 4 (4•) | + | | 4 |
| 3 | O (432) | [4,3]+ | | 24 |
| 4 | 1/6[O×O].2 (O/V;O/V) | [4,3,3]+ | | 192 |
| 5 | | [4,3,3,3]+ | | 1920 |
| 6 | | [4,3,3,3,3]+ | | 23040 |
| ...n | | [4,(3 n−2)+] | ... | 2 n−1 n! |
Another notable index 2 subgroup can be called hyper-pyritohedral symmetry, by dimension: These groups have n orthogonal mirrors in n-dimensions.
| n | Symmetry group | Coxeter notation | Order | Mirrors | Related polytopes |
--- --- --- |
| 2 | D 2 (2•) | [4,1+]= | | 4 | 2 | Rectangle |
| 3 | T h (32) | [4,3+] | | 24 | 3 | snub octahedron |
| 4 | ±1/3[T×T].2 (T/V;T/V) | [4,(3,3)+] | | 192 | 4 | snub 24-cell |
| 5 | | [4,(3,3,3)+] | | 1920 | 5 | |
| 6 | | [4,(3,3,3,3)+] | | 23040 | 6 | |
| ...n | | [4,(3 n−2)+] | ... | 2 n−1 n! | n | |
Homology
[edit]
The group homology of the hyperoctahedral group is similar to that of the symmetric group, and exhibits stabilization, in the sense of stable homotopy theory.
H 1: abelianization
[edit]
The first homology group, which agrees with the abelianization, stabilizes at the Klein four-group, and is given by:
H 1(C n,Z)={0 n=0 Z/2 n=1 Z/2×Z/2 n≥2.{\displaystyle H_{1}(C_{n},\mathbf {Z} )={\begin{cases}0&n=0\\mathbf {Z} /2&n=1\\mathbf {Z} /2\times \mathbf {Z} /2&n\geq 2\end{cases}}.}
This is easily seen directly: the −1{\displaystyle -1} elements are order 2 (which is non-empty for n≥1{\displaystyle n\geq 1}), and all conjugate, as are the transpositions in S n{\displaystyle S_{n}} (which is non-empty for n≥2{\displaystyle n\geq 2}), and these are two separate classes. These elements generate the group, so the only non-trivial abelianizations are to 2-groups, and either of these classes can be sent independently to −1∈{±1},{\displaystyle -1\in {\pm 1},} as they are two separate classes. The maps are explicitly given as "the product of the signs of all the elements" (in the n copies of {±1}{\displaystyle {\pm 1}}), and the sign of the permutation. Multiplying these together yields a third non-trivial map (the determinant of the matrix, which sends both these classes to −1{\displaystyle -1}), and together with the trivial map these form the 4-group.
H 2: Schur multipliers
[edit]
The second homology groups, known classically as the Schur multipliers, were computed in (Ihara & Yokonuma 1965).
They are:
H 2(C n,Z)={0 n=0,1 Z/2 n=2(Z/2)2 n=3(Z/2)3 n≥4.{\displaystyle H_{2}(C_{n},\mathbf {Z} )={\begin{cases}0&n=0,1\\mathbf {Z} /2&n=2\(\mathbf {Z} /2)^{2}&n=3\(\mathbf {Z} /2)^{3}&n\geq 4\end{cases}}.}
Notes
[edit]
^ abcdConway & Smith 2003
^du Val 1964, #47
^du Val 1964, #42
^du Val 1964, #27
^Coxeter 1999, p.121, Essay 5 Regular skew polyhedra
^du Val 1964, #41
References
[edit]
Miller, G. A. (1918). "Groups formed by special matrices". Bull. Am. Math. Soc. 24 (4): 203–6. doi:10.1090/S0002-9904-1918-03043-7.
du Val, P. (1964). Homographies, Quaternions and Rotations. Oxford mathematical monographs. Clarendon Press. OCLC904102141.
Ihara, Shin-ichiro; Yokonuma, Takeo (1965), "On the second cohomology groups (Schur-multipliers) of finite reflection groups", Journal of the Faculty of Science. University of Tokyo. Section IA. Mathematics, 11: 155–171, ISSN0040-8980, MR0190232
Kerber, Adalbert (1971), Representations of permutation groups. I, Lecture Notes in Mathematics, vol.240, Springer-Verlag, doi:10.1007/BFb0067943, ISBN978-3-540-05693-5, MR0325752
Kerber, Adalbert (1975), Representations of permutation groups. II, Lecture Notes in Mathematics, vol.495, Springer-Verlag, doi:10.1007/BFb0085740, ISBN978-3-540-07535-6, MR0409624
Young, Alfred (1930), "On Quantitative Substitutional Analysis 5", Proceedings of the London Mathematical Society, Series 2, 31: 273–288, doi:10.1112/plms/s2-31.1.273, ISSN0024-6115, JFM56.0135.02
Coxeter, H.S.M.; Moser, W.O.J. (2013) . Generators and Relations for Discrete Groups (4th ed.). Springer. p.92 §7.4 Linear fractional groups, p. 122 §9.3 Finite Groups. ISBN978-3-662-21943-0.
Baake, M. (1984). "Structure and representations of the hyperoctahedral group". J. Math. Phys. 25 (11): 3171. Bibcode:1984JMP....25.3171B. doi:10.1063/1.526087.
Stembridge, John R. (1992). "The projective representations of the hyperoctahedral group". J. Algebra. 145 (2): 396–453. doi:10.1016/0021-8693(92)90110-8. hdl:2027.42/30235.
Coxeter, H.S.M. (1999). The Beauty of Geometry: Twelve Essays. Dover. ISBN0-486-40919-8. LCCN99035678.
Conway, John H.; Smith, Derek A. (2003). On Quaternions and Octonions. CRC Press. ISBN978-1-000-68777-4.
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8983 | https://www.analyzemath.com/intermediate_algebra/problems_12.html | Simplifying Algebraic Expressions with Exponents – Step-by-Step Examples
Simplifying Algebraic Expressions with Exponents
Step-by-Step Examples
This collection of intermediate algebra problems focuses on simplifying expressions using the rules and properties of exponents. Detailed solutions are provided at the bottom of the page to help reinforce your understanding.
Basic Exponent Rules for Simplifying Expressions
Product of Powers Rule:
a m⋅a n=a m+n
Quotient of Powers Rule:
a m a n=a m−n(for a≠0)
Power of a Power Rule:
(a m)n=a m⋅n
Power of a Product Rule:
(a b)m=a m⋅b m
Power of a Quotient Rule:
(a b)m=a m b m(for b≠0)
Zero Exponent Rule:
a 0=1(for a≠0)
Negative Exponent Rule:
a−m=1 a m(for a≠0)
Note: all letters (variables) used in the expressions below represent numbers that are not equal to 0.
1. Simplify the following expressions and write them with positive exponents.
a) x 2⋅x 3
b) a−2⋅a−3
c) a 5 a 2
d) a−3 a 7
e) (x 2)3
f) (x−2)5
g) (x 2 y 3)3
h) (x−2 y 3)8
2. Simplify the following expressions and write them with positive exponents.
a) (y 4 y 2)3
b) (y 4 y−3)−2
c) (y 4 x 12 y 2)3
d) (y 12 x 56 y 3)0
e) (x 2 y 4)2(x y−2)3
3. Simplify the following expressions and write them with positive exponents.
a) (x y 2)2(x 2 y 2)2(x y)4
b) ((a b 2)3(a 2 b 2)4)2
c) (x 2 y 3)0(x 4 y 3)4(x y 2)3
d) (x 3 y−2)2(x−2 y 3)4(x−2 y−2)3
Solutions to the Above Questions
1. a) x 2⋅x 3=x 2+3=x 5 b) a−2⋅a−3=a−2−3=a−5=1 a 5 c) a 5 a 2=a 5−2=a 3 d) a−3 a 7=a−3−7=a−10=1 a 10 e) (x 2)3=x 2⋅3=x 6 f) (x−2)5=x−2⋅5=x−10=1 x 10 g) (x 2 y 3)3=x 2⋅3 y 3⋅3=x 6 y 9 h) (x−2 y 3)8=x−2⋅8 y 3⋅8=x−16 y 24=y 24 x 16
2. a) (y 4 y 2)3=y 4⋅3 y 2⋅3=y 12 y 6=y 6 b) (y 4 y−3)−2=y 4⋅(−2)y−3⋅(−2)=y−8 y 6=1 y 14 c) (y 4 x 12 y 2)3=(y 4 x 12)3 y 2⋅3=y 12 x 36 y 6=y 6 x 36 d) (y 12 x 56 y 3)0=1 e) (x 2 y 4)2(x y−2)3=x 4 y 8 x 3 y−6=x y 14
3. a) (x y 2)2(x 2 y 2)2(x y)4=x 2 y 4⋅x 4 y 4 x 4 y 4=x 6 y 8 x 4 y 4=x 2 y 4 b) ((a b 2)3(a 2 b 2)4)2=(a b 2)6(a 2 b 2)8=a 6 b 12⋅a 16 b 16=a 22 b 28 c) (x 2 y 3)0(x 4 y 3)4(x y 2)3=1⋅x 16 y 12 x 3 y 6=x 13 y 6 d) (x 3 y−2)2(x−2 y 3)4(x−2 y−2)3=x 6 y−4⋅x−8 y 12 x−6 y−6=x−2 y 8 x−6 y−6=x 4 y 14
More References and Links
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8984 | https://www.usatoday.com/story/news/world/2017/08/31/united-states-troops-afghanistan/621140001/ | U.S. to send additional troops to Afghanistan
Defense Secretary Jim Mattis said Thursday that the Pentagon will dispatch additional troops to Afghanistan, where U.S.-backed security forces are stalemated in a war against the Taliban and other insurgent groups.
Mattis declined to specify how many additional troops would be heading there, but the top coalition commander in Afghanistan, Gen. John Nicholson, has said a few thousand more troops would be required to break the stalemate in America's longest war.
Currently about 11,000 U.S. forces are in Afghanistan, plus several thousand troops from coalition partner countries.
Mattis said he would hold off on revealing details on the deployment until he has briefed members of Congress next week. He said he has begun signing deployment orders to dispatch the additional forces.
The Pentagon on Wednesday revised the number of U.S. troops it said were in Afghanistan, saying it wanted to be more transparent.
The Pentagon had said 8,400 U.S. troops were in Afghanistan under Obama administration rules that capped the number of troops authorized to be in the country. Those rules did not require the Pentagon to include any troops who were in the country on short assignments.
The Pentagon said it would weigh the efforts to be transparent about troop deployments against security concerns over publicizing troop movements and numbers that might help the enemy.
“We will balance informing the American people (with) maintaining operational security and denying the enemy any advantage,” said Pentagon spokeswoman Dana White.
The Trump administration has given more leeway to the Pentagon to decide how many American troops to send to war zones, such as Afghanistan and Iraq.
It has also allowed field commanders to make decisions without always seeking approval from Washington.
“The current administration has empowered the chain of command to make more decisions on their own,” Lt. Gen, Stephen Townsend, the top coalition commander in Iraq, said at a briefing Thursday.
Mattis has said he wanted to hold off sending additional troops to Afghanistan until the Trump administration had decided on a broad strategy for the region.
In an August speech, Trump announced a strategy that includes supporting Afghan security forces and placing pressure on Pakistan, where many insurgent leaders have sought sanctuary.
The U.S. and coalition troops are not engaging in direct combat, but are serving as advisers and in other ways supporting Afghan forces, which are leading the fight against the Taliban.
As the number of U.S. and coalition troops have declined in Afghanistan, the Taliban has gained ground, particularly in remote parts of the country. |
8985 | https://academic.oup.com/eurjpc/advance-article/doi/10.1093/eurjpc/zwaf182/8097593 | Published Time: 2025-03-27
Occasional smoking is a risk factor for myocardial infarction in the population-based Tromsø Study, 2001–21 | European Journal of Preventive Cardiology | Oxford Academic
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Occasional smoking is a risk factor for myocardial infarction in the population-based Tromsø Study, 2001–21 Open Access
Sweta Tiwari,
Sweta Tiwari
Department of Clinical Medicine, UiT The Arctic University of Norway
,
PO Box 6050 Langnes, N-9037 Tromsø
,
Norway
Research and Innovation Department, Førde Health Trust
,
PO Box 1000, N-6807 Førde
,
Norway
Department of Community Medicine, UiT The Arctic University of Norway
,
PO Box 6050 Langnes, N-9037 Tromsø
,
Norway
Corresponding author. Tel: +47 99 86 94 95, +47 77 64 48 40, Email: sweta.tiwari@uit.no
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Ola Løvsletten,
Ola Løvsletten
Department of Community Medicine, UiT The Arctic University of Norway
,
PO Box 6050 Langnes, N-9037 Tromsø
,
Norway
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Bjarne K Jacobsen,
Bjarne K Jacobsen
Department of Community Medicine, UiT The Arctic University of Norway
,
PO Box 6050 Langnes, N-9037 Tromsø
,
Norway
Center for Sami Health Research, Department of Community Medicine, UiT The Arctic University of Norway
,
Tromsø
,
Norway
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Tom Wilsgaard,
Tom Wilsgaard
Department of Community Medicine, UiT The Arctic University of Norway
,
PO Box 6050 Langnes, N-9037 Tromsø
,
Norway
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Ellisiv B Mathiesen,
Ellisiv B Mathiesen
Department of Clinical Medicine, UiT The Arctic University of Norway
,
PO Box 6050 Langnes, N-9037 Tromsø
,
Norway
Department of Neurology, University Hospital of North Norway
,
Tromsø
,
Norway
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Henrik Schirmer,
Henrik Schirmer
Institute of Clinical Medicine, Campus Ahus, University of Oslo
,
Lørenskog
,
Norway
Department of Cardiology, Medical Division, Akershus University Hospital
,
Lørenskog
,
Norway
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Inger T Gram,
Inger T Gram
Department of Community Medicine, UiT The Arctic University of Norway
,
PO Box 6050 Langnes, N-9037 Tromsø
,
Norway
Norwegian Centre for E-health Research, University Hospital of North Norway
,
Tromsø
,
Norway
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Jan Mannsverk,
Jan Mannsverk
Department of Clinical Medicine, UiT The Arctic University of Norway
,
PO Box 6050 Langnes, N-9037 Tromsø
,
Norway
Department of Cardiology, University Hospital of North Norway
,
Tromsø
,
Norway
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Maja-Lisa Løchen
Maja-Lisa Løchen
Department of Clinical Medicine, UiT The Arctic University of Norway
,
PO Box 6050 Langnes, N-9037 Tromsø
,
Norway
Department of Community Medicine, UiT The Arctic University of Norway
,
PO Box 6050 Langnes, N-9037 Tromsø
,
Norway
Department of Cardiology, University Hospital of North Norway
,
Tromsø
,
Norway
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Conflict of interest: M.-L.L. has received lecture fees from Bayer, Sanofi, and Bristol Myers Squibb (BMS)/Pfizer and is an Associate Editor of the European Journal of Preventive Cardiology. Henrik Schirmer has received lecture fees from Amgen, Boehringer Ingelheim, BMS, Novartis, and Sanofi-Aventis.
Author Notes
European Journal of Preventive Cardiology, zwaf182,
Published:
27 March 2025
Article history
Received:
04 September 2024
Revision received:
07 December 2024
Accepted:
01 March 2025
Published:
27 March 2025
Corrected and typeset:
12 April 2025
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Sweta Tiwari, Ola Løvsletten, Bjarne K Jacobsen, Tom Wilsgaard, Ellisiv B Mathiesen, Henrik Schirmer, Inger T Gram, Jan Mannsverk, Maja-Lisa Løchen, Occasional smoking is a risk factor for myocardial infarction in the population-based Tromsø Study, 2001–21, European Journal of Preventive Cardiology, 2025;, zwaf182, Select Format Download citation Close
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Abstract
Aims
While the association between daily smoking and myocardial infarction (MI) risk is well-established, little is known concerning the relationship between occasional smoking and MI risk. In this Norwegian study, we aimed to investigate the risk of first-time MI among occasional smokers over a 20-year follow-up period.
Methods and results
In this prospective cohort study, data on smoking habits and relevant risk factors were collected from 15 617 participants enrolled in the fifth and/or sixth survey of the Tromsø Study. Participants were followed up for the first MI from 2001 to 2021. Smoking, age, marriage/cohabitant status, body mass index, hypertension, cholesterol, alcohol consumption, education, physical activity, and sex were included as covariates in Cox proportional hazard models. At baseline, 9.2% of participants were occasional smokers. Over the follow-up period, 1297 participants experienced their first MI. Adjusted for covariates, we observed a 41% increased MI risk among occasional smokers compared to never smokers [hazard ratio (HR) 1.41, 95% confidence interval (CI) 1.11–1.80]. We noted a dose–response relationship in the hazards of smoking, categorized as never, former, occasional, and daily. Compared with never smokers, the relative risk of MI in occasional smokers was twice as high in women (HR 2.08, 95% CI 1.43–3.04) and 11% higher in men (HR 1.11, 95% CI 0.81–1.52).
Conclusion
This study highlights an elevated risk of MI among occasional smokers, with higher relative risk in women. The result emphasizes that there is no safe threshold for smoking, underlining the importance of smoking cessation.
Graphical Abstract
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Smoking, Occasional smoking, Myocardial infarction, Cardiovascular disease, Prospective cohort study
Topic:
myocardial infarction
smoking
prospective studies
smokers
Issue Section:
Full Research Paper>Cvd Risk Factors
Lay Summary
The study shows that people who smoke occasionally have a higher risk of myocardial infarction (MI) than never smokers. It emphasizes that no level of smoking is safe, and that occasional as well as daily smoking has detrimental effects on cardiovascular health and increases the risk of MI.
Occasional smokers have a higher risk of MI compared to both never smokers and ex-smokers.
Female occasional smokers have a higher risk than occasional smoking men when compared to non-smokers.
Introduction
The tobacco epidemic is one of the largest threats to public health with more than 8 million deaths a year, including 1.2 million deaths from secondhand smoking.1 Previous studies have shown that smoking is a significant risk factor for myocardial infarction (MI) and that smoking in women is associated with higher relative risk of MI compared to men of similar age groups.2–4 This finding was supported by the study based on data from the Tromsø Study, which also demonstrated that the sex difference extends to passive smoking as well.3 There is no safe threshold of tobacco exposure, and the sharpest increase in the risk for ischaemic heart disease appears to occur from non-smoking to smoking 1–4 cigarettes daily.5 Despite this evidence, there is a misconception that smoking a few cigarettes carries a relatively low risk for cardiovascular disease (CVD).6
In Norway, the prevalence of daily smoking in inhabitants aged 16–74 years has decreased from 17% in 2011 to 7% in 2023. The overall rate of occasional smoking has remained relatively unchanged over the past ten years, at 7–11%.7 However, among young adults (16–24 years old), the proportion of occasional smokers was 15% for women and 18% for men in 2023.7 The constant overall prevalence and growing number among the young population made occasional smoking the main theme of Norwegian Directorate of Health’s smoking cessation programme in 2012–13.8
While smoking attributed CVD mortality and morbidity declined globally from 1990 to 2019,9 a previous study demonstrated increased total and cardiovascular mortality among occasional smoking men.10 Our previous 14-year follow-up from the Tromsø Study found a 38% increase in overall mortality among occasional smokers compared to never smokers.11 Additionally, we observed a higher prevalence of occasional smokers among highly educated individuals.11 In Iranian men, a 12-year follow-up study revealed a higher risk of coronary heart disease (CHD) among occasional smokers compared to non-smokers.12 There is a lack of information on the association between occasional smoking and the risk of MI.13 To our knowledge, the long-term risk of MI among occasional smokers in both women and men in a general population has not been reported previously.
The purpose of this study was to examine the relative risk of incident MI for occasional smokers with a 20-year follow-up period with updated smoking status and other covariates as time-dependent variables. Importantly, we also investigated potential sex differences in this relationship.
Methods
Study population
The Tromsø Study is a Norwegian prospective cohort study conducted in the municipality of Tromsø, Norway.14 Seven surveys referred to as Tromsø1–Tromsø7 have been conducted from 1974 to 2016, and more than 45 000 subjects have participated in one or more surveys. In each survey, invitations to attend the examinations along with a questionnaire (Questionnaire 1) were sent by mail to full birth cohorts and/or representative samples of the registered inhabitants of Tromsø municipality.14 The participants were asked to bring the completed questionnaire when they came for physical examination and biological sampling. Participants attending physical examinations received a second questionnaire (Questionnaire 2), which was to be returned by mail.14 In the present analyses, we included data from participants attending the Tromsø5 (2001) and/or the Tromsø6 (2007–08) surveys. In Tromsø5, 4619 women and 3511 men aged 30–89 years attended with attendance rates of 80.8% and 75.7%, respectively. Similarly, in Tromsø6, 6930 women and 6054 men aged 30–87 years attended with attendance rates of 68.4% and 62.9%, respectively. After exclusion of those without valid written consent, with previous MI, and inconsistent or insufficient data on smoking status, 15 617 participants (8561 women, 7056 men) were included in the analytical cohort (Figure 1).
Figure 1
Flowchart of the analytical cohort, the Tromsø Study 2001–08.
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The Tromsø Study complies with the Declaration of Helsinki and has been approved by the Regional Committee for Medical and Health Research Ethics (REK) and Norwegian Data Protection Authority. All participants gave written informed consent. The current study was approved by REK North (reference 599786) and evaluated by the Norwegian Agency for Shared Services in Education and Research.
Measurements
Data from the Tromsø Study were used to gather information on various factors including marriage/cohabitation status (no/yes), smoking habits, education level, alcohol use, leisure time physical activity level, and anti-hypertensive medication use (no/yes).
In Tromsø5, information on smoking was obtained by two different questionnaires, where the first questionnaire included a question about daily smoking (‘Do you/did you smoke daily?’ with options ‘Yes, now’, ‘Yes, previously’, and ‘Never’). The second questionnaire included a question about occasional smoking (‘Do you smoke?’ with options ‘Yes, daily’, ‘Yes, sometimes’, and ‘No, never’). Participants who reported to be a current daily smoker on the second questionnaire, but a former or never smoker on the first or a never smoker on the second questionnaire and current smoker on the first, were excluded from the analyses. In Tromsø6, information on smoking was obtained from a single questionnaire, including questions about daily smoking (‘Do you/did you smoke daily?’ with options ‘Yes, now’, ‘Yes, previously’, and ‘Never’) and occasional smoking (‘Do you smoke sometimes, but not daily?’ with options ‘Yes’ and ‘No’).
In both surveys, participants were also asked about the duration of smoking, the number of cigarettes smoked daily (for ever-smokers), and the duration of smoking cessation (for former smokers), which was used to check for inconsistencies. For participants attending both surveys, smoking status on both surveys was also used to check for inconsistencies. Participants with missing smoking status information in both Tromsø5 and Tromsø6 or those reporting current daily smoking in Tromsø5 but never smoking in Tromsø6 were excluded from the analyses. Thus, a total of 79 subjects were excluded from the analytical cohort because of missing or inconsistent information concerning smoking habits (Figure 1).
Based on reported smoking habits, participants were categorized into four groups (never smokers, former smokers, occasional smokers, and daily smokers). If there was a change in smoking habits from Tromsø5 to Tromsø6, the smoking status was updated. Additionally, pack-years were calculated for ever-smokers using information on the number of cigarette consumption per day and duration of smoking (years) [(number of cigarettes per day × duration of smoking) / 20].
Information concerning marital status/cohabitation, education, physical activity, and alcohol use was obtained by questionnaires. Education level was categorized as primary/secondary school, upper secondary or vocational, university or college education <4 years, and university or college education ≥4 years. Alcohol use was assessed as intake per drinking session based on Alcohol Use Disorder Identification Test15 and categorized as no alcohol, 1–2 units, 3–4 units, and 5 or more units, respectively. The Saltin–Grimby leisure time physical activity questionnaire was used to assess physical activity level16 categorized as sedentary, light activity, and moderate-to-vigorous activity.
Body mass index (BMI) was calculated as measured body weight/height 2 (kg/m 2). Measurements of height and weight were performed in light clothing and without footwear. Blood pressure was recorded three times with 1 min intervals after the participants had 2 min seated rest, using an automatic device (Dinamap Vital Signs Monitor 1846; Citrikon). The mean of the last two measurements was used in the statistical analyses. Hypertension was defined as systolic blood pressure (SBP) ≥ 140 mmHg and/or diastolic blood pressure (DBP) ≥ 90 mmHg and/or the use of anti-hypertensive medication. Non-fasting serum total cholesterol and serum high-density lipoprotein (HDL) cholesterol were analyzed using enzymatic colorimetric methods. The analyses were performed at the Department of Clinical Chemistry, University Hospital of North Norway in Tromsø.
Follow-up and statistical analysis
Diagnoses of MI were identified from several sources. All residents in Norway have a personal identification number that allows exact matching of population registries. MI events from baseline through 2014 were identified through linkage to the diagnosis registries at the University Hospital of North Norway and the National Cause of Death Registry. First-ever incident MI cases were ascertained by reviewing medical records with ICD-10 discharge diagnosis of I20–I25. Other diagnostic codes such as I46–I48, I50, I60–I69, G45, G46, G81, R96, R98, or R99 were also used for validation. Events from 2015 to 2021 were obtained from the Norwegian Myocardial Infarction Registry (NORMI) and the National Cause of Death Registry. NORMI was established in 2012 as a national quality registry of all MI events hospitalized in Norwegian hospitals. Reporting to the registry is mandatory and the registry has high accuracy and completeness.17
Participants were followed from enrolment until the date of first documented MI or until censoring due to death, migration, or the end of follow-up, 31 December 2021, whichever occurred first. Participants who died or emigrated during follow-up were identified through the Population Registry of Norway. The mean follow-up time for the 15 617 included participants was 11.0 years (from 0.01 to 20.8 years).
We calculated percentage (%) and means with standard deviation (SD) for the distribution of selected characteristics of the study population at enrolment either in Tromsø5 or Tromsø6. We also estimated crude and age-adjusted MI incidence rates per 10 000 person-years overall, according to updated smoking status and for women and men separately, by dividing the number of cases by the total number of persons-years.
Hazard ratios (HRs) for MI were estimated by multivariable Cox proportional hazards regression models, with 95% confidence intervals (CIs). We used age as the underlying time scale with smoking and sex in the simplest model (Model 1). In Model 2, we also included education, marriage/cohabitant status, BMI, hypertension, cholesterol, alcohol consumption, and physical activity as covariates. Except for education (and age and sex), the covariates were treated as time-dependent, i.e. updated at Tromsø6 for those who attended both surveys. Never smokers constituted the reference category. The data were restructured from a wide format to a long format in STATA, allowing repeated measures across the time points to be analyzed appropriately. These restructured data were also used to calculate crude and age-adjusted MI incidence rate per 10 000 person-years overall, according to updated smoking status. We tested for sex interactions by including two-way cross-product terms between sex and each indicator variable of smoking status level in the model. Although sex interaction was not statistically significant (P> 0.05), we also performed separate analyses for women and men. In addition, we conducted a sub-analysis among ever-smokers, incorporating pack-years as a covariate, to reduce the confounding due to differences in exposure levels within the same smoking status. In addition, an analysis restricted to participants aged 79 years or below at baseline and during follow-up was performed. Follow-up information after 80 years was disregarded, as this population is more likely to experience competing risks. The proportional hazard assumption was validated with visual inspection of log-log plots and Schoenfeld residuals. A two-sided P value <0.05 was considered statistically significant. Statistical analysis was conducted using STATA V.18 (Stata, College Station, TX, USA).
Results
The mean age of participants was ∼55 years. In both sexes, occasional smokers tended to be the younger age group. Table 1 shows selected characteristics at enrolment in women and men, stratified, by smoking status. The proportion with higher education level was higher in occasional smokers than in daily smokers. Additionally, a higher proportion of current smokers, including both occasional and daily smokers, were single, consumed more alcohol per drinking session, were physically inactive, and had (in women) lower BMI compared to never or former smokers. The prevalence of hypertension was lower among occasional smokers. Selected characteristics at enrolment are presented separately for Tromsø5 and Tromsø6 in Supplementary files; Supplementary material online, Tables S1 and S2.
Table 1
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Distribution of selected characteristics at enrolment given as percentage and mean (SD) by smoking and sex, in Tromsø5 (2001) or Tromsø6 (2007–08)
| . | Never smokers . | Former smokers . | Occasional smokers . | Daily smokers . |
--- ---
| Women, n (%) | 3321 (38.8) | 2529 (29.5) | 723 (8.5) | 1988 (23.2) |
| Number of pack-years, mean (SD) Median (IQR) | 0 | 10.0 (10.0) 7 (11) | 10.2 (9.9) 7 (12) | 16.8 (10.5) 15 (11.6) |
| Age, mean (SD), years | 56.5 (13.8) | 56.2 (12.2) | 51.2 (12.0) | 53.4 (12.3) |
| Education >12 years, % | 41.5 | 35.9 | 40 | 25.5 |
| Married/cohabitant, % | 69.8 | 72.4 | 66.2 | 65.2 |
| Body mass index, mean (SD), kg/m 2 | 26.7 (4.7) | 27.1 (4.6) | 25.7 (4.5) | 25.4 (4.4) |
| Total/HDL cholesterol ratio, mean (SD) | 3.88 (1.21) | 3.87 (1.19) | 3.76 (1.21) | 4.20 (1.32) |
| Systolic blood pressure (mmHg), mean (SD) | 134.7 (23.8) | 132.9 (22.6) | 125.3 (21.8) | 128.8 (22.2) |
| Hypertension, % | 31.4 | 22.8 | 11.9 | 24.0 |
| Alcohol (per drinking session), % No alcohol 1–2 units 3–4 units 5 or more units | 22.9 63.9 11.3 1.9 | 9.6 67.4 19.7 3.3 | 7.2 53.0 32.4 7.4 | 8.7 53.1 29.0 9.2 |
| Leisure time physical activity, % Sedentary Light activity Moderate-to-vigorous activity | 17.8 67.6 14.6 | 17.8 69.0 13.2 | 20.2 67.5 12.3 | 27.2 62.7 10.1 |
| Men, n (%) | 2199 (31.2) | 2642 (37.4) | 718 (10.2) | 1497 (21.2) |
| Number of pack-years, mean (SD) Median (IQR) | 0 | 16.4 (16.2) 12 (17) | 14.3 (13.3) 11.2 (13.3) | 21.7 (13.6) 19.8 (16.1) |
| Age, mean (SD), years | 51.8 (12.0) | 59.7 (11.8) | 51.7 (11.7) | 54.3 (12.6) |
| Education >12 years, % | 53.3 | 33.2 | 39.7 | 28.4 |
| Married/cohabitant, % | 82.7 | 83.8 | 78.3 | 75.5 |
| Body mass index, mean (SD), kg/m 2 | 26.9 (3.5) | 27.5 (3.7) | 27.3 (3.8) | 26.1 (3.8) |
| Total/HDL cholesterol ratio, mean (SD) | 4.50 (1.37) | 4.50 (1.34) | 4.70 (1.36) | 4.88 (1.52) |
| Systolic blood pressure (mmHg), mean (SD) | 134.9 (19.2) | 141.1 (20.1) | 133.2 (18.0) | 134.8 (19.3) |
| Hypertension, % | 16.1 | 30.5 | 12.7 | 25.3 |
| Alcohol (per drinking session), % No alcohol 1–2 units 3–4 units 5 or more units | 9.2 52.2 27.7 10.9 | 7.1 51.3 29.9 11.7 | 3.6 35.7 34.0 26.7 | 4.6 35.1 31.9 28.4 |
| Leisure time physical activity, % Sedentary Light activity Moderate-to-vigorous activity | 17.2 49.5 33.3 | 19.3 56.5 24.2 | 23.4 53.4 23.2 | 32.8 48.3 18.9 |
| . | Never smokers . | Former smokers . | Occasional smokers . | Daily smokers . |
--- ---
| Women, n (%) | 3321 (38.8) | 2529 (29.5) | 723 (8.5) | 1988 (23.2) |
| Number of pack-years, mean (SD) Median (IQR) | 0 | 10.0 (10.0) 7 (11) | 10.2 (9.9) 7 (12) | 16.8 (10.5) 15 (11.6) |
| Age, mean (SD), years | 56.5 (13.8) | 56.2 (12.2) | 51.2 (12.0) | 53.4 (12.3) |
| Education >12 years, % | 41.5 | 35.9 | 40 | 25.5 |
| Married/cohabitant, % | 69.8 | 72.4 | 66.2 | 65.2 |
| Body mass index, mean (SD), kg/m 2 | 26.7 (4.7) | 27.1 (4.6) | 25.7 (4.5) | 25.4 (4.4) |
| Total/HDL cholesterol ratio, mean (SD) | 3.88 (1.21) | 3.87 (1.19) | 3.76 (1.21) | 4.20 (1.32) |
| Systolic blood pressure (mmHg), mean (SD) | 134.7 (23.8) | 132.9 (22.6) | 125.3 (21.8) | 128.8 (22.2) |
| Hypertension, % | 31.4 | 22.8 | 11.9 | 24.0 |
| Alcohol (per drinking session), % No alcohol 1–2 units 3–4 units 5 or more units | 22.9 63.9 11.3 1.9 | 9.6 67.4 19.7 3.3 | 7.2 53.0 32.4 7.4 | 8.7 53.1 29.0 9.2 |
| Leisure time physical activity, % Sedentary Light activity Moderate-to-vigorous activity | 17.8 67.6 14.6 | 17.8 69.0 13.2 | 20.2 67.5 12.3 | 27.2 62.7 10.1 |
| Men, n (%) | 2199 (31.2) | 2642 (37.4) | 718 (10.2) | 1497 (21.2) |
| Number of pack-years, mean (SD) Median (IQR) | 0 | 16.4 (16.2) 12 (17) | 14.3 (13.3) 11.2 (13.3) | 21.7 (13.6) 19.8 (16.1) |
| Age, mean (SD), years | 51.8 (12.0) | 59.7 (11.8) | 51.7 (11.7) | 54.3 (12.6) |
| Education >12 years, % | 53.3 | 33.2 | 39.7 | 28.4 |
| Married/cohabitant, % | 82.7 | 83.8 | 78.3 | 75.5 |
| Body mass index, mean (SD), kg/m 2 | 26.9 (3.5) | 27.5 (3.7) | 27.3 (3.8) | 26.1 (3.8) |
| Total/HDL cholesterol ratio, mean (SD) | 4.50 (1.37) | 4.50 (1.34) | 4.70 (1.36) | 4.88 (1.52) |
| Systolic blood pressure (mmHg), mean (SD) | 134.9 (19.2) | 141.1 (20.1) | 133.2 (18.0) | 134.8 (19.3) |
| Hypertension, % | 16.1 | 30.5 | 12.7 | 25.3 |
| Alcohol (per drinking session), % No alcohol 1–2 units 3–4 units 5 or more units | 9.2 52.2 27.7 10.9 | 7.1 51.3 29.9 11.7 | 3.6 35.7 34.0 26.7 | 4.6 35.1 31.9 28.4 |
| Leisure time physical activity, % Sedentary Light activity Moderate-to-vigorous activity | 17.2 49.5 33.3 | 19.3 56.5 24.2 | 23.4 53.4 23.2 | 32.8 48.3 18.9 |
There were missing values for some variables.
IQR, interquartile range.
Table 1
Open in new tab
Distribution of selected characteristics at enrolment given as percentage and mean (SD) by smoking and sex, in Tromsø5 (2001) or Tromsø6 (2007–08)
| . | Never smokers . | Former smokers . | Occasional smokers . | Daily smokers . |
--- ---
| Women, n (%) | 3321 (38.8) | 2529 (29.5) | 723 (8.5) | 1988 (23.2) |
| Number of pack-years, mean (SD) Median (IQR) | 0 | 10.0 (10.0) 7 (11) | 10.2 (9.9) 7 (12) | 16.8 (10.5) 15 (11.6) |
| Age, mean (SD), years | 56.5 (13.8) | 56.2 (12.2) | 51.2 (12.0) | 53.4 (12.3) |
| Education >12 years, % | 41.5 | 35.9 | 40 | 25.5 |
| Married/cohabitant, % | 69.8 | 72.4 | 66.2 | 65.2 |
| Body mass index, mean (SD), kg/m 2 | 26.7 (4.7) | 27.1 (4.6) | 25.7 (4.5) | 25.4 (4.4) |
| Total/HDL cholesterol ratio, mean (SD) | 3.88 (1.21) | 3.87 (1.19) | 3.76 (1.21) | 4.20 (1.32) |
| Systolic blood pressure (mmHg), mean (SD) | 134.7 (23.8) | 132.9 (22.6) | 125.3 (21.8) | 128.8 (22.2) |
| Hypertension, % | 31.4 | 22.8 | 11.9 | 24.0 |
| Alcohol (per drinking session), % No alcohol 1–2 units 3–4 units 5 or more units | 22.9 63.9 11.3 1.9 | 9.6 67.4 19.7 3.3 | 7.2 53.0 32.4 7.4 | 8.7 53.1 29.0 9.2 |
| Leisure time physical activity, % Sedentary Light activity Moderate-to-vigorous activity | 17.8 67.6 14.6 | 17.8 69.0 13.2 | 20.2 67.5 12.3 | 27.2 62.7 10.1 |
| Men, n (%) | 2199 (31.2) | 2642 (37.4) | 718 (10.2) | 1497 (21.2) |
| Number of pack-years, mean (SD) Median (IQR) | 0 | 16.4 (16.2) 12 (17) | 14.3 (13.3) 11.2 (13.3) | 21.7 (13.6) 19.8 (16.1) |
| Age, mean (SD), years | 51.8 (12.0) | 59.7 (11.8) | 51.7 (11.7) | 54.3 (12.6) |
| Education >12 years, % | 53.3 | 33.2 | 39.7 | 28.4 |
| Married/cohabitant, % | 82.7 | 83.8 | 78.3 | 75.5 |
| Body mass index, mean (SD), kg/m 2 | 26.9 (3.5) | 27.5 (3.7) | 27.3 (3.8) | 26.1 (3.8) |
| Total/HDL cholesterol ratio, mean (SD) | 4.50 (1.37) | 4.50 (1.34) | 4.70 (1.36) | 4.88 (1.52) |
| Systolic blood pressure (mmHg), mean (SD) | 134.9 (19.2) | 141.1 (20.1) | 133.2 (18.0) | 134.8 (19.3) |
| Hypertension, % | 16.1 | 30.5 | 12.7 | 25.3 |
| Alcohol (per drinking session), % No alcohol 1–2 units 3–4 units 5 or more units | 9.2 52.2 27.7 10.9 | 7.1 51.3 29.9 11.7 | 3.6 35.7 34.0 26.7 | 4.6 35.1 31.9 28.4 |
| Leisure time physical activity, % Sedentary Light activity Moderate-to-vigorous activity | 17.2 49.5 33.3 | 19.3 56.5 24.2 | 23.4 53.4 23.2 | 32.8 48.3 18.9 |
| . | Never smokers . | Former smokers . | Occasional smokers . | Daily smokers . |
--- ---
| Women, n (%) | 3321 (38.8) | 2529 (29.5) | 723 (8.5) | 1988 (23.2) |
| Number of pack-years, mean (SD) Median (IQR) | 0 | 10.0 (10.0) 7 (11) | 10.2 (9.9) 7 (12) | 16.8 (10.5) 15 (11.6) |
| Age, mean (SD), years | 56.5 (13.8) | 56.2 (12.2) | 51.2 (12.0) | 53.4 (12.3) |
| Education >12 years, % | 41.5 | 35.9 | 40 | 25.5 |
| Married/cohabitant, % | 69.8 | 72.4 | 66.2 | 65.2 |
| Body mass index, mean (SD), kg/m 2 | 26.7 (4.7) | 27.1 (4.6) | 25.7 (4.5) | 25.4 (4.4) |
| Total/HDL cholesterol ratio, mean (SD) | 3.88 (1.21) | 3.87 (1.19) | 3.76 (1.21) | 4.20 (1.32) |
| Systolic blood pressure (mmHg), mean (SD) | 134.7 (23.8) | 132.9 (22.6) | 125.3 (21.8) | 128.8 (22.2) |
| Hypertension, % | 31.4 | 22.8 | 11.9 | 24.0 |
| Alcohol (per drinking session), % No alcohol 1–2 units 3–4 units 5 or more units | 22.9 63.9 11.3 1.9 | 9.6 67.4 19.7 3.3 | 7.2 53.0 32.4 7.4 | 8.7 53.1 29.0 9.2 |
| Leisure time physical activity, % Sedentary Light activity Moderate-to-vigorous activity | 17.8 67.6 14.6 | 17.8 69.0 13.2 | 20.2 67.5 12.3 | 27.2 62.7 10.1 |
| Men, n (%) | 2199 (31.2) | 2642 (37.4) | 718 (10.2) | 1497 (21.2) |
| Number of pack-years, mean (SD) Median (IQR) | 0 | 16.4 (16.2) 12 (17) | 14.3 (13.3) 11.2 (13.3) | 21.7 (13.6) 19.8 (16.1) |
| Age, mean (SD), years | 51.8 (12.0) | 59.7 (11.8) | 51.7 (11.7) | 54.3 (12.6) |
| Education >12 years, % | 53.3 | 33.2 | 39.7 | 28.4 |
| Married/cohabitant, % | 82.7 | 83.8 | 78.3 | 75.5 |
| Body mass index, mean (SD), kg/m 2 | 26.9 (3.5) | 27.5 (3.7) | 27.3 (3.8) | 26.1 (3.8) |
| Total/HDL cholesterol ratio, mean (SD) | 4.50 (1.37) | 4.50 (1.34) | 4.70 (1.36) | 4.88 (1.52) |
| Systolic blood pressure (mmHg), mean (SD) | 134.9 (19.2) | 141.1 (20.1) | 133.2 (18.0) | 134.8 (19.3) |
| Hypertension, % | 16.1 | 30.5 | 12.7 | 25.3 |
| Alcohol (per drinking session), % No alcohol 1–2 units 3–4 units 5 or more units | 9.2 52.2 27.7 10.9 | 7.1 51.3 29.9 11.7 | 3.6 35.7 34.0 26.7 | 4.6 35.1 31.9 28.4 |
| Leisure time physical activity, % Sedentary Light activity Moderate-to-vigorous activity | 17.2 49.5 33.3 | 19.3 56.5 24.2 | 23.4 53.4 23.2 | 32.8 48.3 18.9 |
There were missing values for some variables.
IQR, interquartile range.
Table 2 shows the smoking habits in Tromsø5 and Tromsø6 in participants attending both surveys. Among participants attending both studies, there was an increase in the prevalence of occasional smoking from 6.2% in 2001 to 9.7% in 2007–08, indicating a shift in smoking habits. Conversely, the prevalence of daily smoking decreased from 24.7% to 12.1% among these participants.
Table 2
Open in new tab
Change in smoking status from Tromsø5 to Tromsø6 in 4305 subjects who attended both Tromsø5 and Tromsø6
| Smoking status in Tromsø5 . | Smoking status in Tromsø6 . |
--- |
| Never . | Previous . | Occasional . | Daily . | Total (Tromsø5) . |
| Never | 1458 | 42 | 7 | 0 | 1507 (35.0) |
| Previous | 0 | 1398 | 39 | 32 | 1469 (34.1) |
| Occasional | 0 | 124 | 120 | 22 | 266 (6.2) |
| Daily | 0 | 343 | 253 | 467 | 1063 (24.7) |
| Total (Tromsø6) | 1458 (33.9) | 1907 (44.3) | 419 (9.7) | 521 (12.1) | 4305 (100) |
| Smoking status in Tromsø5 . | Smoking status in Tromsø6 . |
--- |
| Never . | Previous . | Occasional . | Daily . | Total (Tromsø5) . |
| Never | 1458 | 42 | 7 | 0 | 1507 (35.0) |
| Previous | 0 | 1398 | 39 | 32 | 1469 (34.1) |
| Occasional | 0 | 124 | 120 | 22 | 266 (6.2) |
| Daily | 0 | 343 | 253 | 467 | 1063 (24.7) |
| Total (Tromsø6) | 1458 (33.9) | 1907 (44.3) | 419 (9.7) | 521 (12.1) | 4305 (100) |
The Tromsø Study 2001–08.
Values are numbers (percent).
Participants with missing values for smoking variables are not included.
Table 2
Open in new tab
Change in smoking status from Tromsø5 to Tromsø6 in 4305 subjects who attended both Tromsø5 and Tromsø6
| Smoking status in Tromsø5 . | Smoking status in Tromsø6 . |
--- |
| Never . | Previous . | Occasional . | Daily . | Total (Tromsø5) . |
| Never | 1458 | 42 | 7 | 0 | 1507 (35.0) |
| Previous | 0 | 1398 | 39 | 32 | 1469 (34.1) |
| Occasional | 0 | 124 | 120 | 22 | 266 (6.2) |
| Daily | 0 | 343 | 253 | 467 | 1063 (24.7) |
| Total (Tromsø6) | 1458 (33.9) | 1907 (44.3) | 419 (9.7) | 521 (12.1) | 4305 (100) |
| Smoking status in Tromsø5 . | Smoking status in Tromsø6 . |
--- |
| Never . | Previous . | Occasional . | Daily . | Total (Tromsø5) . |
| Never | 1458 | 42 | 7 | 0 | 1507 (35.0) |
| Previous | 0 | 1398 | 39 | 32 | 1469 (34.1) |
| Occasional | 0 | 124 | 120 | 22 | 266 (6.2) |
| Daily | 0 | 343 | 253 | 467 | 1063 (24.7) |
| Total (Tromsø6) | 1458 (33.9) | 1907 (44.3) | 419 (9.7) | 521 (12.1) | 4305 (100) |
The Tromsø Study 2001–08.
Values are numbers (percent).
Participants with missing values for smoking variables are not included.
Table 3 shows the person-years, MI cases, and crude and age-adjusted incidence of MI with updated smoking status for all participants. A total of 1297 participants experienced MI during the study period. Table 3 shows that the age-adjusted incidence rate per 10 000 person-years ranged from 33.6 (never smokers) to 64.6 (daily smokers). Supplementary material online, Table S4 shows similar results when the analyses were stratified by sex. Smoking and MI status for the total number of participants attending Tromsø5 and Tromsø6 are presented separately in Supplementary files; Supplementary material online, Table S3.
Table 3
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Crude and age-adjusted incidence rates and multivariable-adjusted hazard ratio (95% confidence interval) for myocardial infarction according to smoking status
| Smoking status . | Person-years . | MI cases, n . | MI incidence per 10 000 person-years . | Age-adjusted MI incidence per 10 000 person-years . | Model 1 . | Model 2 . |
--- --- ---
| Never | 77 996.5 | 375 | 48.1 | 33.6 | 1 (reference) | 1 (reference) |
| Former | 75 730.2 | 492 | 65.0 | 40.2 | 1.05 (0.91–1.21) | 1.02 (0.86–1.21) |
| Occasional | 21 331.5 | 114 | 53.4 | 52.4 | 1.42 (1.15–1.76) | 1.41 (1.11–1.80) |
| Daily | 42 897.7 | 316 | 73.7 | 64.6 | 1.87 (1.60–2.18) | 1.70 (1.40–2.06) |
| Smoking status . | Person-years . | MI cases, n . | MI incidence per 10 000 person-years . | Age-adjusted MI incidence per 10 000 person-years . | Model 1 . | Model 2 . |
--- --- ---
| Never | 77 996.5 | 375 | 48.1 | 33.6 | 1 (reference) | 1 (reference) |
| Former | 75 730.2 | 492 | 65.0 | 40.2 | 1.05 (0.91–1.21) | 1.02 (0.86–1.21) |
| Occasional | 21 331.5 | 114 | 53.4 | 52.4 | 1.42 (1.15–1.76) | 1.41 (1.11–1.80) |
| Daily | 42 897.7 | 316 | 73.7 | 64.6 | 1.87 (1.60–2.18) | 1.70 (1.40–2.06) |
The Tromsø Study 2001–21.
Model 1: adjusted for age and sex.
Model 2: as Model 1 + married/cohabitant, education, BMI, total/HDL cholesterol ratio, hypertension, alcohol, physical activity.
There were missing values for some variables in Model 2.
Table 3
Open in new tab
Crude and age-adjusted incidence rates and multivariable-adjusted hazard ratio (95% confidence interval) for myocardial infarction according to smoking status
| Smoking status . | Person-years . | MI cases, n . | MI incidence per 10 000 person-years . | Age-adjusted MI incidence per 10 000 person-years . | Model 1 . | Model 2 . |
--- --- ---
| Never | 77 996.5 | 375 | 48.1 | 33.6 | 1 (reference) | 1 (reference) |
| Former | 75 730.2 | 492 | 65.0 | 40.2 | 1.05 (0.91–1.21) | 1.02 (0.86–1.21) |
| Occasional | 21 331.5 | 114 | 53.4 | 52.4 | 1.42 (1.15–1.76) | 1.41 (1.11–1.80) |
| Daily | 42 897.7 | 316 | 73.7 | 64.6 | 1.87 (1.60–2.18) | 1.70 (1.40–2.06) |
| Smoking status . | Person-years . | MI cases, n . | MI incidence per 10 000 person-years . | Age-adjusted MI incidence per 10 000 person-years . | Model 1 . | Model 2 . |
--- --- ---
| Never | 77 996.5 | 375 | 48.1 | 33.6 | 1 (reference) | 1 (reference) |
| Former | 75 730.2 | 492 | 65.0 | 40.2 | 1.05 (0.91–1.21) | 1.02 (0.86–1.21) |
| Occasional | 21 331.5 | 114 | 53.4 | 52.4 | 1.42 (1.15–1.76) | 1.41 (1.11–1.80) |
| Daily | 42 897.7 | 316 | 73.7 | 64.6 | 1.87 (1.60–2.18) | 1.70 (1.40–2.06) |
The Tromsø Study 2001–21.
Model 1: adjusted for age and sex.
Model 2: as Model 1 + married/cohabitant, education, BMI, total/HDL cholesterol ratio, hypertension, alcohol, physical activity.
There were missing values for some variables in Model 2.
Table 3 also displays HRs for MI according to smoking status. Both occasional smoking and daily smoking were associated with an increased risk of MI. In age- and sex-adjusted Cox proportional hazards regression analysis (Model 1), occasional smoking was associated with a 42% increased risk of MI [hazard ratio (HR) 1.42, 95% confidence interval (CI) 1.15–1.76] compared to never smoking. The results were similar when adjusting for covariates [HR 1.41, 95% CI 1.11–1.80 (Model 2)]. Excluding individuals with missing values in covariates did not change the result for Model 1. In sex-specific analyses, occasional smoking was associated with a doubling in risk of MI in women (HR 2.08, 95% CI 1.43–3.04) and an 11% increase (HR 1.11, 95% CI 0.81–1.52) in men, as compared with never smokers (see Supplementary files; Supplementary material online, Table S4). The sex difference was, however, not statistically significant neither for the overall smoking status variable (P = 0.18) nor for the HR comparing occasional smoking with no smoking (P = 0.06).
In a sub-group of 8789 ever-smokers with complete information on pack-years, the age- and sex-adjusted risk for MI in occasional smokers was higher than in former smokers (HR 1.35, 95% CI 1.08–1.70). Adjusting for pack-years gave essentially the same result (HR 1.35, 95% CI 1.07–1.69).
We also conducted an analysis restricted to 15 271 participants aged 79 years or below at baseline and during follow-up, disregarding follow-up information after 80 years. The results were like those obtained when including all participants with follow-up after 79 years, though the estimates were slightly higher. Likewise, when restricting the cohort to ever-smokers aged 79 years or below at baseline and during follow-up and adjusting for pack-years (n = 8650), the results remained consistent.
Discussion
In this population-based study, we observed an increased risk of MI among occasional smokers compared to both never- and ex-smokers. This association remained consistent also when the analysis was restricted to a sub-cohort consisting of individuals aged 79 years or younger at follow-up. Additionally, sex-specific analysis revealed a significantly higher relative risk of MI among women who were occasional smokers compared to never smokers, while no significant association was observed in men. The larger impact of smoking in women than in men is in accordance with other studies.2–4 To the best of our knowledge, there are no previous studies examining the association between occasional smoking and the risk of incident fatal or non-fatal MI in men and women in the general population and with a long follow-up period.
The prevalence of occasional smoking in our study was 9.2%, which was similar to official statistics for Norway.7 There has been a shift in smoking habits among the Norwegian population, with a decline in daily smoking and a shift from daily to occasional smoking. The occasional smoking rates have remained relatively unchanged, but with an increase in the younger population.7 This may be attributed to an increased awareness of the adverse health effects of smoking, prompting many smokers to attempt quitting. Those unsuccessful in quitting may transition to occasional smoking as an alternative.18
We found that both occasional smoking and daily smoking were associated with a higher risk of MI, consistent with previous findings demonstrating an increased risk of MI among daily smokers.2,3,19 Although extensive research has established daily smoking as a significant risk factor for MI, the impact of occasional smoking has been less studied. A 12-year follow-up study among men in Iran showed a higher risk of CHD among occasional smokers’ compared to never smokers’ HR of 1.98 (95% CI 1.00–3.92).12 Other studies, including the Tromsø Study, have found an association between occasional smoking and overall mortality11 and cardiovascular mortality.10,20
Tobacco smoke contains over 7000 chemicals that contribute to various cardiovascular complications, including increased heart rate and myocardial contractility, inflammation, endothelial impairment, and thrombus formation in veins and arteries, all of which can lead to CVD, including MI.21,22 Although the risk for MI was lower in occasional smokers compared to daily smokers, the risk was higher than in never or former smokers. A recent multinational study including cohorts from the USA, UK, Norway, and Canada found that quitting smoking substantially and quickly reduces the excess risk of cardiovascular premature death.23 Previous studies have shown that there is no safe limit for smoking, with a non-linear relationship between low levels of smoking and CVD.5,6 A pooled analysis of nine cohort studies in Japan found that low-intensity smokers had higher cardiovascular mortality compared to never smokers, with the risk increasing in a dose–response manner as the number of cigarettes per day increased.24 Despite this, most regular smokers hold the misconception that occasional smoking carries little or no risk.6,25 Therefore, quitting smoking completely is, by far, the best approach to reduce the risk of MI and improve overall health.
We also conducted a sex-specific analysis as an additional investigation, given the known sex difference in smoking patterns and the risk for CVD. Our study revealed a higher relative risk of MI among women who were occasional smokers compared to never smokers. Other studies have also found a higher relative MI risk among women who smoke, compared to men.3,26 Furthermore, a meta-analysis revealed a pooled adjusted female-to-male relative risk ratio of 1.25 (P< 0.0001) for smoking compared to non-smoking in relation to heart disease.27 The observed sex difference in the risk for heart disease could be attributed to various factors, including biological differences and disparities in smoking behaviours between women and men. The absorption of or the pathophysiological effects of toxic chemicals from cigarette smoke might be higher in women who smoke compared to men who smoke.28
The main limitation of this study is that, despite nearly 1300 incident MI cases, the number of cases was low for all subset analyses, especially for occasional smokers. Due to the relatively low number of incident MI cases among occasional smokers and the non-significant interaction with sex, we combined the results for women and men in the main analysis and adjusted for sex. The lack of statistical significance of the interaction with sex is likely a function of limited statistical power rather than the strength of the associations. Additionally, we lack information on passive smoking, which has been associated with an increased risk of MI in this population.3 Thus, the presence of passive smoking in the reference group might have attenuated the result. Studies have also shown an association between electronic cigarette use and increased risk of MI.29 At the time of Tromsø5 (2001) and Tromsø6 (2007–08), vaping products were nonexistent. However, vaping products have gained popularity in Norway, particularly among the younger population and smokers or former smokers who tend to see vaping products as a tool to quit or reduce smoking.30 Therefore, we recommend including vaping products in future studies if possible.
There is always a risk of selection bias in population studies, as individuals with health issues and unhealthy lifestyles may be less likely to participate in the study. Consequently, the study participants may not be fully representative of the entire population. This bias could influence our findings if the association between smoking and MI in invited subjects differs between those who attend and those who do not, a comparison that is not possible to conduct. A more relevant bias is probably misclassification, as the information about smoking habits was obtained from self-reported questionnaires. We lack objective measures for tobacco exposure such as cotinine or thiocyanate. A previous Norwegian study demonstrated a strong relationship between self-reported smoking habits and the measure of serum thiocyanate if the questioning occurred in an unbiased environment.31 Thus, we believe our study demonstrated reasonable validity. However, less socially acceptable habits are often underreported, which may also be the case in our study. We attempted to minimize this issue by cross-checking multiple smoking questionnaires for inconsistencies. In addition, Tromsø7 survey conducted in 2015–16 reported a decrease in daily smoking prevalence within this population.32 The shift in smoking prevalence may have led to an underestimation of the true effects of smoking on MI. Another limitation is the lack of information regarding the number of cigarettes per smoking occasion for occasional smokers.
As the identification of MI cases included non-fatal and fatal hospitalized cases and fatal non-hospitalized cases, it is likely that we have missed cases of non-fatal non-hospitalized MI in the population. The external validity is limited to a Caucasian population and may not be generalizable.
The major strength of our study was the use of longitudinal data with a large sample size and high attendance rate, which provided robust and reliable findings. Additionally, we have a 20-year-long follow-up period for the subjects who attended Tromsø5. We utilized longitudinal data collected 6 years apart and updated changes in smoking habits and other covariates during the follow-up period. Another significant strength was the availability of endpoint information with high validity obtained from local and national registries, ensuring accuracy and completeness in our analysis.
Conclusion
Our study shows an increased risk of MI among occasional smokers. It underscores the notion that no level of smoking is safe. Any amount of smoking can have detrimental effects on cardiovascular health and increase the risk of MI. Consequently, public health initiatives should prioritize smoking cessation efforts to mitigate risk and enhance overall health.
Supplementary material
Supplementary material is available at European Journal of Preventive Cardiology.
Author contributions
S.T.: methodology, software, validation, formal analysis, investigation, data curation, writing—original draft, visualization, project administration. M.-L.L.: data collection, conceptualization, supervision, writing—critical review and editing, project administration, funding acquisition. B.K.J.: conceptualization, methodology, validation, writing—critical review and editing. T.W., O.L.: methodology, software, validation, writing—critical review and editing. E.B.M.: data collection, writing—critical review and editing. H.S., I.T.G., J.M.: writing—critical review and editing. All authors have reviewed and approved the final version of the manuscript and agree to be accountable for all aspects of work ensuring integrity and accuracy.
Funding
S.T.’s research was funded by the Department of Clinical Medicine, UiT The Arctic University of Norway.
Data availability
The data underlying this article are not publicly available but are available upon reasonable request to the Tromsø Study. More information can be found at
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Author notes
Conflict of interest: M.-L.L. has received lecture fees from Bayer, Sanofi, and Bristol Myers Squibb (BMS)/Pfizer and is an Associate Editor of the European Journal of Preventive Cardiology. Henrik Schirmer has received lecture fees from Amgen, Boehringer Ingelheim, BMS, Novartis, and Sanofi-Aventis.
© The Author(s) 2025. Published by Oxford University Press on behalf of the European Society of Cardiology.
This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted reuse, distribution, and reproduction in any medium, provided the original work is properly cited.
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How to determine if 2 points are on opposite sides of a line
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How can I determine whether the 2 points (a x,a y)(a x,a y) and (b x,b y)(b x,b y) are on opposite sides of the line (x 1,y 1)→(x 2,y 2)(x 1,y 1)→(x 2,y 2)?
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edited Jun 25, 2012 at 5:47
Zev Chonoles
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asked Jun 25, 2012 at 5:41
Craig DayCraig Day
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Find the equation of the line. It should be of the form a x+b y+c=0 a x+b y+c=0. Given two points (x 1,y 2)(x 1,y 2) and (x 2,y 2)(x 2,y 2), plug these into that equation. They are on opposite side of the line if a x 1+b y 1+c<0 a x 1+b y 1+c<0 and a x 2+b y 2+c>0 a x 2+b y 2+c>0, or visa-versa.
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answered Jun 25, 2012 at 5:48
WilliamWilliam
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Explicitly, they are on opposite sides iff
((y 1−y 2)(a x−x 1)+(x 2−x 1)(a y−y 1))((y 1−y 2)(b x−x 1)+(x 2−x 1)(b y−y 1))<0.((y 1−y 2)(a x−x 1)+(x 2−x 1)(a y−y 1))((y 1−y 2)(b x−x 1)+(x 2−x 1)(b y−y 1))<0.
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answered Jun 25, 2012 at 6:03
copper.hatcopper.hat
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Appreciate the explicitness of your answer. Clearly I am no math geek :) Cheers!Craig Day –Craig Day 2012-06-25 07:42:34 +00:00 Commented Jun 25, 2012 at 7:42
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C D C D are on the opposite sides of A B A B if and only if (A B−→−×A C−→−)⋅(A B−→−×A D−→−)<0(A B→×A C→)⋅(A B→×A D→)<0, where ×× is cross product and ⋅⋅ is dot product.
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answered Jun 25, 2012 at 5:46
Yai0PhahYai0Phah
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2 How is the cross product defined for 2 dimensions?marty cohen –marty cohen 2015-12-01 00:44:05 +00:00 Commented Dec 1, 2015 at 0:44
@martycohen Defined for an arbitrary embedding R 2↪R 3 R 2↪R 3.Yai0Phah –Yai0Phah 2015-12-01 09:35:01 +00:00 Commented Dec 1, 2015 at 9:35
The cross product of two ℝ3 vectors is perpendicular to them. If both vectors are in ℝ2, the result is [0,0,z] and z could be negative, zero, or positive. Often the cross product of two ℝ2's is defined as just an ℝ1 value, x1y2-x2y1.david van brink –david van brink 2022-10-22 16:44:11 +00:00 Commented Oct 22, 2022 at 16:44
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Writing A A and B B for the points in question, and P 1 P 1 and P 2 P 2 for the points determining the line ...
Compute the "signed" areas of the △P 1 P 2 A△P 1 P 2 A and △P 1 P 2 B△P 1 P 2 B via the formula (equation 16 here)
1 2∣∣∣∣x 1 x 2 x 3 y 1 y 2 y 3 1 1 1∣∣∣∣1 2|x 1 y 1 1 x 2 y 2 1 x 3 y 3 1|
with (x 3,y 3)(x 3,y 3) being A A or B B. The points A A and B B will be on opposite sides of the line if the areas differ in sign, which indicates that the triangles are being traced-out in different directions (clockwise vs counterclockwise).
You can, of course, ignore the "1/2 1/2", as it has not affect the sign of the values. Be sure to keep the row-order consistent in the two computations, though.
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answered Jun 25, 2012 at 6:10
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8988 | https://projecteuclid.org/journals/topological-methods-in-nonlinear-analysis/volume-55/issue-2/Borsuk-Ulam-theorems-for-products-of-spheres-and-Stiefel-manifolds/10.12775/TMNA.2019.103.pdf | Topological Methods in Nonlinear Analysis Volume 55, No. 2, 2020, 553–564 DOI: 10.12775/TMNA.2019.103 c ⃝2020 Juliusz Schauder Centre for Nonlinear Studies Nicolaus Copernicus University in Toru´ n BORSUK–ULAM THEOREMS FOR PRODUCTS OF SPHERES AND STIEFEL MANIFOLDS REVISITED Yu Hin Chan — Shujian Chen Florian Frick — J. Tristan Hull Abstract. We give a different and possibly more accessible proof of a gen-eral Borsuk–Ulam theorem for a product of spheres, originally due to Ramos. That is, we show the non-existence of certain (Z/2)k-equivariant maps from a product of k spheres to the unit sphere in a real (Z/2)k-representation of the same dimension. Our proof method allows us to derive Borsuk–Ulam theorems for certain equivariant maps from Stiefel manifolds, from the corresponding results about products of spheres, leading to alter-native proofs and extensions of some results of Fadell and Husseini.
1. Introduction Let X be a compact n-dimensional CW complex with an action by the group G. A fundamental question with a multitude of applications in topological combinatorics is to decide whether an equivariant map X →V (that is, a map commuting with a G-action) into some n-dimensional real G-representation V must have 0 ∈V in its image. Equivalently, one is interested in deciding the ex-istence of an equivariant map X →S(V ) into the unit sphere of V . This method has found applications in hyperplane mass partitions , the “square-peg” prob-lem , Tverberg-type results , and chromatic numbers of hypergraphs , among others; see , . Thus the identification of easily computable obstruc-tions to the existence of such equivariant maps is of fundamental importance.
2020 Mathematics Subject Classification. 55M20, 54H25.
Key words and phrases. Borsuk–Ulam theorem; Stiefel manifold; equivariant map.
553 554 Y.H. Chan — S. Chen — F. Frick — J.T. Hull One incarnation of this problem that has received particular attention is the case that X is a product of spheres Sn1 × . . . × Snk, and G is (Z/2)k where the jth copy of Z/2, generated by εj, acts non-trivially exactly on the jth factor Snj.
The case k = 1 is the classical Borsuk–Ulam theorem, which states that there is no map Sn →Sn−1 that commutes with the antipodal actions. Extensions of this result to products of spheres have been studied also because such maps naturally appear for the problem of equipartitions by hyperplanes; see for example , and .
Here we study a binary-valued obstruction for the existence of equivariant maps: the parity of the sum of degrees of a specially extended map restricted to various submanifolds obstructs the existence of an equivariant map.
This yields a possibly more accessible proof of Ramos’ general Borsuk–Ulam theorem for products of spheres . Moreover, our reasoning extends to Stiefel mani-folds Vn,k of k mutually orthonormal vectors in Rn. A classical result of Fadell and Husseini establishes the non-existence of an equivariant map from Vn,k into S(V ), where V is (Rn−k)k and εj acts non-trivially on the jth copy of Rn−k.
The difference in dimensions of Vn,k and (Rn−k)k is k 2 , which leaves room for improvement of Fadell and Husseini’s result. It seems that the following theorem has not been recorded before, although (as pointed by an anonymous referee) it is implicit in , where the difference of dimensions is also exploited in a different way: Theorem 1.1. Every (Z/2)k-equivariant map Vn,k →Rn−1 ⊕Rn−2 ⊕. . . ⊕Rn−k has a zero. Here εj acts non-trivially precisely on the j-th factor Rn−j and by (x1, . . . , xj, . . . , xn) 7→(x1, . . . , −xj, . . . , xn) on Vn,k.
We prove a more general result for arbitrary (Z/2)k-actions on the codomain; see Theorem 3.3. We also show that the zeros of an equivariant map Vn,k → (Rn−k)k can be restricted to lie in a fixed submanifold of codimension k 2 ; see Corollary 3.5. Ramos’ approach is elementary and technical, depending on equi-variant approximations of sufficiently generic PL maps for appropriately defined triangulations. Some effort has been invested into simplifying his proofs. For example, a special case was proved by Dzedzej, Idzik, and Izydorek . Further, we mention that some results of Ramos have been reproven since gaps have been pointed out in the treatment of non-free actions (but not for the treatment of free actions as in the present manuscript). These results have been salvaged by different methods , and Vre´ cica and ˇ Zivaljevi´ c have proposed a supplement for Ramos’ proof . Several authors have studied the existence of (Z/2)k-equivariant maps from products of spheres to a sphere using the theory of Fadell and Husseini, such as Mani-Levitska, Vre´ cica, and ˇ Zivaljevi´ c , Blagojevi´ c and Borsuk–Ulam Theorems Revisited 555 Ziegler , and Simon . For computations of Fadell and Husseini’s cohomo-logical index of Stiefel manifolds see Inoue and Blagojevi´ c and Karasev .
2. Borsuk–Ulam theorems for products of spheres We denote the standard generators of (Z/2)k by ε1, . . . , εk.
We think of Z/2 = {0, 1} additively and write ⟨α, β⟩= P j αjβj ∈Z/2 for the inner product in (Z/2)k. For α ∈(Z/2)k denote by Vα the vector space R with the action of (Z/2)k where εj acts non-trivially by x 7→−x if ⟨εj, αi⟩= 1 and trivially otherwise.
Denote the closed upper hemisphere of Sni by Bni and let B = Bn1 × . . . × Bnk. If the equivariant map f : Sn1 × . . . × Snk →V is never zero on ∂B, then f induces a map b f : ∂B →S(V ), x 7→f(x)/|f(x)|. We will show that the parity of the degree of this map is independent of f and only depends on the module V and the numbers n1, . . . , nk. We denote the degree of b f modulo 2 by r(n1, . . . , nk; V ) ∈Z/2.
The degree of b f equivalently counts the number of zeros of f in B counted with signs and multiplicities. The notion of sign and multipicity here is captured by the local degree: Let X and Y be oriented closed n-dimensional manifolds, x ∈X, and f : X →Y a continuous map. Then f induces a map f∗: Hn(X, X \ {x}) →Hn(Y, Y \ {f(x)}).
Both the domain and codomain of this homomorphism are isomorphic to Z, and thus f∗is uniquely determined by d = f∗(1), the local degree deg f|x of f around x. We refer to Outerelo and Ruiz for the basics of mapping degree theory . For example, they prove (see [15, Proposition 4.5]): Lemma 2.1. Let W be a compact, oriented (n+1)-manifold with boundary X.
Let f : W →Rn+1 be continuous with f −1(0) finite and disjoint from X. Then the degree of the map b f : X →Sn, x 7→f(x)/|f(x)| is the sum of local degrees of f around its zeros: deg b f = X x∈f −1(0) deg f|x.
While in this lemma is stated in the smooth category for maps with regular value 0, it is simple to see that the lemma holds in this slightly more general setting. See for instance [2, Proof of Lemma 5.6] for a proof.
As a second ingredient we need that if the action on the domain is free, any two G-equivariant maps have congruent degrees modulo the order of G; see Kushkuley and Balanov [10, Corollary 2.4]: Theorem 2.2. Let X and Y be closed oriented n-dimensional manifolds with actions by the finite group G, such that the G-action on X is free. Then for any 556 Y.H. Chan — S. Chen — F. Frick — J.T. Hull two equivariant maps f1, f2 : X →Y their degrees are congruent modulo |G|: deg f1 ≡deg f2 mod |G|.
For x ∈Sn ⊂Rn+1 we denote the ith coordinate of x by e∗ i (x). We fix the upper hemisphere Bn ⊂Sn as the set of x ∈Sn with e∗ n+1(x) ≥0. We can now prove that r(n1, . . . , nk; V ) is indeed independent of f: Lemma 2.3. Let f : Sn1 × . . . × Snk →V be a (Z/2)k-equivariant map that is never zero on ∂B and has only isolated zeros. Then the degree modulo 2 of the induced map b f : ∂B →S(V ) is independent of f.
Proof. The degree of b f is equal to the sum of zeros of f in B counted with sign and multiplicity by Lemma 2.1. Denote by M = Sn1−1×Sn2×. . .×Snk. The degree modulo 2k = (Z/2)k of the map f|M is independent of f by Theorem 2.2.
This degree counts the number of zeros of f in W = Bn1 × Sn2 × . . . × Snk with signs and multiplicities, again by Lemma 2.1. Every zero of f in B occurs 2k−1 times in W by symmetry. If these 2k−1 symmetric copies of a zero in B all have the same sign – this is the case if each εj, j ≥2, preserves orientation on M if and only if it preserves the orientation of S(V ) – then deg f|M is the sum of zeros of f in B multiplied by 2k−1. Since deg f|M modulo 2k is independent of f, so is the parity of deg b f.
We now induct on the number of generators εj that act in opposite ways on the orientation of M and S(V ). Suppose we have already shown that no matter how the generators εj, 2 ≤j ≤ℓ−1 act on the orientations of M and S(V ), the parity of the degree deg b f is independent of f. Further, assume that εℓacts orientation-preservingly on M and orientation-reversingly on S(V ) or vice versa.
Let M ′ = Sn1+1 ×. . .×Snk and B′ = Bn1+1 ×. . .×Bnk. Extend f equivariantly to a map f ′ : M ′ →V . In this extension process ensure that f ′ has only finitely many zeros x = (x1, . . . , xk) with e∗ 1(x1)e∗ 1(xℓ) = 0. This is possible since f has no zeros with e∗ 1(x1) = 0 or e∗ 1(xℓ) = 0, and any map Sn−1 →Sn−1 can be extended to the entire ball Bn →Rn such that it only has finitely many zeros.
Now consider the map F : M ′ →V ⊕Vα, (x1, . . . , xk) 7→ f ′(x1, . . . , xk), e∗ 1(x1)e∗ 1(xℓ) , where α = ε1 + εℓ.
The map F is equivariant and has only isolated zeros.
Further, by perhaps slightly rotating some of the spheres in the domain M ′, we can guarantee that F has no zeros in ∂B′, since F has only finitely many zeros.
Now εℓacts in the same way on the orientations of M ′ and S(V ⊕Vα), since εℓacts in opposite ways on S(V ) and S(V ⊕Vα). Thus by induction the parity of the degree of the induced map b F : ∂B′ →S(V ⊕Vα) does not depend on F. This counts the zeros of F in B′ with signs and multiplicities. These zeros are precisely the zeros of f in B and the zeros of F in Bn1+1 × Bn2 × . . . × Bnℓ−1 × . . . × Bnk.
Borsuk–Ulam Theorems Revisited 557 But the parity of the latter number of zeros does not depend on F by induction.
Thus the parity of the number of zeros of f in B does not depend on f either.□ That the parity of deg b f is independent of f could also be derived as a con-sequence of elementary obstruction theory. We briefly sketch this argument and refer to tom Dieck for the basics of (equivariant) obstruction theory. Let X be an n-dimensional CW complex with a free cellular G-action. Denote the k-skeleton of X by X(k). Let Y be an (n−2)-connected, (n−1)-simple G-space.
Then there is a G-map h: X(n−1) →Y . Whether h can be extended (up to homotopy on X(n−2)) to a G-map defined on all of X is captured by the ob-struction cocycle o ∈Hn G(X; πn−1Y ). In the situation described here – a primary obstruction problem – the cohomology class o is independent of the map h, and a G-map X →Y exists if and only if the cohomology class o vanishes.
If Y = Sn−1 then the value (of a representative) of o on an n-cell σ of X is the degree of h restricted to ∂σ. A sphere has a Z/2-equivariant CW structure with two cells in each dimension. This induces a CW complex structure on a product of spheres that is equivariant with respect to (Z/2)k. This CW structure has one orbit of top-dimensional n-cells, and thus deg b f determines o. Each orbit of (n −1)-cells intersects the boundary of a fixed n-cell in an even number of cells.
Since each equivariant (n−1)-cochain has the same value up to signs on the cells of the same orbit, the parity of the value of o on each n-cell is well-defined (and thus independent of the map h).
Lemma 2.4. Let f : Sn1 × . . . × Snk →V be an equivariant map that is never zero on ∂B, has only isolated zeros, and is a local homeomorphism around zeros. Then r(n1, . . . , nk; V ) = 1 if and only if f has an odd number of zeros in B. Moreover, if an equivariant map Sn1 × . . . × Snk →S(V ) exists, then r(n1, . . . , nk; V ) = 0.
Proof. Since f is a local homeomorphism around zeros, every local degree deg f|x for x with f(x) = 0 is ±1. By Lemma 2.1 the sum P x∈f −1(0)∩B deg f|x is odd if and only if r(n1, . . . , nk; V ) = 1. This is the case precisely if f −1(0)∩B has an odd number of elements. The second statement is an immediate consequence of Lemma 2.1.
□ It is now an elementary exercise to prove the following result, originally due to Ramos .
Theorem 2.5 (Ramos [16, Section 3]). The value of r(n1, . . . , nk; V ) can be computed recursively via r n1, . . . , nk; n M i=1 Vαi = k X j=1 ⟨αn, εj⟩r n1, . . . , nj −1, . . . , nk; n−1 M i=1 Vαi .
(2.1) 558 Y.H. Chan — S. Chen — F. Frick — J.T. Hull Proof. Let f : Sn1 × . . . × Snk → n−1 L i=1 Vαi be an equivariant map such that for (x1, . . . , xk) ∈Sn1 × . . . × Snk with e∗ 1(xj)e∗ nj+1(xj) = 0 = e∗ 1(xℓ) for j ̸= ℓ, f(x1, . . . , xk) ̸= 0. Such a map exists by obstruction theory since S n−1 L i=1 Vαi is (n −3)-connected. In the next step of the construction of f, where we define f on the (n −1)-skeleton, we can moreover ensure that f has finitely many zeros x with e∗ 1(xj) = 0 for some j. This is because cell-by-cell a map Sn−2 →Sn−2 can be extended to a map Bn−1 →Rn−1 with finitely many zeros. Now define F : Sn1 × . . . × Snk → n M i=1 Vαi, (x1, . . . , xk) 7→ f(x1, . . . , xk), Y j:⟨αn,εj⟩=1 e∗ 1(xj) .
The product in the last coordinate ensures that F is also equivariant in the Vαn-component. Now observe that the zeros of F in Bn1 × . . . × Bnk are in bijection with zeros of f in Bn1 × . . . × Bnk that satisfy e∗ 1(xj) = 0, where j ranges over indices with ⟨αn, εj⟩= 1. Using Lemma 2.4 and reducing modulo 2 finishes the proof.
□ Remark 2.6. The value of r(n1, . . . , nk; V ) is well-defined even in the case that some ni are 0. In this case the manifold X = Sn1 × . . . × Snk splits into several connected components and thus Hn−1(∂B) ∼ = Zc, where c is the number of components, that is, c = 2t with t the number of S0-factors in X. In this case we define the degree of the map b f : ∂B →S(V ) to be b f∗(1, . . . , 1). Since this counts the parity of the number of zeros of f in B with signs and multiplicities, our results hold in the same way, even if some ni = 0. Further, an equivariant map Sn1 × . . . × Snk−1 × S0 →S(V ) exists if and only if an equivariant map Sn1 × . . . × Snk−1 →S(V ) exists, where we forget the action of εk to define the induced (Z/2)k−1-actions. This is true simply because Sn1 × . . . × Snk−1 × S0 consists of two disjoint copies of Sn1 ×. . .×Snk−1. For the nontrivial Z/2-action on R we have that r(1; R) = 1; this is the Intermediate Value Theorem. Using these observations and Theorem 2.5, we can compute the value of the obstruction r(n1, . . . , nk; V ) by induction.
Other Borsuk–Ulam results for products of spheres are corollaries of Theo-rem 2.5: Remark 2.7. An immediate consequence of Theorem 2.5 is r(n1, . . . , nk; W ⊕Vεj) = r(n1, . . . , nj −1, . . . , nk; W) (2.2) and thus r n1, . . . , nk; V ⊕n1 ε1 ⊕. . . ⊕V ⊕nk εk = 1. This implies that any (Z/2)k-map Sn1 ×. . .×Snk →V ⊕n1 ε1 ⊕. . .⊕V ⊕nk εk has a zero – a proof of this special case Borsuk–Ulam Theorems Revisited 559 of Ramos’ result using the cohomological index theory of Fadell and Husseini is due to Dzedzej, Idzik and Izydorek .
Remark 2.8. As another special case we remark that r(n1, n2; W ⊕Vε1+ε2) = r(n1 −1, n2; W) + r(n1, n2 −1; W), (2.3) and thus r n1, n2; W ⊕Vε1+ε2 = 1 if and only if r(n1 −1, n2; W) ̸= r(n1, n2 −1; W).
This is the same recursion that computes binomial coefficients n1+n2 n1 , and thus r n1, n2; V ⊕(n1+n2) ε1+ε2 = 1 if and only if n1+n2 n1 is odd. This is the case if and only if the binary expansions of n1 and n2 do not share a common 1.
Remark 2.9. Combining the observations of the previous two remarks shows that r 3·2t−1, 3·2t−2; (Vε1 ⊕Vε2 ⊕Vε1+ε2)⊕(2t+1−1) = r 2t, 2t−1; V ⊕(2t+1−1) ε1+ε2 = 1, obstructing the existence of a (Z/2)2-equivariant map S3n−1 × S3n−2 →S((Vε1 ⊕Vε2 ⊕Vε1+ε2)⊕(2n−1)) for n = 2t.
This is a result of Mani-Levitska, Vre´ cica, and ˇ Zivaljevi´ c .
3. Borsuk–Ulam theorems for Stiefel manifolds The purpose of this section is to derive our main result, Theorem 1.1, from the methods developed in the preceding section. Recall that for positive integers k ≤n, Vn,k denotes the Stiefel manifold of k pairwise orthonormal vectors in Rn.
In particular, Vn,k ⊂(Sn−1)k and Vn,k is invariant under the action of (Z/2)k and thus inherits an action of (Z/2)k. We will strengthen the following result of Fadell and Husseini: Theorem 3.1 (Fadell and Husseini ). Every equivariant map Vn,k →(Vε1 ⊕. . . ⊕Vεk)⊕(n−k) has a zero.
The dimension of Vn,k exceeds the dimension of the codomain (Vε1 ⊕. . .
⊕Vεk)⊕(n−k) by k 2 .
We give two strengthenings of Theorem 3.1, where the dimensions of domain and codomain coincide. This is possible by restricting the domain to an invariant submanifold of codimension k 2 (this is achieved by Corollary 3.5), or by mapping to a larger codomain instead while still guarantee-ing the existence of a zero – this is Theorem 1.1, which is already implicit in .
560 Y.H. Chan — S. Chen — F. Frick — J.T. Hull Moreover, we prove a generalized result that applies to arbitrary (Z/2)k-actions on the codomain; see Theorem 3.3.
For α ∈(Z/2)k denote by |α| the ℓ1-norm of α, that is, the number of non-zero entries. The dimension of L |α|=2 Vα is k 2 = k−1 P i=0 i. We will need the following: Lemma 3.2. r k −1, k −2, . . . , 1, 0; L |α|=2 Vα = 1.
We provide two short proofs of this lemma. The first proof using the recursive formula of Theorem 2.5, the second by exhibiting an appropriate equivariant map with an odd number of zeros in a fundamental domain of the (Z/2)k-action.
Proof 1 of Lemma 3.2. Let W = M α∈(Z/2)k−1, |α|=2 Vα and U = k−1 M j=1 Vεj.
By Remark 2.6 and using equation (2.2) above k −1 times, r k −1, k −2, . . . , 1, 0; M |α|=2 Vα = r(k −1, k −2, . . . , 1; W ⊕U) = r(k −2, k −3, . . . , 1, 0; W).
Thus the lemma follows by induction on k. The base case is the Intermediate Value Theorem.
□ Proof 2 of Lemma 3.2. Consider a filtration S0 ⊂. . . ⊂Sk−1 obtained by successively intersecting Sk−1 ⊂Rk by coordinate hyperplanes. Then it is simple to check that the map Sk−1 × . . . × S1 × S0 → M |α|=2 Vα, (x1, . . . , xk) 7→(⟨xi, xj⟩)i<j is equivariant and has exactly one zero up to symmetry.
□ We can adapt the reasoning above to show a Borsuk–Ulam type theorem for Stiefel manifolds. The proof of the following theorem uses the same reasoning as [2, Prop. 3.3], which was previously used to derive Tverberg-type results in .
Theorem 3.3. Let k ≤n be integers and m = k(n−1)− k 2 . Let α1, . . . , αm ∈ (Z/2)k, and denote m L i=1 Vαi by V . If r(n −1, . . . , n −1 | {z } k times ; V ⊕ M |α|=2 Vα) = 1 then there is no equivariant map Vn,k →S(V ).
Borsuk–Ulam Theorems Revisited 561 Proof. We prove the contrapositive. Given an equivariant map f : Vn,k → S(V ), extend it to an equivariant map f ′ : (Sn−1)k →V . We could argue that this extension exists by exhibiting a cell complex for (Sn−1)k that respects the (Z/2)k-action and has Vn,k as a subcomplex. Then by contractibility of V and the action of (Z/2)k being free, there is no obstruction to the existence of this exten-sion. Here we instead give an explicit formula for f ′: let (x1, . . . , xk) ∈(Sn−1)k.
To define f ′(x1, . . . , xk), first inductively define points y1, . . . , yk ∈Rn. Think of Sn−1 as the unit sphere in Rn. Set y1 = x1, and having defined y1, . . . , yj−1, let yj = πj(xj), where πj : Rn →Rn is the orthogonal projection onto the orthog-onal complement of the subspace spanned by x1, . . . , xj−1. Further, whenever yj ̸= 0, let y′ j = yj/|yj|. Now, if all yj are non-zero, define f ′(x1, . . . , xk) = f(y′ 1, . . . , y′ k) · Y |yj|, whereas if some yj = 0, then let f ′(x1, . . . , xk) = 0. This map is continuous: the map f as a continuous map on the compact manifold Vn,k is bounded and so are the |yj|. As (x1, . . . , xk) approaches a point where xj is in the subspace spanned by x1, . . . , xj−1 and thus yj approaches zero, the value of f ′(x1, . . . , xk) approaches zero as well. Moreover, f ′ is equivariant: flipping xj to −xj leaves yℓfor ℓ̸= j unchanged and flips y′ j to −y′ j (so long as yj ̸= 0). The equivariance of f ′ now follows from the equivariance of f.
Now define the equivariant map F : (Sn−1)k →V ⊕ M |α|=2 Vα, (x1, . . . , xk) 7→f ′(x1, . . . , xk) ⊕(⟨xi, xj⟩)i<j.
The map F does not have any zeros, since F(x1, . . . , xk) = 0 implies that f ′(x1, . . . , xk) = 0 and the xi are mutually orthogonal, thus (x1, . . . , xk) ∈Vn,k is a zero of f. As F does not have any zeros, r n −1, . . . , n −1; V ⊕ M |α|=2 Vα = 0.
□ Example 3.4. We provide an example of a consequence of Theorem 3.3. Let k = 2 and let n −1 be a power of two. By Remark 2.8 we have r n −1, n −2; (Vε1+ε2)⊕(2n−3) = 1 and thus, in particular, there is no (Z/2)2-map Vn,2 →S V ⊕(2n−4) ε1+ε2 .
We can now derive Theorem 1.1 that any equivariant map Vn,k →V ⊕(n−1) ε1 ⊕. . . ⊕V ⊕(n−k) εk has a zero.
562 Y.H. Chan — S. Chen — F. Frick — J.T. Hull Proof of Theorem 1.1. According to Theorem 3.3, we need to show that r n −1, . . . , n −1 | {z } k times ; V ⊕ M |α|=2 Vα = 1, where V = V ⊕(n−1) ε1 ⊕. . . ⊕V ⊕(n−k) εk . By Theorem 2.5 r n −1, . . . , n −1; V ⊕ M |α|=2 Vα = r k −1, k −2, . . . , 1, 0; M |α|=2 Vα , which is equal to 1 by Lemma 3.2.
□ Lastly, we can strengthen Theorem 3.1 by showing that there always is a zero of an equivariant map from the Stiefel manifold Vn,k that lies in some proper fixed submanifold. Let M = {(x1, . . . , xk) ∈Sn−k × Sn−k+1 × . . . × Sn−1 : ⟨xi, xj⟩= 0 for all i ̸= j}.
In particular, M is a (Z/2)k-invariant submanifold of Vn,k of codimension k 2 .
Corollary 3.5. Any equivariant map M →(Vε1 ⊕. . . ⊕Vεk)⊕(n−k) has a zero.
Proof. Given any such equivariant map f : M →(Vε1 ⊕. . . ⊕Vεk)⊕(n−k) extend it to an equivariant map f ′ : Vn,k →(Vε1 ⊕. . . ⊕Vεk)⊕(n−k).
This extension exists for the same reason as in the proof of Theorem 3.3, but now the explicit construction is simpler: think of Sn−k ⊂Sn−k+1 ⊂. . . ⊂Sn−1 ⊂Rn as a filtration of the unit sphere in Rn by successive equatorial spheres. Let (y1, . . . , yk) ∈Vn,k such that yj is orthogonal to all points in Sn−k+j. Declare f ′(y1, . . . , yk) to be zero and extend linearly along geodesics to Sn−k+j.
There is an equivariant map h: Vn,k →V ⊕(k−1) ε1 ⊕. . . ⊕V ⊕2 εk−2 ⊕Vεk−1 with h−1(0) = M: explicitly, the V ⊕(k−ℓ) εℓ -component, ℓ< k, of h(x1, . . . , xk) is given by (e∗ n(xℓ), e∗ n−1(xℓ), . . . , e∗ n−(k−ℓ)+1(xℓ)).
Then define the equivariant map F : Vn,k →(Vε1 ⊕. . . ⊕Vεk)⊕(n−k) ⊕V ⊕(k−1) ε1 ⊕V ⊕(k−2) ε2 ⊕. . . ⊕V ⊕2 εk−2 ⊕Vεk−1, x 7→(f ′(x), h(x)).
By Theorem 1.1 the map F has a zero. Now, as before, if F(x) = 0, then both f ′(x) = 0 and h(x) = 0. The latter implies x ∈M, while f ′(x) = 0 means that f(x) = 0.
□ Borsuk–Ulam Theorems Revisited 563 Acknowledgements. We thank the two referees for their thorough reading of our manuscript and numerous thoughtful comments, which improved the ex-position substantially. These results were obtained during the Summer Program for Undergraduate Research 2017 at Cornell University. The authors are grate-ful for the excellent research conditions provided by the program. The authors would like to thank Maru Sarazola for many insightful conversations and Pavle Blagojevi´ c for pointing out an additional relevant reference.
References P.V.M. Blagojevi´ c, F. Frick, A. Haase and G.M. Ziegler, Hyperplane mass partitions via relative equivariant obstruction theory, Doc. Math. 21 (2016), 735–771.
P.V.M. Blagojevi´ c, F. Frick, A. Haase and G.M. Ziegler, Topology of the Gr¨ unbaum–Hadwiger–Ramos hyperplane mass partition problem, Trans. Amer. Math.
Soc. 370 (2018), no. 10, 6795–6824.
P.V.M. Blagojevi´ c, F. Frick and G.M. Ziegler, Tverberg plus constraints, Bull. Lond.
Math. Soc. 46 (2014), no. 5, 953–967.
P.V.M. Blagojevi´ c and R. Karasev, Extensions of theorems of Rattray and Makeev, Topol. Methods Nonlinear Anal. 40 (2012), no. 1, 189–213.
P.V.M. Blagojevi´ c and G.M. Ziegler, The ideal-valued index for a dihedral group action, and mass partition by two hyperplanes, Topology Appl. 158 (2011), no. 12, 1326– 1351.
P.V.M. Blagojevi´ c and G.M. Ziegler, Beyond the Borsuk–Ulam theorem: The topolog-ical Tverberg story, A Journey Through Discrete Mathematics, Springer, 2017, pp. 273– 341.
Z. Dzedzej, A. Idzik and M. Izydorek, Borsuk–Ulam type theorems on product spaces II, Topol. Methods Nonlinear Anal. 14 (1999), no. 2, 345–352.
E. Fadell and S. Husseini, An ideal-valued cohomological index theory with applications to Borsuk–Ulam and Bourgin–Yang theorems, Ergodic Theory Dynam. Systems 8 (1988), no. 8, 73–85.
A. Inoue, Borsuk–Ulam type theorems on Stiefel manifolds, Osaka J. Math. 43 (2006), no. 1, 183–191.
A.M. Kushkuley and Z.I. Balanov, Geometric Methods in Degree Theory for Equivari-ant Maps, Springer, 2006.
P. Mani-Levitska, S.T. Vre´ cica and R.T. ˇ Zivaljevi´ c, Topology and combinatorics of partitions of masses by hyperplanes, Adv. Math. 207 (2006), no. 1, 266–296.
J. Matouˇ sek, Using the Borsuk–Ulam Theorem. Lectures on Topological Methods in Combinatorics and Geometry, second ed., Universitext, Springer–Verlag, Heidelberg, 2008.
J. Matouˇ sek and G.M. Ziegler, Topological lower bounds for the chromatic number: A hierarchy, Jahresber. Dtsch. Math.-Ver. 106 (2004), 71–90.
B. Matschke, A survey on the square peg problem, Notices Amer. Math. Soc. 61 (2014), no. 4, 346–352.
E. Outerelo and J.M. Ruiz, Mapping Degree Theory, Graduate Studies in Mathematics, vol. 108, Amer. Math. Soc., 2009.
E.A. Ramos, Equipartition of mass distributions by hyperplanes, Discrete Comput. Geom.
15 (1996), no. 2, 147–167.
564 Y.H. Chan — S. Chen — F. Frick — J.T. Hull S. Simon, Hyperplane equipartitions plus constraints, J. Combin. Theory, Ser. A 161 (2019), 29–50.
T. tom Dieck, Transformation Groups, vol. 8, Walter de Gruyter, 2011.
S.T. Vre´ cica and R.T. ˇ Zivaljevi´ c, Hyperplane mass equipartition problem and the shielding functions of Ramos (2015), arXiv: 1508.01552.
R.T. ˇ Zivaljevi´ c, Topological Methods, Handbook of Discrete and Computational Geome-try (J. O’Rourke, J.E. Goodman and C. Toth, eds.), Chapman & Hall/CRC, 2017.
Manuscript received February 13, 2019 accepted November 6, 2019 Yu Hin Chan Department of Mathematics University of California at Davis Davis, CA, USA E-mail address: yuhchan@math.ucdavis.edu Shujian Chen Department of Mathematics Brandeis University Waltham, MA, USA E-mail address: shujianchen@brandeis.edu Florian Frick Department of Mathematical Sciences Carnegie Mellon University Pittsburgh, PA, USA E-mail address: frick@cmu.edu J. Tristan Hull Department of Mathematics University of California at Berkeley Berkeley, CA, USA E-mail address: jth242@berkeley.edu TMNA : Volume 55 – 2020 – No 2 |
8989 | https://medium.com/@srinivas_59338/overcoming-the-burden-of-responsibility-b74b913f41b7 | Overcoming The Burden of Responsibility | by Seena | Medium
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Overcoming The Burden of Responsibility
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Do you feel that you're taking responsibility in your life?
A lot of people tell us that we should do that and I think it's one of those concepts that we really need to re-examine. Because there's this typical kind of - yeah I should be doing that, I should take responsibility for my life.
But there is kind of this secretive version to it too - Kind of a hesitation about taking responsibility for our lives. And I think it's probably a lot of it is to do with what do we mean when we say take responsibility for our lives?
Now, I'm going to talk about what I think it means to take responsibility for ourselves in life. And the first thing we need to know about it is it's kind of like a subject or a concept that has two levels to it.
1. Psychological Responsibility
2. Worldly Responsibility
There's to take Psychological responsibility for your life Because there's this other thing of like worldly responsibility. You know, what the world talks about is being taken responsibility – Are you legally responsible for this? Are you professionally responsible for this? These are all kind of worldly forms of responsibility.
I think it's much more important to talk about psychological responsibility. Because that's much more of an intimate personal choice that we make to take responsibility for our thoughts, our feeling and our emotions.
So really what I'm focused on more here is to forget this whole thing about Worldly responsibility and focus more on Psychological responsibility.
See the problem with it is the way the world typically thinks about responsibility.
Here's an example - You know, there's an office meeting taking place and somebody barges through the door and says who's responsible for this?
Now immediately everybody's head drops, Nobody's making eye contact anymore.
There's this dirty word, It's called responsibility that's come in here and nobody wants to take it. because typically and this is why we hesitant around responsibility. Responsibility secretly means blame. Okay who's to blame? Which is why it can be a very heavy word to use.
Well, you're saying I should take responsibility for my life? Are you saying everything is my fault? Am I to blame here? And if that's the way we're thinking about responsibility, who would want to take it? Okay, so like I agree with that.
I think this is my definition of responsibility
Responsibility without blame
Okay, so your responsibility like – I'm not blaming other people for this, but I'm not going to take responsibility and then therefore start to blame myself. Responsibility and blame have become unfortunately synonymous with each other and what I've taken responsibility for our life means I completely take ownership of this experience but I'm not going to blame myself for that experience or for what's happened here.
What's actually transpiring in my life whatsoever. Just more of an openness and a curiosity to what's happening is much more appropriate when we're taking responsibility.
One of the ways to think about why we're so hesitant about this responsibility is whenever I ask people this question in any given situation in life which of these two would you rather be?
Would you rather be the victim or would you rather be the victimizer? Would you rather be the person who gets hurt or the person doing the hurting?
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Most of us really do not like the idea of being the victimizer or the person who’s doing the hurting.
So, to avoid that, we sometimes find ourselves attracted to being the victim in any given situation. But it’s really just to avoid being the bad guy quite often, right?
It’s just, it’s not necessarily that. We want to be in victim consciousness, We just can not handle the idea that we’re going to be perceived as the bad guy or the wrong person or the wrong one in this situation.
So if we're going to actually move into responsibility in life, one of the things we can think about is how is this whole victim victimizer concept or paradigm actually helping me in my life?
Is it useful at all? Seeing good and bad, right and wrong, victim victimizer, in everything in life. What if it's not as simple as that? Or what if that's just not a helpful way to think about this?
A better concept I like to think about is again, responsibility without blame or self-empowerment.
I actually think we should be still using the word responsibility because I think it’s just a word that has lost its true meaning. But if you don’t like that word responsibility, or you have that sort of aversion to it.
Start to say, okay I’m going going to be self-empowered now. I’m going to take responsibility That means self-empowerment.
Once I take ownership of something in my life, I stop seeing myself as either a victimizer or a victim, no blame involved here. Now I can do something about this Ownership, self-empowerment. This is kind of the thing we’re looking at here, This is what we’re shooting for.
I'll tell you just a little quick story about this, what it means to kind of take responsibility to leave behind that victim victimizer thing and step into self-empowerment. It's just a little example but I remember personally a little while ago.
You know have that feeling in the morning when you wake up and you're like, oh my God, the world is just out to get me today. The world is imposing itself on top of me. I don't want to deal with it. I want it to go away. I just want to stay in bed. One of those types of days. And you know, I kind of immediately recognized. okay well, am I taking psychological responsibility here? Or am I kind of feeling a bit like a victim in this situation? And of course I did feel like a victim in that situation.
So I could have went into what's wrong with me? Why am I doing this? Why am I a victim? It's my fault that I'm the victim.
So what I did was I'm going to take responsibility without blaming myself. Am I a victim of this situation? No. Is it my fault? No.
I started to step into self-empowerment, and in self-empowerment what we can do is we can find, okay well, where is my role in what I'm experiencing right now?
I came to realize I feel exhausted, really so I don't want to handle this day. Self-empowerment responsibility was why am I so tired? What have I done to contribute to this tiredness? Not blaming myself, but I realized, you know what? I haven't been to bed early for a long time. Maybe that's something I could look at. Maybe I'm contributing at the very least.
This experience of being the victim of this day, which is imposing itself on me. But as soon as I did that, I began to feel less victimized by the day because I realized I'm actually contributing to this experience.
Did I feel bad about it? No, I felt good about it because now I could actually do something about this situation. So taking responsibility is very liberating for us. And it makes us feel less helpless and certainly less victimized by other people, by the world, you name it.
But the takeaway from this story today is, I want you to really take this away.
You know, we don't go to someone who is in deep pain and feeling victimized. we don't go and say you know you should really take responsibility because that's very cruel and we shouldn't do that.
We should see responsibility as probably the first issue most people have a responsibility or refusing to take it is that they're probably blaming themselves already secretly so responsibility see without blame every single time you use that term I'm going to take responsibility without blaming myself.
If we can do that we start to see solutions where we could only see problems, we start to we stop seeing the external world as this thing imposing itself on us victimizing us bullying us pushing us around and we start to feel less helplessness.
So responsibility without blame is the concept your going to take from today's Story.
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8990 | https://www.reddit.com/r/AskPhysics/comments/d2atdz/why_is_initial_velocity_not_0_when_something_is/?tl=es-419 | ¿Por qué la velocidad inicial no es 0 cuando algo se lanza hacia arriba en el aire? : r/AskPhysics
Skip to main content¿Por qué la velocidad inicial no es 0 cuando algo se lanza hacia arriba en el aire? : r/AskPhysics
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ParticularWar
¿Por qué la velocidad inicial no es 0 cuando algo se lanza hacia arriba en el aire?
Entonces, me estoy confundiendo un poco aquí. Si alguien lanza un objeto directamente hacia arriba en el aire, pensé que la velocidad inicial indicaba que estamos en t=0, lo que significaría que nuestra velocidad inicial es 0 porque en t=0 aún no ha salido de la mano de la persona. ¿No es 0 porque la persona está ejerciendo fuerza para lanzar el objeto al aire?
ej. ¿Cuál es la diferencia entre alguien que simplemente suelta un objeto desde un acantilado versus alguien que lo lanza hacia arriba y lo deja caer desde un acantilado?
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2025 AMC 8 Problems/Problem 1
Contents
1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Solution 6
8 Video Solution 1 (Detailed Explanation) 🚀⚡📊
9 Video Solution 2 Super Easy to understand and efficient!
10 Video Solution 3 by SpreadTheMathLove
11 Video Solution 4
12 Video Solution 5 by Daily Dose of Math
13 Video Solution 6 by Thinking Feet
14 Video Solution 7 by CoolMathProblems
15 Video Solution 8 by Pi Academy
16 Video Solution 9
17 Video Solution(Quick, fast, easy!)
18 See Also
Problem
The eight-pointed star, shown in the figure below, is a popular quilting pattern. What percent of the entire grid is covered by the star?
Solution 1
Each of the unshaded triangles has base length and height , so they all have area . Each of the unshaded unit squares has area . The area of the shaded region is equal to the area of the entire grid minus the area of the unshaded region, or . The star is then , or percent of the entire grid. ~cxsmi
Solution 2
There are total squares in the diagram and each square has triangles whose areas are half the area of a unit square. Thus, the total number of triangles in the diagram is triangles. There are shaded triangles in the diagram, so the area of the star is , or percent. ~Pi_in_da_box
Solution 3
There are squares that are entirely shaded and squares that have no shading. This cancels them out. The rest of the squares are half-half. Therefore the shaded region is percent of the grid.
Solution 4
Note that we can move one triangle from each of the four cells in the middle to each of the four corners. This will leave every cell in the grid with one triangle each, and each triangle has an area of half the area of each cell. Thus, our answer must equal to , and so our answer is ~derekwang2048
Solution 5
The shaded area is a square in the middle of the figure combined with small triangles. Since each small triangle is of a unit square, the star's area is equal to the area of unit squares, which percent of the grid.
-vockey
Solution 6
You can move the extra half a squares and make 4 squares so there are 8/16. The answer is percent of 16.
William
Video Solution 1 (Detailed Explanation) 🚀⚡📊
~ ChillThingz:)
Video Solution 2 Super Easy to understand and efficient!
Video Solution 3 by SpreadTheMathLove
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Video Solution 5 by Daily Dose of Math
~Thesmartgreekmathdude
Video Solution 6 by Thinking Feet
Video Solution 7 by CoolMathProblems
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Video Solution 9
Video Solution(Quick, fast, easy!)
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8992 | https://www.quora.com/If-f-x-2-f-x-f-2-what-is-the-value-of-f-x | If f (x+2) = f(x) + f(2) ,what is the value of f(x)? - Quora
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If f (x+2) = f(x) + f(2) ,what is the value of f(x)?
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·5y
No, no, no… So far all the answers are wrong.
If x x is an even integer, then sure, you can write: f(x)=x 2 f(2)f(x)=x 2 f(2)
Since f(2)f(2) is just a constant, we can set k=f(2)k=f(2), and then have:
f(x)=k 2 x f(x)=k 2 x
You can prove this quite easily with induction (hint: f(2)=f(0)+f(2)⟹f(0)=0 f(2)=f(0)+f(2)⟹f(0)=0, and then you need to both increment and decrement from there).
But what happens when x x is not an even integer? It can be odd, or real, or complex. In those cases we are lacking a base case. So we can pick any offset: f(x)=k 2 x+C(x)f(x)=k 2 x+C(x). All we require is that C(x)C(x) have a period of 2 2, and that C(0)C(0)
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No, no, no… So far all the answers are wrong.
If x x is an even integer, then sure, you can write: f(x)=x 2 f(2)f(x)=x 2 f(2)
Since f(2)f(2) is just a constant, we can set k=f(2)k=f(2), and then have:
f(x)=k 2 x f(x)=k 2 x
You can prove this quite easily with induction (hint: f(2)=f(0)+f(2)⟹f(0)=0 f(2)=f(0)+f(2)⟹f(0)=0, and then you need to both increment and decrement from there).
But what happens when x x is not an even integer? It can be odd, or real, or complex. In those cases we are lacking a base case. So we can pick any offset: f(x)=k 2 x+C(x)f(x)=k 2 x+C(x). All we require is that C(x)C(x) have a period of 2 2, and that C(0)=0 C(0)=0. This gives us a lot of flexibility, and allows us to produce an uncountable infinite number of solutions. For example,
f(x)=k 2 x+A sin(π x)f(x)=k 2 x+A sin(π x)
or
f(x)=k 2 x+B⋅{x/2}f(x)=k 2 x+B⋅{x/2}, where we take the fractional part of x/2 x/2
and where A A and B B can be any complex constants.
Here are plots of these two examples, with k=f(2)=2 k=f(2)=2, A=1 A=1 and B=1 B=1
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Rock Brentwood
Knows German · Author has 124 answers and 97.1K answer views
·3y
More generally, assume f(x + a) = f(x) + f(a) for a given number a. Then g(y) ≡ f(ay) - y f(a) has a period of 1, i.e. g(y + 1) = g(y). In addition, we also have g(0) = g(1) = f(a·1) - 1·f(a) = 0.
The case a = 0 is trivial, since all the condition says, then, is that f(0) = 0. Thus, in that case, the general solution is f(x) = h(x) - h(0), where h(x) is an arbitrary function.
For the other cases, assume a ≠ 0. Then, we may write f(x) = x/a f(a) + g(x/a), where g(y) is any function with period 1 for which g(0) = 0. This can be generalized further to arbitrary functions h(x) of period 1 by writing
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More generally, assume f(x + a) = f(x) + f(a) for a given number a. Then g(y) ≡ f(ay) - y f(a) has a period of 1, i.e. g(y + 1) = g(y). In addition, we also have g(0) = g(1) = f(a·1) - 1·f(a) = 0.
The case a = 0 is trivial, since all the condition says, then, is that f(0) = 0. Thus, in that case, the general solution is f(x) = h(x) - h(0), where h(x) is an arbitrary function.
For the other cases, assume a ≠ 0. Then, we may write f(x) = x/a f(a) + g(x/a), where g(y) is any function with period 1 for which g(0) = 0. This can be generalized further to arbitrary functions h(x) of period 1 by writing f(x) = x/a f(a) + h(x/a) - h(0).
For the case of interest, where a = 2, we can similarly write f(x) = ½x f(2) + h(½x) - h(0), where h(x) is an arbitrary function of period 1.
In Surfese, the most totally egregious example is h(x) = 1 if x is rational and h(x) = 0 if x is not rational and f(2) = 2·10¹⁰⁰, so that f(x) is a google times x plus h(x) minus 1 (since, here, h(x) = h(½x)), so that it only matches a google-x if x is irrational, while rational x’s, it comes short of a google-x by 1.
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Jonathan Devor
PhD in Astronomy, Harvard University (Graduated 2008) · Author has 3.7K answers and 19.4M answer views
·3y
Originally Answered: What if (x+2) =f(x) +f(2)? ·
Option #1: If the OP meant: x+2=f(x)+f(2)x+2=f(x)+f(2)
Then, setting x=2⟹4=2 f(2)⟹f(2)=2 x=2⟹4=2 f(2)⟹f(2)=2
So the OP’s equation becomes: x+2=f(x)+2⟹f(x)=x x+2=f(x)+2⟹f(x)=x
Option #2: If the OP meant: f(x+2)=f(x)+f(2)f(x+2)=f(x)+f(2)
Then, setting x=0⟹f(2)=f(0)+f(2)⟹f(0)=0 x=0⟹f(2)=f(0)+f(2)⟹f(0)=0
Setting x=−2⟹f(0)=f(−2)+f(2)⟹f(−2)=−f(2)x=−2⟹f(0)=f(−2)+f(2)⟹f(−2)=−f(2)
And in general, setting y=−x−2 y=−x−2 gives us:
f(−y)=f(−y−2)+f(2)⟹−f(−(y+2))=−f(−y)+f(2)f(−y)=f(−y−2)+f(2)⟹−f(−(y+2))=−f(−y)+f(2)
In other words, all the functional solutions must be odd, so that f(−x)=−f(x)f(−x)=−f(x)
Perhaps the simplest solution is: f(x)=x f(x)=x
But there are an infinite number of other functions that can also solve the O
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Option #1: If the OP meant: x+2=f(x)+f(2)x+2=f(x)+f(2)
Then, setting x=2⟹4=2 f(2)⟹f(2)=2 x=2⟹4=2 f(2)⟹f(2)=2
So the OP’s equation becomes: x+2=f(x)+2⟹f(x)=x x+2=f(x)+2⟹f(x)=x
Option #2: If the OP meant: f(x+2)=f(x)+f(2)f(x+2)=f(x)+f(2)
Then, setting x=0⟹f(2)=f(0)+f(2)⟹f(0)=0 x=0⟹f(2)=f(0)+f(2)⟹f(0)=0
Setting x=−2⟹f(0)=f(−2)+f(2)⟹f(−2)=−f(2)x=−2⟹f(0)=f(−2)+f(2)⟹f(−2)=−f(2)
And in general, setting y=−x−2 y=−x−2 gives us:
f(−y)=f(−y−2)+f(2)⟹−f(−(y+2))=−f(−y)+f(2)f(−y)=f(−y−2)+f(2)⟹−f(−(y+2))=−f(−y)+f(2)
In other words, all the functional solutions must be odd, so that f(−x)=−f(x)f(−x)=−f(x)
Perhaps the simplest solution is: f(x)=x f(x)=x
But there are an infinite number of other functions that can also solve the OP’s formula. All we need to do is take one solution and add to it any odd function with a period of two. For example, f(x)=x+A sin(π x)f(x)=x+A sin(π x)
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If f(x+1)+f(x+3)=f(x)+f(x+2)f(x+1)+f(x+3)=f(x)+f(x+2), what's f f?
What is f(x)f(x) when f(x+1)−f(x)=x f(x+1)−f(x)=x?
Is f(x) + f(2) = f(x+2)?
What is the value of f(2) in a function f(x) = x +1 if x<1?
If f(x)f(x) is a polynomial satisfying f(x)⋅f(1 x)=f(x)+f(1 x)f(x)⋅f(1 x)=f(x)+f(1 x) and f(3)=28 f(3)=28 then f(x)f(x)=?
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To solve the functional equation f(x+2)=f(x)+f(2)f(x+2)=f(x)+f(2), we can analyze the form of f(x)f(x).
Substitution: Let's make some substitutions to understand the function better. Start by substituting x=0 x=0:
f(2)=f(0)+f(2)f(2)=f(0)+f(2)
This simplifies to f(0)=0 f(0)=0.
Recursive Relation: Now, substitute x=2 x=2:
f(4)=f(2)+f(2)=2 f(2)f(4)=f(2)+f(2)=2 f(2)
Next, substitute x=4 x=4:
f(6)=f(4)+f(2)=2 f(2)+f(2)=3 f(2)f(6)=f(4)+f(2)=2 f(2)+f(2)=3 f(2)
Continuing this process, we can see a pattern forming. Let's denote f(2)=c f(2)=c. We can express the values of f f at even integers:
f(0)=0 f(0)=0
f(2)=c f(2)=c
f(4)=2 c f(4)=2 c
f(6)=3 c f(6)=3 c
From this, we can conjecture that:
f(2n)=nc\quad\text{for f(2n)=nc\quad\text{for
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To solve the functional equation f(x+2)=f(x)+f(2)f(x+2)=f(x)+f(2), we can analyze the form of f(x)f(x).
Substitution: Let's make some substitutions to understand the function better. Start by substituting x=0 x=0:
f(2)=f(0)+f(2)f(2)=f(0)+f(2)
This simplifies to f(0)=0 f(0)=0.
Recursive Relation: Now, substitute x=2 x=2:
f(4)=f(2)+f(2)=2 f(2)f(4)=f(2)+f(2)=2 f(2)
Next, substitute x=4 x=4:
f(6)=f(4)+f(2)=2 f(2)+f(2)=3 f(2)f(6)=f(4)+f(2)=2 f(2)+f(2)=3 f(2)
Continuing this process, we can see a pattern forming. Let's denote f(2)=c f(2)=c. We can express the values of f f at even integers:
f(0)=0 f(0)=0
f(2)=c f(2)=c
f(4)=2 c f(4)=2 c
f(6)=3 c f(6)=3 c
From this, we can conjecture that:
f(2 n)=n c for integers n f(2 n)=n c for integers n
Odd Integers: Now let's look at odd integers. Substitute x=1 x=1:
f(3)=f(1)+f(2)f(3)=f(1)+f(2)
Substitute x=3 x=3:
f(5)=f(3)+f(2)=(f(1)+f(2))+f(2)=f(1)+2 f(2)f(5)=f(3)+f(2)=(f(1)+f(2))+f(2)=f(1)+2 f(2)
Continuing this process, we can see that odd integers depend on the value of f(1)f(1).
General Form: We can assume that f(x)f(x) is a linear function. Let’s assume f(x)=k x f(x)=k x for some constant k k.
Substituting back into the original equation:
f(x+2)=k(x+2)=k x+2 k f(x+2)=k(x+2)=k x+2 k
f(x)+f(2)=k x+k(2)=k x+2 k f(x)+f(2)=k x+k(2)=k x+2 k
Since both sides are equal, our assumption holds.
Conclusion: Thus, the function f(x)f(x) can be expressed as:
f(x)=k x f(x)=k x
where k=f(2)/2 k=f(2)/2. If we set k=c/2 k=c/2 (where c=f(2)c=f(2)), we have:
f(x)=f(2)2 x f(x)=f(2)2 x
In summary, the general solution to the functional equation is:
f(x)=k x for some constant k.f(x)=k x for some constant k.
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Preston Lui
actuarial science from The University of Hong Kong · Author has 66 answers and 44.4K answer views
·5y
While Mr. Jonathan Devor, have an interesting graph, I don’t see why the x x part is required.
I will argue that as long as f(x)f(x) is in the form of
f(x)=G(x)+f l o o r(x/2)f(x)=G(x)+f l o o r(x/2)
it is a solution. With G(x)G(x) be a function with period 2
Unfortunately, I am unable to prove it is a unique solution.
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Drake Way
Mathematics Hobbyist · Upvoted by
Rhys Thomas
, MSc Mathematics & Physics, De Montfort University (2002) and
Erik Bergland
, PhD Mathematics, Brown University (2024) · Author has 4.2K answers and 3.3M answer views
·5y
Related
What is f(x),if f(x+y) =f(x) +f(y) +2xy?
First, let’s find the derivative.
d f d x=lim h→0 f(x+h)−f(x)h d f d x=lim h→0 f(x+h)−f(x)h
=lim h→0 f(x)+f(h)+2 x h−f(x)h=lim h→0 f(x)+f(h)+2 x h−f(x)h
=lim h→0 f(h)+2 x h h=lim h→0 f(h)+2 x h h
=2 x+lim h→0 f(h)h=2 x+lim h→0 f(h)h
Assume this derivative exists, if it does, then the limit as h approaches 0 of f(h)/h must be a constant. (If the function is differentiable, then the limit must exist, and because the derivative must be with respect to x, the limit being not a function of h must be a constant) So we can write this as:
=2 x+C 1=2 x+C 1
Now we integrate this to get our original func
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First, let’s find the derivative.
d f d x=lim h→0 f(x+h)−f(x)h d f d x=lim h→0 f(x+h)−f(x)h
=lim h→0 f(x)+f(h)+2 x h−f(x)h=lim h→0 f(x)+f(h)+2 x h−f(x)h
=lim h→0 f(h)+2 x h h=lim h→0 f(h)+2 x h h
=2 x+lim h→0 f(h)h=2 x+lim h→0 f(h)h
Assume this derivative exists, if it does, then the limit as h approaches 0 of f(h)/h must be a constant. (If the function is differentiable, then the limit must exist, and because the derivative must be with respect to x, the limit being not a function of h must be a constant) So we can write this as:
=2 x+C 1=2 x+C 1
Now we integrate this to get our original function:
f(x)=∫2 x+C 1 d x f(x)=∫2 x+C 1 d x
f(x)=x 2+C 1 x+C 2 f(x)=x 2+C 1 x+C 2
Let’s test this:
(x+y)2+C 1(x+y)+C 2(x+y)2+C 1(x+y)+C 2
=x 2+2 x y+y 2+C 1 x+C 1 y+C 2=x 2+2 x y+y 2+C 1 x+C 1 y+C 2
=[x 2+C 1 x+C 2]+[y 2+C 1 y+C 2]+2 x y−C 2=[x 2+C 1 x+C 2]+[y 2+C 1 y+C 2]+2 x y−C 2
=f(x)+f(y)+2 x y−C 2=f(x)+f(y)+2 x y−C 2
Close, but not quite—unless that C 2=0 C 2=0, in which case it all works, and therefore f(x)=x 2+C x f(x)=x 2+C x where C is an arbitrary constant.
Proofs
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Wan Kang
b.a. in Mathematics, Nanjing University (南京大学) (Graduated 1994) · Author has 631 answers and 268.8K answer views
·5y
possible functions:
f(x)=x f(x)=x
f(2)=0 f(2)=0, and f(.)f(.) is a periodic function. e.g.: f(x)=C sin(π x)f(x)=C sin(π x)
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Choi
Works at Wuhan, Hubei, China · Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979)
·11y
Related
If f[f(x)]=x 2+x f[f(x)]=x 2+x, what is f(x)f(x)?
You can see this paper
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You can see this paper
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Mehdi Khayeche
IMO Silver Medalist · Upvoted by
Hoosain Ebrahim
, BSc Mathematics & Mathematical Statistics (2021) and
Marek Kobera
, Ph. D. Mathematics & Physics, Charles University in Prague (2016) · Author has 75 answers and 1.4M answer views
·5y
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If f(x)f(x) is a polynomial satisfying f(x)⋅f(1 x)=f(x)+f(1 x)f(x)⋅f(1 x)=f(x)+f(1 x) and f(3)=28 f(3)=28 then f(x)f(x)=?
Forget about f(3)f(3). Let’s solve the problem in the general setting.
As others have noticed, we can rewrite the given condition as:
f(x)f(1 x)−f(x)−f(1 x)=0⇔f(x)f(1 x)−f(x)−f(1 x)+1=1 f(x)f(1 x)−f(x)−f(1 x)=0⇔f(x)f(1 x)−f(x)−f(1 x)+1=1
Now we are able to factor the equation:
(f(x)−1)(f(1 x)−1)=1(f(x)−1)(f(1 x)−1)=1
We can simplify it by setting Q(x)=f(x)−1 Q(x)=f(x)−1, which gives:
Q(x)Q(1 x)=1 Q(x)Q(1 x)=1
If Q(0)=0 Q(0)=0, we can factor out x x from the expression of Q Q and get Q(x)=x H 1(x)Q(x)=x H 1(x) for some polynomial H 1 H 1. Again, if H 1(0)=0 H 1(0)=0 we can repeat the process, and get Q(x)=x 2 H 2(x Q(x)=x 2 H 2(x
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Forget about f(3)f(3). Let’s solve the problem in the general setting.
As others have noticed, we can rewrite the given condition as:
f(x)f(1 x)−f(x)−f(1 x)=0⇔f(x)f(1 x)−f(x)−f(1 x)+1=1 f(x)f(1 x)−f(x)−f(1 x)=0⇔f(x)f(1 x)−f(x)−f(1 x)+1=1
Now we are able to factor the equation:
(f(x)−1)(f(1 x)−1)=1(f(x)−1)(f(1 x)−1)=1
We can simplify it by setting Q(x)=f(x)−1 Q(x)=f(x)−1, which gives:
Q(x)Q(1 x)=1 Q(x)Q(1 x)=1
If Q(0)=0 Q(0)=0, we can factor out x x from the expression of Q Q and get Q(x)=x H 1(x)Q(x)=x H 1(x) for some polynomial H 1 H 1. Again, if H 1(0)=0 H 1(0)=0 we can repeat the process, and get Q(x)=x 2 H 2(x)Q(x)=x 2 H 2(x) etc.. So let’s say Q(x)=x k R(x)Q(x)=x k R(x) where R(0)≠0 R(0)≠0 (if it was the case, then we could factor out x x again). Note that k k could be 0 0, in the case where Q(0)≠0 Q(0)≠0.
The intuition behind what I just did might not be obvious, but it will become clear later on.
Here’s the nice thing about our new polynomial R R. Since we have:
Q(x)Q(1 x)=1 Q(x)Q(1 x)=1
We can rewrite it as:
x k R(x)(1 x k R(1 x))=1 x k R(x)(1 x k R(1 x))=1
i.e.
R(x)R(1 x)=1 R(x)R(1 x)=1
So R R satisfies the same equation as Q Q. Isn’t that cute?
Now the good thing about polynomials, is that if two of them are equal, then their coefficients must match. For example if :
R(x)=∑n i=0 a i x i=x 5+3 x 2 R(x)=∑i=0 n a i x i=x 5+3 x 2
We can conclude that a 5=1 a 5=1, a 2=3 a 2=3 and a i=0 a i=0 otherwise.
In our case, however, we can’t directly do this since when we expand R(x)R(1 x)R(x)R(1 x) we will get negative powers of x x, so we wouldn’t have a polynomial.
The way to get around this is to consider n n, the degree of R R, and multiply the equation by x n x n, this would guarantee that we have no negative powers! We’d get:
R(x)⋅(x n R(1 x))=x n R(x)⋅(x n R(1 x))=x n
Now LHS is a polynomial in x x, so we can equate the coefficients with RHS.
The bad news is, expanding the LHS would be a mess as every coefficient would be some ugly-looking sum.
The good news, however, is that we only need to consider one coefficient to get what we need.
Always think of the extreme points, because they usually give the most amount of information. (doesn’t only apply to math ;-) )
The largest power in LHS is x 2 n x 2 n, which can only be achieved by combining a n x n a n x n from R(x)R(x) and a 0 x n a 0 x n from x n R(1 x)x n R(1 x). This tells us that the coefficient of x 2 n x 2 n is a n a 0 a n a 0.
We know that a n≠0 a n≠0, since it’s the largest coefficient, and we also know that a 0≠0 a 0≠0 because R(0)≠0 R(0)≠0 (Now you can see why we factored x k x k out of Q Q!).
However, the coefficient of x 2 n x 2 n in RHS is 0 0… Or is it?
Well, it shouldn’t be, but the only non-zero coefficient in RHS is that of x n x n. This tells us that we must have 2 n=n 2 n=n i.e. n=0 n=0 and R(x)R(x) is constant.
From R(x)R(1 x)=1 R(x)R(1 x)=1 we conclude that R(x)=±1 R(x)=±1, implying that:
f(x)=1+Q(x)=1+x k R(x)=1±x k f(x)=1+Q(x)=1+x k R(x)=1±x k
and we can easily check that it verifies our equation. ■◼
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Hilmar Zonneveld
Translator (1985–present) · Author has 58.5K answers and 19.4M answer views
·3y
Originally Answered: What if (x+2) =f(x) +f(2)? ·
It typically isn’t - just try this out with some common functions.
However, it is possible to design functions with this property.
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Upvote ·
20K 20K
1.6K 1.6K
999 446
Lewis Sue
5y
f(x) isn’t unique in this case.
Upvote ·
9 1
L Viswanathan
Former Member of Technical Staff at Bell Telephone Labs (1977–1999) · Author has 3K answers and 1.4M answer views
·5y
Let y = x + 2. f(y) = f(y-2) + f(2). Replacing y by x,
==> f(x) = f(x -2) + f(2), f(x) = x. f(2) = 2.
Upvote ·
Hrishikesh Deka
Noob at heart
·Updated 3y
Related
If f(f(x)) = 3x + 2, what is f(x)?
If you look at it closely, you will find that f(x) will be a polynomial function of degree 1, i.e., of the form — f(x) = ax + b.
The reason being since f(f(x)) is linear, if you take any other polynomial function of higher degree, f(f(x)) would turn out to have an even higher degree.
f(f(x)) = a(ax + b) + b
f(f(x)) = a²x + ab + b
Comparing, a²= 3 , (a+1)b = 2
If a=√3, b=√3–1 and if a=-√3, b=-(√3 +1).
Hope it helps !!!!
Edit: This is just one of the solutions of f(x), assuming f(x) to be a polynomial function.
Upvote ·
9 4
9 2
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If f(f(x))−x 2=x⋅f(x)f(f(x))−x 2=x⋅f(x) then what is the value of f(−100)f(−100)?
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8993 | https://doc.cgal.org/latest/Surface_mesh_shortest_path/index.html | CGAL 6.0.2 - Triangulated Surface Mesh Shortest Paths: User Manual
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▼CGAL 6.0.2 - Triangulated Surface Mesh Shortest Paths
▼User Manual
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►User Interface Description
►Examples
►Benchmarks
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Design and Implementation History
►Reference Manual
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►Class and Concept List
►Examples
User Manual
1 Introduction
2 User Interface Description
2.1 Surface Mesh Shortest Path Class
2.1.1 Specifying the Input
2.1.2 Specifying the Source Points
2.1.3 Building the Internal Sequence Tree
2.1.4 Shortest Path Queries
2.1.5 Additional Convenience Functionalities
2.2 Kernel Recommendations
3 Examples
3.1 Simple Example
3.2 Example Using Polyhedron_3
3.3 Example Using Polyhedron Items without IDs
3.4 Using Multiple Source Points
3.5 Shortest Path Sequence Visitor
4 Benchmarks
4.1 Single Source Point
4.2 Ten Source Points
4.3 Comparison of Construction and Query Times with Multiple Source Points
5 Implementation Details
5.1 Definitions
5.1.1 Geodesic Paths
5.1.2 Visibility Window
5.1.3 Saddle Vertices
5.1.4 The Sequence Tree
5.2 Algorithm Overview
5.3 Continuous Dijkstra
5.4 One Angle, One Split
5.5 Distance Filtering
5.6 Locating Shortest Paths
5.7 Pseudo-Code
6 Design and Implementation History
Author Stephen Kiazyk, Sébastien Loriot, and Éric Colin de Verdière
This package provides an algorithm to compute geodesic shortest paths on a triangulated surface mesh.
Figure 77.1 Shortest paths on a terrain using one source point represented by a green square.
1 Introduction
The motion planning of a robot across the surface of a 3-dimensional terrain is a typical application of the shortest path computation. Using a 2-dimensional approximation would fail to capture anything interesting about the terrain we are trying to cross, and would give a poor solution. The problem is often called the Discrete Geodesic Problem. Although the more general version of this problem, shortest paths in 3D in the presence of obstacles, is NP-Hard, when the motion is constrained to the 2D surface of an object it can be solved efficiently.
The algorithm implemented in this package builds a data structure to efficiently answer queries of the following form: Given a triangulated surface mesh M, a set of source points S on M, and a target point t also on M, find a shortest path λ between t and any element in S, where λ is constrained to the surface of M.
The algorithm used is based on a paper by Xin and Wang , a fast and practical algorithm for exact computation of geodesic shortest paths. It is an extension of earlier results by Chen and Han and Mitchell, Mount, and Papadimitriou .
This package is related to the package The Heat Method. Both deal with geodesic distances. The geodesic shortest path package computes the exact shortest path between any two points on the surface. The heat method package computes for every vertex of a mesh an approximate distance to one or several source vertices.
2 User Interface Description
2.1 Surface Mesh Shortest Path Class
The main class of this package is Surface_mesh_shortest_path. In the following we describe the typical workflow when using this class
2.1.1 Specifying the Input
The shortest paths are computed on a triangulated surface mesh, represented by a model of the FaceListGraph concept. There is no restriction on the genus, connectivity, or convexity of the input surface mesh.
For efficiency reason, index property maps for vertices, halfedges and faces are internally used. For each simplex type the property map must provide an index between 0 and the number of simplices. We recommend to use the class CGAL::Surface_mesh as model of FaceListGraph. If you use the class CGAL::Polyhedron_3, you should use it with the item class CGAL::Polyhedron_items_with_id_3, for which default property maps are provided. This item class associates to each simplex an index that provides a O(1) time access to the indices. Note that the initialization of the property maps requires a call to set_halfedgeds_items_id().
The access to the embedding of each vertex is done using a point vertex property map associating to each vertex a 3D point. Defaults are provided for CGAL classes.
If the traits class used holds some local state, it must also be passed to the class when constructing it (the default one provided does not).
2.1.2 Specifying the Source Points
The set of source points for shortest path queries can be populated one by one or using a range. A source point can be specified using either a vertex of the input surface mesh or a face of the input surface mesh with some barycentric coordinates. Given a point p that lies inside a triangle face (A,B,C), its barycentric coordinates are a weight triple (b 0,b 1,b 2) such that p=b 0⋅A+b 1⋅B+b 2⋅C, and b 0+b 1+b 2=1.
For convenience, a function Surface_mesh_shortest_path::locate() is provided to construct face locations from geometric inputs:
given a point p living in 3D space, this function computes the point closest to p on the surface, and returns the face containing this point, as well as its barycentric coordinates;
given a ray r living in 3D space, this function computes the intersection of the ray with the surface, and (if an intersection exists) returns the face containing this point, as well as its barycentric coordinates;
Usage of this function is illustrated in the example Surface_mesh_shortest_path/shortest_path_with_locate.cpp.
2.1.3 Building the Internal Sequence Tree
A time consuming operation for shortest path queries consists in building an internal data structure used to make the queries. This data structure is called the sequence tree. It will be built automatically when the first shortest path query is done and will be reused for any subsequent query as long as the set of source points does not change. Each time the set of source points is changed the sequence tree needs to be rebuilt (if already built). Note that it can also be built manually by a call to Surface_mesh_shortest_path::build_sequence_tree().
2.1.4 Shortest Path Queries
As for specifying the source points, the target point for a shortest path query can be specified using either a vertex of the input surface mesh or a face of the input surface mesh and some barycentric coordinates.
There are three different kinds of query functions that can be called using the class Surface_mesh_shortest_path. Given a target point, all these functions compute the shortest path between that target point and the set of source points:
Surface_mesh_shortest_path::shortest_distance_to_source_points() provides the closest source point to the target point together with the length of the shortest path.
Surface_mesh_shortest_path::shortest_path_points_to_source_points() provides all the intersection points of the shortest path with the edges and vertices of the input surface mesh (including the source and the target point). This function is useful for visualization purposes.
Surface_mesh_shortest_path::shortest_path_sequence_to_source_points gives access to the complete sequence of simplices crossed by the shortest path using a visitor object model of the concept SurfaceMeshShortestPathVisitor.
2.1.5 Additional Convenience Functionalities
Some convenience functions are provided to compute:
the point on the input surface mesh specified as a face of the input surface mesh and some barycentric coordinates.
the closest point on the input surface mesh (specified as a face of the input surface mesh and some barycentric coordinates) to a given 3D point. Those function are using the class CGAL::AABB_tree.
2.2 Kernel Recommendations
In short, we recommend to use a CGAL kernel with exact predicates such as CGAL::Exact_predicates_inexact_constructions_kernel.
If you need the constructions to be exact (for the shortest path point computation for example), you should use a kernel with exact constructions. Although the algorithm uses square root operations, it will also work on geometry kernels which do not support them by first converting the kernel's number type to double, using the std::sqrt, and converting it back. Note that it would be preferable to use a kernel with directly supports square roots to get the most precision of the shortest path computations.
Using a kernel such as CGAL::Exact_predicates_exact_constructions_kernel_with_sqrt with this package will indeed provide the exact shortest paths, but it will be extremely slow. Indeed, in order to compute the distance along the surface, it is necessary to unfold sequences of faces, edge-to-edge, out into a common plane. The functor SurfaceMeshShortestPathTraits::Construct_triangle_3_to_triangle_2_projection provides an initial layout of the first face in a sequence, by rotating a given face into the xy-plane. SurfaceMeshShortestPathTraits::Construct_triangle_3_along_segment_2_flattening unfolds a triangle into the plane, using a specified segment as a base. Since this results in a chain of constructed triangles in the plane, the exact representation types used with this kernel (either CORE::Expr or leda_real) will process extremely slow, even on very simple inputs. This is because the exact representations will effectively add an O(n) factor to every computation.
3 Examples
3.1 Simple Example
The following example shows how to get the shortest path to every vertex from an arbitrary source point on a surface. The shortest path class needs to have an index associated to each vertex, halfedge and face, which is naturally given for the class Surface_mesh.
FileSurface_mesh_shortest_path/shortest_paths.cpp
include
include
include <CGAL/Surface_mesh_shortest_path.h>
include
include
include
include
include
typedefCGAL::Exact_predicates_inexact_constructions_kernelKernel;
typedefCGAL::Surface_mesh Triangle_mesh;
typedefCGAL::Surface_mesh_shortest_path_traits Traits;
typedefCGAL::Surface_mesh_shortest_path Surface_mesh_shortest_path;
typedef boost::graph_traits Graph_traits;
typedef Graph_traits::vertex_iterator vertex_iterator;
typedef Graph_traits::face_iterator face_iterator;
int main(int argc, char argv)
{
const std::string filename = (argc>1) ? argv : CGAL::data_file_path("meshes/elephant.off");
Triangle_mesh tmesh;
if((filename, tmesh) ||
(tmesh))
{
std::cerr <<"Invalid input file."<< std::endl;
return EXIT_FAILURE;
}
// pick up a random face
const unsigned int randSeed = argc > 2 ? boost::lexical_cast(argv) : 7915421;
CGAL::Random rand(randSeed);
const int target_face_index = rand.get_int(0, static_cast(num_faces(tmesh)));
face_iterator face_it = faces(tmesh).first;
std::advance(face_it,target_face_index);
// ... and define a barycentric coordinates inside the face
Traits::Barycentric_coordinates face_location = {{0.25, 0.5, 0.25}};
// construct a shortest path query object and add a source point
Surface_mesh_shortest_path shortest_paths(tmesh);
shortest_paths.add_source_point(face_it, face_location);
// For all vertices in the tmesh, compute the points of
// the shortest path to the source point and write them
// into a file readable using CGAL Lab
std::ofstream output("shortest_paths_with_id.polylines.txt");
vertex_iterator vit, vit_end;
for ( boost::tie(vit, vit_end) = vertices(tmesh);
vit != vit_end; ++vit)
{
std::vector points;
shortest_paths.shortest_path_points_to_source_points(vit, std::back_inserter(points));
// print the points
output << points.size() <<" ";
for (std::size_t i = 0; i < points.size(); ++i)
output <<" "<< points[i];
output << std::endl;
}
return 0;
}
Surface_mesh_shortest_path.h
Convenience header file only including CGAL/Surface_mesh_shortest_path/Surface_mesh_shortest_path....
CGAL::Exact_predicates_inexact_constructions_kernel
CGAL::Surface_mesh_shortest_path_traits
A model of the concept SurfaceMeshShortestPathTraits as required by the Surface_mesh_shortest_path cl...
Definition: Surface_mesh_shortest_path_traits.h:43
CGAL::Surface_mesh_shortest_path
Computes shortest surface paths from one or more source points on a surface mesh.
Definition: Surface_mesh_shortest_path.h:73
CGAL::Surface_mesh
CGAL::is_triangle_mesh
bool is_triangle_mesh(const FaceGraph &g)
CGAL::IO::read_polygon_mesh
bool read_polygon_mesh(const std::string &fname, Graph &g, const NamedParameters &np=parameters::default_values())
Kernel
3.2 Example Using Polyhedron_3
The following example shows how to get the shortest path to every vertex from an arbitrary source point on the surface. Note that this example uses the Polyhedron_items_with_id_3 item class. The shortest path class needs to have an index associated to each vertex, halfedge and face. Using this item class provide an efficient direct access to the required indices.
FileSurface_mesh_shortest_path/shortest_paths_with_id.cpp
include
include
include
include <CGAL/Surface_mesh_shortest_path.h>
include
include
include
include
include
include
typedefCGAL::Exact_predicates_inexact_constructions_kernelKernel;
typedefCGAL::Polyhedron_3 Triangle_mesh;
typedefCGAL::Surface_mesh_shortest_path_traits Traits;
typedefCGAL::Surface_mesh_shortest_path Surface_mesh_shortest_path;
typedef boost::graph_traits Graph_traits;
typedef Graph_traits::vertex_iterator vertex_iterator;
typedef Graph_traits::face_iterator face_iterator;
int main(int argc, char argv)
{
// read input polyhedron
Triangle_mesh tmesh;
std::ifstream input((argc>1)?argv:CGAL::data_file_path("meshes/elephant.off"));
input >> tmesh;
// initialize indices of vertices, halfedges and faces
CGAL::set_halfedgeds_items_id(tmesh);
// pick up a random face
const unsigned int randSeed = argc > 2 ? boost::lexical_cast(argv) : 7915421;
CGAL::Random rand(randSeed);
const int target_face_index = rand.get_int(0, static_cast(num_faces(tmesh)));
face_iterator face_it = faces(tmesh).first;
std::advance(face_it,target_face_index);
// ... and define a barycentric coordinates inside the face
Traits::Barycentric_coordinates face_location = {{0.25, 0.5, 0.25}};
// construct a shortest path query object and add a source point
Surface_mesh_shortest_path shortest_paths(tmesh);
shortest_paths.add_source_point(face_it, face_location);
// For all vertices in the tmesh, compute the points of
// the shortest path to the source point and write them
// into a file readable using CGAL Lab
std::ofstream output("shortest_paths_with_id.polylines.txt");
vertex_iterator vit, vit_end;
for ( boost::tie(vit, vit_end) = vertices(tmesh);
vit != vit_end; ++vit)
{
std::vector points;
shortest_paths.shortest_path_points_to_source_points(vit, std::back_inserter(points));
// print the points
output << points.size() <<" ";
for (std::size_t i = 0; i < points.size(); ++i)
output <<" "<< points[i];
output << std::endl;
}
return 0;
}
CGAL::Polyhedron_3
CGAL::set_halfedgeds_items_id
void set_halfedgeds_items_id(Polyhedron_with_id &P)
3.3 Example Using Polyhedron Items without IDs
Although it is better to have an index built into each simplex, you can also use a surface mesh without internal indices by using external indices. The following example shows how to proceed in this case.
FileSurface_mesh_shortest_path/shortest_paths_no_id.cpp
include
include
include
include <CGAL/Surface_mesh_shortest_path.h>
include
include
include
include
include
typedefCGAL::Exact_predicates_inexact_constructions_kernelKernel;
typedefCGAL::Polyhedron_3 Triangle_mesh;
typedefCGAL::Surface_mesh_shortest_path_traits Traits;
// default property maps
typedef boost::property_map<Triangle_mesh,
boost::vertex_external_index_t>::const_type Vertex_index_map;
typedef boost::property_map<Triangle_mesh,
CGAL::halfedge_external_index_t>::const_type Halfedge_index_map;
typedef boost::property_map<Triangle_mesh,
CGAL::face_external_index_t>::const_type Face_index_map;
typedefCGAL::Surface_mesh_shortest_path<Traits,
Vertex_index_map,
Halfedge_index_map,
Face_index_map> Surface_mesh_shortest_path;
typedef boost::graph_traits Graph_traits;
typedef Graph_traits::vertex_iterator vertex_iterator;
typedef Graph_traits::halfedge_iterator halfedge_iterator;
typedef Graph_traits::face_iterator face_iterator;
int main(int argc, char argv)
{
Triangle_mesh tmesh;
std::ifstream input((argc>1)?argv:CGAL::data_file_path("meshes/elephant.off"));
input >> tmesh;
// pick up a random face
const unsigned int randSeed = argc > 2 ? boost::lexical_cast(argv) : 7915421;
CGAL::Random rand(randSeed);
const int target_face_index = rand.get_int(0, static_cast(num_faces(tmesh)));
face_iterator face_it = faces(tmesh).first;
std::advance(face_it,target_face_index);
// ... and define a barycentric coordinates inside the face
Traits::Barycentric_coordinates face_location = {{0.25, 0.5, 0.25}};
// construct a shortest path query object and add a source point
// Note that the external index property map are automatically initialized
Surface_mesh_shortest_path shortest_paths(tmesh,
get(boost::vertex_external_index, tmesh),
get(CGAL::halfedge_external_index, tmesh),
get(CGAL::face_external_index, tmesh),
get(CGAL::vertex_point, tmesh));
shortest_paths.add_source_point(face_it, face_location);
// For all vertices in the tmesh, compute the points of
// the shortest path to the source point and write them
// into a file readable using CGAL Lab
std::ofstream output("shortest_paths_no_id.polylines.txt");
vertex_iterator vit, vit_end;
for ( boost::tie(vit, vit_end) = vertices(tmesh);
vit != vit_end; ++vit)
{
std::vector points;
shortest_paths.shortest_path_points_to_source_points(vit, std::back_inserter(points));
// print the points
output << points.size() <<" ";
for (std::size_t i = 0; i < points.size(); ++i)
output <<" "<< points[i];
output << std::endl;
}
return 0;
}
3.4 Using Multiple Source Points
This example shows how to compute the sequence tree from multiple source points, using an iterator range of Surface_mesh_shortest_path::Face_location objects generated at random.
FileSurface_mesh_shortest_path/shortest_paths_multiple_sources.cpp
include
include
include
include <CGAL/Surface_mesh_shortest_path.h>
include
include
include
include
include
typedefCGAL::Exact_predicates_inexact_constructions_kernelKernel;
typedefCGAL::Surface_mesh Triangle_mesh;
typedefCGAL::Surface_mesh_shortest_path_traits Traits;
typedefCGAL::Surface_mesh_shortest_path Surface_mesh_shortest_path;
typedefSurface_mesh_shortest_path::Face_location Face_location;
typedef boost::graph_traits Graph_traits;
typedef Graph_traits::vertex_iterator vertex_iterator;
typedef Graph_traits::face_iterator face_iterator;
typedef Graph_traits::face_descriptor face_descriptor;
int main(int argc, char argv)
{
const std::string filename = (argc>1) ? argv : CGAL::data_file_path("meshes/elephant.off");
Triangle_mesh tmesh;
if((filename, tmesh) ||
(tmesh))
{
std::cerr <<"Invalid input file."<< std::endl;
return EXIT_FAILURE;
}
// pick up some source points inside faces,
const unsigned int randSeed = argc > 2 ? boost::lexical_cast(argv) : 7915421;
CGAL::Random rand(randSeed);
// by copying the faces in a vector to get a direct access to faces
std::size_t nb_faces=num_faces(tmesh);
face_iterator fit, fit_end;
boost::tie(fit, fit_end) = faces(tmesh);
std::vector face_vector(fit, fit_end);
// and creating a vector of Face_location objects
const std::size_t nb_source_points = 30;
Traits::Barycentric_coordinates face_location = {{0.25, 0.5, 0.25}};
std::vector faceLocations(nb_source_points, Face_location(face_descriptor(), face_location));
for (std::size_t i = 0; i < nb_source_points; ++i)
{
faceLocations[i].first=face_vector[rand.get_int(0, static_cast(nb_faces))];
}
// construct a shortest path query object and add a range of source points
Surface_mesh_shortest_path shortest_paths(tmesh);
shortest_paths.add_source_points(faceLocations.begin(), faceLocations.end());
// For all vertices in the tmesh, compute the points of
// the shortest path to the source point and write them
// into a file readable using the CGAL Tmesh demo
std::ofstream output("shortest_paths_multiple_sources.polylines.txt");
vertex_iterator vit, vit_end;
for ( boost::tie(vit, vit_end) = vertices(tmesh);
vit != vit_end; ++vit)
{
std::vector points;
shortest_paths.shortest_path_points_to_source_points(vit, std::back_inserter(points));
// print the points
output << points.size() <<" ";
for (std::size_t i = 0; i < points.size(); ++i)
output <<" "<< points[i];
output << std::endl;
}
return 0;
}
CGAL::Surface_mesh_shortest_path::Face_location
std::pair< face_descriptor, Barycentric_coordinates > Face_location
An ordered pair specifying a location on the surface of the Triangle_mesh.
Definition: Surface_mesh_shortest_path.h:151
3.5 Shortest Path Sequence Visitor
This example shows how to implement a model of the SurfaceMeshShortestPathVisitor concept to get detailed information about the sequence of simplicies crossed by a shortest path.
FileSurface_mesh_shortest_path/shortest_path_sequence.cpp
include
include
include
include <CGAL/Surface_mesh_shortest_path.h>
include
include
include
include
include
typedefCGAL::Exact_predicates_inexact_constructions_kernelKernel;
typedefCGAL::Surface_mesh Triangle_mesh;
typedefCGAL::Surface_mesh_shortest_path_traits Traits;
typedefCGAL::Surface_mesh_shortest_path Surface_mesh_shortest_path;
typedef Traits::Barycentric_coordinates Barycentric_coordinates;
typedef boost::graph_traits Graph_traits;
typedef Graph_traits::vertex_iterator vertex_iterator;
typedef Graph_traits::face_iterator face_iterator;
typedef Graph_traits::vertex_descriptor vertex_descriptor;
typedef Graph_traits::face_descriptor face_descriptor;
typedef Graph_traits::halfedge_descriptor halfedge_descriptor;
// A model of SurfacemeshShortestPathVisitor storing simplicies
// using std::variant
struct Sequence_collector
{
typedef std::variant< vertex_descriptor,
std::pair,
std::pair> Simplex;
std::vector< Simplex > sequence;
void operator()(halfedge_descriptor he, double alpha)
{
sequence.push_back(std::make_pair(he, alpha));
}
void operator()(vertex_descriptor v)
{
sequence.push_back(v);
}
void operator()(face_descriptor f, Barycentric_coordinates alpha)
{
sequence.push_back(std::make_pair(f, alpha));
}
};
// A visitor to print what a variant contains using std::visit
struct Print_visitor
{
int i;
Triangle_mesh& g;
Print_visitor(Triangle_mesh& g) :i(-1), g(g) {}
void operator()(vertex_descriptor v)
{
std::cout <<"#"<< ++i <<" Vertex: "<< get(boost::vertex_index, g)[v];
std::cout <<" Position: "<< Surface_mesh_shortest_path::point(v, g) <<"\n";
}
void operator()(const std::pair& h_a)
{
std::cout <<"#"<< ++i <<" Edge: "<< get(CGAL::halfedge_index, g)[h_a.first] <<" , ("
<< 1.0 - h_a.second <<" , "
<< h_a.second <<")";
std::cout <<" Position: "<< Surface_mesh_shortest_path::point(h_a.first, h_a.second, g) <<"\n";
}
void operator()(const std::pair& f_bc)
{
std::cout <<"#"<< ++i <<" Face: "<< get(CGAL::face_index, g)[f_bc.first] <<" , ("
<< f_bc.second <<" , "
<< f_bc.second <<" , "
<< f_bc.second <<")";
std::cout <<" Position: "<< Surface_mesh_shortest_path::point(f_bc.first, f_bc.second, g) <<"\n";
}
};
int main(int argc, char argv)
{
const std::string filename = (argc>1) ? argv : CGAL::data_file_path("meshes/elephant.off");
Triangle_mesh tmesh;
if((filename, tmesh) ||
(tmesh))
{
std::cerr <<"Invalid input file."<< std::endl;
return EXIT_FAILURE;
}
// pick up a random face
const unsigned int randSeed = argc > 2 ? boost::lexical_cast(argv) : 7915421;
CGAL::Random rand(randSeed);
const int target_face_index = rand.get_int(0, static_cast(num_faces(tmesh)));
face_iterator face_it = faces(tmesh).first;
std::advance(face_it,target_face_index);
// ... and define a barycentric coordinates inside the face
Barycentric_coordinates face_location = {{0.25, 0.5, 0.25}};
// construct a shortest path query object and add a source point
Surface_mesh_shortest_path shortest_paths(tmesh);
std::cout <<"Add source: "<<Surface_mesh_shortest_path::point(face_it, face_location, tmesh) << std::endl;
shortest_paths.add_source_point(face_it, face_location);
// pick a random target point inside a face
face_it = faces(tmesh).first;
std::advance(face_it, rand.get_int(0, static_cast(num_faces(tmesh))));
std::cout <<"Target is: "<<Surface_mesh_shortest_path::point(face_it, face_location, tmesh) << std::endl;
// collect the sequence of simplicies crossed by the shortest path
Sequence_collector sequence_collector;
shortest_paths.shortest_path_sequence_to_source_points(face_it, face_location, sequence_collector);
// print the sequence using the visitor pattern
Print_visitor print_visitor(tmesh);
for (size_t i = 0; i < sequence_collector.sequence.size(); ++i)
std::visit(print_visitor, sequence_collector.sequence[i]);
return 0;
}
CGAL::Surface_mesh_shortest_path::point
Point_3 point(const face_descriptor f, const Barycentric_coordinates &location) const
returns the 3-dimensional coordinates at the barycentric coordinates of the given face.
Definition: Surface_mesh_shortest_path.h:2742
4 Benchmarks
These benchmarks were run using randomly generated source and destination points over multiple trials. The measurements were executed using CGAL 4.5, under Cygwin 1.7.32, using the Gnu C++ compiler version 4.8.3 with options -O3 -DNDEBUG. The system used was a 64bit Intel Core i3 2.20GHz processor with 6GB of RAM
4.1 Single Source Point
| Model | Number of Vertices | Average Construction Time (s) | Average Queries Per Second | Peak Memory Usage (MB) |
--- ---
| ellipsoid.off | 162 | 0.00258805 | 1.21972e+06 | 0.39548 |
| anchor.off | 519 | 0.0580262 | 230461 | 3.88799 |
| rotor.off | 600 | 0.0386633 | 326175 | 3.10571 |
| spool.off | 649 | 0.0418305 | 299766 | 3.75773 |
| handle.off | 1165 | 0.0976167 | 227343 | 7.66706 |
| couplingdown.off | 1841 | 0.138467 | 246833 | 10.1731 |
| bones.off | 2154 | 0.0101125 | 1.31834e+06 | 0.865896 |
| mushroom.off | 2337 | 0.206034 | 202582 | 22.5804 |
| elephant.off | 2775 | 0.136177 | 313785 | 14.0987 |
| cow.off | 2904 | 0.259104 | 206515 | 17.4796 |
| knot1.off | 3200 | 0.279455 | 207084 | 25.314 |
| retinal.off | 3643 | 0.255788 | 247617 | 29.8031 |
| femur.off | 3897 | 0.25332 | 264825 | 21.4806 |
| knot2.off | 5760 | 0.295655 | 309593 | 22.5549 |
| bull.off | 6200 | 0.513506 | 209994 | 34.983 |
| fandisk.off | 6475 | 0.609507 | 198768 | 71.3617 |
| lion-head.off | 8356 | 1.23863 | 145810 | 86.6908 |
| turbine.off | 9210 | 2.23755 | 93079.5 | 172.072 |
| man.off | 17495 | 1.59015 | 187519 | 148.358 |
4.2 Ten Source Points
| Model | Number of Vertices | Average Construction Time (s) | Average Queries Per Second | Peak Memory Usage (MB) |
--- ---
| ellipsoid.off | 162 | 0.00321017 | 911025 | 0.245674 |
| anchor.off | 519 | 0.03601 | 353062 | 3.19274 |
| rotor.off | 600 | 0.015864 | 805416 | 1.97554 |
| spool.off | 649 | 0.0165743 | 802701 | 2.09675 |
| handle.off | 1165 | 0.0294564 | 646057 | 4.62122 |
| couplingdown.off | 1841 | 0.126045 | 272465 | 7.80517 |
| bones.off | 2154 | 0.055434 | 536646 | 4.0203 |
| mushroom.off | 2337 | 0.139285 | 290425 | 11.462 |
| elephant.off | 2775 | 0.167269 | 285076 | 11.2743 |
| cow.off | 2904 | 0.15432 | 328549 | 13.0676 |
| knot1.off | 3200 | 0.114051 | 454640 | 16.1735 |
| retinal.off | 3643 | 0.233208 | 287869 | 18.6274 |
| femur.off | 3897 | 0.128097 | 457112 | 16.8295 |
| knot2.off | 5760 | 0.413548 | 260195 | 33.484 |
| bull.off | 6200 | 0.371713 | 297560 | 30.522 |
| fandisk.off | 6475 | 0.545929 | 223865 | 39.5607 |
| lion-head.off | 8356 | 0.70097 | 229449 | 59.6597 |
| turbine.off | 9210 | 1.35703 | 157301 | 90.7139 |
| man.off | 17495 | 1.75936 | 185194 | 122.541 |
4.3 Comparison of Construction and Query Times with Multiple Source Points
The following figures track the construction time, query time, and peak memory usage for the various test models as the number of source points increases. Notice that none of the values increases significantly as the number of source points increases. In fact, in most cases, the running time and memory go down. This is because a larger number of source points tends to result in a more flat sequence tree, which translates to reduced runtime and memory costs.
Figure 77.2 Plot of construction times against different numbers of source points.
Figure 77.3 Plot of query times against different numbers of source points.
Figure 77.4 Plot of peak memory usage against different numbers of source points.
5 Implementation Details
5.1 Definitions
5.1.1 Geodesic Paths
A geodesic curve is a locally shortest path on the surface of some manifold, that is, it cannot be made shorter by some local perturbations. On a surface mesh, this translates to a curve where, when the faces crossed by the curve are unfolded into the plane, the curve forms a straight line. Another way of describing it is that there is exactly π surface angle to both sides at every point along the curve (except possibly at the curve's endpoints).
A geodesic curve between two points is not necessarily a shortest path, but all shortest paths on surface meshes are formed by sequences of one or more geodesic paths, whose junction points are either vertices on the boundary of the mesh, or saddle vertices. We call such a curve on the surface of the mesh a potential shortest path between its two endpoints.
Figure 77.5 A geodesic on the surface of a simple surface mesh.
Figure 77.6 The same geodesic, with its faces unfolded into the plane. Note in the unfolding, the geodesic forms a straight line.
5.1.2 Visibility Window
A visibility window (or visibility cone) is a pair of geodesic curves which share a common source point and enclose a locally flat region of the surface mesh. Locally flat means that between every pair of points inside the window, there is exactly one geodesic path between them which also stays inside the bounds of the window. Thus, operations, such as distance calculations, can be done with normal 2D Euclidean operations while inside the window. When a visibility window encounters a vertex (a non-flat part of the surface), a branch occurs, forming a sub-window to either side.
Figure 77.7 A single visibility window, before it encounters a vertex.
Figure 77.8 After encountering a convex vertex, the visibility window branches to either side (blue on the left, red on the right). Note that the two new windows immediately overlap on the other side of the vertex, since the surrounding surface area is less than 2 π. Points inside this region of overlap might have two possible shortest paths from the origin point.
5.1.3 Saddle Vertices
A saddle vertex on a surface mesh is a vertex v where the sum of surface angles of all faces incident at v is greater than 2 π, or, in simpler terms, one cannot flatten all the faces incident to v into the plane without overlap. Identifying and dealing with saddle vertices are important in shortest path algorithms because they form blind spots which cannot be reached by a single geodesic curve.
Figure 77.9 A visibility window (shaded blue) encounters a saddle vertex; the shaded red region behind the vertex is not reachable by a geodesic curve from the source point (assuming the geodesic must stay inside the initial window).
In order to deal with this, we must create a new set of child visibility windows which branch out around the saddle vertex. The paths through these child windows would first arrive at the saddle vertex, and then follow a new visibility window (forming a kind of poly-line on the surface). Note that similar behavior is required when we reach a boundary vertex of a non-closed surface mesh.
Figure 77.10 In order to see past the blind spot created by the saddle vertex, we create a branching set of visibility windows emanating from the saddle vertex. Note that only the branches which cover the blind spot for the parent visibility window are needed for our algorithm.
5.1.4 The Sequence Tree
In order to compute shortest paths, we build a sequence tree (or cone tree) from each source point. The sequence tree describes the combinatoric structure of all potential shortest paths which originate from a single source point, by organizing them into a hierarchy of visibility windows.
Whenever a vertex of the surface mesh is encountered, a branch occurs in the sequence tree. If the vertex is a non-saddle vertex, then only two children are created, one for each edge incident to that vertex on the current face. If the vertex is a saddle vertex, in addition to the two children mentioned above, a special type of node, called a pseudo-source, is created which branches out from that vertex.
Once a sequence tree is built, the potential shortest paths from the source to every point inside a given visibility window can be computed. The sequence of faces along each branch of the tree are laid out edge to edge, into a common plane, such that the geodesic distance from any point on the surface to its nearest source point can be obtained using a single 2D Euclidean distance computation. Note that if the window belongs to a pseudo-source, the distance is measured from the target to the pseudo-source, and then the distance from the pseudo-source back to its parent is measured, and so on back to the original source.
5.2 Algorithm Overview
The size of the sequence tree from any source point is theoretically infinite, however we only ever care about trees which are of depth at most N, where N is the number of faces in the surface mesh (since no shortest path can cross the same face twice). Even then, the size of this truncated sequence tree is potentially exponential in the size of the surface mesh, thus a simple breadth-first search is not feasible. Rather, we apply techniques to eliminate entire branches which are provably unable to contain shortest paths from the source point(s). The techniques used are given in greater detail in a paper by Xin and Wang , which itself expands an earlier work by Chen and Han and Mitchell, Mount, and Papadimitriou .
Handling multiple source points is simply a matter of constructing multiple sequence trees concurrently, using a method similar to the multi-source Dijsktra's algorithm.
5.3 Continuous Dijkstra
Continuous Dijkstra is simply the application of the graph-search algorithm to a non-discrete setting. As we build the search tree, newly created nodes are tagged with a distance metric, and inserted into a priority queue, such that the shortest distance nodes are always first.
5.4 One Angle, One Split
This observation by Chen and Han states that out of all the branches that occur at any given vertex of the surface mesh, only a limited number have more than one child which can define shortest paths. This is accomplished by maintaining, for each vertex, all nodes of the sequence tree which can contain that vertex inside their visibility window.
For each vertex, only one two-way branch may occur per face incident to that vertex, specifically, that of the nearest node to that vertex which crosses that face. We call that closest node the occupier of that vertex.
If the vertex is a saddle vertex, only one pseudo-source may be established at that vertex, this time by the absolute nearest node to that vertex.
This method alone can decrease the running time for construction of the sequence tree construction to polynomial time.
5.5 Distance Filtering
An additional distance filter proposed by Xin and Wang helps prune the search tree even further by comparing the current node's distance to the closest distance so far of the three vertices on the current face. Details on this method can be found in their paper .
5.6 Locating Shortest Paths
In order to locate the shortest path from a target point to a source point, we must select the correct visibility window. A simple method is to keep track for each face f of all windows which cross f. In practice, at most a constant number of windows will cross any given face, so for simplicity this is the method we employ. An alternative is to construct a Voronoi-like structure on each face, where each cell represents a visibility window. We did not attempt this method, however it would seem likely that it would be of no computational benefit.
5.7 Pseudo-Code
In this section we give a brief outline of the pseudo-code for this algorithm. More details can be found in and .
--
-- Global Values
--
G : FaceGraph(V,E,F)
-- V - the set of vertices
-- E - the set of edges
-- F - the set of planar faces
Q : PriorityQueue
-- A priority queue ordered using the metric given by Xin and Wang
--
-- Types
--
type VisbilityWindow:
f : a face of F, the current face of this window, we say this window 'crosses' face f
s : a point on the surface of F, the source point of this window
d : the 'base distance' to s, only non-zero if s is a pseudo-source
l : the left-side bounding ray of this window, with its origin at s
r : the right-side bounding ray of this window, with its origin at s
p : its parent VisibilityWindow
--
-- Methods
--
method XinWangDistanceFilter:
Input:
w : a VisibilityWindow
Output:
filter : true if w passes the distance filtering metric given by Xin and Wang, false otherwise
method PropagateWindow:
Input:
w : a visibility window
e : an edge on face w.f
Output:
w' : A new visibility window on the face opposite w.f across edge e
Begin:
Let f' be the face on the opposite side of e as w.f
Lay out face f' along e, such that it shares a common plane with w.f
Create a new VisibilityWindow w', with
- w its parent
- the same source point and base distance as w
- its boundary rays clipped to the sub-segment of e covered by w
return w'
method CreateFaceWindow:
Input:
f : a face of F
v : a vertex of f
w : a VisibilityWindow which intersects f and contains v
Output:
w' : a new VisibilityWindow, with
-- w its parent
-- its source point s = v
-- its two bounding rays along the edges incident to v
-- face f as its crossed face
-- its base distance being the distance of window w to v
method CreatePseudoSource:
Input:
w : the parent window
v : a saddle vertex of V
Begin:
For each face f incident to v:
w' = CreateFaceWindow(f, v, w)
Q.insert(w')
method TreeDepth:
Input:
w : a VisibilityWindow in some sequence tree T
Output:
The depth of node w in its current sequence tree (this would typically be cached in w itself)
method ShortestPathTree:
Input:
s[1..n] : a set of source points on the surface of G.
For simplicity (and without loss of generality),
we will assume they are all vertices of G.
Output:
T[1..n] : a set of sequence trees for the source points
Declare:
O : a map of (f,v) => VisibilityWindow, which gives the 'vertex occupier' for (f,v),
that is the window which crosses face f and whose source is nearest to vertex v
S : a map of v => VisibilityWindow, which gives the window whose source is nearest
to v. Note that this is a strict subset of O
Begin:
for i in 1..n:
Let r[i] be the root of T[i], with distance 0 to s[i]
CreatePseudoSource( r[i], s[i] )
While Q is not empty:
w = Q.take()
if XinWangDistanceFilter(w) and TreeDepth(w) <= |F|:
if w contains a vertex v of w.f:
if w is closer to v than O[w.f,v]:
O[w.f,v] = w
if v is a boundary vertex or a saddle vertex, and w is closer to v than S[v]:
S[v] = w
CreatePseudoSource(w, v, w.dist(v))
let {e_0, e_1} be the edges of f incident to v
for i in [0,1]:
w' = PropagateWindow(w, e_i)
Q.insert(w')
else:
let e_n be the edge 'closer' to window w
w' = PropagateWindow(w, e_n)
Q.insert(w')
else:
let e_o be the only edge crossed by window w
w' = PropagateWindow(w, e_o)
Q.insert(w')
return T[1..n]
To perform shortest path distance queries to each vertex, we can simply use the results stored in S after the completion of ShortestPathTree, as it contains a map from each vertex to the VisibilityWindow which has the shortest path to that vertex. Performing shortest path computations to any arbitrary face location is slightly more complex. As eluded to above, after completion of the algorithm, we traverse each of the sequence trees, and for each face f we store all the VisibilityWindows which cross f in a look-up structure. Then, to find the shortest path to a point on the surface of face f, we look up that pre-stored set of VisibilityWindows associated with it, and among those windows we select the one which contains the query point and has the shortest path back to the origin. Though it may seem slow since it involves a linear search but it is efficient in practice since the number of faces crossing any single face is typically limited (this is due to the additional filtering method given by Xin and Wang .
The actual surface paths can be reconstructed by backtracking from the VisibilityWindow, through its parents in the tree up to the root, and keeping track of each face that was crossed.
6 Design and Implementation History
This package is the result of the work of Stephen Kiazyk during the 2014 season of the Google Summer of Code. He has been mentored by Sébastien Loriot and Éric Colin de Verdière who also contributed to the documentation and the API definition.
Generated by 1.9.6 |
8994 | https://math.stackexchange.com/questions/4773901/combinatorial-problem-stacking-books-on-a-shelf | combinatorics - Combinatorial Problem: Stacking Books on a Shelf - Mathematics Stack Exchange
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Combinatorial Problem: Stacking Books on a Shelf
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A shelf has five books. In how many ways can we stack some or all of these books? The stack can contain only one book.
Attempt: Why is my reasoning wrong? 5!×2 5 5!×2 5, with each book having two possibilities: being inside the stack or not being in it.
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asked Sep 22, 2023 at 22:06
Lambert macuseLambert macuse
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That's far too large. To get the books 1,2 1,2 stacked that way you could have taken the orders 12345,12354,12435,12453,12534,12543 12345,12354,12435,12453,12534,12543 and then chosen to suppress the back three.lulu –lulu 2023-09-22 22:22:00 +00:00 Commented Sep 22, 2023 at 22:22
Just sum over the possible choices for the number of books, it's easy.lulu –lulu 2023-09-22 22:22:23 +00:00 Commented Sep 22, 2023 at 22:22
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Writing "The stack can contain only one book", I presume you mean The stack can even contain only one book, because without the additional word, it means that the stack can contain only one book which has the trivial answer of 5 5
I hope you are familiar with the notation for permutations n P r≡n C r×r!n P r≡n C r×r!, or in the notation favored on this forum, (n r)×r!(n r)×r!
So using the simpler permutation notation for this problem, the answer is
5 P 1+5 P 2+5 P 3+5 P 4+5 P 5 5 P 1+5 P 2+5 P 3+5 P 4+5 P 5
or if you want to condense it
∑r=1 5 5 P r∑r=1 5 5 P r
You had two errors in your attempt
2 5 2 5 includes 0 0 books in the stack
You are multiplying all combinations by 5!5! instead of the actual number of books you put in the stack
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edited Sep 22, 2023 at 23:42
answered Sep 22, 2023 at 23:25
true blue aniltrue blue anil
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8995 | https://pi.math.cornell.edu/~mec/2008-2009/TianyiZheng/Conditional.html | Conditional Probability,The Monty Hall Problem
Sometimes we already know the ocurrence of an event A, then the probability of a relevent event B given A is different from P(B) without any information on A. Since the sample space is reducedd from the total space to A and the probability that B will occur given that A has occured is
Example: Suppose we throw two fair dices. Consider the following three events:
A: Dice 1 lands on 3
B: Dice 2 lands on 1
C: The sum is 8
First notice that C= {(2,6),(3,5),(4,4),(5,3),(6,2)}, and the total number of possible outcomes rolling two dices is 66=36, therefore P(C)=5/36. Obviously, P(A)=1/6 and P(B)=1/6. Sometimes this is called the priority probability of each event.
Next, we have 'Both A and C occur'={(3,5)} , 'Both B and C occur'=empty and 'Both A and B occur'={(3,1)}. The probability of events occuring together is called joint probability. We have P(A intersects C)=1/36 , P(B intersects C)=0 and P(A intersects B)=1/36.
Then you can compute all the conditional probabilities using the priority probability and joint probability given above.
Notice that we have
When the above equation holds for two events A and B, we say they are (statistically) independent. Intuitively, this means that the two events don't interfere with each other. Equivalently, independence means that
As the last example may have suggested, the mapping from event B to conditional probability of B given A (A a fixed event) is a probability. You may look up the axioms of probability and check the conditions one by one.
You may wish to try the next problem by yourself:
Problem: Anne and Billy are playing a simple dice game. Each rolls one dice and the one with higher number wins. If the numbers are the same, they roll again. If Anne just won, what is the probability that she rolled a '5'?
Let A denote the event that 'Anne wins'. If we write out the outcomes with Anne's roll first and Billy's roll second, the event A is
There are 5+4+3+2+1=15 outcomes in event A and the event 'Anne rolled a 5 and won' has 4 outcomes as shown clearly in the above table. Therefore the joint probability is 4/36 and P(A)=15/36, then we have the conditional probability is 4/15.
Conditional probability can be very puzzling sometimes, actually it is the sourse of many 'paradoxes' in probability. One of these attracted worldwide attention in 1990 when Marilyn vos Savant discussed it in her weekly column in the Sunday Parade magazine.
The Monty Hall Problem:
The statement of this famous problem in Parade Magazine is as follows:
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, donkey. You pick a door, say No.1, and the host, who knows what's behind the doors, opens another door, say No.3, which has a donkey. He then says to you, "Do you want to pick door No.2?" Is it to your advantage to switch your choice? (Whitaker 1990)
Many people argue that "the two unopened doors are the same so they each will contain the car with probability 1/2, and hence there is no point in switching." As we will now show, this naive reasoning is incorrect. To compute the answer, we will suppose that the host always chooses to show you a donkey (When the problem and the solution appeared in Parade, approximately 10,000 readers, including nearly 1,000 with Ph.D.s, wrote to the magazine claiming the published solution was wrong. Some of the controversy was because the Parade version of the problem is technically ambiguous since it leaves certain aspects of the host's behavior unstated, for example whether the host must open a door and must make the offer to switch).
Assuming that you have picked door No.1, there are 3 cases:
Notice that although it took a number of steps to compute this answer, it is "obvious". When we picked one of the three doors initially we had probability 1/3 of picking the car, and since the host can always open a door with a donkey the new information does not change our chance of winning.
The above argument may easily persuade you in a moment by when you think about it again by yourself, it is still somewhat confusing. So why is it so confusing?
Source of Confusion:
When first presented with the Monty Hall problem an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter (Mueser and Granberg, 1999). Out of 228 subjects in one study, only 13% chose to switch (Granberg and Brown, 1995:713).
In her book The Power of Logical Thinking, vos Savant (1996:15) quotes cognitive psychologist Massimo Piattelli-Palmarini as saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer."
Most statements of the problem, notably the one in Parade Magazine, do not match the rules of the actual game show (Krauss and Wang, 2003:9), and do not fully specify the host's behavior or that the car's location is randomly selected (Granberg and Brown, 1995:712). Krauss and Wang (2003:10) conjecture that people make the standard assumptions even if they are not explicitly stated. Although these issues are mathematically significant, even when controlling for these factors nearly all people still think each of the two unopened doors has an equal probability and conclude switching does not matter (Mueser and Granberg, 1999). This "equal probability" assumption is a deeply rooted intuition (Falk 1992:202). People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not (Fox and Levav, 2004:637).
A competing deeply rooted intuition at work in the Monty Hall problem is the belief that exposing information that is already known does not affect probabilities (Falk 1992:207). This intuition is the basis of solutions to the problem that assert the host's action of opening a door does not change the player's initial 1/3 chance of selecting the car. For the fully explicit problem this intuition leads to the correct numerical answer, 2/3 chance of winning the car by switching, but leads to the same solution for other variants where this answer is not correct (Falk 1992:207).
Another source of confusion is that the usual wording of the problem statement asks about the conditional probability of winning given which door is opened by the host, as opposed to the overall or unconditional probability. These are mathematically different questions and can have different answers depending on how the host chooses which door to open if the player's initial choice is the car (Morgan et al., 1991; Gillman 1992). For example, if the host opens Door 3 whenever possible then the probability of winning by switching for players initially choosing Door 1 is 2/3 overall, but only 1/2 if the host opens Door 3. In its usual form the problem statement does not specify this detail of the host's behavior, making the answer that switching wins the car with probability 2/3 mathematically unjustified. Many commonly presented solutions address the unconditional probability, ignoring which door the host opens; Morgan et al. call these "false solutions" (1991).
There are various version of the Monte Hall problem developed after the original Parade version.
Example:Cognitive dissonance
An economist, M. Keith Chen, has recently uncovered a version of the Monty Hall problem in the theory of cognitive dissonance. For a half-century, experimenters have been using the so-called free choice paradigm to test our tendency to rationalize decisions. In an experiment typical of the genre, Yale psychologists measured monkeys preferences by observing how quickly each monkey sought out different colors of M&Ms.
In the first step, the researchers gave the monkey a choice between say red and blue. If the monkey chose red, then it was given a choice between blue and green. Nearly two-thirds of the time it rejected blue in favor of green, which seemed to jibe with the theory of choice rationalization: once we reject something, we tell ourselves we never liked it anyway.
Putting aside this interpretation it is natural to ask: What would happen if monkeys were acting at random? The six orderings RGB, RBG, GRB, GBR, BGR, and BRG would have equal probability. In the ?rst three cases red is preferred to blue, but in 2/3s of those cases green is preferred to blue. Just as in the Monty Hall problem, we think that the probability of preferring blue to green is 1/2 due to symmetry, but the probability is 1/3. This time however conditioning on red being preferred to green reduced the original probability of 1/2 to 1/3, whereas in the Monty Hall problem the probability was initially 1/3 and did not change. |
8996 | https://hal.science/hal-04940409v1/file/Article_Simus_APC.pdf | Primary and secondary motions in an annular plane Couette flow Rémi Macadré Institut de Mécanique des Fluides de Toulouse (IMFT), CNRS, Université de Toulouse, France Laboratoire de Génie Chimique (LGC), CNRS-INPT-UPS, Université de Toulouse, France TotalEnergies S.E., Pôle d’études et de Recherche de Lacq (PERL), France and FR FERMAT, Université de Toulouse, CNRS, Toulouse, France F. Risso Institut de Mécanique des Fluides de Toulouse (IMFT), CNRS, Université de Toulouse, France and FR FERMAT, Université de Toulouse, CNRS, Toulouse, France O. Masbernat Laboratoire de Génie Chimique (LGC), CNRS-INPT-UPS, Universié de Toulouse, France and FR FERMAT, Université de Toulouse, CNRS, Toulouse, France R. Belt TotalEnergies S.E., Pôle d’études et de Recherche de Lacq (PERL), France (Dated: February 11, 2025) Axisymmetric direct numerical simulations are carried out to study the hydrodynamics of a laminar, sta-tionary, incompressible, Newtonian, single-phase flow in an Annular Plane Couette (APC) channel. This con-figuration is that of a Straight Plane Couette (SPC) flow but curved around itself to form an annulus. These simulations are validated by Particle Image Velocimetry measurements at different Reynolds numbers. The flow is analyzed using three dimensionless parameters: the channel aspect ratio Ac, which controls the effects of sidewall confinement, the channel curvature ratio Cr, which affects the centrifugal forces due to curvature, and the Reynolds number Re. The rotation of the top annular plate generates a main flow in the azimuthal direction, while generating a secondary recirculation flow in the plane of the channel cross-section, due to the presence of a centrifugal force difference. As a result, the vertical profile of the azimuthal velocity deviates from the classical linear profile of the SPC flow, adopting an unexpected S-shape, similar to that observed in a turbulent plane Couette flow. Depending on the values of Ac and Cr, the flow exhibits a wide range of behaviors, from a quasi-2D flow with a homogeneous shear rate at moderate Re with appropriate geometrical parameters (Ac ≳5, Cr ≲0.1), to a complex 3D flow otherwise. Whereas it is laminar, the APC flow shares a strong analogy with a turbulent Taylor Couette (TC) flow. At large Reynolds numbers, the velocity gradients concentrate near the wall and the flow reaches an asymptotic regime where the torque scales as Reα and the flow structure becomes independent of Re. Such properties make the APC flow an interesting configuration for fundamental investiga-tions as a complement of the TC flow when shear and gravity are required to be in the same plane. In particular, it can be used to investigate the rheology of a dispersed two-phase mixture similarly to the work done by Yi et al. [1, 2] in the TC device.
I.
INTRODUCTION Studies in fluid mechanics often use configurations of reference to investigate relevant physical processes. In particular among these reference cases, Couette-type and Poiseuille flows find a wide range of applications. They are widely used to study transitions to turbulence [3–5], turbulence [6–9], rheology [1, 2, 10, 11] or transitions of flow configurations [12–14] in multiphase flows.
Couette-type devices can be classified into two distinct geometries: (1) the Taylor-Couette (TC) flow, characterized by two concentric rotating cylinders, and (2) the Straight Plane Couette flow (SPC), characterized by one static wall and one moving wall. The fluid sandwiched between these walls in both geometries is propelled forward by the shear caused by the wall motion, leading to classify these configurations as shear-driven flows. Poiseuille flows, on the other hand, result from a pressure difference along a cylindrical or rectangular pipe. These flows are referred to as pressure-driven flows.
While Poiseuille flows are predominant in industrial applications, Taylor-Couette flows are more commonly used for research purposes. This preference stems from the ability to conduct long-term observations thanks to their periodicity, as well as maintaining a uniform shear rate. While the SPC flow keeps the advantage of a constant controllable shear rate, it has no periodicity due to inlet and outlet boundary conditions. However, in multiphase flow studies, it is desirable to have a plane of shear comprising the vertical direction, as in the SPC flow, rather than perpendicular to gravity, as in the TC flow, to study the effects of both gravity and shear on the vertical migration of the phases. A configuration that combines both the vertical plane of shear and flow periodicity to allow for long-term observations is the Annular Plane Couette (APC) flow. This geometry is basically that of a SPC flow that would be curved around itself to form a closed loop.
The APC flow has often been used for the study of geophysical flows [15–18]. It is well-known that due to channel curvature, 2 secondary flows of Prandtl’s first kind (generated by centrifugal forces) develop in the cross-section, resulting in a complex 3D flow. Previous studies of the hydrodynamics in this geometry mainly focused on the turbulent regime [20–22] and on the optimal rotation speed ratio between top and bottom walls, in order to minimize the curvature effect on the radial shear-stress distribution at the bottom wall [23–26]. However, the hydrodynamics of this flow remains poorly addressed in the laminar regime. This issue is tackled in the present work for single-phase flows, in view of considering this configuration as a reference case for the study of complex multiphase flows, as already done to characterize the rheology of suspensions . In this work, the flow in the APC is investigated in the laminar regime, with direct numerical simulations using OpenFOAM, and further validated by Particle Image Velocimetry (PIV) in an experimental APC device at various Reynolds numbers.
The present paper is organized as follows: in Section II, the geometry of the APC flow is detailed and the non-dimensional parameters characteristic of the flow hydrodynamics in laminar regime are introduced. Section III describes the numerical method and the experimental set-up used for the simulations and the measurements respectively. The results obtained and their physical interpretation are discussed in Section IV. Then, scaling laws for both the torque exerted by the fluid on the moving wall and the secondary flow are presented in section IV. Finally, section VI summarizes the main results.
II.
ANNULAR PLANE COUETTE FLOW A.
Description of the geometry The Straight Plane Couette (SPC) flow refers to the standard plane Couette configuration, distinguished by a top moving wall that drives the fluid in the longitudinal direction through shearing in an unbound channel. Its laminar flow is characterized by a vertical linearity of the longitudinal velocity profile, resulting in a constant shear rate ˙ γm = Utop/H, with Utop the velocity of the top moving wall and H the channel height. The laminar SPC flow has been shown to be stable to infinitely small disturbances at any Reynolds number , in contrast to what happens in the APC flow (Section IV.B.1).
Adding lateral walls to the straight channel shapes the configuration into a rectangular-section channel, confining the fluid inside and turning the SPC flow into the Straight Confined Plane Couette (SCPC) flow. The analytical solution for the SCPC flow is also well-established and depends on the channel aspect ratio Ac = W/H, with W the channel width. The influence of the channel aspect ratio is illustrated in Fig. 1. When Ac = 1 (Fig. 1a), the channel has a square cross-section and significant wall effects distort the velocity field. In contrast, when Ac = 10 (Fig. 1b), the solution reverts to the classical linear profile of the SPC flow in the majority of the channel, but the flow remains confined at the side-wall regions due to the influence of the no-slip boundary conditions.
(a) Ac = 1 (b) Ac = 10 FIG. 1: Velocity profile in a Straight Confined Plane Couette (SCPC) flow for different aspect ratios Ac = W/H.
Finally, a configuration capable of accommodating long-term observations and the benefits of the SPC flow can be designed. A well-suited geometry that fulfill these requirements is the configuration termed here as Annular Plane Couette (APC) flow, which is schematized in Fig. 2. The channel is shaped as an annular ring of rectangular cross-section, with a height H, inner radius Ri and outer radius Ro, corresponding to a channel width of ∆R = Ro −Ri. The middle radius is defined by Rmid = (Ro +Ri)/2 and the middle height is noted Zmid = H/2. The cylindrical coordinate system (r,θ,z) is used for the analysis. By rotating the 3 top annular plate, the fluid within the channel is propelled forward, generating a primary flow in the azimuthal direction. As in the SCPC configuration, the confinement due to the lateral walls influences the velocity field. Additionally, the curvature of the channel introduces a centrifugal force difference along the radius that further distorts the velocity field, as it will be discussed in section IV.
FIG. 2: Schematic of the Annular Plane Couette (APC) flow. The fluid is driven by the rotation of the top annular plate.
B.
Dimensionless parameters The analysis of the hydrodynamics of the laminar, stationary, axisymmetric, incompressible, Newtonian, single-phase flow within the APC channel is conducted using three different dimensionless parameters: the channel Reynolds number Re, the channel aspect ratio Ac = ∆R/H and the channel curvature ratio Cr = ∆R/Ro.
The Reynolds number Re is based on the channel’s half-height H/2 and its half maximum velocity Utop,max/2 = ΩRo/2 at the top. It writes Re = ρΩRoH 4µ , (1) where Ωis the rotation velocity of the top annular plate, ρ the fluid density and µ its dynamic viscosity.
A criterion for the transition to turbulence was obtained experimentally by Tillmark et al. in a SPC flow. They found that the onset of the turbulence occurs at Re ∼360, while the transition to fully developed turbulence takes place around Re ∼500. However, since the SPC flow is linearly stable at any Reynolds number, that experimental result probably depends on uncontrolled finite disturbances and cannot be considered as universal. We ignore whether a critical Reynolds number beyond which the APC flow becomes unstable exists. Steady laminar solutions of APC numerical simulations for Re ≳360 are thus not irrelevant, but have to be considered with care regarding their application to real flows at large Re.
The geometrical effects are twofold. As in the SCPC configuration, the channel aspect ratio number Ac = ∆R/H influences the velocity field due to the sidewall confinement. In addition, the effect of the channel curvature on the velocity field is analyzed by considering the channel curvature ratio Cr = ∆R/Ro. This ratio can be derived by comparing the centrifugal stress difference across the radius τc = ρΩ2Ro∆R and the viscous stress τµ = µΩRo/H. This results in the product of the Reynolds number and the curvature ratio as τc τµ = ρΩRoH µ ∆R Ro = 4ReCr.
(2) While τc/τµ characterizes the effects of the centrifugal force difference on the flow, we separate the analysis by varying in-dependently the channel Reynolds number (defined in Eq. 1) and Cr, since the results are expected to depend on both Re and Cr.
4 III.
DESCRIPTION OF NUMERICAL SIMULATIONS AND EXPERIMENTAL SET-UP A.
Methods and flow conditions Direct numerical simulations of the APC flow are performed to better understand the relevance of such a configuration as a reference flow. These simulations are done with the open-source CFD software OpenFOAM. The Navier-Stokes equations are solved by the simpleFOAM module, which uses the finite-volume method and the SIMPLE algorithm for velocity-pressure coupling. The equations are discretized using Gauss linear numerical schemes of the second order accuracy in space. A steady, incompressible and Newtonian single-phase flow is assumed here.
Since the flow is axisymmetric, only the radial-vertical section of the channel is simulated to ensure faster computations. The mesh is exclusively composed of hexahedron cells. Regular meshing is applied to the central region of the section while the mesh is refined near the walls to provide more accurate results of the boundary layer in both vertical and radial directions. An example of the numerical mesh in the radial-vertical plane is shown in Fig. 3.
0.0 0.2 0.4 0.6 0.8 1.0 (r −Ri)/∆R [−] 0.0 0.2 0.4 0.6 0.8 1.0 z/H [−] FIG. 3: Example of the numerical mesh in the radial-vertical plane (r −z).
Regarding the boundary conditions, a no-slip condition is enforced on every wall for the velocity. The top annular plate rotates at a fixed angular velocity Ω, causing the azimuthal velocity of the fluid at the top wall to vary radially as Utop(r) = Ωr while the other walls remain stationary. The zero-gradient boundary condition is used for pressure at each wall. Finally, periodic boundary conditions are applied in the azimuthal direction for the velocity and pressure.
Various simulations are performed by varying the inner radius Ri, the height H of the channel and the rotational speed Ωof the top annular plate, leading to the exploration of the range of dimensionless parameters summarized in Table I.
Parameters Re Ac Cr Ranges 5 to 500 1 to 10 0.1 to 0.9 TABLE I: Ranges of dimensionless parameters for the numerical simulations.
B.
Mesh validation To validate the numerical mesh, the most critical case corresponding to the highest Re, Cr and the lowest Ac is examined. The selected parameters for this case are Re = 500, Cr = 0.9 and Ac = 1. Different mesh sizes are compared for the radial-vertical plane: Nr ×Nz = 70×250, 140×500 and 280×1000. Regarding mesh convergence, the vertical and radial profiles of two key quantities are considered for validation. The first quantity is the dimensionless azimuthal velocity Uθ/Ωrval, where rval is the middle radius Rmid for vertical profiles and the outer radius Ro for radial profiles. The second quantity is the dimensionless shear rate ˙ γ/ ˙ γm, with ˙ γ = r 2S : S based on the strain-rate tensor S, and ˙ γm = ΩRmid/H. With this definition, ˙ γ/ ˙ γm is equal to unity in a SPC flow and constitutes a relevant scale for the mesh assessment in an APC flow since it accounts for all velocity-gradient components.
Fig. 4 shows the profiles of the azimuthal velocity Uθ along the vertical direction at the middle radius r = Rmid and in the radial direction at the middle height z = Zmid. The profiles corresponding to the three different meshes exhibit similarity. Convergence 5 is achieved even at the lowest resolution (Nr ×Nz = 70×250), with only a small difference in the vertical profiles in the lower central region.
0.0 0.2 0.4 0.6 0.8 1.0 Uθ/(ΩRmid) [−] 0.0 0.2 0.4 0.6 0.8 1.0 z/H [−] 70 × 250 140 × 500 280 × 1000 (a) Profile of Uθ(z) at r = Rmid.
0.0 0.2 0.4 0.6 0.8 1.0 (r −Ri)/∆R [−] 0.0 0.2 0.4 0.6 0.8 1.0 Uθ/(ΩRo) [−] 70 × 250 140 × 500 280 × 1000 (b) Profile of Uθ(r) at z = Zmid.
FIG. 4: Azimuthal velocity profiles in the most critical case (Re = 500, Ac = 1, Cr = 0.9) for mesh validation. The results of three different mesh sizes are presented: Nr ×Nz = 70×250, 140×500 and 280×1000.
Fig. 5 displays the shear-rate profiles for the same conditions as in Fig. 4. Once more, the simulations present only minute differences across all mesh resolutions. However, for the radial profiles, the boundary layers at the outer and inner radii exhibit steep shear-rate gradients, and increasing the mesh resolution improves the capture of this effect.
In conclusion, the results show that all tested meshes are suitable. To properly capture the gradients of shear rate at the boundary layers, the resolution of the 140×500 mesh grid emerges as the most pertinent choice for this study. Subsequently, to keep a same mesh resolution for each value of Ac or Cr, the number of cells in both radial (Nr) and vertical (Nz) directions will be adjusted to keep the same cell sizes, ∆r and ∆z, as in the 140×500 case.
C.
Experimental set-up To validate the numerical simulations of the APC flow, experiments are carried out using the set-up illustrated in Fig. 6. The geometrical dimensions of the channel are Ri = 0.48 m and Ro = 0.58 m, corresponding to a curvature ratio Cr = 0.17. The 0 10 20 30 40 ˙ γ/ ˙ γm [−] 0.0 0.2 0.4 0.6 0.8 1.0 z/H [−] 70 × 250 140 × 500 280 × 1000 (a) Profile of ˙ γ(z) at r = Rmid.
0.0 0.2 0.4 0.6 0.8 1.0 (r −Ri)/∆R [−] 0 1 2 3 4 5 6 ˙ γ/ ˙ γm [−] 70 × 250 140 × 500 280 × 1000 (b) Profile of ˙ γ(r) at z = Zmid.
FIG. 5: Shear-rate profiles in the most critical case (Re = 500, Ac = 1, Cr = 0.9) for mesh validation. 3 different mesh sizes are plotted: 70×250, 140×500 and 280×1000.
6 FIG. 6: Experimental set-up for the PIV measurement of the azimuthal velocity.
height of the channel is adjustable through the top plate vertical position and is set at H = 0.01 m, thus yielding an aspect ratio of Ac = 10. The channel is made of transparent PMMA to allow for optical measurements. The flow field of a low viscosity oil (n-Dodecane, CAS: 112-40-3, ρ = 749 kg·m−3, µ = 1.34×10−3 Pa·s) is measured using Particle Image Velocimetry (PIV) with a PIV system that uses a double pulsed Nd:YAG laser (532 nm, 2 × 120 mJ, Nanopiv – Litron Lasers). The liquid is seeded using rhodamine-doped polystyrene particles of average diameter dmean = 10 µm (PS-FluoRed from microParticles, Germany). The azimuthal-vertical plane at middle radius r = Rmid is illuminated by a laser sheet (Fig.6) to obtain a vertical profile of the azimuthal velocity Uθ. Recording of instantaneous flow images is achieved using an sCMOS camera (Imager sCMOS from LaVision, Germany) with a definition of 5.5 million pixels and a high-pass filter. Image pairs are captured at a rate of 9 Hz, and for each case, 1000 instantaneous velocity fields are recorded. The images are then processed by DaVis software (LaVision, Germany), using a decreasing interrogation window size (from 64×64 pixels2 to 32×32 pixels2) with 50 % overlap. Standard cross-correlation with Fast Fourier Transform is employed to determine the corresponding spatially-averaged displacement vectors. The spatial resolution of the velocity fields is 600 µm. The uncertainties in the PIV measurements are obtained by using the method of Wieneke .
IV.
RESULTS AND DISCUSSION This section outlines the results of the numerical simulations obtained with meshes of same resolution, ∆r and ∆z, as the 140 × 500 mesh grid previously validated. The hydrodynamic characteristics are initially scrutinized on a representative case, followed by a parametric study within the ranges of dimensionless parameters given in Table I.
A.
Study of a representative case Here, the analysis focuses on a representative case characterized by Re = 280, Cr = 0.5 and Ac = 5, corresponding to in-termediate values of the dimensionless parameters. The rotation of the top annular plate shears the fluid within the channel, creating a flow in the azimuthal direction. This flow is referred to as primary flow (Fig. 7a). Simultaneously, the curvature of the channel induces a centrifugal force difference along the radius that generates a secondary flow in the cross-section. Indeed, this centrifugal force propels the fluid in the top region towards the outer radius, pushing the fluid in the outer region downward, due to the presence of the lateral walls. Consequently, the fluid in the bottom region moves towards the inner radius, and due to mass conservation, the fluid in the inner region ascends, completing the formation of a radial-vertical recirculation cell. That flow, involving radial and vertical velocity components, is referred to as secondary flow (Fig. 7b). One interesting feature is that, in the inner region, the secondary flow spreads upwards over a larger area than in the three other near-wall regions. This results in a lower intensity in the inner region.
The corresponding flow fields are depicted in Fig. 7. For the azimuthal velocity (Fig. 7a), the flow exhibits a high vertical gradient at the bottom and top regions, deviating from the constant shear-rate of the SPC flow. This is caused by the presence of the secondary flow in these regions, which disturbs the linear velocity profile. Moreover, its influence is also visible in the outer region where the secondary flow goes downward. Regarding the secondary flow (Fig. 7b) made of vertical and radial velocities, it is mainly concentrated near the wall regions with a magnitude up to 15% of the primary-flow magnitude, resulting 7 0.0 0.2 0.4 0.6 0.8 1.0 (r −Ri)/∆R [−] 0.0 0.2 0.4 0.6 0.8 1.0 z/H [−] 0.0 0.2 0.4 0.6 0.8 1.0 Uθ ΩRo [−] (a) Primary flow (Uθ).
0.0 0.2 0.4 0.6 0.8 1.0 (r −Ri)/∆R [−] 0.0 0.2 0.4 0.6 0.8 1.0 z/H [−] 0.00 0.03 0.06 0.09 0.12 0.15 Usec ΩRo [−] (b) Secondary-flow (Ur,Uz).
FIG. 7: Velocity fields of the primary and secondary flows in the radial-vertical plane from direct numerical simulations at Re = 280, Cr = 0.5 and Ac = 5. For the secondary flow, the color map represents its magnitude Usec = p U2 r +U2 z and the vector field its velocity vectors.
0.0 0.2 0.4 0.6 0.8 1.0 (r −Ri)/∆R [−] 0.0 0.2 0.4 0.6 0.8 1.0 z/H [−] Ztop Zmid Zbot Rinner Rmid Router FIG. 8: Vertical and radial positions of the profiles in the radial-vertical plane. The colors and line patterns correspond to their respective positions in Figs 9 and 10. For the radial profiles: Ztop (-·-), Zmid (-), Zbot (- -). For the vertical profiles: Router (-·-), Rmid (-), Rinner (- -).
in a significant influence on the primary flow.
Profiles of the azimuthal velocity Uθ and shear rate ˙ γ at various radial (Rinner, Rmid, Router) and vertical positions (Zbot, Zmid, Zbot), defined in Fig. 9, are now presented. The vertical profiles of the azimuthal velocity (Fig. 9a) are normalized by the top-plate tangential velocity at the considered radial position Ωrref, where rref = Rinner, Rmid or Router. They deviate from the typical linear profile associated with the laminar SCPC flow (corresponding to the red curve of Fig. 9a). Unexpectedly, they exhibit a closer resemblance to the S-shaped profile, characteristic of a turbulent SPC flow, with the presence of significant gradients near the walls. Meanwhile, the profiles in the central region remain almost flat (low vertical gradient), and this pattern holds for each considered radius. The influence of the sidewall confinement being already accounted for in the SCPC channel, this distortion may thus be ascribed to the effect of the centrifugal force difference resulting from the curvature of the channel. Additionally, the gap between each plateau in the central region of the velocity does not appear to be proportional to rref and diminishes as rref increases. This suggests that the secondary flow does not distort the profiles in the same way at each radius. Finally, the near-wall regions exhibit distinct linear slopes, especially at the top wall, while the curves for the middle and outer radii coincide at the bottom wall, indicating a variation of the intensity of the secondary flow in these regions.
As for the vertical profiles, the radial profiles of the azimuthal velocity (Fig. 9b) undergo significant changes in the APC flow compared to a SCPC flow. The profile in the central region is distorted into a quasi-linear profile with a slope that does not depend on the vertical position in the channel. However, differences are visible in the near-wall regions. At the outer radius, all profiles drop to zero within almost the same boundary layer thickness, but they peak at different values depending on the vertical position of the considered profile.
The shear-rate profiles, obtained in the same conditions as in Fig. 9, are reported in Fig. 10. Since ˙ γ accounts for local velocity gradients in all directions, it is influenced by the presence of gradients in both the primary and secondary flows. An examination 8 0.0 0.2 0.4 0.6 0.8 1.0 Uθ/(Ωrref) [−] 0.0 0.2 0.4 0.6 0.8 1.0 z/H [−] Router Rmid Rinner SCPC (a) Profile of Uθ(z) at different radii.
0.0 0.2 0.4 0.6 0.8 1.0 (r −Ri)/∆R [−] 0.0 0.2 0.4 0.6 0.8 1.0 Uθ/(ΩRo) [−] Ztop Zmid Zbot SCPC (b) Profile of Uθ(r) at different heights.
FIG. 9: Azimuthal velocity profiles obtained in the simulations at Re = 280, Cr = 0.5 and Ac = 5. The profile positions are given in Fig. 8. The red lines correspond to the analytical solution of the Straight Confined Plane Couette (SCPC) flow taken at the middle radius Rmid for the vertical profiles and middle height Zmid for the radial profiles.
0 2 4 6 8 ˙ γ/ ˙ γm [−] 0.0 0.2 0.4 0.6 0.8 1.0 z/H [−] Router Rmid Rinner SPC (a) Profile of ˙ γ(z) at different radii.
0.0 0.2 0.4 0.6 0.8 1.0 (r −Ri)/∆R [−] 0 1 2 3 4 5 ˙ γ/ ˙ γm [−] Ztop Zmid Zbot (b) Profile of ˙ γ(r) at different heights.
FIG. 10: Computed Shear-rate profiles for Re = 280, Cr = 0.5 and Ac = 5. The profile positions are given in Fig. 8. The red line correspond to the shear rate of the Straight Plane Couette (SPC) flow.
of the vertical profiles (Fig. 10a) reveals several key features. Specifically, at rref = Rinner and Rmid, the shear rate in the central region is constant and consistently lower than the value in a SPC flow (red curve in Fig. 10a). This is consistent with the absence of a secondary flow (Fig. 7b) and the flatness of the vertical profiles of the azimuthal velocity in the central region (Fig. 9a). This is not the case for shear rate profile at rref = Router, where the secondary flow is strong with stiff gradients in the wall region. Near the top and bottom regions, at rref = Rmid and Router, the shear rate exhibits important gradients issued from both primary and secondary flows. At rref = Rinner the profile remains flat in the bottom region, due to the low magnitude of the secondary-flow gradients.
Regarding the radial profiles of the shear rates (Fig. 10b), the central part of the profile at middle height Zmid stays notably low and flat, as a result of the small primary and secondary-flow gradients in this region. The bottom and top-height profiles show a slight increase of the shear rate along the radial coordinate, from the inner to the outer wall. In the vicinity of the outer wall, the shear rate abruptly grows due to the magnitude of secondary-flow gradients in this region.
In conclusion, the velocity vertical profile of the primary flow exhibits an unexpected S-shape as in a turbulent plane Couette flow. The presence of a secondary-flow is evidenced at the bottom, top, and outer wall zones, forming a recirculating loop in the channel section. This secondary flow is impacting the primary velocity field within a boundary layer thickness near the wall. In these regions, significant gradients of both primary and secondary flow are developing, considerably increasing the local shear 9 0.0 0.2 0.4 0.6 0.8 1.0 Uθ/(ΩRmid) [−] 0.0 0.2 0.4 0.6 0.8 1.0 z/H [−] Re SCPC Re = 5 Re = 50 Re = 500 (a) Vertical profiles of the azimuthal velocity Uθ(z) at r = Rmid.
0.0 0.2 0.4 0.6 0.8 1.0 (r −Ri)/∆R [−] 0.0 0.2 0.4 0.6 0.8 1.0 Uθ/(ΩRo) [−] SCPC Re = 5 Re = 50 Re = 500 (b) Radial profiles of the azimuthal velocity Uθ(r) at z = Zmid.
FIG. 11: Azimuthal velocity profiles for Ac = 5 and Cr = 0.5 at different Reynolds number. The red dotted lines represent the corresponding profiles of the SCPC flows.
rate. However, this general flow structure can be modulated by adjusting the channel Reynolds number Re and geometrical parameters Ac and Cr. In the next section, the influence of these flow parameters is explored and discussed.
B.
Parametric study In this section, we present the results obtained from simulations using the ranges of dimensionless parameters given in Table. I.
Only the vertical profiles at the middle radius Rmid and the radial profiles at the half-height Zmid of the azimuthal velocity are focused on.
1.
Influence of the Reynolds number The first parameter of interest is the channel Reynolds number Re. Its influence on the azimuthal velocity is studied in a case of moderate centrifugal and confinement effects, taking Cr = ∆R/R0 = 0.5 and Ac = ∆R/H = 5. Fig. 11a shows the vertical profiles of the azimuthal velocity at various Re. The case at the lowest Reynolds number (Re = 5) exhibits a linear profile similar to that of the SCPC flow, despite the presence of the wall confinement and centrifugal force. The influence of Re, and hence of the centrifugal force (as discussed in Section II.B), becomes more pronounced at higher values, causing the profiles to converge towards a single S-shaped profile. This feature is likely to result from the reduction of the portion occupied by the secondary flow in the channel cross-section as Re increases, as it can be seen by comparing Fig. 7b and Fig. 12. When increasing Re from 50 to 280, the secondary flow compacts towards the walls while increasing in magnitude. A further increase of Re only amplifies the secondary-flow magnitude. This leads to a higher alteration of the primary flow in these regions, ending in the S-shaped profile, which results from the redistribution of momentum by the secondary flow from the central region to the upper and lower regions.
The radial profiles, depicted in Fig. 11b, reveal a similar trend: as Re increases, they converge towards a given radial profile.
At Re = 5, the value of Uθ at middle radius is still equal to that of a SCPC flow, although a significant slope is present. Increasing Re, the central part of the radial profile converges toward an asymptotic linear evolution. In parallel, we observe the development of a velocity peak in the outer region for the radial profiles (Fig. 11b), arising from the increasing magnitude of the secondary flow and its migration to the wall regions.
2.
Effect of the aspect and curvature ratios In order to investigate the influence of the channel aspect ratio Ac and of the curvature ratio Cr, specific values have been selected to cover a wide range of these parameters: Cr = (0.1,0.5,0.9), Ac = (1,2,10), and Re = (5,50,500). Through Ac, the channel spans from a confined square section (Ac = 1) to an almost 2D channel (Ac = 10). The values of 0.1 and 0.9 of Cr respectively corresponds to a weak and strong effect of the centrifugal force on the flow. The range of the channel Reynolds 10 0.0 0.2 0.4 0.6 0.8 1.0 (r −Ri)/∆R [−] 0.0 0.2 0.4 0.6 0.8 1.0 z/H [−] 0.00 0.05 0.10 0.15 Usec ΩRo [−] FIG. 12: Velocity field of the secondary flow in the radial-vertical plane for Re = 50 (Ac = 5 and Cr = 0.5). The double arrow shows the characteristic height Hsec of the secondary flow.
number covers the whole evolution of the flow structure in the laminar regime, starting from a linear profile (Re = 5), shifting to an intermediary curved profile (Re = 50) and ending with the sharp S-shaped profile (Re = 500).
FIG. 13: Vertical profiles of the azimuthal velocity at r = Rmid for different Reynolds numbers Re = (5,50,500). The different subplots correspond to their respective aspect ratio Ac = ∆R/H and curvature ratio Cr = ∆R/Ro.
Fig. 13 shows the evolution of the vertical profiles of the azimuthal velocity Uθ across the parameter map (Ac, Cr). For the lowest confinement and curvature ratios (Ac = 10 and Cr = 0.1, top right-hand corner of Fig. 13), the velocity profile is linear for Re = 5 and Re = 50 (red and yellow curves), and is S-shaped at Re = 500 (blue curve). As the curvature ratio Cr increases (vertically downwards in Fig. 13), the profiles tend to deviate from the linear trend, and the deviation is more pronounced the higher the Reynolds number. A similar effect is observed when increasing the degree of confinement (moving horizontally to the left).
As the confinement rate increases (Ac decreases), the profiles deviate from the linear trend, and the deviation also increases with Re. At the highest Reynolds number (Re = 500), the S-curve tends to flatten in the upper central region as confinement 11 FIG. 14: Radial profiles of the azimuthal velocity at z = Zmid (middle height) for different Reynolds numbers Re = (5,50,500).
The different subplots correspond to their respective aspect ratio Ac = ∆R/H and curvature ratio Cr = ∆R/Ro.
increases (Ac decreases), while gradients increase near the walls. Thus, for the plot at Ac = 1 and Cr = 0.9 (bottom left-hand corner of Fig. 13), the profile at Re = 5 is no longer linear but strongly curved, and the profile at Re = 500 is flattened over 3/4 of the cross-section height. This flattening can be attributed to a significant increase of the influence of the secondary flow and of the confinement in these cases, which causes the increase of the velocity in the entire section. This results in a flat profile in the central to upper central region, reducing the boundary layer thickness close to the top moving wall. This effect is also observable in the lower central region, where the curve is steeper and the boundary layer thickness in the bottom region is decreased close to the bottom immobile wall.
Additionally, as Ac or Cr increases, the asymmetry of the velocity profile with respect to the half-height also increases. That behavior is reflected in the development of a near-wall gradient that is always stronger near the moving wall than near the immobile lower wall. The higher the curvature ratio Cr, the stronger this asymmetry is, which results from the secondary flow.
The radial profiles are reported in Fig. 14 and, as expected, a significant influence of Ac and Cr is also observed. For the case of lowest confinement and curvature effects (Ac = 10 and Cr = 0.1), the profiles are flat for Re = 5 and Re = 50 in the major part of the channel section, while the velocity increases almost linearly between the inner and the outer wall for Re = 500. When the curvature ratio Cr is increased, the profiles tend to adopt this linear trend at all Reynolds numbers. Increasing the confinement (i.e. decreasing Ac) turns out to reduce the linear part of the profile and curves the profiles, by increasing the gradient thickness near the wall in the radial direction.
In conclusion, the influence of the aspect ratio Ac and the curvature ratio Cr on the flow structure is important in an annular plane Couette flow. Up to moderate Re, the vertical profiles are linear and the radial profiles are homogeneous in the channel section, provided that the confinement and curvature effects are weak (high aspect ratio Ac and low curvature ratio Cr). In this configuration, at higher Re, the Uθ profiles tends towards a S-shape in the vertical direction and a linear profile in the radial direction. Increasing the curvature ratio Cr or decreasing the aspect ratio Ac tends to distort the shape of the velocity profile, curving the linear profiles into a S-shape with a more and more flat central part and with steeper wall gradients. The value of Re at which this distortion appears is smaller as the effects of curvature and confinement become more pronounced. The selection of the APC configuration’s geometrical parameters is therefore essential to ensure either a simpler 2D flow similar to SPC flow with a constant shear rate, or a more complex 3D flow. This choice will depend on the study and the intended applications.
12 C.
Experimental validation Using the experimental set-up described in Sec. III C, the numerical simulations are tested against the PIV measurements of a low-viscosity oil flow in the APC channel. The vertical profiles of the azimuthal velocity Uθ at the middle radius r = Rmid are obtained for three different Reynolds numbers: Re = (300,400,500) in order to address the S-shape profile regime. The results are plotted in Fig. 15. The experimental measurements show only a slight deviation from the simulations, mostly in the top region. These results confirm the validity of the numerical simulations for the single-phase flow in the annular plane Couette channel.
0.0 0.2 0.4 0.6 0.8 1.0 Uθ/(ΩRmid) 0.0 0.2 0.4 0.6 0.8 1.0 z/H Simulations Measurements (a) Re = 300 0.0 0.2 0.4 0.6 0.8 1.0 Uθ/(ΩRmid) 0.0 0.2 0.4 0.6 0.8 1.0 z/H Simulations Measurements (b) Re = 400 0.0 0.2 0.4 0.6 0.8 1.0 Uθ/(ΩRmid) 0.0 0.2 0.4 0.6 0.8 1.0 z/H Simulations Measurements (c) Re = 500 FIG. 15: Comparison of the vertical profiles of the azimuthal velocity obtained at r = Rmid by PIV measurement and numerical simulations. The error bars on the measurements are quantified using the method from Wieneke V.
SCALING LAW FOR THE TORQUE AND SECONDARY FLOW In this section, we focus on the torque Γ exerted by the fluid on the rotating top wall, which is a global quantity of primary interest. Firstly, it is directly related to the total viscous dissipation, D = ΓΩ, and its scaling with the Reynolds number can provide insight into the flow dynamics. Secondly, it can be used to determine the effective viscosity of a dispersed two-phase mixture, which is an important issue for predicting the dynamics of suspensions of solid particles as well as droplet emul-sions [1, 2, 10], especially at large Reynolds numbers where experimental results are rare. The detailed description of the velocity field in the APC flow for a wide range of parameters showed a rather complex picture. Thus, obtaining simple laws for Γ might seem unlikely. However, this was achieved for a single-phase Taylor Couette (TC) flow at large-Reynolds-number by Eckhardt et al. [6, 31] by drawing an analogy between momentum transfer in this flow and the heat transfer by convection in a Rayleigh-Bénard flow. Later, this approach was successfully applied to the experimental determination of the effective 13 viscosity of an emulsion in a TC flow by Yi et al. [1, 2], who adjusted the parameters of the scaling law to the multiphase case.
In the following, we adapt the method developed by those authors to find scaling laws for the APC flow. Before doing so, it is worth mentioning the differences and similarities between the TC and the APC configurations. In both cases, the velocity gradients concentrate towards the walls as the Reynolds number increases. However, the transition from the linear solution at small Reynolds number to the nonlinear solution at large Reynolds number is not of the same nature. In the laminar regime for the TC case, only the azimuthal velocity is not zero and the non-linear terms in the azimuthal momentum equation are identically null by symmetry. Beyond a critical Reynolds number, the steady laminar solution becomes unstable and the scaling law found by [1, 2, 6, 31] was obtained in the turbulent regime. In the APC case, the secondary flow in the radial and vertical directions is already present at any Reynolds number and increases regularly with Re until it plays a major role on the shape of the primary flow. Even though the present results show a S-shape profile similar to that observed in a turbulent APC flow, they are obtained in the laminar regime.
A.
Case of the Taylor-Couette flow We begin by recalling the approach introduced for the TC flow by [1, 2, 6, 31]. Assuming a steady axisymmetric flow, an analytical solution can be derived in the laminar regime. When the inner cylinder rotates at a constant angular velocity ωi while the outer cylinder is immobile, the torque is given by ΓTC,0 = 4πµr2 i H ωi 1−(ri/ro)2 , (3) where ri and ro are the inner and outer cylinder radii, H is the cylinder height and µ the dynamic viscosity of the pure fluid . In order to find a scaling law for large Reynolds numbers, it is relevant to rewrite this expression by making the role of the Reynolds number explicit. That can be achieved by introducing a viscous torque scale that is independent of the angular velocity [6, 33], ΓTC,µ = 2πµ2H ρ .
(4) Normalizing the laminar torque by it, it yields Γ∗ TC,0 = ΓTC,0 ΓTC,µ = K0ReTC, (5) where the Taylor Couette Reynolds number writes ReTC = ρωiri(ro −ri) µ , (6) and K0 depends only on the geometry, K0 = 2(ri/ro) (1+ri/ro)(1−ri/ro)2 .
(7) Beyond the laminar regime, the linear evolution of the torque with the Reynolds number is no longer valid. Eckhardt et al.
[6, 31] proposed to generalize Eq. 5 to high Reynolds number turbulent flows as Γ∗= Γ ΓTC,µ = K∞Reα TC , (8) where the prefactor K∞only depends on the geometry. Various works found the power-law exponent α to range from 1.5 to 1.7 in a single-phase flow [1, 2, 6, 32, 33].
Yi et al. [1, 2] applied the same approach to the flow of a droplet emulsion by assuming that the two-phase mixture can be described as an equivalent homogeneous fluid of effective density ρm and viscosity µm, which are used in the definitions of ΓTC,µ and ReTC instead of the pure fluid values. The experimental measurements of the torque led to α = 1.58, a value similar to that of a single-phase flow. Thus, assuming K∞in Eq. 8 are the same in single-phase and two-phase flows, an effective viscosity of the emulsion was determined.
14 B.
Case of the APC flow We now analyze the torque in the annular plane Couette configuration. As opposed to the TC case, because the azimuthal velocity at the top moving wall varies with the radial position (Uθ = Ωr), the velocity field at any finite Reynolds number is three-dimensional and does not admit a trivial analytical solution. Instead, to build a reference torque Γ0, we consider the shear stress τθz = µU/H from the Straight Plane Couette flow in which we substitute the velocity U by ΩRo. Then, by integrating this stress over the surface Stop = π(R2 o −R2 i ) of the top wall and by using Ro as the lever arm, it yields Γ0 = µ ΩRo H π(R2 o −R2 i )Ro .
(9) Dividing by the APC Reynolds number Re, we get the following viscous torque scale Γµ = Γ0 Re = 4πµ2(R2 o −R2 i )R0 ρH2 .
(10) The torque Γ exerted by the flow on the top annular plate can be computed from the numerical results by integrating the local wall shear stress (µ∂Uθ/∂z) multiplied by the radial coordinate over the surface of the top plate. Following the approach developed for the TC flow [6, 33], we can expect the dimensionless torque Γ∗to behaves as Γ∗= Γ Γµ = A0Re, (11) at low Re and as Γ∗= Γ Γµ = A∞Reα , (12) at large Re. Figure 16 presents Γ∗as a function of the Reynolds number Re for various geometries defined by different parameter pairs of (Ac,Cr). The symbols show the torque computed from the numerical results and the plain straight lines its corresponding fit by Eq. 12 for each geometry. The values of α, A0 and A∞are given in the caption.
At low to moderate Re, the linear regime is clearly present in all cases. The value of A0 is independent of Cr, since Cr is associated to the centrifugal effect which is negligible in this regime. Moreover, it also becomes independent of the lateral confinement when Ac becomes large. The transition to the large-Re asymptotic regime takes place in the range from Re = 10 to 100, starting and ending sooner as Cr increases. At large Re, Γ∗is well described by a power law, A∞Reα, with an exponent in the same range as in a TC flow: α = 1.5 for Cr ≥0.5 and α = 1.65 for Cr ≤0.1 while A∞depends on both Ac and Cr. It is out of the scope of this work to perform an exhaustive investigation of the effect of the geometry. However, the present results indicate that, in cases of low confinement (Ac ≥5), the value of A∞increases with Cr, as the centrifugal effect is enhanced.
Increasing the Reynolds number, the velocity gradients concentrate near the walls. It is thus interesting to examine the evolution of the boundary layer thickness δ of the primary flow. Here it is estimated from the local shear strain (∂Uθ/∂z) on the top moving wall (z = H) at the middle of the channel (r = Rmid), i.e. it is estimated as (∂Uθ/∂z)δ = ΩRmid. Figure 17 shows the evolution of δ/H against Re for the same cases as in Fig. 16. Note that if the shear stress was uniform over the entire top wall, there should be a direct relation between δ/H and Γ∗, an evolution of Γ∗as Reα involving an evolution of δ/H as Re1−α. The numerical results show that this relation is satisfied in the linear regime as well as in the large-Re regime. At low Re, δ/H is almost unity, as expected for a constant shear rate along the channel height. At large Re, δ/H indeed scales as Re1−α, with the same value of α as that obtained from the fit of Γ∗. The fact that a local measurement is sufficient to characterize the global feature of the flow has major consequences. Regarding physical mechanisms, it suggests that the entire flow reaches a self-similar asymptotic state.
Practically, it shows that the evolution of the torque and total dissipation can be obtained from the measurement of the velocity profile of the primary flow in a single plane, such as those presented in section IV C.
We examine now the scaling of the secondary flow in the two asymptotic regimes. The secondary flow in the radial-vertical plane consists of a single recirculation cell (Fig. 12). In order to characterize it, we introduce two parameters: its maximal magnitude |Usec,max| and the vertical distance Hsec between the two points located in the lower half of the domain where the secondary-flow magnitude is 0.3|Usec,max| (see Fig. 12). Figures 18 and 19 present the evolution of Hsec and Usec,max against Re. At low Re, all cases show a similar trend: Hsec = 1 2H, which means that the secondary-flow cell occupies the entire cross section, and the velocity Usec,max of the secondary flow is two orders of magnitude smaller than that of the primary flow. At large Re, the three cases with a significant curvature ratio (Cr ≥0.5) behave in the same way. As Re increases, the secondary-flow cell localizes more and more near the walls, Hsec showing the same scaling law in Re1−α as δ. In parallel, the velocity of the secondary flow reaches a plateau (0.13 ≤Usec,max/ΩRo ≤0.15) that slightly depends on the geometry. Thus, the shear rate characteristic of the primary flow (ΩRo/δ) and that of the secondary flow (Usec,max/Hsec) obey the same scaling law with Re, 15 10−1 100 101 102 103 Re 10−4 10−3 10−2 10−1 100 101 Γ∗ 1 Ac = 2, Cr = 0.5 Ac = 5, Cr = 0.5 Ac = 5, Cr = 0.9 Ac = 10, Cr = 0.1 FIG. 16: Dimensionless torque Γ∗= Γ/ΓAPC,µ versus the Reynolds number Re for various pairs of geometrical parameters: (Ac = 2, Cr = 0.5), (Ac = 5, Cr = 0.5), (Ac = 5, Cr = 0.9) and (Ac = 10, Cr = 0.1). Symbols: numerical simulations; straight lines: high-Re fits by A∞Reα. The fitted parameters are respectively A0 = (3.5×10−3,1.6×10−3,1.8×10−3,1.5×10−3), A∞= (3.8×10−4,2.1×10−4,2.3×10−4,0.51×10−4) and α = (1.5,1.5,1.5,1.65).
10−1 100 101 102 103 Re 10−2 10−1 100 δ/H Ac = 2, Cr = 0.5 Ac = 5, Cr = 0.5 Ac = 5, Cr = 0.9 Ac = 10, Cr = 0.1 FIG. 17: Dimensionless boundary-layer thickness of the primary flow at the top wall. Symbols: numerical simulations for the same pairs of parameters as in Fig. 16; straight lines: high-Re fits by B∞Re1−α with B∞= (1.9,3.18,2.3,21.8) and α = (1.5,1.5,1.5,1.65).
which confirms the close connection between primary and secondary flows in this regime. In the case of a small curvature ratio (Cr ≥0.1), centrifugal inertia is reduced and the transition towards the large-Re asymptotic regime is delayed. The plateau of Usec,max is hardly reached at the largest investigated Re, while the power-law decrease of Hsec is not attained at all.
16 10−1 100 101 102 103 Re 0.1 0.2 0.3 0.4 0.5 0.6 1 Hsec/H - 0.5 Ac = 2, Cr = 0.5 Ac = 5, Cr = 0.5 Ac = 5, Cr = 0.9 Ac = 10, Cr = 0.1 FIG. 18: Dimensionless characteristic height of the secondary flow for the same pairs of parameters as in Fig. 16.
10−1 100 101 102 103 Re 10−2 10−1 Usec,max/(ΩRo) Ac = 2, Cr = 0.5 Ac = 5, Cr = 0.5 Ac = 5, Cr = 0.9 Ac = 10, Cr = 0.1 FIG. 19: Dimensionless characteristic velocity of the secondary flow for the same pairs of parameters as in Fig. 16.
VI.
CONCLUSION The goal of this work was to understand the hydrodynamics of the flow within an Annular Plane Couette (APC), a peculiar configuration involving secondary flows of Prandtl’s first kind that can be used as a reference flow for fundamental studies, such as investigations of the rheology of multiphase media. Direct numerical simulations of a single-phase, laminar, stationary, ax-isymmetric, incompressible, Newtonian flow within an APC channel have been carried out using OpenFOAM. These simulations were further validated by PIV measurements in an experimental APC device at various Reynolds numbers, using a low-viscosity oil seeded by fluorescent particles.
The numerical study focuses on analyzing the effects of three dimensionless parameters: (1) the channel Reynolds number Re, which compares inertial to viscous forces, (2) the channel aspect ratio Ac, which characterizes the lateral wall confinement, 17 and (3) the curvature ratio Cr, which modulates the influence of the centrifugal forces. Initially, the main features of the hydro-dynamics are examined in a single representative case, then the APC flow is scanned by varying the geometric parameters within broad ranges.
The fluid is sheared due to the rotation of the top annular plate, inducing a primary flow in the azimuthal direction, akin to the one observed in a Straight Plane Couette (SPC) flow. However, the vertical profile of the azimuthal velocity deviates from the classical linear profile, showing a S-shaped form reminiscent of a turbulent SPC flow. A secondary flow, in the form of a single recirculation cell in the radial-vertical plane, is also generated by the centrifugal force difference across the radial direction. This secondary flow has a magnitude up to 15% of the primary flow. It thus exerts a significant influence on the primary flow by increasing the shear rate close to the walls and decreasing it in the central region.
Regarding the parametric study, several key observations can be drawn. The results show that as Re increases, the flow evolves from a linear profile to a S-shape profile, due to the increased localization and magnitude of the secondary flow in the near-wall regions. The effects of lateral confinement are significant for a high confinement (Ac = 1), but they notably diminish once the section becomes wider (Ac ≥2), and become negligible when the section is almost 2D (Ac = 10). Meanwhile, increasing the curvature ratio Cr leads to a transformation of the vertical profile of the azimuthal velocity into a S-shape, while lowering it results in an almost 2D flow, fairly similar to the SPC flow. Especially, for the case of low confinement and centrifugal forces (Ac = 10 and Cr = 0.1), the linear profile is maintained at higher Reynolds numbers, while the velocity profile is kept uniform in the radial direction.
Then, the torque exerted on the top moving wall has been examined. The dimensionless torque Γ∗is a global quantity that characterizes the flow as a whole. Its evolution with the Reynolds number clearly exhibits the existence of two asymptotic regimes. At low Re, Γ∗is a linear function of Re, with a slope A0 depending on the lateral confinement through the value of Ac. At large Re, it is a power law, Γ∗= A∞Reα, with an exponent α close to 1.5, a value in the same range as those measured in a turbulent Taylor Couette flow of either a single fluid [6, 31] or a droplet emulsion [1, 2]. Such similarities, between a laminar APC flow and either the turbulent TC flow regarding the torque scaling or the turbulent SPC flow regarding the S-shaped velocity profile, suggest an analogy of the underlying mechanisms. As Re increases, our results show that the velocity gradients concentrate near the wall, in a boundary layer whose thickness δ scales as Re1−α. An important result is that various global or local quantities of either the primary or the secondary flow show all the same scaling with the Reynolds number. This indicates that, for a given geometry, the structure of the APC flow becomes self-similar at large Reynolds numbers. Practically, it means that a local measurement, such as the velocity gradient at the top wall in the channel middle can be enough to characterize the evolution of the torque, as well as the total viscous dissipation by the flow.
In conclusion, the annular plane Couette flow can be used as a reference flow despite the centrifugal effects due to curvature.
Indeed, by enabling long-duration studies, a shear plane parallel to gravity and a controllable shear rate, this configuration seems to open up new possibilities for the investigation of more complex processes, especially in emulsion flows, with slow evolution of the interfacial area. By adjusting the Reynolds number (Re) and the geometry (Ac and Cr), the flow field within the channel can extend from an almost 2D flow, close to the classical plane Couette flow and its constant shear rate, to a complex 3D flow, more representative of industrial applications where the shear rate in pipe flows is not uniform.
ACKNOWLEDGMENTS The authors would like to thank the technical department of the LGC and of IMFT for their help to construct the experimental set-up, Thomas Ménager for the construction and tuning of the experimental set-up and Emmanuel Cid for his help on the development of the optical metrology.
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R. C. Chin, R. Vinuesa, R. Örlü, J. I. Cardesa, A. Noorani, M. S. Chong, and P. Schlatter, Backflow events under the effect of secondary flow of prandtl’s first kind, Phys. Rev. Fluids 5, 0746060 (2020).
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Couette flow, Wikipedia Rectangular channel.
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B. Eckhardt, S. Grossmann, and D. Lohse, Torque scaling in turbulent taylor–couette flow between independently rotating cylinders, Journal of Fluid Mechanics 581, 221–250 (2007).
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8997 | https://www.myeloma.org/resource-library/understanding-mgus-smoldering-myeloma | Understanding Myeloma Precursors| Intnt’l Myeloma Fnd
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Monoclonal gammopathy of undetermined significance (MGUS) is the earliest disease state associated with the subsequent development of myeloma, and smoldering multiple myeloma (SMM) is an asymptomatic precursor to active disease. If you are a patient with MGUS or SMM, it is important and helpful to learn as much as possible about your diagnosis in order to play an active role in your own medical care and to make good decisions about your care with your doctor.
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8998 | https://www.youtube.com/watch?v=t4mWCKu44lU | Calculating Percent Increase | Percent Change | Math with Mr. J
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Welcome to Calculating Percent Increase with Mr. J! Need help with how to find a percent increase? You're in the right place!
Whether you're just starting out, or need a quick refresher, this is the video for you if you're looking for help with percent change. Mr. J will go through an example of calculating percent change, specifically a percent increase, and explain the process step-by-step.
✅ Here's a percent decrease example:
✅ Check out more percent change examples here:
✅ Here are even more examples:
About : Math with Mr. J is a math education channel that offers instructional math videos to anyone looking for a little extra help with math!
Email : math5.mrj@gmail.com
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Hopefully this video is what you're looking for when it comes to calculating percent increase.
Have a great rest of your day and thanks again for watching! ✌️✌️✌️
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Transcript:
[Music] welcome to math with Mr J in this video I'm going to cover how to calculate percent change and specifically we're going to take a look at an example of a percent increase let's jump into our example where we have an original value of 16 and then a new value of 28 we had a change the original value 16 increased in increased to 28 so we need to figure out the percent change we have here now we can calculate percent change by taking the new value and subtracting the original value that gives us the amount of change the amount of increase or decrease then we divide that result by the original value and keep in mind always divide by the original value because we are finding the percent change from from that original value that's going to give us a decimal we then multiply that decimal by 100 to convert it to a percent so here for percent change we need to do the new value 28 Minus the original value 16 divided by the original value 16 and then we multiply by 100 let's start by subtracting here 28 - 16 gives us 12 and that's a positive 12 having a positive there tells us we have an increase if we end up with a negative that tells us we have a decrease so we have 12 over 16 100 now let's divide so we have 12 / by 16 that gives us 75 hundredths 75 and we need to multiply that decimal by 100 in order to convert it to a percent now a quick way to multiply by 100 is to move the decimal twice to the right so once twice to the right and that gives us 75 percent and since that's positive that tells us we have an increase we went up 75% so this was a 75% increase so there you have it there's an example of calculating percent change and specifically we looked at a percent increase check the description for an example of a percent decrease I hope that helped thanks so much for watching until next time peace |
8999 | https://www.accessscience.com/content/article/a089400 | Skip to main content
Boiling point
Ira N. Levine
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Boiling point
Key Concepts
The boiling point of a pure substance is the temperature at which the substance transitions from a liquid to the gaseous phase. At this point, the vapor pressure of the liquid is equal to the applied pressure on the liquid.
The boiling point at a pressure of 1 atmosphere is called the normal boiling point. The normal boiling point of water is 100°C, and remains constant until the water has vaporized.
The normal boiling point is high for liquids with strong intermolecular attractions and low for liquids with weak intermolecular attractions.
For a solution boiled at a fixed pressure, the composition of the vapor will usually differ from that of the liquid. In addition, the change in the liquid composition during boiling will change the boiling point.
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The temperature at which the liquid and vapor phases are in equilibrium with each other at a specified pressure. At the boiling point, the transition from the liquid to the gaseous phase occurs in a pure substance. Therefore, the boiling point is the temperature at which the vapor pressure of the liquid is equal to the applied pressure on the liquid. The boiling point at a pressure of 1 atmosphere is called the normal boiling point (see figure). See also: Boiling; Gas; Liquid; Pressure; Temperature
Phase diagram for water. (1 atm = 101.325 kPa)
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For a pure substance at a particular pressure P, the stable phase is the vapor phase at temperatures immediately above the boiling point and is the liquid phase at temperatures immediately below the boiling point (see figure). The liquid-vapor equilibrium line on the phase diagram of a pure substance gives the boiling point as a function of pressure. Alternatively, this line gives the vapor pressure of the liquid as a function of temperature. The vapor pressure of water is 1 atm (101.325 kilopascals) at 100°C, the normal boiling point of water. The vapor pressure of water is 3.2 kPa (0.031 atm) at 25°C, so the boiling point of water at 3.2 kPa is 25°C. The liquid-vapor equilibrium line on the phase diagram of a pure substance begins at the triple point (where solid, liquid, and vapor coexist in equilibrium) and ends at the critical point, where the densities of the liquid and vapor phases have become equal. For pressures below the triple-point pressure or above the critical-point pressure, the boiling point is meaningless. Carbon dioxide has a triple-point pressure of 5.11 atm (518 kPa), so carbon dioxide has no normal boiling point. See also: Carbon dioxide; Phase equilibrium; Triple point; Vapor pressure; Water
The normal boiling point is high for liquids with strong intermolecular attractions and low for liquids with weak intermolecular attractions. Helium has the lowest normal boiling point, 4.2 kelvin (−268.9°C). Some other normal boiling points are 111.1 K (−162°C) for methane (CH4), 450°C for triacontane (n-C30H62), 1465°C for sodium chloride (NaCl), and 5555°C for tungsten (W). See also: Helium; Intermolecular force; Methane; Tungsten
The rate of change of the boiling-point absolute temperature Tb of a pure substance with pressure is given by the equation below.
dTbdP=TbΔVvap,mΔHvap,m
ΔHvap,m is the molar enthalpy (heat) of vaporization, and ΔVvap,m is the molar volume change on vaporization. See also: Enthalpy
The quantity ΔHvap,m/Tb is ΔSvap,m, the molar entropy of vaporization. The molar entropy of vaporization at the normal boiling point (nbp) is given approximately by Trouton's rule: ΔSvap,m,nbp ≈ 87 J/mol K (21 cal/mol K). Trouton's rule fails for highly polar liquids (especially hydrogen-bonded liquids). It also fails for liquids boiling at very low or very high temperatures, because the molar volume of the vapor changes with temperature and the entropy of a gas depends on its volume. See also: Entropy
When a pure liquid is boiled at fixed pressure, the temperature remains constant until all the liquid has vaporized. When a solution is boiled at fixed pressure, the composition of the vapor usually differs from that of the liquid, and the change in liquid composition during boiling changes the boiling point. Thus the boiling process occurs over a range of temperatures for a solution. An exception is an azeotrope, which is a solution that boils entirely at a constant temperature because the vapor in equilibrium with the solution has the same composition as the solution. In fractional distillation, the variation of boiling point with composition is used to separate liquid mixtures into their components. See also: Azeotropic mixture; Distillation
Related Primary Literature
Q. Li et al., Prediction of critical properties and boiling point of fluorine/chlorine-containing refrigerants, Int. J. Refrig. , 143:28–36, 2022
O. D. Samuel et al., Performance comparison of empirical model and Particle Swarm Optimization & its boiling point prediction models for waste sunflower oil biodiesel, Case Stud. Therm. Eng., 33:101947, 2022
Additional Reading
P. Atkins, J. de Paula, and J. Keeler, Atkins' Physical Chemistry, 12th ed., Oxford University Press, 2022
J. Rumble, CRC Handbook of Chemistry and Physics, 104th ed., CRC Press, 2023
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